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Source: Silk Road Mathematics Competition 2009

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Ulanbek_Kyzylorda KTL

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Ulanbek_Kyzylorda KTL

#1 h Apr 1, 2009, 12:24 PM • 2 Y

Y by Adventure10, Mango247

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.

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Ahiles

#2 h Apr 1, 2009, 7:21 PM • 2 Y

Y by Adventure10, Mango247

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...

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#3 h Apr 2, 2009, 8:20 PM • 2 Y

Y by Adventure10, Mango247

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 $\boxed {\ \frac {XF}{XE} = \frac {DB}{DC}\cdot \frac {AF}{AE}\cdot \frac {AC}{AB}\ }$ .
Proof.

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 $\boxed {\ \frac {XF}{XE} = \frac {BF}{BA}\cdot\frac {CA}{CE}\ }$ .
Proof.

Lemma III. Let $ABCD$ be a convex quadrilateral for which denote $I\in AC\cap BC$ .

Consider $X\in (BC)$ , $Y\in (AD)$ . Then $\boxed {\ I\in XY\ \Longleftrightarrow\ \frac {IA}{IC}\cdot\frac {IB}{ID} = \frac {XB}{XC}\cdot\frac {YA}{YD}\ }$ .
Proof.

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#4 h Jul 23, 2019, 4:55 AM • 3 Y

Y by top1csp2020, Adventure10, Mango247

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

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