Difference between revisions of "2021 AMC 12B Problems/Problem 11"
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− | Using Stewart's Theorem we find <math>BP = 8\sqrt{2}</math>. From the similar triangles <math>BPA | + | Using Stewart's Theorem we find <math>BP = 8\sqrt{2}</math>. From the similar triangles <math>BPA\sim DPC</math> and <math>BPC\sim EPA</math> we have |
<cmath>DP = BP\cdot\frac{PC}{PA} = \frac{1}{2} BP</cmath> | <cmath>DP = BP\cdot\frac{PC}{PA} = \frac{1}{2} BP</cmath> | ||
<cmath>EP = BP\cdot\frac{PA}{PC} = 2BP</cmath> | <cmath>EP = BP\cdot\frac{PA}{PC} = 2BP</cmath> |
Revision as of 20:17, 11 February 2021
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 (cheese)
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.