Difference between revisions of "2021 AMC 12B Problems/Problem 11"
(→Solution 2) |
(→Solutions) |
||
Line 12: | Line 12: | ||
===Solution 2=== | ===Solution 2=== | ||
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 | 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} = | + | <cmath>DP = BP\cdot\frac{PC}{PA} = 2BP</cmath> |
− | <cmath>EP = BP\cdot\frac{PA}{PC} = | + | <cmath>EP = BP\cdot\frac{PA}{PC} = \frac12 BP</cmath> |
So | So | ||
<cmath>DE = \frac{3}{2}BP = \boxed{\textbf{(D) }12\sqrt2}</cmath> | <cmath>DE = \frac{3}{2}BP = \boxed{\textbf{(D) }12\sqrt2}</cmath> | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Let <math>x</math> be the length <math>PE</math>. From the similar triangles <math>BPA\sim DPC</math> and <math>BPC\sim EPA</math> we have | ||
+ | <cmath>BP = \frac{PA}{PC}x = \frac12 x</cmath> | ||
+ | <cmath>PD = \frac{PA}{PC}BP = \frac14 x</cmath> | ||
+ | Therefore <math>BD = DE = \frac{3}{4}x</math>. Now extend line <math>CD</math> to the point <math>Z</math> on <math>AE</math>, forming parallelogram <math>ZABC</math>. As <math>BD = DE</math> we also have <math>EZ = ZC = 13</math> so <math>EC = 26</math>. | ||
+ | |||
+ | We now use the Law of Cosines to find <math>x</math> (the length of <math>PE</math>): | ||
+ | <cmath>x^2 = EC^2 + PC^2 - 2(EC)(PC)\cos{(PCE)} = 26^2 + 10^2 - 2\cdot 26\cdot 10\cos(\angle PCE)</cmath> | ||
+ | As <math>\angle PCE = \angle BAC</math>, we have (by Law of Cosines on triangle <math>BAC</math>) | ||
+ | <cmath>\cos(\angle PCE) = \frac{13^2 + 15^2 - 14^2}{2\cdot 13\cdot 15}.</cmath> | ||
+ | Therefore | ||
+ | <cmath>\begin{align*} | ||
+ | x^2 &= 26^2 + 10^2 - 2\cdot 26\cdot 10\cdot\frac{198}{2\cdot 13\cdot 15}\ | ||
+ | &= 776 - 264\ | ||
+ | &= 512 | ||
+ | \end{align*}</cmath> | ||
+ | And <math>x = 16\sqrt2</math>. The answer is then <math>\frac34x = \boxed{\textbf{(D) }12\sqrt2}</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021|ab=B|num-b=10|num-a=12}} | {{AMC12 box|year=2021|ab=B|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:40, 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 (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
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
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.