369. Mixtilinear circles of ABC.

by Virgil Nicula, Jan 19, 2013, 5:52 PM

Lemma (own). Let $\triangle ABC$ with the incircle $i=C(I,r)$ , the exincircle $i_a=C(I_a,r_a)$ and the circumcircle $e=C(O,R)$ . Denote

$\left\{\begin{array}{ccc} M\in AB\ ,\ IM\perp IA\ ; & I_1\in AI\ ,\ I_1M\perp AB\ ; & I_1M=r_1\\\\ N\in AB\ ,\ I_aN\perp IA\ ; & I_2\in AI\ ,\ I_2N\perp AB\ ; & I_2N=r_2\end{array}\right|$ . Prove that:

$1.\blacktriangleright\ s\cdot AM=(s-a)\cdot AN=bc\ \ ;\ \ \frac {r}{r_1}=\frac {r_a}{r_2}=\cos^2\frac A2$ .

$2.1\blacktriangleright$ The circle $c_1=C(I_1,r_1)$ is tangent to the rays $[AB$ , $[AC$ and interior tangent to $e$ .

$2.2\blacktriangleright$ The circle $c_2=C(I_2,r_2)$ is tangent to the rays $[AB$ , $[AC$ and exterior tangent to $e$ .

$3.\blacktriangleright$ If $T_1\in AC\cap i$ , $T_2\in AC\cap i_a$ , then $MC\parallel BT_2$ and $NC\parallel BT_1$ ( the construction of of the circles $c_1$ , $c_2$).

Proof.

$1.\blacktriangleright$ If $T\in AB\ ,\ IT\perp AB$ , then $IT=r\ ,\ AT=s-a=$ $AI\cdot \cos\frac A2=$ $AM\cdot \cos^2\frac A2=$ $AM\cdot \frac {s(s-a)}{bc}\implies$ $\boxed{AM=\frac {bc}{s}}$ . If $S\in AB\ ,\ I_aS\perp AB$ ,

then $I_aS=r_a\ ,\ AS=s=$ $AI_a\cdot \cos\frac A2=$ $AN\cdot \cos^2\frac A2=$ $AN\cdot \frac {s(s-a)}{bc}\implies$ $\boxed{AN=\frac {bc}{s-a}}$ . In conclusion, $s\cdot AM=(s-a)\cdot AN=bc$ . Observe that

$\frac {r}{r_1}=\frac {AT}{AM}=\frac {s-a}{\frac {bc}{s}}\ \ \wedge\ \ \frac {r_a}{r_2}=$ $\frac {AS}{AN}=\frac {s}{\frac {bc}{s-a}}$ . In conclusion, $\frac {r}{r_1}=\frac {r_a}{r_2}=\frac {s(s-a)}{bc}=\cos^2\frac A2$ .

$2.1\blacktriangleright$ The circle $c_1$ is interior tangent to the circle $e\iff$ $OI_1+r_1=R\iff$ $OI_1^2=(R-r_1)^2\iff$ $AI_1^2+R^2-2R\cdot AI_1\cdot\cos \frac {B-C}{2}=$ $R^2-2Rr_1+r_1^2$ .

Since $r_1=AI_1\sin\frac A2$ obtain that $r_1^2-2Rr_1\sin\frac A2\cos\frac {B-C}{2}=\left(-2Rr_1+r_1^2\right)\sin^2\frac A2\iff$ $r_1-R(\cos B+\cos C)=r_1\sin^2\frac A2-R(1-\cos A)\iff$

$r_1\cos^2\frac A2=R(\cos A+\cos B+\cos C-1)\iff$ $\boxed{r_1=\frac {r}{\cos^2\frac A2}}$ . Since $r_1=\frac {bc}{s}\cdot \tan \frac A2$ obtain that $\frac {bc}{s}\cdot\sin\frac A2\cos\frac A2=r\iff$ $\frac {bc}{2s}\cdot\sin A=r\iff$ $bc\sin A=2sr$ , true.

$2.2\blacktriangleright$ $c_2$ is exterior tangent to $e\iff$ $OI_2=R+r_2\iff$ $OI_2^2=(R+r_2)^2\iff$ $AI_2^2+R^2-2R\cdot AI_2\cdot\cos \frac {B-C}{2}=$ $R^2+2Rr_2+r_2^2$ . Since $r_2=AI_2\sin\frac A2$ obtain

$r_2^2-2Rr_2\sin\frac A2\cos\frac {B-C}{2}=\left(2Rr_2+r_2^2\right)\sin^2\frac A2\iff$ $r_2-R(\cos B+\cos C)=r_2\sin^2\frac A2+R(1-\cos A)\iff$ $r_2\cos^2\frac A2=R(\cos A+\cos B-\cos A+1)$ . Since

$r_2=\frac {bc}{s-a}\cdot \tan \frac A2$ obtain that $\frac {bc}{s-a}\cdot\sin\frac A2\cos\frac A2=4R\sin\frac A2\cos\frac B2\cos\frac C2\iff$ $\frac {bc}{2(s-a)}\cdot\sin A=\frac {4Rs(s-b)(s-c)}{abc}\iff$ $Sabc=4rS^2\iff$ $4RS=abc$ , true.

$3.\blacktriangleright\ \left\{\begin{array}{ccccccc} NC\parallel BT_1 & \iff & \frac {AB}{AN}=\frac {AT_1}{AC} & \iff & AN\cdot AT_1=bc & \iff & (s-a)\cdot\frac {bc}{s-a}=bc\\\\ MC\parallel BT_2 & \iff & \frac {AM}{AB}=\frac {AC}{AT_2} & \iff & AM\cdot AT_2=bc & \iff & \frac {bc}{s}\cdot s=bc\end{array}\right\|$ , what is truly.


Application 1. Let $\triangle ABC$ with incircle $i=C(I,r)$ and circumcircle $e=C(O,R)$ . Let $s_a=C(\rho_a)$ be the circle which

is tangent to $[AB)$ , $[AC)$ and which is interior tangent to $e$ . Define analogously $s_b$ and $s_c$ . Prove that $\frac{1}{\rho_a}+\frac{1}{\rho_b}+\frac{1}{\rho_c}\leq \frac{9}{4r}$ .

Proof. From the upper lemma obtain that $\rho_a=\frac {r}{\cos^2\frac A2}\ \mathrm{a.s.o.}\implies$ $\frac {1}{\rho_a}+\frac {1}{\rho_b}+\frac {1}{\rho_a}=$ $\frac 1r\sum\cos^2\frac A2=$ $\frac {1}{2r}\sum (1+\cos A)=\frac {1}{2r}\left(4+\frac rR\right)\le\frac {9}{4r}$ . I used the inequality $R\ge 2r$ .

Application 2. Let $\triangle ABC$ with incircle $i=C(I,r)$ and circumcircle $e=C(O,R)$ . Let $w_a=C(R_a)$ be the circle which

is tangent to $[AB)$ , $[AC)$ and which is exterior tangent to $e$ . Define analogously $w_b$ and $w_c$ . Prove that $\frac{1}{R_a}+\frac{1}{R_b}+\frac{1}{R_c}\geq \frac{3}{4r}$ .

Proof. From the upper lemma obtain that $R_a=\frac {r_a}{\cos^2\frac A2}\ \mathrm{a.s.o.}\implies$ $\frac {1}{R_a}+\frac {1}{R_b}+\frac {1}{R_a}=$ $\sum\frac {\cos^2\frac A2}{r_a}=$ $\sum\frac {\frac {s(s-a)}{bc}}{\frac {sr}{s-a}}=$

$\frac 1r\sum\frac {(s-a)^2}{bc}\stackrel{(\mathrm{C.B.S})}{\ge}\frac 1r\frac {s^2}{\sum bc}\stackrel{(*)}{\ge}$ $ \frac {3s^2}{r(a+b+c)^2}=\frac {3}{4r}$ . I used and the inequality $(a+b+c)^2\ge 3(ab+bc+ca)\ \ (*)$ .
This post has been edited 25 times. Last edited by Virgil Nicula, Nov 15, 2015, 6:24 PM

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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 OlympusHero, Dec 30, 2020, 6:08 PM

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    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

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

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

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

  • impressive :D
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    by OlympusHero, May 14, 2020, 8:43 PM

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About Owner
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  • Joined: Jun 22, 2005
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  • Blog created: Apr 20, 2010
  • Total entries: 456
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