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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 11:16 PM
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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. 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)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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
Yesterday at 11:16 PM
0 replies
P(x), integer, integer roots, P(0) =-1,P(3) = 128
parmenides51   3
N 2 minutes ago by Rohit-2006
Source: Nordic Mathematical Contest 1989 #1
Find a polynomial $P$ of lowest possible degree such that
(a) $P$ has integer coefficients,
(b) all roots of $P$ are integers,
(c) $P(0) = -1$,
(d) $P(3) = 128$.
3 replies
parmenides51
Oct 5, 2017
Rohit-2006
2 minutes ago
2017 CGMO P1
smy2012   9
N 6 minutes ago by Bardia7003
Source: 2017 CGMO P1
(1) Find all positive integer $n$ such that for any odd integer $a$, we have $4\mid a^n-1$
(2) Find all positive integer $n$ such that for any odd integer $a$, we have $2^{2017}\mid a^n-1$
9 replies
smy2012
Aug 13, 2017
Bardia7003
6 minutes ago
Euler's function
luutrongphuc   1
N 21 minutes ago by luutrongphuc
Find all real numbers \(\alpha\) such that for every positive real \(c\), there exists an integer \(n>1\) satisfying
\[
\frac{\varphi(n!)}{n^\alpha\,(n-1)!} \;>\; c.
\]
1 reply
luutrongphuc
an hour ago
luutrongphuc
21 minutes ago
Square problem
Jackson0423   1
N 36 minutes ago by maromex
Construct a square such that the distances from an interior point to the vertices (in clockwise order) are
1,2,3,4, respectively.
1 reply
Jackson0423
an hour ago
maromex
36 minutes ago
No more topics!
Question 5
Nima Ahmadi Pour   8
N Jun 5, 2022 by guptaamitu1
Source: Iran TST 2006
Let $ABC$ be a triangle such that it's circumcircle radius is equal to the radius of outer inscribed circle with respect to $A$.
Suppose that the outer inscribed circle with respect to $A$ touches $BC,AC,AB$ at $M,N,L$.
Prove that $O$ (Center of circumcircle) is the orthocenter of $MNL$.
8 replies
Nima Ahmadi Pour
Apr 18, 2006
guptaamitu1
Jun 5, 2022
Source: Iran TST 2006
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Nima Ahmadi Pour
160 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ABC$ be a triangle such that it's circumcircle radius is equal to the radius of outer inscribed circle with respect to $A$.
Suppose that the outer inscribed circle with respect to $A$ touches $BC,AC,AB$ at $M,N,L$.
Prove that $O$ (Center of circumcircle) is the orthocenter of $MNL$.
This post has been edited 1 time. Last edited by Nima Ahmadi Pour, Apr 19, 2006, 9:57 AM
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Sailor
256 posts
#2 • 1 Y
Y by Adventure10
Nima Ahmadi Pour wrote:
Let $ABC$ be a triangle such that it's circumcircle radius is equal to the radius of outer inscribed circle with respect to $A$.
Suppose that the outer inscribed circle with respect to $A$ touches $BC,AC,BC$ at $M,N,L$.
Prove that $O$ (Center of circumcircle) is the orthocenter of $MNL$.

I think you meant that it touches $AB$ at $L$...

Anyway, let $BI_a$ meet the circumcircle of $\triangle{ABC}$ at $K$. Then $K$ is the midpoint of the arc $\widehat{ABC}$ , thus $OK\perp{AC}$. Since $I_aN\perp{AC}$ and $OK=I_aN$ we conclude that $KI_a||ON$, but $LM\perp{KI_a}$ $\Longleftrightarrow$ $LM\perp{ON}$.

Using similar arguments we obtain the conclusion.
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Nima Ahmadi Pour
160 posts
#3 • 2 Y
Y by Adventure10, Mango247
Sorry, I just made it correct.
Anyway, thanks for the notification.
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Albanian Eagle
1693 posts
#4 • 2 Y
Y by Adventure10, Mango247
I thought of
$R=r_a \rightarrow R=p tg \frac{A}{2} \rightarrow |OM|^2-|ON|^2=|LM|^2-|LN|^2$ ...
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Virgil Nicula
7054 posts
#5 • 2 Y
Y by Adventure10, Mango247
An interesting remark. Denote the projection $D$ of the vertex $A$ on the opposite side $[BC]$. Then $\boxed {R=r_a\Longleftrightarrow D\in OI}\ .$
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epitomy01
240 posts
#6 • 2 Y
Y by Adventure10, Mango247
this problem is way too easy for IRAN tst.
a solution very similar to sailor's but even simpler:
M on BC, N on AC, L on AB. Suffices to prove OL perpendicular to MN (because then the similar argument that OM perpendicular to LN finishes off the problem neatly). MN is perpendicular to IC with I = excentre of A-excircle. So it suffices to prove a stronger result, i.e. ICOL is an iscoceles trapezium. OC = IL (since R=r_a), so it suffices to prove <OCI=<LIC. Simple angle chasing yields if <A = 2a,etc; then <OCI=<OCB+<ICB=90-2a+90-c=2b+c; <LIC=<LIB+<BIC = b + (b+c) = 2b+c, so we are done.
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epitomy01
240 posts
#7 • 1 Y
Y by Adventure10
hey virgil nicula, do you have a good proof of the 'interesting remark' about D,I,O collinear when R= r_a? i cant do it without side-bashing or trig-bashing.
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thecong33
6 posts
#8 • 2 Y
Y by Adventure10, Mango247
I have a solution:
Let X be the excentre opposite A and D be the point (#A) where AX intersect (ABC). Because OD // MX ( perpendicular BC),
OD=MX so ODXM is a parallelogram,AX perpendicular LN so OM perpen dicular LN.
Let the internal angle bisector of BAC intersect (ABC) at E, EBX = 90, EA = EC. Let T the point of intersection of BX and(ABC) so TA = TC and TE passes O.Let B' the foot of the perpendicular on AC from T. TB' passes O
Because OT // NX and OT= NX so TONX is a parallelogram so TX // ON hence ON perpendicular NM
Hence O is the ortheocentre of MNL
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guptaamitu1
656 posts
#9 • 1 Y
Y by Mango247
Let $I_A$ be the $A$-excenter ; lines $I_AA,I_AB,I_AC$ meet $\odot(ABC)$ again at points $A',B',C'$, respectively. It isn't hard to note that:
  • $I_A$ is the orthocenter of $\triangle A'B'C'$.
  • corresponding sides of $\triangle A'B'C'$ and $\triangle MNL$ are parallel to each other and the two triangles have opposite orientations.
Using the fact that $\triangle A'B'C'$ and $\triangle MNL$ have equal circumradius, we obtain there is a point $S$ such that reflection in $S$ (which we denote by $\mathcal R$) swaps $\{A',M\},\{B',N\},\{C',L\}$. Then as $O,I_A$ are corresponding cirucmenters of $\triangle A'B'C,\triangle MNL$, so $\mathcal R$ swaps $\{O,I_A\}$. Since $I_A$ was the orthocenter of $\triangle A'B'C'$, hence $O$ is the orthocenter of $\triangle MNL$. $\blacksquare$
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