Three collinear points

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Three collinear points

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Source: Iranian third round geometry problem 4

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824 posts

andria

#1 h Sep 10, 2015, 12:27 PM • 3 Y

Y by Ali3085, Adventure10, Mango247

Let $ABC$ be a triangle with incenter $I$ . Let $K$ be the midpoint of $AI$ and $BI\cap \odot(\triangle ABC)=M,CI\cap \odot(\triangle ABC)=N$ . points $P,Q$ lie on $AM,AN$ respectively such that $\angle ABK=\angle PBC,\angle ACK=\angle QCB$ . Prove that $P,Q,I$ are collinear.

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#2 h Sep 10, 2015, 10:56 PM • 9 Y

Y by andria, ATimo, ali.agh, A_Math_Lover, Mathuzb, amar_04, hakN, Adventure10, Mango247

Let $L \equiv AI \cap BC,$ $D \equiv MN \cap BC,$ $Y \equiv AM \cap BC$ and $U \equiv BP \cap AI.$ As $MN$ is the perpendicular bisector of $AI$ (well-known), then $K \in MN.$ Since $BU,BL$ are the reflections of $BK,BA$ across $BI,$ we deduce that $(L,I,A,K)=(A,K,L,I)=(L,U,A,I)=B(L,U,A,I)=(Y,P,A,M)$ $\Longrightarrow$ $D(L,I,A,K)=D(Y,P,A,M)$ $\Longrightarrow$ $P \in DI.$ By similar reasoning $Q \in DI$ $\Longrightarrow$ $P,Q,I$ are collinear.

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2327 posts

#3 h Sep 11, 2015, 3:15 AM • 2 Y

Y by Adventure10, Mango247

I have seen general problem

Let $ABC$ be a triangle inscribed in circle $(O)$ and $P,Q$ are two isogonal conjugate points. $PB,PC$ cut $(O)$ again at $M,N$ . $QA$ cuts $MN$ at $K$ . $L$ is isogonal conjugate of $K$ . $LB,LC$ cut $AM,AN$ at $S,T$ , resp. Prove that $S,Q,T$ are collinear.

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4147 posts

#4 h Sep 11, 2015, 3:38 AM • 3 Y

Y by buratinogigle, Adventure10, Mango247

buratinogigle wrote:

I have seen general problem

Let $ABC$ be a triangle inscribed in circle $(O)$ and $P,Q$ are two isogonal conjugate points. $PB,PC$ cut $(O)$ again at $M,N.$ $QA$ cuts $MN$ at $K.$ $L$ is isogonal conjugate of $K.$ $LB,LC$ cut $AM,AN$ at $S,T,$ resp. Prove that $S,Q,T$ are collinear.

The proof to this generalization is very similar to what I did in my previous solution. Letting $D \equiv MN \cap BC,$ $Y \equiv AM \cap BC,$ $U \equiv AQ \cap BC$ and $V \equiv AP \cap BC,$ we get $B(Y,S,A,M)=B(V,L,A,P)=(A,K,U,Q)=(U,Q,A,K)$ $\Longrightarrow$ $D(Y,S,A,M)=D(U,Q,A,K)$ $\Longrightarrow$ $S \in DQ$ and likewise $T \in DQ$ $\Longrightarrow$ $S,Q,T$ are collinear.

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532 posts

K.N

#5 h Aug 6, 2016, 12:11 PM • 2 Y

Y by Adventure10, Mango247

Hello
Please give some other solutions to see different ideas
I have got that $BP,CQ$ interest each other on $AI$ At a point like $S$ and we can easily get that $(I_{a}ISK)=-1$
Can this be useful?!

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328 posts

#8 h Aug 5, 2018, 10:38 AM • 2 Y

Y by Adventure10, Mango247

we'll solve the problem using bary coordinates. let $\triangle ABC$ be the refrence triangle and let $k'=(QC,BP)$ and literally $K,K'$ are isogonal conjugates and we have that: $K=(2a+b+c:b:c) \implies K'=(\frac{a^2}{2a+b+c}:b:c)$ and now we can easily calculate $Q$ from cevians $CK',AN$ and already know that: $N=(a:b:\frac{-c^2}{a+b}), M=(a:\frac{-b^2}{a+c}:c) \implies Q=(\frac{a^2}{2a+b+c}:b:\frac{-c^2}{a+b}), P=(\frac{a^2}{2a+b+c}:\frac{-b^2}{a+c}:c)$ and it's pretty much straightforward to show

$\begin{vmatrix} \frac{a^2}{2a+b+c} & b &\frac{-c^2}{a+b} \\ a & b &c \\ \frac{a^2}{2a+b+c} & \frac{-b^2}{a+c} & c \\ \end{vmatrix}=0$

This post has been edited 1 time. Last edited by H.HAFEZI2000, Aug 5, 2018, 10:39 AM

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#9 h Oct 2, 2019, 7:40 PM • 2 Y

Y by Adventure10, Mango247

Iran MO Round 3 2015 G4 wrote:

Let $ABC$ be a triangle with incenter $I$ . Let $K$ be the midpoint of $AI$ and $BI\cap \odot(\triangle ABC)=M,CI\cap \odot(\triangle ABC)=N$ . points $P,Q$ lie on $AM,AN$ respectively such that $\angle ABK=\angle PBC,\angle ACK=\angle QCB$ . Prove that $P,Q,I$ are collinear.

Solution (with Aryan-23): Let $MN \cap BC=S$ , $AN \cap BC=E$ & $AI$ $\cap$ $BC$ $=$ $U$ . Let $K'$ be the isogonal conjugate of $K$ WRT $\Delta ABC$
$(A,Q;N,E) \overset{C}{=} (A,K';I,U) =(U,K;I,A)=(A,I;K,U)$ Hence, $S,P,I$ collinear and similarly, $Q,I,S$ collinear $\qquad \blacksquare$

This post has been edited 2 times. Last edited by Luis González, Oct 6, 2019, 2:47 AM
Reason: Unhiding solution

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Arefe

#11 h Aug 1, 2020, 4:34 PM • 1 Y

Y by AlastorMoody

Let $B' , C'$ be points on $BK , CK$ such that $AB'||BI$ and $AC'||CI$ , so $CK=CK'$ and $BK=BK'$ then $BC'||CB'$ and we can get $\angle{ABC'}+\angle{ACB'}=\angle{BAC}$ $(1)$
It's easy to check that $P , B'$ are isogonal conjucate and $Q , C'$ are , too .
If $S$ is intersection of $QB , CP$ , we can check easily from $(1)$ that $S$ is on the circle of $ABC$ .
So with pascal on $NCSBMA$ we can get $I , P , Q$ are collinear $\blacksquare$

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214 posts

#12 h Oct 8, 2020, 5:34 PM • 1 Y

Y by Stuffybear

isogonality lemma kills it

Let $MN \cap BC =R$
note that $(CK,CQ) , (CA,CR)$ are pairs of isogonal line wrt $\angle BCA$
since $N=AQ \cap KR$ we have that $RQ \cap AK$ is isogonal to $AN$ so $RQ \cap AK ,A,N,I$ are collinear
so $I=RQ \cap AK$
similarly $I=PQ \cap AK$

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