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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[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.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]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
Inspired by Philippine 2025
sqing   1
N 3 minutes ago by sqing
Source: Own
Let $ a,b,c $ be real numbers . Prove that
$$\frac{(a-1)(b-1)(c-1)}{(a^2+1)(b^2+1)(c^2+1)} \ge -\frac{7+5\sqrt 2}{8}$$$$\frac{(a-1)(b-1)(c-1)}{(a^2+2)(b^2+2)(c^2+2)} \ge -\frac{5+3\sqrt 3}{32}$$
1 reply
1 viewing
sqing
11 minutes ago
sqing
3 minutes ago
My Unsolved Problem
ZeltaQN2008   1
N 13 minutes ago by Funcshun840
Source: IDK
Let triangle \(ABC\) be inscribed in circle \((O)\). Let \((I_a)\) be the \(A\)-excircle of triangle \(ABC\), which is tangent to \(BC\), the extension of \(AB\), and the extension of \(AC\). Let \(BE\) and \(CF\) be the angle bisectors of triangle \(ABC\). Let \(EF\) intersect \((O)\) at two points \(S\) and \(T\).

a) Prove that circle \((O)\) bisects the segments \(I_aT\) and \(I_aS\).
b) Prove that \(S\) and \(T\) are the points of tangency of the common external tangents of circles \((O)\) and \((I_a)\) .

1 reply
ZeltaQN2008
Yesterday at 3:07 PM
Funcshun840
13 minutes ago
Random walk
EthanWYX2009   2
N 28 minutes ago by mathematical-forest
As shown in the graph, an ant starts from $4$ and walks randomly. The probability of any point reaching all adjacent points is equal. Find the probability of the ant reaching $1$ without passing through $6.$
2 replies
EthanWYX2009
3 hours ago
mathematical-forest
28 minutes ago
Inspired by old results
sqing   2
N 35 minutes ago by pooh123
Source: Own
Let $a,b,c $ be reals such that $a^2+b^2+c^2=3$ .Prove that
$$(1-a)(k-b)(1-c)+abc\ge -k$$Where $ k\geq 1.$
$$(1-a)(1-b)(1-c)+abc\ge -1$$$$(1-a)(1-b)(1-c)-abc\ge -\frac{1}{2}-\sqrt 2$$
2 replies
sqing
6 hours ago
pooh123
35 minutes ago
No more topics!
Grasshopper jumps in an equilateral triangle
shobber   1
N Oct 6, 2005 by yetti
Source: Pan African 2001
Let $ABC$ be an equilateral triangle and let $P_0$ be a point outside this triangle, such that $\triangle{AP_0C}$ is an isoscele triangle with a right angle at $P_0$. A grasshopper starts from $P_0$ and turns around the triangle as follows. From $P_0$ the grasshopper jumps to $P_1$, which is the symmetric point of $P_0$ with respect to $A$. From $P_1$, the grasshopper jumps to $P_2$, which is the symmetric point of $P_1$ with respect to $B$. Then the grasshopper jumps to $P_3$ which is the symmetric point of $P_2$ with respect to $C$, and so on. Compare the distance $P_0P_1$ and $P_0P_n$. $n \in N$.
1 reply
shobber
Oct 4, 2005
yetti
Oct 6, 2005
Grasshopper jumps in an equilateral triangle
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Source: Pan African 2001
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shobber
3498 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ABC$ be an equilateral triangle and let $P_0$ be a point outside this triangle, such that $\triangle{AP_0C}$ is an isoscele triangle with a right angle at $P_0$. A grasshopper starts from $P_0$ and turns around the triangle as follows. From $P_0$ the grasshopper jumps to $P_1$, which is the symmetric point of $P_0$ with respect to $A$. From $P_1$, the grasshopper jumps to $P_2$, which is the symmetric point of $P_1$ with respect to $B$. Then the grasshopper jumps to $P_3$ which is the symmetric point of $P_2$ with respect to $C$, and so on. Compare the distance $P_0P_1$ and $P_0P_n$. $n \in N$.
Z K Y
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yetti
2643 posts
#2 • 3 Y
Y by Adventure10, Mango247, DiaaSaid
Let a = AB be the side of the equilateral triangle $\triangle ABC$. Since $AP_0 = AP_1,\ BP_1 = BP_2$, AB is a midline of the triangle $\triangle P_0P_1P_2$. Hence, the point $P_2$ is obtained from $P_0$ by a translation in the direction $\vec{AB}$ by the distance 2a. Similarly, $P_4$ is obtained from $P_2$ by a translation in the direction $\vec{CA}$ by the distance 2a and $P_6$ from $P_4$ by a translation in the direction $\vec{BC}$ by the distance 2a. It follows that the points $P_0, P_2, P_4$ form an equilateral triangle with the side lengths 2a and the sides $P_0P_2 \parallel AB,\ P_2P_4 \parallel CA, P_4P_0 \parallel BC$ and that the point $P_6 \equiv P_0$ are identical. For similar reasons, $P_3$ is obtained from $P_1$ by a translation in the direction $\vec{BC}$ by the distance 2a, $P_5$ is obtained from $P_3$ by a translation in the direction $\vec{AB}$ by the distance 2a, and $P_7$ is obtained from $P_5$ by a translation in the direction $\vec{CA}$ by the distance 2a. It follows that the points $P_1, P_3, P_5$ also form an equilateral triangle with the side lengths 2a and the sides $P_1P_3 \parallel BC,\ P_3P_5 \parallel AB, P_5P_1 \parallel CA$ and that the point $P_7 \equiv P_1$ are identical. Thus the equilateral triangles $\triangle P_0P_2P_4 \cong \triangle P_3P_5P_1$ are congruent and have parallel sides and the second is obtained from the first by a translation in the direction $\vec{P_0P_3}$ by the distance $P_0P_3$.

Up to now, we did not need the fact that the triangle $\triangle ABC$ is equilateral or that the triangle $\triangle AP_0C$ is isosceles, only the fact that $\vec{AB} + \vec{BC} + \vec{CA} = 0$, which holds for any triangle. Now we use these facts. Since $P_6 \equiv P_0$ are identical and $P_0A = P_0C, AP_0 = AP_1, CP_0 = CP_5$, the triangle $\triangle P_1P_0P_5$ is also isosceles. This means that the point $P_0$ lies on the altitude of the equilateral triangle $\triangle P_3P_5P_1$ from the vertex $P_3$. Since in addition, the isosceles triangle $\triangle P_1P_0P_5$ is right,

$P_0P_3 = P_1P_5 \dfrac{\sqrt 3}{2} - \dfrac{P_1P_5}{5} = a (\sqrt 3 - 1)$

Obviously, $P_0P_5 = P_0P_1 = 2AP_0 = 2CA \dfrac{\sqrt 2}{2} = a \sqrt 2$ and $P_0P_2 = P_0P_4 = 2a$. As a result,

$P_0P_2 = P_0P_4 = P_0P_1 \sqrt 2$

$P_0P_3 = P_0P_1 \dfrac{\sqrt 3 - 1}{\sqrt 2} = P_0P_1 \dfrac{\sqrt 6 - \sqrt 2}{2}$

$P_0P_5 = P_0P_1$

$P_0P_6 = 0$

$P_0P_n = P_0P_{(n \mod 6)}$
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