Difference between revisions of "2021 AMC 12B Problems/Problem 11"
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Using Stewart's Theorem of <math>man+dad=bmb+cnc</math> calculate the cevian to be <math>8\sqrt{2}</math>. It then follows that the answer must also have a factor of the <math>\sqrt{2}</math>. Having eliminated 3 answer choices, we then proceed to draw a rudimentary semiaccurate diagram of this figure. Drawing that, we realize that <math>6\sqrt2</math> is too small making out answer <math>\boxed{\textbf{(D) }12\sqrt2}</math> ~Lopkiloinm | Using Stewart's Theorem of <math>man+dad=bmb+cnc</math> calculate the cevian to be <math>8\sqrt{2}</math>. It then follows that the answer must also have a factor of the <math>\sqrt{2}</math>. Having eliminated 3 answer choices, we then proceed to draw a rudimentary semiaccurate diagram of this figure. Drawing that, we realize that <math>6\sqrt2</math> is too small making out answer <math>\boxed{\textbf{(D) }12\sqrt2}</math> ~Lopkiloinm |
Revision as of 21:45, 11 February 2021
Contents
[hide]Problem
Triangle has and . Let be the point on such that . There are exactly two points and on line such that quadrilaterals and are trapezoids. What is the distance
Solutions
Solution 1 (fakesolve)
Using Stewart's Theorem of calculate the cevian to be . It then follows that the answer must also have a factor of the . Having eliminated 3 answer choices, we then proceed to draw a rudimentary semiaccurate diagram of this figure. Drawing that, we realize that is too small making out answer ~Lopkiloinm
Solution 2
Using Stewart's Theorem we find . From the similar triangles and we have So
Solution 3
Let be the length . From the similar triangles and we have Therefore . Now extend line to the point on , forming parallelogram . As we also have so .
We now use the Law of Cosines to find (the length of ): As , we have (by Law of Cosines on triangle ) Therefore And . The answer is then
Video Solution by OmegaLearn (Using properties of 13-14-15 triangle)
~ pi_is_3.14
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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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