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

Revision as of 14:44, 14 February 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 $\triangle ABC$ 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 $\boxed{A) 0}$

iron

Solution 2

0 (SuperWill)

Solution 3

Draw segment AB with length 10:

$A-------10-------B$

To have area 100, we need the height to be 20, but there is no way to do this when the sum of the other sides is $50-10=40$ .

--mguempel

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

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

@@ Line 25: / Line 25: @@
 <math>A-------10-------B</math>
-To have area 100, we need the height to be 20, but there is to way to do this when the sum of the other sides is <math>50-10=40</math>.
+To have area 100, we need the height to be 20, but there is no way to do this when the sum of the other sides is <math>50-10=40</math>.
 --mguempel

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