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

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==Problem==
 
==Problem==
  
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In a given plane, points <math>A</math> and <math>B</math> are <math>10</math> units apart. How many points <math>C</math> are there in the plane such that the perimeter of <math>\triangle ABC</math> is <math>50</math> units and the area of <math>\triangle ABC</math> is <math>100</math> square units?
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In a given plane, points <math>A</math> and <math>B</math> are <math>10</math> units apart. How many points <math>C</math> are there in the plane such that the perimeter of <math>\triangle ABC</math> is <math>50</math> units and the area of $\triangle
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<math>\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}</math>
 
  
 
==Solution 1==
 
==Solution 1==

Revision as of 21:26, 22 April 2019

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

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 $(0,10)$, since for any other combination of points, we can just rotate the plane to make them $(0,0)$ and $(0,10)$ 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 $\boxed{\textbf{(A) }0}$.

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 $\boxed{\textbf{(A) }0}$.

See Also

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

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