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Difference between revisions of "1962 AHSME Problems/Problem 37"

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==Solution==
 
==Solution==
 
{{solution}}
 
{{solution}}
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Let <math>AE=AF=x</math>
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<math>[CDFE]=[ABCD]-[AEF]-[EBC]=1-\frac{x^2}{2}-\frac{1-x}{2}</math>
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Or
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<math>[CDFE]=\frac{\frac{5}{4}-(x-\frac{1}{2})^2}{2}\le \frac{5}{8}</math>
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As <math>(x-\frac{1}{2})^2\ge 0</math>
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So <math>[CDFE]\le \frac{5}{8}</math>
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Equality occurs when <math>AE=AF=x=\frac{1}{2}</math>
 +
So maximum value is <math>\frac{5}{8}</math>

Revision as of 07:20, 6 July 2018

Problem

$ABCD$ is a square with side of unit length. Points $E$ and $F$ are taken respectively on sides $AB$ and $AD$ so that $AE = AF$ and the quadrilateral $CDFE$ has maximum area. In square units this maximum area is:

$\textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ \frac{9}{16}\qquad\textbf{(C)}\ \frac{19}{32}\qquad\textbf{(D)}\ \frac{5}{8}\qquad\textbf{(E)}\ \frac{2}3$


Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. Let $AE=AF=x$ $[CDFE]=[ABCD]-[AEF]-[EBC]=1-\frac{x^2}{2}-\frac{1-x}{2}$ Or $[CDFE]=\frac{\frac{5}{4}-(x-\frac{1}{2})^2}{2}\le \frac{5}{8}$ As $(x-\frac{1}{2})^2\ge 0$ So $[CDFE]\le \frac{5}{8}$ Equality occurs when $AE=AF=x=\frac{1}{2}$ So maximum value is $\frac{5}{8}$

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