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?

$0$

$2$

$4$

$8$

$\text{infinitely many}$

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 $20.$

The perimeter condition gives $AC+BC=40,$ so $C$ lies on an ellipse with foci $A,B$ and major axis $2a=40,$ hence $a=20$ and $c=5.$ Its semi-minor axis is $b=\sqrt{a^2-c^2}=\sqrt{400-25}=\sqrt{375}\approx19.36.$

The greatest possible distance from a point of the ellipse to line $AB$ is exactly $b\approx19.36,$ which is less than the required $20.$ So no point $C$ satisfies both conditions.

Thus, the answer is A .

2019 AMC 10B Problem 10

Solution:

Problem 10 in Other Years