Difference between revisions of "2024 AMC 12B Problems/Problem 20"
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<cmath>\implies a^2 = \frac{58^2}{2}-x^2</cmath> | <cmath>\implies a^2 = \frac{58^2}{2}-x^2</cmath> | ||
<cmath>\implies a = \sqrt{\frac{58^2}{2}-x^2}</cmath> | <cmath>\implies a = \sqrt{\frac{58^2}{2}-x^2}</cmath> | ||
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+ | By triangle inequality, <math>2a+40>42 \implies a>1 </math> and <math>40+42>2a \implies a<41</math> | ||
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+ | Let's tackle the first inequality: <cmath>\sqrt{\frac{58^2}{2}-x^2}>1 \implies x^2 < \frac{58^2}{2}-1</cmath> | ||
==Solution #1 == | ==Solution #1 == |
Revision as of 11:22, 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:
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 |
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