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Consider an isosceles triangle $ABC$ ($AB=AC$), with points $D, E$ lying on sides $AB, AC$, respectively. Suppose that $\angle BAC=\alpha, \angle EBC=\beta, \angle DCB=\gamma, \angle DEB=\omega$.

Suppose that the value (in degrees) of each of the angles $\alpha, \beta, \gamma, \omega$ is an integer.

a) Prove that $\alpha$ must be divisble by four!

b) Prove it is impossible that exactly three of the angles $\alpha, \beta, \gamma, \omega$ is divisible by three!

Using lots of trigonometry I proved that $$\cot \omega=\dfrac{2\cos \frac{\alpha}{2}\cos (\alpha-\gamma)}{\sin \gamma(\cos \alpha+\cos2\beta)}-\tan (\alpha+\beta)$$

Can someone help to prove a) and b)? Thans in advance. Maybe with trigonometry (perhaps using my fact)? I would be really thankful if someone would post a solution without trigonometry!

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  • $\begingroup$ Where did you get this problem? As stated, neither of those claims are necessarily true. $\endgroup$ Commented Apr 23, 2018 at 18:50
  • $\begingroup$ It is my own problem I need help with. What are the counterexamples? $\endgroup$ Commented Apr 23, 2018 at 18:51
  • $\begingroup$ For example, take a triangle with $\alpha = 50^\circ$ and choose $D$ and $E$ so that $\beta = \gamma = 30^\circ$. $\endgroup$ Commented Apr 23, 2018 at 18:54
  • $\begingroup$ Than $\omega$ is not an integer, so it doesn't work. $\endgroup$ Commented Apr 23, 2018 at 18:56
  • $\begingroup$ Then $\omega = 30^\circ$. $\endgroup$ Commented Apr 23, 2018 at 19:00

1 Answer 1

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The statements are not true, here is a figure.

enter image description here

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