Stay ahead of learning milestones! Enroll in a class over the summer!

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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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)!

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[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
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April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
Involved conditional geo
Assassino9931   1
N 4 minutes ago by hukilau17
Source: Balkan MO 2024 Shortlist G4
Let $ABC$ be an acute-angled triangle with $AB < AC$, orthocenter $H$, circumcircle $\Gamma$ and circumcentre $O$. Let $M$ be the midpoint of $BC$ and let $D$ be a point such that $ADOH$ is a parallellogram. Suppose that there exists a point $X$ on $\Gamma$ and on the opposite side of $DH$ to $A$ such that $\angle DXH + \angle DHA = 90^{\circ}$. Let $Y$ be the midpoint of $OX$. Prove that if $MY = OA$, then $OA = 2OH$.
1 reply
Assassino9931
39 minutes ago
hukilau17
4 minutes ago
Inversion exercise
Assassino9931   2
N 10 minutes ago by awesomeming327.
Source: Balkan MO Shortlist 2024 G5
Let $ABC$ be an acute scalene triangle $ABC$, $D$ be the orthogonal projection of $A$ on $BC$, $M$ and $N$ are the midpoints of $AB$ and $AC$ respectively. Let $P$ and $Q$ are points on the minor arcs $\widehat{AB}$ and $\widehat{AC}$ of the circumcircle of triangle $ABC$ respectively such that $PQ \parallel BC$. Show that the circumcircles of triangles $DPQ$ and $DMN$ are tangent if and only if $M$ lies on $PQ$.
2 replies
Assassino9931
36 minutes ago
awesomeming327.
10 minutes ago
all functions satisfying f(x+yf(x))+y = xy + f(x+y)
falantrng   26
N 25 minutes ago by awesomeming327.
Source: Balkan MO 2025 P3
Find all functions $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$,
\[f(x+yf(x))+y = xy + f(x+y).\]
Proposed by Giannis Galamatis, Greece
26 replies
falantrng
Today at 11:52 AM
awesomeming327.
25 minutes ago
One more problem defined only with lines
Assassino9931   0
34 minutes ago
Source: Balkan MO 2024 Shortlist G6
Let $ABC$ be a triangle and the points $K$ and $L$ on $AB$, $M$ and $N$ on $BC$, and $P$ and $Q$ on $AC$ be such that $AK = LB < \frac{1}{2}AB, BM = NC < \frac{1}{2}BC$ and $CP = QA < \frac{1}{2}AC$. The intersections of $KN$ with $MQ$ and $LP$ are $R$ and $T$ respectively, and the intersections of $NP$ with $LM$ and $KQ$ are $D$ and $E$, respectively. Prove that the lines $DR, BE$ and $CT$ are concurrent.
0 replies
1 viewing
Assassino9931
34 minutes ago
0 replies
No more topics!
A simple geometry
Rushil   10
N Oct 31, 2024 by namanrobin08
Source: Indian RMO 1999 Problem 3
Let $ABCD$ be a square and $M,N$ points on sides $AB, BC$ respectively such that $\angle MDN = 45^{\circ}$. If $R$ is the midpoint of $MN$ show that $RP =RQ$ where $P,Q$ are points of intersection of $AC$ with the lines $MD, ND$.
10 replies
Rushil
Oct 26, 2005
namanrobin08
Oct 31, 2024
A simple geometry
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G H BBookmark kLocked kLocked NReply
Source: Indian RMO 1999 Problem 3
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Rushil
1592 posts
#1 • 3 Y
Y by Adventure10, Adventure10, Mango247
Let $ABCD$ be a square and $M,N$ points on sides $AB, BC$ respectively such that $\angle MDN = 45^{\circ}$. If $R$ is the midpoint of $MN$ show that $RP =RQ$ where $P,Q$ are points of intersection of $AC$ with the lines $MD, ND$.
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Andreas
578 posts
#2 • 4 Y
Y by adythedaddy, Adventure10, Adventure10, namanrobin08
Maybe...

$\angle PDN = 45^{o}= \angle PCN$. Then $PDCN$ is cyclic and $\angle PCD = \angle PND = 45^{o}$.
So $PN \bot MD$. Since $MR = RN$ and $\angle MPN = 90^{o}$ we have that $MN$ is the diameter of the circumscribed circle of $MPQN$. Then $RM = RN = RP = RQ = r$.
This post has been edited 1 time. Last edited by Andreas, Jul 14, 2006, 10:03 AM
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srulikbd
400 posts
#3 • 2 Y
Y by Adventure10, Mango247
how does $\angle PDC = 45^{o}$ it's said $\angle PDN =45^{o}$?
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Andreas
578 posts
#4 • 2 Y
Y by Adventure10, Mango247
Yes, there's a typo. :)
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sen
361 posts
#5 • 1 Y
Y by Adventure10
prove $\text{AMQD}$ and $\text{DPNC}$ are cyclic. angle bash some more. and then prove $\text{MNQP}$ is cyclic, upon noticing $\angle \text{MPN}= 90$
$\implies \text{MN}$ is diameter. $\text{R}$ is centre hence $\text{RP}=\text{RQ}$ oops sorry. i thought ur soln had an error. but it was actually a typo. sorry :blush:
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jayme
9782 posts
#6 • 1 Y
Y by Adventure10
Dear Mathlinkers,

also at

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=500164

Sincerely
Jean-Louis
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jayme
9782 posts
#7 • 1 Y
Y by Adventure10
Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20I.pdf p. 16-18.

Sincerely
Jean-Louis
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AlastorMoody
2125 posts
#8 • 1 Y
Y by Adventure10
It's already clear that, $DCNP$ is cyclic, which implies that, $\angle MPN=\angle MCN=90^{\circ} \implies BMPN \text{ is cyclic}$ Also, $MR=RN \implies R \text{ is the center of circle passing through } BMPN$, Now since, $$\angle DPN=2\angle DBN \implies PD=BP \implies \angle PBC =\angle MDB +45^{\circ}=\angle NDC+45^{\circ}=180^{\circ}-\angle AQN$$$ \implies Q \text{ also passes through } BMPN \implies RP=RQ=RN=RB=RM$
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vasudevkrishna
57 posts
#9 • 2 Y
Y by Adventure10, Mango247
jayme wrote:
Dear Mathlinkers,

also at

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=500164

Sincerely
Jean-Louis
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bal09
21 posts
#10
Y by
after using power of point we get that DR lies on radical axis of circumcircle of ADP and DQC so Hence RP=RQ
This post has been edited 1 time. Last edited by bal09, May 29, 2024, 12:20 PM
Reason: Spelling mistakes
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namanrobin08
99 posts
#11
Y by
Could we solve it somehow by a coordinate bash?
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