372. The distancies : OI , OI_a , ON a.s.o.

by Virgil Nicula, Apr 26, 2013, 3:39 PM

PP1. Let $\triangle ABC$ with circumcenter $O$ , incenter $I$ and $A$-excenter $I_a$ . Find $OI$ and $OI_a$ .

I used the notations: line $XY$ ; segment $[XY]$ ; the open ray $(XY$ (left) or $XY)$ (right). For example $AB$ is sideline , i.e. $[AB]\subset AB$ and it is the length of $[AB]$ .

Denote $\left\{\begin{array}{ccc} w=C(O,R) & - & \mathrm{circumcircle}\\\\ \rho=C(I,r) & - & \mathrm{incircle}\\\\ w_a=C\left(I_a,r_a\right) & - & A-\mathrm{excircle}\end{array}\right\|$ , the midpoint $M$ of $[BC]$ , $D\in AI\cap BC$ and $\left\{\begin{array}{ccc} A'\in (AI\cap w & ; & N\in (A'O\cap w\\\\ T\in AC\cap \rho & ; & L\in AC\cap w_a\end{array}\right\|$ .

Proof 1 (synthetic - shortest). It is well-known that $A'I=A'I_a=A'B=A'C$ , i.e. $IBI_aC$ is inscribed in circle with the diameter $[II_a]$ and $BC=II_a\cdot\cos\frac A2$ , i.e. $II_a=4R\sin\frac A2$

$\blacktriangleright\ \triangle ATI\sim\triangle NCA' \iff$ $\frac {IA}{A'N}=\frac {IT}{A'C}\iff$ $\frac {IA}{2R}=\frac {r}{IA'}\iff$ $IA\cdot IA'=2Rr\iff$ $p_w(I)=-2Rr\iff$ $OI^2-R^2=-2Rr\iff$ $\boxed{OI^2=R^2-2Rr}\ (1)$.

$\blacktriangleright\ \triangle ALI_a\sim\triangle NCA'\iff$ $\frac {I_aA}{A'N}=\frac {I_aL}{A'C}$ $\iff$ $\frac {I_aA}{2R}=\frac {r_a}{I_aA'}\iff$ $I_aA\cdot I_aA'=2Rr_a\iff$ $p_w(I_a)=2Rr_a\iff$ $OI_a^2-R^2=2Rr_a\iff$ $\boxed{OI_a^2=R^2+2Rr_a}\ (2)$ .

Proof 2. Let $d$ and $S\in d$ . For $X$ and $Y$ let $\left\{\begin{array}{c cc} X_0=\mathrm{pr}_dX & ; & \delta_d(X)=x\\\\ Y_0=\mathrm{pr}_dY & ; & \delta_d(Y)=y\end{array}\right\|$ . Then $\boxed{XY^2=X_0Y_0^2+(x-y)^2}$ . Here are $d:=BC$ , $X:=O$ and $Y\in\{I,I_a\}$ .

$\blacktriangleright\ OI^2=\left(R\cos A-r\right)^2+\left(\frac {b-c}{2}\right)^2=$ $(R-r)^2-R^2\sin^2A+2Rr-2Rr\cos A+\left(\frac {b-c}{2}\right)^2=$ $R^2-2Rr+r^2-\frac {a^2-(b-c)^2}{4}+2Rr\cdot 2\sin^2\frac A2=$

$R^2-2Rr+r^2-(s-b)(s-c)+\frac {4Rr(s-b)(s-c)}{bc}=$ $R^2-2Rr+\frac {(s-a)(s-b)(s-c)}{s}-(s-b)(s-c)+\frac {a(s-b)(s-c)}{s}\implies$ $\boxed{OI^2=R^2-2Rr}$ .

$\blacktriangleright\ OI_a^2=\left(r_a+R\cos A\right)^2+\left(\frac {b-c}{2}\right)^2=$ $(R+r_a)^2-R^2\sin^2A-2Rr_a+2Rr_a\cos A+\left(\frac {b-c}{2}\right)^2=$ $R^2+2Rr_a+r_a^2-\frac {a^2-(b-c)^2}{4}-2Rr_a(1-\cos A)=$

$R^2+2Rr_a+r_a^2-(s-b)(s-c)-\frac {4Rr_a(s-b)(s-c)}{bc}=$ $R^2+2Rr_a+\frac {s(s-b)(s-c)}{s-a}-(s-b)(s-c)-\frac {a(s-b)(s-c)}{s-a}\implies$ $\boxed{OI_a^2=R^2+2Rr_a}$ .

Proof 3. Let a line $d$ and $S\in d$ . For $X$ and $Y$ define $\left\{\begin{array}{c cc} X_0=\mathrm{pr}_dX & ; & \delta_d(X)=x\\\\ Y_0=\mathrm{pr}_dY & ; & \delta_d(Y)=y\end{array}\right\|$ . Then $\boxed{XY^2=X_0Y_0^2+(x-y)^2}$ . Here are $d:=AB$ , $X:=O$ and $Y\in \{I,I_a\}$ .

$\blacktriangleright\ OI^2=\left(r-R\cos C\right)^2+\left[(s-a)-R\sin C\right]^2=$ $r^2+R^2+(s-a)^2-2Rr+2Rr(1-\cos C)-2R(s-a)\sin C=$

$R^2+r^2+(s-a)^2-c(s-a)-2Rr+4Rr\sin^2\frac C2=$ $R^2-2Rr+\frac {(s-a)(s-b)(s-c)}{s}-(s-a)(s-b)+\frac {4Rr(s-a)(s-b)}{ab}=$

$R^2-2Rr-\frac {c(s-a)(s-b)}{s}+\frac {4Rr(s-a)(s-b)}{ab}\implies$ $\boxed{OI^2=R^2-2Rr}$ because $\frac {4Rr}{ab}=\frac cs$ .

$\blacktriangleright\ OI_a^2=\left(r_a-R\cos C\right)^2+\left(s-R\sin C\right)^2=$ $r_a^2+R^2+s^2-2Rr_a\cos C-2Rs\sin C=$ $(R+r_a)^2+s^2-2Rr_a(1+\cos C)-sc=$

$R^2+2Rr_a+r_a^2+s^2-\frac {4Rr_as(s-c)}{ab}-sc=$ $R^2+2Rr_a+r_a^2+s(s-c)-\frac {abc}{s-a}\cdot\frac {s(s-c)}{ab}=$ $R^2+2Rr_a+$ $\frac {s(s-b)(s-c)}{s-a}+s(s-c)-\frac {cs(s-c)}{s-a}=$

$R^2+2Rr_a$ because $\frac {s(s-b)(s-c)}{s-a}+s(s-c)-\frac {cs(s-c)}{s-a}=0$ . In conclusion, $\boxed{OI_a^2=R^2+2Rr_a}$ .

Remark. $IBI_aC\ \mathrm{cyclic}\implies$ $BC=II_a\cdot\sin\widehat{BIC}\implies$ $II_a=\frac {a}{\cos\frac A2}\implies$ $II_a^2=\frac {a^2bc}{(s-a)}=$ $\frac {a\cdot 4Rr_a(s-a)}{s(s-a)}=$ $4R\cdot \frac {ar_a}{s}=$ $4R\cdot \frac {s(r_a-r)}{s}\implies$ $\boxed{II_a^2=4R(r_a-r)}\ (*)$ .

Otherwise, $A'I^2=A'B^2=A'M\cdot A'N=$ $\frac {r_a-r}{2}\cdot 2R=R(r_a-r)\implies$ $A'I^2=R(r_a-r)\implies$ $II_a^2=4\cdot A'I^2\implies$ $II_a^2=4R(r_a-r)$ .

Apply theorem of median in $\triangle OII_a\ : 2\left(OI^2+OI_a^2\right)-II_a^2=$ $4\cdot A'O^2\iff$ $OI^2+OI_a^2\stackrel{(*)}{=}2R^2+2R(r_a-r)\iff$ $\boxed{OI^2+OI_a^2=2R(R+r_a-r)}$ .


PP2. Prove that $(\forall )\ ABC$ with the circumcircle $w=C(O,R)$ and the Nagel point $N$ there is the relation $\boxed{\ ON=R-2r\ }$ .

Proof 1 (metric). Let $\left\{\begin{array}{cc} D\in AN\cap BC\\\\ E\in BN\cap CA\\\\ F\in CN\cap AB\end{array}\right\|$ . Is well-known that $\left\{\begin{array}{c} BF=CE=s-a\\\\ CD=AF=s-b\\\\ AE=BD=s-c\end{array}\right\|$ , $\boxed{\frac {NA}a=\frac {ND}{s-a}=\frac {AD}{s}}$ and identity $\boxed{\sum a^2(s-b)(s-c)=4s^2r(R-r)}\ (*)$ .

I denoted $\{A,L\}=\{A,N\}\cap w$ . Apply the Stewart's relation to the cevian $AD$ in $\triangle ABC\ :\ \boxed{a\cdot AD^2+a(s-b)(s-c)=c^2(s-b)+b^2(s-c)}\ (1)$ .

Since $\boxed{R^2-ON^2=NA\cdot NL}\ (2)$ and $DB\cdot DC=DA\cdot DL$ obtain that $NA\cdot NL=\frac as\cdot AD\cdot(ND+DL)=$ $\frac as\cdot AD\cdot \frac {s-a}{s}\cdot AD+\frac as\cdot DB\cdot DC=$

$\frac {s-a}{s^2}\cdot a\cdot AD^2+\frac {a(s-b)(s-c)}{s}$ . Therefore, $NA\cdot NL\stackrel{(1)}{=}$ $\frac {s-a}{s^2}\cdot \left[c^2(s-b)+b^2(s-c)-a(s-b)(s-c)\right]+$ $\frac {a(s-b)(s-c)}{s}=$ $\frac {1}{s^2}\cdot\sum a^2(s-b)(s-c)\stackrel{(*)}{=}$

$4r(R-r)$ . In conclusion, $NA\cdot NL=4r(R-r)$ and from the relation $(2)$ obtain that $ON^2=R^2-NA\cdot NL=$ $R^2-4r(R-r)=(R-2r)^2\implies$ $ON=R-2r$ .

$(*)$ Remark. $\sum a^2(s-b)(s-c)=\sum a^2\left[bc-s(s-a)\right]=$ $abc\sum a-s\cdot\sum a^2(s-a)=$ $8Rs^2r-s^2\cdot \sum a^2+$ $s\cdot\sum a^3=$ $8Rs^2r-s^2\cdot 2\left(s^2-r^2-4Rr\right)+$

$s\left[8s^3-6s\left(s^2+r^2+4Rr\right)+12Rsr\right]=$ $4Rs^2r-4s^2r^2\implies$ $\boxed{\sum a^2(s-b)(s-c)=4s^2r(R-r)}$ . Otherwise. $\sum a^2(s-b)(s-c)=$ $\sum a^2bc\sin^2\frac A2=$

$abc\sum a\sin^2\frac A2=$ $abc\sum 2R\sin A\cdot \sin^2\frac A2=$ $Rabc\sum \sin A(1-\cos A)=$ $4R^2sr\left(\sum\sin A-\frac 12\sum\sin 2A\right)=$ $4R^2sr\left(\frac sR-2\prod\sin A\right)=$ $4R^2sr\left(\frac sR-\frac S{R^2}\right)=$

$4R^2sr\left(\frac sR-\frac {sr}{R^2}\right)=$ $4s^2r(R-r)\implies$ $\sum a^2(s-b)(s-c)=4s^2r(R-r)\ .$

Proof 2 (metric).
This post has been edited 47 times. Last edited by Virgil Nicula, Nov 30, 2016, 5:38 PM

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129. BX=CY and three properties.
128. Value of IG and two inequalities in neighbourhood of 0.
127. Pagina educativa.
126. Routh's theorem.
125. I like very much this Darij Grinberg's message.
124. A nice and easy problem with a right-angled triangle.
123. My extension and its proof.
122. An identity for an equilateral triangle. Extension.
121. Viete's relations with cos A , cos B , cos C.
120. Two equal neighbouring angles/sides in a quadrilateral.
119. Inspired by a Miculita's problem.
118. An usual remarkable geometrical locus.
117. Some applications of harmonic division/quadrilateral.
116. The length of AE (nine methods !).
115. An difficult equation with one radical.
114. A property of tangential triangle for ABC.
113. Range for a function.
112. The equivalence relation ".s.s."
111. Simple, but interesting ...
110. German TST 2004 - Exam V , Problem 1
109. Two equal areas ==> find A.
108. Old, short and nice proof of Euler's relation.
107. XY parallel BC and X,Y are isogonal/izotomic points.
106. OLM - Bucuresti (geometrie).
105. A problem with three circles.
104. Concursul Revistei de Matematica din Galati (RMG).
103. Metrical usual relations in triangle. Applications.
102. Characterization of a metrical relation in a triangle.
101. O problema cu numere complexe.
100. Bobiller's theorem.
99. Some simple and nice geometry problems.
98. Symmedian & median (metrical relations).
97. Problem of butterfly.
+ August 2010
96. Integrals.
95. Three circles in an angle.
94. Equations of angle bisectors (interior and exterior).
93. The Appolonius' construct problems.
92. CA , CB tangent to parabola - OIMU 2008 Problem 4.
91. JBMO 2010, Problem 3.
90. Nice and difficult inequality in ABC (own).
89. Similarity (parallel/antiparallel).
88. A nice geometric locus conditioned metrically.
87. An interesting vectorial identity.
86. A very nice maxmin.
85. Problem "slicing" : 60-24-12.
84. IRAN national math olympiad - 2010
83. A metrical characterization of concurrence.
82. AA' , BN , CM concur in Gergonne point.
81. Sawayama and Thebault’s theorem.
80. Three concurrent perpendiculars (the Carnot's lemma).
79. A quadrilateral with many perpendicularities.
78. Jakobi's theorem.
77. Two equivalent perpendicularities.
76. A parallelogram and a circle (nice !).
75. Extended Steinbart Theorem.
74. Inegalitatea Erdos-Mordell.
73. A nice problems with a square and a perpendicularity.
+ July 2010
72. Colinearity in a triangle - H , P , M .
70. Colinearity with pole/polar - H\in PQ .
71. An angle questions (synthetic/trigonometric proofs).
69*1. Slicing problem 3.
69. Position of x1, x2 w.r.t. a given real number k.
68. Some nice applications of "pole/polar".
67. Remarkable algebraic/geometrical inequalities (1).
66. Remarkable perpendicularities in a triangle.
65. Parallel to BC through I .
64. Problems of the type "instant" (at most one minute).
63. A nice problem with radical center (livetolove212).
62. Nice and easy problem with a right triangle.
61. IMO Shortlist 2009 (very nice problem and its proof).
60. Mediteranean M.C. 2010 Problem 3.
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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  • Total entries: 456
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