Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Menelaus' Theorem"

(→‎Proof Using Barycentric coordinates)
(→‎Proof Using Barycentric coordinates)
Line 46: Line 46:
 
Suppose we give the points <math>P, Q, R</math> the following coordinates:
 
Suppose we give the points <math>P, Q, R</math> the following coordinates:
  
−
<math>P: (0,P,1-P)</math>
+
<math>P: (0, P, 1-P)</math>
  
−
<math>R: (R,1-R,0)</math>
+
<math>R: (R , 1-R, 0)</math>
  
−
<math>Q: (1-Q,0,Q)</math>
+
<math>Q: (1-Q ,0 , 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:

Revision as of 23:32, 28 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 $P, Q, R$ on the respective sides $BC, CA, AB$ (or their extensions) of a triangle $ABC$ to be collinear is that

$BP\cdot CQ\cdot AR = -PC\cdot QA\cdot RB$

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]

Proof

Draw a line parallel to $QP$ through $A$ to intersect $BC$ at $K$:

[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]

$\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}$

$\triangle QCP \sim \triangle ACK \implies \frac{QC}{QA}=\frac{PC}{PK}$

Multiplying the two equalities together to eliminate the $PK$ factor, we get:

$\frac{AR}{RB}\cdot\frac{QC}{QA}=-\frac{PC}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{PC}=-1$

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 $P, Q, R$ the following coordinates:

$P: (0, P, 1-P)$

$R: (R , 1-R, 0)$

$Q: (1-Q ,0 , Q)$

The line through $R$ and $P$ is given by:

$\begin{vmatrix} X & 0 & R \\ Y & P & 1-R\\ Z & 1-P & 0 \end{vmatrix} = 0$

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 $Q$ yields:


$(Q-1)(R-1)(P-1) = QPR$

QED

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Menelaus%27_Theorem&oldid=52880"