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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
Integer polynomial commutes with sum of digits
cjquines0   40
N 15 minutes ago by ihategeo_1969
Source: 2016 IMO Shortlist N1
For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geq 2016$, the integer $P(n)$ is positive and $$S(P(n)) = P(S(n)).$$
Proposed by Warut Suksompong, Thailand
40 replies
cjquines0
Jul 19, 2017
ihategeo_1969
15 minutes ago
easy geo
ErTeeEs06   3
N 18 minutes ago by NicoN9
Source: BxMO 2025 P3
Let $ABC$ be a triangle with incentre $I$ and circumcircle $\Omega$. Let $D, E, F$ be the midpoints of the arcs $\stackrel{\frown}{BC}, \stackrel{\frown}{CA}, \stackrel{\frown}{AB}$ of $\Omega$ not containing $A, B, C$ respectively. Let $D'$ be the point of $\Omega$ diametrically opposite to $D$. Show that $I, D'$ and the midpoint $M$ of $EF$ lie on a line.
3 replies
ErTeeEs06
Yesterday at 11:13 AM
NicoN9
18 minutes ago
2024 IMO P1
EthanWYX2009   103
N 26 minutes ago by ashwinmeena
Source: 2024 IMO P1
Determine all real numbers $\alpha$ such that, for every positive integer $n,$ the integer
$$\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor$$is a multiple of $n.$ (Note that $\lfloor z\rfloor$ denotes the greatest integer less than or equal to $z.$ For example, $\lfloor -\pi\rfloor =-4$ and $\lfloor 2\rfloor= \lfloor 2.9\rfloor =2.$)

Proposed by Santiago Rodríguez, Colombia
103 replies
EthanWYX2009
Jul 16, 2024
ashwinmeena
26 minutes ago
Benelux fe
ErTeeEs06   8
N an hour ago by NicoN9
Source: BxMO 2025 P1
Does there exist a function $f:\mathbb{R}\to \mathbb{R}$ such that $$f(x^2+f(y))=f(x)^2-y$$for all $x, y\in \mathbb{R}$?
8 replies
ErTeeEs06
Yesterday at 11:05 AM
NicoN9
an hour ago
Another binomial coefficients sum
aether01   7
N 2 hours ago by fungarwai
Prove that $$\sum_{k=0}^{n}{{\left(-1\right)}^{k}\binom{n}{k} \binom{2n-k}{n-k}} = 1$$
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aether01
Mar 3, 2022
fungarwai
2 hours ago
đề hsg toán
akquysimpgenyabikho   0
2 hours ago
làm ơn giúp tôi giải đề hsg

0 replies
akquysimpgenyabikho
2 hours ago
0 replies
Number theory
MathsII-enjoy   0
4 hours ago
$Find$ $all$ $integers$ $n$ $such$ $that$ $n-1$ $and$ $\frac{n(n+1)}{2}$ $is$ $a$ $perfect$ $number$.
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MathsII-enjoy
4 hours ago
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inquality
Doanh   2
N 4 hours ago by anduran
Given that \( x, y, z \) are positive real numbers satisfying the condition \( xy + yz + zx = 1 \),
find the maximum value of the expression:
\[
P = \frac{1}{1+x^2} + \frac{1}{1+y^2} + \frac{z}{1+z^2}
\]
2 replies
Doanh
Today at 2:15 AM
anduran
4 hours ago
A problem with a rectangle
Raul_S_Baz   5
N 5 hours ago by Raul_S_Baz
On the sides AB and AD of the rectangle ABCD, points M and N are taken such that MB = ND. Let P be the intersection of BN and CD, and Q be the intersection of DM and CB. How can we prove that PQ || MN?
IMAGE
5 replies
Raul_S_Baz
Yesterday at 11:13 AM
Raul_S_Baz
5 hours ago
Combinatoric
spiderman0   3
N 6 hours ago by MathBot101101
Let $ S = \{1, 2, 3, \ldots, 2024\}.$ Find the maximum positive integer $n \geq 2$ such that for every subset $T \subset S$ with n elements, there always exist two elements a, b in T such that:

$|\sqrt{a} - \sqrt{b}| < \frac{1}{2} \sqrt{a - b}$
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spiderman0
Apr 22, 2025
MathBot101101
6 hours ago
set of sum of three or fewer powers of 2, 2024 TMC AIME Mock #13
parmenides51   4
N 6 hours ago by maromex
Let $S$ denote the set of all positive integers that can be expressed as a sum of three or fewer powers of $2$. Let $N$ be the smallest positive integer that cannot be expressed in the form $a-b$, where $a, b \in S$. Find the remainder when $N$ is divided by $1000$.
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Yesterday at 8:16 PM
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6 hours ago
Inequalities
sqing   9
N Today at 2:17 AM by sqing
Let $ a,b \in [0 ,1] . $ Prove that
$$\frac{a}{ 1-ab+b }+\frac{b }{ 1-ab+a } \leq 2$$$$ \frac{a}{ 1+ab+b^2 }+\frac{b }{ 1+ab+a^2 }+\frac{ab }{2+ab }  \leq 1$$$$\frac{a}{ 1-ab+b^2 }+\frac{b }{ 1-ab+a^2 }+\frac{1 }{1+ab  }\leq \frac{5}{2}$$$$\frac{a}{ 1-ab+b^2 }+\frac{b }{ 1-ab+a^2 }+\frac{1 }{1+2ab  }\leq \frac{7}{3}$$$$\frac{a}{ 1+ab+b^2 }+\frac{b }{ 1+ab+a^2 } +\frac{ab }{1+ab }\leq \frac{7}{6 }$$
9 replies
sqing
Apr 25, 2025
sqing
Today at 2:17 AM
20 fair coins are flipped, N of them land heads 2024 TMC AIME Mock #6
parmenides51   2
N Today at 2:14 AM by MelonGirl
$20$ fair coins are flipped. If $N$ of them land heads, find the expected value of $N^2$.
2 replies
parmenides51
Yesterday at 8:05 PM
MelonGirl
Today at 2:14 AM
Polynomial with integer ciefficient
girishpimoli   9
N Today at 12:44 AM by jasperE3
Let $P(x)$ be a polynomial with integer coefficient, It is known that $P(x)$ takes the values $2015$ for $4$ distinct integers , Then the number of integer values of $x$ for which $ P(x)=2022$
9 replies
girishpimoli
Yesterday at 12:57 AM
jasperE3
Today at 12:44 AM
really nice
darij grinberg   1
N Dec 11, 2004 by darij grinberg
Source: VMO 02-03
The author of this posting is : nttu
____________________________________________________________________

Let a triangle ABC . M , N , P are the midpoints of BC, CA, AB .
a) $d_1, d_2, d_3$ are lines throughing M, N, P and dividing the perimeter of triangle ABC into halves . Prove that : $d_1, d_2, d_3$ are concurrent at K .
b) Prove that : among the ratios : $\frac{KA}{BC}, \frac{KB}{AC}, \frac{KC}{AB}$, there exists at least one ratio $\geq \frac{1}{\sqrt3}$.
1 reply
darij grinberg
Dec 11, 2004
darij grinberg
Dec 11, 2004
really nice
G H J
G H BBookmark kLocked kLocked NReply
Source: VMO 02-03
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darij grinberg
6555 posts
#1 • 2 Y
Y by Adventure10, Mango247
The author of this posting is : nttu
____________________________________________________________________

Let a triangle ABC . M , N , P are the midpoints of BC, CA, AB .
a) $d_1, d_2, d_3$ are lines throughing M, N, P and dividing the perimeter of triangle ABC into halves . Prove that : $d_1, d_2, d_3$ are concurrent at K .
b) Prove that : among the ratios : $\frac{KA}{BC}, \frac{KB}{AC}, \frac{KC}{AB}$, there exists at least one ratio $\geq \frac{1}{\sqrt3}$.
Z K Y
The post below has been deleted. Click to close.
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darij grinberg
6555 posts
#2 • 2 Y
Y by Adventure10, Mango247
See Hyacinthos messages #9627, #9628, #9629 and some later messages. Since some people have troubles with displaying these pages, here is a copy of the relevant contents:
Hyacinthos message #9540 wrote:
From: "orl_ml" (Orlando Döhring)
Subject: a ratio

On the sides of triangle ABC take the points M_1, N_1, P_1 such
that each line MM_1, NN_1, PP_1 divides the perimeter of ABC in two
equal parts (M, N, P are respectively the midpoints of the sides BC,
CA, AB).

1/ Prove that the lines MM_1, NN_1, PP_1 are concurrent at a point K.

2/ Prove that among the ratios KA/BC, KB/CA, KC/AB there exist at
least a ratio which is not less than 1/?ã3.
Hyacinthos message #9627 wrote:
From: "ben_goss_ro" (aka Grobber)
Subject: Re: a ratio

--- In "Hyacinthos", "orl_ml" (Orlando Döhring) wrote:
> On the sides of triangle ABC take the points M_1, N_1, P_1 such
> that each line MM_1, NN_1, PP_1 divides the perimeter of ABC in two
> equal parts (M, N, P are respectively the midpoints of the sides BC,
> CA, AB).
>
> 1/ Prove that the lines MM_1, NN_1, PP_1 are concurrent at a point K.
>
> 2/ Prove that among the ratios KA/BC, KB/CA, KC/AB there exist at
> least a ratio which is not less than 1/?ã3.

(1): The intersection point is the centroid of the perimeter of
triangle ABC. It's the well-known Spieker center (X(10)). The fact
that it's the centroid of the perimeter is a good-enough argument to
show the concurrence: the centroid of the perimeter of ABC must lie
on MM_1, on NN_1 and on PP_1, so they must be concurrent.

What's the ratio for (2)? I see gibberish, but maybe it's my
browser's fault.
Hyacinthos message #9628 wrote:
From: Darij Grinberg
Subject: Re: a ratio

Dear "ben_goss_ro" (= grobber at MathLinks?),

In Hyacinthos message #9627, you wrote:

>> (1): The intersection point is the centroid of
>> the perimeter of triangle ABC. It's the
>> well-known Spieker center (X(10)). The fact
>> that it's the centroid of the perimeter is a
>> good-enough argument to show the concurrence:
>> the centroid of the perimeter of ABC must lie
>> on MM_1, on NN_1 and on PP_1, so they must be
>> concurrent.

This puzzles me a bit. In fact, the lines MM_1,
NN_1, PP_1 do bisect the perimeter of triangle
ABC, but the cevians of the Nagel point (X(8))
do it as well; however, the former lines pass
through the centroid of the perimeter of
triangle ABC, whereas the latter ones don't!
I guess this is a consequence of my bad
physics knowledge (in fact, must a centroid of
a perimeter always lie on every perimeter
bisector?), but I am still searching a better
explanation why X(10) is the centroid of the
perimeter, while X(8) is not.

>> What's the ratio for (2)? I see gibberish,

Actually it's because Orlando used a non-ASCII
sign. If I understand him correctly, he means
"[...] a ratio which is not less than
1/sqrt(3)". Actually, the problem is quite
simple; K doesn't need to be the Spieker point
in order to have this satisfied.

We will prove that for every point K in the
plane of triangle ABC, at least one of the
ratios KA / BC, KB / CA, KC / AB is not less
than 1/sqrt(3).

Well, assume the contrary: Let all ratios
KA / BC, KB / CA, KC / AB be less than
1/sqrt(3). If the sidelengths BC, CA, AB are
denoted by a, b, c, then we have

KA^2 < a^2 / 3,
KB^2 < b^2 / 3,
KC^2 < c^2 / 3.

Hence,

KA^2 + KB^2 + KC^2 < (a^2 + b^2 + c^2) / 3.

But the well-known Leibniz formula gives

KG^2 = (KA^2 + KB^2 + KC^2) / 3
- (a^2 + b^2 + c^2) / 9,

where G is the centroid of triangle ABC.

[Actually, this formula is also known in the
form

KG^2 = (KA^2 + KB^2 + KC^2 - GA^2 - GB^2 - GC^2) / 3.

But this is equivalent to the above, as some
formulas for the length of a median show.]

Since KG^2 >= 0, we thus have

KA^2 + KB^2 + KC^2 >= (a^2 + b^2 + c^2) / 3,

contradicting our assumption. Hence, at
least one of the ratios KA / BC, KB / CA,
KC / AB must be >= 1/sqrt(3). Qed..

Sincerely,
Darij Grinberg
Hyacinthos message #9629 wrote:
From: "ben_goss_ro" (aka Grobber)
Subject: Re: a ratio

--- In "Hyacinthos", Darij Grinberg wrote:
> Dear "ben_goss_ro" (= grobber at MathLinks?),

> This puzzles me a bit. In fact, the lines MM_1,
> NN_1, PP_1 do bisect the perimeter of triangle
> ABC, but the cevians of the Nagel point (X(8))
> do it as well; however, the former lines pass
> through the centroid of the perimeter of
> triangle ABC, whereas the latter ones don't!
> I guess this is a consequence of my bad
> physics knowledge (in fact, must a centroid of
> a perimeter always lie on every perimeter
> bisector?), but I am still searching a better
> explanation why X(10) is the centroid of the
> perimeter, while X(8) is not.
>

Hi! Yes, I am grobber (:)).

About the first part of the problem:
I apologize for not completing it. No, not all the lines which bisect
the perimeter pass through Spieker's point, but these ones do because
they also pass through the midpoints of the sides, and here's why:

Let's look at MM_1. Assume AB<AC, and take X on [AC] s.t. AB=CX. It's
obvious that M_1 is the midpoint of AX. Now look at
the "quadrilateral" ABCX and let's try to find the centroid of its
perimeter. First of all, we take the centroid G_1 of the system
formed by the sides AX and BC. To do this we take the midpoints of AX
and BC (which are M_1 and M respectively), and G_1 must lie on MM_1
s.t. MG_1*BC=M_1G_1*AX (basic rod equilibrium physics). Now take G_2
to be the centroid of the system formed by the sides AB and XC. We do
pretty much the same as before, taking G_2 on the line connecting the
midpts of AB and XC (call them S and T respectively) s.t.
G_2X*AB=G_2B*XC, only this time AB=XC, so SG_2=BG_2. This means that
G_2 is also the centroid of the system of points A,B,X,C and that
it's also situatd on the line connecting the midpts of AX and BC,
namely MM_1.

We have 2 things: G_1 is on MM_1 and G_2 is on MM_1, but G_1 and G_2
are the centroids of 2 disjoint parts of our system (AB, BC, CX, XA)
the union of which is the entire system, so the centroid of the whole
system must lie on G_1G_2=MM_1, so the centroid of the perimeter of
ABC is on MM_1. I'd like to stress that G_1 is never equal to G_2,
because that would mean that AX=BC iff AC-AB=BC, which is clearly
false (triangle inequality). X(10) is on MM_1 so it must, of course,
be on NN_1 and PP_1 as well, so we're done.

Darij
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