HMMT February February 16, 2019

Solution:

Let $A'B'C'$ be the medial triangle of $ABC$, where $A'$ is the midpoint of $BC$ and so on. Notice that the midpoint of $OH$, which is the nine-point-center $N$ of triangle $ABC$, is also the circumcenter of $A'B'C'$ (since the midpoints of the sides of $ABC$ are on the nine-point circle). Thus, if $N$ is on $BC$, then $NA'$ is parallel to $B'C'$, so by similarity, we also know that $OA$ is parallel to $BC$.

Next, $AB < AC$, so $B$ is on the minor arc $AC$. This means that $\angle OAC = \angle OCA = \angle C$, so $\angle AOC = 180 - 2\angle C$. This gives us the other two angles of the triangle in terms of angle $C$: $\angle B = 90 + \angle C$ and $\angle A = 90 - 2\angle C$. To find the area, we now need to find the height of the triangle from $A$ to $BC$, and this is easiest by finding the circumradius $R$ of the triangle.

We do this by the Extended Law of Sines. Letting $AC = x$ and $AB = 5 - x$,
$$\frac{1}{\sin(90 - 2C)} = \frac{x}{\sin(90 + C)} = \frac{5 - x}{\sin C} = 2R$$
which can be simplified to
$$\frac{1}{\cos 2C} = \frac{x}{\cos C} = \frac{5 - x}{\sin C} = 2R.$$
This means that
$$\frac{1}{\cos 2C} = \frac{x + (5 - x)}{\cos C + \sin C} = \frac{5}{\cos C + \sin C}$$
and the rest is an easy computation:
$$\cos C + \sin C = 5 \cos 2C = 5(\cos^2 C - \sin^2 C)$$
$$\frac{1}{5} = \cos C - \sin C$$
Squaring both sides,
$$\frac{1}{25} = \cos^2 C - 2\sin C \cos C + \sin^2 C = 1 - \sin 2C$$
so $\sin 2C = \frac{24}{25}$, implying that $\cos 2C = \frac{7}{25}$. Therefore, since $\frac{1}{\cos 2C} = 2R$ from above, $R = \frac{25}{14}$. Finally, viewing triangle $ABC$ with $BC = 1$ as the base, the height is
$$\sqrt{R^2 - \left(\frac{BC}{2}\right)^2} = \frac{12}{7}$$
by the Pythagorean Theorem, yielding an area of $\frac{1}{2} \cdot 1 \cdot \frac{12}{7} = \frac{6}{7}$.

Solution:

The midpoint of $OH$ is the nine-point center $N$. We are given $N$ lies on $BC$, and we also know $N$ lies on the perpendicular bisector of $EF$, where $E$ is the midpoint of $AC$ and $F$ is the midpoint of $AB$. The main observation is that $N$ is equidistant from $M$ and $F$, where $M$ is the midpoint of $BC$.

Translating this into coordinates, we pick $B(-0.5, 0)$ and $C(0.5, 0)$, and arbitrarily set $A(a, b)$ where (without loss of generality) $b > 0$. We get $E\left(\frac{a + 0.5}{2}, \frac{b}{2}\right)$, $F\left(\frac{a - 0.5}{2}, \frac{b}{2}\right)$, $M(0, 0)$. Thus $N$ must have $x$-coordinate equal to the average of those of $E$ and $F$, or $\frac{a}{2}$. Since $N$ lies on $BC$, we have $N\left(\frac{a}{2}, 0\right)$.

Since $MN = EN$, we have $\frac{a^2}{4} = \frac{1}{16} + \frac{b^2}{4}$. Thus $a^2 = b^2 + \frac{1}{4}$. The other equation is $AB + AC = 5$, which is just
$$\sqrt{(a + 0.5)^2 + b^2} + \sqrt{(a - 0.5)^2 + b^2} = 5$$
This is equivalent to
$$\begin{gathered} \sqrt{2a^2 + a} + \sqrt{2a^2 - a} = 5 \\ \sqrt{2a^2 + a} = 5 - \sqrt{2a^2 - a} \\ 2a^2 + a = 25 - 10\sqrt{2a^2 - a} + 2a^2 - a \\ 25 - 2a = 10\sqrt{2a^2 - a} \\ 625 - 100a + 4a^2 = 200a^2 - 100a \\ 196a^2 = 625 \end{gathered}$$
Thus $a^2 = \frac{625}{196}$, so $b^2 = \frac{576}{196}$. Thus $b = \frac{24}{14} = \frac{12}{7}$, so $[ABC] = \frac{b}{2} = \frac{6}{7}$.