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 (A bit tedious)

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. ~hashbrown2009

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

Solution 5 (kinda bashy quadratics)

Since $AB$ has a length of $10$ , the height must be $10$ as well for the area to be $50$ , so we can drop an altitude from $C$ that has length $10$ , and call the foot of the altitude on $AB$ as $D$ . Let $AD$ have length $x$ , so $BD$ must have length $10-x$ . From the Pythagorean Theorem, $AC$ has length $\sqrt{x^2+100}$ and $BC$ has length $\sqrt{(10-x)^2+100}$ . For the perimeter to be $50$ , $10+\sqrt{x^2+100}+\sqrt{(10-x)^2+100}=50$ , so $40-\sqrt{x^2+100}=\sqrt{(10-x)^2+100}$ . Squaring both sides and simplifying, $1600-80\sqrt{x^2+100}=-20x+100$ . We can rearrange this and divide all sides by $10$ to get $150+2x=8\sqrt{x^2+100}$ . Squaring both sides yet again, rearranging, and simplifying some more, we get the quadratic $3x^2-30x-805=0$ . So, from the quadratic formula, $x = \frac{30+\sqrt{900+4\cdot3\cdot805}}{6}$ or $x = \frac{30-\sqrt{900+4\cdot3\cdot805}}{6}$ . Immediately we can rule out the negative solution (it's negative because the square root of the discriminant is greater than $30$ , and we'll use this later too), as $x$ must be a positive number since it is a length. However, for the positive solution, we see that the discriminant is greater than $900$ , so the square root is greater than $30$ , so the entire expression when divided by $6$ for $x$ is greater than $10$ , but this is not possible because $x$ should be less than $10$ . Therefore, the number of possibilities is $\boxed{\textbf{(A) }0}$ .

~vaishnav

Video Solution

https://youtu.be/MNVKkjVvBUU

~Education, the Study of Everything

Video Solution

https://youtu.be/7xf_g3YQk00

~IceMatrix

https://youtu.be/INvRdwQzC-w

~savannahsolver

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search

2019 AMC 10B Problems/Problem 10

Contents

Problem

Solution 1

Solution 2

Solution 3 (A bit tedious)

Solution 4 (graphing)

Solution 5 (kinda bashy quadratics)

Video Solution

Video Solution

See Also