191. Six nice problems with the incircle of a triangle.

by Virgil Nicula, Dec 14, 2010, 8:25 AM

Proposed problem 1. Let $ABC$ be a triangle with the incircle $w=C(I,r)$ which touches its sides at $D\in BC$ ,

$E\in CA$ , $F\in AB$ . The rays $[BE$ , $[CF$ meet again $w$ at $X$ , $Y$ respectively. Prove that $BY\cap CX\cap AD\ne\emptyset$ .

Proof. Observe that $\Gamma\in AD\cap BE\cap CF$ (Gergogne's point), $\Gamma X\cdot \Gamma E=\Gamma Y\cdot\Gamma F\ ,\ \left\|\begin{array}{c} \frac {\Gamma B}{b(s-b)}=\frac {\Gamma E}{(s-a)(s-c)}=\frac {BE}{(s-a)(s-c)+b(s-b)}\\\ \frac {\Gamma C}{c(s-c)}=\frac {\Gamma F}{(s-a)(s-b)}=\frac {CF}{(s-a)(s-b)+c(s-c)}\end{array}\right\|$

and $\left\|\begin{array}{c} BX\cdot BE=(s-b)^2\\\ CY\cdot CF=(s-c)^2\end{array}\right\|$ . Apply the Ceva's theorem to $\triangle B\Gamma C$ for $D\in BC$ , $X\in \Gamma B$ and $Y\in \Gamma C\ :\ \frac {DB}{DC}\cdot\frac {YC}{Y\Gamma}\cdot\frac {X\Gamma}{XB}=1\iff$

$\frac {s-b}{s-c}\cdot\frac {\frac {(s-c)^2}{CF}}{Y\Gamma}\cdot \frac {X\Gamma}{\frac {(s-b)^2}{BE}}=1\iff$ $\frac {s-c}{s-b}\cdot\frac {BE}{CF}\cdot\frac {\Gamma F}{\Gamma E}=1\iff$ $\frac {s-c}{s-b}\cdot \frac {(s-a)(s-c)+b(s-b)}{(s-a)(s-c)}\cdot\frac {(s-a)(s-b)}{(s-a)(s-b)+c(s-c)}=1\iff$

$(s-a)(s-b)+c(s-c)=(s-a)(s-c)+b(s-b)\iff$ $(s-a)[(s-b)-(s-c)]=s(b-c)-(b-c)(b+c)$ $\iff$

$(s-a)(c-b)=(c-b)(b+c-s)$ $\iff$ $s-a=$b+c-s $\iff$ $2s=a+b+c$, what is truly.


Proposed problem 2. The incircle $w=C(I,r)$ of $\triangle ABC$ touches its sides at $D\in BC$ , $E\in CA$ and $F\in AB$ respectively. Consider the point $A'\in w$

for which the circumcircle $w_a$ of $\triangle BA'C$ is tangent in $A'$ to the circle $w$ . Define analogously the points $B'$ and $C'$ . Prove that $AA'\cap BB'\cap CC'\ne\emptyset$ .

Proof 1. From the well-known property obtain that the ray $[A'D$ is the bisector of the angle $\widehat {BA'C}$ . Denote the intersection $L\in BC\cap EF$. The division $(L,B,D,C)$ is harmonically, i.e. $\frac{LB}{LC}=\frac{DB}{DC}$ . From a characterization of the harmonical division obtain that $A'D\perp A'L$ . With other words, the point $A'$ is the intersection between the circle with the diameter $DL$ and the incircle $w$ for which $(AA')\cap BC=\emptyset$ . Then the point $A'$ is the reflection of the point $D$ w.r.t. the line $SI$, where the point $S$ is the middlepoint of the segment $[LD]$ , i.e. the lines $SD$ and $SA'$ is the tangents from $S$ to $w$ , $\frac{SB}{SC}=\left(\frac{DB}{DC}\right)^2=\left(\frac{p-b}{p-c}\right)^2$, $SD^2=SB\cdot SC$ , $\frac {SB}{(p-b)^2}=$ $\frac {SC}{(p-c)^2}=$ $\frac {a}{a(b-c)}=\frac {1}{b-c}$ , $SD=SB+BD=$ $\frac {(p-b)^2}{b-c}+(p-b)=$ $\frac {(p-b)(p-c)}{b-c}$ and $\tan \widehat {IDA'}=\frac{r}{SD}=\frac{r|b-c|}{(p-b)(p-c)}$ a.s.o.

Proof 2. Denote $N\in I_aD\cap w\ ,\ T\in BC$ for which $TN$ is tangent to $w$ and $R\in TI\cap ND$ . Observe that the pentagon $BRICI_a$ is inscribed in the circle with

diameter $[II_a]$ and $TN^2=TR\cdot TI=TB\cdot TC$ . Therefore, $TN$ is tangent to the circumcircle of $\triangle BNC$ , i.e. $w$ is tangent to the circumcircle of $\triangle BNC$ in $N$ .


Proposed problem 3. The incircle $w$ of $\triangle ABC$ touches $BC$ at $K$. Let $M$ be the midpoint of $A$-altitude $AD$ of $\triangle ABC$ , where $D\in BC$ .

The line $MK$ meets the incircle of $\triangle ABC$ at $N$ (apart from $K$) . Show that the circumcircle of $\triangle BNC$ is tangent to $w$ at $N$ .

The proof. Suppose w.l.o.g. $c<b$ . Denote the A-exincircle $w=C(I_{a},r_{a})$ , $D'\in BC\cap w$, the intersection $P$ between $BC$ and the tangent to $w$ in $N$ and the intersection $R\in NK\cap IP$ . Prove easily that $KD'=b-c$, $KD=\frac{(b-c)(p-a)}{a}$ and $\frac{KD}{KD'}=\frac{p-a}{a}$ . But $\frac{DM}{D'I_{a}}=$ $\frac{h_{a}}{2r_{a}}=$ $\frac{ah_{a}}{2ar_{a}}=$ $\frac{S}{ar_{a}}=$ $\frac{r_{a}(p-a)}{ar_{a}}=$ $\frac{p-a}{a}$ . Therefore, $\frac{DM}{D'I_{a}}=\frac{KD}{KD'}$, i.e. $\boxed{\ K\in MI_{a}\ }$ . The points $B,C,R$ belong to the circle with the diameter $II_{a}$ . But $IP\perp NK$ $\Longrightarrow$ $PN^{2}=$ $PR\cdot PI=$ $PB\cdot PC$ . Thus $PN^{2}=PB\cdot PC$ , i.e. the line $PN$ is tangent to the circumcircle of the triangle $BNC$ .


Proposed problem 4. Let $ ABC$ be a triangle with incircle $w$ which touches its sides at $ D\in BC$ , $E\in CA$ ,

$F\in AB$ . The line $CF$ meet again $w$ in $P$ . Prove that $ABPE$ is a cyclic quadrilateral $\implies DP\parallel AB$ .

Proof. Denote $M\in DE\cap AB$ . From a well-known property $MI\perp CF$ obtain that $MP$ is tangent to $w$ . Denote $K\in PM\cap AC$ and $H\in PB\cap DE$ .

Since $ABPE$ is cyclically obtain that $\widehat{PBM}\equiv \widehat{KEP}\equiv \widehat{PDE}\equiv \widehat{KPE}$ $\implies$ $PDBM$ is cyclically $\implies$ $\widehat{BDE}=\widehat{BPM}=\widehat{DEA}$ $\implies$

$KPHE$ is cyclically $\implies$ $\widehat{PHD}\equiv \widehat{PKE}$ $\implies$ $\widehat{HPD}=\widehat{HDP}$ $\implies$ $ PD\parallel AB$ .


Proposed problem 5. Let $ ABC$ be a triangle ($ b\ne c$) for which denote the circumcircle $ w = C(O,R)$ , the incircle $ C(I,r)$ , the midpoint

$M$ of $ [BC]$ and the point $ N\in w$ so that $ MN\perp BC$ and $ BC$ doesn't separate $ A$ , $ N$ . Prove that $ \widehat {ANI}\equiv \left\{\begin{array}{ccc} \widehat {IMC} & \mathrm{if} & b < c \\ \ \widehat {IMB} & \mathrm{if} & b > c\end{array}\right\|\ .$

Proof I. Suppose $ b<c$ . Denote the diameter $ [NS]$ of circle $ w$ , $ T\in BC$ , $ IT\perp IA$ and $ R\in NI\cap ST$ .

Prove easily that $ R\in w$ and the quadrilateral $ MITS$ is cyclically. Thus, $ IT\parallel NA$ and $ \widehat {IMC}\equiv\widehat {ISR}\equiv\widehat {ANI}\ .$

Proof 2. Suppose $ b < c$ . Denote $ D\in BC$ for which $ ID\perp BC$ . Observe that $ MD = \frac {c - b}{2}$ and $ \tan\widehat {IMC} = \tan\widehat {IMD} = \frac {ID}{MD} = \frac {2r}{c - b}$ $ \implies$

$ \boxed {\ \tan\widehat {IMC} = \frac {2r}{c - b}\ }\ .$ Since $ AN\perp AI$ obtain $ \tan \widehat {ANI} = \frac {AI}{AN} = \frac {b + c}{2p}\cdot\frac {2bc\cdot\cos\frac A2}{b + c}\cdot\frac {1}{2R\sin\frac {C - B}{2}} =$ $ \frac {bc\cdot\cos\frac A2\sin\frac A2}{2pR\sin\frac {C - B}{2}\cos\frac {B + C}{2}} =$

$ \frac {bc\sin A}{2pR(\sin C - \sin B)} = \frac {2pr}{p(c - b)}$ $ \implies$ $ \boxed {\ \tan\widehat {ANI} = \frac {2r}{c - b}\ }$ . Thus, $b < c\implies \tan\widehat {ANI} = \tan\widehat {IMC} = \frac {2r}{c - b}\implies \widehat {ANI}\equiv\widehat {IMC}\ .$

You can prove analogously $b > c\implies \tan\widehat {ANI} = \tan\widehat {IMB} = \frac {2r}{b - c}\implies \widehat {ANI}\equiv\widehat {IMB}\ .$

Proposed problem 6. Let $ \triangle ABC$ with incircle $ w = C(I,r)$ and the Nagel's point $ N$ . Denote $ D\in BC\cap w$ , the diameter $ [MD]$ of $ w$ and $ D'\in AN\cap BC$ .

Prove that $ AM = ND'$ and $ N\in w\Longleftrightarrow$$ \prod (b + c - 3a) = 0 \Longleftrightarrow IG\perp BC\ \vee\ IG\perp CA\ \vee\ IG\perp AB$ $ \Longleftrightarrow s^2 + 4r^2 = 16Rr$ .

Proof. Prove easily that $ AM = ND'$ (nice property !). Therefore, $ N\in w$ $ \Longleftrightarrow$ $ IA = ID'\Longleftrightarrow DD' = s - a\Longleftrightarrow$ $ |b - c| = s - a$ $ \Longleftrightarrow$

$a + b = 3c\ \vee\ a + c = 3b$ a.s.o. Prove easily that $ IG\perp BC\Longleftrightarrow b + c = 3a$ . Using the well-known relation $ IN^2 = s^2 + 5r^2 - 16Rr$

obtain $ N\in w\Longleftrightarrow IN = r\Longleftrightarrow$ $ s^2 + 4r^2 = 16Rr$ .
This post has been edited 54 times. Last edited by Virgil Nicula, Nov 22, 2015, 5:36 PM

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+ 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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