Triangles, flagpoles and heights, oh my! Here is a math question i got from school:
On a horizontal plane, there are two flagpoles. One is 20m, and the other is 10m. There is a wire connected from the top of each flagpole, to the bottom of the other one so that they cross each other . How high is the point they cross from the ground?

How do I solve it without using pythagoras
 A: Let $h$ be the desired height, and let $a$ and $b$ be the horizontal distances from the point where the wires cross to the left and right flagpoles, respectively. Then using similar triangles, we have:
$$
\frac{h}{a} = \frac{10}{a + b} \qquad\text{and}\qquad
\frac{h}{b} = \frac{20}{a + b}
$$
Solving for $h$, we can combine the above to get that:
$$
\frac{10a}{a+b} = h = \frac{20b}{a + b} \iff a = 2b
$$
Substituting into the second equation, we get:
$$
\frac{h}{b} = \frac{20}{(2b) + b} \iff \boxed{h = \dfrac{20}{3}}
$$
A: This is the definition of the harmonic mean.
Let's call the left flag pole $h_1$ and the the right one $h_2$.
The distance from $h_1$ to $x$ is $a$, from $x$ to $h_2$ is $b$.
Then using similar right triangles:
$$\begin{align}
\frac{a+b}{h_1} &= \frac{b}{x}\\
\frac{a+b}{h_2} &= \frac{a}{x}\\
\text{Adding them together}\\
\frac{a+b}{h_1} + \frac{a+b}{h_2} &= \frac{a+b}{x}\\
\text{Divide through by $a+b$}\\
\frac{1}{h_1} + \frac{1}{h_2} &= \frac{1}{x}\\
\end{align}
$$
Plug in 10 and 20 for $h_1$ and $h_2$.
