Play with triangle

Geometry Level 5

If the bisector of the angle A of triangle ABC meets BC in D, then a a = ( b + c ) [ 1 − A D x b c ] y (b+c){[1-\frac{{AD}^{x}}{bc}]}^{y} where a a , b b & c c are the length of sides opposite to angle A, B & C respectively and x x & y y are rational numbers. Find the value x y xy .

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The answer is 1.

Purushottam Abhisheikh by Purushottam Abhisheikh , 24, India

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1 solution

By Stewart's Theorem we know that the length A D AD is:

A D = b c ( ( b + c ) 2 − a 2 ) b + c AD=\dfrac{\sqrt{bc((b+c)^2-a^2)}}{b+c}

So all we have to do is solve for a a :

A D ( b + c ) = b c ( ( b + c ) 2 − a 2 ) A D 2 ( b + c ) 2 = b c ( ( b + c ) 2 − a 2 ) A D 2 ( b + c ) 2 b c = ( b + c ) 2 − a 2 a 2 = ( b + c ) 2 − A D 2 ( b + c ) 2 b c a 2 = ( b + c ) 2 ( 1 − A D 2 b c ) a = ( b + c ) 1 − A D 2 b c a = ( b + c ) [ 1 − A D 2 b c ] 1 2 AD(b+c)=\sqrt{bc((b+c)^2-a^2)} \\ AD^2(b+c)^2=bc((b+c)^2-a^2) \\ \dfrac{AD^2(b+c)^2}{bc}=(b+c)^2-a^2 \\ a^2=(b+c)^2-\dfrac{AD^2(b+c)^2}{bc} \\ a^2=(b+c)^2\left(1-\dfrac{AD^2}{bc}\right) \\ a=(b+c)\sqrt{1-\dfrac{AD^2}{bc}} \\ a=(b+c)\left[1-\dfrac{AD^2}{bc}\right]^{\frac{1}{2}}

Comparing we get x = 2 x=2 and y = 1 2 y=\dfrac{1}{2} , so x y = 2 ( 1 2 ) = 1 xy=2\left(\dfrac{1}{2}\right)=\boxed{1} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

you need to solve only half.....by dimensional analysis x is 2

Rushikesh Joshi - 5 years, 11 months ago

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Nothing good in your approach of dimensions who will do the other half.

Spandan Senapati - 4 years, 3 months ago

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