HMMT February, 2016

Solution:

We present five different solutions and outline a sixth and seventh one. In what follows, let $a = BC$, $b = CA$, $c = AB$ 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
$$\angle AIO = 90^\circ \Longrightarrow b + c = 2a$$
Let us see how this solves the problem. This lemma implies that $s = 216$. If we let $E$ be the foot of $I$ on $AB$, then $AE = s - BC = 72$, consequently the inradius is $r = \sqrt{97^2 - 72^2} = 65$. Finally, the area is $sr = 216 \cdot 65 = 14040$.

First Solution.
Since $OI \perp DA$, $AI = DI$. Now, it is a well-known fact that $DI = DB = DC$ (this is occasionally called "Fact 5"). Then by Ptolemy's Theorem,
$$DB \cdot AC + DC \cdot AB = DA \cdot BC \Longrightarrow AC + AB = 2BC$$

Second Solution.
As before note that $I$ is the midpoint of $AD$. Let $M$ and $N$ be the midpoints of $AB$ and $AC$, and let the reflection of $M$ across $BI$ be $P$; thus $BM = BP$. Also, $MI = PI$, but we know $MI = NI$ as $I$ lies on the circumcircle of triangle $AMN$. Consequently, we get $PI = NI$; moreover by angle chasing we have
$$\angle INC = \angle AMI = 180^\circ - \angle BPI = \angle IPC$$
Thus triangles $INC$ and $PIC$ are congruent ($CI$ is a bisector) so we deduce $PC = NC$. Thus,
$$BC = BP + PC = BM + CN = \frac{1}{2}(AB + AC)$$

Third Solution.
We appeal to Euler's Theorem, which states that $IO^2 = R(R - 2r)$.
Thus by the Pythagorean Theorem on $\triangle AIO$ (or by Power of a Point) we may write
$$(s - a)^2 + r^2 = AI^2 = R^2 - IO^2 = 2Rr = \frac{abc}{2s}$$
with the same notations as before. Thus, we derive that
$$\begin{aligned} abc & = 2s\left((s - a)^2 + r^2\right) \\ & = 2(s - a)(s(s - a) + (s - b)(s - c)) \\ & = \frac{1}{2}(s - a)\left((b + c)^2 - a^2 + a^2 - (b - c)^2\right) \\ & = 2bc(s - a) \end{aligned}$$
From this we deduce that $2a = b + c$, and we can proceed as in the previous solution.

Fourth Solution.
From Fact 5 again ($DB = DI = DC$), drop perpendicular from $I$ to $AB$ at $E$; call $M$ the midpoint of $BC$. Then, by $AAS$ congruency on $AIE$ and $CDM$, we immediately get that $CM = AE$. As $AE = \frac{1}{2}(AB + AC - BC)$, this gives the desired conclusion.

Fifth Solution.
This solution avoids angle-chasing and using the fact that $BI$ and $CI$ are angle bisectors. Recall the perpendicularity lemma, where
$$WX \perp YZ \Longleftrightarrow WY^2 - WZ^2 = XY^2 - XZ^2$$
Let $B'$ be on the extension of ray $CA$ such that $AB' = AB$. Of course, as in the proof of the angle bisector theorem, $BB' \parallel AI$, meaning that $BB' \perp IO$. Let $I'$ be the reflection of $I$ across $A$; of course, $I'$ is then the incenter of triangle $AB'C'$. Now, we have $B'I^2 - BI^2 = B'O^2 - BO^2$ by the perpendicularity and by power of a point $B'O^2 - BO^2 = B'A \cdot B'C$. Moreover $BI^2 + B'I^2 = BI^2 + BI'^2 = 2BA^2 + 2AI^2$ by the median formula. Subtracting, we get $BI^2 = AI^2 + \frac{1}{2}(AB)(AB - AC)$. We have a similar expression for $CI$, and subtracting the two results in $BI^2 - CI^2 = \frac{1}{2}(AB^2 - AC^2)$. Finally,
$$BI^2 - CI^2 = \frac{1}{4}\left[(BC + AB - AC)^2 - (BC - AB + AC)^2\right]$$
from which again, the result $2BC = AB + AC$ follows.

Sixth Solution, outline.
Use complex numbers, setting $I = ab + bc + ca$, $A = -a^2$, 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 $v = bc$. 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 $v$: $2u + v + 1 = 0$ and $u^2 = \left(\frac{130}{97}\right)^2 v$ 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 $(1 - b^2)(1 - c^2) = (1 + bc)^2 - (b + c)^2 = (1 + v)^2 - u^2$ from the area homogenizes perfectly... because $AB \cdot AC = AI \cdot AI_A$, where $I_A$ is the $A$-excenter, and of course the way the problem is set up $AI_A = 3AI$), 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 write everything in terms of the angles and solve the equations; for instance, the $\angle AIO = 90^\circ$ 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 $\cos \frac{A}{2}$.