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Difference between revisions of "Carnot's Theorem"

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==Proof==
 
==Proof==
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'''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.
  
 
==Problems==
 
==Problems==

Revision as of 07:51, 3 July 2011

Carnot's Theorem states that in a triangle $ABC$ with $A_1\in BC$, $B_1\in AC$, and $C_1\in AB$, perpendiculars to the sides $BC$, $AC$, and $AB$ at $A_1$, $B_1$, and $C_1$ are concurrent if and only if $A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2$.

Proof

Only if: Assume that the given perpendiculars are concurrent at $M$. Then, from the Pythagorean Theorem, $A_1B^2=BM^2-MA_1^2$, $C_1A^2=AM^2-MC_1^2$, $B_1C^2=CM^2-MB_1^2$, $A_1C^2=MC^2-MA_1^2$, $C_1B^2=MB^2-MC_1^2$, and $B_1A^2=AM^2-MB_1^2$. Substituting each and every one of these in and simplifying gives the desired result.

If: Consider the intersection of the perpendiculars from $A_1$ and $B_1$. Call this intersection point $N$, and let $C_2$ be the perpendicular from $N$ to $AB$. From the other direction of the desired result, we have that $A_1B^2+C_2A^2+B_1C^2=A_1C^2+C_2B^2+B_1A^2$. We also have that $A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2$, which implies that $C_1A^2-C_1B^2=C_2A^2-C_2B^2$. This is a difference of squares, which we can easily factor into $(C_1A-C_1B)(C_1A+C_1B)=(C_2A-C_2B)(C_2A+C_2B)$. Note that $C_1A+C_1=C_2A+C_2B=AB$, so we have that $C_1A-C_1B=C_2A-C_2B$. This implies that $C_1=C_2$, which gives the desired result.

Problems

Olympiad

$\triangle ABC$ is a triangle. Take points $D, E, F$ on the perpendicular bisectors of $BC, CA, AB$ respectively. Show that the lines through $A, B, C$ perpendicular to $EF, FD, DE$ respectively are concurrent. (Source)

See also

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