Question

A point on the hypotenuse of a right triangle is at distances a and b from the sides making right angle. (a, b constant). Prove that the hypotenuse has minimum length(a²/³+b²/³)³/².

Answer 1

Let ∆ABC is a right - angled triangle at B where AB = x and BC = y.

from Pythagoras theorem,

AC = $\bf{\sqrt{x^2+y^2}}$

Let P is the point on the hypotenuse of triangle ABC in such a way that P is at a distance of a and b from the sides AB and BC respectively.

Let angle BCA = θ

see figure, PC = bcosecθ

AP = asecθ

also, AC = AP + PC

AC = asecθ + bcosecθ

differentiate AC with respect to θ,

d(AC)/dθ = asecθ.tanθ + bcosecθ(-cotθ)

= asecθ.tanθ - bcosecθ.cotθ

now, if d(AC)/dθ = 0

then, asecθ.tanθ - bcosecθ.cotθ = 0

a(sinθ/cos²θ) - b(cosθ/sin²θ) = 0

asin³θ - bcos³θ = 0

asin³θ = bcos³θ

sin³θ/cos³θ = b/a

tan³θ = b/a

tanθ = $\bf{[\frac{b}{a}]^{1/3}}$

so, sinθ = $\bf{\frac{a^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}}$

and cosθ = $\bf{\frac{b^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}}$

it is also clear that $\frac{d^2AC}{d\theta^2}$ < 0 when tanθ = (b/a)^{1/3}

Therefore, by second derivative test, the length of the hypotenuse is the maximum when tanθ = (b/a)^{1/3}

now, when tanθ = (b/a)^{1/3} we get ,

AC = bsecθ + acosecθ

= b/cosθ + a/sinθ

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

solve it , we get,

AC = $\bf{[a^{2/3}+b^{2/3}]^{3/2}}$

hence proved