Carnot's Theorem

by SomeonecoolLovesMaths, Mar 17, 2024, 6:56 AM

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.

geometry

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Carnot's Theorem

by SomeonecoolLovesMaths, Mar 17, 2024, 6:56 AM

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