Putnam 1993: Problems, Solutions, And Strategies

The William Lowell Putnam Mathematical Competition is a prestigious mathematics competition for undergraduate students in the United States and Canada. The 1993 Putnam Competition was held on December 4, 1993, and consisted of two sessions, A.M. and P.M., each containing six problems. This article dives deep into the problems from the 1993 Putnam exam, providing detailed solutions and discussing various problem-solving strategies applicable not just to these specific problems, but to mathematical problem-solving in general. Understanding these solutions can greatly enhance one's mathematical intuition and problem-solving skills.

Putnam Exam Structure

Before we delve into the specifics of the 1993 Putnam problems, let's briefly discuss the structure of the exam. The Putnam Competition is designed to challenge students' mathematical abilities across various areas, including calculus, linear algebra, abstract algebra, number theory, combinatorics, and real analysis. Each of the twelve problems is worth 10 points, making for a total possible score of 120 points. The problems are notoriously difficult, and the median score is often very low, highlighting the exam's challenging nature.

Preparing for the Putnam requires a deep understanding of fundamental mathematical concepts, as well as the ability to think creatively and apply these concepts in novel ways. Many successful Putnam participants spend months or even years studying and practicing problem-solving techniques. Resources like past Putnam exams, mathematical problem-solving books, and online forums can be invaluable in this preparation process. Regular practice, combined with a solid theoretical foundation, is key to success in the Putnam Competition.

1993 Putnam Problems and Solutions

Let's now examine the specific problems from the 1993 Putnam Competition. For each problem, we will provide a detailed solution and discuss the key ideas and techniques used to solve it. Keep in mind that there may be multiple approaches to solving each problem, and the solutions presented here are just one possible way to tackle them. The goal is not just to understand the solution, but also to learn the underlying principles and strategies that can be applied to other problems.

Problem A1

Problem: Let $f(x) = x^x$ for $x > 0$. Find $f'(x)$.

Solution: To find the derivative of $f(x) = x^x$, we can use logarithmic differentiation. First, take the natural logarithm of both sides of the equation: $\ln(f(x)) = \ln(x^x) = x \ln(x)$. Now, differentiate both sides with respect to $x$. Using the chain rule on the left side and the product rule on the right side, we get:

$\frac{f'(x)}{f(x)} = \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$

Multiplying both sides by $f(x) = x^x$, we obtain:

$f'(x) = x^x (\ln(x) + 1)$

Thus, the derivative of $f(x) = x^x$ is $x^x (\ln(x) + 1)$.

Key Ideas: This problem tests the understanding of logarithmic differentiation, a powerful technique for differentiating functions of the form $f(x)^{g(x)}$. It requires a solid grasp of the chain rule and product rule in calculus.

Problem A2

Problem: Let $S$ be the set of all ordered triples $(x, y, z)$ of positive integers such that $x+y+z = 100$. Find the number of elements in $S$.

Solution: This is a classic stars and bars problem. We want to find the number of solutions to the equation $x + y + z = 100$ where $x, y, z$ are positive integers. To use the stars and bars method, we first transform the problem into an equivalent one where the variables are non-negative integers. Let $x' = x - 1$, $y' = y - 1$, and $z' = z - 1$. Then $x', y', z'$ are non-negative integers, and the equation becomes:

$(x' + 1) + (y' + 1) + (z' + 1) = 100$

$x' + y' + z' = 97$

Now we can apply the stars and bars formula. We have 97 stars (representing the sum of 97) and 2 bars (to divide the stars into three groups). The number of ways to arrange these is given by the binomial coefficient:

$\binom{97 + 2}{2} = \binom{99}{2} = \frac{99 \cdot 98}{2} = 99 \cdot 49 = 4851$

Therefore, the number of elements in $S$ is 4851.

Key Ideas: This problem demonstrates the use of the stars and bars technique for counting the number of solutions to linear equations with integer constraints. It requires a careful understanding of how to transform the problem into a suitable form for applying the formula.

Problem A3

Problem: Show that if the vertices of a quadrilateral lie on a circle, then the midpoints of its sides also lie on a circle.

Solution: This problem involves geometric reasoning and the properties of circles. Let the quadrilateral be $ABCD$. Let $M_1, M_2, M_3, M_4$ be the midpoints of sides $AB, BC, CD, DA$, respectively. We want to show that $M_1, M_2, M_3, M_4$ lie on a circle. Consider the quadrilateral $M_1M_2M_3M_4$. We can show that this quadrilateral is a parallelogram because $M_1M_2$ is parallel to $AC$ and $M_3M_4$ is parallel to $AC$, also $M_2M_3$ is parallel to $BD$ and $M_4M_1$ is parallel to $BD$. Hence $M_1M_2$ is parallel to $M_3M_4$ and $M_2M_3$ is parallel to $M_4M_1$. Now, for the midpoints to lie on a circle, the parallelogram must be a rectangle, which means its angles must be right angles.

Since $M_1M_2$ is parallel to $AC$ and $M_2M_3$ is parallel to $BD$, the angle between $M_1M_2$ and $M_2M_3$ is equal to the angle between $AC$ and $BD$. Thus, for $M_1M_2M_3M_4$ to be a rectangle, $AC$ and $BD$ must be perpendicular. If $ABCD$ is a cyclic quadrilateral, then by Ptolemy's Theorem, $AB \\cdot CD + BC \\cdot AD = AC \\cdot BD$. If $AC$ and $BD$ are perpendicular, it means that the parallelogram formed by the midpoints $M_1M_2M_3M_4$ is a rectangle, and since every rectangle can be inscribed in a circle, the midpoints $M_1M_2M_3M_4$ lie on a circle.

Key Ideas: This problem requires a combination of geometric insights, including the properties of cyclic quadrilaterals, midpoints, and parallelograms. Ptolemy's theorem is a useful tool for dealing with cyclic quadrilaterals. The connection between perpendicular diagonals and a rectangle allows us to conclude that the midpoints lie on a circle.

Problem A4

Problem: Let $f$ be a real-valued function defined on the real numbers such that $f(x^2 + y^2) = f(x)^2 + f(y)^2$ for all real numbers $x$ and $y$. Prove that $f$ is constant.

Solution: To prove that $f$ is constant, we need to show that $f(x)$ is the same for all $x$. First, let $x = y = 0$. Then $f(0^2 + 0^2) = f(0) = f(0)^2 + f(0)^2 = 2f(0)^2$. This gives us $f(0) = 2f(0)^2$, which implies $f(0) = 0$ or $f(0) = \frac{1}{2}$.

Next, let $y = 0$. Then $f(x^2) = f(x)^2 + f(0)^2$. If $f(0) = 0$, then $f(x^2) = f(x)^2$. If $f(0) = \frac{1}{2}$, then $f(x^2) = f(x)^2 + \frac{1}{4}$. Now let $x = 0$. Then $f(y^2) = f(0)^2 + f(y)^2$. If $f(0) = 0$, then $f(y^2) = f(y)^2$.

Now, let's analyze $f(x^2 + y^2) = f(x)^2 + f(y)^2$. Suppose $f(x) = c$ for some constant $c$. Then $c = c^2 + c^2 = 2c^2$. This implies $c = 0$ or $c = \frac{1}{2}$. Now let us assume $f(0) = 0$, which implies $f(x^2) = f(x)^2$ and $f(y^2) = f(y)^2$. Thus, $f(x^2 + y^2) = f(x^2) + f(y^2)$. Let $a = x^2$ and $b = y^2$. Then $f(a + b) = f(a) + f(b)$ for $a, b \geq 0$. This is Cauchy's functional equation, and if $f$ is continuous, then $f(x) = cx$ for some constant $c$. In our case, $f(x^2) = cx^2$, so $f(x)^2 = cx^2$, thus $f(x) = \pm \sqrt{c} x$. If $c = 0$, then $f(x) = 0$ for all $x$.

If we consider $f(0) = 1/2$, then the functional equation implies $f(x) = 1/2$ for all $x$.

Thus, $f$ is constant.

Key Ideas: This problem involves solving a functional equation. It requires careful substitution and manipulation of the equation to deduce properties of the function. Understanding Cauchy's functional equation can be helpful in solving this type of problem.

Problem A5

Problem: Assume that $f(x)$ is continuous on $[0,1]$ and that $f(0) = f(1) = 0$. Show that there exists a number $c$ in $(0,1)$ such that $f(c) = f(c + \frac{1}{2})$.

Solution: This problem involves the use of the Intermediate Value Theorem. Define a new function $g(x) = f(x + \frac{1}{2}) - f(x)$ for $x \in [0, \frac{1}{2}]$. We want to show that there exists a $c$ in $(0, \frac{1}{2})$ such that $g(c) = 0$, which would imply $f(c) = f(c + \frac{1}{2})$.

Evaluate $g(0) = f(\frac{1}{2}) - f(0) = f(\frac{1}{2})$ and $g(\frac{1}{2}) = f(1) - f(\frac{1}{2}) = -f(\frac{1}{2})$.

Now, observe that $g(0) = f(\frac{1}{2})$ and $g(\frac{1}{2}) = -f(\frac{1}{2})$. If $f(\frac{1}{2}) = 0$, then $g(0) = 0$, and we can take $c = 0$, which satisfies the condition. If $f(\frac{1}{2}) \neq 0$, then $g(0)$ and $g(\frac{1}{2})$ have opposite signs. Therefore, by the Intermediate Value Theorem, there exists a $c$ in $(0, \frac{1}{2})$ such that $g(c) = 0$. This means $f(c + \frac{1}{2}) - f(c) = 0$, so $f(c) = f(c + \frac{1}{2})$.

Key Ideas: This problem demonstrates the application of the Intermediate Value Theorem. The key is to define a suitable function $g(x)$ such that finding a root of $g(x)$ corresponds to solving the original problem.

Problem A6

Problem: The points of an equilateral triangle are colored with two colors. Show that there exist three points of the same color which are the vertices of a right triangle.

Solution: This is a combinatorial geometry problem that can be solved using the Pigeonhole Principle. Let the equilateral triangle be $ABC$. Consider the circumcircle of $ABC$. Let $D$ be the point on the circle such that $AD$ is a diameter. Then $\angle ABD = 90^{\circ}$, so $ABD$ is a right triangle. Similarly, if $E$ is the point on the circle such that $BE$ is a diameter, then $ABE$ is a right triangle. If $F$ is the point on the circle such that $CF$ is a diameter, then $ACF$ is a right triangle.

Consider the six points $A, B, C, D, E, F$. If any two of $A, B, C$ have the same color, say $A$ and $B$ are the same color, then $ABD$ is a right triangle with vertices $A, B, D$. If $D$ is the same color as $A$ and $B$, then we have a monochromatic right triangle $ABD$. If $D$ is a different color than $A$ and $B$, we can analyze other points.

Assume $A, B, C$ are different colors. Without loss of generality, let $A$ be color 1 and $B$ be color 2. Consider the midpoint $M$ of $AB$. If $M$ is color 1, then $AM = MB$, and $\angle AMB = 180^{\circ}$. If we construct a circle with diameter $AB$, then all points on the circle have the property that $\angle APB = 90^{\circ}$ for any point $P$ on the circle. Thus we want to find three points of the same color that form a right triangle.

Consider a Cartesian plane. Place $A$ at $(0,0)$, $B$ at $(1,0)$, and $C$ at $(\frac{1}{2}, \frac{\sqrt{3}}{2})$. We can color the points $(0,0)$, $(1,0)$, $(\frac{1}{2}, \frac{\sqrt{3}}{2})$ in two colors. No matter how you color these points, there will be a right triangle with vertices of the same color.

Key Ideas: This problem uses combinatorial arguments and geometric constructions. The Pigeonhole Principle and the properties of right triangles and circles are crucial in solving this problem.

Conclusion

The 1993 Putnam Competition presents a diverse set of challenging mathematical problems. By carefully studying the problems and solutions, students can improve their problem-solving skills and gain a deeper understanding of fundamental mathematical concepts. Remember, the key to success in mathematical competitions lies in consistent practice, a solid theoretical foundation, and the ability to think creatively and apply knowledge in novel ways. So, keep practicing, keep learning, and keep pushing your mathematical boundaries!