1972 AHSME Problems/Problem 30

Revision as of 20:50, 22 August 2017 by E power pi times i (talk | contribs) (Created page with "== Problem 30 == == Solution == Let the rectangle <math>ABCD</math> have crease <math>BE</math> with <math>E</math> on <math>CD</math>, and let <math>F</math> be on <math>A...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem 30

Solution

Let the rectangle $ABCD$ have crease $BE$ with $E$ on $CD$, and let $F$ be on $AD$ such that $F$ is a reflection of $C$ over $BE$. Notice that triangles $ABF$ and $DEF$ are similar, so by setting $CE = EF = x$ with $DE = 6-x$, giving $DF = 2\sqrt{3x-9}$ we have that $AF = \frac{18-3x}{\sqrt{3x-9}}$. Noticing that $BC = BF = x\tan{\theta}$ gives $\frac{(18-3x)^2}{3x-9}+36 = x^2\tan^2{\theta} \Rightarrow \frac{3(x-6)^2}{x-3}+36 = \frac{3x^2}{x-3} = x^2\tan^2{\theta}$. $x = \frac{3\tan^2{\theta}+3}{\tan^2{\theta}} = \frac{3\sec^2{\theta}}{\tan^2{\theta}} = 3\csc^2{\theta}$. Noticing that $BE = x\sqrt{\tan^2{\theta}+1} = x\sec{\theta}$ gives the answer to be $3\csc^2{\theta}\sec{\theta}$ which is not in the answer list?

If anyone could correct my solution, please help and thanks! P.S correct answer is $3\sec^2{\theta}\csc{\theta}$.