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A rectangle is inscribed by a triangle with sides of $5$, $8$, and $\sqrt{41}$. Find the largest area of the rectangle!

I'm not exactly sure how to compute this area. I'm looking for the fastest possible way to solve, for this type of problem appears quite often in math competitions.

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Let $\Delta ABC$ be our triangle and $KLMN$ be our rectangle.

There are three cases:

1) $KN\subset AC$;

2) $KN\subset AB$ and

3) $KN\subset BC$.

In the first case let $L\in AB$ and $M\in BC$.

By the Heron's formula $$S_{\Delta ABC}=\frac{1}{4}\sqrt{2(5^28^2+5^2(\sqrt{41})^2+8^2(\sqrt{41})^2)-5^4-(\sqrt{41})^4-8^4}=16.$$ Let $BD$ be an altitude of $\Delta ABC$.

Thus, $$\frac{8BD}{2}=16,$$ which gives $$BD=4.$$ Now, let $MN=x$ and since $\Delta LBM\sim\Delta ABC,$ we obtain $$\frac{4-x}{4}=\frac{ML}{8},$$ which gives $$ML=2(4-x).$$ Thus, by AM-GM $$S_{KLMN}=2x(4-x)\leq2\left(\frac{x+4-x}{2}\right)^2=8.$$ The equality occurs for $x=2$, which says that $8$ is a maximal value in this case.

By the same way we can consider another cases and choose the maximal value.

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  • $\begingroup$ In general, can this problem also be solved with geometric applications? $\endgroup$ – A Piercing Arrow Nov 21 '17 at 5:47
  • $\begingroup$ I don't know to draw in the net. Sorry. Try to draw it by yourself. $\endgroup$ – Michael Rozenberg Nov 21 '17 at 5:50

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