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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
Inequalities
sqing   8
N 6 minutes ago by cj13609517288
Let $ a,b> 0 ,\frac{a}{2b+1}+\frac{b}{3}+\frac{1}{2a+1} \leq 1.$ Prove that
$$  a^2+b^2 -ab\leq 1$$$$ a^2+b^2 +ab \leq3$$Let $ a,b,c> 0 , \frac{a}{2b+1}+\frac{b}{2c+1}+\frac{c}{2a+1} \leq 1.$ Prove that
$$    a +b +c +abc \leq 4$$
8 replies
sqing
May 24, 2025
cj13609517288
6 minutes ago
Original Problem (Complex Numbers)
qrxz17   0
9 minutes ago
Problem: Given that for a complex number \(z\), the value of \(z \cdot \overline{z} +(z+\overline{z})\) is \(19\). Find all complex numbers \(z\).

Answer: Click to reveal hidden text

Solution: Writing the given equation in standard form, we get
\begin{align*}
        (a+bi)\cdot(a-bi) +[(a+bi)+(a-bi)] =a^2+b^2+2a.
    \end{align*}
Then we have
\begin{align*}
        a^2+b^2+2a &= 19 \\
        (a+1)^2+b^2 &=20
    \end{align*}
This is the standard equation of a circle. So, \(r^2=20\) and \(r=2\sqrt{5}\).

Therefore, the combined value of the sum and product of a complex number and its conjugate is \(19\) when the complex numbers lie on the circle centered at \((-1,0)\) with radius \(2\sqrt{5}\).
0 replies
qrxz17
9 minutes ago
0 replies
Sipnayan 2025 SHS Orals Final Round VD-FriedChicken
qrxz17   0
13 minutes ago
Problem: Let \( a, b, c \) be complex numbers. For a complex number \( z = p + qi \) where \( i = \sqrt{-1} \), define the norm \( |z| \) to be the distance of \( z \) from the origin, or \( |z| = \sqrt{p^2 + q^2} \). Let \( m \) be the minimum value and \( M \) be the maximum value of
\[
\frac{ |a + b| + |b + c| + |c + a| }{ |a| + |b| + |c| }
\]for all complex numbers \( a, b, c \) where \( |a| + |b| + |c| \ne 0 \). Find \( M + m \).

Answer: Click to reveal hidden text

Solution: Let us first get the maximum value \(M\). When two complex numbers \(a\) and \(b\) are added, their sum \(a + b\) is also a vector. Geometrically, this is represented by placing the tail of \(b\) at the head of \(a\) or vice versa, following the parallelogram rule. The vector \(a + b\) then extends from the origin to the head of the translated \(b\), forming a triangle with the origin and the head of \(a\).

Then, we can apply the triangle inequality. We have
\begin{align*}
        |a+b| & \leq |a|+|b| \\
        |b+c| & \leq |b|+|c| \\
        |c+a| & \leq |c| + |a| 
    \end{align*}
The given equation can then be rewritten to
\begin{align*}
        \frac{ |a + b| + |b + c| + |c + a| }{ |a| + |b| + |c| } \leq \frac{ |a| + |b| + |b| + |c| + |c| + |a| }{ |a| + |b| + |c| } \leq \frac{ 2(|a| + |b| + |c |) }{ |a| + |b| + |c| } \leq 2
    \end{align*}
Thus, the maximum value M is equal to 2.

Now, let us determine the minimum value \(m\). To find the minimum value, we can look for configurations where ``cancellation" in the vector sums is maximized. This typically happens when the complex numbers are collinear but point in opposite directions. To do this, let \(a\) and \(b\) be at the same point, and let \(c\) be the same distance from the origin as \(a\) and \(b\), but in the opposite direction.

We have
\begin{align*}
        a =\text{ } &b \text{ } =-c \\
        |a| =|&b|=|c|.
    \end{align*}Substituting this in the given equation, we get

\begin{align*}
        \frac{ |a + b| + |b + c| + |c + a| }{ |a| + |b| + |c| } = \frac{ |a+a| + |a + -a| + |-a + a| }{ |a| + |a| + |a| } =\frac{ |2a| }{ 3|a| }=\frac{2}{3}.
    \end{align*}
Thus, the minimum value \(m\) is equal to \(\frac{2}{3}\).

Therefore, \(M+m=2+\frac{2}{3}=\boxed{\frac{8}{3}}\).
0 replies
qrxz17
13 minutes ago
0 replies
Daily Problem Writing Practice
KSH31415   0
15 minutes ago
I'm trying to get better at writing problems so I decided to challenge myself to write one problem for every day this month (June 2025). I will post them in this thread as well as edit this post with all of them in hide tags. If I can, I'll include a difficulty level in the form of AIME placement. If anybody wants to solve them and give feedback on the problem and/or my difficulty rating, please do!

June 1 (AIME P4)

June 1 (AIME P4)
A bag contains $6$ red balls and $6$ blue balls. A draw consists of randomly selecting $2$ of the remaining balls from bag without replacement, and then setting them aside. A draw is called a match if the two balls have the same color. Compute the expected number of draws until either a match occurs or the bag is empty.
0 replies
KSH31415
15 minutes ago
0 replies
One of those hand-waving proofs
math_explorer   0
Sep 18, 2010
[quote]46337
$A, B, C$ are points on a line in that order. A line through $B$ intersects the circle with diameter $AC$ at $P$ and $Q$, the circle with diameter $AB$ at $R$, and the circle with diameter $BC$ at $S$. Prove that $PR$ = $QS$. (Note that it doesn't matter if $P$ and $Q$ are swapped.)[/quote]

Click to reveal hidden text

The next problem was pseudopseudopseudorandomly selected from Geometry Unsolved Problems, because I skipped the first two numbers, which were both larger than 5000. Ha.
213436
I don't want to type the whole thing, so a link will have to do. My goodness, more terminology.
0 replies
math_explorer
Sep 18, 2010
0 replies
First spam
math_explorer   2
N Sep 5, 2010 by phiReKaLk6781
I used barycentric coordinates to solve half of a problem. I derived the incenter coordinates and made a bunch of lines and it worked. I read the solution for the other half, not really because I gave up, but because the problem and the solution were next to each other and it was too easy to read.

TRIANGLE CENTERS
Incenter, centroid, circumcenter, orthocenter, nine-point center, Gergonne point, symmedian point, Fermat point. Rooting through Wikipedia references for the Gergonne point you find a paper with 43 theorems, like "The Gergonne point is the perspector of the intouch triangle and the triangle of the reflections of the internal center of similitude of the incenter and the circumcenter in the sides of the excentral triangle." And the scariest bit is I now understand every word there.

THEOREMS I KEEP FORGETTING TO USE
Ptolemy's theorem

Actually, that's it.

STUFF I NEVER GET TO USE
Desargues' theorem
Steiner-Lehmus
Complex number bash
Arrow's theorem (*)

* Yes, it has nothing to do with geometry, or really with anything in the range of high-school contests, but it's awesome.

SOMETHING ELSE
I received my abstract algebra textbook the class I'm in will use. Nothing related to AoPS there, unfortunately. Anyway, much friendlier than last year's.

My math teacher also reports that some of my classmates want to take the AMC 8.

I really wonder too often how much of my personal information a stalker could deduce from this site.

If you read all the way here, please comment just for the heck of it.
2 replies
math_explorer
Sep 4, 2010
phiReKaLk6781
Sep 5, 2010
No more topics!
<A=45^o, CD \perp AB, P\perp BC iff |AP| = |BC|.
parmenides51   1
N Aug 4, 2019 by AlastorMoody
Let $ABC$ be an acute triangle such that $\angle BAC = 45^o$. Let $D$ a point on $AB$ such that $CD \perp AB$. Let $P$ be an internal point of the segment $CD$. Prove that $AP\perp BC$ if and only if $|AP| = |BC|$.
1 reply
parmenides51
Aug 4, 2019
AlastorMoody
Aug 4, 2019
<A=45^o, CD \perp AB, P\perp BC iff |AP| = |BC|.
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parmenides51
30653 posts
#1 • 1 Y
Y by Adventure10
Let $ABC$ be an acute triangle such that $\angle BAC = 45^o$. Let $D$ a point on $AB$ such that $CD \perp AB$. Let $P$ be an internal point of the segment $CD$. Prove that $AP\perp BC$ if and only if $|AP| = |BC|$.
This post has been edited 3 times. Last edited by parmenides51, Jan 10, 2020, 1:57 PM
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AlastorMoody
2125 posts
#2 • 1 Y
Y by Adventure10
Refer here. Let $P$ $\in$ $CD$, such $AP \perp BC$ $\implies$ $AP=2R \cos A$ $=$ $\sqrt{2}a \cdot \frac{1}{\sqrt{2}}$ $=$ $a$. The iff part follows from phantom points
This post has been edited 3 times. Last edited by AlastorMoody, Aug 4, 2019, 11:38 PM
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