Doesn't the announcement say to wait til 8:00 PM EST? @below ok.

lrjr24 wrote: Doesn't the announcement say to wait til 8:00 PM EST? The problems have been posted on the official website, and thus discussion should be allowed.

lrjr24 wrote: Doesn't the announcement say to wait til 8:00 PM EST? I actually don't think discussion should be allowed. It's not 23:59 UTC yet.

This problem was proposed by Burii.

naman12 wrote: Consider the convex quadrilateral $ABCD$. The point $P$ is in the interior of $ABCD$. The following ratio equalities hold: \[\angle PAD:\angle PBA:\angle DPA=1:2:3=\angle CBP:\angle BAP:\angle BPC\]Prove that the following three lines meet in a point: the internal bisectors of angles $\angle ADP$ and $\angle PCB$ and the perpendicular bisector of segment $AB$.

Definitely European geometry all right. I'll edit in a diagram at some point, but for now: Choose points $S$ on $\overline{BC}$ and $T$ on $\overline{AD}$ so that $\triangle BSP$ and $\triangle ATP$ are both isosceles. The given angle conditions imply $\angle PSC = \angle SPC$ and $\angle DPT=\angle DTP$, so the given angle bisectors are actually the perpendicular bisectors of $\overline{PS}$ and $\overline{PT}$. But also $S$ and $T$ are the midpoints of the arcs $\widehat{PB}$ and $\widehat{PA}$ of $\odot(PAB)$, so these bisectors concur at the center of $\odot(PAB)$, which, by definition, lies on the perpendicular bisector of $\overline{AB}$.

Let the circumcircle of $PAB$ intersect $AD,BC$ at $X$ and $Y$ respectively. It can be seen by angle chase that $XD=PD, YC=PC$ so the angle bisectors in the problem are just bisectors of $PX$ and $PY$. Now the three lines concur since they are perpendiculars of chords of a common circle ($ABYPX$).

What a disaster.

Truly the pinnacle of olympiad geometry. Let $O$ be the circumcenter of $(PAB).$ Now using the angle conditions note that $(POAD)$ is cyclic, so an angle chase shows $DO$ bisects $\angle ADP$. Similarly $CO$ bisects $\angle BCP$.

itslumi wrote: This isn't for a p1,What a disaster(very easy). Wait, I actually don't think this is a bad problem (though perhaps not the pinnacle of Euclidean geometry). While the diagram is definitely... artificial, you actually need to understand how to interpret all the angle conditions in order to make any sort of progress on this problem. In my case, I drew all the angle bisectors and trisectors necessary to create a bunch of equal angles, then realized that many of those lines were not necessary once I figured out what claims to make. The problem actually seems pretty instructive in this regard. Also, obligatory easy ≠ bad, and certainly nobody's going to coordinate bash this! xD

Continuing off of dj, I thought this problem was actually quite difficult for a 1 (around IMO 2018 1), just because of how unbelievably strange the given condition was. I think it's hard to get started on the problem if you're not experienced with geometry - the key after all to solving this problem is to focus less on the condition and trying to find a way to construct the diagram accurately, which is why I think this is not an easy $1$. I also don't particularly like it, because of how contrived it was. Again, easy $\neq$ bad.

If no one can coordbash it, then it is a joy for some and a horror for the others.

I agree, I thought that this problem was ok for P1, probably harder than last year's P1 so that's good. Now that I see that just adding the circumcenter is enough instead of what I did changes my opinion slightly but it's still not too easy. It's like 2018 P1.

Hardest part of the problem by far was getting the diagram right... Let $O$ be the circumcenter of $PAB$. Let $\angle PAD = \alpha$, then $\angle PBA = 2\alpha, \angle DPA = 3\alpha$ and $\angle AOP$. By looking at triangle $APD$ we have \[ \angle ADP = 180^{\circ} - \angle DAP - \angle DPA = 180^{\circ} - 4 \alpha = 180^{\circ} - \angle AOP \] And therefore $AOPD$ is cyclic. As $OA = OP$ it follows that $DO$ is the internal angle bisector of $\angle ADP$. Similarly, $O$ lies on the internal angle bisector of $\angle PCD$. Since $OA = OB$ it follows that this point also lies on the perpendicular bisector of $AB$, as desired.

I'm convinced that no matter how easy/low quality last year's IMO problem 1 was, this year's IMO problem 1 forum will consist of people saying how trivial it is compared to the previous year.

people please dont say this is trivial i think it's pretty fit for its position

After guessing the circumcenter $O$ of $PAB$, the distances from $P$ to $DA$ and $DP$ are both $R\sin(90-\angle PAD)$, and similar for the other one, implying the result. bah guessing the point took so long though

Let $O$ be the circumcenter of $PAB$. We can prove that $O$ lies on the angle bisector of $\angle PDA$, which is enough by symmetry. We have $\angle POA = 4\angle{PAD}$ and $\angle PDA = 180^{\circ} - 4\angle{PAD}$, so $PDAO$ is cyclic. Then it follows that $O$ is the arc midpoint of $DA$, so we are done.

well, Let $O$ be circumcenter of $\triangle APB$ We will prove $ADPO$ and $BCPO$ cyclic if we connect $AO,BO,PO$ we can easyly deduce that.

Let $O$ be the circumcenter of triangle $APB$. Then the quadrilaterals $CPOB$ and $AOPD$ are cyclic as $\angle POB + \angle PCB = 180^\circ$ given the angle condition. It follows that $\overline{DO}$ and $\overline{CO}$ bisect their respective angles, implying the result.

construct a circle around $ABP$ let it intersect $AD$ at $Q$, and $BC$ at $R$ a simple angle chase yields that $DQP=2*QAP$ and $QAP=QPA$, and $QPD=DQP$ therefore, $QDP$ is isosceles and since $QP$ is a chord, the angle bisector of $ADP$ is the perpendicular bisector of $QP$ and intersects this center of the circle $(ABP)$ similarly, the angle bisector of $BCP$ is the perpendicular bisector of $PR$ and intersects this center of the circle $(ABP)$ then, since $AB$ is a chord, it also passes through the center of the circle $(ABP)$

Let $\angle PAD = \alpha$ and $\angle PBC = \beta$. Denote O as the circumcenter of $\triangle APB$. Now $\angle AOB = 360 - 2\angle APB = 360 - 2(180 - 2\alpha - 2\beta) = 4\alpha + 4\beta$ $\Rightarrow$ $\angle OAB = \angle OBA = \frac{180 - 4\alpha - 4\beta}{2} = 90 - 2\alpha - 2\beta$ $\Rightarrow$ $\angle OAP = \angle OAB + \angle BAP = 90 - 2\alpha - 2\beta + 2\beta = 90 - 2\alpha$. Now from AO = OP, we have that $\angle OAP = \angle OPA = 90 - 2\alpha$ $\Rightarrow$ $\angle OAD + \angle OPD = \angle OAP + \angle PAD + \angle OPA + \angle APD = 90 - 2\alpha + \alpha + 90 - 2\alpha + 3\alpha = 180^\circ$ $\Rightarrow$ AOPD is cyclic. Since AOPD is cyclic we have that $\angle OAP = \angle ODP = 90 - 2\alpha$. Since $\angle ADP = 180 - \angle DAP - \angle APD = 180 - 4\alpha$, then $\angle ADO = 180 - 4\alpha - \angle ODP = 180 - 4\alpha - (90 - 2\alpha) = 90 - 2\alpha$ $\Rightarrow$ $\angle ADO = \angle ODP = 90 - 2\alpha$ $\Rightarrow$ DO is the angle bisector of $\angle ADP$. Similarly we will show BOPC is cyclic. From OP = OB, $\angle OPB = \angle OBP = \angle OBA + \angle ABP = 90 - 2\alpha - 2\beta + 2\alpha = 90 - 2\beta$. Now $\angle OPC + \angle OBC = \angle OPB + \angle BPC + \angle OBP + \angle PBC = 90 - 2\beta + 3\beta + 90 - 2\beta + \beta = 180^\circ$ $\Rightarrow$ BOPC is cyclic $\Rightarrow$ $\angle OPB = \angle OCB = 90 - 2\beta$. Also $\angle PCO = \angle PCB - \angle BCO = 180 - \angle CPB - \angle CBP - \angle BCO = 180 - 3\beta - \beta - (90 - 2\beta) = 90 - 2\beta$ $\Rightarrow$ $\angle PCO = \angle BCO = 90 - 2\beta$ $\Rightarrow$ CO is the angle bisector of $\angle PCB$. From this it follows that $DO \cap CO = O$, where DO and CO are the angle bisectors of $\angle ADP$ and $\angle PCB$, respectively. The last thing we need is, to show that the perpendicular bisector of AB passes trough O, which is obvious since O is the circumcenter of $\triangle APB$ $\Rightarrow$ the internal bisectors of angles $\angle ADP$ and $\angle PCB$ and the perpendicular bisector of segment AB meet in a point, which is explicitly O - the circumcenter of $\triangle APB$ $\Rightarrow$ we are ready.

Let $\alpha=\angle DAP$ and $\beta=\angle CBP$. Let $\omega=(ABP)$ meet $AD$ and $BC$ again at $Q_1$ and $Q_2$, respectively. Then $\angle Q_1BP=\angle Q_1AP=\alpha$, so $BQ_1$ bisects $\angle ABP$. Thus $Q_1$ is the midpoint of arc $AP$. Thus $\angle DQ_1P=2\alpha=\angle DPQ_1$, so the angle bisector of $\angle PDA$ is just the perpendicular bisector of $Q_1P$. Similarly, the angle bisector of $\angle PCB$ is just the perpendicular bisector of $Q_2P$. These three lines clearly meet at the center of $\omega$, so we are done. $\blacksquare$

Consider the circumcenter $O$ of $\triangle ABP$, which we claim is the desired concurrency point: $O$ lies on the perpendicular bisector of $AB$. $O$ lies on $(BCP)$, as $\measuredangle BOP = 2 \measuredangle BAP = \measuredangle CBP + \measuredangle BPC = \measuredangle BCP$. Additionally, $BO = PO$ tells us $O$ is the midpoint of arc $BP$, which means $O$ lies on the bisector of $\angle PCB$. $O$ lies on the bisector of $\angle ADP$ analogously. $\blacksquare$

One of the better things I've drawn in a while (Edit: Oops I forgot CO below but it should be updated in the link!) Denote the given angles by $\alpha,2\alpha,3\alpha$ and $\beta,2\beta,3\beta$ in the order they appear; and let the bisectors of $\angle PAB$ and $\angle PBA$ meet $BC$ and $AD$ at $E$ and $F$, respectively. We have: $\angle PAF=\angle PBF=\alpha\implies(PABF)$ $\angle FAP=\angle FPA=\alpha\implies FA=FP$ $\angle DFP=\angle DPF=2\alpha\implies DF=DP$ and $\angle PBE=\angle PAE=\beta\implies(PABE)$ $\angle EBP=\angle EPB=\beta\implies EB=EP$ $\angle CEP=\angle CPE=2\beta\implies CE=CP$. Then, the given lines are the perpendicular bisectors of $AB$, $EP$, and $FP$, which concur at the circumcenter of $(ABEFP)$.

solved with help from peace09 (: thanks for the problem rec <3 We bisect and trisect the angles that are 2x or 3x another. This is supposed to suggest another center -- the circumcenter of $\triangle ABP$, which we'll name $O$. If we let $\angle PAD = \alpha$ and $\angle CBP = \beta$, we have \[\angle PDA = 180-4\alpha = 180 - 2\angle ABP = 180-\angle AOP.\]Similarly, $\angle BCP = 180 - \angle BOP$. This means quadrilaterals $AOPD$ and $BOPC$ are cyclic. Because $OA=OP$, $OD$ is the angle bisector of $\angle ADP$. Similarly, $OC$ is $\angle BCP$'s bisector. Thus, the two angle bisectors intersect at $O$, which lies on the perpendicular bisector of $AB$. $\square$

Let $O$ be the circumcenter of $\triangle PAD.$ We claim that this works. Notice that $POAD$ and $POBC$ are cyclic. Thus, we are done.

Let $O$ be the circumcenter of $\Delta ABP$. Then easy angle chase gives us that $AOPD$ and $BOPC$ are cyclic. Due to fact $5$, this means that $O$ passes through angle bisectors of $\angle PCB$ and $\angle ADP$ as desired.

Let $O$ be the circumcenter of $\triangle{ABP}$. We will show that these three lines all pass through $O$. First, $O$ clearly lies on the perpendicular bisector of segment $AB$. Claim: $BOPC$ and $AOPD$ are cyclic. Note that $$\angle{BCP}=180-\angle{CBP}-\angle{BPC}=180-2\cdot\angle{BAP}=180-\angle{BOP},$$so $BOPC$ is cyclic. Similarly, $$\angle{ADP}=180-\angle{PAD}-\angle{DPA}=180-2\angle{PBA}=180-\angle{AOP},$$so $AOPD$ is cyclic. Since $O$ lies on the perpendicular bisector of $AP$ and $(DPA)$, we have $\angle{ADO}=\angle{ODP}$. Similarly, $\angle{PCO}=\angle{OCB}$, done.

A classic example of a single construction tearing the question apart. Let the given angle bisectors meet in $X$. Construct $M$, $N$ on $BC,AD$ with $BM=MP$ and $AN=NP$. Let $O$ be the circumcentre of $\triangle ABP$. Observe that $\angle MBP=\angle MPB= \frac{1}{2}\angle BAP$ and $\angle CMP= \angle CPM$ so $M \in \odot(ABP)$. Similarly, $N \in \odot(ABP)$. Note that the angle bisectors of $\angle PCB$ and $\angle PDA$ are the perpendicular bisectors of $MP$, $NP$, which meet, by definition, in $O$, which lies, by definition, on the perpendicular bisector of $AB$.

Let $O$ be the circumcenter of $\triangle ABP$. Angle chasing implies that $\square OPBC$ and $\square OPAD$ are cyclic. The intersection of the internal bisectors of $\angle PDA$ and $\angle PCB$ is therefore $O$, which obviously lies on the perpendicular bisector of $AB$.

itslumi wrote: What a disaster. And here we see the convictions of someone who has no taste.