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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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0 replies
jlacosta
Apr 2, 2025
0 replies
inquequality
ngocthi0101   10
N 14 minutes ago by MS_asdfgzxcvb
let $a,b,c > 0$ prove that
$\frac{a}{b} + \sqrt {\frac{b}{c}}  + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$
10 replies
ngocthi0101
Sep 26, 2014
MS_asdfgzxcvb
14 minutes ago
Functional Equation
AnhQuang_67   5
N 17 minutes ago by jasperE3
Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $$2\cdot f\Big(\dfrac{-xy}{2}+f(x+y)\Big)=xf(y)+yf(x), \forall x, y \in \mathbb{R} $$
5 replies
AnhQuang_67
Yesterday at 4:50 PM
jasperE3
17 minutes ago
Special line through antipodal
Phorphyrion   8
N 21 minutes ago by optimusprime154
Source: 2025 Israel TST Test 1 P2
Triangle $\triangle ABC$ is inscribed in circle $\Omega$. Let $I$ denote its incenter and $I_A$ its $A$-excenter. Let $N$ denote the midpoint of arc $BAC$. Line $NI_A$ meets $\Omega$ a second time at $T$. The perpendicular to $AI$ at $I$ meets sides $AC$ and $AB$ at $E$ and $F$ respectively. The circumcircle of $\triangle BFT$ meets $BI_A$ a second time at $P$, and the circumcircle of $\triangle CET$ meets $CI_A$ a second time at $Q$. Prove that $PQ$ passes through the antipodal to $A$ on $\Omega$.
8 replies
Phorphyrion
Oct 28, 2024
optimusprime154
21 minutes ago
PoP+Parallel
Solilin   1
N an hour ago by sansgankrsngupta
Source: Titu Andreescu, Lemmas in Olympiad Geometry
Let ABC be a triangle and let D, E, F be the feet of the altitudes, with D on BC, E on CA, and F on AB. Let the parallel through D to EF meet AB at X and AC at Y. Let T be the intersection of EF with BC and let M be the midpoint of side BC. Prove that the points T, M, X, Y are concyclic.
1 reply
Solilin
an hour ago
sansgankrsngupta
an hour ago
School Math Problem
math_cool123   6
N 5 hours ago by anduran
Find all ordered pairs of nonzero integers $(a, b)$ that satisfy $$(a^2+b)(a+b^2)=(a-b)^3.$$
6 replies
math_cool123
Apr 2, 2025
anduran
5 hours ago
New geometry problem
titaniumfalcon   3
N Yesterday at 11:16 PM by mathprodigy2011
Post any solutions you have, with explanation or proof if possible, good luck!
3 replies
titaniumfalcon
Thursday at 10:40 PM
mathprodigy2011
Yesterday at 11:16 PM
Geo Mock #10
Bluesoul   2
N Yesterday at 8:26 PM by Bluesoul
Consider acute $\triangle{ABC}$ with $AB=10$, $AC<BC$ and area $135$. The circle $\omega$ with diameter $AB$ meets $BC$ at $E$. Let the orthocenter of the triangle be $H$, connect $CH$ and extend to meet $\omega$ at $N$ such that $NC>HC$ and $NE$ is the diameter of $\omega$. Draw the circumcircle $\Gamma$ of $\triangle{AHB}$, chord $XY$ of $\Gamma$ is tangent to $\omega$ and it passes through $N$, compute $XY$.
2 replies
Bluesoul
Apr 1, 2025
Bluesoul
Yesterday at 8:26 PM
Inequalities
sqing   4
N Yesterday at 3:28 PM by sqing
Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=11.$ Prove that
$$a+ab+abc\leq\frac{49}{6}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=10.$ Prove that
$$a+ab+abc\leq\frac{169}{24}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=14.$ Prove that
$$a+ab+abc\leq\frac{63+5\sqrt 5}{6}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=32.$ Prove that
$$a+ab+abc\leq48+\frac{64\sqrt{2}}{3}$$
4 replies
sqing
Apr 1, 2025
sqing
Yesterday at 3:28 PM
Geo Mock #9
Bluesoul   1
N Yesterday at 3:19 PM by vanstraelen
Consider $\triangle{ABC}$ with $AB=12, AC=22$. The points $D,E$ lie on $AB,AC$ respectively, such that $\frac{AD}{BD}=\frac{AE}{CE}=3$. Extend $CD,BE$ to meet the circumcircle of $\triangle{ABC}$ at $P,Q$ respectively. Let the circumcircles of $\triangle{ADP}, \triangle{AEQ}$ meet at points $A,T$. Extend $AT$ to $BC$ at $R$, given $AR=16$, find $[ABC]$.
1 reply
Bluesoul
Apr 1, 2025
vanstraelen
Yesterday at 3:19 PM
Geo Mock #6
Bluesoul   1
N Yesterday at 1:59 PM by vanstraelen
Consider triangle $ABC$ with $AB=5, BC=8, AC=7$, denote the incenter of the triangle as $I$. Extend $BI$ to meet the circumcircle of $\triangle{AIC}$ at $Q\neq I$, find the length of $QC$.
1 reply
Bluesoul
Apr 1, 2025
vanstraelen
Yesterday at 1:59 PM
Congruence
Ecrin_eren   1
N Yesterday at 1:39 PM by Ecrin_eren
Find the number of integer pairs (x, y) satisfying the congruence equation:

3y² + 3x²y + y³ ≡ 3x² (mod 41)

for 0 ≤ x, y < 41.

1 reply
Ecrin_eren
Thursday at 10:34 AM
Ecrin_eren
Yesterday at 1:39 PM
Probability
Ecrin_eren   1
N Yesterday at 1:38 PM by Ecrin_eren
In a board, James randomly writes A , B or C in each cell. What is the probability that, for every row and every column, the number of A 's modulo 3 is equal to the number of B's modulo 3?

1 reply
Ecrin_eren
Thursday at 11:21 AM
Ecrin_eren
Yesterday at 1:38 PM
Excalibur Identity
jjsunpu   9
N Yesterday at 12:21 PM by fruitmonster97
proof is below
9 replies
jjsunpu
Thursday at 3:27 PM
fruitmonster97
Yesterday at 12:21 PM
.problem.
Cobedangiu   2
N Yesterday at 12:06 PM by Lankou
Find the integer coefficients after expanding Newton's binomial:
$$(\frac{3}{2}-\frac{2}{3}x^2)^n (n \in Z)$$
2 replies
Cobedangiu
Yesterday at 6:20 AM
Lankou
Yesterday at 12:06 PM
a virus problem, playing a game with the doctor
parmenides51   6
N Mar 11, 2021 by dgrozev
Source: 2014 Kurchatov Olympiad 11 p6 - Russia
Vasya, an invisible virus, is sitting on a square plate with a side of $1$ cm. He and the doctor Petya take turns. With the next $n$-th move, Petya draws with the vaccine like ink a segment $1$ micron long, and then Vasya must choose a direction and crawl in this straight line a direction of $1 / n$ micron distance (without going beyond the edge of the plate). If Vasya crawls through any of the points with the vaccine or touches it, he will die. Petya can act with any precision. Can Petya within a finite number of moves for sure kill the virus?

original wording
6 replies
parmenides51
Mar 7, 2021
dgrozev
Mar 11, 2021
a virus problem, playing a game with the doctor
G H J
Source: 2014 Kurchatov Olympiad 11 p6 - Russia
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parmenides51
30629 posts
#1 • 2 Y
Y by samrocksnature, Nofancyname
Vasya, an invisible virus, is sitting on a square plate with a side of $1$ cm. He and the doctor Petya take turns. With the next $n$-th move, Petya draws with the vaccine like ink a segment $1$ micron long, and then Vasya must choose a direction and crawl in this straight line a direction of $1 / n$ micron distance (without going beyond the edge of the plate). If Vasya crawls through any of the points with the vaccine or touches it, he will die. Petya can act with any precision. Can Petya within a finite number of moves for sure kill the virus?

original wording
This post has been edited 2 times. Last edited by parmenides51, Mar 7, 2021, 11:22 PM
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pog
4917 posts
#2 • 6 Y
Y by Inconsistent, L567, Galinagames, snoozingnewt, samrocksnature, GianDR
No matter what happens in the future, here is one thing that I can guarantee will remain constant: if there are two people in a Russian Olympiad problem, there is at least an $80\%$ chance that they are named Vasya and Petya.
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dgrozev
2459 posts
#3 • 3 Y
Y by pog, parmenides51, samrocksnature
Petya can kill the virus. Suppose doctor Petya is at his $2^k$ th move. During his next $2^k$ number of moves he draws $2^{k-1}$ aligned vertical segments at a distance of $2^{-k-2}$ microns between any two consecutive ones, and another $2^{k-1}$ aligned horizontal segments, at a distance of $2^{-k-2}$ microns between any adjacent ones, so that they form a sieve.
Thus, he makes a square $Q$ of side length $\left(2^{k-1}-1\right)\cdot 2^{-k-2}>2^{-4}$ microns full of cells of size $2^{-k-2}\times 2^{-k-2}$ inside it, and all borders are drawn.
Next, consider the $2^{k+1}$ th move of the virus. If it is inside $Q$ he will be killed. In this way, Petya eliminates, one by one, squares with a side length greater than $2^{-4}$ microns, and finally kills the virus.
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Kamran011
678 posts
#4 • 5 Y
Y by samrocksnature, Nofancyname, Mango247, Mango247, Mango247
I think it's a bit troll, anyways idea is similar to that of #3 , with a different bounding.

Consider the non-overlapping squares of size $0.2\times 0.2$, if Vasya the Virus is on one of the diagonals of it and goes on it in his first (or few first) move(s) he'll die obviously as the length of his allowed path would be $0.2\sqrt{2}<1$ microns. So he moves to inner part of a $0.2\times 0.2$ table and we may wlog assume he was inside. Let Petya check these tables, and suppose he checks some of them in $k$ moves without success and eventually he's checking the board where Vasya the Virus is located. Then he draws $k$ rows and $k$ columns in order to partition the table into squares of size $\frac{0.2}{k}\times\frac{0.2}{k}$. If the virus died when running away from one line by hitting some other we're done, so assume he's still alive and now he's inside some of these squares,the next time he'll go $\frac{1}{3k}$ microns away which exceeds $\frac{0.2}{k}$, so he dies .
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Nofancyname
102 posts
#5 • 1 Y
Y by Mango247
@above. I don't understand the first few sentences in your proof , why "its allowed path" instead of just "path" , why is it bounded ?
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Kamran011
678 posts
#6
Y by
Petya is the one who draws these tables , that's why :huh: . These moves are counted in the part
Kamran011 wrote:
Let Petya check these tables, and suppose he checks some of them in $k$ moves without success and eventually he's checking the board where Vasya the Virus is located.
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dgrozev
2459 posts
#7
Y by
Funny, but the problem can be made a bit more complicated. There is no need the virus to be framed in an unit square. So, the virus is somewhere on the plane, no one knows exactly where, and Petya (the doctor) is at the origin. The rest of the rules are the same. Still, Petya can kill the virus after finite amount of time!

With $4\cdot 10^6$ consecutive moves Petya can frame any unit square on the plane. He begins to frame the unit squares on the lattice points, so that he consecutively fills the squares with corners at $(-N,-N), (-N,N), (N,N),(N,-N)$ for $N=1,2,\dots.$ One move for every $4\cdot 10^6$ ones, he saves for another goal. With these moves he begins a process of killing the virus, as described in #3, inside each of the framed unit squares, such that he eliminates them, one after another. It can be easily checked that after $n$ moves, the doctor frames all unit squares at a distance $C\sqrt{n}$ from the origin, where $C$ is a constant. At the same time, at its $n$ th move the virus is at a distance $C_1+1+\frac{1}{2}+\dots+\frac{1}{n}\le C_1+C_2\log n$ from the origin. Since the growth rate of $\sqrt{n}$ is faster than $\log n,$ the virus will be framed after finite amount of time, and later it will be killed. The only drawback is the doctor will not know exactly when this will happen.
This post has been edited 1 time. Last edited by dgrozev, Mar 11, 2021, 12:28 PM
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