Difference between revisions of "2019 AMC 10B Problems/Problem 10"
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==Solution 1== | ==Solution 1== | ||
− | Notice that whatever point we pick for <math>C</math>, <math>AB</math> will be the base of the triangle. Without loss of generality, let points <math>A</math> and <math>B</math> be <math>(0,0)</math> and <math>(0 | + | Notice that whatever point we pick for <math>C</math>, <math>AB</math> will be the base of the triangle. Without loss of generality, let points <math>A</math> and <math>B</math> be <math>(0,0)</math> and <math>(10,0)</math>, since for any other combination of points, we can just rotate the plane to make them <math>(0,0)</math> and <math>(10,0)</math> under a new coordinate system. When we pick point <math>C</math>, we have to make sure that its <math>y</math>-coordinate is <math>\pm20</math>, because that's the only way the area of the triangle can be <math>100</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>. | 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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==Solution 3== | ==Solution 3== | ||
− | + | We have: | |
+ | 1. Area = <math>100</math> | ||
− | + | 2. Perimeter = <math>50</math> | |
− | <math> | + | 3. Semiperimeter <math>s = 50 \div 2 = 25</math> |
+ | We let: | ||
− | + | 1. <math>z = \overline{AB} = 10</math> | |
+ | 2. <math>x = \overline{AC}</math> | ||
− | <math>x^2 - | + | 3. <math>y = 50-10-x = 40-x</math>. |
+ | |||
+ | |||
+ | Heron's formula states that for real numbers <math>x</math>, <math>y</math>, <math>z</math>, and semiperimeter <math>s</math>, the area is <math>\sqrt{(s)(s-x)(s-y)(s-z)}</math>. | ||
+ | |||
+ | Plugging numbers in, we have <math>100 = \sqrt{(25)(25-10)(25-x)(25-(40-x))} = \sqrt{(375)(25-x)(x-15)}</math>. | ||
+ | |||
+ | |||
+ | Square both sides, divide by <math>375</math> and expand the polynomial to get <math>40x - x^2 - 375 = \frac{80}{3}</math>. | ||
+ | |||
+ | |||
+ | <math>x^2 - 40x + \left(375 + \frac{80}{3}\right) = 0</math> and the discriminant is <math>\left((-40)^2 - 4 \times 1 \times 401 \frac{2}{3}\right) < 0</math>. Thus, there are no real solutions. | ||
+ | |||
+ | ==Solution 4 (graphing)== | ||
+ | |||
+ | First, let's assume that A and B are <math>(-5,0)</math> and <math>(5,0)</math> respectively. The graph of "the perimeter is <math>50</math>" means that <math>\overline{AC}+\overline{BC}=50-10=40</math>. So this is the graph of an ellipse (memorize that!). Now let the endpoints of the major axis be <math>(-x,0)</math> and <math>(x,0)</math>. Then <math>(x-5)+(x+5)=40</math> and <math>x=20</math>. So the <math>2</math> endpoints of the major axis are <math>(-20,0)</math> and <math>(20,0)</math>. We can also figure out the endpoints of the minor axis must have a y-coordinate less than <math>20</math>. It is actually <math>\sqrt{395}</math>. | ||
+ | |||
+ | Now, we consider "the area is <math>100</math>". Since the base has length <math>10</math>, then the height must have length <math>20</math>. So the graph of "the area is 100" is <math>2</math> lines, one at <math>y=20</math> and the other at <math>y=-20</math>. However, this graph does NOT intersect the ellipse, as <math>\sqrt{395} < 20</math>. So, there are no intersections and thus no solutions, so the answer is <math>\boxed{\textbf{(A) }0}</math>. | ||
+ | |||
+ | ~Yrock | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/MNVKkjVvBUU | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/7xf_g3YQk00 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | https://youtu.be/INvRdwQzC-w | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 17:14, 27 May 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.