+0

# Circle is the incircle of triangle ABC and is also the circumcircle of triangle XYZ. The point X is on line BC, point Y is on

+1
57
2
+1442

Circle $$\Gamma$$ is the incircle of triangle ABC and is also the circumcircle of triangle XYZ. The point X is on line BC, point Y is on overline AB, and the point Z is on line AC. If angle A=40 degrees, angle B=60 degrees, and angle C=80 degrees, what is the measure of angle AYX?

I posted this one awhile ago but got no response, it's a tough one.  Also, I see this question, or something similar to it, has been answered before with a result of 130, however I have tried that and it is not the solution, at least not according to my program.  Thanks!

AnonymousConfusedGuy  Jun 21, 2018

#1
+19653
+1

Circle  is the incircle of triangle ABC and is also the circumcircle of triangle XYZ. The point X is on line BC, point Y is on overline AB, and the point Z is on line AC. If angle A=40 degrees, angle B=60 degrees, and angle C=80 degrees, what is the measure of angle AYX?

$$\text{Let the center M of the incircle, \\called the incenter,\\can be found as the intersection of the three internal angle bisectors }$$

$$\begin{array}{|rcll|} \hline \angle XBM &=& \dfrac{60^{\circ}}{2} \\ &=& 30^{\circ} \\\\ \angle XMB &=& 90^{\circ}- \angle XBM \\ &=& 90^{\circ}-30^{\circ} \\ &=& 60^{\circ} \\\\ \angle XMY &=& 2\times \angle XMB \\ &=& 2\times 60^{\circ} \\ &=& 120^{\circ} \\\\ \angle MYX &=& \dfrac{180^{\circ}-\angle XMY}{2} \\ &=& \dfrac{180^{\circ}-120^{\circ}}{2} \\ &=& 30^{\circ} \\\\ \angle AYX &=& 90^{\circ} + \angle MYX \\ &=& 90^{\circ} + 30^{\circ} \\ & \mathbf{=}& \mathbf{120^{\circ}} \\ \hline \end{array}$$

heureka  Jun 21, 2018
#1
+19653
+1

Circle  is the incircle of triangle ABC and is also the circumcircle of triangle XYZ. The point X is on line BC, point Y is on overline AB, and the point Z is on line AC. If angle A=40 degrees, angle B=60 degrees, and angle C=80 degrees, what is the measure of angle AYX?

$$\text{Let the center M of the incircle, \\called the incenter,\\can be found as the intersection of the three internal angle bisectors }$$

$$\begin{array}{|rcll|} \hline \angle XBM &=& \dfrac{60^{\circ}}{2} \\ &=& 30^{\circ} \\\\ \angle XMB &=& 90^{\circ}- \angle XBM \\ &=& 90^{\circ}-30^{\circ} \\ &=& 60^{\circ} \\\\ \angle XMY &=& 2\times \angle XMB \\ &=& 2\times 60^{\circ} \\ &=& 120^{\circ} \\\\ \angle MYX &=& \dfrac{180^{\circ}-\angle XMY}{2} \\ &=& \dfrac{180^{\circ}-120^{\circ}}{2} \\ &=& 30^{\circ} \\\\ \angle AYX &=& 90^{\circ} + \angle MYX \\ &=& 90^{\circ} + 30^{\circ} \\ & \mathbf{=}& \mathbf{120^{\circ}} \\ \hline \end{array}$$

heureka  Jun 21, 2018
#2
+1442
+2

Thank you Heureka, must have just been a typo haha!

AnonymousConfusedGuy  Jun 21, 2018