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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

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Solution

Consider that ΔABC is right angled at point B.

Let AB=x and BC=y.

Let P be the point on the hypotenuse in such a way that it is at a distance of a from the side AB and at a distance of b from the side BC.

Consider that C=θ.



Now, by Pythagoras theorem,

AC= x 2 + y 2

Also,

PC=bcscθ

And,

AP=asecθ

Since, AC=AP+PC, then,

AC=bcscθ+asecθ

Differentiate both sides with respect to θ,

d( AC ) dθ =bcscθcotθ+asecθtanθ(1)

Put d( AC ) dθ =0,

asecθtanθbcscθcotθ=0 asecθtanθ=bcscθcotθ a× 1 cosθ × sinθ cosθ =b× 1 sinθ × cosθ sinθ a sin 3 θ=b cos 3 θ

Simplify further,

tan 3 θ= b a tanθ= ( b a ) 1/3

It can be observed that,

sinθ= b 1/3 a 2/3 + b 2/3 (2)

And,

cosθ= a 1/3 a 2/3 + b 2/3 (3)

Differentiate equation (1) with respect to θ, we get

d 2 ( AC ) d θ 2 <0

Thus, it can be concluded that the length of the hypotenuse is maximum when tanθ= ( b a ) 1/3 .

Using equation (2) and (3) in AC=bcscθ+asecθ,

AC=b× a 2/3 + b 2/3 b 1/3 +a× a 2/3 + b 2/3 a 1/3 = a 2/3 + b 2/3 ( a 2/3 + b 2/3 ) = ( a 2/3 + b 2/3 ) 3/2

Hence, it is proved that the maximum length of the hypotenuse is ( a 2/3 + b 2/3 ) 3/2 .


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