IMO Problem 4

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138 posts

#1 h Jul 14, 2003, 5:09 PM • 13 Y

Y by Davi-8191, nguyendangkhoa17112003, TurtleKing123, HWenslawski, Adventure10, centslordm, megarnie, proxima1681, Mahmood.sy, Mango247, Rounak_iitr, and 2 other users

Let $ABCD$ be a cyclic quadrilateral. Let $P$ , $Q$ , $R$ be the feet of the perpendiculars from $D$ to the lines $BC$ , $CA$ , $AB$ , respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$ .

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sebadollahi

1 post

sebadollahi

#2 h Jul 15, 2003, 11:58 AM • 4 Y

Y by Adventure10, centslordm, Mango247, and 1 other user

in two cyclic quadrilatrals APRD & DRCQ ,AD & CD are diameters respectively and we have:
RQ/sin(<RCQ)=CD/2 or RQ/sin(<BCA)=CD/2
PR/sin(<PAR)=AD/2 or PR/sin(<BAC)=AD/2
By dividing:
CD/AD= sin(<BAC)/sin(<BCA)=BC/AB
Since CD/AD=BC/AB thus bisectors of <CBA and <ADC intersect AC in the same poin.
S.Ebadollahi

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galois

400 posts

galois

#3 h Jul 16, 2003, 8:17 PM • 2 Y

Y by Adventure10, Mango247

i must admit that this problem was quite easy by imo standards.my solution is based on a standard trick using pedal triangles and projections which is damn easy and pretty similar to the proof posted here.

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Lagrangia

1326 posts

Lagrangia

#4 h Jul 16, 2003, 8:42 PM • 5 Y

Y by Inconsistent, Adventure10, Mango247, and 2 other users

here is an interesting thing about this problem and other 2 IMO problems in the past! it's about the so called Pedal Triangle Trick!

this was posted on a forum by: fritue2000 and I thought it would be interesting to post it here also!

"The previous two IMO problems solved by the Pedal Triangle Trick.
(1996 IMO) Let P be a point inside the triangle ABC such that angle
APB - angle ACB = angle APC - angle ABC. Let D, E be the incenters of
triangles APB, APC respectively. Show that AP, BD, CE meet at a
point.

As I know, at the jury meeting at IMO 1996, there is a discussion
for this problem becasuse its solution is very similar to problem
2 of the 34th IMO. The techniques include 'inversion' and 'PTT', etc.
For example, http://home1.pacific.net.sg/~slwee/imo96/imo96op.htm )

(1993 IMO) Let D be a point inside the acute-angled triangle ABC such
that angle ADB = angle ACB + 90 degrees, and AC*BD = AD*BC.

(a) Calculate the ratio AB*CD/(AC*BD).

The well-known 'Pedal Triangle Trick' is "For any point D, let
X, Y, Z be feet of the altitudes from D to AB, BC and CA. Then,
YZ = (DA*BC)/2r, etc, where r is the circumradius of ABC."
The proof is very easy, since D, A, Y, Z lies on a circle with
diameter DA, by the law of sines, YZ = DA sin A = DA*(BC/2r).
(2003 IMO) Given is a cyclic quadrilateral ABCD and let P, Q, R
be feet of the altitudes from D to AB, BC and CA respectively.

Prove that if PR = RQ then the interior angle bisectors of the
angles <ABC and <ADC are concurrent on AC.
Solution) By PTT, PR=RQ implies (DA*BC)/2r = (DC/AB)/2r, so,
CD/DA=BC/AB implies the results. q.e.d.

The most noticeable thing is in the above solution, we did not
used the condtion 'ABCD is cyclic'. And as an IMO problem, it
is not so intersting because all three problems 1993, 1996,
2003 solved by exactly the same TWO ways (PTT and inversion).
Well, the PTT is really well-known, for example, it appears
at "Geometry Revisited" by Coexter. It also used for a proof
of Ptolemy's Theorem.

In a cyclic qudrilateral ABCD, the three points are on the
simson line, we have PR+RQ=RP, with the same notation of 2003
imo. Then, (DA*BC)/2r + (DC*AB)/2r = (DB*AC)/2r or
DA*BC + DC*AB = DB*AC. "

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Anonymous

334 posts

Anonymous

#5 h Jul 27, 2003, 7:18 PM • 2 Y

Y by Adventure10, Mango247

A bit of standard angle chasing shows that triangles DPR and DBC are similar and so are DQR and DBA. Thus PR=QR gives DC/CB = DR/RP = DR/RQ = DA/AB, and the result follows from the angle bisector theorem.

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Peter

3615 posts

Peter

#6 h Feb 20, 2004, 10:45 PM • 4 Y

Y by Adventure10, Mango247, Dream6068., ehuseyinyigit

Yeah, I had something like that too, was quite easy

solved in about 30 minutes (which is extremely few for me)

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Philip_Leszczynski

327 posts

Philip_Leszczynski

#7 h Dec 9, 2005, 3:06 AM • 3 Y

Y by Adventure10, Adventure10, Mango247

Lemma: Let $ABCD$ be a cyclic quadrilateral. Let $DR$ and $DP$ be the altitudes from $D$ to $AB$ and $DC$ , respectively. Let $RP$ intersect $AC$ at $Q$ . Then $\angle AQD$ is a right angle.

Proof of Lemma:

$\angle DRB = \angle DPB = \pi / 2$ , so $DRPB$ is cyclic. Then $\angle DRP = \angle DBP$ . From cyclic quadrilateral $ABCD$ , $\angle DBP = \angle DAP$ . $\angle DRP = \angle DAP$ , so $RAQD$ is cyclic. Thus $\angle AQD = \pi - \angle ARD = \pi / 2$ .

Proof:

Let the bisector of $\angle ABC$ meet $AC$ at S. Let the bisector of $\angle ADC$ meet $AC$ at T.
By the Lemma, $R,Q,P$ are collinear.
Let $\angle DRQ = \alpha$ . $\angle APD = \pi - \alpha - \angle RDP = \pi - \alpha - (\pi - \angle ABC) = \angle ABC - \alpha$ .
$\frac{RQ}{\sin \angle RDQ} = \frac{DQ}{\sin \alpha}$ , $\frac{PQ}{\sin \angle PDQ} = \frac{DQ}{\sin (\angle ABC - \alpha)}$ .
$\frac{RQ}{PQ} \cdot \frac{\sin PDQ}{\sin RDQ} = \frac{sin(\angle ABC - \alpha)}{\sin \alpha}$ .
$\angle RDQ = \pi - \angle RAQ = \angle BAC$ . Also $\angle PDQ = \angle BCA$ .
$\frac{RQ}{PQ} \cdot \frac{\sin \angle BCA}{\sin \angle BAC} = \frac{\sin (\angle ABC - \alpha)}{\sin \alpha}$ .
$\frac{RQ}{PQ} \cdot \frac{AB}{BC} = \frac{\sin \angle QCD}{\sin \angle QAD} = \frac{AD}{DC}$ .
So $RQ=PQ$ if and only if $\frac{AB}{BC} = \frac{AD}{DC}$ .

By the Angle Bisector Theorem,
$\frac{AS}{SC} = \frac{AB}{BC}$ and $\frac{AT}{TC} = \frac{AD}{DC}$ .
So $\frac{AB}{BC} = \frac{AD}{DC}$ if and only if $\frac{AS}{SC} = \frac{AT}{TC}$ .
This can happen if and only if $S=T$ .
So $PQ=QR$ if and only if the angle bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$ . QED.

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beta

3001 posts

beta

#8 h Dec 12, 2005, 12:31 AM • 2 Y

Y by Adventure10, Mango247

Yay I exploded this problem with like 20 cyclic quadrilaterals.

Okay first well-know fact: Simson's Line, so we know P, Q, R are collinear.

i'll go with the only if direction
Let X be the intersection of the angle bisectors. Extend BX, DX to Y, Z, where Y and Z lies on the circumcircle of ABCD. Y, Z lies on the perpendicular bisector of AC, hence YZ is a diameter, and it goes through midpoint M of AC. angle AMY = 90, angle ZDY = 90 because YZ is a diameter, so MYDX is cyclic, so angle ZYB = MDX = ADB. DX bisects ADC hence ADM = BDC = BAC = PDQ.
angle QAD=QPD. Hence triangle PQD is similar to triangle AMD, and triangle ADC is simlar to PDR. DM is a median hence DQ is a median => Q is the midpoint of PR as desired.

If direction:
Define X to be on AC such that DX bisects ADC, DX intersect the circumcircle at Z, M is the midpoint of AC, Y is the intersection of MZ and the circumcircle. triangle ADC is simlar to PDR. DM is a median and DQ is a median, so triangle PQD is similar to triangle AMD, angle ADM = PDQ=BAC = BDC. By definition XD bisects ADC hence XD bisects MDB so MDX = XDQ. QMY=90, YZ is still diameter, MYDX is cyclic, ZYX = MDX = ZDB. Now let BX intersect the circumcircle at Y'. XY'Z = XYZ, now Y' is unique(consider the locus of points Y' such that XY'Z = XYZ, which is a circle, and Y' lies on the cirucmcircle of ABCD. Two circle intersects in two points, one of which is Z, hence there's only one such point left and thus it's unique), hence Y = Y', thus B, X, Y are collinear. Y lies on the perpendicular bisector of AC so YA=YC, thus angle ABY = YBC.

QED

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FelixD

588 posts

FelixD

#9 h Sep 23, 2009, 7:46 AM • 2 Y

Y by Adventure10, Mango247

The following will be useful: Let $\triangle ABC$ be a triangle, $D$ an arbitrary point and $X$ , $Y$ , $Z$ the pedal points wrt the sides $BC$ , $CA$ , $AB$ . Then $YZ= \frac{AD \cdot BC}{2R}$ , where $R$ denotes the circrumradius of $\triangle ABC$ . That's very easy to prove, I won't write it down now

.
Using the lemma shown above, we have $PQ= \frac{CD \cdot AB}{2R}$ and $QR= \frac{AD \cdot BC}{2R}$ . Hence, $PQ =QR \Leftrightarrow \frac{AB}{BC} = \frac{AD}{DC}$ . Let $S$ denote the intersection of the angle bisectors of $\angle ABC$ and $\angle CDA$ . If $S \in AC$ , then $\frac{AB}{BC} = \frac{AS}{SC} = \frac{AD}{DC}$ . If $PQ=QR$ , then $\frac{AS_1}{S_1C} = \frac{AB}{BC} = \frac{AD}{DC} = \frac{AS_2}{S_2C}$ , hence $S_2=S_1=S$ , where $S_1$ and $S_2$ denote the intersections of the angle bisectors of angles $\angle ABC$ and $\angle CDA$ with the side $AC$ . Thus, the problem is proved.

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serialk11r

1449 posts

serialk11r

#10 h Dec 21, 2009, 3:44 AM • 1 Y

Y by Adventure10

Hmmm I came up with this, which uses a bit more of Simson line properties.

Let the angle bisector of $\angle{ABC}$ intersect $AC$ at $N$ , let the midpoint of $AC$ be $M$ , let $E$ and $F$ be points diametrically opposite on the circumcircle of $ABCD$ such that $EF$ is the perpendicular bisector of $AC$ , with $E$ on the same side of $AC$ as $B$ . Let $B'$ be the point diametrically opposite to $B$ .

$FDQN$ is cyclic, $FDMN$ is a rectangle (cyclic), so $\angle{FND} = \angle{FQD} = \angle{MDQ}$ .
$\angle{FND} = \angle{FBD} + \angle{BFE}$
$\angle{B'BF} = \angle{BFE}$ , since both are diameters.
Thus $\angle{B'BD} = \angle{BNE} = \angle{MDQ}$ .

$P,Q,R$ are collinear and are the Simson line of $D$ . $AC$ is the Simson line of $B'$ . By a well known property of Simson lines, $\angle{PQA} = \frac12\overarc{B'D}$ , but $\frac12\overarc{B'D} = \angle{B'BD} = \angle{BNE} = \angle{MDQ}$ and $\angle{PQA} = \angle{PDA} = \angle{CDR}$ , since $PDR$ is directly similar to $ADC$ . Thus we have a spiral similarity with center $D$ mapping $A,M,C$ to $P,Q,R$ respectively, and since $M$ is the midpoint of $AC$ , $Q$ is the midpoint of $PR$ so we are done.

The argument should work in reverse as well, with some tweaking here and there.

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hatchguy

555 posts

hatchguy

#11 h Dec 2, 2010, 11:05 AM • 4 Y

Y by AllanTian, Adventure10, Mango247, LeYohan

The concurrent with $AC$ bisectors condition is equivalent to $\frac{AB}{BC}=\frac{AD}{DC}$ $=> \frac{AD*BC}{AB*DC} =1$

Since $\angle DAB+ \angle DCB= 180$ and $\angle DCP+ \angle DCB= 180$ we have $\angle DAB= \angle DCP$ .

Therefore we have that $ARD$ is similar to $CPD$ and therefore $\frac{AR}{PC}=\frac{AD}{DC}$ . (1)

By Menelaus theorem for transversal $CQA$ in triangle $BRP$ we obtain:

$\frac{AB*RQ*PC}{AR*PQ*BC}=1$ . Therefore $\frac{RQ}{QP}=\frac{AR*BC}{AB*PC}=\frac{AD*BC}{AB*DC}$ (because of (1)). The conclusion is now obvious.

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MariusBocanu

429 posts

MariusBocanu

#12 h May 6, 2011, 9:22 AM • 1 Y

Y by Adventure10

Use that a quadrilateral is harmonic if and only if the diagonals are symmedians(easy to prove with polarity and crossed ratio).
Now, the problem is equivalent to this:
$\triangle{ABC}$ , $D$ is on the circumcircle of $\triangle{ABC}$ (on the arc $BC$ ). Denote $X,Y,Z$ the projections of $D$ onto $AB,BC,CA$ . $XY=YZ$ if and only if $D$ is on the symmedian with respect to $A$ .
Proof: See that $BXDY$ and $DYZC$ are inscribed in circles with diameter $BD$ ,respectively $CD$ . We have from Simson's line that $X-Y-Z$ are collinear. In $\triangle{XBY}$ (aplying law of sines) $XY=BDsinB$ , and in $\triangle{YZC}$ we have $YZ=DCsinC$ . So, $XY=YZ$ if and only if $\frac{BD}{CD}=\frac{sinC}{sinB}$ , but it happens only for symmedian(note that $\frac{sinBAM}{sinCAM}=\frac{sinB}{sinC}$ , where $M$ is the midpoint of $BC$ .

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arshakus

769 posts

arshakus

#13 h May 16, 2011, 12:53 PM • 1 Y

Y by Adventure10

hey guys)
I solved this problem in about 10 minutes, but I think there is some thing wrong because I didn't use the fact that $ABCD$ is cyclic.
Can it be so?

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nickthegreek

35 posts

nickthegreek

#14 h May 28, 2011, 7:21 AM • 2 Y

Y by Adventure10, Mango247

A different approach:

To begin with, it is clear that triangles $\bigtriangleup DRP$ and $\bigtriangleup DAC$ similar. (This follows by a very simple angle chasing, as in the above posts). This also gives us the fact that triangles $\bigtriangleup DAR$ and $\bigtriangleup DPC$ are similar, therefore $\frac{DA}{DC}=\frac{RA}{PC} (1)$

Also, points $P,Q,R$ are collinear (Simson's line)

It remains to prove the equivalence $PQ=QR \Leftrightarrow \frac{BA}{BC}=\frac{DA}{DC}$

Let's focus on triangle $\bigtriangleup RBP$ . Points $A,Q,C$ are collinear and lie on the lines $BR,RP,PB$ respectively. Applying Menelaus' Theorem, we have:
$\frac{AR}{AB}\cdot \frac{CB}{CP}\cdot \frac{QP}{QR}=1$

Therefore, $QP=QR\Leftrightarrow \frac{QP}{QR}=1\Leftrightarrow \frac{AB}{BC}=\frac{RA}{PC}\Leftrightarrow \frac{AB}{BC}= \frac{DA}{DC}$ ,

since $\frac{DA}{DC}=\frac{RA}{PC}$ by relation (1). The conclusion follows immediately.

Nick

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StefanS

149 posts

StefanS

#15 h Jun 28, 2011, 12:08 PM • 1 Y

Y by Adventure10

nickthegreek wrote:

A different approach:

Your solution is actually identical to hatchguy's. Take a look:

hatchguy wrote:

The concurrent with $AC$ bisectors condition is equivalent to $\frac{AB}{BC}=\frac{AD}{DC}$ $=> \frac{AD*BC}{AB*DC} =1$

Since $\angle DAB+ \angle DCB= 180$ and $\angle DCP+ \angle DCB= 180$ we have $\angle DAB= \angle DCP$ .

Therefore we have that $ARD$ is similar to $CPD$ and therefore $\frac{AR}{PC}=\frac{AD}{DC}$ . (1)

By Menelaus theorem for transversal $CQA$ in triangle $BRP$ we obtain:

$\frac{AB*RQ*PC}{AR*PQ*BC}=1$ . Therefore $\frac{RQ}{QP}=\frac{AR*BC}{AB*PC}=\frac{AD*BC}{AB*DC}$ (because of (1)). The conclusion is now obvious.

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MagicalToaster53

159 posts

MagicalToaster53

#96 h Feb 28, 2024, 10:00 PM

Y by

Notice that the concurrent angle condition is identical to the condition that $ABCD$ is a harmonic quadrilateral. Observe now that $\frac{PQ}{QD} = \frac{\sin \angle PDQ}{\sin \angle QPD} = \frac{\sin \angle BAC}{\sin \angle DAC},$ and $\frac{QR}{QD} = \frac{\sin \angle QDR}{\sin \angle DRQ} = \frac{\sin \angle ACB}{\sin \angle DBA}.$ Hence, $\frac{PQ}{QR} = \frac{\sin \angle BAC}{\sin \angle DAC} \cdot \frac{\sin \angle DBA}{\sin \angle ACB} = \frac{BC}{BA} \cdot \frac{DA}{DC} = 1 \iff ABCD \text{ is harmonic,}$ as desired. $\blacksquare$

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Markas

150 posts

Markas

#97 h May 5, 2024, 8:28 PM • 1 Y

Y by panche

If the bisectors meet on AC this gives us $\frac{AD}{DC} = \frac{AB}{BC}$ . From ABCD cyclic $\angle BAC = \angle PCD$ , so $\triangle CPD \sim \triangle ARD$ , so $\frac{AR}{CP} = \frac{AD}{DC}$ . From Menelaus on $\triangle BPR$ we get $\frac{BA}{AR}. \frac{RQ}{QP}. \frac{PC}{CB} = 1$ (we can do this since P, Q, R lie on one line aka Simpson line) and $\frac{BA}{AR}. \frac{PC}{CB} = \frac{AD}{DC}. \frac{DC}{AD} = 1$ . This means $\frac{PQ}{QP} = 1$ $\Rightarrow$ PQ = QP. In the other direction is the same because the pair of similar triangles is there, so we get that $\frac{BA}{CB} = \frac{AD}{DC}$ which is equivalent to the fact that the bisectors meet on AC. We are ready.

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BestAOPS

707 posts

BestAOPS

#98 h Jul 16, 2024, 8:25 PM • 1 Y

Y by teomihai

Notice that $P,Q,R$ lie on the Simson line.

Let $X$ be the second intersection of line $DQ$ with the circle $\omega$ circumscribing $ABCD$ . We claim that $\overline{BX} \parallel \overline{RQ}$ . Using directed angles,
$\angle BXD = \angle BAD = \angle RAD = \angle RQD,$ and thus, $\overline{BX}$ is parallel to the Simson line.

Next, note that by the angle bisector theorem, the bisector condition is equivalent to $ABCD$ being harmonic. Additionally, we have
$(A,C;B,D) \stackrel{X}{=} (A,C;\overline{BX} \cap \overline{AC}, Q) \stackrel{B}{=} (R,P;P_\infty,Q),$ and since $(R,P;P_\infty,Q) = -1 \iff PQ=QR$ , we are done.

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ezpotd

1293 posts

ezpotd

#99 h Aug 25, 2024, 3:01 AM • 1 Y

Y by m4thbl3nd3r

Complex bash, let $A = w, C = \overline{w}, B = b, D = d$ . Let $M,N = 1,-1$ be the arc midpoints of $AC$ . Note $BD,MN$ meet a point whose polar is parallel to $AC$ . If $BM,DN,AC$ are concurrent (or $BN, DM, AC$ ), then Brokard tells us that $AC$ is precisely this polar, so the condition can be reduces to the intersection of tangents from $A,C$ being the intersection of $BD, MN$ . The intersection of $BD, MN$ is just $\frac{b + d}{bd + 1}$ , so the desired condition is just $\frac{b + d}{bd + 1} = \frac{2}{w + \overline{w}}$ , rearrange to get $2bd + 2 = (b + d)(w + \overline{w})$ . Now for the condition that $PQ = QR$ . We dilate by two and instead find a condition for $P'Q' = Q'R'$ . By Simson line, it just suffices to show $p + r = 2q$ . The reflection of $D$ over $AB$ is just $w + b - \frac{bw}{d}$ , likewise the reflection of $D$ over $BC$ is just $b + \overline{w} - \frac{b\overline{w}}{d}$ . The reflection of $D$ over $AC$ is just $w + \overline w - \frac 1d$ . Now the condition is $2b + w + \overline{w} - \frac bd (w + \overline{w}) = 2(w + \overline{w} ) - \frac 2d$ . Multiplying by $d$ , an equivalent condition is $2bd + (d -b)(w + \overline{w}) = 2d(w + \overline{w}) - 2$ , rearranging we get $(b + d)(w + \overline{w}) = 2bd + 2$ , as desired. Since the conditions are equivalent, we can conclude that $PQ = QR$ and the concurrency happen at exactly the same configurations, we are done.

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peppapig_

280 posts

peppapig_

#100 h Oct 31, 2024, 7:20 PM

Y by

First, note that if the angle bisectors of $\angle ABC$ and $\angle ADC$ meet on segment $AC$ , we then have
$\frac{AB}{CB}=\frac{AD}{CD} \iff AB*CD=AD*BC \iff (AC;BD)=-1,$ and by the uniqueness of the harmonic conjugate, this means that $B$ is the unique point on $(ACD)$ such that $(AC;BD)=-1$ . By the harmonic quadrilateral configuration, this means that $BD$ is the $D$ -symmedian of $\triangle ACD$ . It then suffices to show that
$PQ=QR\iff BD\text{ is the } D\text{-symmedian of }\triangle ACD.$
To start, we make the following claims.

***

Claim 1. $P$ , $Q$ , $R$ are collinear.

Proof.
This is immediate from the Simson Line at $D$ with respect to $\triangle ABC$ .

***

Claim 2. There exists a spiral similarity centered at $D$ sending $\triangle DRP$ to $\triangle DAC$ .

Proof.
All angles below are directed.

We can prove this through (directed) angle chasing. First, note that since $\angle CQD=\angle CPD=90$ , we have that $C$ , $P$ , $D$ , $Q$ are concyclic, which gives us that
$\angle RPD=\angle QPD=\angle QCD=\angle ACD,$ and since $\angle ARD=\angle AQD=90$ , we have that $A$ , $R$ , $Q$ , $D$ concyclic, meaning that
$\angle PRD=\angle QRD=\angle QAD=\angle CAD,$ and these two combined gives us that $\triangle DRP\sim \triangle DAC$ , as desired. This proves our claim.

***

Now, let $M$ be the midpoint of $AC$ . Notice that $Q$ is the midpoint of $RP$ if and only if the spiral similarity described in Claim 2. sends $Q$ to $M$ . However, note that
$\angle Q\rightarrow M \iff \angle MDQ=\angle CDP \iff 90-\angle MDQ=90-\angle CDP \iff \angle DMQ=\angle DCP,$ and since $\angle DCP=\angle BAD$ by cyclic properties, and $\angle DMQ=\angle DMC$ , this gives us that
$\iff \angle BAD=\angle DMC.$
Now, let us fix $A$ , $C$ , and $D$ . We make the following claim.

***

Claim 3. If $\angle BAD=\angle DMC$ , then $B\neq D$ must be the unique point on $(ACD)$ such that $BD$ is the $D$ -symmedian of $\triangle ACD$ .

Proof.
Notice that since $\angle DMC$ is fixed, $\angle BAD$ must also be fixed, meaning that there is a unique point on $(ACD)$ such that $\angle BAD=\angle DMC$ . Now, we can angle chase. We get that
$\angle ADM=180-\angle AMD-\angle MAD=\angle CMD-\angle MAD=\angle BAD-\angle MAD=\angle BAC=\angle BDC,$ which means that $BD$ is indeed the $D$ -symmedian of $\triangle ABC$ , as desired. This proves our claim.

***

By Claim 3. means that $\angle BAD=\angle DMC$ is equivalent to the condition that $BD$ is the $D$ -symmedian of $ADC$ . However, we also established earlier that the former was equivalent to $Q$ being the midpoint of $PR$ , meaning that
$PQ=QR\iff BD\text{ is the } D\text{-symmedian of }\triangle ACD,$ as desired. Since
$BD\text{ is the } D\text{-symmedian of }\triangle ACD\iff (AC;BD)=-1\iff \text{the angle bisector of }\angle ABC \text{ and } \angle ADC \text{ meet on } AC,$ we have that the angle bisectors meet on $AC$ if and only if $PQ=QR$ , which is what we wished to prove. This completes our proof.

This post has been edited 1 time. Last edited by peppapig_, Nov 1, 2024, 10:11 PM
Reason: \implies but really \iff

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shendrew7

799 posts

shendrew7

#101 h Nov 4, 2024, 3:31 PM

Y by

Notice that
$PQ = QR \iff AD \sin \angle A = CD \sin \angle C \iff AD \cdot BC = CD \cdot AB. \quad \blacksquare$

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lnzhonglp

120 posts

lnzhonglp

#102 h Nov 28, 2024, 3:28 AM

Y by

Let $DP$ intersect $(ABC)$ again at $K$ . By Simson line $P, Q, R$ are collinear, and $\measuredangle QPD = \measuredangle PCD = \measuredangle AKD$ so $AK \parallel PQ$ . Then $(AC;BD) \overset{P}{=} (AP \cap (ABC), B; C, K) \overset{A}{=} (PR; Q\infty),$ which concludes by angle bisector theorem.

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DakshAggarwalRedsurgance

10 posts

DakshAggarwalRedsurgance

#103 h Jan 7, 2025, 2:02 PM

Y by

arshakus wrote:

hey guys)
I solved this problem in about 10 minutes, but I think there is some thing wrong because I didn't use the fact that $ABCD$ is cyclic.
Can it be so?

Yes. (same here) it is not a requirement at all . The fact that it is cyclic helps in the creation of simson line and just makes the diagram easy to construct plus it gets rid of nasty configuration issues. I was able to see some possible issues(which were not quite problematic are so) But there could have been some config that I missed. I asked a few people too and all of them agree. IMO committee wanted the problem to own its own merit instead of relying on configs so yippee

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Perelman1000000

136 posts

Perelman1000000

#104 h Jan 13, 2025, 4:50 PM

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$\boxed{nice problem}$

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Mquej555

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Mquej555

#105 h Jan 14, 2025, 11:42 AM

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How can it be to IMO?

$PQ = QR \iff AD \sin \angle A = CD \sin \angle C \iff AD \cdot BC = CD \cdot AB. \quad \blacksquare$

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LeYohan

56 posts

LeYohan

#106 h Feb 10, 2025, 10:12 PM

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My first IMO #4 sol :blush:

We notice that second condition is equivalent to showing that $ABCD$ is an harmonic quadrilateral because of the angle bisector theorem, so we want to proof that $\frac{CB}{AB}=\frac{CD}{AD}$ .

Assuming that $PQ=QR$ , it's well known that $P-Q-R$ , so we proceed by applying Menelaus's theorem on $\triangle PRB$ with transversal line $AC$ , getting that $\frac{CB}{AB}=\frac{CP}{AR}$ , so now we want to show that $\frac{CP}{AR}=\frac{CD}{AD}$ , but just notice that $\triangle DPC \sim \triangle DRA$ which gives the proportion.

Assuming that $\frac{CB}{AB}=\frac{CD}{AD}$ , by the previous proportion we know $\frac{CP}{AR}=\frac{CD}{AD}$ so by applying Menelaus's theorem again on $\triangle PRB$ with transversal line $AC$ , we obtain $PQ=QR$ , so we're done. $\square$

This post has been edited 1 time. Last edited by LeYohan, Feb 10, 2025, 10:12 PM

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Maximilian113

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Maximilian113

#107 h Feb 27, 2025, 12:36 AM

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By the Angle Bisector Theorem, the second condition is equivalent to $\frac{AB}{BC} = \frac{AD}{CD}.$
Now, note that $DQAR$ is a cyclic quadrilateral as $\angle DQA = \angle DRA = 90^\circ.$ Therefore by the Law of Sines $\frac{QR}{\sin \angle QDR} = \frac{DR}{\sin \angle DQR} \iff \frac{QR}{\sin \angle BAC} = \frac{DR}{\sin \angle DAC}.$ Similarly $\frac{PQ}{\sin \angle BCA} = \frac{DR}{\sin \angle DCA}.$ Dividing these equations, it follows that $PQ=QR \iff \frac{\sin \angle BAC}{\sin \angle BCA} = \frac{\sin \angle DAC}{\sin \angle DCA} \iff \frac{AB}{BC} = \frac{AD}{CD}$ by the Law of Sines. QED

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cj13609517288

1923 posts

cj13609517288

#108 h Apr 13, 2025, 2:46 AM

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The first condition, after multiplying by $2$ , subtracting from $a+b+c+d$ , then multiplying by $d$ , is equivalent to
$2ac+2bd=ab+bc+cd+da$ in complex. The second condition is equivalent to $ABCD$ being harmonic, which is
$(a-b)(c-d)=-(c-b)(a-d)\Longleftrightarrow 2ac+2bc=ab+bc+cd+da.$ Wow. $\blacksquare$

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cj13609517288

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cj13609517288

#109 h Apr 13, 2025, 3:08 AM • 1 Y

Y by Maximilian113

Okay fine here's a synthetic solution.

Let's prove the forwards direction. Let $B'$ be on $(ABCD)$ such that $BB'\parallel AC$ , and similarly for $D'$ . We have
$-1=(P,R;Q,\infty_{PR}) \stackrel{D}{=}(\infty_{\perp BA},\infty_{\perp BC};\infty_{\perp AC},\infty_{PR}) \stackrel{\text{rotate }90^\circ}{=}(\infty_{BA},\infty_{BC};\infty_{AC},\infty_{\perp PR}) \stackrel{B}{=}(A,C;B',B\infty_{\perp PR}\cap(ABCD)).$ But we claim that the unique such point $B\infty_{\perp PR}\cap(ABCD)$ is exactly $D'$ . Indeed, $BD$ is the $B$ -circumcenter cevian of triangle $BPR$ (since $D$ is the antipode), so $BD'$ , being isogonal to $BD$ , is perpendicular to $PR$ .

Thus $(A,C;B,D)=-1$ and the angle bisector thing follows by the angle bisector theorem. Everything in this proof was reversible so we get the other direction too. $\blacksquare$

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alexanderchew

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alexanderchew

#110 h Yesterday at 8:04 AM

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not so hard
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