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Source: Iran TST 2008

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1275 posts

#1 h May 20, 2008, 9:29 AM • 7 Y

Y by TRYTOSOLVE, itslumi, cadaeibf, Adventure10, Mango247, Mango247, and 1 other user

Suppose that $I$ is incenter of triangle $ABC$ and $l'$ is a line tangent to the incircle. Let $l$ be another line such that intersects $AB,AC,BC$ respectively at $C',B',A'$ . We draw a tangent from $A'$ to the incircle other than $BC$ , and this line intersects with $l'$ at $A_1$ . $B_1,C_1$ are similarly defined. Prove that $AA_1,BB_1,CC_1$ are concurrent.

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316 posts

#2 h May 21, 2008, 2:06 PM • 1 Y

Y by Adventure10

I think that we could reduce the problem to the following:

Given triangle $ABC$ and $(I)$ is its incircle. Let $A', B'$ be points on $BC, CA$ and $d$ is a tangent of $(I)$ . Tangents of $(I)$ from $A', B'$ (different from $BC, CA$ ) intersect $d$ at $A_1, B_1$ respectively. Then $AA_1, BB_1, A'B'$ are concurrent.

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99 posts

#3 h May 21, 2008, 6:06 PM • 2 Y

Y by Adventure10, Mango247

Lets incircle tachs $AB$ , $BC$ , $CA$ in $F$ , $E$ , $D$ respectively.
Points $F'$ , $E'$ , $D'$ lie on incircle. $D' \in A_{1}A'$ , $E' \in B_{1}B'$ , $F' \in C_{1}C'$ .
$X = EF \cap LD'$ $Y = DF \cap LE'$ $Z = DE \cap LF'$
Let $P$ be a pole of line $l$ . Then $P = DD' \cap EE' \cap FF'$
$AA_{1}$ , $BB_{1}$ , $CC_{1}$ are concurent. $\Longleftrightarrow$ $X$ , $Y$ , $Z$ are colinear. ( $X$ is a pole of $AA_{1}$ )
From Pascal teorem for $DEFF'LD'$ we get that $X$ , $Y$ , $P$ are colinear. Similarly $X$ , $Z$ , $P$ are colinear. So $X$ , $Y$ , $Z$ , $P$ are colinear.

Attachments:

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316 posts

#4 h May 22, 2008, 7:23 AM • 2 Y

Y by Adventure10, ehuseyinyigit

My solution is similar to yours.

Let $D, E$ be intersections of $B'B_1, AB$ and $l, AC$ . We easily note that the hexagon $B'EA_1A'BD$ circumscribes the incircle of triangle $ABC$ . It follows from Brianchon's Theorem that $A'B', BE, A_1D$ are concurrent at $K$ . Hence, applying Pappus' Theorem to the 2 triples $(B_1, A_1, E)$ and $(A, B, D)$ , we have the 3 points $B', K$ and the intersection $H$ of $AA_1, BB_1$ are collinear. Hence, $H$ is on $A'B'$ .
This means $AA_1, BB_1, A'B'$ are concurrent.

The problem then follows from this.

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2860 posts

#5 h Mar 18, 2014, 9:44 PM • 2 Y

Y by Adventure10, Mango247

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#6 h Oct 22, 2014, 11:09 AM • 15 Y

Y by navi_09220114, Eray, Fermat_Theorem, AlastorMoody, amar_04, Kamran011, mijail, SenatorPauline, Nymoldin, parola, mathleticguyyy, Adventure10, Mango247, EpicBird08, and 1 other user

My solution:

From Brianchon theorem ( for $A'CAC'C_1A_1$ and $B'ABA'A_1B_1$ )
we get $AA_1, BB_1, CC_1, l$ are concurrent .

Q.E.D

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sayantanchakraborty

505 posts

sayantanchakraborty

#7 h Nov 30, 2014, 12:52 PM • 2 Y

Y by Adventure10, Mango247

Before proving the statement I need a prerequisite lemma:

Lemma:With respect to a given circle $\omega$ ,the polars are concurrent if and only if their respective poles are collinear.

Proof:Consider the intersection points of the polars and use La Hire's theorem.

Now back to our problem.I shall first name the points: $D,E,F$ are the tangency points of the incircle with $BC,CA,AB$ respectively.Let $X$ be the point on $(I)$ such that $B'X$ is tangent to $(I)$ (other than $E$ ), $Y$ be the point where $l'$ touches $(I)$ , $Z$ be the point where $A'A_1$ touches $(I)$ and $L$ be the point where $C'C_1$ touches $(I)$ .In the rest of my proof the poles and polars,etc will be considered with respect to the incircle.

Note that the polars of $A',B',C'$ are $DZ,EX,FL$ and as $A',B',C'$ are collinear,by our lemma $DZ,EX,FL$ concur.Set $YX \cap DF=P,EF \cap YZ=Q,YL \cap DE=R$ .Applying Pascal's theorem on the cyclic hexagon $XYZDFE$ we see that $P,Q,EX \cap DZ$ are collinear.Similarly Pascal's theorem in $XYLFDE$ yeilds that $P,R,XE \cap LF$ are collinear.But $XE \cap DZ=XE \cap LF$ as $DZ,EX,FL$ concur so $P,Q,R$ are collinear.

Now see that $DF$ is the polar of $B$ ,while $XY$ is the polar of $B_1$ .Hence by La Hire's theorem $P=DF \cap XY$ is the pole of $BB_1$ .Similarly $Q$ is the pole of $AA_1$ and $R$ is the pole of $CC_1$ .Finally by our lemma,as $P,Q,R$ are collinear, $AA_1,BB_1,CC_1$ concur,as desired.

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1916 posts

#8 h Mar 6, 2020, 6:23 AM • 7 Y

Y by GeoMetrix, mueller.25, Hexagrammum16, BinomialMoriarty, Bumblebee60, A-Thought-Of-God, Mathematicsislovely

Wonderful Problem!!! I guess there is a solution with Dual of Desargues Involution too, can anyone provide that solution too.

Iran TST 2008 P2 wrote:

Suppose that $I$ is incenter of triangle $ABC$ and $l'$ is a line tangent to the incircle. Let $l$ be another line such that intersects $AB,AC,BC$ respectively at $C',B',A'$ . We draw a tangent from $A'$ to the incircle other than $BC$ , and this line intersects with $l'$ at $A_1$ . $B_1,C_1$ are similarly defined. Prove that $AA_1,BB_1,CC_1$ are concurrent.

Let the Tangency Points made by $\odot(I)$ with $\{BC,C,A,AB\}$ be $\{D,E,F\}$ respectively. Let $\ell'$ be tangent to $\odot(I)$ at a point $X$ andlet $\{A'P,B'Q,C'R\}$ be the other tangents where $\{P,Q,R\}\in\odot(I)$ .

Notice that the Polars of $\{A',B',C'\}$ are $\{PD,QE,RF\}$ respectively WRT $\odot(I)$ and as $\overline{A'-B'-C'}$ . Hence, $PD,QE,RF$ are concurrent. Now by Pascal on $XQPDFR$ we get that $XQ\cap FD,QP\cap FR,PD\cap RX$ are collinear. Now Pascal on $XPDEFR$ we get that $XP\cap EF,PD\cap FR,DE\cap RX$ are collinear. So combining both the Collinearities we get that $XP\cap EF, DE\cap XR,FD\cap XQ$ are collinear. Now the Polar of $XP\cap EF$ is $AA_1$ , the Polar of $XQ\cap FD$ is $BB_1$ and the Polar of $DE\cap RX$ is $CC_1$ . So, $AA_1,BB_1,CC_1$ are concurrent.

This post has been edited 3 times. Last edited by amar_04, Mar 6, 2020, 6:25 AM

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238 posts

#9 h Sep 16, 2020, 12:49 AM

Y by

Suppose that $\ell$ touches the incircle at $P$ .

We will prove that $AA_1,B_1,CC_1,\ell$ are concurrent.

Firstly, we will show that $BB_1,CC_1,\ell$ are concurrent, and similarly, we would have that $\{AA_1,CC_1,\ell\}; \{AA_1,BB_1,\ell\}$ are concurrent, implying the desired result.

Now, we use Moving Points. Fix $A_1,B_1,C_1$ and move $P$ projectively on the incircle. Let $T_A=C'C_1 \cap B'B_1$ .

$\implies I$ is the incenter of $\Delta C_1T_AB_1$ $\implies \angle ZIY=90º+\frac{\angle FT_AE}{2}$ (fixed).

Hence, the composed map $C'C_1 \mapsto \mathcal{C}_I \mapsto \mathcal{C}_I \mapsto B'B_1$ given by $C_1 \mapsto IC_1 \mapsto IB_1 \mapsto B_1$ is projective $(*)$ , since it's a projection followed by a rotation with fixed angle through $I$ , followed by another projection.

Now, define $CC_1 \cap \ell= Q_C, BB_1 \cap \ell= Q_B$ .

Thus, the composed map $C_1C' \mapsto \mathcal{C}_C \mapsto \mathcal{C}_C \mapsto \ell$ given by $A_1 \mapsto CA_1 \mapsto CQ_C \mapsto Q_C$ is projective, since all maps are projections, which are projective maps.

Similarly, $Y \mapsto Q_B$ is projective $\implies$ from $(*)$ , the map $Q_B \mapsto B_1 \mapsto C_1 \mapsto Q_C$ is projective. Hence, since $Q_B,Q_C \in \ell$ , which is a fixed line, by the Moving Points Lemma, in order to prove that $Q_B=Q_C$ , it is sufficient to verify the problem for $3$ choices of $P$ .

Choosing $P$ to be the points such that the incircle touches $AB,BC,CA$ , we get the desired result easily.

Hence, we are done.

$\blacksquare$

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1610 posts

#10 h Oct 23, 2021, 1:58 PM

Y by

We will henceforth work in the projective plane. First of all send $\ell'$ to the line at infinity, so that $A', B', C'$ are the points at infinity for $BC, CA, AB$ respectively. Under this transformation, however the incircle becomes an ellipse. In order to fix this; we take an affine homography converting the ellipse into a circle again. Since affine transforms preserve the line at infinity, $\ell'$ remains the line at infinity. Now considering the projective dual of the problem, it can be seen to be equivalent to the following:

Reduced Problem:
Given a triangle $ABC$ with circumcircle $\gamma$ , an arbitrary point $P$ on $\gamma$ and $A', B', C'$ as the respective $A, B, C$ antipodes with respect to $\gamma$ prove that $A'P \cap BC, B'P \cap CA$ and $C'P \cap AB$ are collinear.

Now this is easy to prove by spamming Pascal's Hexagram Theorem. Using Pascal on $APCA'BC'$ yields that centre of $\gamma$ , $A'P \cap BC$ , $C'P \cap AB$ are collinear. Reiterating this result for other pairs of intersections yields that $A'P \cap BC, B'P \cap CA$ and $C'P \cap AB$ are collinear and thus we are done.

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93 posts

#11 h Nov 7, 2023, 6:47 PM • 3 Y

Y by popop614, dolphinday, ohiorizzler1434

Polar reciprocate the entire diagram to get the equivalent problem statement.

New Problem Statement wrote:

Let $ABC$ be a triangle with point $X$ in the plane and point $P$ on circumcircle $\Gamma$ . Let $XA, XB, XC$ intersect $\Gamma$ at $D, E, F$ . Then let $DY, EY, FY$ intersect $BC, AC, AB$ at $G, H, I$ . Show that $G, H, I$ are collinear.

Now, take a homography in $\mathbb{CP}^2$ that maps $X$ to a point of infinity. Let $L$ be one of the midpoint of arcs $AD, BE, CF$ . Oh wait, this is just USAMO 2012/5 with $\gamma = PL$ .

This post has been edited 1 time. Last edited by TheHazard, Nov 7, 2023, 6:52 PM

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1747 posts

OronSH

#12 h Aug 9, 2024, 1:52 AM • 6 Y

Y by centslordm, ihatemath123, GeoKing, ohiorizzler1434, ehuseyinyigit, EpicBird08

brianchon on $ABDXYE,ACDXZF$ gives $AX,BY,CZ,DEF$ concurrent done

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#13 h Aug 26, 2024, 5:35 PM

Y by

We apply Brianchon on $A_1A'BAB'B_1$ so we obtained that $AA_1,BB_1,l$ are concurrent.
Analogous $BB_1,CC_1,l$ are concurrent so $AA_1,BB_1,CC_1$ are concurrent. So the problem is proved.

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cursed_tangent1434

636 posts

cursed_tangent1434

#14 h Dec 4, 2024, 1:53 PM

Y by

Denote by $\omega$ the incircle of $\triangle ABC$ . We claim that in particular $AA_1,BB_1,CC_1$ are concurrent on line $\ell$ .

Now, note that each side of $AC'C_1A_1A'C$ and $BC'C_1B_1B'C$ are tangent to $\omega$ . Thus both these hexagons are tangential, which implies that by Brianchon's Theorem, $AA_1$ and $CC_1$ intersect on $\ell$ and also $BB_1$ and $CC_1$ intersect on $\ell$ , which finishes the problem.

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658 posts

#15 h May 16, 2025, 1:29 PM

Y by

Brianchon applied on $AFZXDB$ implies that $AX\cap FD\cap BX$ are concurrent. Symmetry finishes.

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