2019 AMC 10B Problem 10

The perimeter condition gives $AC + BC = 50 - 10 = 40$. The area condition gives $\frac12 \cdot 10 \cdot h = 100$, so $C$ must be at height $h = 20$ above line $AB$.

Among all points $C$ with $AC + BC = 40$, the one farthest from line $AB$ is the symmetric one with $AC = BC = 20$ (moving $C$ off the perpendicular bisector while keeping the sum fixed lowers it, since the locus is an ellipse with foci $A$ and $B$ and its highest point is on the minor axis). In that position the height is
$$\sqrt{20^2 - 5^2} = \sqrt{375} < 20.$$

So no point with $AC + BC = 40$ reaches height $20$, and the maximum possible area is $\frac12 \cdot 10 \cdot \sqrt{375} \approx 96.8 < 100$. No such $C$ exists.

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

Two constraints on $C$ (a perimeter and an area) normally intersect in finitely many points, but only if each is achievable given the other. Fixing $AC + BC$ bounds how far $C$ can be from $AB$, and the isosceles configuration is the extreme case. When a counting question offers $0$ as a choice, check feasibility before counting.

• Assuming a triangle with the given perimeter and area exists and answering 2 or 4 by symmetry, without checking whether height 20 is reachable.
• Comparing the required height $20$ with $AC = 20$ and concluding it works, forgetting that the height is a leg, shorter than the hypotenuse $AC$.