The Napoleon Point and More
Date: 09/04/98 at 03:10:25 From: Adi Schulthess Subject: Fermat On each side of a triangle you put on triangles having three sides of the same length. You take the centers of them and draw a line to the opposite vertex in the original triangle. Does anybody know the proof why the three lines all go to the same point?
Date: 09/05/98 at 12:46:58 From: Doctor Floor Subject: Re: Fermat Hi Adi, Thank you for sending your question to Dr. Math. I wonder why you write "Fermat" in your subject line, because the point of concurrence you are describing is the "Napoleon point" of the triangle. The "Fermat point" is constructed by taking the 'new' vertices of the triangles constructed at the sides, not their centers. For the Napoleon point, visit Clark Kimberling's page: http://www.evansville.edu/~ck6/tcenters/class/napoleon.html From this page you can look for other triangle centers too. Both points are special cases of the following theorem: Let ABC be a triangle, and let A', B' and C' be points such that: angle(ABC') = angle(CBA') = angle(BCA') = angle(ACB') = angle(CAB') = angle(BAC') = t (where the angles are only equal when they are all outside or all inside the triangle). Then the lines AA', BB' and CC' concur in one point. To prove this, let's for simplicity reasons take all angles outside triangle ABC. Let's name the sides of triangle ABC in the standard way, i.e. AB = c, BC = a and CA = b. Let's name the angles angle(CAB) = angle(A), angle(ABC) = angle(B) and angle(BCA) = angle(C). Here is a picture for your reference: Then for AA' we have the following picture (where of course BC'=AC'): C' A X X is intersection of AB and CC'. B C Area(BCC') = a*BC'*sin(B+t) and area(ACC') = b*AC'*sin(A+t). So: area(ACC') b*sin(A+t) ---------- = ---------- area(BCC') a*sin(B+t) Because ACC' and BCC' have the same base CC', the altitudes from B and from C to CC' must have the same ratio as the areas, and thus BX and AX too. So: AX b*sin(A+t) -- = ---------- BX a*sin(B+t) We can compute in the same way ratios for BY/CY (where Y is the point of intersection of AA' and BC) and AZ/CZ (Z intersection of BB' and AC). When we multiply them we get: b*sin(A+t) c*sin(B+t) a*sin(C+t) ---------- * ---------- * --------- = 1 a*sin(B+t) b*sin(C+t) c*sin(A+t) And then from Ceva's theorem (which I hope you know - if not, I've included a statement of it below), it follows that XC, YA and ZB concur in one point. But then AA', BB' and CC' concur in one point too. So this proves your question. In fact there is a more powerful theorem, but the proof needs the use of trilinear coordinates as far as I know. Then it is only needed that angle(BAC') = angle(CAB'), angle(CBA') = angle(ABC'), and angle(ACB') = angle(BCA'). Here is Ceva's Theorem: Given a triangle with vertices A, B, and C and points along the sides D, E, and F: AD, BE, and CF are concurrent (intersect in a single point) if and only if BD * CE * AF = DC * EA * FB. This theorem was first published by Giovanni Cevian in 1678. If you have another math question, please send it to Dr. Math. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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