## Perpendicular Medians

#### Problem

In triangle $ABC$, $A'$ is the midpoint of $BC$, $B'$ is the midpoint of $AC$, and the line segments $AA'$ and $BB'$ are perpendicular. Let $AC = b$, $BC = a$, and $AB = c$.

1. Find $c$ in terms of $a$ and $b$.
2. Determine the values of $b$ in terms of $a$ for which the triangle exists.
3. Calculate the size of the maximum interior angle at vertex $C$.

#### Solution

As any median splits each other median in the ratio 2:1 (see Triangle Median), we can label the following lengths.

1. Using the Pythagorean Theorem:

$x^2 +(2y)^2 = \left( \dfrac{a}{2} \right )^2 \implies x^2 + 4y^2 = \dfrac{a^2}{4} (\Delta BPA')$
$(2x)^2 + y^2 = \left( \dfrac{b}{2} \right )^2 \implies 4x^2 + y^2 = \dfrac{b^2}{4} (\Delta APB')$

$\therefore 5x^2 + 5y^2 = \dfrac{a^2 + b^2}{4} \implies x^2 + y^2 = \dfrac{a^2 + b^2}{20}$

In $\Delta APB$, $(2x)^2 + (2y)^2 = c^2 \implies 4x^2 + 4y^2 = c^2$

$\therefore c^2 = 4(x^2 + y^2) = \dfrac{a^2 + b^2}{5} \implies c = \sqrt{ \dfrac{a^2 + b^2}{5} }$

2. Using the cosine rule in $\Delta ABC$:

\begin {align}\cos(C) &= \dfrac{a^2 + b^2 - c^2}{2ab}\\&= \dfrac{a^2 + b^2 - \frac{a^2 + b^2}{5}}{2ab}\\&= \dfrac{5a^2 + 5b^2 - (a^2 + b^2)}{10ab}\\&= \dfrac{4a^2 + 4b^2}{10ab}\\&= \dfrac{2a^2 + 2b^2}{5ab}\end {align}

As $a$ and $b$ are both positive, $0 \lt \cos(C) \lt 1$.

$\therefore \dfrac{2^2 + 2b^2}{5ab} \lt 1$

Solving at equality with 1 in $b$:

\begin{align}2a^2 + 2b^2 &= 5ab\\\therefore 2b^2 - 5ab + 2a^2 &= 0\end{align}

\begin{align}\therefore b &= \dfrac{5a \pm \sqrt{25a^2 - 16a^2}}{4}\\&= \dfrac{5a \pm 3a}{4}\\&= \dfrac{a}{2},2a\end{align}

That is, $\dfrac{a}{2} \lt b \lt 2a$ for the triangle to exist.

3. By writing $\cos(C) = \dfrac{2a^2 + 2b^2}{5ab} = r$, we get $2b^2 - 5abr + 2a^2 = 0$.

Solving this quadratic in $b$ once again, we get,

$$b = \dfrac{5ar \pm \sqrt{25a^2 r^2 - 16a^2}}{4}$$

However, for a real solution the discriminant, $25a^2 r^2 - 16a^2 \ge 0$.

Solving equal to zero, $a^2(25r^2 - 16) = 0 \implies a = 0$ (in which case the triangle does not exist) or $25r^2 - 16 = 0 \implies r^2 = \dfrac{16}{25} \implies r = \dfrac{4}{5}$.

Hence $r = \cos(C) \ge \dfrac{4}{5}$; that is, $\dfrac{4}{5} \le \cos(C) \le 1$. But as $\cos(0) = 1$ and $\cos(C)$ decreases as $C$ increases the greatest interior angle at $C$ will be $\cos^{-1}(4/5) \approx 36.9^o$.

(Why must angle C be acute?)

Problem ID: 296 (10 Dec 2006)     Difficulty: 4 Star

Only Show Problem