Difference between revisions of "2002 AMC 12B Problems/Problem 23"
Greenturtle (talk | contribs) (→Solution 3) |
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=== Solution 3 === | === Solution 3 === | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | unitsize(4cm); | ||
+ | |||
+ | pair A, B, C, D, M; | ||
+ | |||
+ | A = (1.768,0.935); | ||
+ | B = (1.414,0); | ||
+ | C = (0,0); | ||
+ | D = (1.768,0); | ||
+ | M = (0.707,0); | ||
+ | |||
+ | draw(A--B--C--cycle); | ||
+ | draw(A--D); | ||
+ | draw(D--B); | ||
+ | draw(A--M); | ||
+ | |||
+ | label("$A$",A,N); | ||
+ | label("$B$",B,S); | ||
+ | label("$C$",C,S); | ||
+ | label("$D$",D,S); | ||
+ | label("$x$",(A+D)/2,E); | ||
+ | label("$y$",(B+D)/2,S); | ||
+ | label("$a$",(C+M)/2,S); | ||
+ | label("$a$",(M+B)/2,S); | ||
+ | label("$2a$",(A+M)/2,SE); | ||
+ | label("$1$",(A+B)/2,SE); | ||
+ | label("$2$",(A+C)/2,NW); | ||
+ | |||
+ | draw(rightanglemark(B,D,A,3)); | ||
+ | |||
+ | </asy> | ||
Let <math>D</math> be the foot of the altitude from <math>A</math> to <math>\overline{BC}</math> extended past <math>B</math>. Let <math>AD = x</math> and <math>BD = y</math>. | Let <math>D</math> be the foot of the altitude from <math>A</math> to <math>\overline{BC}</math> extended past <math>B</math>. Let <math>AD = x</math> and <math>BD = y</math>. |
Revision as of 20:24, 27 November 2017
Problem
In , we have and . Side and the median from to have the same length. What is ?
Solution
Solution 1
Let be the foot of the median from to , and we let . Then by the Law of Cosines on , we have
Since , we can add these two equations and get
Hence and .
Solution 2
From Stewart's Theorem, we have Simplifying, we get
Solution 3
Let be the foot of the altitude from to extended past . Let and . Using the Pythagorean Theorem, we obtain the equations
Subtracting equation from and , we get
Then, subtracting from and rearranging, we get , so
~greenturtle 11/26/2017
See also
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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.