2013 AMC 10A Problem 15

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 $\boxed{\textbf{(D)}\ 12}$.

Altitudes are not independent quantities: each is $2K$ over its side. Any statement about altitudes therefore translates into a statement about reciprocals of sides, with the area as a common factor that cancels. Here the third side is the harmonic-type combination $\frac{2}{1/10 + 1/15}$, which is why it is not the arithmetic mean $12.5$.

• Averaging the two side lengths (getting 12.5) or assuming the altitudes are proportional to the sides rather than inversely proportional.
• Dropping the factor $\frac12$ in the average, which gives $\frac{2}{c} = \frac{1}{3}$ and the wrong answer $6$.