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> | ||
+ | |||
+ | <asy> | ||
+ | pair a,b,c,O,i,d,f,g; | ||
+ | a=(0,0); | ||
+ | b=(4,0); | ||
+ | c=(1,3); | ||
+ | O=circumcenter(a,b,c); | ||
+ | i=incenter(a,b,c); | ||
+ | draw(a--b--c--cycle); | ||
+ | draw(circumcircle(a,b,c)); | ||
+ | draw(incircle(a,b,c)); | ||
+ | dot(i); | ||
+ | dot(O); | ||
+ | label("$A$",a,W); | ||
+ | label("$B$",b,E); | ||
+ | label("$C$",c,N); | ||
+ | label("$I$",i,N); | ||
+ | label("$O$",O,N); | ||
+ | d=foot(O,b,c); | ||
+ | dot(d); | ||
+ | draw(O--d); | ||
+ | label("$O_A$",d,N); | ||
+ | draw(rightanglemark(O,d,b)); | ||
+ | f=foot(O,a,b); | ||
+ | dot(f); | ||
+ | draw(O--f); | ||
+ | draw(rightanglemark(O,f,a)); | ||
+ | label("$O_C$",f,S); | ||
+ | g=foot(O,c,a); | ||
+ | dot(g); | ||
+ | draw(O--g); | ||
+ | draw(rightanglemark(O,g,a)); | ||
+ | label("$O_B$",g,W); | ||
+ | </asy> | ||
+ | |||
+ | where r is the [[inradius]] and R is the [[circumradius]]. The sign of the distance is chosen to be negative iff the entire segment <math>OO_i</math> lies outside the triangle. | ||
+ | Explicitly, | ||
+ | |||
+ | <math>OO_A+OO_B+OO_C=\frac{abc(|\cos{A}|+|\cos{B}|+|\cos{C}|)}{4|\Delta|}</math> | ||
+ | |||
+ | where <math>\Delta</math> is the area of triangle <math>\Delta ABC</math>. | ||
+ | |||
+ | |||
+ | Weisstein, Eric W. "Carnot's Theorem." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/CarnotsTheorem.html | ||
+ | |||
+ | |||
+ | =Carnot's Theorem= | ||
+ | |||
'''Carnot's Theorem''' states that in a [[triangle]] <math>ABC</math> with <math>A_1\in BC</math>, <math>B_1\in AC</math>, and <math>C_1\in AB</math>, [[perpendicular]]s to the sides <math>BC</math>, <math>AC</math>, and <math>AB</math> at <math>A_1</math>, <math>B_1</math>, and <math>C_1</math> are [[concurrent]] [[iff|if and only if]] <math>A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2</math>. | '''Carnot's Theorem''' states that in a [[triangle]] <math>ABC</math> with <math>A_1\in BC</math>, <math>B_1\in AC</math>, and <math>C_1\in AB</math>, [[perpendicular]]s to the sides <math>BC</math>, <math>AC</math>, and <math>AB</math> at <math>A_1</math>, <math>B_1</math>, and <math>C_1</math> are [[concurrent]] [[iff|if and only if]] <math>A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2</math>. | ||
− | ==Proof== | + | ====Proof==== |
'''Only if:''' Assume that the given perpendiculars are concurrent at <math>M</math>. Then, from the Pythagorean Theorem, <math>A_1B^2=BM^2-MA_1^2</math>, <math>C_1A^2=AM^2-MC_1^2</math>, <math>B_1C^2=CM^2-MB_1^2</math>, <math>A_1C^2=MC^2-MA_1^2</math>, <math>C_1B^2=MB^2-MC_1^2</math>, and <math>B_1A^2=AM^2-MB_1^2</math>. Substituting each and every one of these in and simplifying gives the desired result. | '''Only if:''' Assume that the given perpendiculars are concurrent at <math>M</math>. Then, from the Pythagorean Theorem, <math>A_1B^2=BM^2-MA_1^2</math>, <math>C_1A^2=AM^2-MC_1^2</math>, <math>B_1C^2=CM^2-MB_1^2</math>, <math>A_1C^2=MC^2-MA_1^2</math>, <math>C_1B^2=MB^2-MC_1^2</math>, and <math>B_1A^2=AM^2-MB_1^2</math>. Substituting each and every one of these in and simplifying gives the desired result. | ||
+ | ''' If:''' Consider the intersection of the perpendiculars from <math>A_1</math> and <math>B_1</math>. Call this intersection point <math>N</math>, and let <math>C_2</math> be the perpendicular from <math>N</math> to <math>AB</math>. From the other direction of the desired result, we have that <math>A_1B^2+C_2A^2+B_1C^2=A_1C^2+C_2B^2+B_1A^2</math>. We also have that <math>A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2</math>, which implies that <math>C_1A^2-C_1B^2=C_2A^2-C_2B^2</math>. This is a difference of squares, which we can easily factor into <math>(C_1A-C_1B)(C_1A+C_1B)=(C_2A-C_2B)(C_2A+C_2B)</math>. Note that <math>C_1A+C_1=C_2A+C_2B=AB</math>, so we have that <math>C_1A-C_1B=C_2A-C_2B</math>. This implies that <math>C_1=C_2</math>, which gives the desired result. | ||
+ | |||
+ | =Carnot Extended= | ||
+ | Let <math>P,Q,R</math> be points in the plane of triangle <math>ABC</math>. Then the perpendiculars from <math>P,Q,R</math> to <math>BC,CA,AB</math> respectively are concurrent if and only if <cmath>PB^2-PC^2+QC^2-QA^2+RA^2-RB^2=0</cmath> | ||
− | + | ====Proof==== | |
+ | 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! | ||
+ | |||
+ | |||
+ | =Problems= | ||
− | |||
===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]]) | ||
Line 14: | Line 72: | ||
==See also== | ==See also== | ||
+ | [[Carnot's Polygon Theorem]] | ||
+ | [[Japanese Theorem]] | ||
[[Category:Geometry]] | [[Category:Geometry]] | ||
[[Category:Theorems]] | [[Category:Theorems]] |
Latest revision as of 01:45, 18 February 2021
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 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)