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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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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
Asymmetric FE
sman96   15
N a minute ago by youochange
Source: BdMO 2025 Higher Secondary P8
Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that$$f(xf(y)-y) + f(xy-x) + f(x+y) = 2xy$$for all $x, y \in \mathbb{R}$.
15 replies
sman96
Feb 8, 2025
youochange
a minute ago
Mount Inequality erupts in all directions!
BR1F1SZ   3
N 3 minutes ago by sqing
Source: Austria National MO Part 1 Problem 1
Let $a$, $b$ and $c$ be pairwise distinct nonnegative real numbers. Prove that
\[
(a + b + c) \left( \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2} \right) > 4.
\](Karl Czakler)
3 replies
1 viewing
BR1F1SZ
May 5, 2025
sqing
3 minutes ago
Looks like power mean, but it is not
Nuran2010   4
N 36 minutes ago by sqing
Source: Azerbaijan Al-Khwarizmi IJMO TST 2025
For $a,b,c$ positive real numbers satisfying $a^2+b^2+c^2 \geq 3$,show that:

$\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+\frac{a+b+c}{9} \geq \frac{4}{3}$.
4 replies
+1 w
Nuran2010
2 hours ago
sqing
36 minutes ago
ISI UGB 2025 P6
SomeonecoolLovesMaths   1
N 40 minutes ago by ronitdeb
Source: ISI UGB 2025 P6
Let $\mathbb{N}$ denote the set of natural numbers, and let $\left( a_i, b_i \right)$, $1 \leq i \leq 9$, be nine distinct tuples in $\mathbb{N} \times \mathbb{N}$. Show that there are three distinct elements in the set $\{ 2^{a_i} 3^{b_i} \colon 1 \leq i \leq 9 \}$ whose product is a perfect cube.
1 reply
SomeonecoolLovesMaths
3 hours ago
ronitdeb
40 minutes ago
Real parameter equation
L.Lawliet03   1
N an hour ago by Mathzeus1024
For which values of the real parameter $a$ does only one solution to the equation $(a+1)x^{2} -(a^{2} + a + 6)x +6a = 0$ belong to the interval (0,1)?
1 reply
L.Lawliet03
Nov 3, 2019
Mathzeus1024
an hour ago
a inequality problem
Polus425   1
N 2 hours ago by Mathzeus1024
$x_1,x_2\; are\; such\; two\; different\; real\; numbers:\; $
$(x_1 ^2 -2x_1 +4ln\, x_1)+(x_2 ^2 -2x_2 +4ln\, x_2)- x_1 ^2 x_2 ^2=0$
$prove\; that:\; x_1+x_2\ge 3$
1 reply
Polus425
Dec 19, 2019
Mathzeus1024
2 hours ago
Find the range of 'f'
agirlhasnoname   1
N 2 hours ago by Mathzeus1024
Consider the triangle with vertices (1,2), (-5,-1) and (3,-2). Let Δ denote the region enclosed by the above triangle. Consider the function f:Δ-->R defined by f(x,y)= |10x - 3y|. Then the range of f is in the interval:
A)[0,36]
B)[0,47]
C)[4,47]
D)36,47]
1 reply
agirlhasnoname
May 14, 2021
Mathzeus1024
2 hours ago
Function of Common Area [China HS Mathematics League 2021]
HamstPan38825   1
N 3 hours ago by Mathzeus1024
Define the regions $M, N$ in the Cartesian Plane as follows:
\begin{align*}
M &= \{(x, y) \in \mathbb R^2 \mid 0 \leq y \leq \text{min}(2x, 3-x)\} \\
N &= \{(x, y) \in \mathbb R^2 \mid t \leq x \leq t+2 \}
\end{align*}for some real number $t$. Denote the common area of $M$ and $N$ for some $t$ be $f(t)$. Compute the algebraic form of the function $f(t)$ for $0 \leq t \leq 1$.

(Source: China National High School Mathematics League 2021, Zhejiang Province, Problem 5)
1 reply
HamstPan38825
Jun 29, 2021
Mathzeus1024
3 hours ago
Functions
Entrepreneur   2
N 3 hours ago by alexheinis
Let $f(x)$ be a polynomial with integer coefficients such that $f(0)=2020$ and $f(a)=2021$ for some integer $a$. Prove that there exists no integer $b$ such that $f(b) = 2022$.
2 replies
Entrepreneur
Aug 18, 2023
alexheinis
3 hours ago
Inequalities
sqing   3
N 4 hours ago by sqing
Let $ a,b,c\geq 0 , (a+8)(b+c)=9.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{38}{23}$$Let $ a,b,c\geq 0 , (a+2)(b+c)=3.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2(2\sqrt{3}+1)}{5}$$
3 replies
sqing
Yesterday at 12:50 PM
sqing
4 hours ago
Plz help
Bet667   1
N 4 hours ago by Mathzeus1024
f:R-->R for any integer x,y
f(yf(x)+f(xy))=(x+f(x))f(y)
find all function f
(im not good at english)
1 reply
Bet667
Jan 28, 2024
Mathzeus1024
4 hours ago
Minimum value of 2 variable function
girishpimoli   6
N 4 hours ago by Mathzeus1024
Minimum value of $x^2+y^2-xy+3x-3y+4$ , Where $x,y\in\mathbb{R}$
6 replies
girishpimoli
Apr 1, 2024
Mathzeus1024
4 hours ago
Function prob
steven_zhang123   4
N 5 hours ago by Mathzeus1024
If the function $f(x)=x^2+ax+b$ has a maximum value of $M$ and a minimum value of $m$ in the interval $[0,1]$. Confirm whether the value of $M-m$ depends on $a$ or $b$.
4 replies
steven_zhang123
Sep 22, 2024
Mathzeus1024
5 hours ago
Angle Formed by Points on the Sides of a Triangle
xeroxia   4
N Today at 8:03 AM by jainam_luniya

In triangle $ABC$, points $D$, $E$, and $F$ lie on sides $BC$, $CA$, and $AB$, respectively, such that
$BD = 20$, $DC = 15$, $CE = 13$, $EA = 8$, $AF = 6$, $FB = 22$.

What is the measure of $\angle EDF$?


4 replies
xeroxia
Yesterday at 10:28 AM
jainam_luniya
Today at 8:03 AM
Strange angle condition and concyclic points
lminsl   127
N May 3, 2025 by reni_wee
Source: IMO 2019 Problem 2
In triangle $ABC$, point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$, respectively, such that $PQ$ is parallel to $AB$. Let $P_1$ be a point on line $PB_1$, such that $B_1$ lies strictly between $P$ and $P_1$, and $\angle PP_1C=\angle BAC$. Similarly, let $Q_1$ be the point on line $QA_1$, such that $A_1$ lies strictly between $Q$ and $Q_1$, and $\angle CQ_1Q=\angle CBA$.

Prove that points $P,Q,P_1$, and $Q_1$ are concyclic.

Proposed by Anton Trygub, Ukraine
127 replies
lminsl
Jul 16, 2019
reni_wee
May 3, 2025
Strange angle condition and concyclic points
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2019 Problem 2
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lminsl
544 posts
#1 • 31 Y
Y by e_plus_pi, Hexagrammum16, abdelkrim, cauchyguess, char2539, Mr.Chagol, Seicchi28, a_simple_guy, Carpemath, FadingMoonlight, AlastorMoody, samuel, RAMUGAUSS, Davrbek, Vietjung, JustKeepRunning, poplintos, sabrinamath, OlympusHero, k12byda5h, bariboaa, math31415926535, HWenslawski, megarnie, TFIRSTMGMEDALIST, ImSh95, SADAT, GeoKing, Adventure10, deplasmanyollari, Rounak_iitr
In triangle $ABC$, point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$, respectively, such that $PQ$ is parallel to $AB$. Let $P_1$ be a point on line $PB_1$, such that $B_1$ lies strictly between $P$ and $P_1$, and $\angle PP_1C=\angle BAC$. Similarly, let $Q_1$ be the point on line $QA_1$, such that $A_1$ lies strictly between $Q$ and $Q_1$, and $\angle CQ_1Q=\angle CBA$.

Prove that points $P,Q,P_1$, and $Q_1$ are concyclic.

Proposed by Anton Trygub, Ukraine
This post has been edited 1 time. Last edited by djmathman, Jul 18, 2019, 4:40 AM
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lminsl
544 posts
#4 • 20 Y
Y by TUGUMEANDREW, Assassino9931, jeteagle, Wizard_32, pavel kozlov, JustKeepRunning, HolyMath, Gumnaami_1945, jnzm02, myh2910, Kagebaka, k.ata.tkm, AllanTian, Illuzion, ImSh95, rayfish, Adventure10, Mango247, ihatemath123, Rounak_iitr
Let $R$ be the intersection of $AA_1$, $BB_1$. Let $\Omega$ denote the circumcircle of triangle $ABC$, and let $A_0=AA_1 \cap \Omega, B_0=BB_1 \cap \Omega$.

Step 1. The points $P,Q, A_0, B_0$ are concyclic.
Since $PQ \parallel BC$, $\angle PQR=\angle ABB_0=\angle PA_0B_0$. The conclusion follows.

Step 2. The points $P,Q, B_0, P_1$ are concyclic, hence $P_1$ and $Q_1$ lies in $\odot(PQA_0B_0)$.

Note that $\angle B_1B_0C=\angle BAC=\angle B_1P_1C$, or $(B_0B_1P_1C)$ are concyclic. Hence $$\angle B_0P_1P=\angle B_1CB_0=\angle ABB_0=\angle PQB_0,$$or $P, Q, B_0, P_1$ are cyclic. This concludes the proof. $\square$
This post has been edited 3 times. Last edited by lminsl, Jul 16, 2019, 3:51 PM
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mofumofu
179 posts
#5 • 14 Y
Y by Flash_Sloth, zhangruichong, jeteagle, Wizard_32, JustKeepRunning, HolyMath, Not_real_name, ImSh95, ike.chen, Quidditch, CyclicISLscelesTrapezoid, SADAT, Adventure10, Rounak_iitr
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(13cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -7.63900015685, xmax = 14.779588377, ymin = -2.38665690592, ymax = 8.57946862605;  /* image dimensions */

 /* draw figures */
draw((1.,4.96)--(-0.798898966671,-0.708008219337)); 
draw((1.,4.96)--(7.68847429913,-0.833514663193)); 
draw((7.68847429913,-0.833514663193)--(-0.798898966671,-0.708008219337)); 
draw(circle((5.34391751537,3.21734977311), 4.68043258616), green); 
draw(circle((0.279066234699,2.06933899636), 2.97920568808), green); 
draw(circle((1.75406921626,4.18630229154), 4.07948863865), blue); 
draw(circle((3.47180131167,1.0560363319), 4.62068545199), red); 
draw((3.88543194857,7.66473878747)--(2.88060830443,-0.762418678059)); 
draw((-2.28184506812,3.59163983096)--(1.27443840444,-0.738667551996)); 
draw((-1.14173984034,1.31288664799)--(7.68847429913,-0.833514663193)); 
draw((-0.798898966671,-0.708008219337)--(5.73321976419,5.08551776613)); 
draw((0.386107411678,0.343008384736)--(3.00779285736,0.304240392932)); 
 /* dots and labels */
dot((1.,4.96),dotstyle); 
label("$C$", (1.0680093857,5.04959989258), NE * labelscalefactor); 
dot((7.68847429913,-0.833514663193),dotstyle); 
label("$B$", (7.75122752106,-0.739384830301), NE * labelscalefactor); 
dot((-0.798898966671,-0.708008219337),dotstyle); 
label("$A$", (-0.736145744742,-0.613878386444), NE * labelscalefactor); 
dot((3.3230046287,2.94782789959),dotstyle); 
label("$A_1$", (3.38987859704,3.04149679088), NE * labelscalefactor); 
dot((-0.22800366956,1.09078053776),dotstyle); 
label("$B_1$", (-0.171366747388,1.19027674399), NE * labelscalefactor); 
dot((0.386107411678,0.343008384736),dotstyle); 
label("$P$", (0.456165471895,0.437238080855), NE * labelscalefactor); 
dot((3.00779285736,0.304240392932),dotstyle); 
label("$Q$", (3.0761124874,0.405861469891), NE * labelscalefactor); 
dot((1.27443840444,-0.738667551996),dotstyle); 
label("$R$", (1.33471057889,-0.645254997408), NE * labelscalefactor); 
dot((2.88060830443,-0.762418678059),dotstyle); 
label("$S$", (2.95060604354,-0.66094330289), NE * labelscalefactor); 
dot((-2.28184506812,3.59163983096),dotstyle); 
label("$P_1$", (-2.22653476554,3.68471731564), NE * labelscalefactor); 
dot((3.88543194857,7.66473878747),dotstyle); 
label("$Q_1$", (3.9546575944,7.76367674098), NE * labelscalefactor); 
dot((-1.14173984034,1.31288664799),dotstyle); 
label("$E$", (-1.08128846535,1.40991302074), NE * labelscalefactor); 
dot((5.73321976419,5.08551776613),dotstyle); 
label("$D$", (5.7901893358,5.17510633644), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

Let $AA_1,BB_1$ intersect the circumcircle of $\triangle ABC$ again at $D,E$ respectively, and $PB_1,QA_1$ meet $BC$ again at $R,S$ respectively. It is clear that $A,R,C,P_1$ and $B,S,C,Q_1$ are cyclic. By Reim's theorem, we know that $P,Q,D,E$ are cyclic. Taking power of a point at $B_1$, we get $B_1B\cdot B_1E=B_1A\cdot B_1C=B_1R\cdot B_1P_1$. Also, since $\triangle B_1PQ\sim \triangle B_1RB$, we get $B_1B\cdot B_1P=B_1R\cdot B_1Q$. Combining the two gives us $B_1P_1\cdot B_1P= B_1Q\cdot B_1E$, whence $P_1\in (PQDE)$. Similarly $Q_1\in (PQDE)$ are we are done.
This post has been edited 4 times. Last edited by mofumofu, Jul 17, 2019, 12:39 PM
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nguyenhaan2209
111 posts
#6 • 5 Y
Y by top1csp2020, jeteagle, ImSh95, Pranav1056, Adventure10
$PB_1$ cuts $CB$,$AB$ at $E$,$D$,$QA_1$ cuts $CA$,$AB$ at $G$,$F$. Apply Pappus for ${B_1GA}$ and ${A_1BE}$ we have $B_1B$-$A_1G$=$Q$, $B_1E$-$A_1A$=$P$, $GE$-$AB$ collinear or parallel but $PQ$//$AB$ thus $GE//AB$. It's obvious that $P_1$,$Q_1$ also lies on $(AGE)$ thus by Reim we have the conclusion!
This post has been edited 1 time. Last edited by nguyenhaan2209, Jul 16, 2019, 12:48 PM
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Seicchi28
252 posts
#7 • 6 Y
Y by a_simple_guy, jeteagle, Wizard_32, CrazyMathMan, ImSh95, Adventure10
Nice problem :)

Let $A_1Q \cap AC = X$, $B_1P \cap BC = Y$, and extension of $PQ$ meets $AC,BC$ at $M,N$ respectively.
From the condition of the problem, $MN||AB$.

Lemma $XY||AB||MN$.
Proof

Construct the circumcircle $\omega$ of $\bigtriangleup CXY$. Denote $P'$ as the second intersection of $YP$ and $\omega$. Since \[ \angle PP'C = \angle YP'C = \angle YXC = \angle BAC, \]therefore $P'=P_1$. Similarly, $Q'=Q_1$. So, $C,X,Y,P_1,Q_1$ are concyclic and
\[  \angle Q_1P_1P = \angle Q_1P_1Y = \angle Q_1XY = \angle Q_1QP.  \]
Done. $\blacksquare$
This post has been edited 2 times. Last edited by Seicchi28, Jul 16, 2019, 1:00 PM
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JustKeepRunning
2958 posts
#9 • 2 Y
Y by ImSh95, Adventure10
mofumofu wrote:
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(15cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -7.63900015685, xmax = 14.779588377, ymin = -2.38665690592, ymax = 8.57946862605;  /* image dimensions */

 /* draw figures */
draw((1.,4.96)--(-0.798898966671,-0.708008219337)); 
draw((1.,4.96)--(7.68847429913,-0.833514663193)); 
draw((7.68847429913,-0.833514663193)--(-0.798898966671,-0.708008219337)); 
draw(circle((5.34391751537,3.21734977311), 4.68043258616), green); 
draw(circle((0.279066234699,2.06933899636), 2.97920568808), green); 
draw(circle((1.75406921626,4.18630229154), 4.07948863865), blue); 
draw(circle((3.47180131167,1.0560363319), 4.62068545199), red); 
draw((3.88543194857,7.66473878747)--(2.88060830443,-0.762418678059)); 
draw((-2.28184506812,3.59163983096)--(1.27443840444,-0.738667551996)); 
draw((-1.14173984034,1.31288664799)--(7.68847429913,-0.833514663193)); 
draw((-0.798898966671,-0.708008219337)--(5.73321976419,5.08551776613)); 
draw((0.386107411678,0.343008384736)--(3.00779285736,0.304240392932)); 
 /* dots and labels */
dot((1.,4.96),dotstyle); 
label("$C$", (1.0680093857,5.04959989258), NE * labelscalefactor); 
dot((7.68847429913,-0.833514663193),dotstyle); 
label("$B$", (7.75122752106,-0.739384830301), NE * labelscalefactor); 
dot((-0.798898966671,-0.708008219337),dotstyle); 
label("$A$", (-0.736145744742,-0.613878386444), NE * labelscalefactor); 
dot((3.3230046287,2.94782789959),dotstyle); 
label("$A_1$", (3.38987859704,3.04149679088), NE * labelscalefactor); 
dot((-0.22800366956,1.09078053776),dotstyle); 
label("$B_1$", (-0.171366747388,1.19027674399), NE * labelscalefactor); 
dot((0.386107411678,0.343008384736),dotstyle); 
label("$P$", (0.456165471895,0.437238080855), NE * labelscalefactor); 
dot((3.00779285736,0.304240392932),dotstyle); 
label("$Q$", (3.0761124874,0.405861469891), NE * labelscalefactor); 
dot((1.27443840444,-0.738667551996),dotstyle); 
label("$R$", (1.33471057889,-0.645254997408), NE * labelscalefactor); 
dot((2.88060830443,-0.762418678059),dotstyle); 
label("$S$", (2.95060604354,-0.66094330289), NE * labelscalefactor); 
dot((-2.28184506812,3.59163983096),dotstyle); 
label("$P_1$", (-2.22653476554,3.68471731564), NE * labelscalefactor); 
dot((3.88543194857,7.66473878747),dotstyle); 
label("$Q_1$", (3.9546575944,7.76367674098), NE * labelscalefactor); 
dot((-1.14173984034,1.31288664799),dotstyle); 
label("$E$", (-1.08128846535,1.40991302074), NE * labelscalefactor); 
dot((5.73321976419,5.08551776613),dotstyle); 
label("$D$", (5.7901893358,5.17510633644), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

Let $AA_1,BB_1$ intersect the circumcircle of $\triangle ABC$ again at $D,E$ respectively, and $PB_1,QA_1$ meet $BC$ again at $R,S$ respectively. It is clear that $A,R,C,P_1$ and $B,S,C,Q_1$ are cyclic. By Reim's theorem, we know that $P,Q,D,E$ are cyclic. Taking power of a point at $B_1$, we get $B_1B\cdot B_1E=B_1A\cdot B_1C=B_1R\cdot B_1P_1$. Also, since $\triangle B_1PQ\sim \triangle B_1RB$, we get $B_1B\cdot B_1P=B_1R\cdot B_1Q$. Combining the two gives us $B_1P_1\cdot B_1P= B_1Q\cdot B_1E$, whence $P_1\in (PQDE)$. Similarly $Q_1\in (BQDE)$ are we are done.

How do you import a graph form geogebra?
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Wictro
119 posts
#10 • 3 Y
Y by ImSh95, Adventure10, Mango247
Seicchi28 wrote:
Nice problem :)
Let $A_1Q$ meet $AC$ at $X$, $B_1P$ meet $BC$ at $Y$, extension of $PQ$ meets $AC,BC$ at $M,N$ respectively.
From the question, $MN||AB$.

Lemma

Now by the lemma, create the circumcircle $\omega$ of $\bigtriangleup CXY$. Denote $P'$ as the second intersection of $YP$ and $\omega$. Since \[ \angle PP'C = \angle YP'C = \angle YXC = \angle BAC, \]therefore $P'=P_1$. Similarly, $Q'=Q_1$. So, $C,X,Y,P_1,Q_1$ are concyclic and
\[  \angle Q_1P_1P = \angle Q_1P_1Y = \angle Q_1XY = \angle Q_1QP.\]Done.

Exactly how I did it. Though I proved the lemma with Pappus and then Desargues on perspective triangles.
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TUGUMEANDREW
5 posts
#13 • 2 Y
Y by ImSh95, Adventure10
Since Problem 2 was geometry and problem 2 is medium level, is there a chance that Problem 4 won't be geometry?
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richrow12
411 posts
#14 • 2 Y
Y by ImSh95, Adventure10
Seicchi28 wrote:
Nice problem :)

Let $A_1Q \cap AC = X$, $B_1P \cap BC = Y$, and extension of $PQ$ meets $AC,BC$ at $M,N$ respectively.
From the condition of the problem, $MN||AB$.

Lemma $XY||AB||MN$.
Proof

Construct the circumcircle $\omega$ of $\bigtriangleup CXY$. Denote $P'$ as the second intersection of $YP$ and $\omega$. Since \[ \angle PP'C = \angle YP'C = \angle YXC = \angle BAC, \]therefore $P'=P_1$. Similarly, $Q'=Q_1$. So, $C,X,Y,P_1,Q_1$ are concyclic and
\[  \angle Q_1P_1P = \angle Q_1P_1Y = \angle Q_1XY = \angle Q_1QP.  \]
Done. $\blacksquare$

Just the same but my proof of the lemma as follows:

Consider two lines $(P, B_1, Y)$ and $(Q, A_1, X)$. By Pappus's theorem points $S=PA_1\cap QB_1$, $T=PX\cap QY$ and $C=B_1Y\cap A_1X$ are collinear. Now, consider trinagles $APX$ and $BQY$. We claim that they are perspective. Indeed, $AP\cap BQ = S$, $AX\cap BY=C$ and $PX\cap QY=T$ are colinear. Hence, by Desargues's theorem lines $AB$, $PQ$ and $XY$ are concurrent. Since $AB\parallel PQ$ we have $AB\parallel PQ\parallel XY$.
TUGUMEANDREW wrote:
Since Problem 2 was geometry and problem 2 is medium level, is there a chance that Problem 4 won't be geometry?

Usually (at least in recent years) problems 1,2,4,5 have different topics (ACGN).
This post has been edited 1 time. Last edited by richrow12, Jul 16, 2019, 1:56 PM
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-[]-
97 posts
#15 • 5 Y
Y by richrow12, ImSh95, carefully, Adventure10, Mango247
Finally, the first medium geometry problem since the introduction of the new protocol! So surprised that they didn't vote for G1 this time.
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Math-wiz
6107 posts
#16 • 3 Y
Y by ImSh95, Adventure10, Mango247
TUGUMEANDREW wrote:
Since Problem 2 was geometry and problem 2 is medium level, is there a chance that Problem 4 won't be geometry?

P6 can be a Geo. P4,5 would be NT and Combi
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MS_Kekas
275 posts
#17 • 65 Y
Y by a_simple_guy, Wictro, richrow12, MathPassionForever, Mr.Chagol, rmtf1111, RudraRockstar, SHREYAS333, FadingMoonlight, AlastorMoody, UlanKZ, 62861, MNJ2357, GourmetSalad, Illuzion, Systematicworker, anantmudgal09, juckter, niyu, samuel, Anzoteh, Kayak, AlbertEinstein1905, jeteagle, Euler1728, Wizard_32, myh2910, Gumnaami_1945, Bassiskicking, Geronimo_1501, Abbas11235, aa1024, MarkBcc168, Aryan-23, amar_04, Atpar, k12byda5h, CHLORG1, CrazyMathMan, blackbluecar, 554183, tigerzhang, Bubu-Droid, Seicchi28, Wizard0001, rayfish, ImSh95, ike.chen, Pranav1056, Quidditch, parmenides51, IMUKAT, sabkx, wenwenma, CyclicISLscelesTrapezoid, kamatadu, Adventure10, Mango247, LoloChen, EpicBird08, Kingsbane2139, khina, ihatemath123, farhad.fritl, Sedro
Lol, this is my problem! Didn't expect to see this at P2
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AngleChasingXD
109 posts
#18 • 5 Y
Y by jeteagle, ImSh95, DashTheSup, Adventure10, Mango247
Beautiful problem!!! :)
My solution is kinda long, especially because I used Moving Points during it.

Let $B_1P_1$ cut $AB$ at $U$ and $Q_1A_1$ cut $AB$ at $V$. Moreover, let $L$ be the intersection point of lines $B_1P_1$ and $BC$ and $K$ the intersection point of lines $Q_1A_1$ and $AC$.

Since $\angle PP_1C\equiv \angle BAC$, we conclude that $AP_1CU$ is cyclic. Similarly, since $\angle QQ_1C\equiv \angle ABC$, $BQ_1CV$ is cyclic, too. Keep these in mind as (1).

The conclusion is equivalent to $\angle QPP_1\equiv \angle QQ_1P_1$. However, since $PQ\parallel AB$, we have $\angle QPP_1\equiv \angle AUP_1$, so it suffices to prove $\angle AUP_1\equiv \angle QQ_1P_1$, which is equivalent to $UVP_1Q_1$ being cyclic.

$\textbf{Claim.}$ We have that $KL\parallel AB$.

$\textbf{Proof:}$ We show this by the Moving Points method. Fix $\Delta ABC$, points $A_1$, $B_1$ and animate $P$ on $AA_1$. Consider $L'$ on $AB$ such that $KL'\parallel AB$. We prove that the map $$L\to P\to Q\to K\to L'~~by~~\overline{BC}\to \overline{AA_1}\to \overline{BB_1}\to \overline{AC}\to \overline{BC}$$is projective, thus it would ve enough to check $L=L'$ for $3$ distinct positions of $P$.
$\textbf{(1)}$ $L\to P$ by $\overline{BC}\to \overline{AA_1}$ is projective since it is a perspectivity through $B_1$ from $BC$ to $AA_1$.
$\textbf{(2)}$ $P\to Q$ by $\overline{AA_1}\to \overline{BB_1}$ is projective since it preserves linear motion.
$\textbf{(3)}$ $Q\to K$ by $\overline{BB_1}\to \overline{AC}$ is projective since it is a perspectivity through $A_1$ from $BB_1$ to $AC$.
$\textbf{(4)}$ $K\to L'$ by $ \overline{AC}\to \overline{BC}$ is projective since it preserves linear motion.

Now it suffices to check for $3$ distinct positions of $P$ the desired parallelism.
$\textbf{(1)}$ When $P\in AA_1\cap BB_1$, we actually have $Q=P$, $L=B$ and $K=A$ so there is nothing to prove.
$\textbf{(2)}$ When $P=A$, $Q=B$ so $C=K=L$, again nothing to prove.
$\textbf{(3)}$ When $P$ is the point at infinity on $\overline{AA_1}$, $Q$ is the point at infinity on $\overline{BB_1}$ so $B_1L\parallel AA_1$ and $A_1K\parallel BB_1$. Then apply Thales' Lemma to obtain $CB_1\cdot CA_1=CK\cdot CB$ and $CA_1\cdot CB_1=CL\cdot CA$, hence $CK\cdot CB=CL\cdot CA$ so indeed $KL\parallel BC$.$\blacksquare$

Now we are ready to finish. From (1) and Power of a Point Theorem we get $PB_1\cdot B_1U=AB_1\cdot CB_1$. By our lemma and Thales' Lemma, $B_1K\cdot B_1U=B_1L\cdot B_1A$. Divide these two last relations in order to get $PB_1\cdot B_1L=CB_1\cdot B_1K$, which again by Power of a Point leads to $CP_1KL$ being cyclic. In a similar manner one proves the cyclicity of $CQ_1KL$, so the pentagon $CP_1KLQ_1$ is cyclic. But then $\angle KQ_1P_1\equiv \angle KP_1C$, and since by (1) we have $\angle KCP_1\equiv \angle AUP_1$, we arrive at $\angle AUP_1\equiv \angle VQ_1P_1$, hence $UVQ_1P_1$ is cyclic and done.
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prague123
230 posts
#19 • 2 Y
Y by ImSh95, Adventure10
-[]- wrote:
Finally, the first medium geometry problem since the introduction of the new protocol! So surprised that they didn't vote for G1 this time.

This probably just means that G1 and G2 got eliminated as known problems, and that G3 contained traces of combinatorial geometry.
This post has been edited 1 time. Last edited by prague123, Jul 16, 2019, 2:13 PM
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Math-wiz
6107 posts
#20 • 2 Y
Y by ImSh95, Adventure10
MS_Kekas wrote:
Lol, this is my problem! Didn't expect to see this at P2

Wowww. Where are you from?
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