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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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[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
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[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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
Infinite Pairs of Integers
steven_zhang123   0
8 minutes ago
Source: 2025 Spring NSMO P6
Given a positive integer \( k \), prove that there exist infinitely many pairs of positive integers \((m, n)\) (\(m < n\)) such that
\[
\tau(m^k) \tau(m^k + 1) \cdots \tau(n^k - 1) \tau(n^k)
\]is a perfect square, where \(\tau(n)\) denotes the number of positive divisors of \(n\).
Proposed by Dong Zhenyu
0 replies
steven_zhang123
8 minutes ago
0 replies
power sum system of equations in 3 variables
Stear14   0
12 minutes ago
Given that
$x^2+y^2+z^2=8\ ,$
$x^3+y^3+z^3=15\ ,$
$x^5+y^5+z^5=100\ .$

Find the value of $\ x+y+z\ .$
0 replies
Stear14
12 minutes ago
0 replies
Game on 6 by 6 grid
billzhao   26
N an hour ago by Sleepy_Head
Source: USAMO 2004, problem 4
Alice and Bob play a game on a 6 by 6 grid. On his or her turn, a player chooses a rational number not yet appearing in the grid and writes it in an empty square of the grid. Alice goes first and then the players alternate. When all squares have numbers written in them, in each row, the square with the greatest number in that row is colored black. Alice wins if she can then draw a line from the top of the grid to the bottom of the grid that stays in black squares, and Bob wins if she can't. (If two squares share a vertex, Alice can draw a line from one to the other that stays in those two squares.) Find, with proof, a winning strategy for one of the players.
26 replies
1 viewing
billzhao
Apr 29, 2004
Sleepy_Head
an hour ago
USA GEO 2003
dreammath   21
N an hour ago by lpieleanu
Source: TST USA 2003
Let $ABC$ be a triangle and let $P$ be a point in its interior. Lines $PA$, $PB$, $PC$ intersect sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Prove that
\[ [PAF]+[PBD]+[PCE]=\frac{1}{2}[ABC]  \]
if and only if $P$ lies on at least one of the medians of triangle $ABC$. (Here $[XYZ]$ denotes the area of triangle $XYZ$.)
21 replies
dreammath
Feb 16, 2004
lpieleanu
an hour ago
No more topics!
Polygon with minimum internal angle 120^\circ
Kunihiko_Chikaya   1
N Apr 28, 2025 by Mathzeus1024
How many sides can have the polygon with minimum internal angle of $ 120^\circ$ by adding every $ 5^\circ$?
Sorry for my poor English.
1 reply
Kunihiko_Chikaya
Aug 9, 2007
Mathzeus1024
Apr 28, 2025
Polygon with minimum internal angle 120^\circ
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Kunihiko_Chikaya
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#1 • 2 Y
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How many sides can have the polygon with minimum internal angle of $ 120^\circ$ by adding every $ 5^\circ$?
Sorry for my poor English.
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Mathzeus1024
919 posts
#2
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For any convex polygon with $N$ sides, the total internal angle sum (in degrees) computes to $180^{\circ}(N-2)$ (i). The polygon in question has internal angles whose sum is represented by the arithmetic series: $\sum_{k=1}^{N} 120^{\circ} + 5^{\circ}(k-1)$ (ii). Equating (i) with (ii) yields a quadratic in $N$:

$180^{\circ}(N-2) = \sum_{k=1}^{N} 120^{\circ} + 5^{\circ}(k-1)$;

or $180^{\circ}(N-2) = \sum_{k=1}^{N} 115^{\circ} + 5^{\circ}k$;

or $36(N-2) = \sum_{k=1}^{N} 23 + k$;

or $36(N-2) = 23N +\frac{N(N+1)}{2}$;

or $N^2-25N+144=0$;

or $(N-9)(N-16)=0$;

or $N=9, 16$.

In order for the polygon in question to be convex we require all of its interior angles $< 180^{\circ}$. At $N=9$ the maximum angle is $120^{\circ}+5^{\circ}(9) = 165^{\circ}$ (admissible), and $N=16$ yields $120^{\circ}+5^{\circ}(16) = 200^{\circ}$ (not admissible). Hence, our desired polygon has $\boxed{9}$. sides.
This post has been edited 1 time. Last edited by Mathzeus1024, Apr 29, 2025, 9:57 AM
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