Difference between revisions of "2023 AIME II Problems/Problem 12"
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<cmath>\frac{AQ}{AM}=\frac{12-x}{x}</cmath>. Plugging in <math>x=\frac{147}{37}</math> and <math>AM=\sqrt{148}</math>, we find that <cmath>AQ=\frac{99}{\sqrt{148}}</cmath> Thus, our final answer is <math>99+148=\boxed{247}</math>. | <cmath>\frac{AQ}{AM}=\frac{12-x}{x}</cmath>. Plugging in <math>x=\frac{147}{37}</math> and <math>AM=\sqrt{148}</math>, we find that <cmath>AQ=\frac{99}{\sqrt{148}}</cmath> Thus, our final answer is <math>99+148=\boxed{247}</math>. | ||
~sigma | ~sigma | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2023|num-b=11|num-a=13|n=II}} | ||
+ | {{MAA Notice}} |
Revision as of 19:10, 16 February 2023
Solution
Because is the midpoint of , following from the Steward's theorem, .
Because , , , are concyclic, , .
Denote .
In , following from the law of sines,
Thus,
In , following from the law of sines,
Thus,
Taking , we get
In , following from the law of sines,
Thus, Equations (2) and (3) imply
Next, we compute and .
We have
We have
Taking (5) and (6) into (4), we get . Therefore, the answer is . ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2
Define to be the foot of the altitude from to . Furthermore, define to be the foot of the altitude from to . From here, one can find , either using the 13-14-15 triangle or by calculating the area of two ways. Then, we find and using Pythagorean theorem. Let . By AA similarity, and are similar. By similarity ratios, Thus, . Similarly, . Now, we angle chase from our requirement to obtain new information. Take the tangent of both sides to obtain By the definition of the tangent function on right triangles, we have , , and . By abusing the tangent angle addition formula, we can find that By substituting , and using tangent angle subtraction formula we find that Finally, using similarity formulas, we can find . Plugging in and , we find that Thus, our final answer is . ~sigma
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.