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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
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Greek gods flood the world
a1267ab   16
N 7 minutes ago by awesomeming327.
Source: USA Winter TST for IMO 2020, Problem 3, by Nikolai Beluhov
Let $\alpha \geq 1$ be a real number. Hephaestus and Poseidon play a turn-based game on an infinite grid of unit squares. Before the game starts, Poseidon chooses a finite number of cells to be flooded. Hephaestus is building a levee, which is a subset of unit edges of the grid (called walls) forming a connected, non-self-intersecting path or loop*.

The game then begins with Hephaestus moving first. On each of Hephaestus’s turns, he adds one or more walls to the levee, as long as the total length of the levee is at most $\alpha n$ after his $n$th turn. On each of Poseidon’s turns, every cell which is adjacent to an already flooded cell and with no wall between them becomes flooded as well. Hephaestus wins if the levee forms a closed loop such that all flooded cells are contained in the interior of the loop — hence stopping the flood and saving the world. For which $\alpha$ can Hephaestus guarantee victory in a finite number of turns no matter how Poseidon chooses the initial cells to flood?
-----
*More formally, there must exist lattice points $\mbox{\footnotesize \(A_0, A_1, \dotsc, A_k\)}$, pairwise distinct except possibly $\mbox{\footnotesize \(A_0 = A_k\)}$, such that the set of walls is exactly $\mbox{\footnotesize \(\{A_0A_1, A_1A_2, \dotsc , A_{k-1}A_k\}\)}$. Once a wall is built it cannot be destroyed; in particular, if the levee is a closed loop (i.e. $\mbox{\footnotesize \(A_0 = A_k\)}$) then Hephaestus cannot add more walls. Since each wall has length $\mbox{\footnotesize \(1\)}$, the length of the levee is $\mbox{\footnotesize \(k\)}$.

Nikolai Beluhov
16 replies
+1 w
a1267ab
Dec 16, 2019
awesomeming327.
7 minutes ago
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f.e with finite number of f(t)=-t
jjkim0336   2
N 42 minutes ago by jasperE3
Source: own
f:R->R
f(xf(y)+y)=yf(x)+f(f(y)) and there are finite number of t such that f(t)= - t
2 replies
jjkim0336
3 hours ago
jasperE3
42 minutes ago
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inequalities
Cobedangiu   1
N an hour ago by pooh123
Source: UCT
Let $a,b,c>0$ and $a+b+c=3$ Prove that:
$\sum \dfrac{\sqrt{a^2+a+1}}{2a^2+10a+9}\ge \dfrac{\sqrt{3}}{7}$
1 reply
Cobedangiu
Yesterday at 4:14 PM
pooh123
an hour ago
J
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Solve $\sin(17x)+\sin(13x)=\sin(7x)$
Speed2001   0
2 hours ago
How to solve the equation:
$$ \sin(17x)+\sin(13x)=\sin(7x),\;0<x<24^{\circ} $$Approach: I'm trying to factor $\sin(18x)$ to get $x=10^{\circ}$.

Any hint would be appreciated.
0 replies
Speed2001
2 hours ago
0 replies
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the four altitudes are concurrent
Fermat -Euler   6
N Sep 19, 2022 by levifb
Source: brazil 2001
An altitude of a convex quadrilateral is a line through the midpoint of a side perpendicular to the opposite side. Show that the four altitudes are concurrent iff the quadrilateral is cyclic.
6 replies
Fermat -Euler
Sep 5, 2005
levifb
Sep 19, 2022
J
the four altitudes are concurrent
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Source: brazil 2001
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Fermat -Euler
444 posts
Fermat -Euler
#1 h  Sep 5, 2005, 9:09 AM • 2 Y
Y by Adventure10, Mango247
An altitude of a convex quadrilateral is a line through the midpoint of a side perpendicular to the opposite side. Show that the four altitudes are concurrent iff the quadrilateral is cyclic.
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mecrazywong
606 posts
mecrazywong
#2 h  Sep 5, 2005, 2:06 PM • 1 Y
Y by Adventure10
This is well known. The lines concur at the anticenter of quadrilateral $ABCD$.
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Arne
3660 posts
Arne
#3 h  Sep 5, 2005, 2:11 PM • 2 Y
Y by vsathiam, Adventure10
mecrazywong wrote:
The lines concur at the anticenter of quadrilateral $ABCD$.

More information here:

http://www.cut-the-knot.org/Curriculum/Geometry/Brahmagupta2.shtml
http://www.pballew.net/cycquad.html

etcetera
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vsathiam
201 posts
vsathiam
#4 h  Aug 19, 2017, 9:12 PM • 2 Y
Y by Adventure10, Mango247
Or you can complex bash this :)

Once you've proved the cyclic part, you can use that to disprove noncyclic quads.
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jayme
9787 posts
jayme
#5 h  Aug 30, 2017, 1:19 PM • 4 Y
Y by vsathiam, levifb, Adventure10, Mango247
Dear Mathlinkers,
we can speak instead altitude, maltitude...
see

http://jl.ayme.pagesperso-orange.fr/Docs/A%20propos%20de%20l'anticentre%20d'un%20quadrilatere%20cyclique.pdf

Sincerely
Jean-Louis
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jestrada
515 posts
jestrada
#6 h  Aug 30, 2017, 2:09 PM • 1 Y
Y by Adventure10
The midpoints of a quadrilateral form a parallelogram (even better known). Let $X$ be the center of the parallelogram. Rotate $ABCD$ 180 degrees through $X$.
If $ABCD$ is cyclic, let $O$ be the circumcenter. The lines from $O$ to the midpoints are the perpendicular bisectors of the respective sides; the rotation takes the midpoint of each side to the midpoint of the opposite side, and the images of the lines from $O$ to the midpoints are precisely the altitudes of $ABCD$, which meet in the image of $O$. Likewise if the altitudes are concurrent, the rotation takes the altitudes to the perpendicular bisectors of the respective sides, and if these are concurrent then $ABCD$ is cyclic.
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levifb
61 posts
levifb
#7 h  Sep 19, 2022, 5:29 PM
Y by
Just consider the homothety with center in the center of the parallelogram formed by the midpoints of the sides, and radius $-1$. Notice that the altitudes are taken in the perpendicular bisectors of the sides of the quadrilateral. Thus, they concur if and only if this lines meet at an unic point, i.e, ABCD is cyclic.
This post has been edited 1 time. Last edited by levifb, Sep 23, 2022, 10:49 PM
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