Dynamic Geometry: P130

Geometry Level pending

The diagram shows a unit black square. A red quarter circle is drawn using the side of the square as the radius. Two yellow semicircles are drawn. Each center moves on one square's side so the semicircles are growing, shrinking and are tangent to each other at any moment. Then, we draw a cyan circle so it's tangent to the two yellow semicircles and internally tangent to the red quarter circle. Finally we inscribed a blue circle between the cyan circle and the two yellow semicircles. We use the tangency points to draw a green triangle inscribed in the blue circle. When the radius of the cyan circle is equal to 6 35 \dfrac{6}{35} , the area of the green triangle can be expressed as p q \dfrac{p}{q} , where p p and q q are coprime positive integers. Find p + q − 7 p+q-7 .


The answer is 934102.

Valentin Duringer by Valentin Duringer , 30, Belgium

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Chew-Seong Cheong
May 20, 2021

Let the unit square be A B C D ABCD , the centers and radii of the semicircles and circles be O O , P P , S S , and T T , and r 0 r_0 , r 1 r_1 , r 2 r_2 , and r 3 r_3 as shown. By Pythagorean theorem ,

A O 2 + A P 2 = O P 2 ( 1 − r 0 ) 2 + ( 1 − r 1 ) 2 = ( r 0 + r 1 ) 2 1 − r 0 − r 1 = r 0 r 1 ⟹ r 1 = 1 − r 0 1 + r 0 \begin{aligned} AO^2 + AP^2 & = OP^2 \\ (1-r_0)^2 + (1-r_1)^2 & = (r_0+r_1)^2 \\ \blue{1 - r_0 - r_1} & \blue{= r_0r_1} \\ \implies r_1 & = \frac {1-r_0}{1+r_0} \end{aligned}

By Descartes' theorem ,

1 r 0 + 1 r 1 − 1 1 + 2 1 r 0 r 1 − 1 r 0 − 1 r 1 = 1 r 2 r 0 + r 1 r 0 r 1 − 1 + 2 1 − r 0 − r 1 r 0 r 1 = 1 r 2 Note that 1 − r 0 − r 1 = r 0 r 1 1 − r 0 r 1 r 0 r 1 + 1 = 1 r 2 ⟹ r 0 r 1 = r 2 = 6 35 and r 1 = 1 − r 0 1 + r 0 r 0 ( 1 − r 0 ) 1 + r 0 = 6 35 35 r 0 2 − 29 r 0 + 6 = 0 ( 5 r 0 − 2 ) ( 7 r 0 − 3 ) = 0 Since r 0 and r 1 are swappable, ⟹ r 0 , r 1 = 2 5 , 3 7 \begin{aligned} \frac 1{r_0} + \frac 1{r_1} - \frac 11 + 2 \sqrt{\frac 1{r_0r_1} - \frac 1{r_0} - \frac 1{r_1}} & = \frac 1{r_2} \\ \frac {r_0+r_1}{r_0r_1} - 1 + 2 \sqrt{\frac \blue{1-r_0-r_1}{r_0r_1}} & = \frac 1{r_2} & \small \blue{\text{Note that }1-r_0-r_1 = r_0 r_1} \\ \frac {1-r_0r_1}{r_0r_1} + 1 & = \frac 1{r_2} \\ \implies r_0\blue{r_1} & = r_2 = \frac 6{35} & \small \blue{\text{and }r_1 = \frac {1-r_0}{1+r_0}} \\ \frac {r_0(1-r_0)}{1+r_0} & = \frac 6{35} \\ 35r_0^2 - 29r_0+6 & = 0 \\ (5r_0-2)(7r_0-3) & = 0 & \small \blue{\text{Since }r_0 \text{ and }r_1 \text{ are swappable,}} \\ \implies r_0, r_1 & = \frac 25, \frac 37 \end{aligned}

Let r 0 = 2 5 r_0 = \dfrac 25 and r 1 = 3 7 r_1 = \dfrac 37 . Similarly,

1 r 3 = 1 r 0 + 1 r 1 + 1 r 2 + 2 1 r 0 r 1 + 1 r 1 r 2 + 1 r 2 r 0 Note that r 0 r 1 = r 2 = r 0 + r 1 r 2 + 1 r 2 + 2 1 r 2 ( 1 + 1 r 0 + 1 r 1 ) = 1 r 2 − 1 + 1 r 2 + 2 1 r 2 ( 1 + 1 r 2 − 1 ) = 4 r 2 − 1 = 70 3 − 2 = 67 3 ⟹ r 3 = 3 67 \begin{aligned} \frac 1{r_3} & = \frac 1{r_0} + \frac 1{r_1} + \frac 1{r_2} + 2 \sqrt{\frac 1{r_0r_1} + \frac 1{r_1r_2} + \frac 1{r_2r_0}} & \small \blue{\text{Note that }r_0r_1 = r_2} \\ & = \frac {r_0+r_1}{r_2} + \frac 1{r_2} + 2 \sqrt{\frac 1{r_2}\left(1+\frac 1{r_0}+\frac 1{r_1}\right)} \\ & = \frac 1{r_2} - 1 + \frac 1{r_2} + 2 \sqrt{\frac 1{r_2}\left(1+\frac 1{r_2} - 1\right)} \\ & = \frac 4{r_2} - 1 = \frac {70}3 - 2 = \frac {67}3 \\ \implies r_3 & = \frac 3{67} \end{aligned}

To find the area of the triangle we first find the measures of its three central angles by cosine rule and then use 1 2 r 3 2 sin ⁡ θ \frac 12 r_3^2 \sin \theta to find the areas of these three smaller triangle.

cos ⁡ α = ( r 0 + r 3 ) 2 + ( r 1 + r 3 ) 2 − ( r 0 + r 1 ) 2 2 ( r 0 + r 3 ) ( r 1 + r 3 ) = r 3 ( r 0 + r 1 + r 3 ) − r 0 r 1 ( r 0 + r 3 ) ( r 1 + r 3 ) = 3465 5513 ⟹ sin ⁡ α = 1 − cos ⁡ 2 α = 4288 5513 Similarly sin ⁡ β = 5628 6253 sin ⁡ γ = 22780 25181 \begin{aligned} \cos \alpha & = \frac {(r_0+r_3)^2+(r_1+r_3)^2 - (r_0+r_1)^2}{2(r_0+r_3)(r_1+r_3)} \\ & = \frac {r_3(r_0+r_1+r_3)-r_0r_1}{(r_0+r_3)(r_1+r_3)} = \frac {3465}{5513} \\ \implies \sin \alpha & = \sqrt{1-\cos^2 \alpha} = \frac {4288}{5513} & \small \blue{\text{Similarly }} \\ \sin \beta & = \frac {5628}{6253} \\ \sin \gamma & = \frac {22780}{25181} \end{aligned}

The area of the triangle

A = r 3 2 2 ( sin ⁡ α + sin ⁡ β + sin ⁡ γ ) = 1 2 ( 3 67 ) 2 ( 4288 5513 + 5628 6253 + 22780 25181 ) = 2412 931697 A = \frac {r_3^2}2 (\sin \alpha + \sin \beta + \sin \gamma) = \frac 12 \left(\frac 3{67} \right)^2 \left(\frac {4288}{5513} + \frac {5628}{6253} + \frac {22780}{25181} \right) = \frac {2412}{931697}

And the required answer p + q − 7 = 2412 + 931697 − 7 = 934102 p+q-7 = 2412+931697-7 = \boxed{934102} .

1 Helpful 1 Interesting 1 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...