# Difference between revisions of "2019 AMC 10B Problems/Problem 10"

The following problem is from both the 2019 AMC 10B #10 and 2019 AMC 12B #6, so both problems redirect to this page.

## Problem

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? $\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}$

## Solution

Notice that whatever point we pick for C, AB will be the base of the triangle. WLOG, points A and B are (0,0) and (0,10) [notice that for any other combination of points, we can just rotate the plane to be the same thing]. When we pick point C, we have to make sure the y value of C is 20, because that's the only way the area of the triangle can be 100.

We figure that the one thing we need to test to see if there is such a triangle is when the perimeter is minimized, and the value of C is (x, 20). Thus, we put C in the middle, so point C is (5, 20). We can easily see that AC and BC will both be $\sqrt{20^2+5^2} \Rightarrow \sqrt{425}$. The perimeter of this minimized triangle is $2\sqrt{425} + 10$, which is larger than 50. Since the minimized perimeter is greater than 50, there is no triangle that satisfies the condition, giving us $\boxed{A) 0}$.

## Solution 2

WLOG let $AB$ be a horizontal segment of length 10. Now, realize $C$ has to lie on one of the lines parallel to $AB$ and vertically 20 units away from it. But 10+20+20 is already 50, and this doesn't form a triangle. Otherwise, WLOG $AC<20$. Dropping altitude $CD$ we have a right triangle $ACD$ with hypotenuse $AC<20$ and leg $CD=20$ which is clearly impossible, giving us $\boxed{A) 0}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 