Using only elementary geometry, determine angle x.
You may not use trigonometry, such as sines and cosines, the law of sines, the law of cosines, etc.
Using only elementary geometry, determine angle x.
You may not use trigonometry, such as sines and cosines, the law of sines, the law of cosines, etc.
Each node of the regular polygon $P_{18}$ sees all other nodes separated with $10^{\circ}$, making this $P_{18}$ a perfect drawing canvas for this problem (left image):
It is easy to verify that $A$ and $B$ are located on a large diagonal of $P_{18}$.
Now embed the unique, smaller $18$-gon $p_{18}$ inside $P_{18}$, with nodes $A$, $B$ and $C$.
Then we have:
$AB \overset{\text{similarity}}{\parallel} A’B’\overset{\text{regularity}}{\parallel}A’’B’’$
Now using the $10^{\circ}$-property, we see that:
$x = \angle A’’B’’B = 2\times 10^{\circ} = 20^{\circ} \qquad \blacksquare $
This is known as the problem of "adventitious angles". You'll find many references if you search the web for that phrase.
Now, it's easy to prove that $CE=AG$, and $DF=DG=GF$.
Since $AF=CF$, then $EF=GF$.
Then $EF=DF \Rightarrow \angle FED= \angle FDE$.
While $\angle DFE=\angle ABC=80 ^\circ$, so $\angle DEF=50^\circ$.
From $\angle AEB=30^\circ$, we can get $x=\angle DEA=20^\circ$. [Q.E.D]