2015 AMC 10B Problem 13

Setting $y = 0$ gives $x = 5$; setting $x = 0$ gives $y = 12$. So the triangle has vertices $(0,0)$, $(5,0)$, $(0,12)$: a right triangle with legs $5$ and $12$ and hypotenuse $\sqrt{25 + 144} = 13$.

In a right triangle each leg is perpendicular to the other, so the legs are two of the three altitudes: $5$ and $12$.

The third altitude is the one to the hypotenuse. Computing the area two ways, $\frac12 \cdot 5 \cdot 12 = \frac12 \cdot 13 \cdot h$, so $h = \frac{60}{13}$.

The sum is $5 + 12 + \frac{60}{13} = 17 + \frac{60}{13} = \frac{221 + 60}{13} = \frac{281}{13}$.

The answer is $\boxed{\textbf{(E)}\ \frac{281}{13}}$.

A line $ax + by = c$ cuts the axes at $(c/a, 0)$ and $(0, c/b)$, always producing a right triangle at the origin. In a right triangle the legs double as altitudes, and the altitude to the hypotenuse is forced by the area identity $\text{leg}_1 \cdot \text{leg}_2 = \text{hypotenuse} \cdot h$. Recognize the $5$-$12$-$13$ triple instantly to skip the square root.

The altitude to the hypotenuse is the distance from the origin to the line $12x + 5y = 60$: $\frac{|0 + 0 - 60|}{\sqrt{12^2 + 5^2}} = \frac{60}{13}$. Then add the two legs.

• Forgetting that in a right triangle the two legs are themselves altitudes, and reporting only 60/13 or adding the hypotenuse instead.
• Swapping the intercepts (reading $x = 12$, $y = 5$); the sum happens to be the same here, but the habit causes errors elsewhere.