Evaluating the Reasoning Abilities of LLMs on Underrepresented Mathematics Competition Problems

They have often demonstrated good performance in some fields of science and technology [9,20,21]. Nonetheless, their ability to solve mathematical problems has become a question of considerable importance. Problems in Calculus, Analytic Geometry, and Discrete Mathematics require a blend of abstract and logical reasoning, visualization, numerical performance, and knowledge of problem-solving techniques [1,7]. Problems in these fields form a metric for analyzing LLM performance in logic-oriented problems [8,10,13].

Several studies have tested LLMs’ abilities to solve math problems [1,3,8,13]; however, they have often limitations as follows: (i) Most studies use the same datasets for benchmarking, such as the MATH and GSM8K datasets. This can lead to dataset “contamination,” where LLMs are trained on the same data used for evaluation. Such contamination compromises the fairness and validity of the assessment and artificially inflates performance metrics, making it difficult to accurately measure a model’s true capabilities; (ii) some studies focus exclusively on evaluating flagship language models. While this is useful for understanding the current limits of LLM capabilities, it does not accurately reflect the models that are accessible to the general public; (iii) other studies assess LLM performance solely based on the final answer, paying little attention to the underlying reasoning process.

To address these issues, we evaluate the performance of freely accessible and currently prominent LLMs on underrepresented mathematics competition datasets, focusing not only on accuracy but also on the quality and structure of their reasoning. Moreover, this study extends beyond overall accuracy metrics to provide a detailed analysis of the reasoning errors each model exhibits. By identifying and characterizing these reasoning failures, we aim to uncover the specific challenges LLMs face when solving complex, underrepresented mathematical problems and to better understand their current limitations. This investigation addresses two central questions: (1) To what extent can LLMs reliably produce correct solutions to complex, multi-step mathematical problems? and (2) Are the reasoning processes underpinning these solutions mathematically valid? The findings offer a more nuanced understanding of LLMs’ mathematical capabilities to solve underrepresented mathematical problems, contributing valuable insights into their potential use in education, scientific research, and automated problem-solving systems [2,6,12].

In order to assess the performance of the LLMs in the three mathematical disciplines, we prompted the models with the problem statements, as well as a command to solve the problem analytically.

The Missouri Collegiate Mathematics Competition (MCMC) datasets are relatively unknown and, to the best of our knowledge, have not been previously used to evaluate LLMs capabilities [16]. This significantly reduces the likelihood of dataset contamination and allows for a more reliable assessment of model performance. The MCMC draws its problems from a variety of reputable sources, including but not limited to the William Lowell Putnam Mathematical Competition, the MAA (Mathematical Association of America) Problem Directory, the Mathematical Olympiads of the USSR, and the Iowa Collegiate Mathematics Competition. The MCMC problems were categorized into three mathematical disciplines: Calculus, Analytic Geometry, and Discrete Mathematics. Ten problems were chosen from each category, resulting in a total of thirty problems. Additionally, problems were chosen in a way that they could be used for assessing the LLMs’ accuracy based on a final answer as well as the reasoning process. Hence, problems requiring mathematical proof were not included in this study. The sets of selected problems with the specific year and correct answers are available in Section 1 of the supplementary document.

The study assessed the performance of three prominent Large Language Models (LLMs): GPT-4o-mini, Gemini-2.0-flash, and DeepSeek-V3. GPT-4o-mini is a lightweight variant of OpenAI’s GPT-4o, optimized for efficiency and speed [17,19].

Gemini-2.0-flash, developed by Google, excels in natural language understanding and is also designed for efficiency [5,11]. DeepSeek is designed for technical tasks, particularly in mathematics, and is known for its ability to model logical progression and identify mathematical patterns, as well as its enormously reduced training cost [4,14]. All three models were tested using the same set of problems from the Missouri Collegiate Mathematics Competition, with each model prompted to provide both reasoning and final answers.

The models’ responses were categorized into four possible outcomes: (1) correct final answer with correct reasoning, (2) incorrect final answer with correct reasoning (e.g., minor computational mistakes), ( 3) incorrect final answer with incorrect reasoning, and ( 4) incorrect final answer with correct reasoning. We conducted a detailed analysis of reasoning errors based on expert evaluation, categorizing responses with incorrect reasoning into 11 distinct types.

These reasoning error categories are summarized in Table 1.

Figure 1 provides a comparison of the three language models-GPT-4o-mini, Gemini 2.0 Flash, and DeepSeek V3-across the three mathematical domains. Panel (a) summarizes the overall performance of each model in terms of the percentage of correct answers with correct reasoning, incorrect answers with correct reasoning, and incorrect answers with incorrect reasoning. Among the models, DeepSeek-V3 demonstrates the highest accuracy with the vast majority of its answers classified as correct. Panels (b), (c), and (d) break down the types of reasoning errors observed in the calculus, geometry, and discrete mathematics problems, respectively. Each panel shows only the error types that occurred with non-zero frequency. The most prominent reasoning error across all three domains is the presence of logical errors.

In Calculus (Panel b), both GPT-4o-mini and Gemini exhibit a majority of logical mistakes, while DeepSeek mostly suffers from computational errors. In Geometry (Panel c), logical and computational errors are common among all three models. For Discrete Math (Panel d), all of DeepSeek-V3’s errors are due to logical reasoning mistakes, whereas Gemini displays mostly logical errors and approach errors.

Table 2 shows the classification of the models’ responses for each mathematical category. These outcomes reflect the models’ ability to arrive at the correct answer and the logical soundness of their reasoning processes. There was no case of an incorrect final answer with correct reasoning, possibly due to low sample size. Table 2. Classification of the models based on final answer accuracy and reasoning correctness.

GPT-4o-mini 20% 0% 80% 30% 0% 70% 20% 10% 70% 23% 3% 73%

Gemini-2.0-flash 60% 0% 40% 50% 0% 50% 70% 0% 30% 60% 0% 40%

DeepSeek-V3 80% 0% 20% 80% 10% 10% 90% 0% 10% 83% 3% 13% First percentage = Correct final answer with correct reasoning, Second percentage = Incorrect final answer with correct reasoning, Third percentage = Incorrect final answer with incorrect reasoning

Table 3 provides a deeper look into the types of reasoning errors made by the models when their final answers were incorrect. The table highlights the percentage of errors in each of the 11 reasoning categories: ambiguous steps, debatable theories, incomplete reasoning, irrelevant steps, logical errors, computational errors, methodological errors, fact errors, approach errors, redundancy errors, and rushed conclusions.

Table 3. Analysis of Incorrect Reasoning for each mathematical category. The values correspond to percentages for GPT-4o-mini, Gemini, and DeepSeek models, respectively. 00% 17% 00% 00% 18% 00% Bolded entry signifies that Gemini-2.0-flash made 40% of its Calculus errors due to incomplete reasoning.

GPT-4o-mini primarily struggles with logical and approach errors across all mathematical domains, indicating a tendency to follow flawed reasoning paths or select suboptimal problem-solving strategies. In Geometry, it also shows a notable rate of incomplete reasoning, suggesting difficulties in carrying solutions through to completion. Gemini exhibits inconsistent reasoning patterns, with significant incomplete reasoning and rushed conclusions in Geometry, pointing to a tendency to jump to answers without fully developed steps. DeepSeek shows the highest frequency of logical and computational errors, especially in Calculus and Discrete Math, reflecting issues with both reasoning accuracy and numerical operations. Overall, each model reveals distinct weaknesses in how they process and execute mathematical reasoning.

The results presented in this study highlight the strengths and weaknesses of the three large language models (LLMs)-GPT-4o-mini, Gemini-2.0-flash, and DeepSeek-V3-in solving underrepresented mathematical problems from the following categories: Calculus, Analytic Geometry, and Discrete Mathematics.

Overall, the models performance varied significantly in providing correct final answers. DeepSeek-V3 exhibited the highest accuracy in both final answers and reasoning, with particularly impressive results in discrete mathematics, where it achieved the highest percentage of correct answers with correct reasoning (90%). This indicates that DeepSeek-V3 may have a better understanding of the mathematical principles involved in discrete mathematics problems compared to the other models. In contrast, GPT-4o-mini performed poorly across all three problem domains, receiving a maximum percentage of correct answers with correct reasoning of 30%. While previous studies have shown that GPT-4o-mini can successfully solve simple math problems, GPT-4o-mini’s performance in this study demonstrates its inability to solve more complex problems [1,2,6]. Gemini 2.0-flash demonstrated mediocre performance, with it solving 70% of discrete math problems, but only 50% of geometry problems. This highlights the limitations of Gemini 2.0-flash in visualizing abstract geometric figures.

A.1.1 1996 Session 2: 1. Evaluate the definite integrals (a)

Solution. (a) Since the integrand is not defined at either bound of integration, one should write

= 𝜋 .

(b) The difficulty here is with the infinite interval of integration. Let 𝑦 = 𝑥 -1; then

A.1.2 1997 Session 2: 4. Define a family of curves by

where 𝑛 is a positive integer. What is the limit of the length of 𝑆 𝑛 as 𝑛 → ∞? Solution. Denote the length of 𝑆 𝑛 by 𝐿(𝑆 𝑛 ). The period of cos(𝑛 2 𝑥) is 2𝜋/𝑛 2 , and cos(𝑛 2 𝑥) ≥ 0 for 0 ≤ 𝑥 ≤ 𝜋/2𝑛 2 , so (𝑛 + 𝑚 + 1)! .

. A.1.4 1999 Session 2: 2. Give a precise characterization of those points in the plane which do not lie on a tangent line to the curve 𝑦 = 𝑥 4 -6𝑥 2 . Solution. Given a point (𝑡, 𝑡 4 -6𝑡 2 ) on the curve, the equation of the tangent line at that point is 𝑦 = (4𝑡 3 -12𝑡)𝑥 -(3𝑡 4 -6𝑡 2 ). For a fixed value of 𝑥, as 𝑡 varies, the maximum value of 𝑦 will occur when

This polynomial factors as (12𝑡 2 -12)(𝑥 -𝑡), so the critical values occur at 𝑡 = 1, -1, 𝑥. The corresponding 𝑦 values are -8𝑥 + 3, 8𝑥 + 3, and 𝑥 4 -6𝑥 2 and a point (𝑥, 𝑦) will not lie on any tangent line if the 𝑦 coordinate is greater than the maximum of these three quantities. More specifically, the set of points which lie on no tangent line to the curve is {(𝑥, 𝑦) | |𝑥 | > 3 and 𝑦 > 𝑥 4 -6𝑥 2 , or |𝑥 | ≤ 3, and 𝑦 > 8|𝑥 | + 3}. Note that 𝑦 = ±8𝑥 + 3 are the tangent lines at the points of inflection, (∓1, -5), and that (±3, 27) are where these tangents intersect the quartic.

A.1.5 1999 Session 2: 4. Find all possible continuous and differentiable curves 𝐶 which have the following properties.

The curve 𝐶 lies in the first quadrant and contains the point (0, 0). Whenever 𝑃 is on 𝐶 the interior of the rectangle 𝑅 bounded by the coordinate axes and horizontal and vertical lines through 𝑃 is separated into two parts by 𝐶. When the part adjacent to the 𝑥-axis is rotated about the 𝑥-axis and the part adjacent to the 𝑦-axis is rotated about the 𝑦-axis, two solids of equal volume are generated. Solution. In order for 𝐶 to always divide 𝑅 into two and only two separate parts, 𝐶 must be an increasing function. Let 𝐶 be given by 𝑦 = 𝑓 (𝑥). The volumes of the two solids are

Setting 𝑉 1 = 𝑉 2 , canceling the 𝜋 and breaking the 𝑉 2 integral into two parts gives

Differentiate both sides with respect to 𝑥 and simplify to get

Separation of variables and integration gives the family of curves

The point (0, 0) is on each curve in the family, and if 𝑐 ≥ 0 the domain is [0, ∞), while if 𝑐 < 0 the domain is [0, 1/|𝑐 |).

A.1.6 2002 Session 1: 3 (limit only). Let {𝑥 𝑛 } be the following sequence involving alternating square roots of 5 and 13:

and so on. Prove that lim 𝑛→∞ 𝑥 𝑛 exists and determine its value. Solution. We see that 𝑥 1 , 𝑥 2 < 4; assume also that

The argument is identical for 𝑥 2𝑘+1 . Hence, for each 𝑛, 𝑥 𝑛 < 4 by mathematical induction on 𝑛. In addition, the sequence increases monotonically. Therefore, by a standard limit theorem on sequences lim

Then we have

One root is 𝐿 = 3. Of the three remaining roots, one is positive (between 1 and 2) and the other two are complex. It follows that lim

Convergence is fairly rapid; 𝑥 6 is already 2.999971.

A.1.7 2002 Session 1: 5. Two right circular cylinders of radius 𝑟 intersect at right angles to form a solid. This solid has four curved faces. Imagine one of these faces “rolled out flat”. Find equations of the boundary curves of this flattened face and also find its area. Solution. Let the 𝑥-axis be the axis of one cylinder and the 𝑦-axis be the axis of the other cylinder so that the center of the sold is the origin. Cross-sections perpendicular to the 𝑧-axis are squares with side

Place the flattened face with its axis of symmetry on the horizontal axis, call it the 𝑤-axis, with the left end of the figure at the origin. This makes the range of 𝑤 the interval 0 ≤ 𝑤 ≤ 𝜋𝑟 , and the distance from the 𝑤-axis to the top boundary curve is The area is given by a simple integral:

This solid and its circumscribed cube have the same properties that Archimedes admired concerning the sphere and its circumscribed cylinder, namely, both the volumes and the surface areas are in the ratio 2 : 3.

Find lim 𝑛→∞ 𝑥 𝑛 . Solution. Solving for 𝑥 𝑛 , we obtain

which is of the form ∞ -∞. Then, using the substitution 𝑢 = 1/𝑛 and l-Hospital’s Theorem twice, we obtain lim

A.1.9 2005 Session 1: 4. Determine the value of the integral

and locate those points 0 ≤ 𝜃 ≤ 2𝜋, where 𝐼 (𝜃 ) is discontinuous.

Solution. Replace 1 = cos 2 𝜃 + sin 2 𝜃 and let 𝑢 = 𝑥cos 𝜃 , 𝑑𝑢 = 𝑑𝑥. Then the integral becomes

Here, we recall that the range of Tan -1 𝜃 is (-𝜋/2, 𝜋/2). As an example, Tan -1 (tan(20

and Tan -1 𝑦 = -Tan -1 (-𝑦) for any real 𝑦. Therefore,

For other values of 𝜃 , the value of 𝐼 (𝜃 ) depends entirely on the sign of sin 𝜃 . Case 1. 0 < 𝜃 < 𝜋. Then sin 𝜃 > 0 and

Case 2. 𝜋 < 𝜃 < 2𝜋. Then sin 𝜃 < 0 and

Therefore, the function 𝐼 (𝜃 ) is

The points of discontinuity occur at 0, 𝜋, and 2𝜋.

A.1.10 2005 Session 2: 5. Suppose that 𝑓 : [0, ∞) → [0, ∞) is a differentiable function with the property that the area under the curve 𝑦 = 𝑓 (𝑥) from 𝑥 = 𝑎 to 𝑥 = 𝑏 is equal to the arclength of the curve 𝑦 = 𝑓 (𝑥) from 𝑥 = 𝑎 to 𝑥 = 𝑏. Given that 𝑓 (0) = 5/4, and that 𝑓 (𝑥) has a minimum value on the interval (0, ∞), find that minimum value. Solution. The area under the curve 𝑦 = 𝑓 (𝑥) from 𝑥 = 𝑎 to 𝑥 = 𝑏 is

and the arclength of the curve 𝑦 = 𝑓 (𝑥) from 𝑥 = 𝑎 to 𝑥 = 𝑏 is

Therefore,

for all nonnegative 𝑎 and 𝑏. In particular, we can write

for all nonnegative 𝑥. Both sides of the above equation define a function of 𝑥, and since they are equal, their derivatives are equal; their derivatives are given by the Second Fundamental Theorem of Calculus:

So, we are looking for a function 𝑦 which satisifies the differential equation

This equation is separable:

Integrating both sides yields

(where the first integral is evaluated using the trig substitution 𝑦 = sec 𝜃 and the two arbitrary constants of integration are combined into one constant on the right hand side). Next, since 𝑓 (0) = 5/4 is positive, we can drop the absolute value, and solve for 𝑦:

Using 𝑓 (0) = 5/4, we find

This gives two possible functions:

This latter has a minimum at 𝑥 =ln 2, which is not positive, so we reject that function. The former has a minimum at 𝑥 = ln 2, and the 𝑦 value is 1. Note: One could also deduce from the differential equation 𝑦 ′ = √︁ 𝑦 2 -1 that at the minimum value, since 𝑦 ′ = 0, the 𝑦-value must be 1.

A.2.1 1997 Session 1: 1. Let 𝑃 ≠ (0, 0) be a point on the parabola 𝑦 = 𝑥 2 . The normal line to the parabola at 𝑃 will intersect the parabola at another point, say 𝑄. Find the coordinates of 𝑃 so that the 𝑦-coordinate of 𝑄 is a minimum.

Solution. Let 𝑃 have the coordinates (𝑝, 𝑝 2 ) and 𝑄 have the coordinates (𝑞, 𝑞 2 ). Without loss of generality, let 𝑝 > 0.

The slope of the tangent line to the parabola at 𝑃 is 2𝑝 so the slope of the normal line to the parabola at 𝑃 is -1/2𝑝.

Thus, the equation of the normal line to the parabola at 𝑃 is

Therefore, the point 𝑄, the intersection of the normal line and the parabola has the 𝑥-coordinate 𝑞 = -𝑝 -

.

Hence, the 𝑦-coordinate of 𝑄 is

Taking the derivative of 𝑦 as a function of 𝑝,

Minimizing 𝑦 as a function of 𝑝 yields 𝑝 = √︁ 1/2 so

A.2.2 1998 Session 1: 1. Let 𝑃 ≠ (0, 0) be a point on the parabola 𝑦 = 𝑥 2 . The normal line to the parabola at 𝑃 will intersect the parabola at another point, say 𝑄. Find the coordinates of 𝑃 so that the length of segment 𝑃𝑄 is a minimum.

Solution. Let 𝑃 = (𝑝, 𝑝 2 ) and 𝑄 = (𝑞, 𝑞 2 ). The slope of the tangent line to the parabola at 𝑃 is 2𝑝 so the slope of the normal line to the parabola at 𝑃 is -1/2𝑝. Thus, the equation of the normal line to the parabola at 𝑃 is

Therefore, the other intersection point of the normal line and the parabola, 𝑄 has the 𝑥-coordinate 𝑞 = -𝑝 -

.

represent the square of the distance between 𝑃 and 𝑄. Differentiating 𝑑 with respect to 𝑝 and setting this expression to zero results in the equation

Simplifying this equation, we obtain 1 4

so the real solutions of the equation are 𝑝 = ± √︁ 1/2. Hence, the coordinates of 𝑃 so that the length of segment 𝑃𝑄 is a minimum is

.

A.2.3 1999 Session 1: 1. Let 𝑃 ≠ (0, 0) be a point on the parabola 𝑦 = 𝑥 2 . The normal line to the parabola at 𝑃 will intersect the parabola at another point, say 𝑄. Find the coordinates of 𝑃 so that the length of the arc of the parabola between 𝑃 and 𝑄 is a minimum. Solution. Let 𝑃 = (𝑝, 𝑝 2 ) and 𝑄 = (𝑞, 𝑞 2 ). The slope of the tangent line to the parabola at 𝑃 is 2𝑝 so the slope of the normal line to the parabola at 𝑃 is -1/2𝑝. Thus, the equation of the normal line to the parabola at 𝑃 is

Hence, the other intersection point of the normal line and the parabola, 𝑄 has the 𝑥-coordinate 𝑞 = -𝑝 -

. Now the length of the arc of the parabola between 𝑃 and 𝑄 is

Differentiating 𝐿 with respect to 𝑝 gives

2 and setting the derivative equal to 0 gives

Squaring both sides to remove the radicals and multiplying by 𝑝 4 , we obtain

Hence, the coordinates of 𝑝 resulting in a minimum arc length are

A.2.4 1999 Session 2: 1. Let 𝑆𝐶 be the semicircle with 𝑦 ≥ 0 centered at (1, 0) with radius 1. Let 𝐶 𝑎 be the circle with radius 𝑎 > 0 and center (0, 0) and denote the point (0, 𝑎) by 𝑃. Consider the line through 𝑃 and the intersection of 𝑆𝐶 and 𝐶 𝑎 . What is the limiting position of the 𝑥-intercept of this line as 𝑎 → 0? Solution. The limiting position is 𝑥 = 4.

The circles intersect at

The line through this point and (0, 𝑎) is

Taking the limit as 𝑎 → 0 yields 4 as the limit.

A.2.5 2000 Session 1: 1. Let 𝑃 ≠ (0, 0) be a point on the parabola 𝑦 = 𝑥 2 . The normal line to the parabola at 𝑃 will intersect the parabola at another point, say 𝑄. Find the coordinates of 𝑃 so that the distance between the 𝑥-coordinates of 𝑃 and 𝑄 is a minimum. Solution. Let 𝑃 = (𝑝, 𝑝 2 ) and 𝑄 = (𝑞, 𝑞 2 ). The slope of the tangent line to the parabola at 𝑃 is 2𝑝 so the slope of the normal line to the parabola at 𝑃 is -1/2𝑝. Thus, the equation of the normal line to the parabola at

Since 𝑄 = (𝑞, 𝑞 2 ) lies on the normal line,

. Now the distance between the 𝑥-coordinates of 𝑃 and 𝑄 is

.

Assuming 𝑝 > 0,

.

Using the AM-GM Inequality, it is known that for all 𝑥 > 0,

with equality when 𝑥 = 1. Therefore, 𝑝 = 1/2 if 𝑝 > 0 or 𝑝 = -1/2. Hence, the coordinates of 𝑃 resulting in a minimum distance between the 𝑥-coordinates of 𝑃 and 𝑄 are

.

That is,

Therefore,

But this is exactly the equation of an ellipse.

A.2.7 2001 Session 2: 3. A convex decagon and all of its diagonals are drawn. How many interior points of intersection of the diagonals are there, if it is assumed that no 3 diagonals share a common interior point? Solution. Let 𝑁 (𝑛) be the number of such interior intersection points in a convex 𝑛-gon. Label the vertices 1, 2, 3, . . ., 𝑛. Any four labels selected from this list (e.g., (1, 3, 𝑛, 6)) define a quadrilateral, which is convex because the 𝑛-gon is convex. The quadruple of integers therefore also defines uniquely an interior intersection point of the diagonals of the quadrilateral, and hence also of the 𝑛-gon.

The total number of interior intersection points of diagonals of the 𝑛-gon (assuming no 3 of the diagonals meet at a common point) is thus the number of ways that four integers can be selected (without regard to order) from a list of 𝑛 distinct integers. Therefore,

, which for the decagon gives 𝑁 (10) = 210.

A.2.8 2002 Session 1: 1. Let 𝑃 ≠ (0, 0) be a point on the parabola 𝑦 = 𝑥 2 . The normal line to the parabola at 𝑃 will intersect the parabola at another point, say 𝑄. Find the coordinates of 𝑃 so that the average of the 𝑦-coordinates of 𝑃 and 𝑄 is a minimum. Solution. Let 𝑃 = (𝑝, 𝑝 2 ) and 𝑄 = (𝑞, 𝑞 2 ). The slope of the tangent line to the parabola at 𝑃 is 2𝑝, so the slope of the normal line to the parabola at 𝑃 is -1/2𝑝. Thus, the equation of the normal line to the parabola at 𝑃 is

Since 𝑄 = (𝑞, 𝑞 2 ) lies on the normal line,

.

Solving for 𝑞, we obtain

. Now the average of the 𝑦-coordinates of 𝑃 and 𝑄 is

Differentiating this quantity with respect to 𝑝, we obtain

Setting this quantity equal to 0, we arrive at

Therefore, the coordinates of 𝑃 so that the average of the 𝑦-coordinates of 𝑃 and 𝑄 is a minimum are (2 -3/4 , 2 -3/2 ).

A.2.9 2007 Session 1: 4. A triangle is Pythagorean if it is a right triangle and the lengths of all of its sides are integers.

Suppose that △𝐴𝐵𝐶 is Pythagorean; for concreteness assume that the lengths of the three sides satisfy 𝑐 > 𝑎 > 𝑏. The median and the altitude are now drawn from 𝐶 to the hypotenuse, where they meet the latter at 𝑃, 𝑄, respectively. Determine simple conditions upon 𝑎, 𝑏, 𝑐 so that △𝐶𝑄𝑃 will also be Pythagorean.

We have

Next,

.

Finally,

.

Summarizing:

Equivalent formulations are possible.

A.2.10 2007 Session 2: 2. Find the point on a given line such that the sum of its distances from two fixed points is a minimum. Assume the two fixed points and the given line are in the same plane. Solution. If one of the two points is on the line, then the solution is clearly that point. It is also clear that if the two points are on opposite sides of the line, then the solution is the intersection of the line with the segment connecting the points. Now assume the two points are on the same side of the line. Because we are interested in the sum of distances from two fixed points, consider the family of ellipses with foci at the two points, and denote the foci 𝑃 1 and 𝑃 2 . If an ellipse in this family intersects the line in two points, say 𝐴 and 𝐵, then from the definition of an ellipse, the sum of the distances from 𝐴 to the foci equals the sum of the distances from 𝐵 to the foci. If a smaller ellipse in the family also intersects the line in two points, the same thing happens, except that the sum of the distances is less for the smaller ellipse. Therefore, if we continue to shrink the ellipse until it is tangent to the line, the point of tangency will be where the required minimum occurs.

A. 𝑥 0 = 1, and for all 𝑖 ≥ 0, 𝑥 𝑖+1 ≤ 𝑥 𝑖 .

(a) Prove that for every such sequence, there is an 𝑛 ≥ 1 such that

(b) Find such a sequence for which

Solution. (a) We will prove the series (1)

has sum ≥ 4 (with the obvious convention that this holds if the series diverges). This clearly implies that some partial sum of the series is ≥ 3.999.

Let 𝐿 be the inf ( = greatest lower bound) of the sums of all series of the form (1). Clearly 𝐿 > 1 since the first term

For any 𝜖 > 0, we can find a sequence {𝑥 𝑛 } such that

The series on the right side of (2) can be written in the form 1

By the definition of 𝐿, the series in parentheses has sum ≥ 𝐿. Hence from (2) we have

Solution. If 𝑆 is the required sum, then we have

, we obtain successively

,

In attempting to see how sequences of length 𝑛 might be derived from shorter ones, one might notice that a sequence of length 𝑛 can be obtained by attaching either 𝑏𝑎 or 𝑐𝑎 at the end of any sequence of length 𝑛 -2. For example,

Of course, all the sequences generated in this way have the letter 𝑎 in the third-last position. Conversely, a sequence of length 𝑛 whose third-last term is the letter 𝑎 yields an acceptable sequence of length 𝑛 -2 when its last two terms are dropped. Thus the number of sequences of length 𝑛 in which the third-last term is the letter 𝑎 is 2𝑡 𝑛-2 .

For the rest of the sequences of length 𝑛, the third-last term is either 𝑏 or 𝑐, and each sequence of this kind provides a single sequence of length 𝑛 -1 by simply deleting its second-last term:

(This is not allowed when the third-last term is the letter 𝑎.) Conversely, there is only one possible letter that can be inserted between the last two terms of a sequence, and doing so clearly extends one of length 𝑛 -1 to one of length 𝑛.

Therefore there are 𝑡 𝑛-1 sequences of length 𝑛 in which the third-last term is 𝑏 or 𝑐, and we have altogether that

with initial conditions 𝑡 1 = 1 and 𝑡 2 = 0. Now assume 𝑡 𝑛 = 𝑟 𝑛 . Substituting into the difference equation yields the characteristic equation 𝑟 2 -𝑟 -2 = 0, from which 𝑟 = 2 or 𝑟 = -1. So the general solution is

Using the initial conditions lets us find 𝑐 1 = 1/6 and 𝑐 2 = -2/3. Therefore,

A. A Lagrange multiplier solution requires solving the five equations

The second and third equations have only the trivial solution unless 𝜆 2 = 4, and the first and fourth equations have no solution if 𝜆 2 = 4, so we must have 𝑦 = 𝑧 = 0 and either 𝑥 = 0 or 𝑡 = 0. The two possible solutions are (1, 0, 0, 0) and (0, 0, 0, 2). Because 𝐹 (1, 0, 0, 0) = 4 and 𝐹 (0, 0, 0, 2) = 1, the minimum distance is 1 and occurs when 𝑋 = 0 0 0 2 .

A.3.9 2006 Session 1: 4. Find all positive integers 𝑐 such that 𝑛(𝑛 + 𝑐) is never a perfect square for any positive integer 𝑛. Solution. Write 𝑛(𝑛 + 𝑐) = 𝑚 2 ; then 𝑛 2 + 𝑐𝑛 -𝑚 2 = 0, and

.

For 𝑛 to be integral, it is necessary that 𝑐 2 + 4𝑚 2 be a perfect square. Let us consider the situation when then

so no members in this class are acceptable answers. Case 2. 𝑐 = 2 𝑟 , 𝑟 ≥ 3. Choose 𝑛 = 2 𝑟 -3 ; then

so no members in this class are acceptable answers. Case 3. 𝑐 = 2 𝑟 𝑠, 𝑠 odd, 𝑠 > 1. Choose

, so no members in this class are acceptable answers. In summary, the set of integers 𝑐 requested is {1, 2, 4}.

(b) Evaluation of 𝐽 . We define:

Introduce the substitution:

Rewriting the integral:

Using the change of variables:

This shows that:

By adding the two integrals:

Since the denominators are identical, the sum of the fractions equals 1:

B.1.2 1997 Session 2: 4 Correct Reasoning, Correct Final Answer. To determine the limit of the length of 𝑆 𝑛 as 𝑛 → ∞, we analyze the arc length of the curve

Step 1: Compute the Arc Length Formula. The arc length of a function 𝑦 = 𝑓 (𝑥) from 𝑥 = 𝑎 to 𝑥 = 𝑏 is given by:

For our function,

we compute its derivative:

Thus,

Step 2: Evaluate the Arc Length Integral. The arc length integral becomes:

For large 𝑛, observe that cos 2 (𝑛 2 𝑥) oscillates between 0 and 1 rapidly. However, we approximate:

.

This approximation is justified by averaging over oscillations, replacing cos 2 (𝑛 2 𝑥) by its mean value 1 2 . This gives:

.

Step 3: Find the Limit. For large 𝑛,

Thus,

.

Since this expression grows indefinitely as 𝑛 → ∞, we conclude:

B.1.3 1998 Session 1: 3 Correct Reasoning, Incorrect Final Answer. We aim to find a formula for the integral:

Step 1: Change of Variables. Introduce a change of variables:

Then,

Rewriting the powers in terms of 𝑡:

Thus, our integral transforms as:

Rearrange the factors:

Step 2: Recognizing the Beta Function. The integral

is the Beta function:

Since Γ(𝑘) = (𝑘 -1)! for positive integers 𝑘, we get:

Canceling 𝑚!𝑛!:

Step 3: Evaluating ∫ 1 0 (1 -𝑥 2 ) 𝑛 𝑑𝑥. We substitute 𝑎 = 0, 𝑏 = 1, and let 𝑚 = 𝑛, so we need:

Using our formula with 𝑚 = 𝑛:

Thus, the final result for the given integral is:

B.1.4 1999 Session 2: 2 Incorrect Reasoning, Incorrect Final Answer. We seek to evaluate the limit: lim

Step 1: Partial Fraction Decomposition. We factor the denominator:

We seek a partial fraction decomposition:

Multiplying both sides by (4𝑛 -1)(4𝑛 + 1), we obtain:

Setting 𝑛 = 1 4 and 𝑛 = -1 4 , we solve:

Solving for 𝐴 and 𝐵, we find:

.

Thus, we rewrite the fraction as:

Step 2: Evaluating the Sum. Rewriting the sum:

This forms a telescoping series, so most terms cancel, leaving:

Step 3: Taking the Limit. Multiplying by -2 and adding 1:

Taking 𝑁 → ∞, we note that 1 4𝑁 +1 → 0, leaving:

.

Conclusion. Thus, the desired limit is: Find all possible continuous and differentiable curves 𝐶 which have the following properties: 1. The curve 𝐶 lies in the first quadrant and contains the point (0, 0). 2. Whenever 𝑃 is on 𝐶, the interior of the rectangle 𝑅 bounded by the coordinate axes and the horizontal and vertical lines through 𝑃 is separated into two parts by 𝐶. 3. When the part adjacent to the 𝑥-axis is rotated about the 𝑥-axis and the part adjacent to the 𝑦-axis is rotated about the 𝑦-axis, two solids of equal volume are generated.

Let the curve 𝐶 be given by 𝑦 = 𝑓 (𝑥), where 𝑓 (𝑥) is a continuous and differentiable function in the first quadrant and passes through the origin, i.e., 𝑓 (0) = 0.

Step 1: Interpretation of the Condition. For any point 𝑃 (𝑎, 𝑏) on 𝐶, consider the rectangle 𝑅 with sides along the coordinate axes and the lines 𝑥 = 𝑎 and 𝑦 = 𝑏. The area is divided into two regions:

(1) Region 1: Bounded by the 𝑥-axis, the vertical line 𝑥 = 𝑎, and the curve.

(2) Region 2: Bounded by the 𝑦-axis, the horizontal line 𝑦 = 𝑏, and the curve.

Each of these regions is rotated about its respective axis to form solids of revolution, which have equal volumes.

Step 2: Volumes of Revolution. 1. Region 1 rotated about the 𝑥-axis:

Using the shell method, the volume is:

where 𝑥 (𝑦) is the inverse function of 𝑓 (𝑥).

The condition that these two volumes are equal gives:

Canceling 𝜋, we obtain:

Step 3: Change of Variables. Let 𝑦 = 𝑓 (𝑥), so that 𝑑𝑦 = 𝑓 ′ (𝑥) 𝑑𝑥. The equation becomes:

By the inverse function property, if 𝑥 = 𝑔(𝑦), then 𝑔(𝑦) = 𝑥 (𝑦) and 𝑔(𝑓

Thus, we rewrite:

Step 4: Differentiation. Differentiating both sides with respect to 𝑎 using the Fundamental Theorem of Calculus:

Dividing by 𝑓 (𝑎) (assuming 𝑓 (𝑎) ≠ 0 for 𝑎 > 0):

Setting 𝑦 = 𝑓 (𝑎), we obtain 𝑦 = 2𝑓 -1 (𝑦), or

.

Taking inverses on both sides:

Step 5: Verification. If 𝑓 (𝑥) = 2𝑥, then:

.

Similarly, computing 𝑉 𝑦 using 𝑥 = 𝑦 2 :

.

Since 𝑉 𝑥 = 𝑉 𝑦 , this function satisfies the given condition.

Thus, the unique solution is:

B.1.6 2002 Session 1: 3 Incorrect Reasoning, Incorrect Final Answer. We are given the sequence {𝑥 𝑛 } where the terms alternate between square roots of 5 and 13, starting with 𝑥 1 = √ 5. The goal is to find the limit of this sequence as 𝑛 → ∞. To approach this problem, we aim to find a pattern or recurrence that can help us compute the limit.

Step 1: Define the Sequence Recursively We start by defining the sequence more explicitly:

, and so on.

The sequence alternates between involving square roots of 5 and 13, and we observe that the structure of each term involves nested square roots. This suggests the possibility of the sequence converging to a fixed point.

Step 2: Investigate the Limit Let us assume that the sequence converges to some limit 𝐿 as 𝑛 → ∞. If the sequence converges, then the terms of the sequence for large 𝑛 should approach 𝐿. Hence, we assume that for large 𝑛, 𝑥 𝑛 ≈ 𝐿.

Now consider the recurrence relation for the even and odd terms separately.

-For the odd-indexed terms 𝑥 2𝑘 -1 , we have:

-For the even-indexed terms 𝑥 2𝑘 , we have:

Step 3: Solve the System of Equations Assume that the sequence converges to 𝐿. Then, for large 𝑘, both 𝑥 2𝑘 -1 and 𝑥 2𝑘 should be close to 𝐿. Thus, we can set up the following system of equations:

We now solve these equations.

Solve 𝐿 = √ 5 + 𝐿: Squaring both sides:

Rearrange the terms:

Solve this quadratic equation using the quadratic formula:

Thus, the two possible solutions are:

.

Since 𝐿 must be positive, we take the positive root:

13 + 𝐿: Squaring both sides:

Rearrange the terms:

Solve this quadratic equation using the quadratic formula:

Thus, the two possible solutions are:

Since 𝐿 must be positive, we take the positive root:

Step 4: Conclude the Limit We find that the two equations give different values for 𝐿, which means there is no single solution that satisfies both equations simultaneously. Therefore, the sequence does not converge to a single limit in the traditional sense. Thus, we conclude that: Understanding the Geometry. We have two cylinders, both with radius 𝑟 , intersecting at right angles. Let’s assume the cylinders are aligned along the 𝑥-and 𝑦-axes in 3D space. The equation of the first cylinder, which is aligned along the 𝑥-axis, is:

The equation of the second cylinder, aligned along the 𝑦-axis, is:

These two cylinders intersect in a shape known as a Bicylinder.

Flattening One of the Curved Faces. We need to imagine one of the curved faces of the intersection being “unwrapped” or “flattened” onto a plane. We will consider the face corresponding to the portion of the first cylinder 𝑦 2 + 𝑧 2 = 𝑟 2 in the region where the second cylinder intersects it. To do this, we will work in the 𝑦𝑧-plane, and observe that the boundary curve of this face will be part of the intersection of the two cylinders. By substituting the equation of the second cylinder 𝑥 2 + 𝑧 2 = 𝑟 2 into the first cylinder’s equation, we will determine the boundary of the flattened face.

Step 1: Find the Equations of the Boundary Curves. The boundary of the face we are flattening corresponds to the projection of the curve of intersection onto the 𝑦𝑧-plane. From the geometry of the cylinders, we can see that the boundary of the curved face involves the intersection of the two cylinder equations. We already have the two cylinder equations:

At the boundary of the flattened face, we are dealing with the region where 𝑥 = 0, since this is the part of the intersection of the cylinders projected onto the 𝑦𝑧-plane.

When 𝑥 = 0, the second cylinder’s equation becomes:

This gives the boundary for 𝑧.

Substituting this into the first cylinder’s equation, we get:

Thus, the boundary curve of the flattened face lies along the lines where 𝑦 = 0 and 𝑧 = ±𝑟 .

Step 2: Find the Area of the Flattened Face. The area of the flattened face is simply the area of a quarter circle, because the boundary corresponds to a quarter of a circle with radius 𝑟 .

The area of a full circle is 𝜋𝑟 2 . Since we have a quarter circle, the area is:

Final Answer. The equations of the boundary curves of the flattened face are 𝑦 = 0 and 𝑧 = ±𝑟 . The area of the flattened face is

tep 1: Express 𝑥 𝑛 explicitly First, we isolate 𝑥 𝑛 in the given equation:

Step 2: Investigate the behavior of

As 𝑛 → ∞, we know that

Using the approximation for small 𝑧, 2 𝑧 ≈ 1 + 𝑧 ln 2 for 𝑧 close to 0, we get

Step 3: Substitute the asymptotic approximation into the equation for 𝑥 𝑛 Substitute this approximation into the expression for 𝑥 𝑛 :

Step 4: Conclusion Therefore, the limit of 𝑥 𝑛 as 𝑛 → ∞ is:

B.1.9 2005 Session 1: 4 Incorrect Reasoning, Incorrect Final Answer. To solve the integral

let’s begin by examining the integrand.

Step 1: Simplifying the denominator. The denominator 1 -2𝑥 cos 𝜃 + 𝑥 2 can be rewritten as a perfect square:

Thus, the integral becomes:

Step 2: Substitution for simpler form. A standard substitution to simplify this form is:

When 𝑥 = -1, 𝑢 = -1cos 𝜃 , and when 𝑥 = 1, 𝑢 = 1cos 𝜃 . This gives the transformed integral:

Step 3: Evaluating the integral. Now we recognize the integral as a standard form. The integral of 1 𝑢 2 +𝑎 2 is:

, where 𝑎 = sin 𝜃 . Applying this to the limits of integration, we have:

Simplifying, we get:

Step 4: Using the arctangent subtraction identity. The difference of two arctangents can be simplified using the identity:

Applying this identity, we get:

.

Simplifying the numerator:

Simplifying the denominator:

First, expand the product:

Thus, the denominator becomes:

Therefore, the integral simplifies to:

Step 5: Discontinuities. The function 𝐼 (𝜃 ) will be discontinuous at points where the argument inside the arctangent function undergoes a jump, particularly where the denominator in the original integral 1 -2𝑥 cos 𝜃 + 𝑥 2 = 0 for some

This occurs when the quadratic equation has a double root. Specifically, for the equation (𝑥cos 𝜃 ) 2 + sin 2 𝜃 = 0, we need:

which has a solution only when sin 𝜃 = 0, or 𝜃 = 0, 𝜋, 2𝜋.

Thus, 𝐼 (𝜃 ) is discontinuous at 𝜃 = 0, 𝜋, 2𝜋.

Final Answer. The value of the integral is: 4 , and 𝑓 (𝑥) has a minimum value on the interval (0, ∞). To solve this problem, we first express the two given quantities (area under the curve and arclength) in mathematical terms:

1. Area under the curve:

The area under the curve from 𝑥 = 𝑎 to 𝑥 = 𝑏 is given by the integral:

The arclength of the curve from 𝑥 = 𝑎 to 𝑥 = 𝑏 is given by the formula:

We are told that the area under the curve is equal to the arclength, so we have the equation:

To understand the relationship between 𝑓 (𝑥) and its derivative, differentiate both sides of the equation with respect to 𝑏:

5. Solve for 𝑓 ′ (𝑏) :

Squaring both sides of the equation:

Rearranging:

Taking the square root:

At the minimum point of 𝑓 (𝑥), the derivative 𝑓 ′ (𝑥) is zero. Therefore, we set 𝑓 ′ (𝑥) = 0 in the equation:

When 𝑓 ′ (𝑥) = 0, we have:

which implies:

Thus, the minimum value of 𝑓 (𝑥) is 1.

The minimum value of 𝑓 (𝑥) on the interval (0, ∞) is 1 . Thus, the slope of the tangent line at 𝑃 (𝑎, 𝑎 2 ) is 2𝑎. The normal line is perpendicular to the tangent, so its slope is:

.

Using the point-slope form of a line:

which simplifies to:

.

Finding the Intersection with the Parabola. Setting this equal to 𝑦 = 𝑥 2 :

.

Multiplying by 2𝑎 to clear fractions:

Finding the Other Intersection Point. Since one root of the quadratic equation is 𝑥 = 𝑎, let the other root be 𝑥 = 𝑏. Using Vieta’s formulas:

.

Solving for 𝑏:

The corresponding 𝑦-coordinate of 𝑄 is:

Expanding:

Minimizing 𝑦 𝑄 . Define:

Solving numerically, we find:

Thus, the coordinates of 𝑃 that minimize 𝑦 𝑄 are:

.

1 Incorrect Reasoning, Incorrect Final Answer. Let 𝑃 (𝑎, 𝑎 2 ) be a point on the parabola 𝑦 = 𝑥 2 , where

Step 1: Equation of the Normal Line. The derivative of the parabola is:

Thus, the slope of the tangent line at 𝑃 (𝑎, 𝑎 2 ) is 2𝑎, so the slope of the normal line (which is perpendicular to the tangent) is:

The equation of the normal line passing through 𝑃 (𝑎, 𝑎 2 ) is given by the point-slope form:

which simplifies to:

Step 2: Finding the Intersection Point 𝑄. To find the second intersection 𝑄 (𝑏, 𝑏 2 ), substitute 𝑦 = 𝑏 2 into the normal line equation:

Rearrange to form a quadratic equation:

Since 𝑏 = 𝑎 is a known root, factorizing the quadratic gives:

Solving for the second root using Vieta’s formulas:

Solving for 𝑏 2 :

Step 3: Length of 𝑃𝑄. The distance between 𝑃 (𝑎, 𝑎 2 ) and 𝑄 (𝑏 2 , 𝑏 2 2 ) is given by:

Squaring:

The total squared distance is:

Minimizing 𝑃𝑄 2 as a function of 𝑎, differentiation and solving 𝑑 𝑑𝑎 𝑃𝑄 2 = 0 gives 𝑎 = 1 2 .

Step 4: Conclusion. Substituting 𝑎 = 1 2 , we get the optimal point:

B.2.3 1999 Session 1: 1 Incorrect Reasoning, Incorrect Final Answer. We will analyze the problem by finding the normal line to the parabola 𝑦 = 𝑥 2 at a given point 𝑃, determining its second intersection 𝑄 with the parabola, and then minimizing the arc length of the parabola between these two points.

Step 1: Find the Normal Line at 𝑃 Let 𝑃 = (𝑎, 𝑎 2 ) be a point on the parabola.

-The derivative of 𝑦 = 𝑥 2 is:

So, the slope of the tangent line at 𝑃 is 2𝑎. -The normal line is perpendicular to this, meaning its slope is:

-The equation of the normal line at 𝑃 is:

Simplifying:

Step 2: Find the Second Intersection 𝑄 The second intersection 𝑄 satisfies the system:

Multiplying everything by 2𝑎 to clear the fraction:

This quadratic equation in 𝑥 has roots corresponding to 𝑃 and 𝑄. Since 𝑃 has 𝑥-coordinate 𝑎, the other root satisfies:

-𝑎

Step 3: Find the Arc Length The arc length of the parabola from 𝑃 to 𝑄 is:

To minimize this, we apply symmetry arguments. The integral form suggests a natural minimization when 𝑃 and 𝑄 are symmetric about the origin, meaning 𝑎 should satisfy:

which has no real solutions.

Instead, using symmetry and known results, the minimum arc length occurs when 𝑃 is at (1, 1), leading to 𝑄 = (-1, 1). Thus, the optimal 𝑃 is:

(1, 1)

B.2.4 1999 Session 2: 1 Incorrect Reasoning, Incorrect Final Answer. We are tasked with finding the limiting position of the 𝑥-intercept of a line through the point 𝑃 = (0, 𝑎) on the circle 𝐶 𝑎 with radius 𝑎 and center at (0, 0), and the intersection of the semicircle 𝑆𝐶 and the circle 𝐶 𝑎 , as 𝑎 → 0.

Step 1: Equation of the semicircle 𝑆𝐶. The semicircle 𝑆𝐶 is centered at (1, 0) with radius 1. Its equation, for 𝑦 ≥ 0, is:

This describes the upper half of a circle with center at (1, 0) and radius 1.

Step 2: Equation of the circle 𝐶 𝑎 . The circle 𝐶 𝑎 has radius 𝑎 and is centered at (0, 0). Its equation is:

Step 3: Finding the intersection points of 𝑆𝐶 and 𝐶 𝑎 . We now solve the system of equations formed by the semicircle and the circle to find their intersection points.

1. The equation of the semicircle: (𝑥 -1) 2 + 𝑦 2 = 1 2. The equation of the circle:

Substitute 𝑦 2 = 𝑎 2 -𝑥 2 from the circle’s equation into the semicircle’s equation:

Expand the terms:

Substitute 𝑥 = 𝑎 2 2 into the equation of the circle to find 𝑦:

Thus, the intersection points of the semicircle and the circle 𝐶 𝑎 are 𝑎 2 2 , √︃ 𝑎 2 -𝑎 4 4 and its reflection across the 𝑥-axis

Step 4: The line through 𝑃 = (0, 𝑎) and the intersection point. The line through 𝑃 = (0, 𝑎) and the intersection point

The equation of the line is then

Step 5: Finding the 𝑥-intercept. The 𝑥-intercept occurs when 𝑦 = 0. Setting 𝑦 = 0 in the equation of the line:

Step 6: Taking the limit as 𝑎 → 0. As 𝑎 → 0, we have

Thus, the expression for the 𝑥-intercept becomes 𝑥 ≈ -

Therefore, the limiting position of the 𝑥-intercept of the line as 𝑎 → 0 is -∞.

1 Correct Reasoning, Correct Final Answer. We are tasked with finding the point 𝑃 = (𝑎, 𝑎 2 ) on the parabola 𝑦 = 𝑥 2 such that the distance between the 𝑥-coordinates of 𝑃 and 𝑄 is minimized, where 𝑄 is the other point of intersection of the normal line to the parabola at 𝑃.

Step 1: Equation of the normal line at point 𝑃. The equation of the parabola is 𝑦 = 𝑥 2 . The derivative of this equation, 𝑑𝑦 𝑑𝑥 , gives the slope of the tangent line at any point on the parabola. We compute the derivative:

At the point 𝑃 = (𝑎, 𝑎 2 ), the slope of the tangent line is 2𝑎. The slope of the normal line is the negative reciprocal of the tangent slope, which is:

.

Using the point-slope form of the equation of the line, the equation of the normal line at 𝑃 = (𝑎, 𝑎 2 ) is:

Simplifying this equation:

Thus, the equation of the normal line is:

.

Step 2: Finding the other point of intersection 𝑄. To find the other point of intersection 𝑄, we substitute the equation of the normal line into the equation of the parabola 𝑦 = 𝑥 2 . Set:

.

Rearranging terms:

This is a quadratic equation in 𝑥:

Let the roots of this quadratic equation be 𝑥 = 𝑎 and 𝑥 = 𝑏, where 𝑏 is the 𝑥-coordinate of the other intersection point 𝑄. The quadratic equation can be factored as:

Expanding this product gives:

Comparing the coefficients of this equation with the original quadratic equation 𝑥 2 + 1 2𝑎 𝑥 -𝑎 2 + 1 2 = 0, we obtain the following system of equations:

and 𝑎𝑏 = -𝑎 2 + 1 2

.

Step 3: Solving for the distance between the 𝑥-coordinates of 𝑃 and 𝑄. We are interested in minimizing the distance between the 𝑥-coordinates of 𝑃 and 𝑄, which is |𝑎 -𝑏|. From the equation 𝑎 + 𝑏 = -1 2𝑎 , we solve for 𝑏:

Thus, the distance between the 𝑥-coordinates of 𝑃 and 𝑄 is:

.

We want to minimize 2𝑎 + 1 2𝑎 .

Step 4: Minimizing the expression. Let 𝑓 (𝑎) = 2𝑎+ 1 2𝑎 . To minimize |𝑓 (𝑎)|, we first find the critical points by differentiating 𝑓 (𝑎):

Setting 𝑓 ′ (𝑎) = 0 to find the critical points:

Step 5: Conclusion. The distance is minimized when 𝑎 = 1 2 or 𝑎 = -1 2 . Therefore, the coordinates of 𝑃 are 1 2 , 1 4 or - 1 2 , 1 4 .

1 Incorrect Reasoning, Incorrect Final Answer. To find the locus of points that are equidistant from two concentric circles, we proceed as follows.

Let the two circles, 𝐴 and 𝐵, have the following properties:

• Circle 𝐴 has center 𝑂 𝐴 and radius 𝑟 𝐴 ,

• Circle 𝐵 has center 𝑂 𝐵 and radius 𝑟 𝐵 ,

• Circle 𝐵 lies wholly inside Circle 𝐴.

The goal is to find the locus of points that are equidistant from both Circle 𝐴 and Circle 𝐵. A point 𝑃 on the locus will satisfy the condition that its distance to the boundary of Circle 𝐴 equals its distance to the boundary of Circle 𝐵.

Step 1: General coordinates and distances. Let the center of both circles be the same point, 𝑂, which simplifies the problem to concentric circles. Denote the distance from point 𝑃 to the center 𝑂 as 𝑟 𝑃 , and consider the distances from 𝑃 to the circumferences of circles 𝐴 and 𝐵:

• The distance from 𝑃 to the boundary of Circle 𝐵 is |𝑟 𝑃 -𝑟 𝐵 |.

Step 2: Equating the distances. For 𝑃 to be equidistant from the two circles, we set these two distances equal:

Step 3: Solving the equation. We now analyze this equation in two cases based on the absolute value condition:

Case 1: 𝑟 𝑃 ≥ 𝑟 𝐴 and 𝑟 𝑃 ≥ 𝑟 𝐵 . In this case, the equation becomes:

Simplifying:

This is a contradiction since the problem states that Circle 𝐵 is inside Circle 𝐴, implying 𝑟 𝐴 > 𝑟 𝐵 .

Case 2: 𝑟 𝑃 < 𝑟 𝐴 and 𝑟 𝑃 < 𝑟 𝐵 . In this case, the equation becomes:

Simplifying:

Again, this is a contradiction.

Case 3: 𝑟 𝐴 ≤ 𝑟 𝑃 < 𝑟 𝐵 . This is the key case. The solution simplifies as follows:

Rearranging:

.

Step 4: Conclusion. The locus of points equidistant from the two circles is a circle with radius 𝑟 𝐴 +𝑟 𝐵 Step 1: Understanding the Problem. We are given a convex decagon, which has 10 sides, and all of its diagonals are drawn. The problem specifies that no three diagonals share a common interior point. This means that each interior intersection is formed by exactly two diagonals intersecting inside the decagon.

Step 2: Determining the Number of Diagonals. For any convex polygon with 𝑛 sides, the number of diagonals can be calculated using the formula:

Number of diagonals = 𝑛(𝑛 -3) 2 For a decagon, where 𝑛 = 10:

Thus, there are 35 diagonals in the decagon.

Step 3: Finding the Number of Interior Intersection Points. The number of interior points of intersection of the diagonals in a convex polygon is given by the combination formula for choosing 4 vertices from the 𝑛 vertices of the polygon. This is because each interior intersection point is formed by the diagonals connecting four distinct vertices.

The number of ways to choose 4 vertices from 10 vertices is:

Thus, the number of interior points of intersection of the diagonals in a convex decagon is 210 . where 𝑎 is the 𝑥-coordinate of 𝑃. The slope of the tangent line at 𝑃 is 2𝑎, and the slope of the normal line is -1 2𝑎 . The equation of the normal line at 𝑃 is:

(𝑥 -𝑎)

Simplifying:

To find the point 𝑄 where the normal line intersects the parabola, we set the equation of the normal line equal to the equation of the parabola 𝑦 = 𝑥 2 :

Rearranging:

The roots of this quadratic equation are 𝑥 1 = 𝑎 and 𝑥 2 , the 𝑥-coordinate of 𝑄. By Vieta’s formulas, the sum of the

The average of the 𝑦-coordinates of 𝑃 and 𝑄 is:

To minimize this expression, we take the derivative with respect to 𝑎 and solve for 𝑎.

B.2.9 2005 Session 1: 4 Incorrect Reasoning, Incorrect Final Answer. We are given a Pythagorean triangle △𝐴𝐵𝐶 with side lengths satisfying 𝑐 > 𝑎 > 𝑏, where 𝑐 is the hypotenuse. The triangle is right-angled at 𝐵, so 𝑎 2 + 𝑏 2 = 𝑐 2 . The median and altitude are drawn from 𝐶 to the hypotenuse 𝐴𝐵, meeting the hypotenuse at points 𝑃 and 𝑄, respectively.

We seek conditions on 𝑎, 𝑏, 𝑐 such that △𝐶𝑄𝑃 is also a Pythagorean triangle.

Step 1: Geometric Setup. -Let △𝐴𝐵𝐶 be a right triangle with right angle at 𝐵, so 𝑐 is the hypotenuse. -The median 𝐶𝑃 joins 𝐶 to the midpoint 𝑃 of 𝐴𝐵. -The altitude 𝐶𝑄 is perpendicular from 𝐶 to 𝐴𝐵, meeting it at 𝑄.

Using known formulas:

We aim to satisfy the Pythagorean theorem for △𝐶𝑄𝑃:

Step 2: Analyze the Side Lengths of △𝐶𝑄𝑃. Using coordinate geometry, position the points as follows:

Finding 𝑃 (Midpoint of 𝐴𝐵):

Finding 𝑄 (Foot of Perpendicular from 𝐶 to 𝐴𝐵):

.

Step 3: Checking the Pythagorean Condition. For △𝐶𝑄𝑃 to be a Pythagorean triangle, we solve:

After simplification, the necessary condition for this to hold is:

Conclusion. Thus, △𝐶𝑄𝑃 is a Pythagorean triangle if and only if:

B.2.10 2005 Session 2: 5 Correct Reasoning, Incorrect Final Answer.

Problem Statement. Given two fixed points 𝐴 and 𝐵 in a plane and a given line 𝐿, find the point 𝑃 on 𝐿 such that the sum of distances 𝑃𝐴 + 𝑃𝐵 is minimized.

To solve this problem, we employ the method of reflection.

Step 1: Reflection of One Point. We reflect one of the fixed points, say 𝐵, across the given line 𝐿. Let the reflection of 𝐵 be 𝐵 ′ . The point 𝐵 ′ is the image of 𝐵 such that the line 𝐿 is the perpendicular bisector of the segment 𝐵𝐵 ′ .

Step 2: Minimizing the Sum of Distances. Since the shortest path between two points is a straight line, the minimal path from 𝐴 to 𝐵 via a point 𝑃 on 𝐿 must be the direct path from 𝐴 to 𝐵 ′ , intersecting 𝐿 at some point 𝑃. Thus, the optimal point 𝑃 is the intersection of the line segment 𝐴𝐵 ′ with the given line 𝐿.

Therefore, the required point 𝑃 that minimizes 𝑃𝐴 + 𝑃𝐵 is given by the intersection of the line passing through 𝐴 and 𝐵 ′ with the line 𝐿.

B.3.1 1996 Session 1: 3 Correct Reasoning, Incorrect Final Answer. We are given that each of the numbers 𝑥 1 , 𝑥 2 , . . . , 𝑥 𝑛 can be 1, 0, or -1, and we are tasked with finding the minimum possible value of the sum of all products of pairs of these numbers. Specifically, we are asked to minimize the following sum:

Step 1: Expression for the sum The given sum 𝑆 involves all distinct pairs of indices 𝑖 and 𝑗 where 𝑖 < 𝑗. We can express this sum as:

We aim to minimize 𝑆. To do this, let’s explore the relationship between the values of the 𝑥 𝑖 ’s and the sum 𝑆.

Step 2: Total sum of squares Consider the sum of the squares of the 𝑥 𝑖 ’s:

Since each 𝑥 𝑖 can be 1, 0, or -1, we have

Therefore, the sum of the squares is:

where 𝑘 is the number of non-zero values of 𝑥 𝑖 (the number of 𝑥 𝑖 ’s that are either 1 or -1).

Thus, we can rewrite the equation for the total sum of squares as:

Step 3: Minimizing the sum Let 𝑆 denote the sum we wish to minimize:

From the above equation, we can express 𝑆 as:

To minimize 𝑆, we need to minimize the expression 𝑛 𝑖=1 𝑥 𝑖 2 , which is the square of the sum of the 𝑥 𝑖 ’s.

The sum 𝑛 𝑖=1 𝑥 𝑖 will be smallest when the values of 𝑥 𝑖 are as balanced as possible, meaning that the number of 1’s and -1’s should be as equal as possible, and the sum should approach 0. Thus, the sum 𝑛 𝑖=1 𝑥 𝑖 is minimized when it is 0, which happens when half of the 𝑥 𝑖 ’s are 1 and half are -1, and the remaining 𝑥 𝑖 ’s are 0.

Therefore, the minimum value of 𝑆 is achieved when the sum of the 𝑥 𝑖 ’s is 0, and the minimum value of 𝑆 is:

3.2 1997 Session 2: 1 Incorrect Reasoning, Incorrect Final Anwer. We are tasked with finding positive integers 𝑛 and 𝑎 1 , 𝑎 2 , . . . , 𝑎 𝑛 such that

and the product 𝑎 1 𝑎 2 • • • 𝑎 𝑛 is as large as possible.

Step 1: To maximize the product, we aim to make the integers 𝑎 1 , 𝑎 2 , . . . , 𝑎 𝑛 as equal as possible. The product of numbers is maximized when the terms are as close as possible. We start by dividing 1997 by 3, since 3 is the integer that maximizes the product for a given sum.

1997 ÷ 3 = 665 remainder 2.

Thus, we can express 1997 as:

Step 2: We can now construct the integers 𝑎 1 , 𝑎 2 , . . . , 𝑎 𝑛 by setting:

Therefore, 𝑛 = 667, and the integers are composed of 665 threes and two ones.

Step 3: The product of these integers is:

Thus, the maximum product is 3 665 , and the solution is:

B.3.3 1997 Session 2: 5 (b only) Incorrect Reasoning, Incorrect Final Answer. We are tasked with finding a sequence {𝑥 𝑛 } of positive real numbers satisfying the following properties:

• For all 𝑖 ≥ 0, 𝑥 𝑖+1 ≤ 𝑥 𝑖 (the sequence is non-increasing),

• The inequality

𝑥 𝑛 < 4 for all 𝑛.

We will attempt to find a sequence that satisfies all these conditions.

Step 1: Understanding the Inequality

The inequality that we need to satisfy is:

< 4 for all 𝑛.

Since the sequence is non-increasing, each term in the sum is positive.

Step 2: Considering the Form of the Sequence Let’s assume the sequence follows a power law form, i.e., 𝑥 𝑛 = 𝐶 𝑛+1 for some constant 𝐶. This form has several advantages: -It satisfies the condition 𝑥 0 = 1 when 𝐶 = 1. -It is non-increasing, as 𝑥 𝑛 = 1 𝑛+1 decreases as 𝑛 increases.

Step 3: Substituting the Proposed Sequence

Substituting 𝑥 𝑖 = 1 𝑖+1 into the sum, we get:

This simplifies to:

Step 4: Analyzing the Series

We can split the sum into two parts:

The first sum is the harmonic series:

where 𝐻 𝑛 is the 𝑛-th harmonic number. The second sum is a known series:

This sum converges to 𝜋 2 6 as 𝑛 → ∞. Thus, the inequality becomes:

For large 𝑛, the harmonic number 𝐻 𝑛 behaves asymptotically as 𝐻 𝑛 ∼ ln(𝑛) + 𝛾, where 𝛾 is the Euler-Mascheroni constant.

Thus, the sum will not exceed 4 for sufficiently large 𝑛, implying that the sequence 𝑥 𝑛 = 1 𝑛+1 satisfies the inequality for all 𝑛. B.3.4 1998 Session 2: 3 Incorrect Reasoning, Incorrect Final Answer. We are asked to sum the series

Step 1: Simplify the General Term The general term is 36𝑖 2 + 1 (36𝑖 2 -1) 2 . The denominator can be factored as 36𝑖 2 -1 = (6𝑖 -1)(6𝑖 + 1), so the term becomes 36𝑖 2 + 1 ((6𝑖 -1)(6𝑖 + 1)) 2 .

Step 2: Decompose Using Partial Fractions Attempting partial fraction decomposition could simplify the sum further, but solving this by hand is quite complex. Instead, we can look up known results or use numerical methods.

Step 3: Final Result After applying known summation results, we find that the sum converges to 𝑆 = 𝜋 2 6 . Thus, the value of the infinite series is 𝑆 = 𝜋 2 6 . B.3.5 1999 Session 2: 5 Incorrect Reasoning, Incorrect Final Answer. Let 𝐴 𝑛 denote the 𝑛 × 𝑛 matrix whose (𝑖, 𝑗)-th entry is GCD(𝑖, 𝑗). We aim to compute det(𝐴 𝑛 ).

Step 1: Structure of the Matrix

The matrix 𝐴 𝑛 is given by:

This matrix is symmetric, as GCD(𝑖, 𝑗) = GCD( 𝑗, 𝑖), and each entry is a non-negative integer.

Step 2: Pattern for Small Values of 𝑛 For 𝑛 = 1, we have:

For 𝑛 = 2, we have:

, det(𝐴 2 ) = 1.

For 𝑛 = 3, we have:

Step 3: Generalization

From the small examples, we observe that the determinant of 𝐴 𝑛 appears to be 1 for all 𝑛. In fact, it is a known result that for any 𝑛, we have: det(𝐴 𝑛 ) = 1.

We are tasked with finding the number of sequences 𝑥 1 𝑥 2 • • • 𝑥 𝑛 where each 𝑥 𝑖 is one of 𝑎, 𝑏, or 𝑐, such that:

• The sequence begins and ends with the letter 𝑎,

• Adjacent terms are always different letters.

Step 1: Fix the first and last terms The sequence begins and ends with 𝑎, so we have:

Thus, the only terms we need to determine are 𝑥 2 , 𝑥 3 , . . . , 𝑥 𝑛-1 .

Step 2: Constraints on adjacent terms Adjacent terms must differ, so:

Since 𝑥 1 = 𝑎, the second term 𝑥 2 must be either 𝑏 or 𝑐.

Step 3: Alternating pattern After determining 𝑥 2 , each subsequent term alternates between 𝑏 and 𝑐, because adjacent terms must differ. Therefore, for each 𝑖 where 2 ≤ 𝑖 ≤ 𝑛 -1, 𝑥 𝑖 can be either 𝑏 or 𝑐, depending on the previous term.

Step 4: Number of valid sequences -For 𝑛 = 2, there is exactly one sequence: (𝑎, 𝑎). -For 𝑛 ≥ 3, the number of valid sequences is 2 𝑛-2 , because each of the 𝑛 -2 intermediate terms can be either 𝑏 or 𝑐.

Thus, the number of valid sequences is: Let the number of occurrences of -2, 1, and 2 in the sequence be 𝑎, 𝑏, and 𝑐 respectively. Thus, the sequence contains 𝑎 terms equal to -2, 𝑏 terms equal to 1, and 𝑐 terms equal to 2. The sequence has 𝑎 + 𝑏 + 𝑐 = 𝑛 elements in total.

Step 2: Use the Sum Constraints

From the conditions 𝑖 𝑥 𝑖 = 29 and 𝑖 𝑥 2 𝑖 = 59, we can write the following equations based on the values of 𝑥 𝑖 :

Step 3: Solve the System of Equations

We now solve the system of two equations: 1. -2𝑎 + 𝑏 + 2𝑐 = 29, 2. 4𝑎 + 𝑏 + 4𝑐 = 59.

First, subtract the first equation from the second:

Thus, we have:

Substitute this expression for 𝑐 into the first equation:

Step 4: Constraints on 𝑎, 𝑏, and 𝑐

We require that 𝑎, 𝑏, and 𝑐 be non-negative integers, so we need to check the possible values of 𝑎 such that both 𝑏 = 8𝑎 -1 and 𝑐 = 15 -3𝑎 are non-negative integers.

-For 𝑏 = 8𝑎 -1 ≥ 0, we must have 𝑎 ≥ 1. -For 𝑐 = 15 -3𝑎 ≥ 0, we must have 𝑎 ≤ 5.

Thus, 𝑎 can take values in the range 1 ≤ 𝑎 ≤ 5. For each 𝑎, we compute 𝑏 and 𝑐:

• We seek the minimum distance from 𝐴 to Σ, and a matrix 𝑆 ∈ Σ that realizes this minimum.

Step 1: Distance from 𝐴 to a general matrix 𝑋

Then,

Step 2: Constraint from Σ

The constraint for 𝑋 ∈ Σ is:

Step 3: Try a candidate matrix 𝑆 ∈ Σ

The minimum distance from 𝐴 to Σ is We want to determine for which values of 𝑐, this expression is never a perfect square for any positive integer 𝑛.

Suppose 𝑛(𝑛 + 𝑐) is a perfect square for some integer 𝑛. Then we have:

This leads to the equation:

We treat this as a quadratic equation in 𝑛:

The discriminant of this quadratic equation is:

For 𝑛 to be an integer, the discriminant must be a perfect square. Let’s set the discriminant equal to another perfect square, say 𝑚 2 , so we have:

This is a Diophantine equation of the form 𝑚 2 -4𝑘 2 = 𝑐 2 , which is a difference of squares:

Now, let’s consider the case where 𝑐 = 4. Substituting 𝑐 = 4 into the equation 𝑛(𝑛 + 𝑐) = 𝑛 2 + 4𝑛:

Rewriting this as:

the discriminant becomes:

For 𝑛 to be an integer, the discriminant must be a perfect square, so we need:

for some integer 𝑚. This is a difference of squares:

which factors as:

The integer factor pairs of 4 are (1, 4), (-1, -4), (2, 2), and (-2, -2). Solving each pair gives:

-𝑚 -𝑘 = 1 and 𝑚 + 𝑘 = 4 gives 𝑚 = 5 2 , which is not an integer. -𝑚 -𝑘 = -1 and 𝑚 + 𝑘 = -4 gives 𝑚 = -5 2 , which is not an integer. -𝑚 -𝑘 = 2 and 𝑚 + 𝑘 = 2 gives 𝑚 = 2 and 𝑘 = 0. -𝑚 -𝑘 = -2 and 𝑚 + 𝑘 = -2 gives 𝑚 = -2 and 𝑘 = 0. Thus, the only solution is 𝑘 = 0, which implies:

This only has the solution 𝑛 = 0, which is not a positive integer. Therefore, 𝑛(𝑛 + 4) cannot be a perfect square for any positive integer 𝑛.

Thus, the only solution is:

and we are tasked with finding the largest integer that does not belong to 𝑆. This is a classic application of the Frobenius coin problem. For two coprime integers 𝑚 and 𝑛, the largest integer that cannot be expressed as a nonnegative linear combination of 𝑚 and 𝑛 is given by the formula:

In this case, we have 𝑚 = 5 and 𝑛 = 503, which are coprime. Applying the formula, we compute:

𝑛 sin(𝑛 2 𝑥), 0 ≤ 𝑥 ≤ 𝜋 }. The arc length 𝐿 𝑛 is given by:

We have

As 𝑛 → ∞, 𝑛 2 cos 2 (𝑛 2 𝑥) dominates, so Thus,

Repeating this process 𝑚 times, we get

Part 2: Evaluating the Integral. We want to evaluate

Using the formula with 𝑎 = 0 and 𝑏 = 1, 𝑚 = 𝑛, and 𝑛 = -1/2, we have

C.1.4 1999 Session 2: 2 Correct Reasoning, Correct Final Answer. We want to find the limit:

First, we factor the denominator:

Using partial fraction decomposition:

The limit becomes:

We know that:

Therefore,

C.1.5 1999 Session 2: 4 Incorrect Reasoning, Incorrect Final Answer. Let the curve 𝐶 be defined by 𝑦 = 𝑓 (𝑥), where 𝑓 (𝑥) is a continuous and differentiable function. Since 𝐶 lies in the first quadrant and contains the point (0, 0), we have 𝑓 (0) = 0 and 𝑓 (𝑥) ≥ 0 for 𝑥 ≥ 0. Let 𝑃 = (𝑥, 𝑦) be a point on the curve 𝐶. The rectangle 𝑅 has vertices (0, 0), (𝑥, 0), (𝑥, 𝑦), and (0, 𝑦).

The volume of the solid generated by rotating the region adjacent to the 𝑥-axis about the 𝑥-axis is given by:

The volume of the solid generated by rotating the region adjacent to the 𝑦-axis about the 𝑦-axis is given by:

Since the volumes are equal, we have 𝑉 𝑥 = 𝑉 𝑦 . Thus,

Since 𝑦 = 𝑓 (𝑥), we can rewrite the second integral by substituting 𝑡 = 𝑓 (𝑢), 𝑑𝑡 = 𝑓 ′ (𝑢) 𝑑𝑢, and 𝑓 -1 (𝑡) = 𝑢:

Differentiating both sides with respect to 𝑥 using the Fundamental Theorem of Calculus:

Let 𝑦 = 𝑓 (𝑥), so 𝑦 ′ = 𝑓 ′ (𝑥). Then we have:

Integrating both sides:

Since 𝑓 (0) = 0, we consider the limit as 𝑥 approaches 0. As 𝑥 → 0, we have 𝑦 → 0, which is consistent with the initial condition.

For the volumes to be equal in the first quadrant, we need 𝐶 = 0, thus 𝑦 = 𝑥.

If 𝑦 = 𝑥, then 𝑓 (𝑥) = 𝑥 and 𝑓 -1 (𝑥) = 𝑥.

Therefore, the curve is 𝑦 = 𝑥.

C.1.6 2002 Session 1: 3 (limit only) Correct Reasoning, Correct Final Answer. Let {𝑥 𝑛 } be the following sequence:

and so on. We want to determine the value of lim 𝑛→∞ 𝑥 𝑛 .

Let 𝐿 = lim 𝑛→∞ 𝑥 𝑛 . If the limit exists, then we have:

Squaring both sides, we get:

Rearranging, we have:

Squaring again:

We observe that 𝐿 = 3 is a root:

Thus, (𝐿 -3) is a factor. Dividing 𝐿 4 -10𝐿 2 -𝐿 + 12 by (𝐿 -3), we get:

We check that 𝐿 = 3 satisfies the original equation:

Thus, lim 𝑛→∞ 𝑥 𝑛 = 3.

C.1.7 2002 Session 1: 5 Incorrect Reasoning, Incorrect Final Answer. Let the equations of the two cylinders be:

The intersection of the cylinders occurs when 𝑥 2 = 𝑦 2 , so 𝑦 = ±𝑥.

The intersection curve can be parameterized as:

for 0 ≤ 𝑡 ≤ 𝜋 2 in the first octant. When one of the curved faces is flattened, the boundaries are:

The area of one flattened face is:

Since there are four such faces, the total surface area of the intersection solid is 4

C.

Find lim 𝑛→∞ 𝑥 𝑛 .

We have

We know that 𝑛 √ 2 = 2 1/𝑛 = 𝑒 ln(2)/𝑛 . Using the Taylor series expansion for 𝑒 𝑥 around 𝑥 = 0, we have 𝑒 𝑥 = 1+𝑥+ 𝑥 2 2! + 𝑥 3 3! +• • • . Applying this to 𝑒 ln(2)/𝑛 , we get

Now, we can substitute this into the expression for 𝑥 𝑛 :

As 𝑛 → ∞, the term 𝑂 (1/𝑛) goes to 0, so lim

C.1.9 2005 Session 1: 4 Correct Reasoning, Correct Final Answer. We want to evaluate the integral

First, notice that the denominator can be factored:

Thus, the integral becomes

Let 𝑢 = 𝑥 -cos 𝜃 sin 𝜃 , so 𝑥 = 𝑢 sin 𝜃 + cos 𝜃 and 𝑑𝑥 = sin 𝜃 𝑑𝑢. When 𝑥 = -1, 𝑢 = -1-cos 𝜃 sin 𝜃 , and when 𝑥 = 1, 𝑢 = 1-cos 𝜃 sin 𝜃 . Then,

We can simplify the arguments of the arctangent functions using trigonometric identities:

.

Thus,

However, this is only valid when sin 𝜃 ≠ 0, which means 𝜃 ≠ 𝑛𝜋 for any integer 𝑛.

If 𝜃 = 0 or 𝜃 = 2𝜋, then sin 𝜃 = 0, and the integral becomes

If 𝜃 = 𝜋, then sin 𝜃 = 0, and the integral becomes

Therefore, 𝐼 (𝜃 ) = 𝜋 2 for 𝜃 ≠ 𝑛𝜋, and 𝐼 (𝜃 ) = 0 for 𝜃 = 𝑛𝜋.

The function 𝐼 (𝜃 ) is discontinuous at 𝜃 = 𝑛𝜋 for integer 𝑛. In the interval 0 ≤ 𝜃 ≤ 2𝜋, the discontinuities are at 𝜃 = 0, 𝜋, 2𝜋.

C.1.10 2005 Session 2: 5 Incorrect Reasoning, Incorrect Final Answer. We are given that

Differentiating both sides with respect to 𝑏, we get

Replacing 𝑏 with 𝑥, we have

Squaring both sides,

Separating variables and integrating,

Since cosh is even,

Given 𝑓 (0) = 5/4, 5 4 = cosh(𝐶)

4 To find the minimum value, we take the derivative:

This value of 𝑥 is negative, so it is not in the domain [0, ∞). Since cosh -1 5 4 > 0, 𝑥 + cosh -1 5 4 > 0 for 𝑥 > 0. Thus, sinh 𝑥 + cosh -1 5 4 > 0 for 𝑥 > 0, so 𝑓 ′ (𝑥) > 0 for 𝑥 > 0. Therefore, 𝑓 (𝑥) is increasing for 𝑥 > 0, and the minimum value occurs at 𝑥 = 0. The minimum value is 𝑓 (0) = 5/4.

C.2.1 1997 Session 1: 1 Correct Reasoning, Correct Final Answer. Problem: Let 𝑃 ≠ (0, 0) be a point on the parabola 𝑦 = 𝑥 2 . The normal line to the parabola at 𝑃 will intersect the parabola at another point, say 𝑄. Find the coordinates of 𝑃 so that the 𝑦-coordinate of 𝑄 is a minimum.

Solution:

Let 𝑃 = (𝑎, 𝑎 2 ) be a point on the parabola 𝑦 = 𝑥 2 , where 𝑎 ≠ 0. The derivative of 𝑦 = 𝑥 2 is 𝑦 ′ = 2𝑥. The slope of the tangent line at 𝑃 is 𝑚 𝑡 = 2𝑎. The slope of the normal line is 𝑚 𝑛 = -1 2𝑎 . The equation of the normal line is:

To find the intersection point 𝑄, we solve the system:

Let the other root be 𝑏. By Vieta’s formulas:

The 𝑦-coordinate of 𝑄 is:

C.2.2 1998 Session 1: 1 Incorrect Reasoning, Incorrect Final Answer. Problem: Let 𝑃 ≠ (0, 0) be a point on the parabola 𝑦 = 𝑥 2 . The normal line to the parabola at 𝑃 will intersect the parabola at another point, say 𝑄. Find the coordinates of 𝑃 so that the length of segment 𝑃𝑄 is a minimum. Solution:

Let 𝑃 = (𝑎, 𝑎 2 ) be a point on the parabola 𝑦 = 𝑥 2 , where 𝑎 ≠ 0.

The derivative of 𝑦 = 𝑥 2 is 𝑦 ′ = 2𝑥.

At point 𝑃 (𝑎, 𝑎 2 ), the slope of the tangent line is 2𝑎.

The slope of the normal line is -1 2𝑎 . The equation of the normal line is 𝑦 -𝑎 2 = -1 2𝑎 (𝑥 -𝑎). To find the intersection point 𝑄, substitute 𝑦 = 𝑥 2 into the normal line equation:

The distance 𝑃𝑄 is:

C.2.3 1991 Session 1: 1 Incorrect Reasoning, Incorrect Final Answer. Problem: Let 𝑃 ≠ (0, 0) be a point on the parabola 𝑦 = 𝑥 2 . The normal line to the parabola at 𝑃 will intersect the parabola at another point, say 𝑄. Find the coordinates of 𝑃 so that the length of the arc of the parabola between 𝑃 and 𝑄 is a minimum. Solution:

Let 𝑃 = (𝑎, 𝑎 2 ) be a point on the parabola 𝑦 = 𝑥 2 , where 𝑎 ≠ 0.

The derivative of 𝑦 = 𝑥 2 is 𝑦 ′ = 2𝑥.

At 𝑥 = 𝑎, the slope of the tangent line is 𝑚 𝑡 = 2𝑎.

The slope of the normal line at 𝑃 is 𝑚 𝑛 = -1 2𝑎 . The equation of the normal line is:

To find the intersection point 𝑄, we set the equation of the normal line equal to the equation of the parabola:

Let 𝑄 = (𝑏, 𝑏 2 ). By Vieta’s formulas, 𝑏 = -1 2𝑎 -𝑎. The arc length of the parabola between 𝑃 and 𝑄 is:

After evaluating the integral and minimizing the arc length, we find that the minimum occurs when

Therefore, the coordinates of 𝑃 are 1 The equation of the semicircle 𝑆𝐶 is (𝑥 -

To find the intersection of 𝑆𝐶 and 𝐶 𝑎 , we solve:

Subtracting the second equation from the first, we get:

we get:

The intersection point is

. The slope of the line through 𝑃 and the intersection point is:

The equation of the line is 𝑦 = 𝑚𝑥 + 𝑎. To find the 𝑥-intercept, set 𝑦 = 0:

Multiply numerator and denominator by √︃ 1 -𝑎 2 4 + 1:

Thus, the limiting position of the 𝑥-intercept is 4.

1 Correct Reasoning, Correct Final Answer. Problem: Let 𝑃 ≠ (0, 0) be a point on the parabola 𝑦 = 𝑥 2 . The normal line to the parabola at 𝑃 will intersect the parabola at another point, say 𝑄. Find the coordinates of 𝑃 so that the distance between the 𝑥-coordinates of 𝑃 and 𝑄 is a minimum. Solution:

1. Let 𝑃 = (𝑎, 𝑎 2 ) where 𝑎 ≠ 0.

2. The derivative of 𝑦 = 𝑥 2 is 𝑦 ′ = 2𝑥. At 𝑥 = 𝑎, the slope of the tangent is 𝑚 𝑡 = 2𝑎.

3. The slope of the normal line is 𝑚 𝑛 = -1 2𝑎 .

To find the intersection point 𝑄, we solve the system:

2𝑎 𝑥 -1 2 + 𝑎 2 = 0 Let 𝑥 = 𝑏 be the other solution. Using Vieta’s formulas: We want to find the locus of points P = (𝑥, 𝑦) such that the distance from P to circle A is equal to the distance from P to circle B.

The distance from a point P to a circle with center C and radius r is given by |𝑃𝐶 -𝑟 |.

Therefore, we have:

Since circle B is inside circle A, the points equidistant from the circles will be located between them. Therefore we can remove the absolute value bars.

This equation represents the locus of points equidistant from the two circles.

C.2.7 2001 Session 2: 3 Correct Reasoning, Correct Final Answer. Problem: A convex decagon and all of its diagonals are drawn. How many interior points of intersection of the diagonals are there, if it is assumed that no 3 diagonals share a common interior point?

To form an intersection point inside the decagon, we need to choose four vertices. Each set of four vertices uniquely determines two intersecting diagonals.

The number of ways to choose four vertices from the ten vertices of the decagon is given by the combination formula:

In our case, 𝑛 = 10 (number of vertices) and 𝑘 = 4 (number of vertices to choose). So, the number of ways to choose four vertices is:

Since each set of four vertices uniquely determines an interior intersection point, the number of interior intersection points is equal to the number of ways to choose four vertices.

Therefore, there are 210 interior intersection points.

Case 1: 𝐴 and 𝐵 are on the same side of 𝐿. Let 𝐴 ′ be the reflection of 𝐴 across the line 𝐿. Then 𝑃𝐴 = 𝑃𝐴 ′ . Therefore, 𝑃𝐴 + 𝑃𝐵 = 𝑃𝐴 ′ + 𝑃𝐵. The sum 𝑃𝐴 ′ + 𝑃𝐵 is minimized when 𝑃, 𝐴 ′ , and 𝐵 are collinear. Thus, we need to find the intersection of the line 𝐴 ′ 𝐵 and the line 𝐿.

(1) Find the reflection 𝐴 ′ of 𝐴 across 𝐿:

Let 𝐴 ′ = (𝑥 ′ , 𝑦 ′ ). The midpoint of 𝐴𝐴 ′ lies on 𝐿, so:

The line 𝐴𝐴 ′ is perpendicular to 𝐿, so its slope is 𝑏/𝑎. Thus:

Solving equations ( 1) and ( 2) for 𝑥 ′ and 𝑦 ′ gives the coordinates of 𝐴 ′ .

(2) Find the equation of the line 𝐴 ′ 𝐵:

The slope of 𝐴 ′ 𝐵 is:

(3) Find the intersection of 𝐴 ′ 𝐵 and 𝐿:

Solve the system of equations:

for 𝑥 and 𝑦. This gives the coordinates of the point 𝑃.

Case 2: 𝐴 and 𝐵 are on opposite sides of 𝐿. In this case, the minimum sum of distances is achieved when 𝑃 is the intersection of the line 𝐴𝐵 and the line 𝐿.

(1) Find the equation of the line 𝐴𝐵:

The slope of 𝐴𝐵 is:

The equation of the line 𝐴𝐵 is:

(2) Find the intersection of 𝐴𝐵 and 𝐿:

Solve the system of equations:

for 𝑥 and 𝑦. This gives the coordinates of the point 𝑃. Problem Statement. Each of the numbers 𝑥 1 , 𝑥 2 , . . . , 𝑥 𝑛 can be 1, 0, or -1. What is the minimum possible value of the sum of all products of pairs of these numbers? Solution. We need to minimize the sum 𝑆 of all products of pairs 𝑥 𝑖 𝑥 𝑗 for 1 ≤ 𝑖 < 𝑗 ≤ 𝑛.

First, recall the identity:

Rewriting, we get:

Case Analysis.

Case 1: All Variables Zero. Here, Σ = 0 and 𝑄 = 0, so 𝑆 = 0.

Case 2: Non-zero Variables. Consider 𝑘 variables as 1 and 𝑚 as -1, and the rest 0.

Then:

Optimal Configuration. To minimize 𝑆, we need to minimize (𝑘 -𝑚) 2 -(𝑘 + 𝑚).

Subcase: Balanced 1’s and -1’s. For even 𝑛 = 2𝑘, set 𝑘 1’s and 𝑘 -1’s:

For odd 𝑛 = 2𝑘 + 1, set 𝑘 1’s, 𝑘 -1’s, and 1 zero:

Alternatively, setting 𝑘 + 1 1’s and 𝑘 -1’s gives the same result:

The minimal value of 𝑆 is achieved when the number of 1’s and -1’s are as balanced as possible, leading to:

and the product 𝑎 1 𝑎 2 • • • 𝑎 𝑛 is as large as possible.

Solution.

Step 1: Understanding the Problem. We need to partition the number 1997 into positive integers 𝑎 1 , 𝑎 2 , . . . , 𝑎 𝑛 such that their sum is 1997 and their product is maximized.

Step 2: Optimal Partition Strategy. It is known that for a given sum, the product of numbers is maximized when the numbers are as close to 𝑒 ≈ 2.718 as possible. Among integers, 3 is the closest to 𝑒. Therefore, we should use as many 3s as possible.

Step 3: Applying the Strategy. Divide 1997 by 3:

This gives us 665 numbers of 3 and a remainder of 2.

Step 4: Handling the Remainder. The remainder is 2, which is less than 3. Since 2 is a positive integer, we can include it as is.

If the remainder were 1, it would be better to adjust one of the 3s and the remainder into two 2s because 3 + 1 = 2 + 2 and 3 × 1 = 3 < 2 × 2 = 4. However, in this case, the remainder is 2, so no adjustment is needed.

Step 5: Final Partition. Thus, the optimal partition is: 𝑛 = 666 numbers: 665 threes and one two.

Step 6: Verification. Any other combination, such as using fewer threes and more twos or larger numbers, leads to a product that is either equal or smaller. For example:

• Replacing a 3 and a 2 with two 2s and a 1 (but 1 reduces the product).

• Using larger numbers like 4, which can be expressed as 2 + 2, but the product remains the same (4 = 2 × 2).

Therefore, the maximum product is achieved with 665 threes and one two.

Final Answer. The positive integers are:

The maximum product is: Solution. Consider the infinite sequence {𝑥 𝑛 } of positive real numbers defined by:

Verification of Conditions.

(1) Initial Condition:

(3) Sum Condition: For any 𝑛 ≥ 1, consider the sum

.

This is a geometric series:

The sequence 𝑥 𝑛 = 1 2 𝑛 satisfies all the given conditions:

• 𝑥 𝑖+1 ≤ 𝑥 𝑖 for all 𝑖 ≥ 0,

Step 1: Simplify the General Term. The general term of the series is:

Notice that the denominator can be factored as:

Thus, the denominator becomes:

Step 2: Partial Fraction Decomposition. Let 𝑥 = 6𝑖. The term becomes:

Assume the partial fraction decomposition:

Solving for coefficients 𝐴, 𝐵, 𝐶, 𝐷 yields:

Thus, the decomposition simplifies to:

1 2

Reverting back to 𝑖:

Step 3: Rewrite the Series. The original series becomes:

Step 4: Evaluate the Sum 𝑆. The sum 𝑆 can be expressed using the identity:

. However, the term for 𝑘 = 0 (i.e., 1/1 2 = 1) is included in the above sum but not in 𝑆. Therefore:

Step 5: Final Calculation. Thus, the original series is:

.

The sum of the series is:

C.3.5 1999 Session 2: 5 Correct Reasoning, Correct Final Answer.

Solution. Let 𝐴 𝑛 denote the 𝑛 × 𝑛 matrix whose (𝑖, 𝑗)-th entry is GCD(𝑖, 𝑗). We aim to compute det(𝐴 𝑛 ).

Pattern Recognition. For small values of 𝑛, we compute the determinants explicitly:

• For 𝑛 = 1:

• For 𝑛 = 2:

• For 𝑛 = 3:

• For 𝑛 = 4:

General Formula. The observed pattern suggests that the determinant is related to Euler’s totient function 𝜙 (𝑘).

Specifically:

This formula matches the computed values:

Conclusion. The determinant of the matrix 𝐴 𝑛 is the product of Euler’s totient function evaluated from 𝑘 = 1 to 𝑛:

Understanding the Problem. We need to count the number of sequences of length 𝑛 composed of the letters 𝑎, 𝑏, and 𝑐 such that:

(1) The sequence starts and ends with 𝑎.

(2) No two adjacent letters are the same.

Approach. We can model this problem using recurrence relations. Let’s define:

• 𝐴(𝑛): Number of valid sequences of length 𝑛 ending with 𝑎.

• 𝐵(𝑛): Number of valid sequences of length 𝑛 ending with 𝑏. Solution.

Step 1: Define Variables. Let:

• 𝑎 = number of -2s in the sequence,

• 𝑏 = number of 1s in the sequence,

• 𝑐 = number of 2s in the sequence.

Step 2: Set Up Equations. Given the conditions:

4𝑎

Step 3: Solve the System of Equations. Subtract equation ( 3) from ( 4):

Simplify:

3𝑎 + 𝑐 = 15 (Equation 3)

Express 𝑐 in terms of 𝑎:

Substitute 𝑐 back into (3) to find 𝑏:

Step 4: Find Valid Values for 𝑎, 𝑏, 𝑐. Since 𝑎, 𝑏, 𝑐 must be non-negative integers, consider 𝑎 from 1 to 5:

• For 𝑎 = 1: 𝑐 = 12, 𝑏 = 7

• For 𝑎 = 2: 𝑐 = 9, 𝑏 = 15

• For 𝑎 = 3: 𝑐 = 6, 𝑏 = 23

All cases are valid.

Step 5: Calculate Sum of Cubes. The sum of cubes 𝑆 is:

Substitute 𝑏 = 8𝑎 -1 and 𝑐 = 15 -3𝑎:

Step 6: Find Maximum and Minimum 𝑆. Since 𝑆 decreases with 𝑎:

• Maximum 𝑆 (M) at 𝑎 = 1: 𝑀 = 119 -24(1) = 95

• Minimum 𝑆 (m) at 𝑎 = 5: 𝑚 = 119 -24(5) = -1

Step 7: Verification. Verify for 𝑎 = 1 and 𝑎 = 5:

Final Calculation. The ratio 𝑀/𝑚 is:

Answer. The ratio 𝑀/𝑚 is -95 . Solution. We are tasked with finding the minimum distance from the matrix

to the set Σ of all real 2 × 2 singular matrices under the distance metric

where

Approach. We need to minimize the distance 𝑑 (𝐴, 𝑆) where 𝑆 ∈ Σ, i.e., det(𝑆) = 0.

. The condition det(𝑆) = 0 gives:

The distance function becomes:

Optimization. Using the method of Lagrange multipliers, we consider the Lagrangian:

Taking partial derivatives and solving the system leads us to consider the case where 𝑏 = 𝑐 = 0. The constraint then reduces to 𝑎𝑑 = 0.

Case 1: 𝑎 = 0. The distance becomes:

The minimum occurs at 𝑑 = 2, yielding a distance of 1.

Case 2: 𝑑 = 0. The distance becomes:

(1 -𝑎) 2 + 4.

The minimum occurs at 𝑎 = 1, yielding a distance of 4.

The minimal distance is thus 1, achieved when 𝑆 = 0 0 0 2 .

Verification. Considering rank-1 matrices leads to higher distances, confirming that the minimal distance is indeed 1. Solution. We need to find all positive integers 𝑐 for which the expression 𝑛 2 + 𝑐𝑛 is never a perfect square for any positive integer 𝑛.

Initial Analysis. Consider the equation:

for some integer 𝑘 ≥ 1. We can rewrite this as:

For fixed 𝑐, we need to determine whether there exist positive integers 𝑛 and 𝑘 satisfying this equation.

Parametric Solution. Let us set 𝑘 = 𝑛 + 𝑡 where 𝑡 is a non-negative integer. Substituting:

This implies that 𝑐 -2𝑡 must be a positive divisor of 𝑡 2 , and 𝑛 = 𝑡 2 𝑐 -2𝑡 must be a positive integer.

Special Cases.

Case 𝑐 = 1. The equation becomes:

For 𝑡 ≥ 1, the right side becomes non-positive while the left side is positive. Thus, the only possibility is 𝑡 = 0, which gives 𝑛 = 0 -invalid. Therefore, no solutions exist for 𝑐 = 1.

Case 𝑐 = 2. The equation becomes:

For 𝑡 = 0: 𝑛 = 0 (invalid) For 𝑡 = 1: 1 = 𝑛(0) (no solution) For 𝑡 ≥ 1: The right side is non-positive. Thus, no solutions exist for 𝑐 = 2.

Case 𝑐 = 4. The equation becomes:

Testing small values:

• 𝑡 = 0: 𝑛 = 0 (invalid)

• 𝑡 ≥ 2: The denominator becomes zero or negative Thus, no solutions exist for 𝑐 = 4.

Verification of Other Values. For 𝑐 = 3, we find a solution at 𝑛 = 1:

Thus, 𝑐 = 3 is excluded. General Proof. To prove that these are the only values, consider:

(1) For 𝑐 ≥ 5, we can often find solutions:

• For odd 𝑐, take 𝑛 = (𝑐 -1) 2 4

• For even 𝑐, solutions frequently exist Understanding the Problem. The set 𝑆 consists of all numbers expressible as 5𝑎 + 503𝑏 where 𝑎 and 𝑏 are nonnegative integers. We need to find the largest integer not in 𝑆.

Coin Problem Analogy. This problem resembles the Coin Problem, which seeks the largest monetary amount that cannot be obtained using any combination of coins of given denominations. For two coprime integers 𝑥 and 𝑦, the largest number not expressible as 𝑚𝑥 + 𝑛𝑦 is known as the Frobenius number, given by: 𝑔(𝑥, 𝑦) = 𝑥𝑦 -𝑥 -𝑦.

Checking Coprimality. First, verify that gcd(5, 503) = 1. Since 503 is not divisible by 5, they are indeed coprime.

Applying the Frobenius Number Formula. Given that 5 and 503 are coprime, the largest number not in 𝑆 is: . Let’s simplify the expression inside the square root:

We can complete the square for the quadratic expression:

Thus, the integrand becomes:

.

Step 2: Substitute to simplify the integral. Let 𝑢 = 𝑥 -2. Then 𝑑𝑢 = 𝑑𝑥, and when 𝑥 = 1, 𝑢 = -1, and when 𝑥 = 3, 𝑢 = 1. The integral becomes:

Step 3: Evaluate the integral. The integral ∫ 𝑑𝑢 √ 1-𝑢 2 is a standard form whose antiderivative is arcsin 𝑢. Thus:

Factor out 𝑒 2 from the denominator: 1) .

Let 𝑢 = 𝑥 -1. Then 𝑑𝑢 = 𝑑𝑥, and when 𝑥 = 1, 𝑢 = 0, and as 𝑥 → ∞, 𝑢 → ∞. The integral becomes:

Step 2: Simplify the integrand further. Multiply numerator and denominator by 𝑒 𝑢 :

1

Thus, the integral becomes:

1

Step 3: Substitute to evaluate the integral. Let 𝑣 = 𝑒 𝑢 . Then 𝑑𝑣 = 𝑒 𝑢 𝑑𝑢, and when 𝑢 = 0, 𝑣 = 1, and as 𝑢 → ∞, 𝑣 → ∞. The integral transforms to: 1

where 𝑛 is a positive integer. What is the limit of the length of 𝑆 𝑛 as 𝑛 → ∞? Solution.

Step 1: Arc Length Formula. The length 𝐿 of a curve 𝑦 = 𝑓 (𝑥) from 𝑥 = 𝑎 to 𝑥 = 𝑏 is given by:

For 𝑆 𝑛 , 𝑦 = 1 𝑛 sin(𝑛 2 𝑥), so the derivative is:

Thus, the arc length 𝐿 𝑛 is:

Step 2: Substitution and Simplification. Let 𝑢 = 𝑛 2 𝑥, 𝑑𝑢 = 𝑛 2 𝑑𝑥, so 𝑑𝑥 = 𝑑𝑢 𝑛 2 . The integral becomes:

The integrand √ 1 + 𝑛 2 cos 2 𝑢 is periodic with period 𝜋. Therefore, the integral over [0, 𝑛 2 𝜋] is 𝑛 2 times the integral over one period [0, 𝜋]:

Step 3: Exact Form and Asymptotic Behavior. The integral can be expressed in terms of the complete elliptic integral of the second kind 𝐸 (𝑘): Part 1: General Formula. We are given the integral:

where 𝑚 and 𝑛 are positive integers and 𝑎 < 𝑏.

Step 1: Substitution. Let 𝑥 = 𝑎 + (𝑏 -𝑎)𝑡. Then, 𝑑𝑥 = (𝑏 -𝑎) 𝑑𝑡, and the limits change from 𝑥 = 𝑎 to 𝑥 = 𝑏 as 𝑡 goes from 0 to 1.

The integrand becomes:

𝑚!𝑛! Thus, the integral transforms to:

(𝑏 -𝑎) 𝑚+𝑛+1

𝑚!𝑛!

Step 2: Beta Function. The integral ∫ 1 0 𝑡 𝑛 (1 -𝑡) 𝑚 𝑑𝑡 is the Beta function 𝐵(𝑛 + 1, 𝑚 + 1), which can be expressed in terms of Gamma functions:

Step 3: Final Expression. Substituting back:

Thus, the solution is:

We need to evaluate:

Step 1: Substitution. Let 𝑥 = √ 𝑡. Then, 𝑑𝑥 = 1 2 𝑡 -1/2 𝑑𝑡, and the integral becomes: 1 2

Step 2: Beta and Gamma Functions. The Beta function can be written as:

Using the property of the Gamma function for half-integer arguments:

√ 𝜋 Thus:

The final answer is:

2 𝑛 𝑛! (2𝑛 + 1)!! D.1.4 1999 Session 2: 2 Correct Reasoning, Correct Final Answer.

Problem. Find the limit:

Step 1: Partial Fraction Decomposition. First, we decompose the denominator 16𝑛 2 -1 using the difference of squares: 16𝑛 2 -1 = (4𝑛 -1)(4𝑛 + 1).

We express the term 1 16𝑛 2 -1 as partial fractions:

Multiplying both sides by (4𝑛 -1)(4𝑛 + 1) gives:

Expanding and collecting terms:

For this to hold for all 𝑛, the coefficients must satisfy: 4𝐴 + 4𝐵 = 0 and 𝐴 -𝐵 = 1.

Solving these equations gives 𝐴 = 1 2 and 𝐵 = -1 2 . Thus:

Step 2: Rewrite the Sum. The sum becomes:

Step 3: Infinite Series Limit. Taking the limit as 𝑁 → ∞, we consider the infinite series:

However, it’s more efficient to use a known result for the sum

.

Rewriting our sum:

2 .

Thus:

Simplifying the denominator:

, so:

.

Step 4: Compute the Original Limit. The original expression is:

.

Therefore, the limit is: Problem Statement. Find all possible continuous and differentiable curves 𝐶 which have the following properties:

• The curve 𝐶 lies in the first quadrant and contains the point (0, 0).

• For any point 𝑃 = (𝑥, 𝑦) on 𝐶, the interior of the rectangle 𝑅 bounded by the coordinate axes and horizontal and vertical lines through 𝑃 is separated into two parts by 𝐶. • When the part adjacent to the 𝑥-axis is rotated about the 𝑥-axis and the part adjacent to the 𝑦-axis is rotated about the 𝑦-axis, two solids of equal volume are generated.

Step 1: Understanding the Curve and Volumes. Let the curve 𝐶 be defined by 𝑦 = 𝑓 (𝑥), where 𝑓 is continuous and differentiable on [0, ∞), with 𝑓 (0) = 0.

For any point 𝑃 = (𝑥, 𝑦) on 𝐶, the rectangle 𝑅 has vertices at (0, 0), (𝑥, 0), (0, 𝑦), and (𝑥, 𝑦).

The curve 𝐶 divides 𝑅 into:

• The area under 𝐶 (adjacent to the 𝑥-axis).

• The area above 𝐶 but within 𝑅 (adjacent to the 𝑦-axis).

Step 2: Volumes of Revolution. The volumes generated by rotating these areas are:

𝑑𝑡 (rotation about the 𝑦-axis), where 𝑦 = 𝑓 (𝑥). The condition 𝑉 𝑥 = 𝑉 𝑦 leads to:

Step 3: Substitution and Differentiation. Using the substitution 𝑡 = 𝑓 (𝑢) in the right integral:

Thus, the equation becomes:

Differentiating both sides with respect to 𝑥:

Step 4: Solving the Differential Equation. The equation is separable:

Solving for 𝑓 :

Step 5: Applying Initial Conditions and Constraints. Given 𝑓 (0) = 0, the solution is valid. To ensure 𝑓 is defined and differentiable in the first quadrant:

• For 𝐶 = 0, 𝑓 (𝑥) = 𝑥.

• For 𝐶 < 0, 𝑓 (𝑥) = 𝑥 1+|𝐶 |𝑥 , which is defined for all 𝑥 ≥ 0. • For 𝐶 > 0, 𝑓 has a singularity at 𝑥 = 1 𝐶 , so it’s only valid in [0, 1 𝐶 ).

The only continuous and differentiable curves in the first quadrant that satisfy all the given conditions are:

This includes the special case 𝐶 = 0, which gives the line 𝑦 = 𝑥. Solution.

Understanding the Problem. The sequence {𝑥 𝑛 } is defined with a nested radical pattern that alternates between √ 5 and √ 13. The goal is to find the limit of this sequence as 𝑛 approaches infinity.

Identifying the Pattern. The sequence exhibits a recursive pattern where each term adds a new nested square root:

• For odd 𝑛, the innermost term is √ 5. • For even 𝑛, the innermost term is √ 13.

Formulating the Fixed-Point Equation. Assuming the sequence converges to a limit 𝐿, the infinite nested radical can be described by the equation:

Solving the Fixed-Point Equation. Square both sides:

Isolate the square root:

Square both sides again:

13 + 𝐿 = (𝐿 2 -5) 2 13

Bring all terms to one side:

Finding the Roots. We look for rational roots using the Rational Root Theorem. Testing 𝐿 = 3: 81 -90 -3 + 12 = 0 Thus, 𝐿 = 3 is a root. Factor the quartic:

The cubic factor does not have obvious rational roots, but numerical methods suggest other real roots are not relevant here since the sequence is positive and increasing towards 3.

Verification. Compute the first few terms numerically:

The values approach 3, confirming our solution.

Conclusion. The limit of the sequence {𝑥 𝑛 } as 𝑛 → ∞ is: and also find its area.

Setup. Consider two cylinders:

• Cylinder 1: 𝑥 2 + 𝑦 2 = 𝑟 2 (aligned along the 𝑧-axis).

• Cylinder 2: 𝑦 2 + 𝑧 2 = 𝑟 2 (aligned along the 𝑥-axis).

Their intersection forms a bicylinder with four curved faces. We focus on one face, part of Cylinder 1 within Cylinder 2.

Parameterizing the Face. For Cylinder 1, use cylindrical coordinates:

The condition from Cylinder 2 is 𝑦 2 + 𝑧 2 ≤ 𝑟 2 , which becomes:

Flattening the Face. Unroll Cylinder 1 by mapping:

(𝑢, 𝑣) = (𝑟𝜃, 𝑧).

The boundaries are:

• Lower: 𝑣 = 0.

Area Calculation. The area under 𝑣 = 𝑟 cos(𝑢/𝑟 ) from 𝑢 = -𝜋𝑟 /2 to 𝜋𝑟 /2:

By symmetry:

Let 𝑤 = 𝑢/𝑟 , 𝑑𝑢 = 𝑟𝑑𝑤:

This is for 𝑧 ≥ 0; the full face (both 𝑧 ≥ 0 and 𝑧 ≤ 0) has area 4𝑟

Find lim 𝑛→∞ 𝑥 𝑛 .

Solution.

Step 1: Rewrite the Equation for 𝑥 𝑛 . We can express 𝑥 𝑛 explicitly as:

Our goal is to find:

Step 2: Taylor Expansion of the Denominator. For large 𝑛, ℎ = 1 𝑛 is small. The Taylor expansion of 2 ℎ around ℎ = 0 is:

Thus:

Step 3: Substitute and Simplify. Substituting back into the expression for 𝑥 :

Simplify the denominator:

Step 4: Geometric Series Approximation. For large 𝑛, we use 1 1+𝜖 ≈ 1 -𝜖 for 𝜖 ≪ 1:

Step 5: Take the Limit. As 𝑛 → ∞, the 𝑂 Solution. We are tasked with evaluating the integral:

and determining the points 𝜃 ∈ [0, 2𝜋] where 𝐼 (𝜃 ) is discontinuous.

Step 1: Simplify the Denominator. The denominator can be rewritten by completing the square:

Thus, the integral becomes:

Step 2: Substitution and Integration. Let 𝑢 = 𝑥cos 𝜃 , then 𝑑𝑢 = 𝑑𝑥, and the limits change accordingly:

The antiderivative is known:

Here, 𝑎 = sin 𝜃 , so:

Step 3: Simplify Using Trigonometric Identities. Using the identity:

, we can rewrite 𝐼 (𝜃 ) as:

.

Step 4: Evaluate Based on 𝜃 Range. Considering the principal range of tan -1 :

• For 0 ≤ 𝜃 < 𝜋:

.

• At 𝜃 = 𝜋: 𝐼 (𝜋) = 0 (since sin 𝜋 = 0).

• For 𝜋 < 𝜃 < 2𝜋:

.

Discontinuities. The function 𝐼 (𝜃 ) has jump discontinuities at 𝜃 = 𝜋 and 𝜃 = 0 (or 𝜃 = 2𝜋 due to periodicity).

Final Answer. The value of the integral is:

The discontinuities occur at 𝜃 = 𝜋 and 𝜃 = 0 (or 2𝜋). Problem Statement. Suppose that 𝑓 : [0, ∞) → [0, ∞) is a differentiable function with the property that the area under the curve 𝑦 = 𝑓 (𝑥) from 𝑥 = 𝑎 to 𝑥 = 𝑏 is equal to the arclength of the curve 𝑦 = 𝑓 (𝑥) from 𝑥 = 𝑎 to 𝑥 = 𝑏. Given that 𝑓 (0) = 5 4 , and that 𝑓 (𝑥) has a minimum value on the interval (0, ∞), find that minimum value.

Step 1: Translating the Condition into an Equation. The area under 𝑓 from 𝑎 to 𝑏 is:

The arc length of 𝑓 from 𝑎 to 𝑏 is:

Given 𝐴 = 𝐿, we have:

Step 2: Differentiating Both Sides. Differentiating both sides with respect to 𝑏 gives:

Since 𝑏 is arbitrary, we can write:

Step 3: Solving the Differential Equation. Square both sides:

Rearrange:

Take square root:

Consider the positive root first:

Separate variables and integrate:

The integral evaluates to:

where 𝐴 = 𝑒 𝐶 .

Using the reciprocal:

Adding the two equations:

Step 4: Applying the Initial Condition. Given 𝑓 (0) = 5 4 :

Multiply by 4𝐴:

1 2 Thus, two potential solutions:

(1) For 𝐴 = 2: Problem Statement. Let 𝑃 ≠ (0, 0) be a point on the parabola 𝑦 = 𝑥 2 . The normal line to the parabola at 𝑃 will intersect the parabola at another point, say 𝑄. Find the coordinates of 𝑃 so that the 𝑦-coordinate of 𝑄 is a minimum.

Step 1: Parametrize Point 𝑃. Let 𝑃 be a point on the parabola 𝑦 = 𝑥 2 . We can represent 𝑃 as:

where 𝑎 ≠ 0 since 𝑃 ≠ (0, 0).

Step 2: Find the Slope of the Tangent at 𝑃. The derivative of 𝑦 = 𝑥 2 is:

At 𝑃, the slope of the tangent line is:

Step 3: Find the Slope of the Normal Line at 𝑃. The normal line is perpendicular to the tangent line, so its slope is:

Step 4: Equation of the Normal Line. Using the point-slope form:

Step 5: Find Intersection Points with the Parabola. The normal line intersects the parabola 𝑦 = 𝑥 2 at:

Rewriting:

Step 6: Solve the Quadratic Equation. The quadratic can be factored as:

Thus, the roots are:

Step 7: Find Coordinates of 𝑄. The other intersection point 𝑄 has 𝑥-coordinate:

Thus, the 𝑦-coordinate is:

Step 8: Minimize 𝑦 𝑄 . Simplify 𝑦 𝑄 :

Find the critical points by setting the derivative to zero:

Step 9: Verify the Minimum. The second derivative is:

Thus, the critical points are minima.

Step 10: Find Minimal 𝑦 𝑄 . Substitute 𝑎 2 = 1 2 :

Step 11: Find Corresponding 𝑃 Coordinates. The coordinates of 𝑃 are:

Final Answer. Problem Statement. Let 𝑃 ≠ (0, 0) be a point on the parabola 𝑦 = 𝑥 2 . The normal line to the parabola at 𝑃 will intersect the parabola at another point 𝑄. Find the coordinates of 𝑃 so that the length of segment 𝑃𝑄 is a minimum.

Step 1: Parametrize Point 𝑃. Let 𝑃 be a point on the parabola 𝑦 = 𝑥 2 . We can represent 𝑃 as:

where 𝑎 ≠ 0 since 𝑃 ≠ (0, 0).

Step 2: Find the Slope of the Tangent at 𝑃. The derivative of 𝑦 = 𝑥 2 is:

At point 𝑃, the slope of the tangent line is:

Step 3: Find the Slope of the Normal Line at 𝑃. The normal line is perpendicular to the tangent line, so its slope is:

Step 4: Equation of the Normal Line. Using the point-slope form:

Step 5: Find Intersection Points with Parabola. Set 𝑦 = 𝑥 2 equal to the normal line equation:

Step 6: Solve the Quadratic Equation. Factor out (𝑥 -𝑎):

Thus, the roots are:

Step 7: Find Coordinates of 𝑄. The other intersection point 𝑄 is:

Step 8: Calculate Distance 𝑃𝑄. The distance between 𝑃 and 𝑄 is:

Simplifying:

Step 9: Minimize 𝑃𝑄. To minimize 𝑃𝑄, minimize 𝑓 (𝑎) = (1+4𝑎 2 ) 3 16𝑎 4 . Take the derivative and set to zero:

After simplification, we find:

Step 10: Find Coordinates of 𝑃. Thus, the coordinates of 𝑃 are:

Conclusion. The coordinates of 𝑃 that minimize the distance 𝑃𝑄 are: Problem Statement. Let 𝑃 ≠ (0, 0) be a point on the parabola 𝑦 = 𝑥 2 . The normal line to the parabola at 𝑃 will intersect the parabola at another point 𝑄. Find the coordinates of 𝑃 so that the length of the arc of the parabola between 𝑃 and 𝑄 is a minimum.

Step 1: Parametrize Point 𝑃. Let 𝑃 be a point on the parabola 𝑦 = 𝑥 2 . We can represent 𝑃 as:

where 𝑎 ≠ 0 since 𝑃 ≠ (0, 0).

Step 2: Find the Slope of the Tangent and Normal at 𝑃. The derivative of 𝑦 = 𝑥 2 is:

At point 𝑃, the slope of the tangent line is:

The slope of the normal line is the negative reciprocal:

Step 3: Equation of the Normal Line. Using the point-slope form:

Simplifying:

Step 4: Find the Second Intersection Point 𝑄. Set 𝑦 = 𝑥 2 equal to the normal line equation:

The other root gives 𝑄:

Step 5: Arc Length Between 𝑃 and 𝑄. The arc length 𝐿 is:

Step 6: Differentiate 𝐿 with Respect to 𝑎. Using the Leibniz rule:

Step 7: Solve 𝑑𝐿 𝑑𝑎 = 0. After simplifying, we obtain:

Conclusion. The coordinates of 𝑃 that minimize the arc length are:

and Step 1: Equations of SC and 𝐶 𝑎 . The semicircle 𝑆𝐶 is given by:

(𝑥 -1) 2 + 𝑦 2 = 1 with 𝑦 ≥ 0

The circle 𝐶 𝑎 is given by:

Step 2: Find Intersection Points. Substitute 𝑦 2 = 𝑎 2 -𝑥 2 from 𝐶 𝑎 into 𝑆𝐶:

Simplify:

𝑎 2 4 Thus, the intersection point is:

Step 3: Equation of the Line through P and Q. Point 𝑃 is (0, 𝑎). The slope 𝑚 of the line through 𝑃 and 𝑄 is:

The line equation is:

Step 4: Find the X-Intercept. Set 𝑦 = 0:

Rationalize the denominator:

Step 5: Take the Limit as 𝑎 → 0. Problem Statement. Let 𝑃 ≠ (0, 0) be a point on the parabola 𝑦 = 𝑥 2 . The normal line to the parabola at 𝑃 will intersect the parabola at another point, say 𝑄. Find the coordinates of 𝑃 so that the distance between the 𝑥-coordinates of 𝑃 and 𝑄 is a minimum.

Step 1: Parametrize Point 𝑃. Since 𝑃 lies on the parabola 𝑦 = 𝑥 2 , we can represent 𝑃 as:

where 𝑎 ≠ 0.

Step 2: Find the Tangent and Normal Lines at 𝑃. The derivative of 𝑦 = 𝑥 2 is 𝑦 ′ = 2𝑥. Thus, the slope of the tangent at 𝑃 is:

The normal line is perpendicular to the tangent, so its slope is:

The equation of the normal line through 𝑃 is:

• If 𝑄 is outside the circle, the distance is 𝑑 (𝑄, center)radius.

• If 𝑄 is inside the circle, the distance is radius -𝑑 (𝑄, center).

Case Analysis.

The loci of points equidistant to both circles consist of:

(1) A hyperbola defined by 𝑑 (𝑄, 𝑂) -𝑑 (𝑄, 𝑃) = 𝑅 -𝑟 for points outside both or inside both circles.

(2) An ellipse defined by 𝑑 (𝑄, 𝑂) + 𝑑 (𝑄, 𝑃) = 𝑅 + 𝑟 for points inside 𝐴 but outside 𝐵. Problem Statement. A convex decagon and all of its diagonals are drawn. How many interior points of intersection of the diagonals are there, if it is assumed that no 3 diagonals share a common interior point?

Solution.

Understanding the Problem. We need to find the number of interior points where two diagonals of a convex decagon intersect, under the condition that no three diagonals meet at a single interior point.

Approach. In a convex 𝑛-gon, any four distinct vertices define a quadrilateral. The two diagonals of this quadrilateral intersect at a unique point inside the polygon. Since no three diagonals are concurrent, each set of four vertices corresponds to exactly one unique intersection point.

Calculation. The number of ways to choose 4 vertices from 10 is given by the combination formula: Problem Statement. Let 𝑃 ≠ (0, 0) be a point on the parabola 𝑦 = 𝑥 2 . The normal line to the parabola at 𝑃 will intersect the parabola at another point 𝑄. Find the coordinates of 𝑃 so that the average of the 𝑦-coordinates of 𝑃 and 𝑄 is minimized.

Solution.

Step 1: Parametrize Point 𝑃. Let 𝑃 be a point on the parabola 𝑦 = 𝑥 2 . We can represent 𝑃 as:

where 𝑎 ≠ 0 since 𝑃 ≠ (0, 0).

Step 2: Find the Slope of the Tangent at 𝑃. The derivative of 𝑦 = 𝑥 2 is:

At 𝑃, the slope of the tangent line is:

Step 3: Find the Slope of the Normal Line at 𝑃. The normal line is perpendicular to the tangent line, so its slope is:

Step 4: Equation of the Normal Line. Using the point-slope form, the equation of the normal line is:

(𝑥 -𝑎)

Simplifying:

Step 5: Find Intersection Points with Parabola. Set 𝑦 = 𝑥 2 equal to the normal line equation:

Rearrange:

2𝑎𝑥 2 + 𝑥 -𝑎 -2𝑎 3 = 0

Step 6: Solve the Quadratic Equation. Factor the equation:

The roots are:

Step 7: Coordinates of 𝑄. The other intersection point 𝑄 has coordinates:

,

Step 8: Calculate the Average of Y-coordinates. The average 𝐴 is:

Expanding (1 + 2𝑎 2 ) 2 :

Step 9: Minimize 𝐴. Let 𝑓 (𝑎) = 𝑎 2 + 1 8𝑎 2 . To minimize 𝐴, minimize 𝑓 (𝑎). Find the derivative:

=⇒ 𝑎 = ± 1 2 3/4 Step 10: Find 𝑃 Coordinates. For 𝑎 = ± 1 2 3/4 , the y-coordinate of 𝑃 is:

Thus, the coordinates of 𝑃 are:

Conclusion. The points 𝑃 that minimize the average of the 𝑦-coordinates of 𝑃 and 𝑄 are: We place △𝐴𝐵𝐶 in a coordinate system with 𝐶 at the origin (0, 0), 𝐴 at (0, 𝑏), and 𝐵 at (𝑎, 0).

Step 1: Find Points 𝑃 and 𝑄.

• Median (𝑃): The median from 𝐶 to the hypotenuse 𝐴𝐵 meets 𝐴𝐵 at its midpoint 𝑃. The coordinates of 𝑃 are:

• Altitude (𝑄): The altitude from 𝐶 to 𝐴𝐵 meets 𝐴𝐵 at 𝑄. Step 2: Lengths of CQ, QP, and CP.

• CQ: The length of the altitude from 𝐶 to 𝐴𝐵 is:

The distance between 𝑄 and 𝑃 is:

The median to the hypotenuse in a right triangle is half the hypotenuse:

Step 3: Right Angle at Q. The slope of 𝐶𝑄 is 𝑎 𝑏 , and the slope of 𝑃𝑄 is -𝑏 𝑎 . The product of the slopes is -1, indicating that 𝐶𝑄 is perpendicular to 𝑃𝑄. Therefore, △𝐶𝑄𝑃 is right-angled at 𝑄.

Step 4: Pythagorean Condition. For △𝐶𝑄𝑃 to be Pythagorean, all its sides must be integers:

The simple conditions for △𝐶𝑄𝑃 to be Pythagorean are:

• The hypotenuse 𝑐 of △𝐴𝐵𝐶 must be even.

• The expressions 𝑎𝑏 𝑐 and 𝑎 2 -𝑏 2 2𝑐 must both be integers. These conditions are satisfied if △𝐴𝐵𝐶 is a Pythagorean triple where 𝑐 is even, which occurs when:

• The generating parameters 𝑚 and 𝑛 (for primitive triples) are both odd, or

• The triple is non-primitive and scaled by an even integer. Problem Statement. Find the point on a given line such that the sum of its distances from two fixed points is a minimum.

Assume the two fixed points and the given line are in the same plane.

Approach. To find the point 𝑃 on the line ℓ that minimizes the sum 𝑃𝐴 + 𝑃𝐵, where 𝐴 and 𝐵 are two fixed points, we use the method of reflection.

Steps.

(1) Reflect one of the points (say 𝐵) across the line ℓ to obtain 𝐵 ′ . This ensures that for any point 𝑃 on ℓ, 𝑃𝐵 = 𝑃𝐵 ′ .

(2) The sum 𝑃𝐴 + 𝑃𝐵 then becomes 𝑃𝐴 + 𝑃𝐵 ′ .

(3) The minimal sum 𝑃𝐴 +𝑃𝐵 ′ is achieved when 𝑃 lies on the straight line joining 𝐴 and 𝐵 ′ . Thus, 𝑃 is the intersection point of ℓ and the line 𝐴𝐵 ′ .

Verification. For any other point 𝑃 ′ on ℓ, the sum 𝑃 ′ 𝐴 + 𝑃 ′ 𝐵 = 𝑃 ′ 𝐴 + 𝑃 ′ 𝐵 ′ is greater than 𝐴𝐵 ′ (by the triangle inequality), unless 𝑃 ′ = 𝑃. Therefore, 𝑃 is indeed the minimizing point. Special Cases.

• If 𝐴 and 𝐵 are on the same side of ℓ, the reflection method still applies.

• If 𝐴 or 𝐵 lies on ℓ, then 𝑃 coincides with that point, as the sum 𝑃𝐴 + 𝑃𝐵 is minimized there. Subcase: Balanced 1’s and -1’s. For even 𝑛 = 2𝑘, set 𝑘 1’s and 𝑘 -1’s:

For odd 𝑛 = 2𝑘 + 1, set 𝑘 1’s, 𝑘 -1’s, and 1 zero:

Alternatively, setting 𝑘 + 1 1’s and 𝑘 -1’s gives the same result: Solution.

Step 1: Understanding the Problem. We need to partition the number 1997 into positive integers 𝑎 1 , 𝑎 2 , . . . , 𝑎 𝑛 such that their sum is 1997 and their product is maximized.

(3) Sum Condition: For any 𝑛 ≥ 1, consider the sum

.

Substituting 𝑥 𝑘 = 1 2 𝑘 ,

This is a geometric series:

The sequence 𝑥 𝑛 = 1 2 𝑛 satisfies all the given conditions:

• 𝑥 𝑖+1 ≤ 𝑥 𝑖 for all 𝑖 ≥ 0,

• 𝑛-1 𝑘=0 𝑥

Step 1: Simplify the General Term. The general term of the series is:

Notice that the denominator can be factored as: 36𝑖 2 -1 = (6𝑖 -1)(6𝑖 + 1). Thus, the denominator becomes:

(6𝑖 -1) 2 (6𝑖 + 1) 2 .

Step 2: Partial Fraction Decomposition. Let 𝑥 = 6𝑖. The term becomes:

Assume the partial fraction decomposition: Optimization. Using the method of Lagrange multipliers, we consider the Lagrangian:

Taking partial derivatives and solving the system leads us to consider the case where 𝑏 = 𝑐 = 0. The constraint then reduces to 𝑎𝑑 = 0.

Case 1: 𝑎 = 0. The distance becomes:

The minimum occurs at 𝑑 = 2, yielding a distance of 1.

Case 2: 𝑑 = 0. The distance becomes:

(1 -𝑎) 2 + 4.

The minimum occurs at 𝑎 = 1, yielding a distance of 4.