Difference between revisions of "2002 AMC 12B Problems/Problem 23"
Greenturtle (talk | contribs) (→Solution 3) |
Greenturtle (talk | contribs) (→Solution 3) |
||
Line 54: | Line 54: | ||
label("$C$",C,S); | label("$C$",C,S); | ||
label("$D$",D,S); | label("$D$",D,S); | ||
− | label("$M$",M,S) | + | label("$M$",M,S); |
label("$x$",(A+D)/2,E); | label("$x$",(A+D)/2,E); | ||
label("$y$",(B+D)/2,S); | label("$y$",(B+D)/2,S); |
Revision as of 20:27, 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.