Difference between revisions of "2024 AMC 12B Problems/Problem 20"
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\textbf{(E) }913\qquad | \textbf{(E) }913\qquad | ||
</math> | </math> | ||
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+ | ==Solution== | ||
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+ | Let the midpoint of <math>BC</math> be <math>M</math>, and let the length <math>BM = CM = a</math>. We know there are limits to the value of <math>x</math>, and these limits can probably be found through the triangle inequality. But the triangle inequality relates the third side length <math>BC</math> to <math>AC</math> and <math>AB</math>, and doesn't contain any information about the median. Therefore we're going to have to write the side <math>BC</math> in terms of <math>x</math> and then use the triangle inequality to find bounds on <math>x</math>. | ||
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+ | We use Stewart's theorem to relate <math>BC</math> to the median <math>AM</math>: <math>man + dad = bmb + cnc</math>. In this case <math>m = a</math>, <math>n=a</math>, <math>a = m+n</math>, <math>d = x</math>, <math>b = 42</math>, <math>c = 40</math>. | ||
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+ | Therefore we get the equation <math>2a^3 + 2ax^2 = a \cdot 42^2 + a \cdot 40^2</math> | ||
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+ | <math>2a^2 + 2x^2 = 42^2 + 40^2</math>. | ||
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+ | Notice that since <math>20-21-29</math> is a pythagorean triple, this means <math>2a^2 + 2x^2 = 58^2</math>. | ||
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+ | <cmath>\implies a^2 = \frac{58^2}{2}-x^2</cmath> | ||
+ | <cmath>\implies a = \sqrt{\frac{58^2}{2}-x^2}</cmath> | ||
==Solution #1 == | ==Solution #1 == |
Revision as of 11:19, 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 .
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.