Prove the following:

$$ \frac{1}{3} = \frac{1+3}{5+7} = \frac{1+3+5}{7+9+11} = \cdots. $$

Winner: Chris Howard

Runner-Up: Andrew Sansom

Winner: Arnav Iyer

Runner-Up: Aryan Agarwal

Let $0<x<1$. Simplify as much as possible:

$$ \ln(x) \ln(1-x) + \sum_{n=1}^{\infty} \frac{x^{n}}{n^2} +\sum_{n=1}^{\infty} \frac{(1-x)^{n}}{n^2}. $$

Winner: Weilun Sun

Runner-Up: Hubert Yang

Sept. 2019

Let $a$ and $b$ be real numbers with $b\neq 0$. Determine:

$$ \int_{0}^{2\pi} \frac{d\theta}{\cos^{2}(\theta) + 2a\cos(\theta)\sin(\theta) + (a^{2}+b^{2})\sin^{2}(\theta)}. $$

Apr. 2019

Determine all positive integer solutions to the system of equations:

$$ xy=z+w$$

$$ zw=x+y.$$

Winner: Rhythm Garg

Runner-Up: Jingyi Dai

Winner: Edoardo Luna

Runner-Up: Rhythm Garg

Take a square piece of paper with vertices labeled $A, B, C, D$ when proceeding in the counterclockwise direction with vertex $A$ in the upperleft corner. Now take the side $AB$ and fold it in such a way that $A$ gets folded to a point $A'$ and $B$ to $B'$ where $B'$ is on $CD$ and $A'$ is outside the square $ABCD$. Then $A'B'$ intersects $AD$ at $E$. Next draw a circle that is tangent to $CD$, tangent to $AD$ and tangent to $A'B'$. Denote the radius of this circle by $r$. Prove that $r= A'E.$

Hint: A well-known fact from geometry states that $r= \frac{2 \textrm{Area}(T)}{\textrm{Perimeter}(T)} $ where $T$ is the triangle $EB'D$.

Winner: Rhythm Garg

Runner-Up: Fernando Moreno

Winner: Rhythm Garg

Runner-Up: Trenton Hicks

Let $0<a<b$. Determine:

$$ \int_{a}^{b} \cos^{-1}\left( \frac{x}{\sqrt{(a+b)x - ab}} \right)\, dx. $$

Winner: Tien Le

Runner-Up: Rhythm Garg

Investigate the convergence of:

$$ \sum_{n=3}^{\infty} \frac{1}{\ln(\ln(n))^{\ln(n)}} $$

and:

$$ \sum_{n=3}^{\infty} \frac{1}{\ln(\ln(n))^{\ln(\ln(n))}} $$

Let $a>0$ and $b>0$. Determine:

$$ \int_{0}^{2\pi} \frac{ab}{a^2 \cos^{2}(t) + b^2\sin^{2}(t)} \, dt. $$

Determine:

$$ \sin\left(\frac{\pi}{n}\right) \sin\left(\frac{2\pi}{n}\right) \sin\left(\frac{3\pi}{n}\right) \cdots \sin\left(\frac{(n-1)\pi}{n}\right). $$

Hint: Note that: $$ z^{n}-1 = (z-1)(z - e^{\frac{2\pi i}{n}}) (z - e^{\frac{4\pi i}{n}}) (z - e^{\frac{6\pi i}{n}}) \cdots (z - e^{\frac{2(n-1)\pi i}{n}}). $$

Let $x>1$ and define $f(e)=e.$ For $x>1$ and $x\neq e$ let $f(x)>0$ be the unique number such that $f(x)\neq x$ and $x^{f(x)} = f(x)^{x}.$Sketch the graph of $f(x)$ and calculate $f'(x)$ (you may leave this answer in terms of $x$ and $f(x)$). Finally calculate $f'(e)$.

Jan. 2018

Let $a, b, c$ be real numbers such that $ax^{2} + 2bxy +cy^2>0$ for all $(x,y)\neq (0,0)$. Determine: $$ \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \left( e^{-(ax^2 + 2bxy + cy^2)} \right) \, dx \, dy.$$ You may use the well-known fact that for $H>0$: $$ \int_{-\infty}^{\infty} e^{-Ht^2} \, dt = \sqrt\frac{\pi}{H}. $$

Determine: $$ \sum_{n=2}^{\infty} (-1)^{n}\frac{\ln(n)}{n}. $$ Express your answer in terms of Euler's constant, $\gamma$.Note that: $\gamma = \lim\limits_{n \to \infty} \left( 1 + \frac{1}{2} + \cdots \frac{1}{n} - \ln(n) \right). $Hint: You may assume there exists a constant $A$ such that: $$ \lim\limits_{n \to \infty} (\frac{\ln(2)}{2} + \frac{\ln(3)}{3} + \cdots + \frac{\ln(n)}{n} - \frac{1}{2}\ln^{2}(n) ) = A. $$

Winner: Linda Yu

Runner-Up: Austin McGregor

Winner: Brandon Ohl

Runner-Up: Ethan Seal

Consider the sum:
$$ \begin{array}{1 2 3 4 5 6 7 }
\ & \ & \textrm{M} & \textrm{A} & \textrm{J} & \textrm{O} & \textrm{R} \\
\ & + & \textrm{M} & \textrm{I} & \textrm{N} & \textrm{O} & \textrm{R} \\
= & \textrm{R} & \textrm{E} & \textrm{S} & \textrm{U} & \textrm{M} & \textrm{E} \end{array}
$$

where each letter represents a different nonnegative integer 0, 1, .... 9. What is the largest number that RESUME can represent and still have the equation be true?

Note: There are 10 different letters in this sum so all of the nonnegative integers from 0 to 9 will be used exactly once.

Winner: Ethan Seal

Runner-Up: Brandon Ohl

Let $A=(0,2)$, $B = (3,0)$. Find a point $C$ on the unit circle centered at the origin so that triangle $ABC$ is of largest area.

Winner: William Liu

Runner-Up: Xiangyu Kong

November 2016

Winner: William Liu

Runner-Up: Aoxue Chen

October 2016

a. Let $A$ and $B$ be two linear transformations from ${\mathbb R}^N \to {\mathbb R}^N.$ Show that it is impossible for $AB-BA=I$ where $I$ is the identity map.

b. On the other hand show that it is possible to find two linear transformations (defined on infinite dimensional spaces) with $AB-BA=I$.

Winner: William Liu

Determine $$ \lim_{n \to \infty} \frac{1}{n^4} \prod_{i=1}^{2n} (n^2 + i^2)^{\frac{1}{n}}. $$

Note: The $\prod$ sign is a short-hand for product. That is: $\prod\limits_{i=1}^{2n} b_i = b_{1}b_{2} \cdots b_{2n-1} b_{2n}.$

Winner: William Liu

Runner-Up: Christopher Lee

Winner: William Liu

Runner-Up: Jagath Vytheeswaran

$$ \min_{p} \max_{x \in [a,b]} \frac{|x-p|}{x}. $$

Winner: William Liu

Runner-Up Brandon Ohl

Let $f_{n}$ be the Fibonacci sequence. Determine $$\sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{f_{2n+1}} \right). $$

Hint: Use a trig identity and the following identity which holds for the Fibonacci sequence: $f_{n+1}f_{n+2} -f_{n}f_{n+3} = (-1)^n. $

Determine all nonnegative continuous functions which satisfy:

$$ f(x+t) = f(x) + f(t) + 2 \sqrt{f(x)}\sqrt{f(t)} \textrm{ for } x \geq 0, t \geq 0. $$

December 2014

Winner: Murray Lee

Runner Up: Steven Grigsby

Winner: Steven Grigsby

Runner Up: Mark Fincher

September 2014

Simplify: $$ \frac{1}{\sqrt[3]{1} + \sqrt[3]{2} + \sqrt[3]{4}} + \frac{1}{\sqrt[3]{4} + \sqrt[3]{6}+\sqrt[3]{9}} + \frac{1}{\sqrt[3]{9} + \sqrt[3]{12} + \sqrt[3]{16}} $$

Winner: Tamaki Ueno

Runner Up: Murray Lee

April 2014

Squared Matrices and Characteristic Polynomials. Let $A$ be a real $4\times4$-matrix with characteristic polynomial $p(\lambda)=\det(\lambda{I}-A)=\lambda^4-s_1\lambda^3+s_2\lambda^2-s_3\lambda+s_4.$ Show that if $A$ admits real square roots, in the sense that there is a real $4\times4$-matrix $S$ such that $S^2=A,$ then $s_1+s_2+s_3+s_4\ge-1.$

Winner: Mark Fincher

Runner-Up: Chia-Ting Han

Winner: Kevin Lin

Runner-Up: Chia-Ting Han

Winner: Chia-Ting Han

Runner-Up: Kevin Lin

Winner: Chia-Ting Han

Runner-Up: Kevin Lin

Winner: Kevin Lin

Runner-Up: Chris James

Integration Bee Warm-Up. Find an anti-derivative of the function $$f(x)=\sqrt{\sqrt{\sqrt{x}+1}-\sqrt{\sqrt{x}-1}},\;x>1.$$

Winner: Heather Olney

Runner-Up: Alyssa Sylvester