Let $A B C$ be an acute triangle with incenter $I$ and circumcenter $O$. Assume that $\angle O I A=90^{\circ}$. Given that $A I=97$ and $B C=144$, compute the area of $\triangle A B C$.
We present five different solutions and outline a sixth and seventh one. In what follows, let $a=B C$, $b=C A, c=A B$ as usual, and denote by $r$ and $R$ the inradius and circumradius. Let $s=\frac{1}{2}(a+b+c)$. In the first five solutions we will only prove that \[ \angle A I O=90^{\circ} \Longrightarrow b+c=2 a \] Let us see how this solves the problem. This lemma implies that $s=216$. If we let $E$ be the foot of $I$ on $A B$, then $A E=s-B C=72$, consequently the inradius is $r=\sqrt{97^{2}-72^{2}}=65$. Finally, the area is $s r=216 \cdot 65=14040$. First Solution. Since $O I \perp D A, A I=D I$. Now, it is a well-known fact that $D I=D B=D C$ (this is occasionally called "Fact 5"). Then by Ptolemy's Theorem, \[ D B \cdot A C+D C \cdot A B=D A \cdot B C \Longrightarrow A C+A B=2 B C \] Second Solution. As before note that $I$ is the midpoint of $A D$. Let $M$ and $N$ be the midpoints of $A B$ and $A C$, and let the reflection of $M$ across $B I$ be $P$; thus $B M=B P$. Also, $M I=P I$, but we know $M I=N I$ as $I$ lies on the circumcircle of triangle $A M N$. Consequently, we get $P I=N I$; moreover by angle chasing we have \[ \angle I N C=\angle A M I=180^{\circ}-\angle B P I=\angle I P C \] Thus triangles $I N C$ and $P I C$ are congruent ( $C I$ is a bisector) so we deduce $P C=N C$. Thus, \[ B C=B P+P C=B M+C N=\frac{1}{2}(A B+A C) \] Third Solution. We appeal to Euler's Theorem, which states that $I O^{2}=R(R-2 r)$. Thus by the Pythagorean Theorem on $\triangle A I O$ (or by Power of a Point) we may write \[ (s-a)^{2}+r^{2}=A I^{2}=R^{2}-I O^{2}=2 R r=\frac{a b c}{2 s} \] with the same notations as before. Thus, we derive that \[ \begin{aligned} a b c & =2 s\left((s-a)^{2}+r^{2}\right) \\ & =2(s-a)(s(s-a)+(s-b)(s-c)) \\ & =\frac{1}{2}(s-a)\left((b+c)^{2}-a^{2}+a^{2}-(b-c)^{2}\right) \\ & =2 b c(s-a) \end{aligned} \] From this we deduce that $2 a=b+c$, and we can proceed as in the previous solution. Fourth Solution. From Fact 5 again $(D B=D I=D C)$, drop perpendicular from $I$ to $A B$ at $E$; call $M$ the midpoint of $B C$. Then, by $A A S$ congruency on $A I E$ and $C D M$, we immediately get that $C M=A E$. As $A E=\frac{1}{2}(A B+A C-B C)$, this gives the desired conclusion. Fifth Solution. This solution avoids angle-chasing and using the fact that $B I$ and $C I$ are anglebisectors. Recall the perpendicularity lemma, where \[ W X \perp Y Z \Longleftrightarrow W Y^{2}-W Z^{2}=X Y^{2}-X Z^{2} \] Let $B^{\prime}$ be on the extension of ray $C A$ such that $A B^{\prime}=A B$. Of course, as in the proof of the angle bisector theorem, $B B^{\prime} \| A I$, meaning that $B B^{\prime} \perp I O$. Let $I^{\prime}$ be the reflection of $I$ across $A$; of course, $I^{\prime}$ is then the incenter of triangle $A B^{\prime} C^{\prime}$. Now, we have $B^{\prime} I^{2}-B I^{2}=B^{\prime} O^{2}-B O^{2}$ by the perpendicularity and by power of a point $B^{\prime} O^{2}-B O^{2}=B^{\prime} A \cdot B^{\prime} C$. Moreover $B I^{2}+B^{\prime} I^{2}=B I^{2}+B I^{\prime 2}=2 B A^{2}+2 A I^{2}$ by the median formula. Subtracting, we get $B I^{2}=A I^{2}+\frac{1}{2}(A B)(A B-A C)$. We have a similar expression for $C I$, and subtracting the two results in $B I^{2}-C I^{2}=\frac{1}{2}\left(A B^{2}-A C^{2}\right)$. Finally, \[ B I^{2}-C I^{2}=\frac{1}{4}\left[(B C+A B-A C)^{2}-(B C-A B+A C)^{2}\right] \] from which again, the result $2 B C=A B+A C$ follows. Sixth Solution, outline. Use complex numbers, setting $I=a b+b c+c a, A=-a^{2}$, etc. on the unit circle (scale the picture to fit in a unit circle; we calculate scaling factor later). Set $a=1$, and let $u=b+c$ and $v=b c$. Write every condition in terms of $u$ and $v$, and the area in terms of $u$ and $v$ too. There should be two equations relating $u$ and $v: 2 u+v+1=0$ and $u^{2}=\frac{130}{9}^{2} v$ from the right angle and the 144 to 97 ratio, respectively. The square area can be computed in terms of $u$ and $v$, because the area itself is antisymmetric so squaring it suffices. Use the first condition to homogenize (not coincidentally the factor $\left(1-b^{2}\right)\left(1-c^{2}\right)=(1+b c)^{2}-(b+c)^{2}=(1+v)^{2}-u^{2}$ from the area homogenizes perfectly. $\ldots$ because $A B \cdot A C=A I \cdot A I_{A}$, where $I_{A}$ is the $A$-excenter, and of course the way the problem is set up $A I_{A}=3 A I$.), and then we find the area of the scaled down version. To find the scaling factor simply determine $|b-c|$ by squaring it, writing in terms again of $u$ and $v$, and comparing this to the value of 144 . Seventh Solution, outline. Trigonometric solutions are also possible. One can you write everything in terms of the angles and solve the equations; for instance, the $\angle A I O=90^{\circ}$ condition can be rewritten as $\frac{1}{2} \cos \frac{B-C}{2}=2 \sin \frac{B}{2} \sin \frac{C}{2}$ and the 97 to 144 ratio condition can be rewritten as $\frac{2 \sin \frac{B}{2} 2 \sin \frac{C}{2}}{\sin A}=\frac{97}{144}$. The first equation implies $\sin \frac{A}{2}=2 \sin \frac{B}{2} \sin \frac{C}{2}$, which we can plug into the second equation to get $\cos \frac{A}{2}$. $\fbox{14040}$.