Difference between revisions of "Menelaus' Theorem"
(→Proof Using Barycentric coordinates) |
(→Proof Using Barycentric coordinates) |
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<math>Q: (1-Q ,0 , Q)</math> | <math>Q: (1-Q ,0 , Q)</math> | ||
| + | |||
| + | Note that this says the following: | ||
| + | |||
| + | <math>\frac{CP}{PB}=\frac{1-P}{P}</math> | ||
| + | |||
| + | <math>\frac{BR}{AR}=\frac{1-R}{R}</math> | ||
| + | |||
| + | <math>\frac{QA}{QC}=\frac{1-Q}{Q}</math> | ||
The line through <math>R</math> and <math>P</math> is given by: | The line through <math>R</math> and <math>P</math> is given by: | ||
| − | + | <math>\begin{vmatrix} X & 0 & R \\ Y & P & 1-R\\ Z & 1-P & 0 \end{vmatrix} = 0</math> | |
| + | |||
| − | |||
Which yields, after simplification, | Which yields, after simplification, | ||
| − | <center> <math>-X\cdot(R-1)(P-1)+Y\cdotR(1-P)-Z\cdotPR = 0</math> | + | <center> <math>-X\cdot(R-1)(P-1)+Y\cdotR(1-P)-Z\cdotPR = 0</math> |
| Line 69: | Line 77: | ||
<math>(Q-1)(R-1)(P-1) = QPR</math> | <math>(Q-1)(R-1)(P-1) = QPR</math> | ||
| − | </ | + | |
| + | |||
| + | We know that, since | ||
| + | |||
| + | |||
| + | <math>\frac{CP}{PB}=\frac{1-P}{P}</math>, | ||
| + | |||
| + | |||
| + | <math>P=\frac{(1-P)\cdot PB}{CP}</math>. Likewise, | ||
| + | |||
| + | |||
| + | <math>R=\frac{(1-R)\cdot AR}{BR}</math>, and | ||
| + | |||
| + | |||
| + | <math>Q=\frac{(1-Q)\cdot QC}{QA}</math>. | ||
| + | |||
| + | |||
| + | Substituting these values yields: | ||
| + | |||
| + | |||
| + | <math>(Q-1)(R-1)(P-1) = \frac{(1-Q)\cdot QC \cdot (1-P) \cdot PB \cdot (1-R) \cdot AR}{QA\cdotCP\cdotBR}</math> | ||
| + | |||
| + | Which simplifies to: | ||
| + | |||
| + | |||
| + | <math>QA\cdot CP \cdot BR = -QC \cdot AR \cdot PB</math> | ||
QED | QED | ||
Revision as of 17:59, 29 May 2013
This article is a stub. Help us out by expanding it.
Menelaus's Theorem deals with the collinearity of points on each of the three sides (extended when necessary) of a triangle. It is named for Menelaus of Alexandria.
Statement
A necessary and sufficient condition for points 
on the respective sides 
(or their extensions) of a triangle 
to be collinear is that
where all segments in the formula are directed segments.
![[asy] defaultpen(fontsize(8)); pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R; draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75); draw((7,6)--(6,8)--(4,0)); R=intersectionpoint(A--B,Q--P); dot(A^^B^^C^^P^^Q^^R); label("A",A,(1,1));label("B",B,(-1,0));label("C",C,(1,0));label("P",P,(0,-1));label("Q",Q,(1,0));label("R",R,(-1,1)); [/asy]](http://latex.artofproblemsolving.com/8/6/a/86a254a29864a11196f5bbd900856e311e41be6b.png)
Proof
Draw a line parallel to 
through 
to intersect 
at 
:
![[asy] defaultpen(fontsize(8)); pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R, K=(5.5,0); draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75); draw((7,6)--(6,8)--(4,0)); draw(A--K, dashed); R=intersectionpoint(A--B,Q--P); dot(A^^B^^C^^P^^Q^^R^^K); label("A",A,(1,1));label("B",B,(-1,0));label("C",C,(1,0));label("P",P,(0,-1));label("Q",Q,(1,0));label("R",R,(-1,1)); label("K",K,(0,-1)); [/asy]](http://latex.artofproblemsolving.com/f/b/d/fbd7bf9ba4bcb21c083f9f6aef5d813523b36c8f.png)
Multiplying the two equalities together to eliminate the 
factor, we get:
Proof Using Barycentric coordinates
Disclaimer: This proof is not nearly as elegant as the above one. It uses a bash-type approach, as barycentric coordinate proofs tend to be.
Suppose we give the points the following coordinates:
Note that this says the following:
The line through 
and 
is given by:
Which yields, after simplification,
$Z\cdotPR = -X\cdot(R-1)(P-1)+Y\cdotR(1-P)$ (Error compiling LaTeX. Unknown error_msg)
Plugging in the coordinates for 
yields:
We know that, since
,
. Likewise,
, and
.
Substituting these values yields:
$(Q-1)(R-1)(P-1) = \frac{(1-Q)\cdot QC \cdot (1-P) \cdot PB \cdot (1-R) \cdot AR}{QA\cdotCP\cdotBR}$ (Error compiling LaTeX. Unknown error_msg)
Which simplifies to:
QED
