Day 10: Astonishing Geometry

It can be proven that angle APO + angle AQO = 180. In order for the circle to intersect AB at points P and B, the angle APO has to be < 90. The same is true for the point Q. Therefore the conditions stated for the circle are met only when the circle is tangent to AB at B and to AC at C. In that case point P = B and Q = C. Also AB = AC.

When I was making the problem, I discovered more and more interesting properties of this diagram. Geometry really is astonishing! Here is the solution:

By circle theorems $\angle BQC = 108^{\circ}$ . Now notice that $\angle PIQ = 90^{\circ} + \frac{A}{2}$ . Also, $BPIQ$ is cyclic so $\angle PBQ = 90^{\circ} - \frac{A}{2}$ . Now by the exterior angle theorem on $BQA$ , $90^{\circ} + \frac{A}{2} = 108^{\circ}$ . Therefore $A = 36^{\circ}$ .

Therefore $ABOC$ is cyclic. Now $BCO$ is isosceles so $\angle BCO = 18^{\circ}$ , then by angles in the same segment $\angle BAO = \boxed{18^{\circ}}$ as required.