Difference between revisions of "2024 AMC 12B Problems/Problem 20"
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triangle <math>ACD</math> has <math>3</math> sides <math>(40,42,2x)</math> | triangle <math>ACD</math> has <math>3</math> sides <math>(40,42,2x)</math> | ||
− | as such,<cmath> | + | |
− | <cmath>1 | + | for 3 sides a>b>c in any triangle <cmath> a - b < c < a + b </cmath> |
+ | |||
+ | as such,<cmath> 42 - 40 < 2x < 42 + 40 </cmath> | ||
+ | <cmath>1 < x < 41</cmath> | ||
so <cmath>p = 1, q=41</cmath> | so <cmath>p = 1, q=41</cmath> | ||
Revision as of 22:09, 15 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 1
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
~KingRavi
Solution 2
Let midpoint of as , extends to and ,
triangle has sides
for 3 sides a>b>c in any triangle
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 |
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