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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
Help me this problem. Thank you
illybest   3
N 13 minutes ago by jasperE3
Find f: R->R such that
f( xy + f(z) ) = (( xf(y) + yf(x) )/2) + z
3 replies
illybest
Today at 11:05 AM
jasperE3
13 minutes ago
line JK of intersection points of 2 lines passes through the midpoint of BC
parmenides51   4
N 31 minutes ago by reni_wee
Source: Rioplatense Olympiad 2018 level 3 p4
Let $ABC$ be an acute triangle with $AC> AB$. be $\Gamma$ the circumcircle circumscribed to the triangle $ABC$ and $D$ the midpoint of the smallest arc $BC$ of this circle. Let $E$ and $F$ points of the segments $AB$ and $AC$ respectively such that $AE = AF$. Let $P \neq A$ be the second intersection point of the circumcircle circumscribed to $AEF$ with $\Gamma$. Let $G$ and $H$ be the intersections of lines $PE$ and $PF$ with $\Gamma$ other than $P$, respectively. Let $J$ and $K$ be the intersection points of lines $DG$ and $DH$ with lines $AB$ and $AC$ respectively. Show that the $JK$ line passes through the midpoint of $BC$
4 replies
parmenides51
Dec 11, 2018
reni_wee
31 minutes ago
AGM Problem(Turkey JBMO TST 2025)
HeshTarg   3
N 32 minutes ago by ehuseyinyigit
Source: Turkey JBMO TST Problem 6
Given that $x, y, z > 1$ are real numbers, find the smallest possible value of the expression:
$\frac{x^3 + 1}{(y-1)(z+1)} + \frac{y^3 + 1}{(z-1)(x+1)} + \frac{z^3 + 1}{(x-1)(y+1)}$
3 replies
HeshTarg
an hour ago
ehuseyinyigit
32 minutes ago
Shortest number theory you might've seen in your life
AlperenINAN   8
N an hour ago by HeshTarg
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.
8 replies
AlperenINAN
Yesterday at 7:51 PM
HeshTarg
an hour ago
No more topics!
concyclic with two vertices and the circumcenter
danepale   7
N Mar 18, 2024 by anantmudgal09
Source: MEMO 2016 I3
Let $ABC$ be an acute triangle such that $\angle BAC > 45^{\circ}$ with circumcenter $O$. A point $P$ is chosen inside triangle $ABC$ such that $A, P, O, B$ are concyclic and the line $BP$ is perpendicular to the line $CP$. A point $Q$ lies on the segment $BP$ such that the line $AQ$ is parallel to the line $PO$.

Prove that $\angle QCB = \angle PCO$.
7 replies
danepale
Aug 24, 2016
anantmudgal09
Mar 18, 2024
concyclic with two vertices and the circumcenter
G H J
Source: MEMO 2016 I3
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danepale
99 posts
#1 • 4 Y
Y by doxuanlong15052000, pisgood, Adventure10, Mango247
Let $ABC$ be an acute triangle such that $\angle BAC > 45^{\circ}$ with circumcenter $O$. A point $P$ is chosen inside triangle $ABC$ such that $A, P, O, B$ are concyclic and the line $BP$ is perpendicular to the line $CP$. A point $Q$ lies on the segment $BP$ such that the line $AQ$ is parallel to the line $PO$.

Prove that $\angle QCB = \angle PCO$.
Z K Y
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Dukejukem
695 posts
#2 • 4 Y
Y by pisgood, Tawan, Adventure10, Mango247
Lemma 1: Let $P, Q$ be points in the plane of $\triangle ABC$ and denote $R \equiv BP \cap CQ$ and $S \equiv BQ \cap CP.$ Let $\ell$ be the bisector of $\angle BAC.$ If $AP, AQ$ are symmetric in $\ell$, then $AR, AS$ are symmetric in $\ell.$

Proof of Lemma: By the Dual of Desargues' Involution Theorem, there exists an involution that swaps $(AB, AC), (AP, AQ), (AR, AS).$ Since $AB, AC$ and $AP, AQ$ are symmetric in $\ell$, this involution is reflection in $\ell.$ Hence $AR, AS$ are symmetric in $\ell.$ $\blacksquare$

Under inversion at $A$, we obtain the following corollary.

Corollary: Let $P, Q$ be points in the plane of $\triangle ABC$ and denote by $R$ and $S$ the second intersections of $\odot(ABP), \odot(ACQ)$ and $\odot(ABQ), \odot(ACP).$ If $AP, AQ$ are symmetric in $\ell$, then $AR, AS$ are symmetric in $\ell.$

Lemma 2: Let $P, Q$ be isogonal conjugates WRT $\triangle ABC.$ Then $\measuredangle BPC + \measuredangle BQC = \measuredangle BAC.$

Proof of Lemma: By examining $\triangle BQC$, we have \[ \measuredangle BPC + \measuredangle BQC = \measuredangle BPC + \measuredangle QBC + \measuredangle BCQ = \measuredangle BPC + \measuredangle ABP + \measuredangle PCA = \measuredangle BAC, \]where the last step follows from examining quadrilateral $ABPC.$ $\blacksquare$
_______________________________________________________________________________________________________________________________________________

Main Proof: Let $Q^*$ denote the second intersection of $\odot(BOC)$ and $\odot(AB)$ and let $R$ be the projection of $B$ onto $CA.$
[asy]
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label("$B$", (1015.5883543077671,1035.434690063097), NE * labelscalefactor); 
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dot((1010.1406864706545,1019.3916953613821),linewidth(3.pt) + dotstyle); 
label("$O$", (1009.1074013264267,1019.3796019956849), NW * labelscalefactor); 
dot((1017.3999315177143,1010.5378928206159),linewidth(3.pt) + dotstyle); 
label("$P$", (1017.5326402021692,1011.0132808743178), NE * labelscalefactor); 
dot((1007.1756474018246,1015.7560398396774),linewidth(3.pt) + dotstyle); 
label("$Q^*$", (1006.691773397018,1014.636722768431), NW * labelscalefactor); 
dot((1016.9487027379275,1016.2120584526925),linewidth(3.pt) + dotstyle); 
label("$Q$", (1017.0612981671627,1016.3747965225178), NE * labelscalefactor); 
dot((1015.4920652065816,1008.9736305721863),linewidth(3.pt) + dotstyle); 
label("$R$", (1015.1759300271364,1007.8317221380233), E * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
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[/asy]
Since $BR, BO$ are symmetric in the bisector of $\angle ABC$, the corollary for points $R, O$ and $\triangle ABC$ implies that $BP, BQ^*$ are symmetric in the bisector of $\angle ABC.$ Moreover, from $QA \parallel PO$, we infer that \[ \measuredangle BQ^*A + \measuredangle BQA = 90^{\circ} + \measuredangle BPO = 90^{\circ} + \measuredangle BAO = \measuredangle BCA. \]Hence, by Lemma 2, $Q^*$ is uniquely determined as the isogonal conjugate of $Q$ WRT $\triangle ABC.$

Thus, using $\measuredangle BQ^*C = \measuredangle BOC = 2\measuredangle BAC$, we obtain $\measuredangle BQC = -\measuredangle BAC$ by Lemma 2. $(\star)$

By $(\star)$ we have $\measuredangle PCQ = \measuredangle BQC - 90^{\circ} = \measuredangle OCB.$ Hence, $\measuredangle PCO = \measuredangle QCB$, as desired.
This post has been edited 1 time. Last edited by Dukejukem, Aug 24, 2016, 11:05 PM
Reason: Added diagram
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FabrizioFelen
241 posts
#3 • 4 Y
Y by pisgood, ILIILIIILIIIIL, Adventure10, Mango247
Let $\measuredangle ABP=\beta$ and $\measuredangle QAB=\theta$ $\Longrightarrow$ $\measuredangle  PAQ$ $=$ $\beta+\theta$, so from $PO\parallel AQ$ and $ABPO$ is cyclic we get $\measuredangle PAQ$ $=$ $\measuredangle PQA$ $=$ $\measuredangle OAB$ $=$ $\measuredangle OBA$ $=$ $\beta+\theta...(\star)$. Let $R\equiv BP\cap \odot (ABC)$ $\Longrightarrow$ $\measuredangle ARB$ $=$ $\measuredangle ACB$ $=90^{\circ}$ $-\beta$ $-\theta$, so from $\measuredangle AQB$ $=$ $\beta$ $+$ $\theta$ we get $\measuredangle QAR$ $=$ $90^{\circ}$, but by $(\star)$ we get $PA$ $=$ $PQ$, since $\measuredangle QAR$ $=$ $90^{\circ}$ and $PA$ $=$ $PQ$ we get $PA$ $=$ $PQ$ $=$ $PR$, since $CP\perp QR$ we get $ CQ$ $=$ $CR$ $\Longrightarrow$ $\measuredangle BAC$ $=$ $\measuredangle CRQ$ $=$ $\measuredangle CQR...(\star\star)$. Since $O$ is the circumcenter of $\odot (ABC)$ we get $\measuredangle OCB$ $=$ $90^{\circ}-\measuredangle BAC$. On the other hand, so from $(\star\star)$ we get $\measuredangle PCQ$ $=$ $90^{\circ}-\measuredangle PQC$ $=$ $90^{\circ}-\measuredangle BAC$ hence $\measuredangle OCB$ $=$ $\measuredangle PCQ$ $\Longrightarrow$ $\measuredangle PCO=\measuredangle QCB.$
[asy]  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
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label("$B$", (17.94000000000068,12.51999999999999), NE * labelscalefactor); dot((20.52000000000077,18.76000000000002),linewidth(0.7500000000000009pt) + dotstyle + invisible,UnFill(0)); 
label("$A$", (20.60000000000078,18.88000000000000), NE * labelscalefactor); dot((25.42000000000095,12.88000000000002),linewidth(0.7500000000000009pt) + dotstyle + invisible,UnFill(0)); 
label("$C$", (25.50000000000097,13.00000000000000), NE * labelscalefactor); dot((21.76466970387325,14.81555808656031),linewidth(0.7500000000000009pt) + dotstyle + invisible,UnFill(0)); 
label("$O$", (21.84000000000083,14.93999999999999), NE * labelscalefactor); dot((21.96813861016855,16.51463181832704),linewidth(0.7500000000000009pt) + dotstyle + invisible,UnFill(0)); 
label("$P$", (21.98000000000084,16.63999999999999), NE * labelscalefactor); dot((20.03077221953434,14.67468747841660),linewidth(0.7500000000000009pt) + dotstyle + invisible,UnFill(0)); 
label("$Q$", (19.74000000000075,14.75999999999999), NE * labelscalefactor); dot((23.90550500080287,18.35457615823758),linewidth(0.7500000000000009pt) + dotstyle + invisible,UnFill(0)); 
label("$R$", (23.98000000000091,18.47999999999999), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */  [/asy]
This post has been edited 1 time. Last edited by FabrizioFelen, Aug 27, 2016, 5:14 AM
Reason: add graphic
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randomusername
1059 posts
#4 • 2 Y
Y by pisgood, Adventure10
From the conditions, $A,P,O,B$ lie on a circle in this order ($\alpha>45^\circ$).

Need: $\angle PCQ=\angle BCO=90^\circ-\alpha$, same as $\frac{CP}{QP}=\tan \alpha$. $(1)$

$M$ and $N$: centers of $BPC$ and $APOB$, resp.
$MN$ is perpendicular to their radical axis, $BP$.
Also $OM\perp BC$ and $ON\perp AB$, so $\angle ONM=\angle ABP=\varphi$, $\angle OMN=\angle CBP=\theta$.
From $\triangle OMN$, we have $\frac{OM}{ON}=\frac{\sin \varphi}{\sin \theta}$. $(2)$

Let $\rho$ be the radius of circle $APOB$ and $R$ be the circumradius.
$\rho=\frac{AB}{2\sin AOB}=R\cdot \frac1{2\cos \gamma}$.
Since $\angle PQA=\angle QPO=\angle BPO=\angle BAO=90^\circ-\gamma$, and $\angle BPA=2\gamma$, we get $QP=AP$, so
$QP=AP=2\rho \sin \varphi=R\frac1{\cos\gamma}\sin\varphi$ $(3)$
but also
$CP=BC\sin \theta=2R\sin \alpha\sin \theta$, $(4)$
$\frac{OM}{ON}=\frac{R\cos\alpha}{\rho}=2\cos\alpha\cos\gamma$. $(5)$

Now $(2-5)$ imply $(1)$, so done.
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rkm0959
1721 posts
#5 • 4 Y
Y by bgn, K.titu, Adventure10, Mango247
Finally got around to solving this problem.

Angle chasing gives $\angle AQP = 90 - \angle C$ and $\angle APB = \angle AOB = 2 \angle C$.
This gives $PA=PQ$. Now if we set $R = BP \cap \omega$ ($\omega$ is the circumcircle of $\triangle ABC$) we have $\angle ARB = \angle C$, so $\triangle QAR$ is a right triangle. This gives $PQ=PR$, and it is easy to see that $CP \perp RQ$, so $\angle PCQ = 90 - \angle PRC = 90-\angle A = \angle OCB$, which gives the desired.
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MarkBcc168
1595 posts
#6 • 2 Y
Y by Adventure10, Mango247
A bit difficult.

Let $M$ be the midpoint of $BC$ and let $O'$ be the circumcenter of $\triangle BQC$. Let $S$ and $T$ be the circumcenters of $\triangle APB, \triangle AQB$ respectively. As
$$\angle AQB = 180^{\circ}-\angle BPO = 90^{\circ} + \angle C$$, we get that $\angle TAB =\angle C$ or $S$ is midpoint of $OT$. But $MS$ and $O'T$ are both perpendicular to $CP$, thus $M$ is the midpoint of $OO'$.

Hence $\angle BQC = 180^{\circ}-\angle A$ so $\angle PCQ=90^{\circ}-\angle A$, implying the desired isogonality.
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Pinionrzek
54 posts
#7
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$\textbf{Lemma}$ Define the projections of $P$ on $AB, AC$ as $X, Y$ and as $Z, W$ the projections of $Q$. Then $P, Q$ are isogonal conjugates wrt $\angle BAC$ if and only if $X, Y, Z, W$ are concyclic. Moreover the midpoint of $PQ$ is the center of this circle.



The proof is a simple angle chasing exercise. The problem statement is equivalent with the isogonal conjugacy of $Q, O$ wrt $\angle PCB$. Let $M$ be the midpoint of $BC$ and $D$ be another intersection of $AQ$ with the circumcircle of $A, P, O, B$. Then $PQDO$ is a parallelogram. Denote $N$ as the intersection of its diagonals. By our lemma the statement is equivalent with showing that $ND = NP = NM$ which is equivalent to $\angle PMD = 90^{\circ}$. However, $PM= MC= MB$ and $PD = OB$ since $PODB$ is an isosceles trapezoid. By this observation also $\angle DPM = \angle OBM$, thus $\triangle OBM \equiv \triangle DPM$, hence $\angle PMD = \angle OMB = 90^{\circ }$.
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anantmudgal09
1980 posts
#8
Y by
Elegant conjuring problem! Posting for storage.
danepale wrote:
Let $ABC$ be an acute triangle such that $\angle BAC > 45^{\circ}$ with circumcenter $O$. A point $P$ is chosen inside triangle $ABC$ such that $A, P, O, B$ are concyclic and the line $BP$ is perpendicular to the line $CP$. A point $Q$ lies on the segment $BP$ such that the line $AQ$ is parallel to the line $PO$.

Prove that $\angle QCB = \angle PCO$.

Observe that $\angle QAO=\angle POA = \angle PBA$ and $\angle PAO=\angle PBO$ hence $\angle PAQ = \angle OBA$. Likewise, $\angle PQA = \angle QAB+\angle QBA=\angle OAB-\angle OAQ+\angle PBA=\angle OAB$ since $\angle OAQ=\angle PBA$. Thus, $PA=PQ$, and so if we reflect $Q$ across $P$, we get point $R$ such that $\angle RAQ=90^{\circ}$, and $\angle ARB=\angle ARQ = 90^{\circ}-\angle AQP=90^{\circ}-\angle OAB=\angle C$, hence $R$ lies on the circumcircle of $ABC$. Since $CP \perp QR$, it follows that $\angle CQB=180^{\circ}-\angle CQP=180^{\circ}-\angle CRP=180^{\circ}-\angle CAB$ hence $\angle QCP=(180^{\circ}-\angle CAB)-90^{\circ} = 90^{\circ}-\angle CAB=\angle OCB$ so $CO, CQ$ are isogonal in angle $PCB$, which proves our claim.
This post has been edited 1 time. Last edited by anantmudgal09, Mar 18, 2024, 11:02 PM
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