2013 AMC 10A Problem 15

We do not know the area, but it links every side to its altitude. Use it and hope it cancels.

1. Pick an area to see the pattern. Suppose the area is $30$. Area $= \tfrac12 \times \text{side} \times \text{altitude}$, so altitude $= \dfrac{2 \times \text{area}}{\text{side}} = \dfrac{60}{\text{side}}$.
- Altitude to the side $10$: $60 \div 10 = 6$.
- Altitude to the side $15$: $60 \div 15 = 4$.
- Their average: $(6 + 4) \div 2 = 5$.
- The altitude to the third side is $5$, so the third side is $60 \div 5 = 12$.
2. Why any area gives the same answer. Call the area $K$ and the third side $c$. The altitudes are $\dfrac{2K}{10}$, $\dfrac{2K}{15}$ and $\dfrac{2K}{c}$. The rule says
$$\frac{2K}{c} = \frac12\left(\frac{2K}{10} + \frac{2K}{15}\right) = \frac{K}{10} + \frac{K}{15} = \frac{3K + 2K}{30} = \frac{K}{6}.$$
3. Solve. Divide both sides by $K$: $\dfrac{2}{c} = \dfrac16$, so $c = 12$.
4. Check that such a triangle exists. Sides $10, 12, 15$: the two shorter sides add to $22$, more than $15$. Good.

(An area of $30$ may not be the real area of this triangle. That does not matter, because Step 2 shows the area cancels.)

The answer is $\boxed{\textbf{(D)}\ 12}$.

Altitudes are not free to be anything: each one is twice the area divided by its side. So a rule about altitudes is really a rule about "one over each side", and the area cancels. That is why the answer is $12$ and not the plain average $12.5$ of the sides.

Let the triangle have area $K$ and third side $c$. Reading $K = \frac12 (\text{side})(\text{altitude})$ three ways, the altitudes are
$$h_{10} = \frac{2K}{10}, \qquad h_{15} = \frac{2K}{15}, \qquad h_c = \frac{2K}{c}.$$

The given condition $h_c = \frac{h_{10} + h_{15}}{2}$ becomes
$$\frac{2K}{c} = \frac12\left(\frac{2K}{10} + \frac{2K}{15}\right) = K\left(\frac{1}{10} + \frac{1}{15}\right) = K \cdot \frac{3 + 2}{30} = \frac{K}{6}.$$

Cancel $K$: $\frac{2}{c} = \frac{1}{6}$, so $c = 12$.

Check that such a triangle exists: $10, 12, 15$ satisfies the triangle inequality. The answer is $\boxed{\textbf{(D)}\ 12}$.