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, asecθtanθ−bcscθcotθ=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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