2002 AMC 10A Problem 13

Every altitude is tied to the area, so first find the area, then find the altitudes from it.

1. Spot the right triangle. Divide each side by $5$: $3, 4, 5$, the famous right triangle. Check directly: $15^2 + 20^2 = 225 + 400 = 625 = 25^2$. So the angle between the sides $15$ and $20$ is a right angle.

2. Area. In a right triangle the two legs are a base and its height:
$$\text{area} = \frac12 \times 15 \times 20 = 150.$$

1. Which altitude is shortest? From area $= \frac12 \times \text{base} \times \text{height}$, the height to any side is $\frac{2 \times \text{area}}{\text{side}} = \frac{300}{\text{side}}$. Dividing $300$ by a bigger side gives a smaller answer, so the shortest altitude goes to the longest side, $25$.

4. Compute it.
$$\frac{300}{25} = 12.$$
(The other two altitudes are $\frac{300}{20} = 15$ and $\frac{300}{15} = 20$, which are just the legs.)

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

One triangle has one area, and each side with its altitude gives that same area. So altitudes and sides go in opposite directions: the longest side has the shortest altitude. Spotting a scaled $3$-$4$-$5$ triangle makes the area immediate.