matinyousefi wrote: In a trapezoid $ABCD$ with $AB$ parallel to $BC$ points $E, F$ are on sides $AB, CD$ respectively. $A_1, C_1$ are on $AD,BC$ such that $A_1, E, F, A$ lie on a circle and so do $C_1, E, F, C$. Prove that lines $A_1C_1, BD, EF$ are concurrent. Do you mean $B$ is at $P_{\infty}$? How can be $AB\| BC$ P.S: A typo in topic name too! It should be Segment

Thanks. Fixed.

We seek to show that $\triangle A_1DF, \triangle C_1BE$ are perspective. By Desargues' Theorem, it suffices to prove that $DF \cap BE, A_1F \cap C_1E, A_1D \cap C_1B$ are collinear. As $A_1D \cap C_1B$ is the point at infinity along $AD$, we just need to show that $X = AB \cap CD$ and $Y = C_1E \cap A_1F$ satisfy $XY || AD.$ Note that $\angle YFE = \angle A_1FE = \angle XAD = \angle ABC$ and $\angle YEF = \angle FCB$, from the conditions of the problem. This means that $\angle EYF = 180 - \angle YFE - \angle YEF = \angle BXC.$ As $\angle EXF = \angle EYF$, we've $EXYF$ is cyclic. This implies that $\angle YXF = \angle YEF = \angle ECC_1$, and so $XY || BC$ as desired. $\square$

We restate the problem in terms of the quadrilateral $AEFC_1$ as follows:- Restated Problem wrote: Let $\omega_1$ and $\omega_2$ be two circles meeting at $E,F$. Let $A$ be a point on $\omega_1$ and $C_1$ be a point on $\omega_2$. Let $B$ be a variable point on line $AE$, and suppose $BC_1$ meets $\omega_2$ again at $C$. Take $D \in CF$ such that $AD \parallel C_1C$, and denote $A_1=AD \cap \omega_1$ $(A_1 \neq D)$. Then show that lines $A_1C_1,BD,EF$ are concurrent. Animate $B$ linearly on line $AE$. Note that $$\measuredangle ADF=\measuredangle ADC=\measuredangle C_1CD=\measuredangle C_1CF=\measuredangle C_1EF$$which is constant, as $B$ moves. Thus, $D$ moves on a fixed circle $\gamma$ through $A$ and $F$. Now, let $\infty$ be the common point at infinity on lines $BC_1$ and $AD$. Then we have that $$B \mapsto C_1B \mapsto \infty \mapsto A\infty \cap \omega_1=A_1$$is a sequence of projective maps. Also, perspectivity at $A$ from the line at infinity to $\gamma$ gives the projective map $\infty \mapsto D$. Since points $D$ and $A_1$ move on circles, both projectively, so we must have $\text{deg}(D)=\text{deg}(A_1)=2$. Also, as $\text{deg}(B)=1$, so we get $\text{deg}(BD)=3$. Thus, to prove that $A_1C_1,BD,EF$ are concurrent, it suffices to check $(2+3+0)+1=6$ positions of point $B$. That's not that hard, as shown below: Take $B=E$. Then $C=E$, and so $D \in EF$. Thus, lines $BD$ and $EF$ coincide. Take $B=A$. Then both $A_1$ and $D$ lie on line $AC_1$. Thus, in this case, lines $BD$ and $A_1C_1$ coincide. Consider $B=AE \cap \omega_2$. Then $C=B$ gives $D \in BF$. Thus, lines $BD$ and $EF$ meet at $F$. But, by Reim's Theorem, since $AA_1 \parallel CC_1$, and $C \in AE$, so we must have $F,A_1,C_1$ collinear. This gives that lines $BD,A_1C_1,EF$ meet at $F$. When $B=C_1 \infty_{AF} \cap AE$, then $D=F$, as well as $A_1=F$. So lines $BD,A_1C_1,EF$ meet at $F$. When $B=C_1(AF \cap \omega_2) \cap AE$, then $C=AF \cap \omega_2$. This gives $D \in AF$, or equivalently, $D=A$. Thus, lines $BD$ and $EF$ meet at $E$. But, by Reim's Theorem, since $AA_1 \parallel CC_1$, and $C \in AF$, so we must have $E,A_1,C_1$ collinear. This gives that lines $BD,A_1C_1,EF$ meet at $E$. Finally consider $B=\infty_{AE}$. This translates to $\infty=\infty{AE}$, and so $D \in AE$. Thus, lines $BD$ and $EF$ meet at $E$. But, we also have $A_1=E$, and so lines $BD,A_1C_1,EF$ meet at $E$. REMARK: After trying a number of setups, I was able to reduce the problem condition to a degree $6$ statement. I would love to see a proof which reduces the degree to say $5$ or $4$ (restatements might help)

RMM Shortlist 2017 G1

Let $EF $ cuts $BC $, $AD $ at $H $, $I $. $\frac {HC_1}{HB}=\frac {HE\cdot HF}{HB\cdot HC}$ and $\frac {IA_1}{ID}=\frac {IE\cdot IF}{IA\cdot ID}$. Thales theorem $\implies\frac{HE}{HB}=\frac {IE}{IA}$, $\frac {HF}{HC}=\frac {IF}{ID}\implies\frac {HC_1}{HB}=\frac {IA_1}{ID} $ Let $BD $ cuts $EF $ at $G $. By Thales theorem $\frac {DG}{GB}=\frac {ID}{HB}=\frac {IA_1}{HC_1}\implies\overline {G,A_1,C_1} $

Pathological wrote: We seek to show that $\triangle A_1DF, \triangle C_1BE$ are perspective. By Desargues' Theorem, it suffices to prove that $DF \cap BE, A_1F \cap C_1E, A_1D \cap C_1B$ are collinear. As $A_1D \cap C_1B$ is the point at infinity along $AD$, we just need to show that $X = AB \cap CD$ and $Y = C_1E \cap A_1F$ satisfy $XY || AD.$ Note that $\angle YFE = \angle A_1FE = \angle XAD = \angle ABC$ and $\angle YEF = \angle FCB$, from the conditions of the problem. This means that $\angle EYF = 180 - \angle YFE - \angle YEF = \angle BXC.$ As $\angle EXF = \angle EYF$, we've $EXYF$ is cyclic. This implies that $\angle YXF = \angle YEF = \angle ECC_1$, and so $XY || BC$ as desired. $\square$ I think it should be 180°-∠YXF=∠YEF=∠ECC1 typo

Here's a completely different solution. Iranian RMM TST Day 2 P4 wrote: In a trapezoid $ABCD$ with $AD$ parallel to $BC$ points $E, F$ are on sides $AB, CD$ respectively. $A_1, C_1$ are on $AD,BC$ such that $A_1, E, F, A$ lie on a circle and so do $C_1, E, F, C$. Prove that lines $A_1C_1, BD, EF$ are concurrent. For sake of Simplicity Rephrase the Problem as follows. Rephrased Problem wrote: Two Circles $\omega_1,\omega_2$ intersects at $E,F$ respectively. Let $\{A,A_1\}$ be any two points on $\omega_2$ and $\{C,C_1\}$ be any two points on $\omega_1$ such that $AA_1\|EF\|CC_1$ such that $\{A_1,A,E,C_1,C,F\}$ are in this order and if $CF\cap AA_1=D$ and $AE\cap CC_1=B$. Then $A_!C_1,BD,EF$ concurs. Key Claim:- $\{A_1,F,C,C_1,E,A\}$ lies on a unique conic $\mathcal C$. Proof:- Notice that $\{A_1F\cap AE,FC\cap EC_1,AC\cap A_1C_1\}$ all lie on the Perpendicular bisector of $EF$ as $CC_1EF,AA_1FE,AA_1CC_1$ are Isosceles Trapezoids. So, By Converse of Pascal's Theorem on $A_1FCAEC_1$ we get that $\{A_1,A,E,F,C,C_1\}$ lies on a unique conic $\mathcal C$. To Finish off apply Pascal on $AEFCC_1A_1$ to get that $A_1C_1,BD,EF$ are concurrent. $\blacksquare$