# 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 1

Notice that whatever point we pick for $C$, $AB$ will be the base of the triangle. Without loss of generality, let points $A$ and $B$ be $(0,0)$ and $(10,0)$, since for any other combination of points, we can just rotate the plane to make them $(0,0)$ and $(10,0)$ under a new coordinate system. When we pick point $C$, we have to make sure that its $y$-coordinate is $\pm20$, because that's the only way the area of the triangle can be $100$.

Now when the perimeter is minimized, by symmetry, we put $C$ in the middle, at $(5, 20)$. We can easily see that $AC$ and $BC$ will both be $\sqrt{20^2+5^2} = \sqrt{425}$. The perimeter of this minimal triangle is $2\sqrt{425} + 10$, which is larger than $50$. Since the minimum perimeter is greater than $50$, there is no triangle that satisfies the condition, giving us $\boxed{\textbf{(A) }0}$.

~IronicNinja

## Solution 2

Without loss of generality, let $AB$ be a horizontal segment of length $10$. Now realize that $C$ 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, without loss of generality, $AC<20$. Dropping altitude $CD$, we have a right triangle $ACD$ with hypotenuse $AC<20$ and leg $CD=20$, which is clearly impossible, again giving the answer as $\boxed{\textbf{(A) }0}$.

## Solution 3

We have:

1. Area = $100$

2. Perimeter = $50$

3. Semiperimeter $s = 50 \div 2 = 25$

We let:

1. $z = \overline{AB} = 10$

2. $x = \overline{AC}$

3. $y = 50-10-x = 40-x$.

Heron's formula states that for real numbers $x$, $y$, $z$, and semiperimeter $s$, the area is $\sqrt{(s)(s-x)(s-y)(s-z)}$.

Plugging numbers in, we have $100 = \sqrt{(25)(25-10)(25-x)(25-(40-x))} = \sqrt{(375)(25-x)(x-15)}$.

Square both sides, divide by $375$ and expand the polynomial to get $40x - x^2 - 375 = \frac{80}{3}$.

$x^2 - 40x + \left(375 + \frac{80}{3}\right) = 0$ and the discriminant is $\left((-40)^2 - 4 \times 1 \times 401 \frac{2}{3}\right) < 0$. Thus, there are no real solutions.

## Solution 4 (graphing)

First, let's assume that A and B are $(-5,0)$ and $(5,0)$ respectively. The graph of "the perimeter is $50$" means that $\overline{AC}+\overline{BC}=50-10=40$. So this is the graph of an ellipse (memorize that!). Now let the endpoints of the major axis be $(-x,0)$ and $(x,0)$. Then $(x-5)+(x+5)=40$ and $x=20$. So the $2$ endpoints of the major axis are $(-20,0)$ and $(20,0)$. We can also figure out the endpoints of the minor axis must have a y-coordinate less than $20$. It is actually $\sqrt{395}$.

Now, we consider "the area is $100$". Since the base has length $10$, then the height must have length $20$. So the graph of "the area is 100" is $2$ lines, one at $y=20$ and the other at $y=-20$. However, this graph does NOT intersect the ellipse, as $\sqrt{395} < 20$. So, there are no intersections and thus no solutions, so the answer is $\boxed{\textbf{(A) }0}$.

~Yrock

## Video Solution

~Education, the Study of Everything

~IceMatrix

~savannahsolver