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Re:[EMHL] Perimeter things!

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  • Milorad Stevanovic
    Dear TigerSandre, you wrote ... Let be s1=ks,R1=kR,r1=kr. Sides of the first triangle are the roots of equation x^3-2s(x^2)+(s^2+4Rr+r^2)x-4Rrs=0
    Message 1 of 1 , Sep 30, 2003
      Dear TigerSandre,
      you wrote
      >1)Prove that two triangles are alike if and only if the perimeters,
      >the radius of the incircles and the radius of the circumcircles are
      >proportional. Show that two triangles incriebed in the same circle
      >which have the same perimeters and areas are congruent.
      Let be s1=ks,R1=kR,r1=kr.
      Sides of the first triangle are the roots of equation
      x^3-2s(x^2)+(s^2+4Rr+r^2)x-4Rrs=0 (1).
      Sides of the second triangle are the roots of equation
      x^3-2ks(x^2)+(k^2)(s^2+4Rr+r^2)x-4(k^3)Rrs=0 (2).
      If we put x-->a in (1) and x-->a1 in (2) and take
      (-k^3)(1)+(2) then we have
      (a1-ka)[(a1)^2+kaa1+(ka)^2-2ks(a1+ka)+(k^2)(s^2+4Rr+r^2)]=0
      or (a1-ka)[(a1)^2+ka1(a-2s)+(k^2)bc]=0 or
      (a1-ka)(a1-kb)(a1-kc)=0.
      Analogously for sides b,b1 and c,c1.
      Thus we have that triangles are similar.
      The second statement is the special case of the first part when k=1.
      Maybe the easier proof is known?
      Best regards
      Milorad R.Stevanovic


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