In the regular hexagon $ABCDEF$ on the line $AF$, the point $X$ is taken so that the angle $XCD$ is $45^o$. Find the angle $\angle FXE$. (Kiev Olympiad)
Problem
Source: 2008 Oral Moscow Geometry Olympiad grades 8-9 p3
Tags: geometry, angles, hexagon
franzliszt
18.10.2020 01:43
parmenides51 wrote:
In the regular hexagon $ABCDEF$ on the line $AF$, the point $X$ is taken so that the angle $XCD$ is $45^o$. Find the angle $FXE$.
(Kiev Olympiad)
my first attempt wrote:
Let $CX$ meet $AF$ at $G$. Let $AF$ meet $DE$ at $H$.
Note that $\angle GFX = 180^\circ-\angle GFA = 180^\circ - 120^\circ = 60^\circ$. Also note that $AF \parallel CD$ so $\angle AXC=\angle XCD=45^\circ$. Then we deduce that $\angle FGX=180^\circ-60^\circ-45^\circ=75^\circ$.
$\angle XGE = 180^\circ-75^\circ=105^\circ$.
By obvious supplementary angles, $\triangle EFH$ is equilangular.
Now, draw in segment $BD$ and let it meet $SC$ at $I$. Obviously, $\angle BDC=30^\circ$...
. Here is a much nicer solution. Define $X'$ as the unique point on line $AF$ such that $\angle EXF=75^\circ$. In order to prove that $X=X'$ we need to prove that $\angle X'CD=45^\circ$. $\angle X'AC=\angle FAC-\angle CAB=120^\circ-30^\circ=90^\circ$. Note that $\angle AEX'=180^\circ-75^\circ-30^\circ=75^\circ$ so $AX'=AE$. But then note that $\triangle EAC$ is isosceles. So $AX'=AC$ and $\angle \angle ACX'=45^\circ$. Since $\angle ACB=30^\circ$, we see that $\angle X'CD=45^\circ$. So now we can safely deduce that $\angle FXE=75^\circ$. $\blacksquare$
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