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Let $ABC$ be a triangle with inscribed radius $r$ and circumscribed radius $R$. Let $A′B′C′$ be the triangle for which $A′B′$ is the perpendicular to $OC$ through $C$ and so on. Let $r_1$ be the inscribed radius of $\triangle A'BC$, $r_2$ be the inscribed radius of $\triangle AB'C$ and so on. Find $r_1+r_2+r_3$ in terms of $r$ and $R$.

Although it seems easy, I haven't been able to make any progress. I understand that the inscribed radius radius of $\triangle A'B'C'$ is $R$, but not much else. Can someone explain me how to do this? :)

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    $\begingroup$ @DavidHolden: and so? $\endgroup$ – Jack D'Aurizio Aug 11 '14 at 12:55
  • $\begingroup$ quite right, @Jack. a non-sequitur. i will delete the comment. sorry 125a8owp $\endgroup$ – David Holden Aug 11 '14 at 13:00
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Since the circumcircle of $ABC$ is the incircle of $A'B'C'$, $O$ is the incenter of $A'B'C'$, hence $A'B',B'C'$ and $C'A'$ are the exterior angle bisectors of $ABC$, and $A'B'C'$ is the excentral triangle of $ABC$. Due to relations $(3)$ and $(4)$ given here:

$$ r_1+r_2+r_3 = 4R+r. $$

Proof: Since $4R\Delta=abc,r_1\Delta=\frac{\Delta^2}{p-a},r\Delta=\frac{\Delta^2}{p}$, by exploiting Heron's formula we only need to show that:

$$ abc+(p-a)(p-b)(p-c)=p(p-b)(p-c)+p(p-a)(p-c)+p(p-a)(p-b).\tag{1}$$ that is trivial since both the RHS and the LHS can be reduced to the form: $$ -p^3+p(ab+ac+bc) = p(ab+ac+bc-p^2).$$

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  • $\begingroup$ Its still not clear to me :| ... the Carnot theorem says something different, how come its about $r_1, r_2,...$? $\endgroup$ – Theluqfzhluswcpzflabucheecatne Aug 11 '14 at 13:08
  • $\begingroup$ @125a8owp: I modified my proof, now it just depends on simple algebraic identities. $\endgroup$ – Jack D'Aurizio Aug 11 '14 at 13:33

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