An Olympiad Problem

The semiperimeter of $ABC$ is $\large\ s = \frac{20 + 21 + 22}{2} = \frac{63}{2}$ . By Heron's formula, the area of the whole triangle is $\large\ A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{21\sqrt{1311}}{4}$ . Using the formula $A = rs$ , we find that the inradius is $\large\ r = \frac{A}{s} = \frac{\sqrt{1311}}6$ . Since $\triangle ADE \sim \triangle ABC$ , the ratio of the heights of triangles $ADE$ and $ABC$ is equal to the ratio between sides $DE$ and $BC$ . From $\large\ A=\frac{1}{2}bh$ , we find $\large\ h_{ABC} = \frac{21\sqrt{1311}}{40}$ . Thus, we have

$\huge\ \frac{h_{ADE}}{h_{ABC}} = \frac{h_{ABC}-r}{h_{ABC}} = \frac{21\sqrt{1311}/40-\sqrt{1311}/6}{21\sqrt{1311}/40}=\frac{DE}{20}.$

Solving for $DE$ gives $DE= \large\ \frac{860}{63}$ , so the answer is $m+n=\boxed{923}$ .

Area of a triangle = r * s, where r is the in radius and s semi-parameter.
Area of a triangle = (1/2)base * a , a=altitude.
$\therefore \ r*31.5=0.5*20*a. \ \implies \dfrac ra=\dfrac{20}{63}.\\ Since \ DE || BC, \dfrac{DE}{BC}=\dfrac{a-r}a=1- \dfrac r a=1-\dfrac{20}{63}=\dfrac{43}{63}.\\ DE=\dfrac{43}{63}*20=\dfrac m n. \ \ \therefore m+n=\large \ \ 923$

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