2006 AIME I Problems/Problem 8

Problem

Hexagon $ABCDEF$ is divided into four 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 $K.$

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

Solution

Let $x$ denote the common side length of the rhombi. Let $y$ denote one of the smaller interior angles of rhombus $\mathcal{P}$ . Then $x^2sin(y)=\sqrt{2006}.$ $K=x^2\sin(2y) \Rightarrow K=2x^2siny\cdot cosy \Rightarrow K= 2\sqrt{2006}\cdot cosy.$ Thus $K$ is any positive integer on ( $0, 2\sqrt{2006}$ ). $2\sqrt{2006}\approx 89.58$ . Hence, the number of positive values for $K$ is ${089}$ .

Solution provided by 1337h4x

Page

Toolbox

Search

2006 AIME I Problems/Problem 8

Problem

Solution

See also