# 12) A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle.Show that the minimum length of the hypotenuse is$( a ^{\frac{2}{3}}+ b ^\frac{2}{3}) ^ \frac{3}{2}$

It is given that
A point on the hypotenuse of a triangle is at a distance a and b from the sides of the triangle

Let the angle between  AC and BC is $\theta$
So, the angle between AD and ED is also $\theta$
Now,
CD = $b \ cosec\theta$
And
AD = $a \sec\theta$
AC = H = AD + CD
=  $a \sec\theta$ + $b \ cosec\theta$
$\frac{dH}{d\theta} = a \sec\theta\tan\theta - b\cot\theta cosec \theta\\ \frac{dH}{d\theta} = 0\\ a \sec\theta\tan\theta - b\cot\theta cosec \theta =0\\ a \sec\theta\tan\theta = b\cot\theta cosec \theta\\ a\sin^3\theta = b\cos^3\theta\\ \tan^3\theta = \frac{b}{a} \\ \tan\theta = \left ( \frac{b}{a} \right )^\frac{1}{3}$
Now,
$\frac{d^2H}{d\theta^2} > 0$
When  $\tan\theta = \left ( \frac{b}{a} \right )^\frac{1}{3}$
Hence, $\tan\theta = \left ( \frac{b}{a} \right )^\frac{1}{3}$ is the point of minima
$\sec \theta = \frac{a\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{a^\frac{1}{3}}$ and  $cosec \theta = \frac{b\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{b^\frac{1}{3}}$

AC = $\frac{a\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{a^\frac{1}{3}} +$ $\frac{b\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{b^\frac{1}{3}}$ = $(a^\frac{2}{3}+b^\frac{2}{3})^\frac{3}{2}$
Hence proved

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