HMMT-Feb 2019 geo, Problem 8

In triangle $A B C$ with $A B<A C$, let $H$ be the orthocenter and $O$ be the circumcenter. Given that the midpoint of $O H$ lies on $B C, B C=1$, and the perimeter of $A B C$ is 6 , find the area of $A B C$.

Solution: Solution 1. Let $A^{\prime} B^{\prime} C^{\prime}$ be the medial triangle of $A B C$, where $A^{\prime}$ is the midpoint of $B C$ and so on. Notice that the midpoint of $O H$, which is the nine-point-center $N$ of triangle $A B C$, is also the circumcircle of $A^{\prime} B^{\prime} C^{\prime}$ (since the midpoints of the sides of $A B C$ are on the nine-point circle). Thus, if $N$ is on $B C$, then $N A^{\prime}$ is parallel to $B^{\prime} C^{\prime}$, so by similarity, we also know that $O A$ is parallel to $B C$. Next, $A B<A C$, so $B$ is on the minor arc $A C$. This means that $\angle O A C=\angle O C A=\angle C$, so $\angle A O C=$ $180-2 \angle C$. 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 $B C$, and this is easiest by finding the circumradius $R$ of the triangle. We do this by the Extended Law of Sines. Letting $A C=x$ and $A B=5-x$, \[ \frac{1}{\sin (90-2 C)}=\frac{x}{\sin (90+C)}=\frac{5-x}{\sin C}=2 R \] which can be simplified to \[ \frac{1}{\cos 2 C}=\frac{x}{\cos C}=\frac{5-x}{\sin C}=2 R \] This means that \[ \frac{1}{\cos 2 C}=\frac{(x)+(5-x)}{(\cos C)+(\sin C)}=\frac{5}{\cos C+\sin C} \] and the rest is an easy computation: \[ \cos C+\sin C=5 \cos 2 C=5\left(\cos ^{2} C-\sin ^{2} C\right) \] \[ \frac{1}{5}=\cos C-\sin C \] Squaring both sides, \[ \frac{1}{25}=\cos ^{2} C-2 \sin C \cos C+\sin ^{2} C=1-\sin 2 C \] so $\sin 2 C=\frac{24}{25}$, implying that $\cos 2 C=\frac{7}{25}$. Therefore, since $\frac{1}{\cos 2 C}=2 R$ from above, $R=\frac{25}{14}$. Finally, viewing triangle $A B C$ with $B C=1$ as the base, the height is \[ \sqrt{R^{2}-\left(\frac{B C}{2}\right)^{2}}=\frac{12}{7} \] by the Pythagorean Theorem, yielding an area of $\frac{1}{2} \cdot 1 \cdot \frac{12}{7}=\frac{6}{7}$. Solution 2. The midpoint of $O H$ is the nine-point center $N$. We are given $N$ lies on $B C$, and we also know $N$ lies on the perpendicular bisector of $E F$, where $E$ is the midpoint of $A C$ and $F$ is the midpoint of $A B$. The main observation is that $N$ is equidistant from $M$ and $F$, where $M$ is the midpoint of $B C$. 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 $B C$, we have $N\left(\frac{a}{2}, 0\right)$. Since $M N=E N$, 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 $A B+A C=5$, which is just \[ \sqrt{(a+0.5)^{2}+b^{2}}+\sqrt{(a-0.5)^{2}+b^{2}}=5 \] This is equivalent to \[ \begin{gathered} \sqrt{2 a^{2}+a}+\sqrt{2 a^{2}-a}=5 \\ \sqrt{2 a^{2}+a}=5-\sqrt{2 a^{2}-a} \\ 2 a^{2}+a=25-10 \sqrt{2 a^{2}-a}+2 a^{2}-a \\ 25-2 a=10 \sqrt{2 a^{2}-a} \\ 625-100 a+4 a^{2}=200 a^{2}-100 a \\ 196 a^{2}=625 \end{gathered} \] Thus $a^{2}=\frac{625}{196}$, so $b^{2}=\frac{576}{196}$. Thus $b=\frac{24}{14}=\frac{12}{7}$, so $[A B C]=\frac{b}{2}=\frac{6}{7}$. $\fbox{\frac{6}{7}}$.

Answer: \frac{6}{7}