121. Viete's relations with cos A , cos B , cos C.

by Virgil Nicula, Sep 13, 2010, 1:01 AM

Quote:
Let $ABC$ be a triangle. Prove that $\cos A\cos B+\cos B\cos C+\cos C\cos A = \frac{p^2-4R^{2}+r^{2}}{4R^2}$ .
Proof. $\sum\cos A=1+\frac{r}{R}\ \ \ ,\ \ \sum\cos^{2}A=\frac{1}{2}\cdot\sum\left(1+\cos 2A\right)$ $\mathrm{\ \ \ ,\ \ \ }\sum\cos 2A=-1-4\cos A\cos B\cos C$ . The power $p_{w}H$ of the orthocenter $H$

w.r.t. the circumcircle $w=C(O,R)$ is given by the relation $p_{w}H=-8R^{2}\cos A\cos B\cos C$. The power $p_{w}G$ of the centroid $G$ w.r.t. the circle $w$ is given by

the relation $p_{w}G=-\frac{a^{2}+b^{2}+c^{2}}{9}$ . $HO^{2}=R^{2}+p_{w}H=R^{2}-8R^{2}\cos A\cos B\cos C$ and $GO^{2}=R^{2}+p_{w}(G)=R^{2}-\frac{a^{2}+b^{2}+c^{2}}{9}$. Therefore,

$HO=3\cdot GO$ $\implies$ $HO^{2}=9\cdot GO^{2}$ $\implies$ $R^{2}-8R^{2}\prod\cos A=9\left(R^{2}-\frac{a^{2}+b^{2}+c^{2}}{9}\right)$ $\implies$ $\boxed{\ 8R^{2}\cos A\cos B\cos C=a^{2}+b^{2}+c^{2}-8R^{2}\ }$

and $ab+bc+ca=p^{2}+r^{2}+4Rr$ . Observe that $\left\{\begin{array}{c}\cos A+\cos B+\cos C=1+\frac{r}{R}\\\\ \cos A\cos B+\cos B\cos C+\cos C\cos A=\frac{p^{2}++r^{2}-4R^{2}}{4R^{2}}\\\\ \cos A\cos B\cos C=\frac{p^{2}-\left(2R+r\right)^{2}}{4R^{2}}\end{array}\right\|\ \ \Longleftrightarrow$

the equation $4R^{2}\cdot x^{3}-4R(R+r)\cdot x^{2}+\left(p^{2}+r^{2}-4R^{2}\right)\cdot x+(2R+r)^{2}-p^{2}=0$ has the roots $\left\{\cos A,\cos B,\cos C\right\}$ .


The shortest method. $\left\{\begin{array}{c}ab+bc+ca=p^{2}+r^{2}+4Rr\\\\ \cos A=-\cos (B+C)=-\cos B\cos C+\sin B\sin C\\\\ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\\\ \cos A+\cos B+\cos C=1+\frac{r}{R}\end{array}\right\|$ $\implies$ $\sum\cos B\cos C=$ $\sum\sin B\sin C-\sum\cos A=$

$\sum\frac{bc}{4R^{2}}-\left(1+\frac{r}{R}\right)=$ $\frac{p^{2}+r^{2}+4Rr}{4R^{2}}-\frac{R+r}{R}$ $\implies$ $\boxed{\mathcal \sum\cos B\cos C=\frac{p^{2}+r^{2}-4R^{2}}{4R^{2}}}$ .

Consequence. $\frac{(R-2r)(7R+2r)}{4R^{2}}+\frac{3}{4}\le \cos A\cos B+\cos B\cos C+\cos C\cos A\le\left(\frac{R+r}{r\sqrt 3}\right)^{2}$.

You can use above identity and the remarkable chain of inequalities $3r(4R+r)\le p^{2}\le\frac{(4R+r)^{2}}{3}$ .

Another short proof. $\cos B\cos C+\cos A=\sin B\sin C=\frac{S}{aR}$ $\implies$ $\sum\cos B\cos C=\frac{S}{R}\cdot\sum \frac{1}{a}-\left(1+\frac{r}{R}\right)$ $\implies$

$\sum\cos B\cos C=\frac{S}{R}\cdot \frac{ab+bc+ca}{4RS}-\frac{R+r}{R}$ $=$ $\frac{p^{2}+r^{2}+4Rr}{4R^{2}}-\frac{R+r}{R}$ $\implies$ $\sum\cos B\cos C=\frac{p^{2}-4R^{2}+r^{2}}{4R^{2}}$ .

Examples. $\left\{\begin{array}{c} \tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}=\frac{r_{a}+r_{b}+r_{c}}{p}=\frac{4R+r}{p}\\\\ \tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}=\frac{r_{a}r_{b}+r_{b}r_{c}+r_{c}r_{a}}{p^{2}}=1\\\\ \tan\frac{A}{2}\tan\frac{B}{2}\tan\frac{C}{2}=\frac{r}{p-a}\cdot\frac{r}{p-b}\cdot\frac{r}{p-c}=\frac{r^{3}}{pr^{2}}=\frac{r}{p}\end{array}\right\|$ .

Therefore, the equation $f(X)\equiv pX^{3}-(4R+r)X^{2}+pX-r=0$ has the roots $\left\{\ \tan\frac{A}{2}\ ,\ \tan\frac{B}{2}\ ,\ \tan\frac{C}{2}\ \right\}$ .

For example, $\left\{\begin{array}{ccccc} 90^{\circ}\in \left\{\ A\ ,\ B\ ,\ C\right\} & \iff & f(1)=0 & \iff & p=2R+r\\\\ 60^{\circ}\in\left\{\ A\ ,\ B\ ,\ C\ \right\} & \iff & f\left(\frac{\sqrt 3}{3}\right)=0 & \iff & p=(R+r)\sqrt 3\\\\ 30^{\circ}\in\left\{\ A\ ,\ B\ ,\ C\ \right\} & \iff & f\left(2-\sqrt 3\right)=0 & \iff & p=R+r(2+\sqrt 3)\end{array}\right\|$ .

From $\left\{\begin{array}{c} r_{a}+r_{b}+r_{c}=4R+r\\\\ r_{a}r_{b}+r_{b}r_{c}+r_{c}r_{a}=p^{2}\\\\ r_{a}r_{b}r_{c}=p^{2}r\end{array}\right\|$ obtain that the equation $r(X)=X^{3}-(4R+r)X^{2}+p^{2}X-p^{2}r=0$ has the roots $\left\{\ r_{a}\ ,\ r_{b}\ ,\ r_{c}\right\}$ .


Piece of poetry :

\[ \underline {\overline {\left| \ cos A,\ \cos B,\ \cos C\ ...\ \right| }} \]
\[ \blacksquare \]
\[ 2\cos \frac{A+B}{2}\cos \frac{A-B}{2}+\cos C \]\[ 2\sin \frac C2\cos \frac{A-B}{2}+1-2\sin ^2\frac C2 \]\[ 1+2\sin \frac C2\left( \cos \frac{A-B}{2}-\cos \frac{A+B}{2}\right) \]\[ 1+4\sin \frac A2\sin \frac B2\sin \frac C2 \]\[ 1+4\sqrt{\frac{(s-b)(s-c)}{bc}\cdot \frac{(s-c)(s-a)}{ca}\cdot \frac{(s-a)(s-b)}{ab}} \]\[ 1+\frac{4(s-a)(s-b)(s-c)}{abc} \]\[ 1+\frac{4sr^2}{4Rrs} \]\[ 1+\frac rR. \]
\[ \blacksquare \]
Exercise (practice). Let $ABC$ be a acute triangle, its circumcircle $w=C(O,R)$ and the middlepoints $D\in [BC]$,

$E\in [CA]$, $F\in [AB]$ of the sides of $\triangle ABC$. Then $OD+OE+OF=R+r$ (the Euler's relation).

1. A trigonometrical solution. Add the relations $OD=R\cos A\ \ \wedge\ \ OE=R\cos B\ \ \wedge\ \ OF=R\cos C$

$\implies$ $OD+OE+OF=R(\cos A+\cos B+\cos C)=R\left(1+\frac rR\right)=R+r$ .

2. A solution without the previous trigonometrical relation. We apply the Ptolemeu's theorem in the cyclic quadrilaterals $AEOF$, $BFOD$, $CDOE$ :

$b\cdot OF+c\cdot OE=$ $a\cdot R\ \ \wedge\ \ c\cdot OD+$ $a\cdot OF=b\cdot R\ \ \wedge\ \ a\cdot OE+$ $b\cdot OD=c\cdot R$. We add these relations to the wellknown relation

$a\cdot OD+b\cdot OE+c\cdot OF=2sr$ $\implies$ $(a+b+c)\cdot (OD+OE+OF)=(a+b+c)\cdot (R+r)$, i.e. $OD+OE+OF=R+r$ .
This post has been edited 32 times. Last edited by Virgil Nicula, Nov 23, 2015, 7:41 AM

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59. IMO 2010, problem 4.
58. A cevian and two incircles.
57. Opinii si comentarii relativ la inv. rom. II
56. A extremum problem with areas.
55. Two important circles - mixtilinear incircle/excircle.
54. Cinci grade de libertate.
53. An remarkable concurrency.
52. Harmonical division.
51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.
51. A nice problem with perpendicularity from Jaime.
50. IMAC 2010 seniors, the first day.
49. The midpoint of a cevian in an acute triangle.
+ June 2010
48. An inequality in a triangle : h_a+max{b,c} ...
47. Interesting and difficult identities in a triangle.
46. Outside of a quadrilateral.
45. \cos B+\cos C=1 <==> nice property.
44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.
43. A "slicing" problem with a parameter.
42. IMAC 2010 Problema 2.
41. ONM 2010 Calarasi Problema 3.
+ May 2010
40. Lema lui Haruki.
39. A nice and remarkable problem with Brocard's point.
38. Minimum area of triangle touching with semicircle.
37. Own extension of the Steiner-Lehmus' problem.
36. ONM (juniori) - 2010 Iasi, problema 4.
35. Nice problem from Arqady (Eduard_Tur).
34. IMO Shortlist 2002 geometry problem 7.
+ April 2010
32. Linkuri diverse.
31.Some nice metrical problems.
29. Inegalitate integrala de la Kunny.
28. A fost odată ...
27.Problem of Darij Grinberg (I - centroid of triangle DEF).
26. Outside of a triangle.
25. Geometric equations.
24. A fine and cool inequalities.
23. A very nice exponential inequality (own - from CRUX).
22. Some interesting limits (sequence/function).
21 bis. Some problems with complex numbers.
21. Probleme de geometrie (Yetty & I).
20. O ineg.cu suma de AG/AI .
19. Own proposed problems in CRUX Mathematicorum - Canada.
18. O problema a lui Toshio Seimiya.
17. O inegalitate "tare" intr-un triunghi.
16. Length of the (interior) angle bisector (10 proofs).
15. Inequalities stronger than Panaitopol's.
14. Opinii si comentarii relativ la inv. rom.
13. Inegalitatea de la Sorin Borodi.
12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

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    by piphi, Jun 10, 2020, 11:44 PM

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    by CSPAL, May 27, 2020, 4:17 PM

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    by OlympusHero, May 14, 2020, 8:43 PM

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