Three half circles and one mini circle

Geometry Level 3

Given K A = 3 KA=3 and L B = 2 LB=2 . If the radius of circle with center P P can be written as a b \dfrac{a}{b} , where a a and b b are positive coprime integers, what is a + b a+b ?

Source: Syrian math olympiad-regional 2019


The answer is 49.

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2 solutions

Nibedan Norman
May 31, 2019

Solution by: Nibedan Norman Mukherjee

Let the radius of circle with center P P be r r and the center of the biggest semicircle be O ( 0 , 0 ) O(0,0) . Then A ( 2 , 0 ) A(-2,0) and B ( 3 , 0 ) B(3,0) and point P P satisfies the following three equations of circle:

{ ( x + 2 ) 2 + y 2 = ( 3 + r ) 2 . . . ( 1 ) ( x 3 ) 2 + y 2 = ( 2 + r ) 2 . . . ( 2 ) x 2 + y 2 = ( 5 r ) 2 . . . ( 3 ) \begin{cases} (x+2)^2 + y^2 = (3+r)^2 & ...(1) \\ (x-3)^2 + y^2 = (2+r)^2 & ...(2) \\ x^2 + y^2 = (5-r)^2 & ...(3) \end{cases}

{ x 2 + 4 x + 4 + y 2 = 9 + 6 r + r 2 . . . ( 1 ) x 2 6 x + 9 + y 2 = 4 + 4 r + r 2 . . . ( 2 ) x 2 + y 2 = 25 10 r + r 2 . . . ( 3 ) \begin{cases} x^2 + 4x+4 + y^2 = 9 + 6r+ r^2 & ...(1) \\ x^2 - 6x+9 + y^2 = 4 + 4r+ r^2 & ...(2) \\ x^2 + y^2 = 25 -10 r+ r^2 & ...(3) \end{cases}

{ ( 1 ) ( 3 ) : 4 x + 4 = 16 + 16 r x = 4 r 5 . . . ( 4 ) ( 3 ) ( 2 ) : 6 x 9 = 21 14 r 3 x = 15 7 r . . . ( 5 ) \begin{cases} (1) - (3): & 4 x + 4 = - 16 + 16r & \implies x = 4r - 5 & ...(4) \\ (3)-(2): & 6x - 9 = 21 - 14 r & \implies 3x = 15 - 7r & ...(5) \end{cases}

3 × ( 4 ) ( 5 ) : 19 r 30 = 0 r = 30 19 \begin{aligned} 3\times (4) - (5): \quad 19r - 30 & = 0 \\ \implies r & = \frac {30}{19} \end{aligned}

Therefore, a + b = 19 + 30 = 49 a+b = 19+30 = \boxed{49} .

I think you meant (3)-(2) instead of (3)-(1) :)

Mehdi K. - 2 years ago

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