geometras 4 all

Geometry Level 3

The figure shows two circles of different radii, the circle with centre at A has radius r =a, the circle with centre at B has

radius r = b , The two circles are tangential to each other at point G. there is another small circle with centre at C and

radius r = c which is in turn tangential to both circles at points H and I , the three circles are all tangential to the line DFE .

find the value of c in terms of a and b .

If a = 16 , b = 9 , then c = m n \frac{m}{n} , where m and n are positive integers , find m + 16 n − 41 \frac{m +16}{n - 41} .


The answer is 20.

Aziz Alasha by Aziz Alasha , 59, Sudan

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1 solution

Arjen Vreugdenhil
Dec 12, 2017

We will solve the problem in general. Let p = D F p = DF and q = F E q = FE .

The segment A B = a + b AB = a + b may be viewed as the hypotenuse of a right triangle with width p + q p + q and height a − b a - b . Pythagoras gives ( a + b ) 2 = ( a − b ) 2 + ( p + q ) 2 ∴ ( p + q ) 2 = 4 a b , p + q = 2 a b . (a+b)^2 = (a - b)^2 + (p+q)^2\ \ \ \ \therefore\ \ \ \ (p+q)^2 = 4ab,\ \ p + q = 2\sqrt{ab}. Applying a similar strategy to A C AC and B C BC , we obtain ( a + c ) 2 = ( a − c ) 2 + p 2 ∴ p 2 = 4 a c , p = 2 a c ; (a+c)^2 = (a - c)^2 + p^2\ \ \ \ \therefore\ \ \ \ p^2 = 4ac,\ \ p = 2\sqrt{ac}; ( b + c ) 2 = ( b − c ) 2 + q 2 ∴ q 2 = 4 b c , q = 2 b c . (b+c)^2 = (b - c)^2 + q^2\ \ \ \ \therefore\ \ \ \ q^2 = 4bc,\ \ q = 2\sqrt{bc}. Now ( p + q ) 2 = p 2 + q 2 + 2 p q ; (p+q)^2 = p^2 + q^2 + 2pq; 4 a b = 4 a c + 4 b c + 8 a b c = 4 c ( a + b ) 2 ; 4ab = 4ac + 4bc + 8\sqrt{ab}c = 4c\left(\sqrt a + \sqrt b\right)^2; c = a b ( a + b ) 2 . c = \frac{ab}{\left(\sqrt a + \sqrt b\right)^2}.

With the given values, c = 16 ⋅ 9 ( 4 + 3 ) 2 = 144 49 ; c = \frac{16\cdot 9}{(4 + 3)^2} = \frac{144}{49}; we submit the value of 144 + 16 49 − 41 = 160 8 = 20 . \frac{144+16}{49 - 41} = \frac{160}{8} = \boxed{20}.

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