It's all Circles and Trapezoids.

Level pending

Let b > a b > a .

A circle with radius r r is inscribed in a right trapezoid with base lengths a a and b b and the larger circle with radius R R goes thru the vertices P , Q , T P,Q,T of the right trapezoid as shown above.

Let h h be the height of the right trapezoid.

If R 2 h = 2 a b 3 + b 4 4 ( a + b ) 2 R^2 - h = \dfrac{2ab^3 + b^4}{4(a + b)^2} and the area A A of the inscribed circle can be expressed as A = α β π A = \dfrac{\alpha}{\beta}\pi , where α \alpha and β \beta are coprime positive integers, find α + β \alpha + \beta ..


The answer is 89.

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1 solution

Rocco Dalto
Sep 24, 2018

For smaller circle:

( a + b 2 r ) 2 = ( 2 r ) 2 + ( b a ) 2 a 2 + 2 a b + b 2 4 ( a + b ) r + 4 r 2 = 4 r 2 + b 2 2 a b + a 2 4 a b = 4 ( a + b ) r 2 (a + b - 2r)^2 = (2r)^2 + (b - a)^2 \implies a^2 + 2ab + b^2 - 4(a + b)r + 4r^2 = 4r2 + b^2 - 2ab + a^2 \implies 4ab = 4(a + b)r^2

r = a b a + b \implies \boxed{r = \dfrac{ab}{a + b}}

For outer circle ( x x 0 ) 2 + ( y y 0 ) 2 = R 2 (x - x_{0})^2 + (y - y_{0})^2 = R^2 .

Using the three points on the larger circle I chose above:

( 0 , 0 ) : x 0 2 + y 0 2 = R 2 (0,0): x_{0}^2 + y_{0}^2 = R^2

( 0 , 2 r ) : x 0 2 + 4 r 2 4 r y 0 + y 0 2 = R 2 4 r ( r y 0 ) = 0 r 0 y 0 = r = a b a + b (0,2r): x_{0}^2 + 4r^2 - 4ry_{0} + y_{0}^2 = R^2 \implies 4r(r - y_{0}) = 0 \:\ r \neq 0 \implies y_{0} = r = \dfrac{ab}{a + b}

( b , 0 ) : b 2 2 b x 0 + x 0 2 + y 0 2 = R 2 x 0 = b 2 (b,0):b^2 - 2bx_{0} + x_{0}^2 + y_{0}^2 = R^2 \implies x_{0} = \dfrac{b}{2}

R 2 = b 2 4 + a 2 b 2 ( a + b ) 2 \implies \boxed{R^2 = \dfrac{b^2}{4} + \dfrac{a^2b^2}{(a + b)^2}}

The height of the trapezoid h = 2 r = 2 a b a + b h = 2r = \dfrac{2ab}{a + b} \implies

R 2 h = b 2 4 + a 2 b 2 ( a + b ) 2 2 a b a + b = R^2 - h = \dfrac{b^{2}}{4} + \dfrac{a^2b^2}{(a + b)^2} - \dfrac{2ab}{a + b} = 5 a 2 b 2 + 2 a b 3 + b 4 8 a 2 b 8 a b 2 4 ( a + b ) 2 = 2 a b 3 + b 4 4 ( a + b ) 2 \dfrac{5a^2b^2 + 2ab^3 + b^4 - 8a^2b - 8ab^2}{4(a + b)^2} = \dfrac{2ab^3 + b^4}{4(a + b)^2} \implies a b ( 5 a b 8 a 8 b ) = 0 a , b 0 5 a b = 8 ( a + b ) r = a b a + b = 8 5 A = 64 25 π = α β π α + β = 89 ab(5ab - 8a - 8b) = 0 \:\ a,b \neq 0 \implies 5ab = 8(a + b) \implies r = \dfrac{ab}{a + b} = \dfrac{8}{5} \implies A = \dfrac{64}{25}\pi = \dfrac{\alpha}{\beta}\pi \implies \alpha + \beta = \boxed{89} .

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