Question

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

Answer 1

Let $P$ be a point on the hypotenuse $AC$ of right $△ABC$ such that

$PL\perp AB=a$

$PM\perp BC=b$

Let $\angle APL=\angle ACB=\theta$

$AP=a\sec \theta$ and $PC=b\text{cosec}\theta$

Let $l$ be the length of the hypotenuse, then $l=AP+PC$

$\Rightarrow a\sec \theta +b\text{cosec}\theta$, $0<\theta <\frac{\pi }{2}$

Now differentiate $l$ with respect to $\theta$, we get

$\frac{dl}{d\theta }=a\sec \theta .\tan \theta -b\text{cosec}\theta .\cot \theta$

For maxima and minima $\frac{dl}{d\theta }=0$

Thus $a\sec \theta .\tan \theta =b\text{cosec}\theta .\cot \theta$

$\Rightarrow a{\sin }^3\theta =b{\cos }^3\theta$

$\frac{a}{b}=\frac{{\cos }^3\theta }{{\sin }^3\theta }$

Thus $\frac{b}{a}={\tan }^3\theta$

$\Rightarrow \tan \theta ={\left[\frac{b}{a}\right]}^{\frac{1}{3}}$

Now $\frac{d^{2}l}{d{\theta }^2}=a(\sec \theta .{\sec }^2\theta +\tan \theta .\sec \theta .\tan \theta )$

$=-b\left(\text{cosec}\theta \right(-{\text{cosec}}^2\theta )+\cot \theta (-\text{cosec}\theta .\cot \theta \left)\right)$

$=a\sec \theta ({\sec }^2\theta +{\tan }^2\theta )+b\text{cosec}\theta \times b\text{cosec}\theta +{\cot }^2\theta$

Since $0<\theta <\frac{\pi }{2}$, so all $t$ ratios of $\theta$ are positive.

Also $a>0$ and $b<0$

Therefore, $\frac{d^{2}l}{d{\theta }^2}$ is positive.

$\Rightarrow$ $l$ is least when $\tan \theta ={\left[\frac{b}{a}\right]}^{\frac{1}{3}}$

Least value of $l$ $=a\sec \theta +b\text{cosec}\theta$

$=\frac{a.\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{a^{\frac{1}{3}}}+\frac{b.\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{b^{\frac{1}{3}}}$

$=\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}(a^{\frac{2}{3}}+b^{\frac{2}{3}})$

$=(a^{\frac{2}{3}}+b^{\frac{2}{3}}{)}^{\frac{3}{2}}$

Hence, proved.