Difference between revisions of "Carnot's Theorem"
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− | + | '''Carnot's Theorem''' states that in a [[triangle]] <math>ABC</math>, the signed sum of [[perpendicular]] distances from the [[circumcenter]] <math>O</math> to the sides (i.e., signed lengths of the pedal lines from <math>O</math>) is: | |
<math>OO_A+OO_B+OO_C=R+r</math> | <math>OO_A+OO_B+OO_C=R+r</math> |
Revision as of 10:55, 2 August 2017
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
Below is not 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.
Problems
Olympiad
is a triangle. Take points
on the perpendicular bisectors of
respectively. Show that the lines through
perpendicular to
respectively are concurrent. (Source)