lol at the description of the problem ...

Yes that was basically my solution, not_trig.

It's pretty simple, but does anyone know if the concave case will mess up any solutions?

I assumed AB||Q_1Q_2, then translate Q_1AD over so that Q_1 goes to Q_2. After that you angle chase.

But what if $Q_1Q_2$ is not parallel to either $AB$ or $CD$? What if it intersects them? That was the problem I struggled with for about 2 hours.

tenniskidperson3 wrote: But what if $Q_1Q_2$ is not parallel to either $AB$ or $CD$? What if it intersects them? That was the problem I struggled with for about 2 hours. There's no need to consider the intersection of Q_1Q_2, just show that if AB is not parallel to CD then AB, CD, and Q1Q2 concur, solving the problem.

This problem ate up my time like #1 did... but all geometry problems eat up my time because I take 2 hours to make a single trivial observation. I didn't even think of isogonal conjugates. However, I did use the same method (showing that $\angle Q_1XB=\angle Q_2XA=\angle PXD$, but I wasn't able to prove that entirely by angle chase. However, since it's actually evident by isogonal conjugates, I think I'll get some credit on it.

tenniskidperson3 wrote: But what if $Q_1Q_2$ is not parallel to either $AB$ or $CD$? What if it intersects them? That was the problem I struggled with for about 2 hours. ummm. if Q_1Q_2 isn't parallel to either AB or CD, then you don't have to do anything... the question doesn't ask to prove anything in this case.

Never mind... I tried proving that either $AB||CD||Q_1Q_2$, or they are all concurrent. So there was no reason for that first part?

My first line: "If $AB \parallel CD$, then there is nothing to prove."

Hmm obviously I wasn't thinking straight because this is what I decided to come up with:

As far as I can see, the concave case shouldn't screw up the isogonal conjugates solution if you stayed with collinearity (or if you were very careful with saying angles are equal.

hrm you could also try proving "If AB intersects CD then Q_1Q_2 intersects at that point also." darn you could probably trig bash this.

connaissance wrote: hrm you could also try proving "If AB intersects CD then Q_1Q_2 intersects at that point also." I drew a diagram that disproves this.

1=2 wrote: connaissance wrote: hrm you could also try proving "If AB intersects CD then Q_1Q_2 intersects at that point also." I drew a diagram that disproves this. Geogebra says otherwise: [geogebra]37186a7099503831882ed00b754cf1fdd46ca0aa[/geogebra]

1=2 wrote: connaissance wrote: hrm you could also try proving "If AB intersects CD then Q_1Q_2 intersects at that point also." I drew a diagram that disproves this. However it's true, lol. EDIT: zzz real ninja you

1=2 wrote: connaissance wrote: hrm you could also try proving "If AB intersects CD then Q_1Q_2 intersects at that point also." I drew a diagram that disproves this. a lot of people i know (or have heard of, but don't know well) can draw such diagrams. ex1. a certain mall

I think 1=2 was just trying to emphasize his bad diagram-drawing skills, since if you draw a misleading diagram you are going to get pwned.

uuuhh... yeah... that's totally what I did, Yongyi... right. OK I seriously need to become much better at math if I'm to prove Zuming's IMO problem conjecture wrong (2009 AMP Counting Strategies reference)

For the case $AB \parallel CD$ the problem is trivial so assume otherwise. Let $X = AB$ $\cap$ $CD$. By the angle conditions given, $P$ is the isogonal conjugate of $Q_1$ with respect to $\triangle XBC$ and that of $Q_2$ with respect to $\triangle XAD$. Thus, $X, Q_1, Q_2$ are collinear since they lie on the reflection of $XP$ over the angle bisector of $\angle AXD$. Since lines $Q_1Q_2, AB, CD$ concur neither $Q_1Q_2 \parallel AB$ nor $Q_1Q_2 \parallel CD$ hold.

Now if $AB||CD$, the statement is obvious, so assume otherwise. Let $AB$ and $CD$ intersect at $X$. Now note that by the angle condition, we have $P$ and $Q_2$ are isogonal conjugates wrt $XAD$ and $P$ and $Q_1$ are isogonal conjugates wrt $XBC$, so $X,Q_1,Q_2$ are collinear, so we are done.

Case 1: $AB \parallel CD$ This case is obvious as $Q_1Q_2$ cannot be parallel to only one of $AB$ and $CD$ Case 2: $AB \cap CD = E$ Claim: $Q_1-Q_2-E$ Proof: Note that $Q_1$ is the isogonal conjugate of $P$ in triangle $EBC$ Also $Q_2$ is the isogonal conjugate of P in triangle $EAD$ Thus both $Q_1$ and $Q_2$ must lie on the line isogonal to EP Hence $Q_1-Q_2-E$ holds and $Q_1Q_2$ intersects both $AB$ and $CD$ at $E$.

We see if $\overline{AB}\parallel\overline{CD},$ the problem is solved, so we assume otherwise, letting $E=\overline{AB}\cap\overline{CD}.$ Notice $Q_1$ and $Q_2$ are the isogonal conjugates of $P$ with respect to $\triangle EBC$ and $\triangle EAD,$ respectively. Hence, $$\angle Q_1EC=\angle BEP=\angle Q_2ED$$and $\overline{AB},\overline{CD},$ and $\overline{Q_1Q_2}$ are concurrent. Hence, if $\overline{AB}\nparallel\overline{CD},$ it is absurd for $\overline{Q_1Q_2}$ to be parallel to only one of $\overline{AB}$ or $\overline{CD}.$ $\square$

We will solve this problem using casework when $AB || CD$ and when $AB \not{||} CD$. Case 1: $AB || CD$ Note that $Q_1Q_2$ must be either parallel to both $AB, CD$ or neither. So this matches the problem conditions and we are done with this case. Case 2: $AB \not{||} CD$ Let $AB$ intersect $CD$ at $E$. So by the angle conditions in the problem, we know that $P$ is the isogonal conjugate of $Q_1$ with respect to $\triangle{EBC}$. Similarly we know that $P$ is the isogonal conjugate of $Q_2$ with respect to $\triangle{EAD}$. So we see that $$\angle{Q_1ED} = \angle{PEA} = \angle{Q_2ED}.$$This means that $Q_1Q_2, AB, CD$ all concur which means that $Q_1Q_2$ can't be parallel to $AB$ or $CD$ and this satisfies the problems conditions.

Let $M$ be the intersection of $AB$ and $CD$. [asy][asy] import graph; size(8 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(7); defaultpen(dps); pen ds=black; real xmin=-4.5,xmax=6.5,ymin=-0.8,ymax=5.; pen evevfz=rgb(0.8980392156862745,0.8980392156862745,0.9764705882352941), qqqqcc=rgb(0.,0.,0.8); pair A=(-1.,0.), B=(6.,0.), C=(4.,4.), D=(0.,2.), P=(2.,1.), Q_1=(3.91304347826087,2.43478260869565), Q_2=(0.3333333333333333,1.3333333333333333), X=(-4.,0.); filldraw(A--B--C--D--cycle,evevfz,linewidth(0.8)+qqqqcc); draw(A--B,linewidth(0.8)+qqqqcc); draw(B--C,linewidth(0.8)+qqqqcc); draw(C--D,linewidth(0.8)+qqqqcc); draw(D--A,linewidth(0.8)+qqqqcc); draw(P--A,linewidth(0.8)); draw(P--B,linewidth(0.8)); draw(P--C,linewidth(0.8)); draw(P--D,linewidth(0.8)); draw(B--Q_1,linewidth(0.8)); draw(C--Q_1,linewidth(0.8)); draw(Q_2--A,linewidth(0.8)); draw(Q_2--D,linewidth(0.8)); draw(Q_1--Q_2,linewidth(0.8)); draw(C--X,linewidth(0.8)); draw(X--A,linewidth(0.8)); draw(X--Q_1,linewidth(0.8)); draw(X--P,linewidth(0.8)); dot(A,linewidth(3.pt)+ds); label("$A$",(-1.2304077862934957,-0.39306836788223526),NE*lsf); dot(B,linewidth(3.pt)+ds); label("$B$",(5.999939557072708,-0.6367879412541299),NE*lsf); dot(C,linewidth(3.pt)+ds); label("$C$",(4.090802898992868,4.115743739497815),NE*lsf); dot(D,linewidth(3.pt)+ds); label("$D$",(0.08973990280426616,2.1253672236273426),NE*lsf); dot(P,linewidth(3.pt)+ds); label("$P$",(1.9379466675411328,0.5818099256053433),NE*lsf); dot(Q_1,linewidth(3.pt)+ds); label("$Q_1$",(4.1314228278881835,2.470636619237527),NE*lsf); dot(Q_2,linewidth(3.pt)+ds); label("$Q_2$",(0.4959391917574237,1.0286291434538168),NE*lsf); dot(X,linewidth(3.pt)+ds); label("$M$",(-4.114422737860914,-0.3727584034345774),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Claim: $Q_2$ lies on $MQ_1$. Proof: Since $Q_1$ and $Q_2$ are isogonal conjugates of $\Delta MCB$ and $\Delta MDA$, we have \begin{align*} \angle PMB=\angle CMQ_1\\ \angle PMA=\angle DMQ_2=\angle CMQ_2 \end{align*}Since $\angle PMB=\angle PMA$, we have $\angle CMQ_1=\angle CMQ_2$. As $Q_1$ and $Q_2$ are located within quadrilateral $ABCD$, we have $Q_2\in MQ_1$. $\blacksquare$ Since $AB$, $CD$ and $Q_1Q_2$ concur at $M$, then $Q_1Q_2\parallel AB$, if and only if $CD\parallel Q_1Q_2$, as no two of them to be parallel at one time. [asy][asy] import graph; size(8cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(7); defaultpen(dps); pen ds=black; real xmin=-2.,xmax=6.5,ymin=-0.8,ymax=5.; pen evevfz=rgb(0.8980392156862745,0.8980392156862745,0.9764705882352941), qqqqcc=rgb(0.,0.,0.8); pair A=(-1.,0.), B=(6.,0.), C=(4.,4.), D=(0.,4.), P=(2.,1.), Q_1=(3.42857142857143,3.), Q_2=(0.9090909090909097,3.); filldraw(A--B--C--D--cycle,evevfz,linewidth(0.8)+qqqqcc); draw(A--B,linewidth(0.8)+qqqqcc); draw(B--C,linewidth(0.8)+qqqqcc); draw(C--D,linewidth(0.8)+qqqqcc); draw(D--A,linewidth(0.8)+qqqqcc); draw(P--A,linewidth(0.8)); draw(P--B,linewidth(0.8)); draw(P--C,linewidth(0.8)); draw(P--D,linewidth(0.8)); draw(B--Q_1,linewidth(0.8)); draw(C--Q_1,linewidth(0.8)); draw(Q_2--A,linewidth(0.8)); draw(Q_2--D,linewidth(0.8)); draw(Q_1--Q_2,linewidth(0.8)); dot(A,linewidth(3.pt)+ds); label("$A$",(-1.2304077862934957,-0.39306836788223526),NE*lsf); dot(B,linewidth(3.pt)+ds); label("$B$",(5.999939557072708,-0.6367879412541299),NE*lsf); dot(C,linewidth(3.pt)+ds); label("$C$",(4.090802898992868,4.115743739497815),NE*lsf); dot(D,linewidth(3.pt)+ds); label("$D$",(0.08973990280426616,4.115743739497815),NE*lsf); dot(P,linewidth(3.pt)+ds); label("$P$",(1.9379466675411328,0.5818099256053433),NE*lsf); dot(Q_1,linewidth(3.pt)+ds); label("$Q_1$",(3.6439836811443946,3.0393156237719476),NE*lsf); dot(Q_2,linewidth(3.pt)+ds); label("$Q_2$",(1.0646181962918442,2.6940462281617634),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

Suppose FTSOC that $Q_1Q_2$//AB and AB intersects CD at $E\ne P_{\infty}$. Notice $Q_1,Q_2$ are isogonal conjuates of P wrt BCX,ADX, respectively, so $Q_1Q_2$ must be the isogonal line of XP, which implies X is on $Q_1Q_2$, meaning X is $P_{\infty}$, contradiction! The other way for $Q_1Q_2$//CD follows immediately.

Isogonal conjugates go brrrrr. Assume $T = AB \cap CD$, and we are given that $CD \parallel Q_1Q_2$. Then noting that $P$ and $Q_2$ are isogonal conjugates in $\triangle ATD$, we must have $\angle PTD = \angle Q_2TA$. Similarly we find that $Q_1$ and $P$ are isogonal conjugates in $\triangle BCT$, hence $\angle PTC = \angle Q_1TB$. However noting $\angle PTD = \angle PTC$ we find, \begin{align*} \angle Q_2TA = \angle Q_1TB \end{align*}Clearly then we have $Q_1$, $Q_2$ and $T$ collinear, implying that $T = Q_1Q_2 \cap CD$, which is the point at infinity, hence $AB \parallel CD$.

Suppose $\overline{Q_1Q_2}\parallel\overline{AB}$. I will show that $\overline{AB}\parallel \overline{CD}$. Assume for contradiction $AB$ and $CD$ intersect at a point $X$. Note that $Q_1, Q_2$ are the isogonal conjugates of $P$ wrt $\triangle BCX$ and $\triangle ADX$, respectively. Hence, $XQ_1$ is isogonal to $XP$ wrt $\angle AXB$, and similarly $XQ_2$ is isogonal to $XP$ wrt $\angle AXB$. Hence $XQ_1$ and $XQ_2$ are the same line. Hence, lines $Q_1Q_2$ and $AB$ both pass through $X$, contradicting $\overline{Q_1Q_2}\parallel\overline{AB}$. Thus we must have $\overline{AB}\parallel \overline{CD}$, which implies $\overline{Q_1Q_2}\parallel \overline{CD}$ as well. The proof above can be repeated by assuming $\overline{Q_1Q_2}\parallel\overline{CD}$ initially, and we get that this implies $\overline{Q_1Q_2}\parallel\overline{AB}$. Thus $$\overline{Q_1Q_2}\parallel\overline{CD}\iff \overline{Q_1Q_2}\parallel\overline{AB},$$as desired.

why did it take two people to propose this :skull: Let $X$ be the intersection of lines $AB$ and $CD$, for the sake of contradiction. Then, by definition, $Q_1$ and $Q_2$ are the isogonal conjugates of $P$ in $\triangle XBC$ and $\triangle XAD$ respectively, so $X$, $Q_1$ and $Q_2$ are collinear. But then, lines $Q_1 Q_2$ and $AB$ intersect at $X$, giving us a contradiction.