# Problem Of The Week #429 August 10th, 2020

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#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

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A regular 12-gon is inscribed in a circle of radius 12. The sum of the lengths of all sides and diagonals of the 12-gon can be written in the form $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$, where $a,\,b,\,c$ and $d$ are positive integers. Find $a+b+c+d$.

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#### anemone

##### MHB POTW Director
Staff member
Congratulations to kaliprasad for his correct solution!

You can find the suggested solution below:
Position the 12-gon in the Cartesian plane with its center at the origin and one vertex at $(12,\,0)$. Compute the sum, $S$, of the lengths of the eleven segments emanating from this vertex. The coordinates of the other vertices are given by $(12\cos kx,\,12\sin kx)$ where $x=30^{\circ}$ and $k=1,\,2,\,\cdots,\,11$. The length of the segment joining $(12,\,0)$ to $(12\cos kx,\,12\sin kx)$ is

$12\sqrt{(\cos kx-1)^2+(\sin kx)^2}=12\sqrt{2-2\cos kx}=24\sin \dfrac{kx}{2}$

Thus the sum of the lengths of the 11 segments from $(12,\,0)$ is

$S=24(\sin 15^{\circ}+\sin 30^{\circ}+\cdots+\sin 150^{\circ})+24\sin 165^{\circ}$

Since $\sin t=\sin (180^{\circ}-t)$, we may write

$S=48(\sin 15^{\circ}+\sin 30^{\circ}+\sin 45^{\circ}+\sin 60^{\circ}+\sin 75^{\circ})+24\sin 90^{\circ}$

Now,

\begin{align*}\sin 15^{\circ}+\sin 75^{\circ}&=\sin (45^{\circ} -30^{\circ})+\sin (45^{\circ} +30^{\circ})\\&=2\sin 45^{\circ} \cos 30^{\circ}\\&=\dfrac{\sqrt{6}}{2}\end{align*}

Thus,

\begin{align*}S&=48\left(\dfrac{\sqrt{6}}{2}+\dfrac{1}{2}+\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{3}}{2}\right)+24\\&=48+24\sqrt{2}+24\sqrt{3}+24\sqrt{6}\end{align*}

The same values, $S$, occurs if we add the lengths of all segments emanating from any other vertex of the 12-gon. Since each segment is counted at two vertices (its endpoints) the total length of all such segment is

$\dfrac{1}{2}(12S)=288+144\sqrt{2}+144\sqrt{3}+144\sqrt{6}$

Hence, $a+b+c+d=288+144+144+144=720$.

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