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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
Sums of products of entries in a matrix
Stear14   0
a few seconds ago
(a) $\ $Each entry of an $\ 8\times 8\ $ matrix equals either $\ 1\ $ or $\ 2.\ $ Let $\ A\ $ denote the sum of eight products of entries in each row. Also, let $\ B\ $ denote the sum of eight products of entries in each column. Find the maximum possible value of $\ A-B.\ $ In other words, find
$$ {\rm max}\ \left[ \sum_{i=1}^8\ \prod_{j=1}^8\ a_{ij} - 
\sum_{j=1}^8\ \prod_{i=1}^8\ a_{ij} \right]
$$
(b) $\ $Same question, but for a $\ 2025\times 2025\ $ matrix.
0 replies
Stear14
a few seconds ago
0 replies
a father and his son are skating around a circular skating rink
parmenides51   2
N 17 minutes ago by thespacebar1729
Source: Tournament Of Towns Spring 1999 Junior 0 Level p1
A father and his son are skating around a circular skating rink. From time to time, the father overtakes the son. After the son starts skating in the opposite direction, they begin to meet five times more often. What is the ratio of the skating speeds of the father and the son?

(Tairova)
2 replies
parmenides51
May 7, 2020
thespacebar1729
17 minutes ago
Sums of n mod k
EthanWYX2009   1
N 19 minutes ago by Martin.s
Source: 2025 May 谜之竞赛-3
Given $0<\varepsilon <1.$ Show that there exists a constant $c>0,$ such that for all positive integer $n,$
\[\sum_{k\le n^{\varepsilon}}(n\text{ mod } k)>cn^{2\varepsilon}.\]Proposed by Cheng Jiang
1 reply
EthanWYX2009
May 26, 2025
Martin.s
19 minutes ago
Easy P4 combi game with nt flavour
Maths_VC   1
N 3 hours ago by p.lazarov06
Source: Serbia JBMO TST 2025, Problem 4
Two players, Alice and Bob, play the following game, taking turns. In the beginning, the number $1$ is written on the board. A move consists of adding either $1$, $2$ or $3$ to the number written on the board, but only if the chosen number is coprime with the current number (for example, if the current number is $10$, then in a move a player can't choose the number $2$, but he can choose either $1$ or $3$). The player who first writes a perfect square on the board loses. Prove that one of the players has a winning strategy and determine who wins in the game.
1 reply
Maths_VC
May 27, 2025
p.lazarov06
3 hours ago
No more topics!
P on altitude and concurrent lines
Snakes   7
N Oct 3, 2022 by P2nisic
Source: Moldova TST 2017, B3
Let $\omega$ be the circumcircle of the acute nonisosceles triangle $\Delta ABC$. Point $P$ lies on the altitude from $A$. Let $E$ and $F$ be the feet of the altitudes from P to $CA$, $BA$ respectively. Circumcircle of triangle $\Delta AEF$ intersects the circle $\omega$ in $G$, different from $A$. Prove that the lines $GP$, $BE$ and $CF$ are concurrent.
7 replies
Snakes
Mar 6, 2017
P2nisic
Oct 3, 2022
P on altitude and concurrent lines
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G H BBookmark kLocked kLocked NReply
Source: Moldova TST 2017, B3
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Snakes
11033 posts
#1 • 3 Y
Y by microsoft_office_word, Adventure10, Mango247
Let $\omega$ be the circumcircle of the acute nonisosceles triangle $\Delta ABC$. Point $P$ lies on the altitude from $A$. Let $E$ and $F$ be the feet of the altitudes from P to $CA$, $BA$ respectively. Circumcircle of triangle $\Delta AEF$ intersects the circle $\omega$ in $G$, different from $A$. Prove that the lines $GP$, $BE$ and $CF$ are concurrent.
This post has been edited 1 time. Last edited by Snakes, Mar 6, 2017, 7:50 PM
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rmtf1111
698 posts
#2 • 1 Y
Y by Adventure10
Outline of the solution
This post has been edited 1 time. Last edited by rmtf1111, Mar 6, 2017, 8:05 PM
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FabrizioFelen
241 posts
#4 • 1 Y
Y by Adventure10
Let $H=AP\cap BC$, since $PFBH$ and $AEPF$ is cyclic we get $\measuredangle ABC=\measuredangle FPA=\measuredangle FEA$ $\Longrightarrow$ $BCEF$ is cyclic, let $A'$ the antipode of $A$ wrt $\omega$ since $PF\parallel A'B$ and $PE\parallel CA'$ we get the perpendicular bisectors of $BF$ and $CE$ intersect in $PA'$ be that point $O$ $\Longrightarrow$ $O$ is the circumcenter of $\odot (BCEF)$ $\Longrightarrow$ $O,P, A'$ are collinear, it so easy note that $G=\overline {OPA'}\cap \omega$. On the other hand by radical axis in $\odot (AFE)$, $\odot (BCEF)$ and $\omega$ we get $AG$, $FE$, $BC$ are concurrent in $X$, let $Y$ $=$ $BE$ $\cap$ $CF$ $\Longrightarrow$ by Brocard's theorem in $\odot (BCEF)$ we get $OY$ $\perp$ $AX$, but $OG$ $\perp$ $AY$ hence $O$ $,$ $G$ $,$ $Y$ are collinear $\Longrightarrow$ $O$ $,$ $P$ $,$ $G$ $,$ $A'$ $,$ $Y$ are collinear hence $GP$, $BE$ and $CF$ are concurrent in $Y$.
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Itticantdomath
25 posts
#5 • 1 Y
Y by Adventure10
This is a very common miquel configuration.
$\angle AEP + \angle AFP = 90^{\circ}+90^{\circ} = 180^{\circ}$ impies $AEPF$ cyclic.
Similarly, we get $PDCE$ and $PDBF$ cyclic where $D$ is the foot of the perpendicular from $A$ to $BC$. From this we get $BCEF$ cyclic and $G$ its miquel point.
Let $BE \cap CF = X$. Also, let the center of $BCEF$ be $O$. From the miquel configuration, we have $G, X, O$ colinear. So, we reduce the problem into showing $X, P, O$ colinear.
$\angle FEP = \angle FEB - \angle PEX = \angle FCB - 90^{\circ} + \angle BEC = \angle FCB - \angle BCO = \angle FCO$. So, $EP$ meets $CO$ on the circle. Let this point be $Y$. Analogously, $FP$ meets $BO$ at $Z$ which is also on the circle.
Now we apply pascal's theorem to hexagon $BZFCYE$. We get our conclusion.
This post has been edited 1 time. Last edited by Itticantdomath, Apr 3, 2017, 6:06 PM
Reason: Mistake in latex
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AleksaS
41 posts
#6
Y by
Cool problem.
Similar solution as above.
Firstly we have that
$$\angle PEA =\angle PFA = 90^\circ$$from where we obtain that $PEFA$ is cyclic. Let $H$ be feet of altitude from $A$. Now from
$$\angle PFB = \angle PHB = 90^\circ$$and $$\angle PEC = \angle PHC = 90^\circ$$we obtain that $PFBH$ and $PEHC$ are cyclic too.
Now we are doing some angle chase.
$$\angle FBC = 180^\circ - \angle FPH =\angle FPA = \angle FEA = 180^\circ - \angle FEC$$With this we get that $BCFE$ is cyclic. Now let $T = BE \cap CF$. Let $O$ be circumcenter of $BCEF$. $G$ is Miquel point of the quadrilateral and the rest is the same as above so there is no point in writing it all over again :)
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dgreenb801
1896 posts
#7 • 1 Y
Y by math31415926535
See my solution on my Youtube channel here. It is very similar to FabrizioFelen's solution, but with a slight difference in the middle (I think his way is actually simpler).

https://www.youtube.com/watch?v=Rl6S_sU7rK4
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WolfusA
1900 posts
#8
Y by
You can work this up using Hong Kong 2001 Test 1 Q1.
$\definecolor{A}{RGB}{0,90,255}\color{A}\fbox{Solution}$
As I proved there, if you denote by $O$ the circumcenter of quadrilateral $BFEC$ then points $G,P,O$ lie on line perpendicular to $AG$. Now take $X$ as the intersection of lines $BE,CF$. By radical axis theorem and Brocard's theorem $OX$ is the line perpendicular to $AG$. There's only one line perpendicular to $AG$ passing through point $O$, thus $X\in GP.\blacksquare$
#1802
This post has been edited 1 time. Last edited by WolfusA, Oct 2, 2020, 11:53 AM
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P2nisic
406 posts
#9
Y by
$BEFC$ is cyclic.So from radical center $AG,EF,BC$ concur at $S$.
If $BE,CF$ concur at $T$ then $GT,GP$ are perpedicular to $SA$ so $G,T,P$ collinear so we are done
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