2006 AIME I Problems/Problem 8: Difference between revisions

Revision as of 19:35, 11 March 2007

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 $K$ .

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 $x^2\sin(y)=\sqrt{2006}$ . We also see that $\displaystyle K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y$ . 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 089.

Revision as of 19:22, 29 November 2006 view source Xantos C. Guin (talk \| contribs) editor 580 edits there are 5 not 4 rhombuses according to the original problem. ← Older edit		Revision as of 19:35, 11 March 2007 view source K1c2h2o (talk \| contribs) 59 edits mNo edit summary Newer edit →
Line 3:		Line 3:


	~~{{image}}~~		[[Image:2006AimeA8.PNG]]

Page

Toolbox

Search

2006 AIME I Problems/Problem 8: Difference between revisions

Revision as of 19:35, 11 March 2007

Problem

Solution

See also