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Can someone help me to find the value of $x$? I wish I could share my attempted solution but I really couldn't develop something interesting to share. I know that the value of the internal angle are $108^°$, because the sum of the internal angles of a convex polygon is $S=180^°(n-2)$, and they are all the same, because the it's a regular polygon. But that's all I figured out. enter image description here

Alv
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Gustavo Gabriel
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  • Hint: the triangle formed by the right line you drew has one angle given as 30° and one angle is an angle of the pentagon. What can you say about the third one (the one on the bottom line of the pentagon)? – Toffomat Sep 11 '23 at 16:47
  • The angle you mentioned is 42º (180º-30º-108º), but how it can help? – Gustavo Gabriel Sep 11 '23 at 16:52
  • Ah, sorry, maybe it doesn't tell you as much as I thought at first... – Toffomat Sep 11 '23 at 17:42
  • @GustavoGabriel: What is the source of this question? Such information can provide readers with a sense of the level of difficulty expected in a solution. ... FYI: A GeoGebra sketch suggests that the answer is $90^\circ$. – Blue Sep 11 '23 at 18:11
  • @Blue, it's a question from a national math olympiad in Brazil for high school students. The question is from the first phase of the test, so I guess it's not something very complex. – Gustavo Gabriel Sep 11 '23 at 18:26

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Consider the figure below, where $ABCDE$ is a regular pentagon, $F \in BC$ and $G \in AB$ with $EG \cong FG$, and $\measuredangle BFG = 30^\circ$. Also, $G'$ is taken in such a way that $\triangle FGG'$ is equilateral.

enter image description here

  1. Show that $\triangle BGG'$ is isosceles, with $\measuredangle BGG' = \measuredangle BG'G = 18^\circ$.
  2. Use Exterior Angle Theorem on $\triangle BGG'$, plus the fact that $\triangle ABE$ is isosceles, to show that $E$, $B$, and $G'$ are aligned.
  3. Observing that $\triangle EGG'$ is isosceles, prove that $$\boxed{\measuredangle DEG = 90^\circ}.$$
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