2002 AMC 12B Problems/Problem 23
Contents
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 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 11/28/2017
Solution 2
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 3
From Stewart's Theorem, we have Simplifying, we get
- awu2014
Solution 4 [Pappus's Median Theorem]
There is a theorem in geometry known as Pappus's Median Theorem. It states that if you have triangle ABC, and you draw a median from point A to side BC (label this as Ma), then: (Ma)^2 = (1/4)(2(b^2) + 2(c^2) - (a^2)). For this specific problem, call where the median from A hits side BC point M. Let MB = MC = x. Then AM = 2x. Now, we can plug into the formula given above: Ma = 2x, b = 2, c = 1, and a = 2x. After some simple algebra, we find x = \sqrt{2}/2. Then, BC = a = \boxed{\sqrt{2}}.$
-Flames
Video Solution by TheBeautyofMath
~IceMatrix
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.