## Re:[EMHL] Perimeter things!

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• 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