Difference between revisions of "Menelaus' Theorem"
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| − | + | which yields, after simplification, | |
| − | + | <cmath>-X\cdot(R-1)(P-1)+Y\cdotR(1-P)-Z\cdotPR = 0</cmath> | |
| + | <cmath>Z\cdotPR = -X\cdot(R-1)(P-1)+Y\cdotR(1-P).</cmath> | ||
| + | Plugging in the coordinates for <math>Q</math> yields <math>(Q-1)(R-1)(P-1) = QPR</math>. From <math>\frac{CP}{PB}=\frac{1-P}{P},</math> we have <cmath>P=\frac{(1-P)\cdot PB}{CP}.</cmath> Likewise, <cmath>R=\frac{(1-R)\cdot AR}{BR}</cmath> and <cmath>Q=\frac{(1-Q)\cdot QC}{QA}.</cmath> | ||
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| − | + | Substituting these values yields <cmath>(Q-1)(R-1)(P-1) = \frac{(1-Q)\cdot QC \cdot (1-P) \cdot PB \cdot (1-R) \cdot AR}{QA\cdotCP\cdotBR}</cmath> which simplifies to <math>QA\cdot CP \cdot BR = -QC \cdot AR \cdot PB.</math> | |
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| − | Substituting these values yields | ||
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| − | <math>QA\cdot CP \cdot BR = -QC \cdot AR \cdot PB</math> | ||
QED | QED | ||
Revision as of 16:48, 8 September 2014
This article is a stub. Help us out by expanding it.
Menelaus' 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,
\[-X\cdot(R-1)(P-1)+Y\cdotR(1-P)-Z\cdotPR = 0\] (Error compiling LaTeX. Unknown error_msg)
\[Z\cdotPR = -X\cdot(R-1)(P-1)+Y\cdotR(1-P).\] (Error compiling LaTeX. Unknown error_msg)
Plugging in the coordinates for 
yields 
. From 
we have ![\[P=\frac{(1-P)\cdot PB}{CP}.\]](http://latex.artofproblemsolving.com/0/1/4/014878e5d4964845c637419c2985c4674b4a7929.png)
Likewise, ![\[R=\frac{(1-R)\cdot AR}{BR}\]](http://latex.artofproblemsolving.com/3/a/7/3a7abff5351fd621f2d4d4fc42ed9ee84e1ae21f.png)
and ![\[Q=\frac{(1-Q)\cdot QC}{QA}.\]](http://latex.artofproblemsolving.com/f/4/4/f448a124aae74987854ffa8a13bd7738f0ef843c.png)
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
