Let $P$ be a given point inside quadrilateral $ABCD$. Points $Q_1$ and $Q_2$ are located within $ABCD$ such that \[\angle Q_1BC=\angle ABP,\quad\angle Q_1CB=\angle DCP,\quad\angle Q_2AD=\angle BAP,\quad\angle Q_2DA=\angle CDP.\] Prove that $\overline{Q_1Q_2}\parallel\overline{AB}$ if and only if $\overline{Q_1Q_2}\parallel\overline{CD}$.
Problem
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Tags: AMC, USA(J)MO, USAMO, geometry, trigonometry, isogonal conjugates
29.04.2011 01:26
lol at the description of the problem ...
29.04.2011 01:29
Yes that was basically my solution, not_trig.
29.04.2011 01:29
It's pretty simple, but does anyone know if the concave case will mess up any solutions?
29.04.2011 01:31
I assumed AB||Q_1Q_2, then translate Q_1AD over so that Q_1 goes to Q_2. After that you angle chase.
29.04.2011 01:33
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.
29.04.2011 01:33
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.
29.04.2011 01:34
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.
29.04.2011 01:36
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.
29.04.2011 01:37
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?
29.04.2011 01:38
My first line: "If $AB \parallel CD$, then there is nothing to prove."
29.04.2011 02:13
Hmm obviously I wasn't thinking straight because this is what I decided to come up with:
29.04.2011 03:19
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.
29.04.2011 03:44
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.
29.04.2011 03:47
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.
29.04.2011 03:54
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]
29.04.2011 03:54
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
29.04.2011 03:55
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
29.04.2011 03:55
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.
29.04.2011 04:42
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)
01.08.2021 01:12
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.
10.08.2021 03:25
04.09.2021 18:43
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.
17.09.2021 09:39
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 $E$.
05.03.2022 22:25
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 $\overline{CD}.$ $\square$
11.07.2022 22:40
19.07.2022 07:03
Attachments:

21.07.2022 19:58
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.
27.07.2022 18:01
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]
07.03.2023 19:52
12.08.2023 02:06
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.
10.12.2023 05:32
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 $AB \parallel CD$.
13.02.2024 20:07
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.
18.05.2024 18:44
28.11.2024 02:27
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.