Question

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

Answer 1

The correct answer is $=(a^\frac{2}{3}+b^\frac{2}{3})^\frac{3}{2}$
Let $∆ABC$ be right-angled at $B$.Let $AB=x$ and $BC=y$.
Let $P$ be a point on the hypotenuse of the triangle such that $P$ is at a distance of $a$ and $b$ from the sides $AB$ and $BC$ respectively.
Let $∴C=θ$.

We have,$AC=\sqrt{x^2+y^2}$
Now,
$PC=b\,cosecθ$
And,$AP=a\,secθ$
$∴AC=AP+PC$
$∴ AC=b\,cosecθ+a\,secθ … (1)$
$∴\frac{d(AC)}{dθ}=-b\,cosecθ\,cotθ+a\,secθ\,tanθ$
$∴\frac{d(AC)}{dθ}=0$
$⇒a\,secθ\,tanθ=b\,cosecθ\,cotθ$
$⇒\frac{a}{cosθ}.\frac{sinθ}{cosθ}=\frac{b}{sinθ}\frac{cosθ}{sinθ}$
$⇒a\,sin^3θ=b\,cos^3θ$
$⇒(a)^{\frac{1}{3}}sinθ=(b)^\frac{1}{3}cosθ$
$⇒tanθ=(\frac{b}{a})^\frac{1}{3}$
$∴sinθ=\frac{(b)^\frac{1}{3}}{\sqrt{a^\frac{2}{3}+d^\frac{2}{3}}} \,and\, cosθ=\frac{(a)^\frac{1}{3}}{\sqrt{a^\frac{2}{3}+d^\frac{2}{3}}} .....(2)$
It can be clearly shown that $\frac{d^2(AC)}{dθ}<0$ when $tanθ=(\frac{b}{a})^\frac{1}{3}$.
Now when $tanθ=(\frac{b}{a})^\frac{1}{3}$,we have
$AC=\frac{b\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{b^\frac{1}{3}}+\frac{a\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{a^\frac{1}{3}}$ [Using(1) and(2)]
$=\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}(b^\frac{2}{3}+a^\frac{2}{3})$
$=(a^\frac{2}{3}+b^\frac{2}{3})^\frac{3}{2}$
Hence, the maximum length of the hypotenuses is $=(a^\frac{2}{3}+b^\frac{2}{3})^\frac{3}{2}$