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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
primitive polyominoes
N.T.TUAN   29
N 5 minutes ago by Disjunction
Source: USAMO 2007
An animal with $n$ cells is a connected figure consisting of $n$ equal-sized cells[1].

A dinosaur is an animal with at least $2007$ cells. It is said to be primitive it its cells cannot be partitioned into two or more dinosaurs. Find with proof the maximum number of cells in a primitive dinosaur.

(1) Animals are also called polyominoes. They can be defined inductively. Two cells are adjacent if they share a complete edge. A single cell is an animal, and given an animal with $n$ cells, one with $n+1$ cells is obtained by adjoining a new cell by making it adjacent to one or more existing cells.
29 replies
N.T.TUAN
Apr 26, 2007
Disjunction
5 minutes ago
Only consecutive terms are coprime
socrates   37
N 20 minutes ago by blueprimes
Source: 7th RMM 2015, Problem 1
Does there exist an infinite sequence of positive integers $a_1, a_2, a_3, . . .$ such that $a_m$ and $a_n$ are coprime if and only if $|m - n| = 1$?
37 replies
socrates
Feb 28, 2015
blueprimes
20 minutes ago
Cyclic roots are not real, they can't hurt you
anantmudgal09   21
N 28 minutes ago by bjump
Source: INMO 2023 P2
Suppose $a_0,\ldots, a_{100}$ are positive reals. Consider the following polynomial for each $k$ in $\{0,1,\ldots, 100\}$:
$$a_{100+k}x^{100}+100a_{99+k}x^{99}+a_{98+k}x^{98}+a_{97+k}x^{97}+\dots+a_{2+k}x^2+a_{1+k}x+a_k,$$where indices are taken modulo $101$, i.e., $a_{100+i}=a_{i-1}$ for any $i$ in $\{1,2,\dots, 100\}$. Show that it is impossible that each of these $101$ polynomials has all its roots real.

Proposed by Prithwijit De
21 replies
1 viewing
anantmudgal09
Jan 15, 2023
bjump
28 minutes ago
Erasing the difference of two numbers
BR1F1SZ   2
N 43 minutes ago by sami1618
Source: Austria National MO Part 1 Problem 3
Consider the following game for a positive integer $n$. Initially, the numbers $1, 2, \ldots, n$ are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers $3,6,11$ and $17$ are on the board, then $3$ can be erased as $6 - 3=3$, or $6$ as $17 - 11=6$, or $11$ as $17 - 6=11$.)

For which values of $n$ is it possible to end with only one number remaining on the board?

(Michael Reitmeir)
2 replies
BR1F1SZ
Monday at 9:48 PM
sami1618
43 minutes ago
primes and perfect squares
Bummer12345   5
N 4 hours ago by Shan3t
If $p$ and $q$ are primes, then can $2^p + 5^q + pq$ be a perfect square?
5 replies
Bummer12345
Monday at 5:08 PM
Shan3t
4 hours ago
trapezoid
Darealzolt   1
N 5 hours ago by vanstraelen
Let \(ABCD\) be a trapezoid such that \(A, B, C, D\) lie on a circle with center \(O\), and side \(AB\) is parallel to side \(CD\). Diagonals \(AC\) and \(BD\) intersect at point \(M\), and \(\angle AMD = 60^\circ\). It is given that \(MO = 10\). It is also known that the difference in length between \(AB\) and \(CD\) can be expressed in the form \(m\sqrt{n}\), where \(m\) and \(n\) are positive integers and \(n\) is square-free. Compute the value of \(m + n\).
1 reply
Darealzolt
Yesterday at 2:03 AM
vanstraelen
5 hours ago
Polynomial Minimization
ReticulatedPython   1
N Yesterday at 5:36 PM by clarkculus
Consider the polynomial $$p(x)=x^{n+1}-x^{n}$$, where $x, n \in \mathbb{R+}.$

(a) Prove that the minimum value of $p(x)$ always occurs at $x=\frac{n}{n+1}.$
1 reply
ReticulatedPython
Yesterday at 5:07 PM
clarkculus
Yesterday at 5:36 PM
Easy one
irregular22104   0
Yesterday at 5:03 PM
Given two positive integers a,b written on the board. We apply the following rule: At each step, we will add all the numbers that are the sum of the two numbers on the board so that the sum does not appear on the board. For example, if the two initial numbers are 2.5, then the numbers on the board after step 1 are 2,5,7; after step 2 are 2,5,7,9,12;...
1) With a = 3; b = 12, prove that the number 2024 cannot appear on the board.
2) With a = 2; b = 3, prove that the number 2024 can appear on the board.
0 replies
irregular22104
Yesterday at 5:03 PM
0 replies
A pentagon inscribed in a circle of radius √2
tom-nowy   3
N Yesterday at 4:53 PM by itsjeyanth
Can a pentagon with all rational side lengths be inscribed in a circle of radius $\sqrt{2}$ ?
3 replies
tom-nowy
Yesterday at 2:37 AM
itsjeyanth
Yesterday at 4:53 PM
This shouldn't be a problem 15
derekli   2
N Yesterday at 4:09 PM by aarush.rachak11
Hey guys I was practicing AIME and came across this problem which is definitely misplaced. It asks for the surface area of a plane within a cylinder which we can easily find out using a projection that is easy to find. I think this should be placed in problem 10 or below. What do you guys think?
2 replies
derekli
Yesterday at 2:15 PM
aarush.rachak11
Yesterday at 4:09 PM
Regular tetrahedron
vanstraelen   7
N Yesterday at 3:46 PM by ReticulatedPython
Given the points $O(0,0,0),A(1,0,0),B(\frac{1}{2},\frac{\sqrt{3}}{2},0)$
a) Determine the point $C$, above the xy-plane, such that the pyramid $OABC$ is a regular tetrahedron.
b) Calculate the volume.
c) Calculate the radius of the inscribed sphere and the radius of the circumscribed sphere.
7 replies
vanstraelen
May 4, 2025
ReticulatedPython
Yesterday at 3:46 PM
[ABCD] = n [CDE], areas in trapezoid - IOQM 2020-21 p1
parmenides51   4
N Yesterday at 3:44 PM by Kizaruno
Let $ABCD$ be a trapezium in which $AB \parallel CD$ and $AB = 3CD$. Let $E$ be then midpoint of the diagonal $BD$. If $[ABCD] = n \times  [CDE]$, what is the value of $n$?

(Here $[t]$ denotes the area of the geometrical figure$ t$.)
4 replies
parmenides51
Jan 18, 2021
Kizaruno
Yesterday at 3:44 PM
Inequalities
sqing   9
N Yesterday at 1:53 PM by sqing
Let $ a,b,c>0 $ and $ a+b\leq 16abc. $ Prove that
$$ a+b+kc^3\geq\sqrt[4]{\frac{4k} {27}}$$$$ a+b+kc^4\geq\frac{5} {8}\sqrt[5]{\frac{k} {2}}$$Where $ k>0. $
$$ a+b+3c^3\geq\sqrt{\frac{2} {3}}$$$$ a+b+2c^4\geq \frac{5} {8}$$
9 replies
sqing
May 4, 2025
sqing
Yesterday at 1:53 PM
geometry
JetFire008   2
N Yesterday at 12:45 PM by sunken rock
Given four concyclic points. For each subset of three points take the incenter. Show that the four incentres form a rectangle.
2 replies
JetFire008
Monday at 4:14 PM
sunken rock
Yesterday at 12:45 PM
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
10855 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
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$BEFC$ 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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