Difference between revisions of "2002 AMC 12B Problems/Problem 23"
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− | === Solution 4 | + | === Solution 4: Pappus's Median Theorem === |
There is a theorem in geometry known as Pappus's Median Theorem. It states that if you have <math>\triangle{ABC}</math>, and you draw a median from point <math>A</math> to side <math>BC</math> (label this as <math>M</math>), then: <math>(AM)^2 = \dfrac{2(b^2) + 2(c^2) - (a^2)}{4}</math>. Note that <math>b</math> is the length of side <math>\overline{AC}</math>, <math>c</math> is the length of side <math>\overline{AB}</math>, and <math>a</math> is length of side <math>\overline{BC}</math>. Let <math>MB = MC = x</math>. Then <math>AM = 2x</math>. Now, we can plug into the formula given above: <math>AM = 2x</math>, <math>b = 2</math>, <math>c = 1</math>, and <math>a = 2x</math>. After some simple algebra, we find <math>x = \dfrac{\sqrt{2}}{2}</math>. Then, <math>BC = \boxed{\sqrt{2}} \implies \boxed{C}</math>. | There is a theorem in geometry known as Pappus's Median Theorem. It states that if you have <math>\triangle{ABC}</math>, and you draw a median from point <math>A</math> to side <math>BC</math> (label this as <math>M</math>), then: <math>(AM)^2 = \dfrac{2(b^2) + 2(c^2) - (a^2)}{4}</math>. Note that <math>b</math> is the length of side <math>\overline{AC}</math>, <math>c</math> is the length of side <math>\overline{AB}</math>, and <math>a</math> is length of side <math>\overline{BC}</math>. Let <math>MB = MC = x</math>. Then <math>AM = 2x</math>. Now, we can plug into the formula given above: <math>AM = 2x</math>, <math>b = 2</math>, <math>c = 1</math>, and <math>a = 2x</math>. After some simple algebra, we find <math>x = \dfrac{\sqrt{2}}{2}</math>. Then, <math>BC = \boxed{\sqrt{2}} \implies \boxed{C}</math>. | ||
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Note: Pappus's Median Theorem is just a special case of Stewart's Theorem, with <math>m = n</math>. ~Puck_0 | Note: Pappus's Median Theorem is just a special case of Stewart's Theorem, with <math>m = n</math>. ~Puck_0 | ||
+ | aka Apollonius' Theorem - Orion 2010 | ||
===Video Solution by TheBeautyofMath=== | ===Video Solution by TheBeautyofMath=== |
Revision as of 03:15, 30 June 2024
Contents
Problem
In , we have
and
. Side
and the median from
to
have the same length. What is
?
Solution
Solution 1: Pythagoras Theorem
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: Law of Cosines
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: Stewart's Theorem
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 , and you draw a median from point
to side
(label this as
), then:
. Note that
is the length of side
,
is the length of side
, and
is length of side
. Let
. Then
. Now, we can plug into the formula given above:
,
,
, and
. After some simple algebra, we find
. Then,
.
-Flames
Note: Pappus's Median Theorem is just a special case of Stewart's Theorem, with . ~Puck_0
aka Apollonius' Theorem - Orion 2010
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.