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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
Perfect Squares, Infinite Integers and Integers
steven_zhang123   0
4 minutes ago
Source: China TST 2001 Quiz 5 P1
For which integer \( h \), are there infinitely many positive integers \( n \) such that \( \lfloor \sqrt{h^2 + 1} \cdot n \rfloor \) is a perfect square? (Here \( \lfloor x \rfloor \) denotes the integer part of the real number \( x \)?
0 replies
steven_zhang123
4 minutes ago
0 replies
f(f(x)+y)+f(x+y)=2x+2f(y)
parmenides51   3
N 6 minutes ago by Burmf
Source: 2015 AGCN Competition p1 by bobthesmartypants https://artofproblemsolving.com/community/c5h1128876p5232794
Find all functions $f:\mathbb{R}_{\ge 0}\to \mathbb{R}_{\ge 0}$ satisfying$$f(f(x)+y)+f(x+y)=2x+2f(y)$$
3 replies
parmenides51
Dec 5, 2023
Burmf
6 minutes ago
Help to prove an inequality
JK1603JK   1
N 8 minutes ago by jawadkaleem
Source: unknown
If a,b,c\ge 0: ab+bc+ca=1 then prove \frac{a\left(b+c+2\right)}{bc+2a}+\frac{b\left(c+a+2\right)}{ca+2b}+\frac{c\left(a+b+2\right)}{ab+2c}\ge 3
* Please help me convert it to latex form. Thank you.
1 reply
1 viewing
JK1603JK
23 minutes ago
jawadkaleem
8 minutes ago
2^a + 3^b + 5^c = n!
togrulhamidli2011   2
N 12 minutes ago by togrulhamidli2011
\[
\text{Find all non-negative integers } (a, b, c, n) \text{ such that}
\]\[
2^a + 3^b + 5^c = n!
\]
2 replies
togrulhamidli2011
21 minutes ago
togrulhamidli2011
12 minutes ago
No more topics!
Back to the origin
semisimplicity   10
N Aug 21, 2023 by KI_HG
Source: Indian IMOTC 2013, Practice Test 2, Problem 3
A marker is placed at the origin of an integer lattice. Calvin and Hobbes play the following game. Calvin starts the game and each of them takes turns alternatively. At each turn, one can choose two (not necessarily distinct) integers $a, b$, neither of which was chosen earlier by any player and move the marker by $a$ units in the horizontal direction and $b$ units in the vertical direction. Hobbes wins if the marker is back at the origin any time after the first turn. Prove or disprove that Calvin can prevent Hobbes from winning.

Note: A move in the horizontal direction by a positive quantity will be towards the right, and by a negative quantity will be towards the left (and similar directions in the vertical case as well).
10 replies
semisimplicity
May 10, 2013
KI_HG
Aug 21, 2023
Back to the origin
G H J
Source: Indian IMOTC 2013, Practice Test 2, Problem 3
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semisimplicity
141 posts
#1 • 4 Y
Y by Smita, Purple_Planet, Adventure10, Mango247
A marker is placed at the origin of an integer lattice. Calvin and Hobbes play the following game. Calvin starts the game and each of them takes turns alternatively. At each turn, one can choose two (not necessarily distinct) integers $a, b$, neither of which was chosen earlier by any player and move the marker by $a$ units in the horizontal direction and $b$ units in the vertical direction. Hobbes wins if the marker is back at the origin any time after the first turn. Prove or disprove that Calvin can prevent Hobbes from winning.

Note: A move in the horizontal direction by a positive quantity will be towards the right, and by a negative quantity will be towards the left (and similar directions in the vertical case as well).
Z K Y
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dibyo_99
487 posts
#2 • 4 Y
Y by alet, Adventure10, Mango247, and 1 other user
Sorry, wrong solution.
This post has been edited 1 time. Last edited by dibyo_99, May 14, 2013, 8:14 AM
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chaotic_iak
2932 posts
#3 • 2 Y
Y by Adventure10, Mango247
What if $b_i = -a_i$?
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alet
363 posts
#4 • 1 Y
Y by Adventure10
What if Calvin in his first move chooses $(a,b)$ and then Hobbes plays $(-a,-b)$?
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chaotic_iak
2932 posts
#5 • 3 Y
Y by Purple_Planet, Adventure10, Mango247
Well, that's why Calvin's first move must be something in the form of $(a,-a)$.
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mavropnevma
15142 posts
#6 • 3 Y
Y by Purple_Planet, Adventure10, Mango247
Or $(a,0)$ (or $(0,b)$), since then $0$ cannot be used anymore ...
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mnbvmar
7 posts
#7 • 6 Y
Y by dibyo_99, siddigss, confusedhexagon, Purple_Planet, Adventure10, Stuffybear
Proof that Calvin can always prevent Hobbes from reaching (0,0):

Let's say that Calvin is about to move now. Let $A$ be the set containing all earlier chosen numbers and $(x,y)$ - current position of the marker. Set $t = 3 \cdot \max\left(\max_{a \in A} |a|,\ |x|,\ 1\right)$. Then translate the marker by a vector $[t,\ -x-t]$ so that the new position is $(x+t,\ y-x-t)$. It's easy to show that the absolute values of the vector coordinates are strictly bigger than the absolute values of every element of $A$ - for every $a \in A$ we have $|t| > |a|$ and $|-x-t| \geq |t|-|x| \geq \frac{2}{3}|t| > |a|$. It follows that this move is possible. Of course x+t>0, so Calvin will never return to (0,0) this way.

Assume Hobbes got back to the origin after described Calvin's move. Then he would have to move the marker horizontally by $-x-t$ points. However, after last Calvin's move this number became forbidden - contradiction.
Z K Y
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neel02
66 posts
#8 • 2 Y
Y by Adventure10, Mango247
Idea only since I am in hurry ! Suppose Hobbes arrives at some point $(m,n)$ . Then Calvin chooses a point
$(d,-d)$ such that absolute value of d > max{ absolute values of x,y } .
Such d easily exists . Since absolute value is increasing it never turns out to be 0 .Again Hobbes can not send it directly to 0 for the condition that none of $(a,b)$ can be repeated . So done !
But note that I have not fixed d for describing only a strategy ! But this strategy is useful for a lattice plain .There are many flaws if this game is played in the real plane ! :coool: :agent: :whistling:
This post has been edited 1 time. Last edited by neel02, May 18, 2018, 6:06 PM
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Smita
514 posts
#9 • 2 Y
Y by Adventure10, Mango247
semisimplicity wrote:
A marker is placed at the origin of an integer lattice. Calvin and Hobbes play the following game. Calvin starts the game and each of them takes turns alternatively. At each turn, one can choose two (not necessarily distinct) integers $a, b$, neither of which was chosen earlier by any player and move the marker by $a$ units in the horizontal direction and $b$ units in the vertical direction. Hobbes wins if the marker is back at the origin any time after the first turn. Prove or disprove that Calvin can prevent Hobbes from winning.

Note: A move in the horizontal direction by a positive quantity will be towards the right, and by a negative quantity will be towards the left (and similar directions in the vertical case as well).

Can u tell whete to find india imotc 2017 problems????
This post has been edited 1 time. Last edited by Smita, Jun 29, 2018, 10:39 AM
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Wizard_32
1566 posts
#10 • 1 Y
Y by Adventure10
@above https://artofproblemsolving.com/community/c3176_india_contests
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KI_HG
157 posts
#11 • 1 Y
Y by winniep008hfi
Yes.

First, Calvin moves (a,0). Then after Hobbes's nth move, let the y-coordinate be $Y_n$. Cleary, $Y_1=\not=0$. Let $ A= \{\textit{the numbers have been used}\}$.After hobbes nth move, we consider the following set number of number $S=\{-(Y_n+t): t\in A\}$. Clearly, there exists an $x_{n+1}\in A$, but $-(Y_n+x_{n+1})\not\in A$. Let Calvin choose $b=-(Y_n+x_{n+1})$. So the y-coordinate of the marker becomes $-x_{n+1}$. Therefore Hoobes cant make it back to origin on the next turn.
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