viernes, 3 de febrero de 2017

1265. Carnot's theorem

    Nina Guindilla has proved Carnot's theorem for acute angled triangles. Will you demonstrate it for other triangles?

    Dear Teacher:
    In an acute angled triangle... If  a ,  b  and  c  are the distances from the circumcenter to the sides, and  r  and  R  are the inradius and the circumradius, then


a + b + c = r + R
    Proof:
    Let  A ,  B  and  C  be the sides of the triangle...
    If I divide the triangle in this way, its area is


area  =  Ar/2 + Br/2 + Cr/2  =  (A+B+C) r/2
    So
r  =  2 · area / (A+B+C)



    But if I divide the triangle in this other way, thus, the area is


area  =  Aa/2 + Bb/2 + Cc/2  =  (Aa+Bb+Cc) / 2
    So
r  =  (Aa+Bb+Cc) / (A+B+C)
    Finally, dividing the triangle again... but now into three cyclic quadrilaterals (they are cyclic because have two opposite right angles), I can apply Ptolemy's theorem:

RA/2  =  bC/2 + cB/2
RB/2  =  aC/2 + cA/2
RC/2  =  aB/2 + bA/2
    Multiplying by 2 and summing:

RA + RB + RC  =  bC + cB + aC + cA + aB + bA
R (A+B+C)  =  bC + cB + aC + cA + aB + bA
R  =  (bC+cB+aC+cA+aB+bA) / (A+B+C)
    Hence
r + R  =
 (Aa+Bb+Cc) / (A+B+C)  +  (bC+cB+aC+cA+aB+bA) / (A+B+C)  =
=  (Aa+Bb+Cc+bC+cB+aC+cA+aB+bA) / (A+B+C) =
=  (a+b+c) (A+B+C) / (A+B+C)  =
=  a + b + c

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