Area of segments of a circle!

Level pending

In the above diagram, if A R = arccos ( α β α ) α β β α A_{R} = \arccos(\sqrt{\dfrac{\alpha - \sqrt{\beta}}{\alpha}}) - \dfrac{\sqrt{\alpha\sqrt{\beta}- \beta}}{\alpha} , where α \alpha and β \beta are coprime positive integers, find α + β \alpha + \beta .


The answer is 5.

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

Rocco Dalto
Feb 24, 2021

2 = B C sin ( 3 0 ) = B C 2 B C = 4 A C = 2 3 2 = \overline{BC}\sin(30^{\circ}) = \dfrac{\overline{BC}}{2} \implies \overline{BC} = 4 \implies \overline{AC} = 2\sqrt{3}

Let A ( 0 , 0 ) , B ( 0 , 2 ) A(0,0), B(0,2) and C ( 2 3 , 0 ) m B C = 1 3 y = 1 3 x + 2 C(2\sqrt{3},0) \implies m_{\overline{BC}} = -\dfrac{1}{\sqrt{3}} \implies y = -\dfrac{1}{\sqrt{3}}x + 2

and

( x 1 ) 2 + ( y 1 ) 2 = 1 x 2 2 x + 1 + 3 2 3 x + x 2 3 = 1 (x - 1)^2 + (y - 1)^2 = 1 \implies x^2 - 2x + 1 + \dfrac{3 - 2\sqrt{3}x + x^2}{3} = 1 \implies

4 x 2 2 ( 3 + 3 ) x + 3 = 0 x 1 = 3 + 3 + 6 3 4 4x^2 - 2(3 + \sqrt{3})x + 3 = 0 \implies x_{1} = \dfrac{3 + \sqrt{3} + \sqrt{6\sqrt{3}}}{4} and

x 2 = 3 + 3 6 3 4 y 1 = 7 3 3 6 3 4 3 x_{2} = \dfrac{3 + \sqrt{3} - \sqrt{6\sqrt{3}}}{4} \implies y_{1} = \dfrac{7\sqrt{3} - 3 - \sqrt{6\sqrt{3}}}{4\sqrt{3}} and y 2 = 7 3 3 + 6 3 4 3 y_{2} = \dfrac{7\sqrt{3} - 3 + \sqrt{6\sqrt{3}}}{4\sqrt{3}}

Δ x = 6 3 2 \Delta{x} = \dfrac{\sqrt{6\sqrt{3}}}{2} and Δ y = 6 3 2 3 D E = 2 3 \Delta{y} = -\dfrac{\sqrt{6\sqrt{3}}}{2\sqrt{3}} \implies \overline{DE} = \sqrt{2\sqrt{3}}

Using law of cosines on D O E 2 3 = 2 ( 1 cos ( θ ) ) = 4 sin 2 ( θ 2 ) \triangle{DOE} \implies 2\sqrt{3} = 2(1 - \cos(\theta)) = 4\sin^2(\dfrac{\theta}{2}) \implies

sin 2 ( θ 2 ) = 3 2 sin ( θ 2 ) = 3 2 cos ( θ 2 ) = 2 3 2 \sin^2(\dfrac{\theta}{2}) = \dfrac{\sqrt{3}}{2} \implies \sin(\dfrac{\theta}{2}) = \sqrt{\dfrac{\sqrt{3}}{2}} \implies \cos(\dfrac{\theta}{2}) = \sqrt{\dfrac{2 - \sqrt{3}}{2}}

and h = cos ( θ 2 ) A D O E = 1 2 ( D E ) h = h = \cos(\dfrac{\theta}{2}) \implies A_{\triangle{DOE}} = \dfrac{1}{2}(\overline{DE})h = 2 3 3 2 \dfrac{\sqrt{2\sqrt{3} - 3}}{2}

and

A s e c t o r ( D O E ) = arccos ( 2 3 2 ) A R = A s e c t o r ( D O E ) A D O E = A_{sector(DOE)} = \arccos(\sqrt{\dfrac{2 - \sqrt{3}}{2}}) \implies A_{R} = A_{sector(DOE)} - A_{\triangle{DOE}} =

arccos ( 2 3 2 ) 2 3 3 2 = arccos ( α β α ) α β β α \arccos(\sqrt{\dfrac{2 - \sqrt{3}}{2}}) - \dfrac{\sqrt{2\sqrt{3} - 3}}{2} = \arccos(\sqrt{\dfrac{\alpha - \sqrt{\beta}}{\alpha}}) - \dfrac{\sqrt{\alpha\sqrt{\beta}- \beta}}{\alpha} \implies

α + β = 5 \alpha + \beta = \boxed{5} .

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...