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How do we prove that for any triangle this holds: $$\frac{r}{R}+1=\cos A+\cos B+\cos C$$ I can use this beautiful identity to prove several geometric inequalities, but I have no idea how to prove the identity itself. Can anyone give me hints?

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  • $\begingroup$ Use Ptolemy Theorem in quadrilateral $AC_1B_1S$ where $S$ is incenter and $C_1$, $B_1$ are tangential points on sides $AB$ and $AC$ and same for other such quadrilaterals. Then you just need some basic trigonometry of right-angled triangle. $\endgroup$ – user101521 Mar 31 '14 at 21:01
  • $\begingroup$ @user101521 Could you elaborate, maybe give it as answer? $\endgroup$ – Sawarnik Mar 31 '14 at 21:13
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    $\begingroup$ What is $r$ and $R$? $\endgroup$ – Cherufe Mar 31 '14 at 21:14
  • $\begingroup$ @Cherufe Inradius and the circumradius respectively, of course. $\endgroup$ – Sawarnik Mar 31 '14 at 21:15
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    $\begingroup$ Inradius formula (Equation 2); circumradius formula (Equation 1); Law of Cosines (Equations 4, 5, 6). $\endgroup$ – Blue Mar 31 '14 at 23:10
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here is mechanical solution:

$\cos A+\cos B+\cos C-1=\dfrac{a^2b+b^2c+c^2a+b^2a+c^2b+a^2c-a^3-b^3-c^3-2abc}{2abc}=\dfrac{(a+b-c)(b+c-a)(c+a-b)}{2abc}=\dfrac{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}{2abc(a+b+c)}=\dfrac{8S^2}{abc(a+b+c)}=\dfrac{\dfrac{S}{s}}{\dfrac{abc}{4S}}=\dfrac{r}{R}$

$S=\sqrt{s(s-a)(s-b)(s-c)},s=\dfrac{a+b+c}{2}$

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  • $\begingroup$ That is the only thing that came to my mind, just brute forcing using the cosine rule, but I was looking for elegant solutions. But anyways, how did you factor the expression in the second step? $\endgroup$ – Sawarnik Apr 1 '14 at 9:07
  • $\begingroup$ $c^2a+c^2b-c^3-a^3-a^2b+a^2c-b^3-b^2c+b^2c+2a^2b+2b^2a-2abc=c^2(a+b-c)-a^2(a+b-c)-b^2(a+b-c)+2ab(a+b-c)=(a+b-c)(c^2-(a-b)^2)$ $\endgroup$ – chenbai Apr 1 '14 at 9:50
  • $\begingroup$ There is a geometry method that I found it was even more complex than this way. $\endgroup$ – chenbai Apr 1 '14 at 9:51
  • $\begingroup$ Ok, no problem :) $\endgroup$ – Sawarnik Apr 1 '14 at 10:03
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Using this, $$\cos A+\cos B+\cos C-1=4\sin\frac A2\sin\frac B2\sin\frac C2$$

Now from this

or using cosine formula & $\displaystyle\cos A=1-2\sin^2\dfrac A2\implies\sin\frac A2=+\sqrt{\frac{1-\cos A}2}$ as $\displaystyle0<\frac A2<\frac\pi2$

$\displaystyle\sin\frac A2=\sqrt{\frac{(s-b)(s-c)}{bc}}$ where $2s=a+b+c$

$$\implies4\sin\frac A2\sin\frac B2\sin\frac C2=4\frac{(s-a)(s-b)(s-c)}{abc}$$

Now, $\displaystyle\triangle =\frac12ab\sin C=\frac{abc}{4R}=\sqrt{s(s-a)(s-b)(s-c)}=r\cdot s$

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This is corollary of Carnot theorem, which is stated that $d(O,BC)+d(O,CA)+d(O,AB)=R+r$. Let $M,N,P$ be the midpoints of $BC,CA,AB$. Then $\cos A+ \cos B+\cos C=\dfrac{OM+ON+OP}{R}=\dfrac{R+r}{R}=1+\dfrac{r}{R}.$

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