Difference between revisions of "Carnot's Theorem"
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==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. |
Revision as of 07:51, 3 July 2011
Carnot's Theorem states that in a triangle with
,
, and
, perpendiculars to the sides
,
, and
at
,
, and
are concurrent if and only if
.
Contents
[hide]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)