Difference between revisions of "1962 AHSME Problems/Problem 37"
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==Solution== | ==Solution== | ||
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Let <math>AE=AF=x</math> | Let <math>AE=AF=x</math> | ||
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So <math>[CDFE]\le \frac{5}{8}</math> | So <math>[CDFE]\le \frac{5}{8}</math> | ||
Equality occurs when <math>AE=AF=x=\frac{1}{2}</math> | Equality occurs when <math>AE=AF=x=\frac{1}{2}</math> | ||
− | So maximum value is <math>\frac{5}{8}</math> | + | So the maximum value is <math>\frac{5}{8}</math> |
==Solution 2== | ==Solution 2== |
Revision as of 22:36, 25 July 2020
Problem
is a square with side of unit length. Points
and
are taken respectively on sides
and
so that
and the quadrilateral
has maximum area. In square units this maximum area is:
Solution
Let
Or
As
So
Equality occurs when
So the maximum value is
Solution 2
Let us first draw a unit square.
We will now pick arbitrary points
and
on
and
respectively. We shall say that
Thus, our problem has been simplified to maximizing the area of the blue quadrilateral.
If we drop an altitude from
to
, and call the foot of the altitude
, we can find the area of
by noting that
.
We can now finish the problem.
Since and
, we have:
To maximize this, we compete the square in the numerator to have:
Finally, we see that , as:
So,
where the first inequality was from the Trivial Inequality, and the second came from dividing both sides by , which does not change the inequality sign.
Thus, the maximum area is
or
when