# 2006 AIME I Problems/Problem 8

## Problem

Hexagon $ABCDEF$ is divided into five rhombuses, $\mathcal{P, Q, R, S,}$ and $\mathcal{T,}$ as shown. Rhombuses $\mathcal{P, Q, R,}$ and $\mathcal{S}$ are congruent, and each has area $\sqrt{2006}.$ Let $K$ be the area of rhombus $\mathcal{T}$. Given that $K$ is a positive integer, find the number of possible values for $K$. $[asy] // TheMathGuyd size(8cm); pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); draw(A--B--C--D--EE--F--cycle); draw(F--G--(2.1,0)); draw(C--H--(2.1,0)); draw(G--(2.1,-3.2)); draw(H--(2.1,-3.2)); label("\mathcal{T}",(2.1,-1.6)); label("\mathcal{P}",(0,-1),NE); label("\mathcal{Q}",(4.2,-1),NW); label("\mathcal{R}",(0,-2.2),SE); label("\mathcal{S}",(4.2,-2.2),SW); [/asy]$

## Solution 1

Let $x$ denote the common side length of the rhombi. Let $y$ denote one of the smaller interior angles of rhombus $\mathcal{P}$. Then $x^2\sin(y)=\sqrt{2006}$. We also see that $K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y$. Thus $K$ can be any positive integer in the interval $(0, 2\sqrt{2006})$. $2\sqrt{2006} = \sqrt{8024}$ and $89^2 = 7921 < 8024 < 8100 = 90^2$, so $K$ can be any integer between 1 and 89, inclusive. Thus the number of positive values for $K$ is $\boxed{089}$.

## Solution 2

Call the side of each rhombus w. w is the width of the rhombus. Call the height h, where $w*h=\sqrt{2006}$. The height of rhombus T would be 2h, and the width would be $\sqrt{w^2-h^2}$. Substitute the first equation to get $\sqrt{\frac{2006}{h^2}-h^2}$. Then the area of the rhombus would be $2h * \sqrt{\frac{2006}{h^2}-h^2}$. Combine like terms to get $2 * \sqrt{2006-h^4}$. This expression equals an integer K. $2006-h^4$ specifically must be in the form $n^2/2$. There is no restriction on h as long as it is a positive real number, so all we have to do is find all the positive possible values of $n^2$ for $2006-h^4$. Now, quick testing shows that $44^2 < 2006$ and $45^2>2006$, but we must also test $44.5^2$, because the product of two will make it an integer. $44.5^2$ is also less than $2006$, so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5 is also a valid value. (notice 0 is not valid because the height must be a positive number) That gives us $44*2+1=$ $\boxed{089}$

-jackshi2006

## Solution 3 $[asy] size(8cm); pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); draw(A--B--C--D--EE--F--cycle); label("A",A,2*N); label("B",B,2*N); label("C",C,2*E); label("D",D,2*S); label("E",EE,2*S); label("F",F,2*W); label("G",(0.47,-1.55),NW); label("H",(3.73,-1.55),NE); label("I",I,2*N); label("J",J,2*S); label("K",K,2*SW); draw(F--C); draw(F--G--(2.1,0)); draw(C--H--(2.1,0)); draw(G--(2.1,-3.2)); draw(H--(2.1,-3.2)); draw((2.1,0)--(2.1,-3.2)); [/asy]$ To determine the possible values of $[GIHJ],$ we must determine the maximum and minimum possible areas.

In the case where the $4$ rhombi are squares, we have $[GIHJ]=0,$ implying the minimum possible positive-integer-valued area is $1.$

Denote the length $HC=a$ and $KH=b.$ We have $$KI=\sqrt{a^2-b^2}$$ by the Pythagorean Theorem, which implies $$[IBCH]=a\sqrt{a^2-b^2}=\sqrt{2006}$$ and $$[GIHJ]=2b\sqrt{a^2-b^2}.$$ The first equation yields $$\sqrt{a^2-b^2}=\frac{\sqrt{2006}}{a}.$$ Plugging into the second, we have $$[GIHJ]=2\sqrt{2006}\frac{b}{a}.$$ The maximal value of $\frac{b}{a}$ occurs when the height of $ABCDEF$ is minimized, which means $$\frac{b}{a}\leq 1.$$ Plugging back up, we have $$[GIHJ]\leq 2\sqrt{2006}=\sqrt{8024}.$$ We have $$\lfloor \sqrt{8024} \rfloor = 89,$$ thus our answer is $$89-1+1=\boxed{089}.$$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 