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

==Solution 1==

Notice that whatever point we pick for$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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}$.

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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2019 AMC 10B Problems/Problem 10

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