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Difference between revisions of "1962 AHSME Problems/Problem 37"
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where the first inequality was from the Trivial Inequality, and the second came from dividing both sides by <math>2</math>, which does not change the inequality sign.
 
where the first inequality was from the Trivial Inequality, and the second came from dividing both sides by <math>2</math>, which does not change the inequality sign.
 
Thus, the maximum area is <math>\boxed{\frac{5}{8}},</math> when <math>\frac{(x-\frac{1}{2})^2}{2} = 0.</math>
 
Thus, the maximum area is <math>\boxed{\frac{5}{8}},</math> when <math>\frac{(x-\frac{1}{2})^2}{2} = 0.</math>
 +
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~ kcbhatraju

Revision as of 11:00, 11 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 $\frac{5}{8}$

Solution 2

Let us first draw a unit square. [asy] draw((0,0)--(0,1)); draw((0,1)--(1,1)); draw((1,1)--(1,0)); draw((1,0)--(0,0)); label("A",(0,1),NW); label("B",(1,1),NE); label("C",(1,0),SE); label("D",(0,0),SW); label("$1$",(0,0)--(0,1),W); [/asy] We will now pick arbitrary points $E$ and $F$ on $\overline{AB}$ and $\overline{AD}$ respectively. We shall say that $AE = AF = x$ [asy] draw((0,0)--(0,1)); draw((0,1)--(1,1)); draw((1,1)--(1,0)); draw((1,0)--(0,0)); label("A",(0,1),NW); label("B",(1,1),NE); label("C",(1,0),SE); label("D",(0,0),SW); label("$1$",(0,0)--(0,1),W); draw((1,0)--(0.5,1),blue); draw((0.5,1)--(0,0.5),blue); draw((0,0.5)--(0,0),blue); draw((0,0)--(1,0),blue); label("E",(0.5,1),N); label("F",(0,0.5),SE); [/asy] Thus, our problem has been simplified to maximizing the area of the blue quadrilateral. If we drop an altitude from $E$ to $\overline{DC}$, and call the foot of the altitude $G$, we can find the area of $EFDC$ by noting that $[EFDC] = [EGDF] + [EGC]$. [asy] draw((0,0)--(0,1)); draw((0,1)--(1,1)); draw((1,1)--(1,0)); draw((1,0)--(0,0)); label("A",(0,1),NW); label("B",(1,1),NE); label("C",(1,0),SE); label("D",(0,0),SW); label("$1$",(0,0)--(0,1),W); draw((1,0)--(0.5,1),blue); draw((0.5,1)--(0,0.5),blue); draw((0,0.5)--(0,0),blue); draw((0,0)--(1,0),blue); label("E",(0.5,1),N); label("F",(0,0.5),SE); draw((0.5,1)--(0.5,0),red); draw((0.5,0.1)--(0.4,0.1),red); draw((0.4,0.1)--(0.4,0),red); label("G",(0.5,0),S); [/asy] We can now finish the problem.

Since $EG = 1$ and $FD = 1-x$, we have: \[[EFDC] = [EGDF] + [EGC] = DG\cdot\frac{EG + FD}{2} + \frac{(CG)(EG)}{2} = \frac{x(2-x)}{2} + \frac{1-x}{2} = \frac{1+x-x^2}{2}\]

To maximize this, we compete the square in the numerator to have: \[[EFDC] = \frac{-(x-\frac{1}{2})^2 + \frac{5}{4}}{2} = \frac{5}{8} - \frac{(x-\frac{1}{2})^2}{2}\]

Finally, we see that $\frac{(x-\frac{1}{2})^2}{2}\geq0$, as: \[\left(x-\frac{1}{2}\right)^2\geq0\] So, \[\frac{(x-\frac{1}{2})^2}{2}\geq0,\]

where the first inequality was from the Trivial Inequality, and the second came from dividing both sides by $2$, which does not change the inequality sign. Thus, the maximum area is $\boxed{\frac{5}{8}},$ when $\frac{(x-\frac{1}{2})^2}{2} = 0.$

~ kcbhatraju

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