2019 AMC 10B Problem 10

Turn each condition into a fact about point $C$, then see whether both facts can hold at once.

1. Perimeter. $AB + AC + BC = 50$ and $AB = 10$, so $AC + BC = 40$.
2. Area. Use $AB$ as the base: $\frac12 \cdot 10 \cdot h = 100$, so the height is $h = 20$. That means $C$ is $20$ units away from line $AB$.
3. The height limits the sides. A slanted path from $C$ down to line $AB$ is never shorter than the straight-down path. So $AC \ge 20$, and $AC = 20$ only if $C$ sits directly above $A$. In the same way, $BC \ge 20$, with $BC = 20$ only if $C$ sits directly above $B$.
4. $C$ cannot be directly above both $A$ and $B$, since they are $10$ units apart. So at least one of these is strictly bigger than $20$, and
$$AC + BC > 20 + 20 = 40.$$
5. That contradicts step 1. No point $C$ satisfies both conditions.

The answer is $\boxed{\textbf{(A)}\ 0}$.

Two conditions on a point usually meet in a few places, but only if they can both be met at all. Before counting, ask whether the numbers are even possible together. Here the perimeter is too small for such a tall triangle. When $0$ is an answer choice in a counting question, take it seriously.

Find the largest area possible with $AC + BC = 40$. The tallest such triangle is the one with $AC = BC = 20$ (sharing the $40$ evenly puts $C$ as high as possible). Its height, from the right triangle with hypotenuse $20$ and half-base $5$, is $\sqrt{400 - 25} = \sqrt{375}$, which is less than $\sqrt{400} = 20$. So the largest possible area is below $\frac12 \cdot 10 \cdot 20 = 100$, and area $100$ cannot happen.