Let $ABCD$ be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that \[ \frac{1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|. \] When does equality hold?

Suppose $a_1, \dots, a_n$ are integers whose greatest common divisor is 1. Let $S$ be a set of integers with the following properties: (a) For $i=1, \dots, n$, $a_i \in S$. (b) For $i,j = 1, \dots, n$ (not necessarily distinct), $a_i - a_j \in S$. (c) For any integers $x,y \in S$, if $x+y \in S$, then $x-y \in S$. Prove that $S$ must be equal to the set of all integers.

For what real values of $k>0$ is it possible to dissect a $1 \times k$ rectangle into two similar, but noncongruent, polygons?

Alice and Bob play a game on a 6 by 6 grid. On his or her turn, a player chooses a rational number not yet appearing in the grid and writes it in an empty square of the grid. Alice goes first and then the players alternate. When all squares have numbers written in them, in each row, the square with the greatest number in that row is colored black. Alice wins if she can then draw a line from the top of the grid to the bottom of the grid that stays in black squares, and Bob wins if she can't. (If two squares share a vertex, Alice can draw a line from one to the other that stays in those two squares.) Find, with proof, a winning strategy for one of the players.

Click for solution Here's a way: Denote columns as c1, c2, c3, c4, c5, c6. Show that Bob has the ability to do one of the following in any row: (the basic strategy for Bob is to always place numbers in the row Alice put hers) #1 Force the black square to be in one of c1, c2, or c3. #2 Force the black square to be in one of c4, c5, or c6. #3 Force the black square to be in one of c1, c2, c5, c6. Now let Bob execute #1 on row 1, #3 on row 2, and #2 on row 3. If, in the second row, black square is in one of c1, c2, then it cannot be connected to row 3. (case 1) If, in the second row, black square is in one of c5, c6, then it cannot be connected to row 1. (case 2) In either case, Bob wins. _: Grid boxes X: Possible locations of black square Case 1: X X X_ _ _ X X _ _ _ _ _ _ _ X X X Case 2: X X X_ _ _ _ _ _ _ X X _ _ _ X X X So in fact, Bob wins in a 6x3 grid.

Let $a, b, c > 0$. Prove that $(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \geq (a + b + c)^3$.

Click for solution This resembles APMO 2004/5 so much. Note that if $x>0$ then $x^5-x^2+3 \geq x^3+2$, which can be seen by factoring into $(x^3-1)(x^2-1) \geq 0$. Then the inequality reduces to \[ \prod (a^3 +2 ) \geq (a+b+c)^3 \] or, \[ \prod (\frac{a^3-1}{3}+1) \geq (\frac{a+b+c}{3})^3 \] If $a,b,c \geq 1$, or $a,b,c \leq 1$, then by Bernoulli's Inequality, we have \[ \prod (\frac{a^3-1}{3}+1) \geq 1 + \sum \frac{a^3-1}{3} =\sum a^3/3 \] Complete by Power-Mean If $a \geq 1 \geq b,c$ or $a \leq 1 \leq b,c$, then by Bernoulli and Holder, we have \[ \prod (\frac{a^3-1}{3}+1) \geq (\frac{a^3+1+1}{3})(1+\frac{b^3-1}{3}+\frac{c^3-1}{3})\geq (\frac{a^3+1+1}{3})(\frac{1+b^3+c^3}{3})(\frac{1+1+1}{3}) \geq (\frac{a+b+c}{3})^3 \] Q.E.D. Note: Bernoulli's inequality (or perhaps an extension of it, anyone know its name?) states that if $a_1, \ldots, a_n > -1$, and that all the {a} have the same sign, then $\prod(1+a) \geq 1+\sum a$.

A circle $\omega$ is inscribed in a quadrilateral $ABCD$. Let $I$ be the center of $\omega$. Suppose that \[ (AI + DI)^2 + (BI + CI)^2 = (AB + CD)^2. \] Prove that $ABCD$ is an isosceles trapezoid.

These problems are copyright $\copyright$ Mathematical Association of America.