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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
interesting geo config (2/3)
Royal_mhyasd   2
N 3 minutes ago by Ilikeminecraft
Source: own
Let $\triangle ABC$ be an acute triangle and $H$ its orthocenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = |\angle ABC-\angle ACB|$. Define $Q$ and $R$ as points on the parallels through $B$ to $AC$ and through $C$ to $AB$ similarly. If $P,Q,R$ are positioned around the sides of $\triangle ABC$ as in the given configuration, prove that $P,Q,R$ are collinear.
2 replies
Royal_mhyasd
6 hours ago
Ilikeminecraft
3 minutes ago
interesting geo config (1\3)
Royal_mhyasd   1
N 8 minutes ago by Ilikeminecraft
Source: own
Let $\triangle ABC$ be an acute triangle with $AC > AB$, $H$ its orthocenter and $O$ it's circumcenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = \angle ABC - \angle ACB$ and $P$ and $C$ are on different sides of $AB$. Denote by $S$ the intersection of the circumcircle of $\triangle ABC$ and $PA'$, where $A'$ is the reflection of $H$ over $BC$, $M$ the midpoint of $PH$, $Q$ the intersection of $OA$ and the parallel through $M$ to $AS$, $R$ the intersection of $MS$ and the perpendicular through $O$ to $PS$ and $N$ a point on $AS$ such that $NT \parallel PS$, where $T$ is the midpoint of $HS$. Prove that $Q, N, R$ lie on a line.

fiy it's 2am and i'm bored so i decided to look further into this interesting config that i had already made some observations on, maybe this problem is trivial from some theorem so if that's the case then i'm sorry lol :P i'll probably post 2 more problems related to it soon, i'd say they're easier than this though
1 reply
Royal_mhyasd
Yesterday at 11:18 PM
Ilikeminecraft
8 minutes ago
Rootiful sets
InternetPerson10   38
N 18 minutes ago by cursed_tangent1434
Source: IMO 2019 SL N3
We say that a set $S$ of integers is rootiful if, for any positive integer $n$ and any $a_0, a_1, \cdots, a_n \in S$, all integer roots of the polynomial $a_0+a_1x+\cdots+a_nx^n$ are also in $S$. Find all rootiful sets of integers that contain all numbers of the form $2^a - 2^b$ for positive integers $a$ and $b$.
38 replies
InternetPerson10
Sep 22, 2020
cursed_tangent1434
18 minutes ago
weird conditions in geo
Davdav1232   2
N an hour ago by teoira
Source: Israel TST 7 2025 p1
Let \( \triangle ABC \) be an isosceles triangle with \( AB = AC \). Let \( D \) be a point on \( AC \). Let \( L \) be a point inside the triangle such that \( \angle CLD = 90^\circ \) and
\[
CL \cdot BD = BL \cdot CD.
\]Prove that the circumcenter of triangle \( \triangle BDL \) lies on line \( AB \).
2 replies
+1 w
Davdav1232
May 8, 2025
teoira
an hour ago
No more topics!
Tangential quadrilateral collinearity on steroids
Justpassingby   2
N Jan 18, 2022 by Justpassingby
Source: 2021 Mexico Center Zone Regional Olympiad, problem 3
Let $W,X,Y$ and $Z$ be points on a circumference $\omega$ with center $O$, in that order, such that $WY$ is perpendicular to $XZ$; $T$ is their intersection. $ABCD$ is the convex quadrilateral such that $W,X,Y$ and $Z$ are the tangency points of $\omega$ with segments $AB,BC,CD$ and $DA$ respectively. The perpendicular lines to $OA$ and $OB$ through $A$ and $B$, respectively, intersect at $P$; the perpendicular lines to $OB$ and $OC$ through $B$ and $C$, respectively, intersect at $Q$, and the perpendicular lines to $OC$ and $OD$ through $C$ and $D$, respectively, intersect at $R$. $O_1$ is the circumcenter of triangle $PQR$. Prove that $T,O$ and $O_1$ are collinear.

Proposed by CDMX
2 replies
Justpassingby
Jan 17, 2022
Justpassingby
Jan 18, 2022
Tangential quadrilateral collinearity on steroids
G H J
Source: 2021 Mexico Center Zone Regional Olympiad, problem 3
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Justpassingby
81 posts
#1
Y by
Let $W,X,Y$ and $Z$ be points on a circumference $\omega$ with center $O$, in that order, such that $WY$ is perpendicular to $XZ$; $T$ is their intersection. $ABCD$ is the convex quadrilateral such that $W,X,Y$ and $Z$ are the tangency points of $\omega$ with segments $AB,BC,CD$ and $DA$ respectively. The perpendicular lines to $OA$ and $OB$ through $A$ and $B$, respectively, intersect at $P$; the perpendicular lines to $OB$ and $OC$ through $B$ and $C$, respectively, intersect at $Q$, and the perpendicular lines to $OC$ and $OD$ through $C$ and $D$, respectively, intersect at $R$. $O_1$ is the circumcenter of triangle $PQR$. Prove that $T,O$ and $O_1$ are collinear.

Proposed by CDMX
This post has been edited 2 times. Last edited by Justpassingby, Jan 20, 2022, 1:31 AM
Reason: Added proposer
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Ru83n05
171 posts
#2
Y by
Let $S$ be the intersection of the lines perpendicular to $OA$ and $OD$ trough $A$ and $D$ respectively.

Claim 1: $PQRS\sim WXYZ$. In fact they are homothetic (through $O_2$ say)
Proof: All sides are parallel. $\blacksquare$

So we define $\Phi : WXYZ\mapsto PQRS $. We also contend

Claim 2: $P-O-R$ are colinear (and so are $Q-O-S$)
Proof: Consider $O_3:=BC\cap AD$. Then $P$ is the incenter of $\triangle O_3AB$, $O$ is the incenter of $\triangle O_3CD$ and $R$ is the $O_3$-excenter of $\triangle O_3CD$. They all lie on the internal bisector of $\angle AO_3B$. $\blacksquare$

We can now finish by homothety at $O_2$. In fact $O_2-T-O-O_1$ are colinear: $\Phi(T)=O$ so $O_2-T-O$ and $\Phi(O)=O_1$ so $O_2-O-O_1$. We are done. $\blacksquare$
This post has been edited 1 time. Last edited by Ru83n05, Jan 17, 2022, 3:31 PM
Z K Y
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Justpassingby
81 posts
#3
Y by
Ru83n05 wrote:
Claim 1: $PQRS\sim WXYZ$. In fact they are homothetic (through $O_2$ say)
Proof: All sides are parallel. $\blacksquare$
Not to be pedantic, but this is not enough to say they are similar/homothetic. It is also necessary for $PR$ to be parallel to $WY$, which is not so hard to prove, but also not trivial, I think.

The proof looks good from there on :).
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