## Problem of the Day #124: Circle Inscribed Within an Isosceles Trapezoid
*July 21, 2011*

*Posted by Saketh in : potd , add a comment*

A circle is inscribed inside isosceles trapezoid $ABCD$. The circle intersects diagonal $\overline{AC}$ twice, once at point $E$ and again at point $F$. Suppose $E$ lies between $A$ and $F$. The value of

$$\frac{AF \cdot EC}{AE \cdot FC}$$

can be expressed in the form $a + \sqrt{b}$, where $a$ and $b$ are positive integers. Compute the value of $ab$.

## Problem of the Day #123: The Alberts’ Triangle Game
*July 20, 2011*

*Posted by Billy in : potd , add a comment*

Albert Sr. (Senior) and Albert Jr. (Junior) are playing a game. Senior picks a random point inside a triangle with sides $5$, $7$, and $9$. He then draws three lines, one through each of the vertices of the triangle, that also go through the point he picked. This divides the triangle into six smaller triangles. Junior then picks the three smaller triangles with the largest areas. If the sum of the areas of the three triangles is more than $\frac{2}{3}$ of the area of the original triangle, Junior wins the game. If Senior and Junior play $10000$ games, what is the closest integer to the number of games Junior can expect to win?

## End of Second Two-Week Period
*July 19, 2011*

*Posted by Sreenath in : announcement , add a comment*

Please join us in congratulating Lewis Chen, the winner of the second two-week period. This period spanned problems $108$ through $121$.

As the top-scoring solver over the past two weeks that hasn’t already won a two-week prize, Lewis wins a $\$10$ gift card of his choice. You can view the full rankings for these two weeks after the jump. (more…)

## Problem of the Day #122: I like pentagons
*July 19, 2011*

*Posted by Mitchell in : potd , add a comment*

Let ABCDE be a pentagon with $AB \parallel CE$, $BC \parallel AD$, $AC \parallel DE$, $\angle ABC=120^\circ$, $AB=6$, $BC=10$, and $DE = 30$. The ratio of the area of $\triangle ABC$ to the area of $\triangle EBD$ can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Compute $a+b$.

## Problem of the Day #121: qwertyasdfzxcvb
*July 18, 2011*

*Posted by Seungln in : potd , 3 comments*

Sreenath likes mashing. (This link might clarify it a little bit.) Sreenath will start from the top row and randomly choose a key to start with and a key to end with. The key to end with can be the same as the key to start with, but cannot be to the left of the starting key, because Sreenath will only mash to the right. He will then repeat the procedure with the home row and then the bottom row. (The top row is the row from ‘$q$’ to ‘]’, the home row is from ‘$a$’ to ‘;’, and the bottom row is from ‘$z$’ to ‘.’.) If Sreenath is to spam Albert with every possible mash that has **more than** $5$ characters, with how many distinct mashes can Sreenath spam Albert? (e.g. The mash on the title, ‘qwertyasdfzxcvb’, is one that Sreenath will send to Albert. The mash ‘qwaz’ is not, because it is too short. The mash ‘asdfghjkl;’, while long enough, does not take inputs from all three rows, and will not be sent. The mash ‘qwertyzxcvbasdf’, while long enough and containing keys in all three rows, is not done in the right order, and will also not be sent.)

## Problem of the Day #120: Fun With Quartics
*July 17, 2011*

*Posted by Albert in : potd , add a comment*

The quartic $$f(x) = x^4 + 17 x^3 + A x^2 + B x + 3600$$ has roots $r_1, r_2, r_3, r_4$. If $r_1 r_2 = r_3 r_4$, find $|B|$.

## Problem of the Day #119: A Triangle with a Missing Point
*July 16, 2011*

*Posted by Alex in : potd , add a comment*

On $\triangle ABC$, point $A$ is located at $(3, 5)$ and point $B$ is located at $(-1, 2)$. Point $C$ is located a distance of $3$ from the origin. Given that the area of $\triangle ABC$ is $4$, the coordinates of point $C$ can be expressed in the form of $(a, b)$, where $a$ and $b$ are real numbers. Find $|1200a – 1600b|$.

You may use a five-function calculator ($+, -, \times, \div, \sqrt{\;}$), but one is not required.

## Problem of the Day #118: Pooh’s Diophantine
*July 15, 2011*

*Posted by Saketh in : potd , 1 comment so far*

Winnie the Pooh is trying to find solutions for the equation below. Each distinct solution $(x,y,z)$ will earn him a jar of honey. $$413x + 792y + z = 327096$$ Being a bear of very little brain, he can only work with non-negative integers. How many jars of honey can he get?

## Problem of the Day #117: Las cajas del Cid
*July 14, 2011*

*Posted by Sreenath in : potd , add a comment*

The legendary Indian national hero Sid has $3$ boxes. He also regularly receives gifts from his many fans. When he receives a gift, he randomly places it in one of his $3$ boxes. Sid has received $10$ gifts so far. The expected value of the number of empty boxes can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Compute $a+b$.

## Problem of the Day #116: Binomial coefficient sum
*July 13, 2011*

*Posted by Mitchell in : potd , add a comment*

Given that \[\sum_{n = 1003}^{2006} \frac{1}{n} \binom{n}{2006 - n} = \frac{p}{q},\] where $p$ and $q$ are relatively prime integers with $q > 0$, find the smallest positive integer $m$ such that $qm – p$ is divisible by $2011$.