# 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 OO_i 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 | ||

+ | |||

+ | |||

+ | =Below is not 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. | ||

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''' 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. | ''' 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. | ||

− | ==Problems== | + | ====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]]) | ||

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==See also== | ==See also== | ||

+ | [[Carnot's Polygon Theorem]] | ||

+ | [[Japanese Theorem]] | ||

[[Category:Geometry]] | [[Category:Geometry]] | ||

[[Category:Theorems]] | [[Category:Theorems]] |

## Revision as of 11:53, 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)