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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
Combinatorics
P162008   3
N 3 minutes ago by P162008
Let $m,n \in \mathbb{N}.$ Let $[n]$ denote the set of natural numbers less than or equal to $n.$

Let $f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$

Compute the sum of the digits of $f(4,4).$
3 replies
P162008
3 hours ago
P162008
3 minutes ago
Find min and max
lgx57   0
36 minutes ago
Source: Own
$x_1,x_2, \cdots ,x_n\ge 0$,$\displaystyle\sum_{i=1}^n x_i=m$. $k_1,k_2,\cdots,k_n >0$. Find min and max of
$$\sum_{i=1}^n(k_i\prod_{j=1}^i x_j)$$
0 replies
lgx57
36 minutes ago
0 replies
Find min
lgx57   0
40 minutes ago
Source: Own
$a,b>0$, $a^4+ab+b^4=60$. Find min of
$$4a^2-ab+4b^2$$
$a,b>0$, $a^4-ab+b^4=60$. Find min of
$$4a^2-ab+4b^2$$
0 replies
lgx57
40 minutes ago
0 replies
Geometry
Lukariman   3
N 2 hours ago by Lukariman
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
3 replies
Lukariman
Yesterday at 12:43 PM
Lukariman
2 hours ago
No more topics!
Problem 5 -- Making A Point Here
RockmanEX3   6
N Oct 18, 2020 by jayme
Source: 46th Austrian Mathematical Olympiad National Competition Part 2 Problem 5
Let I be the incenter of triangle $ABC$ and let $k$ be a circle through the points $A$ and $B$. The circle intersects

* the line $AI$ in points $A$ and $P$
* the line $BI$ in points $B$ and $Q$
* the line $AC$ in points $A$ and $R$
* the line $BC$ in points $B$ and $S$

with none of the points $A,B,P,Q,R$ and $S$ coinciding and such that $R$ and $S$ are interior points of the line segments $AC$ and $BC$, respectively.

Prove that the lines $PS$, $QR$, and $CI$ meet in a single point.

(Stephan Wagner)
6 replies
RockmanEX3
Jul 14, 2018
jayme
Oct 18, 2020
Problem 5 -- Making A Point Here
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G H BBookmark kLocked kLocked NReply
Source: 46th Austrian Mathematical Olympiad National Competition Part 2 Problem 5
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RockmanEX3
274 posts
#1 • 4 Y
Y by PavelMath, Adventure10, Mango247, AlexCenteno2007
Let I be the incenter of triangle $ABC$ and let $k$ be a circle through the points $A$ and $B$. The circle intersects

* the line $AI$ in points $A$ and $P$
* the line $BI$ in points $B$ and $Q$
* the line $AC$ in points $A$ and $R$
* the line $BC$ in points $B$ and $S$

with none of the points $A,B,P,Q,R$ and $S$ coinciding and such that $R$ and $S$ are interior points of the line segments $AC$ and $BC$, respectively.

Prove that the lines $PS$, $QR$, and $CI$ meet in a single point.

(Stephan Wagner)
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PavelMath
185 posts
#2 • 2 Y
Y by Adventure10, Mango247
RockmanEX3 wrote:
Let I be the incenter of triangle $ABC$ and let $k$ be a circle through the points $A$ and $B$. The circle intersects

* the line $AI$ in points $A$ and $P$
* the line $BI$ in points $B$ and $Q$
* the line $AC$ in points $A$ and $R$
* the line $BC$ in points $B$ and $S$

with none of the points $A,B,P,Q,R$ and $S$ coinciding and such that $R$ and $S$ are interior points of the line segments $AC$ and $BC$, respectively.

Prove that the lines $PS$, $QR$, and $CI$ meet in a single point.

(Stephan Wagner)

Very nice problem!
Denote by $K$ and $L$ the intersection points of the line which passes through the point $I$ parallel to $AB$ with the sides $AC$ and $BC$ respectively. By direct counting of the angles it iseasy to prove that
1) points $K$, $Q$, $R$, $I$ lie on one circle $\omega_1$;
2) points $L$, $P$, $S$, $I$ lie on one circle $\omega_2$;
3) points $K$, $R$, $S$, $L$ li on one circle $\omega_3$.
Then point $C$ is the radical center of the circles $\omega_1$, $\omega_2$ and $\omega_3$ and $(CI)$ is the radical axis of the circles $\omega_1$ and $\omega_2$.
Denote the circle which passes through the points $A$, $R$, $S$, $B$ by $\Omega$. Then the radical center of the circles $\omega_1$, $\omega_2$ and $\Omega$ is the intersection point of the lines $(CI)$, $(QR)$ and $(PS)$. Hence these lines are concurrent.
This post has been edited 2 times. Last edited by PavelMath, Jul 14, 2018, 7:26 PM
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Gryphos
1702 posts
#3 • 1 Y
Y by Adventure10
My solution:
It is easy to see by angle chasing that $PQ \perp CI$. Moreover, $\angle PQR = \angle PAR = \angle BAP = \angle IQP$, hence the reflection across the line $PQ$ maps $QI$ to $QR$. Analogously, this reflection maps $PI$ to $PS$ and the line $CI$ to itself. Thus $QR$, $PS$ and $CI$ concur at the reflection of $I$ across the line $PQ$.
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jayme
9792 posts
#4 • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,
just applie the Pascal's theorem...

Sincerely
Jean-Louis
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dikhendzab
108 posts
#5 • 3 Y
Y by Mango247, Mango247, Mango247
Let $QR$ and $PS$ intersect in $X$. Quadrilateral $ABSR$ is cyclic, so $\angle BSR=180^{\circ}- \alpha$, so $\angle RSC=\alpha.$
$\angle ARS=180^{\circ}- \beta$, so $\angle CRS=\beta.$
$\angle RSP=\angle RSX=\angle RAP=\frac{\alpha}{2}$ (angles on the same arc)
$\angle RSX=\angle XSC=\frac{\alpha}{2}$, so $XS$ is the bisector of $\angle RSC$
Now, $\angle QRA=\angle QBA=\angle CRX=\frac{\beta}{2}$ (angles on the same arc)
$\angle CRX=\angle SRX=\frac{\beta}{2}$, so $XR$ is the bisector of $\angle SRC$
Obviously $CX$ is the bisector of $\angle SCR \implies CI, QR, PS$ intersect in one point. :)
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sonny2011
7 posts
#6 • 1 Y
Y by Mango247
The easy way to solve this problem is used by Pascal's Theorem with hexagon $ARQBSP$.
We will get that incenter of $\triangle ABC$ ($I$), the intersection point of $PS$ and $QR$, and point $C$ are collinear.
Thus, $PS$, $QR$, and $CI$ are concurrent.
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jayme
9792 posts
#7 • 2 Y
Y by Mango247, Mango247
Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/02.%201.%20Points%20sur%20une%20cevienne%20du%20triangle.pdf p.19...

Sincerely

Jean-Louis
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