Dynamic Geometry: P81

Geometry Level 4

The diagram shows an orange circle with radius 25 25 and a purple circle with radius 16 16 which is internally tangent to the orange circle. The cyan semicircle is moving freely so it's internally tangent to the orange semicircle and tangent to the purple circle at any moment. Using the three centers, we draw a black triangle. When the perimeter of the black triangle is equal to 56 56 , the area of the cyan semicircle is equal to p q π \dfrac{p}{q}\pi , where p p and q q are coprime positive integers. Find p + q p+q .


The answer is 51.

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1 solution

Chew-Seong Cheong
Mar 18, 2021

Let the centers of the large circle, purple circle, and cyan semicircle be O O , P P , and C C respectively, A B AB and D E DE be diameters passing through centers C C and O O respectively, and O C = k OC = k . By intersecting chords theorem ,

A C C B = D C C E r r = ( 25 k ) ( 25 + k ) r 2 = 625 k 2 k = 625 r 2 \begin{aligned} AC \cdot CB & = DC \cdot CE \\ r \cdot r & = (25-k)(25+k) \\ r^2 & = 625 - k^2 \\ \implies k & = \sqrt{625-r^2} \end{aligned}

The perimeter of O P C \triangle OPC is given by:

C O + O P + P C = 56 k + ( 25 16 ) + 16 + r = 56 625 r 2 = 31 r Squaring both sides 625 r 2 = r 2 62 r + 961 2 r 2 62 r + 336 = 0 r 2 31 r + 168 = 0 ( r 7 ) ( r 24 ) = 0 r = 7 24 is too big. \begin{aligned} CO+OP+PC & = 56 \\ k + (25-16) + 16 + r & = 56 \\ \sqrt{625-r^2} & = 31 - r & \small \blue{\text{Squaring both sides}} \\ 625 - r^2 & = r^2 - 62 r + 961 \\ 2r^2 - 62 r + 336 & = 0 \\ r^2 - 31 r + 168 & = 0 \\ (r - 7)(r-24) & = 0 \\ \implies r & = 7 & \small \blue{\text{24 is too big.}} \end{aligned}

Then the area of the cyan semicircle is 49 2 π \dfrac {49}2\pi and p + q = 49 + 2 = 51 p+q = 49+2 = \boxed{51} .

Well you are just destroying my problems right now, I hope the next problems will oppose more resistance!

Valentin Duringer - 2 months, 3 weeks ago

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