Bisectors of triangle ABC of an angles A and C intersect with BC and AB at points A1 and C1 respectively. Lines AA1 and CC1 intersect circumcircle of triangle ABC at points A2 and C2 respectively. K is intersection point of C1A2 and A1C2. I is incenter of ABC. Prove that the line KI divides AC into two equal parts.
Problem
Source: Silk Road Mathematics Competition 2009
Tags: geometry, circumcircle, incenter, trigonometry, geometry proposed
Ahiles
01.04.2009 22:21
We have $ \dfrac{C_2C_1}{C_1I} = \dfrac{AC_2}{AI}\cdot \dfrac{\sin{\angle C_2AB}}{\sin \angle C_1AI} = \dfrac{2R\sin \frac {C}{2}}{AI}\cdot \dfrac{\sin\frac {C}{2}}{\sin \frac {A}{2}} = \dfrac{2R\sin^2 \frac {C}{2}}{r}$ Similiarly $ \dfrac{A_2A_1}{A_1I} = \dfrac{2R\sin^2 \frac {A}{2}}{r}$ Let $ D \in KI \cap A_2C_2$. By Ceva in $ \triangle C_2A_2I$ we have $ \dfrac{C_2D}{A_2D} = \dfrac{C_2C_1}{C_1I}\cdot \dfrac{IA_1}{A_1A_2} = \dfrac{\sin^2\frac {C}{2}}{\sin^2\frac {A}{2}}$ Then $ \dfrac{\sin{\angle C_2ID}}{\sin\angle A_2ID} = \dfrac{C_2D}{A_2D}\cdot \dfrac{IA_2}{IC_2} = \dfrac{\sin^2\frac {C}{2}}{\sin^2\frac {A}{2}} \cdot \dfrac{\sin \frac {A}{2} }{\sin \frac {C}{2} } = \dfrac{\sin\frac {C}{2}}{\sin\frac {A}{2}}$ If we denote $ M \in KI \cap AC$. Then $ \dfrac{AM}{MC} = \dfrac{AI}{IC}\cdot \dfrac{\sin \angle AIM}{\sin \angle CIM} = \dfrac{AI}{IC}\cdot \dfrac{\sin\angle A_2ID}{\sin{\angle C_2ID}} = \dfrac{AI}{IC}\cdot \dfrac{\sin\frac {A}{2} } {\sin\frac {C}{2} } = \dfrac{r}{r} = 1$ And we are done...
Virgil Nicula
02.04.2009 23:20
Virgil Nicula wrote: An easy extension. Let $ P$ be an interior point of $ \triangle ABC$ and denote the points $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ so that $ P\in AD\cap BE\cap CF$ . Define the second intersections $ M$ , $ N$ of the lines $ BE$ , $ CF$ respectively with the circumcircle $ w$ of $ \triangle ABC$ and $ K\in MF\cap NE$ , $ L\in KP\cap BC$ . Prove that $ AA\cap MN\cap EF\ne\emptyset$ and $ \frac {LB}{LC} = \left(\frac {DB}{DC}\cdot\frac bc\right)^2$ . Remark. Denote $ T\ \in\ (AP\ \cap\ w$ and $ S\ \in\ TT\ \cap\ BC$ . Then the division $ (\ B\ ,\ C\ ;\ S\ ,\ L\ )$ is harmonically ! Virgil Nicula wrote: Lemma I. Let $ ABC$ be a triangle. Consider the points $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ and $ X\in AD\cap EF$ . Then $ \boxed {\ \frac {XF}{XE} = \frac {DB}{DC}\cdot \frac {AF}{AE}\cdot \frac {AC}{AB}\ }$ .
Denote $ \left\|\begin{array}{c} m(\widehat {DAB}) = x \\
\\
m(\widehat {DAC}) = y\end{array}\right\|$ . Then $ \left\|\begin{array}{c} \frac {XF}{XE} = \frac {AF}{AE}\cdot\frac {\sin x}{\sin y} \\
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\frac {DB}{DC} = \frac {AB}{AC}\cdot\frac {\sin x}{\sin y}\end{array}\right\|\ : \downarrow\ \implies$ $ \frac {XF}{XE} = \frac {DB}{DC}\cdot \frac {AF}{AE}\cdot \frac {AC}{AB}$ .
Lemma II. Letr $ ABC$ be a triangle and $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ be three points so that $ AD\cap BE\cap CF\ne\emptyset$ . Denote $ X\in AD\cap EF$ . Then $ \boxed {\ \frac {XF}{XE} = \frac {BF}{BA}\cdot\frac {CA}{CE}\ }$ .
Apply the upper lemma $ \frac {XF}{XE} = \frac {DB}{DC}\cdot\frac {AC}{AB}\cdot \frac {AF}{AE}$ and the Ceva's theorem : $ \frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB} = 1$ .
Thus, $ \frac {XF}{XE} = \left(\frac {EA}{EC}\cdot\frac {FB}{FA}\right)\cdot\frac {AF}{AE}\cdot\frac {AC}{AB}$ , i.e. $ \frac {XF}{XE} = \frac {BF}{BA}\cdot\frac {CA}{CE}$ .
Lemma III. Let $ ABCD$ be a convex quadrilateral for which denote $ I\in AC\cap BC$ . Consider $ X\in (BC)$ , $ Y\in (AD)$ . Then $ \boxed {\ I\in XY\ \Longleftrightarrow\ \frac {IA}{IC}\cdot\frac {IB}{ID} = \frac {XB}{XC}\cdot\frac {YA}{YD}\ }$ .
Denote $ \left\|\begin{array}{ccc} m(\widehat{XIB}) = \alpha & ; & m(\widehat {XIC}) = \beta \\
\\
m(\widehat {YIA}) = \gamma & ; & m(\widehat {YID}) = \delta\end{array}\right\|$ , where $ \alpha + \beta = \gamma + \delta$ . Using the well-known relations $ \left\|\begin{array}{c} \frac {XB}{XC} = \frac {IB}{IC}\cdot\frac {\sin \alpha}{\sin\beta} \\
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\frac {YA}{YD} = \frac {IA}{ID}\cdot\frac {\sin \gamma}{\sin\delta}\end{array}\right\|$
obtain equivalencies : $ \frac {IA}{IC}\cdot\frac {IB}{ID} = \frac {XB}{XC}\cdot\frac {YA}{YD}$ $ \Longleftrightarrow$ $ \frac {IA}{IC}\cdot\frac {IB}{ID} = \frac {IB}{IC}\cdot\frac {\sin \alpha}{\sin\beta}\cdot\frac {IA}{ID}\cdot\frac {\sin \gamma}{\sin\delta}$ $ \Longleftrightarrow$ $ \sin \alpha\cdot\sin \gamma = \sin\beta\cdot\sin \delta$ $ \Longleftrightarrow$
$ \cos (\alpha - \gamma ) - \cos (\alpha + \gamma ) = \cos (\beta - \delta ) - \cos (\beta + \delta )$ $ \Longleftrightarrow$ $ \left\|\begin{array}{c} \alpha + \gamma = \beta + \delta \\
\\
(\ \alpha + \beta = \gamma + \delta\ )\end{array}\right\|$ $ \Longleftrightarrow$ $ \alpha = \delta\ \wedge\ \beta = \gamma$ $ \Longleftrightarrow$ $ I\in XY$ .
nguyenhaan2209
23.07.2019 07:55
Notice A2C2 is perp bisector of BI hence B2C2 cuts BB at F and FI=FB so FI//BC. Pascal for (BAA2C2CB) so FC1A1 collinear so -1=I(FEC1A1)=I(FDAC) but IF//BC so D is midpoint of BC so q.e.d