Difference between revisions of "2021 AMC 12B Problems/Problem 11"
Brendanb4321 (talk | contribs) (added diagram) |
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<math>\textbf{(A) }\frac{42}5 \qquad \textbf{(B) }6\sqrt2 \qquad \textbf{(C) }\frac{84}5\qquad \textbf{(D) }12\sqrt2 \qquad \textbf{(E) }18</math> | <math>\textbf{(A) }\frac{42}5 \qquad \textbf{(B) }6\sqrt2 \qquad \textbf{(C) }\frac{84}5\qquad \textbf{(D) }12\sqrt2 \qquad \textbf{(E) }18</math> | ||
+ | |||
+ | ==Diagram== | ||
+ | <asy> | ||
+ | size(8cm); | ||
+ | pair A = (5,12); | ||
+ | pair B = (0,0); | ||
+ | pair C = (14,0); | ||
+ | pair P = 2/3*A+1/3*C; | ||
+ | pair D = 3/2*P; | ||
+ | pair E = 3*P; | ||
+ | draw(A--B--C--A); | ||
+ | draw(A--D); | ||
+ | draw(C--E--B); | ||
+ | |||
+ | dot("$A$",A,N); | ||
+ | dot("$B$",B,W); | ||
+ | dot("$C$",C,ESE); | ||
+ | dot("$D$",D,N); | ||
+ | dot("$P$",P,W); | ||
+ | dot("$E$",E,N); | ||
+ | |||
+ | defaultpen(fontsize(9pt)); | ||
+ | label("$13$", (A+B)/2, NW); | ||
+ | label("$14$", (B+C)/2, S); | ||
+ | label("$5$",(A+P)/2, NE); | ||
+ | label("$10$", (C+P)/2, NE); | ||
+ | </asy> | ||
==Solution 1 (fakesolve)== | ==Solution 1 (fakesolve)== |
Revision as of 15:05, 12 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
Diagram
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 Punxsutawney Phil
https://YouTube.com/watch?v=yxt8-rUUosI&t=450s
Video Solution by OmegaLearn (Using properties of 13-14-15 triangle)
~ pi_is_3.14
Video Solution by Hawk Math
https://www.youtube.com/watch?v=p4iCAZRUESs
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.