Question
This question comes from Question 10 of the 2020 AIMO
A circle with centre $O$ has diameter $AD$. With $X$ on $AO$ and points $B$ and $C$ on the circle, triangles $ABX$ and $XCO$ are similar isosceles with base angle $\alpha$ as shown. Find, with proof, the value of $\alpha$.
My Attempt
I knew we probably have to use the fact that the isosceles triangles have a base on the diameter and vertex on the circle. Then $\angle ACD=\angle ABD=90^{\circ}$. Since $\angle ABX=180-2\alpha=\angle XBC$, then $AB||XC$.
Now, when constructing $CD$ and $BD$, I notice that $CD$ looked very much like the angle bisector of $\angle XCD$ and $BD$ looked very much like the angle bisector of $\angle CDA$. Let $CD$ and $BD$ intersect at $I$. I would need to prove that $I$ is in fact the incenter of $XCD$ to give a rigorous proof. Note that $\angle XCO=180-2\alpha$ and since $CO=OD$ (both a radius), then $\angle ODC=\angle COD=\frac{\alpha}{2}$. If $I$ were to be the incenter of $XCD$, then $\angle XCO=\angle OCD\implies 180-2\alpha=\frac{\alpha}{2}\implies \alpha=72^{\circ}$. This looked to be the right answer (could confirm with geogebra).
However, this is also the step that I am struggling to prove. I tried constructing the altitude of $\triangle XID$, $\triangle CIX$ and $\triangle DIC$ to be $IF,IG$ and $IH$ repsetively. Note that since $AB||XC$, $\angle ABD=\angle XGD=90^{\circ}$ and thus the points $B,G,I,D$ are colinear. Refer to the following figure:
It would suffice to prove that $\triangle CGI\cong\triangle CHI$ and $\triangle XGI\cong\triangle XFI$.
Any hints or solutions would be appreciated. It would be nice if you would be able to show how to prove the last step in my attempt. Otherwise, giving me a hint to a better approach would also be much appreciated. Thank you in advance!
Best Answer
Let $B'$ the point symmetric to $B$ with respect to $AO$. Then $B', X, C$ are collinear because both $\angle OXC$ and $\angle AXB'$ are equal to $\alpha$. Since the central angle $COA$ subtends the same arc as the inscribed angle $CB'A$, we have $\angle COA = 2\angle CB'A$. This leads to $\alpha = 2(\pi - 2\alpha)$ from which we calculate $\alpha=\dfrac 25 \pi$.