Angles of ABC in decagon

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Angles of ABC in decagon

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#1 h Jan 18, 2017, 4:16 AM • 1 Y

Y by Adventure10

Let $A_1A_2A_3\cdots A_{10}$ be a regular decagon and $A=A_1A_4\cap A_2A_5, B=A_1A_6\cap A_2A_7, C=A_1A_9\cap A_2A_{10}.$ Find the angles of the triangle $ABC$ .

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#2 h Jan 21, 2017, 5:13 PM • 2 Y

Y by Adventure10, Mango247

We will work in argand plane.
Let , $z = cos(\frac{\pi}{5}) + isin(\frac{\pi}{5})$ .
Let , $A_{i} = z^{i-1}$ , $\forall i \in (1,2,...,10)$
Now see that ,
$A =\frac{z^3(z+z^4)-z^5(1+z^3)}{z^3-z^5} = \frac{z(z^3+1)}{z+1}$
$B = 0$
$C =\frac{z^8(z+z^9)-z^{10}(1+z^8)}{z^8-z^{10}}=\frac{z(z^8+1)}{z+1}$
Now,
Let $A' = 0,B'=\frac{1-z^2}{2} , C' = 1$ it's easy to see that $\triangle ABC \sim \triangle A'B'C'$
See that
$1-z^2 = {1 - cos(\frac{2\pi}{5}) - isin(\frac{2\pi}{5})} = -2isin(\frac{\pi}{5})z$
$1+z^2= {1 + cos(\frac{2\pi}{5})+isin(\frac{2\pi}{5})} = 2cos(\frac{\pi}{5})z$
Now,
$\angle B'A'C' = arg (-z) + arg (isin(\frac{\pi}{5})) = \frac{\pi}{2} - \frac{\pi}{5} = \frac{3\pi}{10}$
$\angle A'B'C' = arg (\frac{z^2-1}{z^2+1}) = arg(itan\frac{\pi}{5}) = \frac{\pi}{2}$
$\angle A'C'B' = \frac{\pi}{5}$
So angles of $\triangle ABC$ are $\frac{3\pi}{10},\frac{\pi}{2},\frac{\pi}{5}$

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