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 $A B C$.
Answer:
$\frac{6}{7}$
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, $A B0$. 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
$$
\begin{array}{c}
\sqrt{2 a^{2}+a}+\sqrt{2 a^{2}-a}=5 \\
\sqrt{2 a^{2}+a}=5-\sqrt{2 a^{2}-a} \\
2 a^{2}+a=25-10 \sqrt{2 a^{2}-a}+2 a^{2}-a \\
25-2 a=10 \sqrt{2 a^{2}-a} \\
625-100 a+4 a^{2}=200 a^{2}-100 a \\
196 a^{2}=625
\end{array}
$$

Thus $a^{2}=\frac{625}{196}$, so $b^{2}=\frac{576}{196}$. Thus $b=\frac{24}{14}=\frac{12}{7}$, so $[A B C]=\frac{b}{2}=\frac{6}{7}$.