Addition on the IMO

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1358 posts

#1 h Sep 22, 2020, 6:28 PM • 32 Y

Y by Smileyklaws, ilovepizza2020, OlympusHero, tree_3, dchenmathcounts, amar_04, A-Thought-Of-God, Shrawon360, Aimingformygoal, Aritra12, phxbesx0705, samrocksnature, Jc426, Yugoo, centslordm, brickmaster8, megarnie, tigerzhang, HWenslawski, son7, rayfish, ImSh95, Lamboreghini, crazyeyemoody907, Mogmog8, aidan0626, deplasmanyollari, tigeryong, Rounak_iitr, Sedro, ItsBesi, NicoN9

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 $AB$ .

Proposed by Dominik Burek, Poland

This post has been edited 2 times. Last edited by djmathman, Oct 6, 2020, 5:51 PM

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khina

993 posts

khina

#2 h Sep 22, 2020, 6:30 PM • 4 Y

Y by samrocksnature, Vaddokx, ImSh95, Math_.only.

lol Click to reveal hidden text

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lrjr24

967 posts

lrjr24

#3 h Sep 22, 2020, 6:33 PM • 4 Y

Y by samrocksnature, Math4Life2020, ImSh95, Nuran2010

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

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naman12

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naman12

#4 h Sep 22, 2020, 6:33 PM • 2 Y

Y by samrocksnature, ImSh95

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.

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jj_ca888

2726 posts

jj_ca888

#5 h Sep 22, 2020, 6:34 PM • 2 Y

Y by samrocksnature, ImSh95

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.

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timon92

224 posts

timon92

#6 h Sep 22, 2020, 6:37 PM • 13 Y

Y by 62861, Aryan-23, 508669, Kanep, qubatae, samrocksnature, CrazyMathMan, megarnie, ImSh95, aidan0626, Mango247, ehuseyinyigit, NicoN9

This problem was proposed by Burii.

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Pluto1708

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Pluto1708

#7 h Sep 22, 2020, 6:45 PM • 8 Y

Y by A-Thought-Of-God, samrocksnature, Vaddokx, megarnie, ImSh95, Aryan-23, Mango247, Math_.only.

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 $AB$ .

Solution

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djmathman

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djmathman

#8 h Sep 22, 2020, 6:58 PM • 9 Y

Y by Pluto1708, OlympusHero, karitoshi, samrocksnature, Jc426, HamstPan38825, megarnie, ImSh95, ehuseyinyigit

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 $\overline{AB}$ .

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InkretBear

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InkretBear

#9 h Sep 22, 2020, 7:00 PM • 4 Y

Y by Pluto1708, Steff9, samrocksnature, ImSh95

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 ( $ABYPX$ ).

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itslumi

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itslumi

#10 h Sep 22, 2020, 7:08 PM • 7 Y

Y by RudraRockstar, samrocksnature, ImSh95, Complete_quadrilateral, mistakesinsolutions, ihatemath123, pokpokben

What a disaster.

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mathlogician

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mathlogician

#12 h Sep 22, 2020, 7:09 PM • 6 Y

Y by Ya_pank, samrocksnature, Complete_quadrilateral, Mango247, Mango247, Mango247

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 $\angle BCP$ .

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djmathman

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djmathman

#14 h Sep 22, 2020, 7:17 PM • 20 Y

Y by timon92, OlympusHero, Kanep, IAmTheHazard, crazyeyemoody907, myh2910, Ronin_The_Nameless_Ninja, samrocksnature, Jc426, HamstPan38825, 554183, centslordm, tigerzhang, rayfish, CyclicISLscelesTrapezoid, mathmax12, Mango247, aidan0626, OronSH, NicoN9

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

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khina

993 posts

khina

#15 h Sep 22, 2020, 7:21 PM • 21 Y

Y by timon92, SnowPanda, Aryan-23, danepale, IAmTheHazard, Illuzion, OlympusHero, lneis1, samrocksnature, Corella, CrazyMathMan, 554183, centslordm, tigerzhang, IMUKAT, rayfish, pog, mathmax12, Mango247, aidan0626, NicoN9

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.

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Smileyklaws

447 posts

Smileyklaws

#16 h Sep 22, 2020, 7:22 PM • 2 Y

Y by samrocksnature, Infinityfun

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

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Aryan-23

558 posts

Aryan-23

#17 h Sep 22, 2020, 7:22 PM • 2 Y

Y by A-Thought-Of-God, samrocksnature

Sol

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AngeloChu

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AngeloChu

#146 h Apr 8, 2024, 12:40 PM

Y by

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 $(ABP)$

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Markas

105 posts

Markas

#147 h May 5, 2024, 12:56 PM

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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.

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cj13609517288

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cj13609517288

#148 h May 15, 2024, 3:08 PM

Y by

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. $\blacksquare$

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shendrew7

794 posts

shendrew7

#149 h Jun 5, 2024, 4:18 PM

Y by

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. $\blacksquare$

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peace09

5417 posts

peace09

#150 h Jun 21, 2024, 4:08 PM • 3 Y

Y by bjump, OronSH, LeonidasTheConquerer

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!)

https://cdn.artofproblemsolving.com/attachments/4/d/d724117e3092b7f3b8c40fc5808d6064c5ed33.png

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 $(ABEFP)$ .

This post has been edited 1 time. Last edited by peace09, Jun 21, 2024, 4:11 PM

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gracemoon124

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gracemoon124

#151 h Jul 4, 2024, 12:39 AM • 5 Y

Y by peace09, dolphinday, OronSH, bjump, LeonidasTheConquerer

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 $AB$ . $\square$

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Mr.Sharkman

498 posts

Mr.Sharkman

#152 h Nov 7, 2024, 1:52 PM

Y by

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

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Eka01

204 posts

Eka01

#153 h Nov 13, 2024, 5:35 PM

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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.

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megahertz13

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megahertz13

#154 h Nov 23, 2024, 7:57 PM

Y by

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.

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SimplisticFormulas

98 posts

SimplisticFormulas

#155 h Dec 21, 2024, 3:27 PM

Y by

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 $AB$ .

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CANTBANKAN

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CANTBANKAN

#156 h Jan 12, 2025, 8:34 PM

Y by

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 $AB$ .

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Scilyse

385 posts

Scilyse

#157 h Jan 17, 2025, 10:51 AM

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itslumi wrote:

What a disaster.

And here we see the convictions of someone who has no taste.

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Ilikeminecraft

359 posts

Ilikeminecraft

#158 h Mar 15, 2025, 1:52 AM

Y by

Let $O = (ADP)\cap(BPC).$ We get $\angle AOP = \angle AOD + \angle DOP = \angle APD + \angle DAP = 2\angle ABP.$ Thus, $O$ is circumcenter of $(ABP).$ Clearly, $\angle ODP = \angle OAP = \angle OPA = \angle ODA$ which finishes.

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endless_abyss

41 posts

endless_abyss

#159 h Mar 24, 2025, 5:24 PM

Y by

Nice! This is such a satisfying problem once you identify the intersection point!

Let $Q$ be the circumcentre of $B A P$ , we claim that

$D Q A P$ and $C Q B P$ are concyclic.

This follows from $\angle D A P = \angle D Q P$ and $\angle P B C = \angle P Q C$

Further, note that $\angle A Q P = 2 \angle A B P$
so, $\angle A Q D = 3x$ ,
$\angle Q A B = 90 - 2 x - 2 y$
so, $\angle A D Q = 90 - 2x = \angle Q D P$ and in triangle $C B P$ the angle chase follows similarly, so $Q$ is the desired concurrence point.

:starwars:

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NicoN9

123 posts

NicoN9

#160 h Apr 5, 2025, 1:30 AM

Y by

My solution is similar to others, but I'll post it for fun anyways.

Let the internal bisector of $\angle PAB$ , $\angle PCD$ and segment $CB$ , $AD$ meet at $K$ , $L$ , respectively.

claim. five points $P$ , $A$ , $B$ , $K$ , $L$ are concyclic.
proof. We have $\angle PAK=\frac{1}{2} \angle PAB=\angle CBP=\angle KBP.$

We claim that $CK=CP$ , which follows by $\angle CKP=\angle KBP+\angle KPB = 2\angle KBP=\angle BPC -\angle BPK=\angle CPK.$ similarly $DP=DL$ . Now it is suffice to show that the perpendicular bisector of three segments $AB$ , $PK$ , $PL$ are concurrent, which is obvious by the claim.

This post has been edited 1 time. Last edited by NicoN9, Apr 5, 2025, 1:30 AM

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