Solution to Week 4 Bonus

This post originally appeared on the Brilliant blog on 9/9/2012.

Bonus Challenge: In triangle $ABC$, $BC=16$ and $AC=43$. Let $D$ and $E$ be the midpoints of $BC$ and $AC$, respectively. It is given that $AD$ and $BE$ are perpendicular. What is the value of $|AB|^2$?

Key Technique: Pythagorean Theorem, Parallel Lines

Note: You may have received different values of $BC$ and $AC$ than the ones listed above.

Solution 1: Since $D$ and $E$ are midpoints, it follows from Parallel Lines Property D that $DE \parallel AB$ and $ED = \frac {1}{2} AB$. Let $G$ be the intersection of $AD$ and $BE$. We have the following Pythagorean Theorem:

$\begin{aligned} GD^2 + GE^2 & = & DE^2 = \frac {1}{4} AB^2\\ GD^2 + GB^2 & = & \left( \frac {16}{2} \right) ^2\\ GE^2 + GA^2 & = & \left( \frac {43}{2} \right) ^2 \\ GA^2 + GB^2 & = & AB^2 \\ \end{aligned}$

Adding the first and fourth statements and subtracting the middle two gives $0 = \frac {5}{4} AB^2 - \frac {16^2+43^2}{4} \Rightarrow AB^2 = \frac {16^2+43^2}{5}=421$.

Solution 2: Use vector notation. Let $C$ be the origin, define $\vec{CA}=\vec{a}$ and $\vec{CB} = \vec{b}$. Then, $\vec{AD} = \vec{AC}+\vec{CD} = -\vec{a}+\frac {1}{2} \vec{b}$ and $\vec{BE} = \vec {BC} + \vec{CE} = -\vec{b} +\frac {1}{2} \vec{a}$. Since the 2 medians are perpendicular, their dot product is 0, which gives us $5 \vec{a} \cdot \vec {b} - 2a^2 - 2 b^2 = 0$. Since $a = 16$ and $b=43$, we get

$\vec{a} \cdot \vec{b} = \frac {2 \times 16 ^2 + 2 \times 43 ^2} {5} =842.$

Now, we can calculate that $|AB|^2 = \vec{AB} \cdot \vec{AB} = (\vec{b} - \vec{a})\cdot (\vec{b} - \vec{a}) = b^2 + a^2 - 2 \vec{a} \cdot \vec{b} = 43 ^2 + 16 ^2 - 2 \times 842 = 421$.

Note: The solutions can be simplified slightly if we know that $G$ is the centroid of triangle $ABC$, hence $AG = 2 GD$ and $BG = 2 GE$. This ratio is a direct consequence of Parallel Lines Property D.

Answer: 421