Difference between revisions of "2024 AMC 12B Problems/Problem 20"
(→Solution) |
(→Solution) |
||
Line 46: | Line 46: | ||
The area of this triangle is <math>\frac{1}{2} 42 \cdot 40 \cdot \sin(\theta)</math>. The maximum value of the area occurs when the triangle is right, i.e. <math>\theta = 90^{\circ}</math>. Then the area is <math>\frac{1}{2} \cdot 40 \cdot 42 = 840</math>. The length of the median of a right triangle is half the length of it's hypotenuse, which squared is <math>40^2+42^2 = 58^2</math>. Thus the length of <math>x</math> is <math>29</math>. | The area of this triangle is <math>\frac{1}{2} 42 \cdot 40 \cdot \sin(\theta)</math>. The maximum value of the area occurs when the triangle is right, i.e. <math>\theta = 90^{\circ}</math>. Then the area is <math>\frac{1}{2} \cdot 40 \cdot 42 = 840</math>. The length of the median of a right triangle is half the length of it's hypotenuse, which squared is <math>40^2+42^2 = 58^2</math>. Thus the length of <math>x</math> is <math>29</math>. | ||
− | Our final answer is <math>1+41+840+29 = \boxed{\textbf{911 }C}</math> | + | Our final answer is <math>1+41+840+29 = \boxed{\textbf{911 } (C)}</math> |
==Solution #1 == | ==Solution #1 == |
Revision as of 11:31, 14 November 2024
Contents
[hide]Problem 20
Suppose , , and are points in the plane with and , and let be the length of the line segment from to the midpoint of . Define a function by letting be the area of . Then the domain of is an open interval , and the maximum value of occurs at . What is ?
Solution
Let the midpoint of be , and let the length . We know there are limits to the value of , and these limits can probably be found through the triangle inequality. But the triangle inequality relates the third side length to and , and doesn't contain any information about the median. Therefore we're going to have to write the side in terms of and then use the triangle inequality to find bounds on .
We use Stewart's theorem to relate to the median : . In this case , , , , , .
Therefore we get the equation
.
Notice that since is a pythagorean triple, this means .
By triangle inequality, and
Let's tackle the first inequality:
Here we use the property that .
Therefore in this case, .
For the second inequality,
Therefore we have , so the domain of is .
The area of this triangle is . The maximum value of the area occurs when the triangle is right, i.e. . Then the area is . The length of the median of a right triangle is half the length of it's hypotenuse, which squared is . Thus the length of is .
Our final answer is
Solution #1
Let midpoint of as , extends to and ,
triangle has sides as such, so
so which is achieved when , then
See also
2024 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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.