Myth has already solved some of them, so I guess that if there's any harm to be done here then it's already done . Assume we have shown that $A,B_1,C_1$ are collinear. Then the result would follow, since $\angle B_1C_1P=\angle AC_1P=\angle ABP=\angle B_1QC=\pi-\angle B_1QP$. Now, in order to show that $A,B_1,C_1$ are collinear, we can work in reverse: given $B_1,C_1$ on the circles s.t. $B_1,A,C_1$ are collinear, we construct the paralleles through $C_1$ to $AC$ and through $B_1$ to $AB$ so that they cut $AB,AC$ respectively. Let these intersect at $Q$. From a quick angle chase we find $BC_1\|CB_1$, so the hexagon $ABC_1QB_1C$ is isncribed in a conic. Since this conic passes through $A,B_1,C_1$ which are collinear, and also through $B,C$, it must be the union of $B_1,C_1,BC$, so $Q\in BC$. Now, if we make $B_1C_1$ the tangent to $(APB)$, then $Q=B$, and if we make it the tangent to $(APC)$, then $Q=C$, so $Q$ will go along the entire edge $BC$.

My intuition tells me this problem can be solved by inversion. I wish I had a better handle on this topic (one that seems strange, elusive, and interesting all at the same time).

So the following is not true? Let $\mathcal C_1,\mathcal C_2$ be two circles intersecting in $U,V$. The tangent to $\mathcal C_1$ in $U$ cuts $\mathcal C_2$ again in $S$. Let $M,N$ be points on $\mathcal C_1,\mathcal C_2$ respectively, collinear with $V$. Then $UM\|NS$. I really think it is, and I also have a proof, but I'll let you check it (and check the fact that it's equivalent to what you have quoted) .

Sorry, my ABC was not acute enough. Maj. Pestich

Er...how much would you say the showing collinearity part is worth?

SJS wrote: My intuition tells me this problem can be solved by inversion. Be sure, it may be solved by inversion (in point $P$), we get an easy angle-counting problem.

Construct $B_2$ on the circumcircle of $APB$ so that $B_1$, $A$, and $B_2$ are collinear, and take $R$ the intersection of lines $QC_1$ and $PB_2$. Observe that $RPQB_1$ is cyclic and that $\angle{PRB_1} \cong \angle{PB_2B_1}$. It must be that $R = B_2 = C_1$! (Thanks to Darij for pointing out the [now corrected] typo in the cyclic quadrilateral.)

Consider $M$ the second intersection of the two circles $APBC_1$ and $AQCB_1$. Let $PC_1$ intersect $AM$ at $T$. Then $MPQT$ is cyclic. Hence $TQ // AB$. Therefore $T, Q, B_1$ are collinear. Hence $PC_1B_1Q$ is cyclic.

Mildorf, what did you think when you saw this problem (or more specifically, when you got the solution)? Because I sure wasn't expecting that to work .

Alas, I prevented myself from the joy of solving this problem, when I repeatedly misread the problem text and drew wrong sketches, what finally irritated me to an extent such that I gave up and read the solution. Before I comment on the solutions, let me say a few words about the problem and the characteristic "classical triangle geo": The problem is not classical. Actually, in the problem, two points ($B_1$ and $C_1$) are defined as intersections of lines and circles, so their coordinates (cartesian, let's say) contain quadratic irrationalities; on the other hand, the assertion of the problem, the concyclicity of the points $B_1$, $C_1$, P and Q, yields that the points $B_1$ and $C_1$ are rationally dependent, i. e. if one of them is given, then the other one can be constructed by rational operations only (e. g., if you know $B_1$, then you can find $C_1$ as the point of intersection of the circumcircles of triangles APB and $B_1PQ$ apart from the point P). As a consequence, the conditions that the points $C_1$ and Q lie on opposite sides of the line AB and that the points $B_1$ and Q lie on opposite sides of the line AC are really important as conditions (to distinguish between the two different points of intersection of lines with circles). This makes the problem different from the usual kind of triangle geometry problems, which either don't contain any irrationalities at all, or the irrationalities are independent (i. e., each line-circle intersection can be replaced by the other intersection without that the problem would become wrong). Anyway, enough of rambling for now, let's come to the solution of the problem. If there is a shortest solution for the problem, then it's Mildorf's solution in post #9, although it contains a typo: Mildorf wrote: Observe that $RPQB$ is cyclic This should be "Observe that $RAPB$ is cyclic" in my opinion. Here is also a simpler (or, at least, more elementary) way to state Grobber's solution from post #2: Let the line $AB_1$ intersect the circumcircle of triangle APB at a point $B_2$ (apart from the point A). We will show that $B_2=C_1$. We will work with directed angles modulo 180°. Since the points A, P, C and $B_1$ lie on one circle, $\measuredangle AB_1C=\measuredangle APC$. Since the points A, P, B and $B_2$ lie on one circle, $\measuredangle AB_2B=\measuredangle APB$. But < APC = < APB, and thus $\measuredangle AB_1C=\measuredangle AB_2B$. Since the point $B_2$ lies on the line $AB_1$, this equality yields $CB_1\parallel BB_2$. In other words, the point of intersection of the lines $BB_2$ and $CB_1$ is an infinite point. Also, since $QB_1\parallel BA$, the point of intersection of the lines $QB_1$ and BA is an infinite point. Now, by the Pappus-Pascal theorem, applied to the collinear point triples (B; C; Q) and ($B_1$; $B_2$; A), we see that the point of intersection of the lines CA and $QB_2$, the point of intersection of the lines $QB_1$ and BA, and the the point of intersection of the lines $BB_2$ and $CB_1$ lie on one line. Since we know that the point of intersection of the lines $BB_2$ and $CB_1$ and the point of intersection of the lines $QB_1$ and BA are infinite points, this line is the line at infinity, so that the point of intersection of the lines CA and $QB_2$ must also lie on the line at infinity, i. e. it is an infinite point. In other words, $QB_2\parallel CA$. On the other hand, $QC_1\parallel CA$. Thus, both points $B_2$ and $C_1$ lie on the parallel to the line CA through the point Q. Also, both of these points lie on the circumcircle of triangle APB, and, more exactly, on the arc AB of this circumcircle (the point $C_1$ lies on this arc because the quadrilateral $APBC_1$ is convex and cyclic and the points $C_1$ and Q lie on opposite sides of the line AB). Since the parallel to the line CA through the point Q intersects the arc AB of the circumcircle of triangle APB at only one point, the points $B_2$ and $C_1$ must therefore coincide. Since we knew that the points $B_1$, $B_2$ and A are collinear, we thus obtain that the points $B_1$, $C_1$ and A are collinear. Now, the rest is easy: We have $\measuredangle C_1PQ=\measuredangle C_1PB$. On the other hand, $\measuredangle C_1PB=\measuredangle C_1AB$, since the points $C_1$, B, P and A lie on one circle. Also, $\measuredangle C_1AB=\measuredangle C_1B_1Q$ since $QB_1\parallel BA$ (and since the points $B_1$, $C_1$ and A are collinear). Thus, $\measuredangle C_1PQ=\measuredangle C_1B_1Q$. And it follows that the points $B_1$, $C_1$, P and Q lie on one circle. This solves the problem. Darij

darktreb wrote: Mildorf, what did you think when you saw this problem (or more specifically, when you got the solution)? Because I sure wasn't expecting that to work . Actually, this was probably one of the (I'm guessing) two problems I lost points on due to sloppy writing. I found that solution with literally a couple minutes remaining. I constructed $B_2$ because I knew that it had to be $C_1$, then, after observing that $\angle{B_1PB_2} \cong \angle{BAC} \cong \angle{B_1QC_1}$, it was logical to construct $R$ and proceed to show that three points are the same.

If I could prove that $ BC_1 || B_1C$, then I'd be done. Unfortunately, I can't. Darn.

Fedor Petrov wrote: SJS wrote: My intuition tells me this problem can be solved by inversion. Be sure, it may be solved by inversion (in point $ P$), we get an easy angle-counting problem. I hear a lot of "you just invert, then angle chase" and I never see a concrete solution. Is anyone willing to provide an actual formal inversive solution? Thanks

Now I actually did an inversion in gsp and it's not showing me the easy solution. I must be missing something obvious.

Attachments:

Hmm, this seems too easy to be a #3...Am I missing something? It feels like cheating to use this reasoning: if the problem is true, $A$, $C_1$ and $B_1$ must be collinear by some angle-chasing. Since the problem is most likely true, that motivates us to construct $B_2$ as the intersection of $C_1$ and the circumcircle of $C_1PQ$. By simple angle chasing, $B_2APC$ is cyclic and $B_2Q \parallel AB$, which means $B_2 = B_1$, and we're done.

It's not as nice, but I'd like to point out that there is a solution without "phantoming" (like Darij, I misread this problem for a long time and got a bit frustrated, resorting to a silly ratio orgy)... We will use directed segments throughout this proof. Let $C_2=QC_1\cap(ABP)$, $B_2=QB_1\cap(ACP)$, $U=AB\cap C_1C_2$, and $V=AC\cap B_1B_2$. As the posters above have shown, $P,Q,B_1,C_1$ lie on a circle iff $A,B_1,C_1$ are collinear (by a one-step angle chase), which is true iff \[\frac{UA}{UC_1}=\frac{VB_1}{VA}.\](We have $AUB\|QVB_1$ and $AVC\|QUC_1$.) First, since $AUQV$ is a parallelogram, parallel lines in $\triangle{ABC}$ yield \[r=\frac{QB}{QC}=\frac{VA}{VC}=\frac{UB}{UA}.\]Considering the power of point $Q$ with respect to $(ABP)$ and $(ACP)$, we also have \begin{align*} QP\cdot QB &= QC_1\cdot QC_2 \\ QP\cdot QC &= QB_1\cdot QB_2, \end{align*}and from the power of points $U,V$ with respect to $(ABP),(ACP)$ (respectively), \begin{align*} UA\cdot UB &= UC_1\cdot UC_2 = (QC_1-QU)(QC_2-QU) = (QC_1-VA)(QC_2-VA) \\ VA\cdot VC &= VB_1\cdot VB_2 = (QB_1-QV)(QB_2-QV) = (QB_1-UA)(QB_2-UA). \end{align*}Letting $u=UA$, $v=VC$, and $(p,c,c_1,c_2,b_1,b_2)=(QP,QC,QC_1,QC_2,QB_1,QB_2)$ and plugging in $r$, we have \begin{align} pcr &= c_1c_2\\ pc &= b_1b_2\\ ru^2 &= (c_1-rv)(c_2-rv)\\ rv^2 &= (b_1-u)(b_2-u) \end{align}and it suffices to show that \[\frac{u}{c_1-rv}=\frac{UA}{UC_1}=\frac{VB_1}{VA}=\frac{b_1-u}{rv}\Longleftrightarrow b_1c_1=uc_1+rvb_1\Longleftrightarrow b_1b_2=b_2u+c_2v,\]where we have used $r=\frac{c_1c_2}{b_1b_2}$ from (1)/(2). Plugging this expression for $r$ again into (3) and (4), we get \begin{align*} u^2 &= b_1b_2-(c_1+c_2)v+\frac{c_1c_2}{b_1b_2}v^2 \\ \frac{c_1c_2}{b_1b_2}v^2 &= b_1b_2-(b_1+b_2)u+u^2. \end{align*}Adding these two equations and simplifying, we now have \[2b_1b_2=(b_1+b_2)u+(c_1+c_2)v.\]Solving for $u$ in terms of $v$ and substituting into the former of the two equations, \[u^2=\frac{(2b_1b_2-(c_1+c_2)v)^2}{(b_1+b_2)^2}=\frac{2(b_1b_2-c_1v)(b_1b_2-c_2v)}{2b_1b_2}=\frac{(c_1-c_2)^2v^2}{(b_1-b_2)^2}.\]But $u=UA$ and $b_1-b_2=B_2B_1$ are parallel with the same orientation while $v=VC$ and $c_1-c_2$ are parallel with the opposite orientation (because $Q,C_1$ are on opposite side of $AB$ and $Q,B_1$ are on opposite side of $AC$), so \[u(b_1-b_2)=v(c_2-c_1)\implies ub_1+vc_1=ub_2+vc_2\implies b_1b_2=b_2u+c_2v,\]as desired. (Recall that $2b_1b_2=(b_1+b_2)u+(c_1+c_2)v$.)

Could we instead assume that $C_{1}$ is the intersection of the circumcircles of $PB_{1}Q$ and $ABP$, then prove that $C_{1}Q$ is parallel to $AC$? This task would be trivial, since we have angles $AC_{1}Q=B_{1}PC=B_{1}AC$. Although we have $C_{1}Q$ intersecting circle $ABP$ two times, the inner intersection isn't relevant because it's inside quad $APQB_{1}$?

Dear Mathlinkers, we can also make a link with http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=369400 Sincerely Jean-Louis

Let $B_2$ be the second intersection of $AC_1$ with $(APC)$. Then $\angle B_2PQ=\angle B_2AC=\angle B_2C_1Q$, so $PQB_2C_1$ is cyclic. Now $\angle C_1B_2Q=\angle C_1PB=\angle C_1AB\implies B_2Q\parallel AB\implies B_1=B_2$, so we're done.

kootrapali wrote: We begin with the following claim: Claim: $A,B_1,C_1$ are collinear. Proof: We have $\measuredangle AB_1C=\measuredangle APC=\measuredangle APB=\measuredangle AC_1B$, so $BC_1\parallel CB_1$. By Reim’s, this implies $B_1C_1$ passes through $A$, as desired. Now, we have $\measuredangle B_1PQ=\measuredangle B_1AC=\measuredangle B_1C_1Q$, so $PQB_1C_1$ is cyclic, and we are done. Unfortunately, this is a circular reasoning.

Dunno if this is right. Let $B_1'$ be the point such that $B_1'C\parallel C_1B$, $B_1'Q\parallel AB$, and $B_1'\in AC_1$ (which must exist by Pappus). Now we can just angle chase to show that $B_1'\equiv B_1$ and we're done. (Idea is to use phantom points to turn weak claim -> strong claim into strong claim -> weak claim)

We prove the following restated problem. Restated problem wrote: Let $ABC$ be a triangle and $P$ lie on segment $\overline{BC}$. Let $E,F$ be selected on $(ABP)$ and $(ACP)$ such that $A,E,F$ are collinear. Then the line through $E$ parallel to $\overline{AC}$, the line through $F$ parallel to $\overline{AB}$, and $\overline{BC}$ concur. To prove this, note that by Reim's theorem $\overline{BE} \parallel \overline{CF}$. Let $\infty_1$ be the point at infinity along $\overline{AB}$ and $\overline{CF}$, $\infty_2$ be the point at infinity along $\overline{AC}$ and $\overline{BE}$, and $\infty_3$ the point at infinity along $\overline{BE}$ and $\overline{CF}$. The problem then follows by Pascal's on degenerate hyperbola $\infty_1A\infty_2E\infty_3F$. Now note that if we vary $E$ along minor arc $AB$ of $(ACP)$, $F$ varies along minor arc $AC$ of $(ACP)$ as well, and the concurrency point can lie anywhere on $\overline{BC}$ (consider two limiting cases, use continuity). Thus in the original problem we always have $B_1,C_1,A$ collinear. Then $\measuredangle C_1QB_1=\measuredangle BAC$, and $$\measuredangle C_1PB_1=\measuredangle BPB_1+\measuredangle C_1PC=\measuredangle BAB_1+\measuredangle C_1AC=\measuredangle BAC,$$proving the desired concyclicity. $\blacksquare$

Well, I wonder if I fakesolved my solution isn't as involved as most others. Simply let $B_1' = (APC) \cap \overline{AC_1}$ and $Q' = (C_1PB_1') \cap \overline{BC}$. Then, one notices that \[\measuredangle BQ'C_1 = \measuredangle PQ'C_1 = \measuredangle PB_1'C_1 = \measuredangle PB_1'A = \measuredangle PCA = \measuredangle BCA\]Thus, $C_1Q' \parallel AC$ and we must have $Q'=Q$. Further, \[\measuredangle B_1'QC = \measuredangle B_1'CP = \measuredangle AC_1P = \measuredangle ABP =\measuredangle ABC\]Thus, we also have that $B_1'Q \parallel AB$ and thus $B_1'= B_1$ and indeed we have that points $P,Q,B_1$ and $C_1$ lie on a circle.

Claim. $A, B_1, C_1$ are collinear. Proof. Note that $\angle AC_1P = B$ and $\angle AB_1P = C$, so $\angle C_1PB_1 = A$. Then $\angle C_1AB + \angle B_1AC + \angle A = 180^\circ$. $\blacksquare$ Now let $R = \overline{B_1C_1} \cap \overline{BC}$. Then note $RQ \cdot RA = RC_1 \cdot RC$, so $$RC_1 \cdot RB_1 = RP \cdot RQ \iff RA \cdot RC_1 \cdot RB_1 = RC_1 \cdot RC \cdot RP$$which is obviously true.

Let $B'_1=AC_1\cap (APC)$. Claim: $A,B_1,C_1$ are all collinear. Proof: It suffices to show that $B'_1=B_1$, or that $QB'_1\parallel AB$. Note that $\triangle PB'_1C_1\sim \triangle ACB$ by AA. Therefore: \[\angle C_1QP=180-\angle C_1QB=180-\angle ACB=180-\angle C_1B'_1P,\]so $C_1QPB'_1$ is cyclic. This implies that: \[\angle C_1B'_1Q=\angle C_1PQ=\angle C_1AB,\]showing that $AB\parallel QB'_1$, or $B_1=B'_1$ $\square$ We have already shown that $C_1QPB'_1$ is cyclic, so $C_1QPB_1$ is cyclic, as desired $\blacksquare$.

We consider $B_1$ as define and $C' = B_1A \cap (APB)$ Now $\angle AC'P = \angle ABC = \angle B_1QC$ hence $C',B_1,P,Q$ are cyclic Now from $$\angle ACP = \angle AB_1P = \angle C'B_1P = \angle C'QP$$which give us $C'Q \parallel AC$ so $C'=C_1$ $\blacksquare$

Notice that \begin{align*} \text{Points } A, B_1, C_1 \text{ are collinear} & \iff \angle B_1C_1P = \angle AC_1P \\ & \iff \angle B_1C_1P = \angle ABP \\ & \iff \angle B_1C_iP = \angle B_1QC \\ & \iff B_1C_1PQ \text{ is cyclic.} \end{align*} Thus, it suffices to prove that points $A$, $B_1$, and $C_1$ are collinear. Denote the intersection of $\overline{AC_1}$ and $(ACP)$ as $B_2$. We have \[\angle AB_2P = \angle ACP = \angle C_1QP,\] meaning $B_2C_1PQ$ is cyclic. Then, \[\angle CQB_2 = \angle AC_1P = \angle ABP,\] so $\overline{AB} \parallel \overline{B_2Q}$. Hence, $B_1 \equiv B_2$, so we are done. $\square$

Okay. Note that $\angle AB_1P = \angle C$ and $\angle AC_1P = \angle B$. Also $\angle B_1QC_1 = \angle A$. Then our next claim finishes the problem. Claim: The points $A$, $B_1$ and $C_1$ are collinear. Proof. Let $\overline{C_1A}$ meet $(ACP)$ at $X$. We claim $X \equiv B_1$. Note \[\angle AXP = \angle ACP = \angle C_1QB\]Thus $(C_1PQX)$ is cyclic. However then \[\angle XQC = \angle XC_1P = \angle B\]so $\overline{XQ} \parallel \overline{AB}$ and hence $X \equiv B_1$. $\square$ Then we finish as we find $\angle B_1 PC_1 = A$ and hence $B_1C_1 P Q$ is cyclic. $\blacksquare$

We use a phantom point argument by defining $C_2 = B_1A \cap (APB)$. We first note that \[\measuredangle B_1QP = \measuredangle B_1QC = \measuredangle ABP = \measuredangle B_1C_2P,\] so $B_1C_2PQ$ is cyclic. We're finished if we can show $C_1 = C_2$, which follows from \[\measuredangle PCA = \measuredangle PB_1A = \measuredangle PQC_2 \implies QC_2 \parallel CA. \quad \blacksquare\]

nice Let $B_2 = \overline{AC_1} \cap (APC)$. We have $$\measuredangle B_2PQ = \measuredangle B_2PC = \measuredangle B_2AC = \measuredangle B_2C_1Q,$$so $B_2C_1PQ$ is cyclic. But $$\measuredangle B_2QP = \measuredangle B_2C_1P = \measuredangle AC_1P = \measuredangle ABP = \measuredangle ABQ,$$which implies that $\overline{BA} \parallel \overline{B_2Q}$, so $B_2 = B_1$ and we are done.

Let $B'_1 = AC_1 \cap (ACP)$ so $\angle B'_1PC = \angle B'_1AC = \angle B'_1C_1Q$ so $B'_1 \in (PQC_1)$. Then we have $\angle C_1AB_1 = \angle C_1AB + \angle BAC + \angle B_1AC$. We angle chase to find $\angle C_1AB = \angle C_1PB$ and $\angle B'_1AC = \angle B'_1PC$ and finally $\angle BAC = \angle C_1QB'_1 = \angle C_1PB'_1$(since $A$ and $Q$ are opposite vertices of a parallelogram). It follows from $\angle C_1PB + \angle C_1PB'_1 + \angle B'_1PC = 180^\circ = \angle C_1AB_1$ that $C_1 - A - B'_1$ so $B'_1 = B_1$ from which we get that $B_1C_1PQ$ is cyclic as desired.

Let line $C_1A$ meet $(PAC)$ again at $B_1'$. Then $$\measuredangle C_1QP = \measuredangle ACB =\measuredangle C_1B_1'P,$$so $C_1B_1'PQ$ is cyclic, and $$\measuredangle B_1'QC = \measuredangle B_1'C_1P = \measuredangle ABC,$$so $QB_1' \parallel AB$ and $B_1' = B_1$. Therefore, $C_1B_1QP$ is cyclic.

Let $X = AC_1 \cap (ACP)$. It follows that \[ \measuredangle PXA = \measuredangle PCA \]Hence we get that $APCX$ is cyclic. Finally we get that \[ \measuredangle PBA = \measuredangle PC_1A = \measuredangle PC_1X = \measuredangle PQX \]Which implies that $QX \parallel BA$. From this it follows that $X \equiv B_1$ proving the required result.