Area and orthocenter

Question

I need help with the following proble. Given a triangle ABC with orthocenter H and altitudes $CM= 2 \sqrt 2$ and $AN=3$. If H divides the altitude BP into segments with ratio 5:1, i.e.BH:HP=5:1, find the area of $\triangle ABC$.

Answer 1

Let $CH=x$ and $HM=Y$. Let $AH=u$ and $HN=v$. Let $BH=5t$ and $HP=t$.

Triangles $HPC$ and $HMB$ are similar. It follows that $\frac{t}{x}=\frac{y}{5t}$ and therefore $xy=5t^2$. In a similar way we can show that $uv=5t^2$. So far we have the equations $$x+y=2\sqrt{2},\qquad u+v=3,\qquad xy=uv.$$ We need another equation! One way of getting it goes as follows.

Let $K$ be the area of $\triangle ABC$. Note that $\triangle ABC$ can be decomposed into the $3$ triangles $\triangle AHC$, $\triangle BHA$, and $\triangle CHB$.

Look first at $\triangle AHC$. This has base $CA$ and height $t$, while our full triangle $ABC$ has base $CA$ and height $6t$. So the area of $\triangle AHC$ is $\frac{t}{6t}K$, that is, $\frac{K}{6}$.

Do a similar calculation with $\triangle BHA$. It has area $\frac{y}{x+y}K$, that is, $\frac{Ky}{2\sqrt{2}}$. Similarly, the area of $\triangle CHB$ is $\frac{Kv}{3}$.

The sum of the areas of these three triangles is $K$. That gives the equation $$\frac{K}{6}+\frac{Ky}{2\sqrt{2}}+\frac{Kv}{3}=K,$$ which simplifies to $$3y+2\sqrt{2}v=5\sqrt{2}.$$

The rest is mechanical. We have $3$ linear equations in our variables $x$, $y$, $u$, $v$. Use these to express $x$, $u$, and $v$ in terms of $y$. Then substitute in the equation $xy=uv$. We get a quadratic equation in $y$. Solve, then find $x$, $u$, $v$. Now we can determine $t$, and therefore everything.

Remark: There is a much simpler solution that I cannot recall. The annoying thing is that a number of years ago I gave a version of this problem in an assignment in a "problem solving" course!