Question

In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?

$0$

$2$

$4$

$8$

$\text{infinitely many}$

Answer: A

Concepts:triangle area triangle inequality

Difficulty rating: 1460

Solution:

The area condition fixes the distance from $C$ to line $AB.$ If that distance is $h,$ then $\frac{10h}{2}=100,$ so $h=20.$ Thus $C$ must lie on a line parallel to $AB$ at distance $20.$

Since the perpendicular distance from $C$ to line $AB$ is $20$ , both $AC$ and $BC$ are at least $20$ . They cannot both equal $20$ , because that would require the perpendicular from $C$ to meet line $AB$ at both distinct points $A$ and $B$ . Hence $AC+BC>40$ , contradicting the perimeter requirement $AC+BC=40$ . Therefore no point $C$ works.

Thus, the answer is A .

2019 AMC 10B Problem 10

Answer: A

Difficulty rating: 1460

Solution:

Problem 10 in Other Years