Difference between revisions of "2002 AMC 12B Problems/Problem 23"
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From [[Stewart's Theorem]], we have <math>(2)(1/2)a(2) + (1)(1/2)a(1) = (a)(a)(a) + (1/2)a(a)(1/2)a.</math> Simplifying, we get <math>(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{\sqrt{2}}.</math> | From [[Stewart's Theorem]], we have <math>(2)(1/2)a(2) + (1)(1/2)a(1) = (a)(a)(a) + (1/2)a(a)(1/2)a.</math> Simplifying, we get <math>(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{\sqrt{2}}.</math> | ||
+ | - awu2014 | ||
===Video Solution by TheBeautyofMath=== | ===Video Solution by TheBeautyofMath=== |
Revision as of 16:37, 23 October 2021
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
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.