Let's see how smart Puzzling stack users really are! Try to solve the value of x (get angle value). This is related to the http://thinkzone.wlonk.com/MathFun/Triangle.htm , you can search online how to solve, but that wouldn't be fun at all.


Good luck! and have fun! ;) enter image description here

PS: sorry for poor paint drawing skills :P


for the answer to be accepted, you must explain in a detailed way how did you reach the value of x, it must be a strong answer that consistently proves your point. Good luck!

EDIT: Sorry, I forgot to add some angles :P this pic has now the full info you need to solve!

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    $\begingroup$ BTW answer is spelt like answer, not awnser or anwser (yeah, that w is tricky). $\endgroup$ – boboquack Jun 12 '17 at 9:14
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    $\begingroup$ This looks like the World's Hardest Easy Geometry Problem. If so, please provide attribution. $\endgroup$ – boboquack Jun 12 '17 at 9:34
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    $\begingroup$ This seems more mathematics problem than puzzle... $\endgroup$ – Gareth McCaughan Jun 12 '17 at 10:33
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    $\begingroup$ You need strategy to solve many mathematics problems. (Take a look some time at the various mathematical olympiads...) $\endgroup$ – Gareth McCaughan Jun 12 '17 at 11:29
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    $\begingroup$ @OldBunny2800 Picture isn't drawn to scale.... $\endgroup$ – Tumbler41 Jun 12 '17 at 19:44

This is the original drawn of this question:

enter image description here

You will easily notice that AFE angle is 20. And AFE and FCA triangles are isosceles triangle:

enter image description here

Moreover, if you draw a line through point E as 20 degrees as seen below, we have another isosceles triangle FHE and equillateral triangle AIE:

enter image description here

You will probably already notice that FCA and FHE triangles are the same triangles but reflections. Therefore ACE angle and EHA angles are equal to each other and if we join the point H and C, it has to be parallel to AE line.

As a result, CHI triangle has to be equillateral triangle because HIC angle is 60 degrees and HI and CI are equal to each other

enter image description here

On the other hand, the red lines are equal to each other because ADE and AED angles are equal to each other. Moreover, JDE triangle becomes isosceles since JDE and JED angles are 30 degree:

enter image description here

Lastly, if we connect point A to point K as shown below, we will have another triangle KCA. We already know that KHA angle is 80 degrees from previous drawings, and KDH and ADI are 80 degrees as well. So as a result, |KH|=|KD|, and |AD|=|AI|, ann we know that CHI triangle is equillateral, we can conclude that |KH|=|KD|=|AD|=|AI| and C ray passing just between AK line is bisector and the result becomes:

30 degree

enter image description here

Note: I will fix mathjax part later and explain it more. (out of time right now :))

  • $\begingroup$ ahaha nice! did you figured it out alone? $\endgroup$ – jeyejow Jun 12 '17 at 14:43
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    $\begingroup$ @jeyejow actually yes, i am good at geometry $\endgroup$ – Oray Jun 12 '17 at 14:50
  • $\begingroup$ thats actually impressive! $\endgroup$ – jeyejow Jun 12 '17 at 14:51
  • $\begingroup$ highlighting the original triangles from the question would make the pictures easier to follow. Impressive work. $\endgroup$ – phyatt Jun 12 '17 at 16:55

This is not a full solution, but merely a first step towards one.

Take a look at this picture:

enter image description here

It's a regular 18-gon with some of its diagonals drawn. It is easy to see that the angles I've drawn match that of the problem. It is also obvious that point B lies on the intersection of three lines, because of symmetry. The only surprising thing that needs proof is that three diagonals intersect at point A. Once that is established it is straightforward to show what the angles at A are.

  • $\begingroup$ Very nice! You're being very modest, calling this a first step. I'd say it's nearly a full solution, and potentially a very elegant one, too. All that's missing is finding why these three diagonals intersect at one point. (If they do, this point must be A.) Although I can't see it right now, it can't be too hard to show, can it? $\endgroup$ – Angkor Jun 15 '17 at 16:48
  • $\begingroup$ @Angkor The meat of the proof really is in showing that the three diagonals intersect (i.e. let A be the intersection of 2 of them, and prove that the third goes through A), and that is very tricky to do without trigonometry. It is also easy to get confused and construct a circular argument or to assume the conclusion. To be honest, I did not come up with this myself, but I don't remember where I first saw it. I feel it is interesting to see this bigger picture that shows what's going on, and it deserves to be better known. $\endgroup$ – Jaap Scherphuis Jun 15 '17 at 20:58

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