enter image description here

This is the picture, and we are aiming for the angle $x$

It's easy to see that $\angle DGA = \angle CGB = 100°$, $\angle CGD = \angle AGB = 80°$, $\angle CBG = 50°$, but now i'm missing $\angle GBA = ? $ and $\angle GAB = x$

Any hints?

  • 1
    $\begingroup$ I now have no time to approach the problem, but there might be something that might help. The first thing is that I don't think that these problems are thought to be solved with trigonometry but with nice constructions and better observations. That said, note that if the solution is $x=20°$, then $\Delta AGB$ is isosceles, so you might want to prove that rather than trying to find $\angle BAG$. In competitions you can't use GeoGebra or similar programs, so you'll only guess the length of an angle through nice construction with the ruler and the angle conveyor... $\endgroup$ – Dr. Mathva Feb 24 at 19:04
  • $\begingroup$ If worse comes to worse you can figure out the sides BG and GA via law of sines and express as trig functions. And thus the angles via law of sines again. It may be expresses as arc trig of side ratios... but it is a value.... Or there may be something clever. $\endgroup$ – fleablood Feb 24 at 19:39

By law of sines we obtain: $$\frac{BG}{GD}=\frac{\frac{BG}{AG}}{\frac{GD}{AG}}=\frac{\sin{x}}{\sin(100^{\circ}-x)}.$$ In another hand, $$\frac{BG}{GD}=\frac{\frac{BG}{CG}}{\frac{GD}{CG}}=\frac{\frac{\sin30^{\circ}}{\sin50^{\circ}}}{\frac{\sin70^{\circ}}{\sin30^{\circ}}}=\frac{1}{4\sin50^{\circ}\sin70^{\circ}},$$ which gives $$\frac{\sin{x}}{\sin(100^{\circ}-x)}=\frac{1}{4\sin50^{\circ}\sin70^{\circ}}$$ or $$\sin100^{\circ}\cot{x}-\cos100^{\circ}=4\sin50^{\circ}\sin70^{\circ}$$ or $$\cot{x}=\frac{\cos100^{\circ}+4\sin50^{\circ}\sin70^{\circ}}{\sin100^{\circ}},$$ which gives $x=20^{\circ}.$

Indeed, $$\cot{x}-\cot20^{\circ}=\frac{\cos100^{\circ}+4\sin50^{\circ}\sin70^{\circ}}{\sin100^{\circ}}-\cot20^{\circ}=$$ $$=\frac{\cos100^{\circ}+2(\cos20^{\circ}-\cos120^{\circ})}{\cos10^{\circ}}-\frac{\cos20^{\circ}}{\sin20^{\circ}}=$$ $$=\frac{2\sin10^{\circ}(\cos100^{\circ}+2\cos20^{\circ}+1)-\cos20^{\circ}}{\sin20^{\circ}}=$$ $$=\frac{2\sin10^{\circ}(\cos40^{\circ}+\cos20^{\circ}+1)-\cos20^{\circ}}{\sin20^{\circ}}=$$ $$=\frac{\sin50^{\circ}-\sin30^{\circ}+\sin30^{\circ}-\sin10^{\circ}+2\sin10^{\circ}-\cos20^{\circ}}{\sin20^{\circ}}=$$ $$=\frac{\sin50^{\circ}+\sin10^{\circ}-\cos20^{\circ}}{\sin20^{\circ}}=0$$ and we are done!

  • $\begingroup$ Thanks, i undestand now. Anyways i was looking for a non-trigonometric solution, but this work definitely. $\endgroup$ – Rodrigo Pizarro Feb 25 at 0:42

Let $y$ denote the angle $\angle GBA$. By the law of sines applied to the four triangles with vertex $G$, we obtain the relations $$\frac{AB}{\sin 80^\circ}=\frac{BG}{\sin x}=\frac{AG}{\sin y}.$$ $$\frac{AD}{\sin 100^\circ}=\frac{AG}{\sin 40^\circ}=\frac{DG}{\sin 40^\circ},$$ $$\frac{BC}{\sin 100^\circ}=\frac{BG}{\sin 30^\circ}=\frac{CG}{\sin 50^\circ},$$ $$\frac{CD}{\sin 50^\circ}=\frac{DG}{\sin 70^\circ}=\frac{CG}{\sin 30^\circ}.$$ Dividing the fourth equation by the third one, we get (cancelling out $CG$) $$\frac{\sin^2 30^\circ}{\sin 50^\circ\sin 70^\circ}=\frac{BG}{DG}.$$ Now, from the second, we get that $DG=AG$, so $$\frac{\sin^2 30^\circ}{\sin 50^\circ\sin 70^\circ}=\frac{BG}{AG}.$$ Now, we apply the law of cosines to get $$AB^2=BG^2+AG^2-2BG\cdot AG \cos 80^\circ$$ Hence, by the first equation we gave by the law of sines $$\frac{1}{\sin^2x} =\frac{1}{\sin^2 80^\circ}\frac{AB^2}{BG^2}=\frac{1}{\sin^2 80^\circ}\left(1+\frac{AG^2}{BG^2}-2\frac{AG}{BG} \cos 80^\circ\right)=\frac{1}{\sin^2 80^\circ}\left(1+\left(\frac{\sin 50^\circ\sin 70^\circ}{\sin^2 30^\circ}\right)^2-2\frac{\sin 50^\circ\sin 70^\circ}{\sin^2 30^\circ} \cos 80^\circ\right)$$ Easily you can check that $x=20^\circ$.


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