Difference between revisions of "2019 AMC 10B Problems/Problem 10"
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Now when the perimeter is minimized, by symmetry, we put <math>C</math> in the middle, at <math>(5, 20)</math>. We can easily see that <math>AC</math> and <math>BC</math> will both be <math>\sqrt{20^2+5^2} = \sqrt{425}</math>. The perimeter of this minimal triangle is <math>2\sqrt{425} + 10</math>, which is larger than <math>50</math>. Since the minimum perimeter is greater than <math>50</math>, there is no triangle that satisfies the condition, giving us <math>\boxed{\textbf{(A) }0}</math>. | Now when the perimeter is minimized, by symmetry, we put <math>C</math> in the middle, at <math>(5, 20)</math>. We can easily see that <math>AC</math> and <math>BC</math> will both be <math>\sqrt{20^2+5^2} = \sqrt{425}</math>. The perimeter of this minimal triangle is <math>2\sqrt{425} + 10</math>, which is larger than <math>50</math>. Since the minimum perimeter is greater than <math>50</math>, there is no triangle that satisfies the condition, giving us <math>\boxed{\textbf{(A) }0}</math>. | ||
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+ | ~IronicNinja | ||
==Solution 2== | ==Solution 2== |
Revision as of 23:01, 17 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
[hide]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
.
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.