HMMT-Feb 2023 guts, Problem 14

Acute triangle $A B C$ has circumcenter $O$. The bisector of $\angle A B C$ and the altitude from $C$ to side $A B$ intersect at $X$. Suppose that there is a circle passing through $B, O, X$, and $C$. If $\angle B A C=n^{\circ}$, where $n$ is a positive integer, compute the largest possible value of $n$.

Solution: Solution: We have $\angle X B C=B / 2$ and $\angle X C B=90^{\circ}-B$. Thus, $\angle B X C=90^{\circ}+B / 2$. We have $\angle B O C=2 A$, so \[ 90^{\circ}+B / 2=2 A \] This gives $B=4 A-180^{\circ}$, which gives $C=360^{\circ}-5 A$. In order for $0^{\circ}<B<90^{\circ}$, we need $45^{\circ}<A<67.5^{\circ}$. In order for $0^{\circ}<C<90^{\circ}$, we require $54^{\circ}<A<72^{\circ}$. The largest integer value in degrees satisfying these inequalities is $A=67^{\circ}$. $\fbox{67}$.

Answer: 67