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