Difference between revisions of "Carnot's Theorem"
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Let <math>X,Y,Z</math> be the feet of perpendiculars from <math>P,Q,R</math> to <math>BC,CA,AB</math> respectively. Note that <math>PB^2-PC^2=XB^2-XC^2</math> from the application of pythogorean theorem to triangles <math>PXB,PXC</math>. Now with similar relations for <math>Y</math> and <math>Z</math>, Carnot's theorem finishes the job! | Let <math>X,Y,Z</math> be the feet of perpendiculars from <math>P,Q,R</math> to <math>BC,CA,AB</math> respectively. Note that <math>PB^2-PC^2=XB^2-XC^2</math> from the application of pythogorean theorem to triangles <math>PXB,PXC</math>. Now with similar relations for <math>Y</math> and <math>Z</math>, Carnot's theorem finishes the job! | ||
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===Olympiad=== | ===Olympiad=== | ||
<math>\triangle ABC</math> is a triangle. Take points <math>D, E, F</math> on the perpendicular bisectors of <math>BC, CA, AB</math> respectively. Show that the lines through <math>A, B, C</math> perpendicular to <math>EF, FD, DE</math> respectively are concurrent. ([[1997 USAMO Problems/Problem 2|Source]]) | <math>\triangle ABC</math> is a triangle. Take points <math>D, E, F</math> on the perpendicular bisectors of <math>BC, CA, AB</math> respectively. Show that the lines through <math>A, B, C</math> perpendicular to <math>EF, FD, DE</math> respectively are concurrent. ([[1997 USAMO Problems/Problem 2|Source]]) |
Revision as of 14:01, 10 March 2018
Carnot's Theorem states that in a triangle , the signed sum of perpendicular distances from the circumcenter
to the sides (i.e., signed lengths of the pedal lines from
) is:
where r is the inradius and R is the circumradius. The sign of the distance is chosen to be negative iff the entire segment OO_i lies outside the triangle. Explicitly,
where is the area of triangle
.
Weisstein, Eric W. "Carnot's Theorem." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/CarnotsTheorem.html
Contents
Carnot's Theorem
Carnot's Theorem states that in a triangle with
,
, and
, perpendiculars to the sides
,
, and
at
,
, and
are concurrent if and only if
.
Proof
Only if: Assume that the given perpendiculars are concurrent at . Then, from the Pythagorean Theorem,
,
,
,
,
, and
. Substituting each and every one of these in and simplifying gives the desired result.
If: Consider the intersection of the perpendiculars from and
. Call this intersection point
, and let
be the perpendicular from
to
. From the other direction of the desired result, we have that
. We also have that
, which implies that
. This is a difference of squares, which we can easily factor into
. Note that
, so we have that
. This implies that
, which gives the desired result.
Carnot Extended
Let be points in the plane of triangle
. Then the perpendiculars from
to
respectively are concurrent if and only if
Proof
Let be the feet of perpendiculars from
to
respectively. Note that
from the application of pythogorean theorem to triangles
. Now with similar relations for
and
, Carnot's theorem finishes the job!
Problems
Olympiad
is a triangle. Take points
on the perpendicular bisectors of
respectively. Show that the lines through
perpendicular to
respectively are concurrent. (Source)