Difference between revisions of "2019 AMC 10B Problems/Problem 10"
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==Video Solution== | ==Video Solution== | ||
Latest revision as of 23:26, 3 December 2024
- 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 1
Notice that whatever point we pick for , 
will be the base of the triangle. Without loss of generality, let points and be 
and 
, since for any other combination of points, we can just rotate the plane to make them and under a new coordinate system. When we pick point , we have to make sure that its 
-coordinate is 
, because that's the only way the area of the triangle can be .
Now when the perimeter is minimized, by symmetry, we put in the middle, at 
. We can easily see that 
and 
will both be 
. The perimeter of this minimal triangle is 
, which is larger than . Since the minimum perimeter is greater than , there is no triangle that satisfies the condition, giving us 
.
~IronicNinja
Solution 2
Without loss of generality, let be a horizontal segment of length . Now realize that has to lie on one of the lines parallel to and vertically 
units away from it. But 
is already 50, and this doesn't form a triangle. Otherwise, without loss of generality, 
. Dropping altitude 
, we have a right triangle 
with hypotenuse and leg 
, which is clearly impossible, again giving the answer as .
Solution 3
We have:
1. Area =
2. Perimeter =
3. Semiperimeter 
We let:
1. 
2. 
3. 
.
Heron's formula states that for real numbers 
, , 
, and semiperimeter 
, the area is 
.
Plugging numbers in, we have 
.
Square both sides, divide by 
and expand the polynomial to get 
.
and the discriminant is 
. Thus, there are no real solutions.
Solution 4 (graphing)
First, let's assume that A and B are 
and 
respectively. The graph of "the perimeter is " means that 
. So this is the graph of an ellipse (memorize that!). Now let the endpoints of the major axis be 
and 
. Then 
and 
. So the 
endpoints of the major axis are 
and 
. We can also figure out the endpoints of the minor axis must have a y-coordinate less than . It is actually 
.
Now, we consider "the area is ". Since the base has length , then the height must have length . So the graph of "the area is 100" is lines, one at 
and the other at 
. However, this graph does NOT intersect the ellipse, as 
. So, there are no intersections and thus no solutions, so the answer is .
~Yrock
Video Solution
~Education, the Study of Everything
Video Solution
~IceMatrix
~savannahsolver
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. 
