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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
equal angles
jhz   6
N 19 minutes ago by aidan0626
Source: 2025 CTST P16
In convex quadrilateral $ABCD, AB \perp AD, AD = DC$. Let $ E$ be a point on side $BC$, and $F$ be a point on the extension of $DE$ such that $\angle ABF = \angle DEC>90^{\circ}$. Let $O$ be the circumcenter of $\triangle CDE$, and $P$ be a point on the side extension of $FO$ satisfying $FB =FP$. Line BP intersects AC at point Q. Prove that $\angle AQB =\angle DPF.$
6 replies
jhz
Mar 26, 2025
aidan0626
19 minutes ago
Very easy number theory
darij grinberg   101
N an hour ago by sharknavy75
Source: IMO Shortlist 2000, N1, 6th Kolmogorov Cup, 1-8 December 2002, 1st round, 1st league,
Determine all positive integers $ n\geq 2$ that satisfy the following condition: for all $ a$ and $ b$ relatively prime to $ n$ we have \[a \equiv b \pmod n\qquad\text{if and only if}\qquad ab\equiv 1 \pmod n.\]
101 replies
darij grinberg
Aug 6, 2004
sharknavy75
an hour ago
Looks difficult number mock#5
Physicsknight   1
N an hour ago by Physicsknight
Source: Vadilal factory
Consider $a_1,a_2,\hdots,a_b$ be distinct prime numbers. Let $\alpha_i=\sqrt{a}, \,\mathrm{A}=\mathbb{Q}[\alpha_1,\alpha_2,\hdots,\alpha_b]$
Let $\gamma=\sum\,\alpha_i$
[list]
[*] Prove that $[\mathrm{A}:\mathbb{Q}]=2^b$
[*] Prove that $\mathrm{A}=\mathbb{Q}[\gamma];$ and deduce that the minimum polynomial $f(X)$ of $\gamma$ over $\mathbb{Q}$ has degree $2^b.$
[*] Prove that $f(X)$ factors in $\mathbb{Z}_a[X]$ into a product of polynomials of degree $\leq 4 \,(a\ne 2)$ either of degree $\leq 8\,(a=2)$
[/list]
1 reply
Physicsknight
Yesterday at 1:51 PM
Physicsknight
an hour ago
IMO 2011 Problem 5
orl   83
N an hour ago by Ihatecombin
Let $f$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers $m$ and $n$, the difference $f(m) - f(n)$ is divisible by $f(m- n)$. Prove that, for all integers $m$ and $n$ with $f(m) \leq f(n)$, the number $f(n)$ is divisible by $f(m)$.

Proposed by Mahyar Sefidgaran, Iran
83 replies
orl
Jul 19, 2011
Ihatecombin
an hour ago
No more topics!
IMO Shortlist 2012, Geometry 4
lyukhson   83
N Mar 15, 2025 by KevinYang2.71
Source: IMO Shortlist 2012, Geometry 4
Let $ABC$ be a triangle with $AB \neq AC$ and circumcenter $O$. The bisector of $\angle BAC$ intersects $BC$ at $D$. Let $E$ be the reflection of $D$ with respect to the midpoint of $BC$. The lines through $D$ and $E$ perpendicular to $BC$ intersect the lines $AO$ and $AD$ at $X$ and $Y$ respectively. Prove that the quadrilateral $BXCY$ is cyclic.
83 replies
lyukhson
Jul 29, 2013
KevinYang2.71
Mar 15, 2025
IMO Shortlist 2012, Geometry 4
G H J
Source: IMO Shortlist 2012, Geometry 4
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lyukhson
127 posts
#1 • 12 Y
Y by anantmudgal09, Davi-8191, Jasurbek, donotoven, geometry6, ike.chen, Adventure10, Mango247, Rounak_iitr, and 3 other users
Let $ABC$ be a triangle with $AB \neq AC$ and circumcenter $O$. The bisector of $\angle BAC$ intersects $BC$ at $D$. Let $E$ be the reflection of $D$ with respect to the midpoint of $BC$. The lines through $D$ and $E$ perpendicular to $BC$ intersect the lines $AO$ and $AD$ at $X$ and $Y$ respectively. Prove that the quadrilateral $BXCY$ is cyclic.
This post has been edited 1 time. Last edited by Amir Hossein, Jul 29, 2013, 5:39 PM
Reason: Edited Title and Changed Format of The Problem.
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Burii
137 posts
#2 • 6 Y
Y by giftctb, Mobashereh, Miku3D, donotoven, Adventure10, and 1 other user
Let $H$ be the orthocenter of triangle $ABC$. If $M$ is midpoint of $DY$, let $P$ be the reflection of $E$ with respect point $M$. Of course $BCPY$ is isosceles trapezoid. The bisector of $\angle BAC$ and perpendicular bisector of $BC$ intersect at $M$, so $M$ lies on circumcircle of triangle $ABC$. Thus we have $BD\cdot DC=AD \cdot DM$. Points $H$ and $O$ are isogonal cojugate wrt triangle, so easy angle chasing and we have that $\angle AXD= \angle DMP$. Thus quadrangle $AXMP$ is cyclic, so $DX\cdot DP=AD\cdot DM = BD\cdot DC$ - quadrangle $BSCP$ is cyclic, so points $B, X, C, P, T$ lie on the same circle.
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sunken rock
4381 posts
#3 • 6 Y
Y by Burii, Mobashereh, Miku3D, donotoven, Adventure10, and 1 other user
Nice problem, nice idea for solution; however it could be made shorter:

Clearly $\Delta ADX\sim\Delta AMO$, so $\Delta ADX$ is isosceles, and so is $\Delta DPM$, since $M$ is the center of the rectangle $DEYP$; as $\angle MPD=\angle MDP=\angle XDA=\angle XAD$ we infer $AXMP$ cyclic, and we are done.

Best regards,
sunken rock
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exmath89
2572 posts
#4 • 7 Y
Y by sunny2000, Sx763_, Miku3D, lneis1, donotoven, Adventure10, and 1 other user
Solution
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JuanOrtiz
366 posts
#5 • 4 Y
Y by AlastorMoody, donotoven, Adventure10, H_Taken
Absolutely trivial for a G4. Here is the solution. Call Y' the reflection of Y through the perpendicular bisector of BC. Clearly if we show that BXCY' is cyclic, we're done. So if we show $DX(DY)=DB(DC)$, we're done. So if we show $DX(EY)=BD(DC)$, we're done. Assume $\angle B \textgreater \angle C$. Call $(A/2)+C=\delta$

Now, note that $\angle DAX = \angle DAC - \angle OAC = A/2 - 90 + B = \angle ADX = \delta$, so $DX = \frac{AD}{2sin \delta}$. Also, we can see that $EY=2DMtan\delta$ since $\angle EDY = \angle BDA = \delta$. So we wish to prove $DB(DC)=\frac{DM(DA)}{cos \delta}$. But call $Z$ the midpoint of arc $BC$ that does not contain $A$ in the circumcircle of $ABC$. We see that $DA(DZ)=DB(DC)$ and $DZ = \frac{DM}{cos \delta}$. So we're done.

Horrible problem, nothing interesting about it. Very poor choice for G4, in my opinion.
This post has been edited 2 times. Last edited by JuanOrtiz, Aug 8, 2013, 1:11 AM
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XmL
552 posts
#6 • 4 Y
Y by donotoven, Adventure10, Mango247, and 1 other user
I swear this has been posted a while ago, but I can't find it!!! my solution was the same as sunken rock's
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Math-lover123
304 posts
#7 • 3 Y
Y by donotoven, Adventure10, Mango247
You are right XmL here it is : http://www.artofproblemsolving.com/Forum/resources.php?c=114&cid=75&year=2013&sid=7c75a2addf93750e55d6db969f0c3964
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Andrew64
33 posts
#8 • 2 Y
Y by donotoven, Adventure10
lyukhson wrote:
Let $ABC$ be a triangle with $AB \neq AC$ and circumcenter $O$. The bisector of $\angle BAC$ intersects $BC$ at $D$. Let $E$ be the reflection of $D$ with respect to the midpoint of $BC$. The lines through $D$ and $E$ perpendicular to $BC$ intersect the lines $AO$ and $AD$ at $X$ and $Y$ respectively. Prove that the quadrilateral $BXCY$ is cyclic.
As shown in the figure below.

$H$ is the intersection of $\circ O$ and $AY$.

$F$ is the midpoint of $AD$.

$DG//EY$ and $DG=EY$.

It's obvious that

$B,C,G,Y$ are concyclic. and

$HE=HD$

so $H$ is the midpoint of $YD$.

As
$DF\times DY=\frac{AD}{2}\times DY=AD\times \frac{DY}{2}=AD\times DH=BD\times DC$

$B,F,C,Y$ are concyclic.

Further, as
$\triangle DXF \sim \triangle DYE$

we have

$DF\times DY = DX\times EY = DX\times DG$

Thus

$B,Y,G,C,F,X$ are concyclic.
Attachments:
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v_Enhance
6872 posts
#9 • 18 Y
Y by TheStrangeCharm, Math-Ninja, AlastorMoody, richrow12, Miku3D, Muaaz.SY, lneis1, HamstPan38825, donotoven, Geometry285, Tellocan, Adventure10, Ritwin, shafikbara48593762, Funcshun840, NicoN9, Lcz, and 1 other user
Assume $AB < AC$ and let $L$ be the midpoint of arc $BC$. Denote by $Y'$ the reflection of $E$ over $L$; evidently $BY'YC$ is a isosceles (and hence cyclic) trapezoid. Now $\triangle DLY' \sim \triangle AXD$ because they are both isosceles (since $\triangle AXD \sim \triangle AOL$) and share an angle, so we see that $\angle XAL = \angle XAD = \angle DY'L = \angle XY'l$ and so $AXLY'$ is cyclic; hence $XD \cdot DY' = AD \cdot DL = BD \cdot DC$ so $XBY'C$ is cyclic as well, and we're done.

[asy]/* DRAGON 0.0.9.6 
Homemade Script by v_Enhance. */

import olympiad; 
import cse5; 
size(13cm); 
real lsf=0.8000; 
real lisf=2011.0; 
defaultpen(fontsize(10pt));

/* Initialize Objects */
pair A = (0.64, 0.96);
pair B = (0.4, -1.0);
pair C = (3.0, -1.0);
pair D = IntersectionPoint(Line(incenter(A,B,C),A,lisf),Line(B,C,lisf));
pair E = B+C-D;
pair M = midpoint(B--C);
pair O = circumcenter(A,B,C);
pair X = IntersectionPoint(Line(D,D+O-M,lisf),Line(A,O,lisf));
pair Y = IntersectionPoint(Line((2)*(O)-X,E,lisf),Line(A,D,lisf));
pair L = midpoint(D--Y);
pair Y_prime = (2)*(L)-E;

/* Draw objects */
draw(A--B, rgb(0.2,0.6,0.0));
draw(B--C, rgb(0.2,0.6,0.0));
draw(C--A, rgb(0.2,0.6,0.0));
draw(circumcircle(A,B,C), rgb(0.0,0.4,0.0) + linewidth(1.0) + linetype("4 4"));
draw(circumcircle(B,C,X), rgb(0.6,0.0,0.6));
draw(A--O, rgb(0.2,0.6,0.0));
draw(E--Y, rgb(0.6,0.0,0.6));
draw(Y_prime--L, rgb(0.6,0.0,0.6) + linetype("4 4"));
draw(D--X, rgb(0.0,0.6,0.6));
draw(D--Y_prime, rgb(0.6,0.0,0.6) + linetype("4 4"));
draw((abs(dot(unit(L-Y_prime),unit(D-Y_prime))) < 1/2011) ? rightanglemark(L,Y_prime,D) : anglemark(L,Y_prime,D), rgb(0.6,0.0,0.6));
draw((abs(dot(unit(D-A),unit(X-A))) < 1/2011) ? rightanglemark(D,A,X) : anglemark(D,A,X), rgb(0.6,0.0,0.6));
draw(A--D, rgb(0.2,0.6,0.0));
draw(D--Y, rgb(0.6,0.0,0.6));
draw(O--M, rgb(0.2,0.6,0.0));
draw(M--L, rgb(0.6,0.0,0.6));

/* Place dots on each point */
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(M);
dot(O);
dot(X);
dot(Y);
dot(L);
dot(Y_prime);

/* Label points */
label("$A$", A, lsf * dir(110));
label("$B$", B, lsf * dir(225));
label("$C$", C, lsf * dir(-45));
label("$D$", D, lsf * dir(45));
label("$E$", E, lsf * dir(45));
label("$M$", M, lsf * dir(45));
label("$O$", O, lsf * dir(45));
label("$X$", X, lsf * dir(60));
label("$Y$", Y, lsf * dir(-90));
label("$L$", L, lsf * dir(45));
label("$Y'$", Y_prime, lsf * dir(-90));
[/asy]

In other news, 2048th post!
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Math-lover123
304 posts
#10 • 3 Y
Y by parola, Adventure10, Mango247
Let $F$ be a foot of perpendicular from $X$ to $AD$.Then it it easy to prove that triangles $DEY$ and $XFD$ are similar.
So $EY/DY=EY/2DL=AD/2XD$ so $EY\cdot XD=DL\cdot AD=BD\cdot CD=CE\cdot CD$ ( we have obviously $BD=CE$) so triangles $CEY$ and $CDX$ are similar.Then $\angle XCY=90$, analogously $\angle XBY=90$ so done. :)
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Particle
179 posts
#11 • 2 Y
Y by Adventure10, Mango247
Calculative proof
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AnonymousBunny
339 posts
#12 • 3 Y
Y by Roct-7, Adventure10, Mango247
Sketch, will complete this soon:
Let $M$ be the midpoint of $OM.$ Extend $OM$ to meet the circumcircle of $\triangle ABC$ at $J.$ It is easy to prove that $AJ$ bisects $\angle BAC,$ and as a consequence $A,D,J$ are collinear. Now, note that $\triangle ADX \sim \triangle DJM.$ By some easy ratio computations using PoP of $D$ wrt the circumcircle of $\triangle ABC,$ it follows that $\triangle BDX \sim \triangle BEY$ and similarly $\triangle CDX \sim \triangle CEY.$ The rest is simple angle chasing.
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junioragd
314 posts
#13 • 2 Y
Y by Adventure10, Mango247
Let M be the midpoint of the minor arc BC and let Y' be the reflection of Y across the perpendicular bisector of BC.Now,it is easy to see that that BCYY' is an iscocelles trapezoid and X,D and Y' are collinear and M is the midpoint of DY(simple angle chasing).Now,by angle chasing we get <Y'MA=<AXY',so AXMY' is a cyclic.Also,BMCA is a cyclic so we have DB*DC=DM*DA=DX*DY',so we are finished.
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jayme
9775 posts
#14 • 1 Y
Y by Adventure10
Dear Mathlinkers,
see also
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=530055
Sincerely
Jean-Louis
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utkarshgupta
2280 posts
#15 • 3 Y
Y by AlastorMoody, Adventure10, Mango247
Call the midpoint of $BC$ $M$ and $AD \cap \odot  ABC = P$
Obviously $OMP$ is a straight line

Consider reflection in line $OP$ as $R(OP)$.
$R(OP) : D \to E, B \to C, O \to O, M \to M, P\to P, Y \to Y'$
$R(OP):DPY \to EPY'$
$\implies EPY'$ is a straight line.
Also easily, $XDY'$ is a straight line.
$\angle XAD = \angle \frac{A}{2}-90+ \angle B = 90-\angle C-\angle \frac{A}{2}$

$\angle PY'D = \angle EY'D = 90-\angle DEY' = 90-\angle EDP = 90 - \angle DBP - \angle DPB = 90-\angle C-\angle \frac{A}{2}$

$\implies \angle XAD=\angle PY'D$
$\implies XAY'P$ is cyclic
$\implies DX \cdot DY'=DA \cdot DP$
$\implies DX \cdot DY'=DB \cdot DC$
$\implies BXCY'$ is cyclic.
$\implies \angle BY'C = 180-\angle BXC$
But by $R(OP):\angle BY'C  \to \angle CYB$
$\implies \angle CYB+\angle BXC=180$

$\implies BXCY$ is cyclic and thus we are done
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