Difference between revisions of "2019 AMC 10B Problems/Problem 10"
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Revision as of 17:04, 15 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.
Contents
Problem
In a given plane, points and are units apart. How many points are there in the plane such that the perimeter of is units and the area of is square units?
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 . The perimeter of this minimized triangle is , which is larger than 50. Since the minimized perimeter is greater than 50, there is no triangle that satisfies the condition, giving us .
Solution 2
WLOG let be a horizontal segment of length 10. Now, realize has to lie on one of the lines parallel to and vertically 20 units away from it. But 10+20+20 is already 50, and this doesn't form a triangle. Otherwise, WLOG . Dropping altitude we have a right triangle with hypotenuse and leg which is clearly impossible, giving us .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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