Blue vs Red

Geometry Level 3

A B C D E ABCDE is a regular pentagon where O O is its center. It is given sin ⁡ 1 8 ∘ = 1 4 ( 5 − 1 ) \sin 18^\circ=\dfrac{1}{4}(\sqrt{5}-1) . The ratio of the blue area to the red area, [ B L U E ] [ R E D ] \frac{\color{#3D99F6}[\color{#3D99F6}B\color{#3D99F6}L\color{#3D99F6}U\color{#3D99F6}E\,\color{#3D99F6}]}{{\color{#D61F06}[\,\color{#D61F06}R}{\color{#D61F06}E}{\color{#D61F06}D\,\color{#D61F06}]}} , can be denoted as a − b c \dfrac{\sqrt{a}-b}{c} , where a a , b b , and c c are positive integers. Calculate a + b + c a+b+c .

157 167 277 287

Tunk-Fey Ariawan by Tunk-Fey Ariawan , 35, Indonesia

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1 solution

Ercole Suppa
May 4, 2014

Let r = A O r=AO . Since [ A O B ] = 1 2 r 2 sin ⁡ 7 2 ∘ [AOB]=\frac{1}{2}r^2\sin 72^\circ and [ A O D ] = 1 2 r 2 sin ⁡ 14 4 ∘ [AOD]=\frac{1}{2}r^2\sin 144^\circ we have:

BLUE RED = r 2 sin ⁡ 14 4 ∘ 5 2 r 2 sin ⁡ 7 2 ∘ − r 2 sin ⁡ 14 4 ∘ = sin ⁡ 3 6 ∘ 5 sin ⁡ 3 6 ∘ cos ⁡ 3 6 ∘ − sin ⁡ 3 6 ∘ = 1 5 cos ⁡ 3 6 ∘ − 1 = 125 − 1 31 \frac{\text{BLUE}}{\text{RED}}=\frac{r^2 \sin 144^\circ}{\frac{5}{2}r^2\sin 72^\circ-r^2\sin 144^\circ}=\frac{\sin 36^\circ}{5\sin 36^\circ \cos 36^\circ-\sin 36^\circ}=\frac{1}{5\cos 36^\circ-1}=\frac{\sqrt{125}-1}{31}

Therefore a + b + c + d = 157 a+b+c+d=\boxed{157}

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