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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

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0 replies
jwelsh
Jul 1, 2025
0 replies
Tangent at the orthocenter
Tofa7a._.36   0
an hour ago
Let $ABC$ be a triangle. Consider the points $D,E,F$ as the feet of the altitudes from $A,B,C,$ respectively, and $H$ its orthocenter which we suppose is the midpoint of $CF$. Let $M$ be the midpoint of $BC$, $N$ be the midpoint of $BE$, and $X=(AN)\cap(MF).$
Prove that $\angle HXM=90^\circ$.
0 replies
Tofa7a._.36
an hour ago
0 replies
Nt and chains
Tofa7a._.36   0
an hour ago
Let \( a_0, a_1, \ldots, a_n \) be positive divisors of the number \( 2024^{2025} \) such that:

\(\bullet\) \( a_0 < a_1 < a_2 < \cdots < a_n \)

\(\bullet\) \( a_0 \mid a_1,\, a_1 \mid a_2,\, \ldots,\, a_{n-1} \mid a_n \)

Find the largest possible value of the positive integer \( n \).
0 replies
Tofa7a._.36
an hour ago
0 replies
Nt with casework
Tofa7a._.36   0
2 hours ago
Find all nonnegative integers \( x, y \in \mathbb{N} \) and primes \( p \) such that:
\[
x^2 = p - 2 \qquad \text{and} \qquad y^2 = 2p^2 - 2
\]
0 replies
Tofa7a._.36
2 hours ago
0 replies
A little tedious digits problem (i think)
menseggerofgod   3
N 2 hours ago by menseggerofgod
How many digits must at least a number that is a multiple of 13 have if by exchanging two of its digits it is still a multiple of 13?
a) 4 b) 5 c) 6 d) 7 e) 8 Well, this is my try
Ir a=13k (the number), trying with 4 77≤k≤769
And , trying with the other keys to find some number that verify the rules,
Another, if the number is (abcd) put in mod13 and find values for a,b,c,d .....and other larges tryes
I hope someone can help me......
3 replies
menseggerofgod
Yesterday at 2:26 AM
menseggerofgod
2 hours ago
Divisors Circle
Tofa7a._.36   0
2 hours ago
Mohamed wrote $9$ distinct natural numbers around the circumference of a circle. Each time, he chooses two numbers and writes all the positive divisors of their difference inside the circle. The process ends when all possible pairs have been chosen.
In this case, prove that all numbers less than $9$ have been written inside the circle (not necessarily alone).
0 replies
Tofa7a._.36
2 hours ago
0 replies
Octagon/Rectangle Area
worthawholebean   9
N 4 hours ago by mudkip42
A regular octagon $ ABCDEFGH$ has an area of one square unit. What is the area of the rectangle $ ABEF$?
IMAGE$ \textbf{(A)}\ 1-\frac{\sqrt2}{2} \qquad
\textbf{(B)}\ \frac{\sqrt2}{4} \qquad
\textbf{(C)}\ \sqrt2-1 \qquad
\textbf{(D)}\ \frac12 \qquad
\textbf{(E)}\ \frac{1+\sqrt2}{4}$
9 replies
worthawholebean
Jan 5, 2009
mudkip42
4 hours ago
2003 AMC 12B Question 15 Help
emma_maths   1
N 4 hours ago by mudkip42
This is question 15:
[quote]A regular octagon $ABCDEFGH$ has an area of one square unit. What is the area of the rectangle $ABEF$?

$\text {(A) } 1-\frac{\sqrt{2}}{2} \qquad \text {(B) } \frac{\sqrt{2}}{4} \qquad \text {(C) } \sqrt{2}-1 \qquad \text {(D) } \frac{1}{2} \qquad \text {(E) } \frac{1+\sqrt{2}}{4}$[/quote]

My Solution:
The attachment below is the way I drew the diagram. Since a regular octagon is made of $8$ equilateral triangles, I know the are of each triangle is $1/8$. I also know that the area of an equilateral triangle is $\frac{\sqrt3}{4}a^2$. So, I can solve for $a$: \begin{align*}\frac{1}{8} &= \frac{\sqrt3}{4}a^2 \\ \frac{1}{2\sqrt3} &= a^2 \\ \frac{\sqrt3}{6} &= a^2 \\ a &= \frac{\sqrt{6\sqrt3}}{6}. \end{align*}
If we notice, in the diagram, the height of the rectangle $ABEF$ is twice the height of each triangle. For an equilateral triangle with side length $s$, the height would be $\frac{s}{2}\sqrt{3}$. So, in this case it would be \begin{align*} \frac{\sqrt{6\sqrt3} \cdot \sqrt3}
{2\cdot6} &= \frac{9\sqrt{2\sqrt{3}}}{12} \\ &= \frac{3\sqrt{2\sqrt3}}{4}. \end{align*}So, the height of the rectangle is $$2 \cdot \frac{3\sqrt{2\sqrt3}}{4} =  \frac{3\sqrt{2\sqrt3}}{2}.$$
All together, this makes the area of the rectangle \begin{align*}\frac{\sqrt{6\sqrt3}}{6} \cdot \frac{3\sqrt{2\sqrt3}}{2} &= \frac{\sqrt{12 \cdot 3}}{4} \\ &= \boxed{\frac{3}{2}.} \end{align*}
That isn’t even in the options. After looking at the answers (D), I know my solution is wayyy too complicated, but I am not sure what exactly I am doing wrong. Is there a way I could have tweaked my solution to get the correct result? Thanks!
1 reply
emma_maths
4 hours ago
mudkip42
4 hours ago
Number Theory (Divisibility)
AbdulWaheed   3
N 4 hours ago by KSH31415
Find all triples (a, b, c) of natural numbers such that the numbers $a^2$ + bc, $b^2$ + ac,
and $c^2$ + ab are powers of 2. (source, Sozopol 2023, Grades 8-9)
3 replies
AbdulWaheed
Yesterday at 7:14 AM
KSH31415
4 hours ago
Grade 10 Inequality
BomboaneMentos316   1
N 5 hours ago by aaravdodhia
Prove that for any x,y,z real and positive the following is true:
(PS: It is written correctly)

\begin{align*}
\frac{x^3}{y^2 + y z + z^2}
\;+\;
\frac{y^3}{2 z^2 + y z x}
\;+\;
\frac{z^3}{x^2 + x y + y^2}
\;\ge\;
\frac{x + y + z}{3}.
\end{align*}
1 reply
BomboaneMentos316
Yesterday at 6:41 PM
aaravdodhia
5 hours ago
10 Problems
Sedro   14
N 5 hours ago by Sedro
Title says most of it. I've been meaning to post a problem set on HSM since at least a few months ago, but since I proposed the most recent problems I made to the 2025 SSMO, I had to wait for that happen. (Hence, most of these problems will probably be familiar if you participated in that contest, though numbers and wording may be changed.) The problems are very roughly arranged by difficulty. Enjoy!

Problem 1: An sequence of positive integers $u_1, u_2, \dots, u_8$ has the property for every positive integer $n\le 8$, its $n^\text{th}$ term is greater than the mean of the first $n-1$ terms, and the sum of its first $n$ terms is a multiple of $n$. Let $S$ be the number of such sequences satisfying $u_1+u_2+\cdots + u_8 = 144$. Compute the remainder when $S$ is divided by $1000$.

Problem 2 (solved by fruitmonster97): Rhombus $PQRS$ has side length $3$. Point $X$ lies on segment $PR$ such that line $QX$ is perpendicular to line $PS$. Given that $QX=2$, the area of $PQRS$ can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Problem 3: Positive integers $a$ and $b$ satisfy $a\mid b^2$, $b\mid a^3$, and $a^3b^2 \mid 2025^{36}$. If the number of possible ordered pairs $(a,b)$ is equal to $N$, compute the remainder when $N$ is divided by $1000$.

Problem 4: Let $ABC$ be a triangle. Point $P$ lies on side $BC$, point $Q$ lies on side $AB$, and point $R$ lies on side $AC$ such that $PQ=BQ$, $CR=PR$, and $\angle APB<90^\circ$. Let $H$ be the foot of the altitude from $A$ to $BC$. Given that $BP=3$, $CP=5$, and $[AQPR] = \tfrac{3}{7} \cdot [ABC]$, the value of $BH\cdot CH$ can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Problem 5: Anna has a three-term arithmetic sequence of integers. She divides each term of her sequence by a positive integer $n>1$ and tells Bob that the three resulting remainders are $20$, $52$, and $R$, in some order. For how many values of $R$ is it possible for Bob to uniquely determine $n$?

Problem 6: There is a unique ordered triple of positive reals $(x,y,z)$ satisfying the system of equations \begin{align*} x^2 + 9 &= (y-\sqrt{192})^2 + 4 \\ y^2 + 4 &= (z-\sqrt{192})^2 + 49 \\ z^2 + 49 &= (x-\sqrt{192})^2 + 9. \end{align*}The value of $100x+10y+z$ can be expressed as $p\sqrt{q}$, where $p$ and $q$ are positive integers such that $q$ is square-free. Compute $p+q$.

Problem 7: Let $S$ be the set of all monotonically increasing six-term sequences whose terms are all integers between $0$ and $6$ inclusive. We say a sequence $s=n_1, n_2, \dots, n_6$ in $S$ is symmetric if for every integer $1\le i \le 6$, the number of terms of $s$ that are at least $i$ is $n_{7-i}$. The probability that a randomly chosen element of $S$ is symmetric is $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Compute $p+q$.

Problem 8: For a positive integer $n$, let $r(n)$ denote the value of the binary number obtained by reading the binary representation of $n$ from right to left. Find the smallest positive integer $k$ such that the equation $n+r(n)=2k$ has at least ten positive integer solutions $n$.

Problem 9: Let $p$ be a quadratic polynomial with a positive leading coefficient. There exists a positive real number $r$ such that $r < 1 < \tfrac{5}{2r} < 5$ and $p(p(x)) = x$ for $x \in \{ r,1,  \tfrac{5}{2r} , 5\}$. Compute $p(20)$.

Problem 10: Find the number of ordered triples of positive integers $(a,b,c)$ such that $a+b+c=995$ and $ab+bc+ca$ is a multiple of $995$.
14 replies
Sedro
Jul 10, 2025
Sedro
5 hours ago
Find the min value of function f(x,y)
PhysicsIsChad   2
N 5 hours ago by maromex
Let
\[
f(x, y) = \sqrt{x^2 + y^2} + \sqrt{x^2 + y^2 - 2x + 1} + \sqrt{x^2 + y^2 - 2y + 1} + \sqrt{x^2 + y^2 - 6x - 8y + 25}
\]for all real \( x, y \in \mathbb{R} \).
Find the min value of this function . What are your insights on this problem ?
My insights :
All these quadratics under the square root can be converted into \[
f(x, y) = \sqrt{(x - 0)^2 + (y - 0)^2} 
+ \sqrt{(x - 1)^2 + (y - 0)^2} 
+ \sqrt{(x - 0)^2 + (y - 1)^2} 
+ \sqrt{(x - 3)^2 + (y - 4)^2}
\]This function can be interpreted as the sum of the distances of points (0,0) , (1,0) , (0,1) , (3,4) from (x,y) .
We can prove that for minimising this distance (x,y) must be the mean of the x and y coordinates of the points respectively . Thus we get (x,y) = (1,1.25) . \[
\begin{aligned}
f(x, y) &= \sqrt{(x - 0)^2 + (y - 0)^2} 
+ \sqrt{(x - 1)^2 + (y - 0)^2} \\
&\quad + \sqrt{(x - 0)^2 + (y - 1)^2} 
+ \sqrt{(x - 3)^2 + (y - 4)^2} \\
\\
f(1, 1.25) &= \sqrt{1^2 + 1.25^2} 
+ \sqrt{(1 - 1)^2 + 1.25^2} \\
&\quad + \sqrt{1^2 + (1.25 - 1)^2} 
+ \sqrt{(1 - 3)^2 + (1.25 - 4)^2} \\
\\
&= \sqrt{2.5625} + \sqrt{1.5625} + \sqrt{1.0625} + \sqrt{11.5625} \\
\\
&\approx 1.6 + 1.25 + 1.03078 + 3.40184 \\
\\
&= \boxed{7.2826}
\end{aligned}
\]But apparently this is wrong . I will post the proof for the intermediate i used in the replies .





2 replies
PhysicsIsChad
Yesterday at 4:36 PM
maromex
5 hours ago
Inversive Geo
Tofa7a._.36   1
N Yesterday at 7:44 PM by aaravdodhia
Let $ABC$ be an acute and scalene triangle with circumcircle $\omega$. The perpendicular bisector of the segment $AB$ intersects the lines $BC$ and $AC$ in points $D$ and $E$, respectively, such that $E$ lies outside segment $AC$. The perpendicular to line $BC$ from $D$ intersects $(BCE)$ at a point $X$ outside $\triangle ABC$. Line $DX$ intersects line $AC$ at $Y$ and $\omega$ at points $Z$ and $T$ such that $Z$ lies on the arc $AC$ that does not contain $B$. The circumcircle of triangle $\triangle ZET$ intersects the side $BC$ and the circumcircle of triangle $\triangle YDE$ in $P$ and $Q$, respectively.
Prove that the tangent to $(YZQ)$ from $Z$, the tangent to $(YTQ)$ from $T$, and the line $PX$ meet at one point.
1 reply
Tofa7a._.36
Yesterday at 3:35 PM
aaravdodhia
Yesterday at 7:44 PM
Trivial Fe
Tofa7a._.36   2
N Yesterday at 6:01 PM by Kempu33334
Find all functions $f : \mathbb R \rightarrow \mathbb R$ such that:
$$f(f(2x+y)) + f(x) = 2x + f(x+y)$$For all real numbers $x,y \in \mathbb{R}$.
2 replies
Tofa7a._.36
Yesterday at 5:20 PM
Kempu33334
Yesterday at 6:01 PM
Hexagon geometry
Tofa7a._.36   0
Yesterday at 5:53 PM
Let $ABC$ be a triangle such that $\angle{ABC}=3\angle{ACB}$. In the circumcircle of this triangle, let $D,E$ and $F$ be points such that: $(AD)\parallel (BC)$, $(DE)\parallel (CA)$ and $(EF) \parallel (AB)$.
Let $J$ be the intersection of lines $DF$ and $AC$ and let $\omega$ be the circle that passes through $J$ and is tangent to the line $BD$ at $D$.
Finally, let $L$ be the intersection point of $\omega$ and $(ABC)$.
Prove that points $E,J$ and $L$ are collinear.
0 replies
Tofa7a._.36
Yesterday at 5:53 PM
0 replies
Function and Quadratic equations help help help
Ocean_MathGod   1
N May 20, 2025 by Mathzeus1024
Consider this parabola: y = x^2 + (2m + 1)x + m(m - 3) where m is constant and -1 ≤ m ≤ 4. A(-m-1, y1), B(m/2, y2), C(-m, y3) are three different points on the parabola. Now rotate the axis of symmetry of the parabola 90 degrees counterclockwise around the origin O to obtain line a. Draw a line from the vertex P of the parabola perpendicular to line a, meeting at point H.

1) express the vertex of the quadratic equation using an expression with m.
2) If, regardless of the value of m, the parabola and the line y=x−km (where k is a constant) have exactly one point of intersection, find the value of k.

3) (where I'm struggling the most) When 1 < PH ≤ 6, compare the values of y1, y2, and y3.
1 reply
Ocean_MathGod
Aug 26, 2024
Mathzeus1024
May 20, 2025
Function and Quadratic equations help help help
G H J
G H BBookmark kLocked kLocked NReply
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Ocean_MathGod
20 posts
#1
Y by
Consider this parabola: y = x^2 + (2m + 1)x + m(m - 3) where m is constant and -1 ≤ m ≤ 4. A(-m-1, y1), B(m/2, y2), C(-m, y3) are three different points on the parabola. Now rotate the axis of symmetry of the parabola 90 degrees counterclockwise around the origin O to obtain line a. Draw a line from the vertex P of the parabola perpendicular to line a, meeting at point H.

1) express the vertex of the quadratic equation using an expression with m.
2) If, regardless of the value of m, the parabola and the line y=x−km (where k is a constant) have exactly one point of intersection, find the value of k.

3) (where I'm struggling the most) When 1 < PH ≤ 6, compare the values of y1, y2, and y3.
Z K Y
The post below has been deleted. Click to close.
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Mathzeus1024
1064 posts
#2
Y by
Part (I):

Completing the square on this parabola yields:

$y = \left(x+\frac{2m+1}{2}\right)^2 + m(m-3) - \frac{(2m+1)^2}{4} \Rightarrow y = \left(x+\frac{2m+1}{2}\right)^2 -\frac{16m+1}{4}$;

with vertex $\textcolor{red}{P\left(-\frac{2m+1}{2}, -\frac{16m+1}{4}\right)}$.

Part (II):

If the parabola intersects the line $y=x-km$ in exactly one point, then:

$x^2 + (2m+1)x + m(m-3) = x-km \Rightarrow x^2 +2mx + (m^2-3m+km) = 0 \Rightarrow x = \frac{-2m \pm \sqrt{4m^2-4(1)[m^2+(k-3)m]}}{2}$ (i);

of which we require the discriminant in (i) to equal zero. This occurs $\Leftrightarrow \textcolor{red}{k=3}$.

Part (III):

For $m \in [-1,4]$ we have the points $A(-m-1, -4m); B\left(\frac{m}{2}, \frac{9m^2-10m}{4}\right); C(-m,-4m)$ on the parabola. Checking $-4m=\frac{9m^2-10m}{4} \Rightarrow 9m^2-6m=m(9m+6)=0 \Rightarrow m = -\frac{2}{3}, 0$ gives us the orderings:

$y_{2}>y_{1}=y_{3}$ for $m \in \left[-1,-\frac{2}{3}\right) \cup (0,4]$;

$y_{2}=y_{1}=y_{3}$ for $m = -\frac{2}{3}, 0$;

$y_{2} < y_{1} = y_{3}$ for $m \in \left(-\frac{2}{3}, 0\right)$.

Rotating the parabola's axis of symmetry (i.e. $x= -\frac{2m+1}{2}$) $90^{\circ}$ counterclockwise about the origin gives us the horizontal line $y = -\frac{2m+1}{2}$ from which $|PH| = \left|-\frac{16m+1}{4} - \left(-\frac{2m+1}{2}\right)\right| = \frac{|1-12m|}{4}$.

If $1 < |PH| \le 6$, then $1 < \frac{|1-12m|}{4}\le 6 \Rightarrow m \in \left[-\frac{23}{12},-\frac{1}{4}\right) \cup \left(\frac{5}{12},\frac{25}{12}\right]$. This results in the following orderings:

$\textcolor{red}{y_{2}>y_{1}=y_{2}}$ for $\textcolor{red}{m \in \left[-1, -\frac{2}{3}\right) \cup \left(\frac{5}{12},\frac{25}{12}\right]}$;

$\textcolor{red}{y_{2}=y_{1}=y_{3}}$ for $\textcolor{red}{m = -\frac{2}{3}}$;

$\textcolor{red}{y_{2}<y_{1}=y_{3}}$ for $\textcolor{red}{m \in \left(-\frac{2}{3}, -\frac{1}{4}\right)}$.
This post has been edited 6 times. Last edited by Mathzeus1024, May 31, 2025, 1:03 PM
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