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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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[*]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
Nice "if and only if" function problem
ICE_CNME_4   3
N a minute ago by ICE_CNME_4
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )

Please do it at 9th grade level. Thank you!
3 replies
ICE_CNME_4
Yesterday at 7:23 PM
ICE_CNME_4
a minute ago
IMO 2014 Problem 1
Amir Hossein   134
N 11 minutes ago by Ihatecombin
Let $a_0 < a_1 < a_2 < \dots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n\geq 1$ such that
\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]
Proposed by Gerhard Wöginger, Austria.
134 replies
Amir Hossein
Jul 8, 2014
Ihatecombin
11 minutes ago
Arrange marbles
FunGuy1   2
N 15 minutes ago by Sunshine132
Source: Own?
Anna has $200$ marbles in $25$ colors such that there are exactly $8$ marbles of each color. She wants to arrange them on $50$ shelves, $4$ marbles on each shelf such that for any $2$ colors there is a shelf that has marbles of those colors.
Can Anna achieve her goal?
2 replies
FunGuy1
38 minutes ago
Sunshine132
15 minutes ago
Nice orthocenter config
Rijul saini   12
N 22 minutes ago by Commander_Anta78
Source: India IMOTC 2024 Day 4 Problem 3
Let $ABC$ be an acute-angled triangle with $AB<AC$, and let $O,H$ be its circumcentre and orthocentre respectively. Points $Z,Y$ lie on segments $AB,AC$ respectively, such that \[\angle ZOB=\angle YOC = 90^{\circ}.\]The perpendicular line from $H$ to line $YZ$ meets lines $BO$ and $CO$ at $Q,R$ respectively. Let the tangents to the circumcircle of $\triangle AYZ$ at points $Y$ and $Z$ meet at point $T$. Prove that $Q, R, O, T$ are concyclic.

Proposed by Kazi Aryan Amin and K.V. Sudharshan
12 replies
Rijul saini
May 31, 2024
Commander_Anta78
22 minutes ago
No more topics!
Lines pass through a common point
April   4
N Apr 7, 2025 by Nari_Tom
Source: Baltic Way 2008, Problem 18
Let $ AB$ be a diameter of a circle $ S$, and let $ L$ be the tangent at $ A$. Furthermore, let $ c$ be a fixed, positive real, and consider all pairs of points $ X$ and $ Y$ lying on $ L$, on opposite sides of $ A$, such that $ |AX|\cdot |AY| = c$. The lines $ BX$ and $ BY$ intersect $ S$ at points $ P$ and $ Q$, respectively. Show that all the lines $ PQ$ pass through a common point.
4 replies
April
Nov 23, 2008
Nari_Tom
Apr 7, 2025
Lines pass through a common point
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Source: Baltic Way 2008, Problem 18
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April
1270 posts
#1 • 1 Y
Y by Adventure10
Let $ AB$ be a diameter of a circle $ S$, and let $ L$ be the tangent at $ A$. Furthermore, let $ c$ be a fixed, positive real, and consider all pairs of points $ X$ and $ Y$ lying on $ L$, on opposite sides of $ A$, such that $ |AX|\cdot |AY| = c$. The lines $ BX$ and $ BY$ intersect $ S$ at points $ P$ and $ Q$, respectively. Show that all the lines $ PQ$ pass through a common point.
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minsoens
505 posts
#2 • 4 Y
Y by Mathasocean, Adventure10, Mango247, and 1 other user
[asy]defaultpen(fontsize(8));
pair A=(0,0), B=(10,0), X=(0,10), Y=(0,-5), H, P, Q;
path S = Circle(midpoint(A--B),abs(A-B)/2);
P = intersectionpoints(S, X--B)[1];
Q = intersectionpoints(S, Y--B)[1];
H = extension(P,Q,A,B);
draw(S);
draw(A--Y--B--X--A--P--Q--A--B);
dot(P^^Q^^A^^B^^X^^Y^^H);
label("A",A,(-1,0));label("B",B,(1,0));label("P",P,(1,1));label("Q",Q,(1,-1));label("X",X,(-1,0));label("Y",Y,(-1,0));label("H",H,(1,-1));[/asy]Let $ H=PQ\cap AB$.
Claim: As $ X$ and $ Y$ vary on $ L$ such that $ AX\cdot AY = c$, $ H$ remains fixed on $ AB$.
It is sufficient to show that $ \frac{[PAQ]}{[PBQ]}$ remains constant.
$ \frac{AX}{AB}=\frac{AP}{BP}$, $ \frac{AY}{AB}=\frac{AQ}{BQ}$
$ \implies\frac{c}{AB^2}=\frac{AX\cdot AY}{AB^2}=\frac{AP\cdot AQ}{BP\cdot BQ}=\frac{\frac{1}{2}AP\cdot AQ\sin{PAQ}}{\frac{1}{2}BP\cdot BQ\sin{PBQ}}=\frac{[PAQ]}{[PBQ]}$
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vittasko
1327 posts
#3 • 4 Y
Y by Adventure10, Mango247, and 2 other users
Let $ X',\ Y'$ be, the points on the tangent line $ L$ of the given circle $ (O)$ at point $ A,$ as the problem states and also such that $ AX' = AY'$ $ ,(1)$

So, we have that $ (AX)\cdot (AY) = u^{2} = (AX')^{2} = (AY')^{2}$ $ ,(2)$

We denote the points $ P'\equiv (O)\cap BX',\ Q'\equiv (O)\cap BY'$ and let be the point $ H\equiv AB\cap P'Q'.$

We will prove that the line-segment $ PQ,$ always passes through the point $ H.$

From the right angled triangle $ \bigtriangleup ABX',$ we have that $ (BP')\cdot (BX') = (AB)^{2}$ $ ,(3)$

Similarly from the right angled triangle $ \bigtriangleup ABY',$ we have that $ (BQ')\cdot (BY') = (AB)^{2}$ $ ,(4)$

From $ (3),$ $ (4),$ we conclude that the points $ X',\ P',\ Q',\ Y'$ are concyclic, with their circumcircle so be it $ (O'),$ which intersects the line-segment $ AB,$ at points $ M,\ N$ such that $ (BP')\cdot (BX') = (BM)\cdot (BN) = (AB)^{2}$ $ ,(5)$ and $ (AM)\cdot (AN) = u^{2}$ $ ,(6)$

By the same way as before, we can say that $ (BP)\cdot (BX) = (AB)^{2} = (BQ)\cdot (BY)$ $ ,(7)$ from the right angled triangles $ \bigtriangleup ABX,\ \bigtriangleup ABY.$

From $ (5),$ $ (7),$ we conclude that the points $ X,\ P,\ M,\ N$ are concyclic, with circumcircle so be it $ (O_{1}).$

Because of $ (AM)\cdot (AN) = u^{2} = (AX)\cdot (AY)$ $ ,(8)$ we conclude that the point $ Y,$ lies on $ (O_{1}).$

The point $ Q$ now, lies also on the circle $ (O_{1}),$ because of $ (7).$

That is, the circle $ (O_{1}),$ always passes through the constant points $ M,\ N$ and then, we conclude that the line segment $ PQ,$ always passes through the fixed point $ H\equiv AB\cap P'Q',$ as the radical center of the circles $ (O),\ (O'),\ (O_{1})$ and the proof is completed.

Kostas Vittas.
Attachments:
t=241323.pdf (12kb)
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Lil_flip38
58 posts
#4
Y by
Consider inversion around $A$ with radius $\sqrt{AX\times AY}$
Now, we need to solve the following problem:
Quote:
Let $A$ be a point on line $l$, and let $B$ be a point such that $AB\perp l$. let $X,Y$ be points such that $AX\times AY=c$ for some positive real constant. If circles $(ABY),(ABX)$ intersect the line through $B$ parallel to $l$ at $P,Q$, show that $(APQ)$ passes through a fixed point.

let $Z=(APQ)\cap AB$. We claim $Z$ is the fixed point. Note that $XYPQ$ is a rectangle, so we have that $PB\times BQ=c$. Thus, by power of a point $Z$ is the point such that $AB\times BZ=c$, which is independent of $X,Y$, as desired.
This post has been edited 2 times. Last edited by Lil_flip38, Oct 3, 2024, 11:46 AM
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Nari_Tom
117 posts
#5
Y by
Nice solutions above, mine is most contest like one. We assume here $A,B$ are fixed and $R$ be the radi of circle.

Let's say $\angle PBA=a$ and $\angle QBA=b$. Then we have $tg(a)*tg(b)$ is constant. Let $Z=AB \cap PQ$. It's suffices to prove that length of $CZ$
is constant. In $\triangle CZP$, by the law of sines we have $CZ=\frac{R*sin(90-b-a)}{sin(90-a+b)}=\frac{R*cos(a+b)}{cos(a-b)}=\frac{cos(a)cos(b)-sin(a)sin(b)}{cos(a)cos(b)+sin(a)sin(b)}=\frac{1-tg(a)tg(b)}{1+tg(a)tg(b)}$, which is clearly constant.
This post has been edited 1 time. Last edited by Nari_Tom, Apr 7, 2025, 4:25 PM
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