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a+b+c+d divides abc+bcd+cda+dab
v_Enhance   51
N an hour ago by BossLu99
Source: USA Team Selection Test for IMO 2021, Problem 1
Determine all integers $s \ge 4$ for which there exist positive integers $a$, $b$, $c$, $d$ such that $s = a+b+c+d$ and $s$ divides $abc+abd+acd+bcd$.

Proposed by Ankan Bhattacharya and Michael Ren
51 replies
1 viewing
v_Enhance
Mar 1, 2021
BossLu99
an hour ago
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three discs of radius 1 cannot cover entirely a square surface of side 2
parmenides51   1
N an hour ago by Blast_S1
Source: 2014 Romania NMO VIII p4
Prove that three discs of radius $1$ cannot cover entirely a square surface of side $2$, but they can cover more than $99.75\%$ of it.
1 reply
parmenides51
Aug 15, 2024
Blast_S1
an hour ago
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Floor sequence
va2010   88
N 2 hours ago by heheman
Source: 2015 ISL N1
Determine all positive integers $M$ such that the sequence $a_0, a_1, a_2, \cdots$ defined by \[ a_0 = M + \frac{1}{2} \qquad \textrm{and} \qquad a_{k+1} = a_k\lfloor a_k \rfloor \quad \textrm{for} \, k = 0, 1, 2, \cdots \]contains at least one integer term.
88 replies
va2010
Jul 7, 2016
heheman
2 hours ago
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2025 Caucasus MO Juniors P6
BR1F1SZ   2
N 2 hours ago by IEatProblemsForBreakfast
Source: Caucasus MO
A point $P$ is chosen inside a convex quadrilateral $ABCD$. Could it happen that$$PA = AB, \quad PB = BC, \quad PC = CD \quad \text{and} \quad PD = DA?$$
2 replies
BR1F1SZ
Mar 26, 2025
IEatProblemsForBreakfast
2 hours ago
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Interesting functions with iterations over integers
miiirz30   5
N 2 hours ago by OGMATH
Source: 2025 Euler Olympiad, Round 2
For any subset $S \subseteq \mathbb{Z}^+$, a function $f : S \to S$ is called interesting if the following two conditions hold:

1. There is no element $a \in S$ such that $f(a) = a$.
2. For every $a \in S$, we have $f^{f(a) + 1}(a) = a$ (where $f^{k}$ denotes the $k$-th iteration of $f$).

Prove that:
a) There exist infinitely many interesting functions $f : \mathbb{Z}^+ \to \mathbb{Z}^+$.

b) There exist infinitely many positive integers $n$ for which there is no interesting function
$$ f : \{1, 2, \ldots, n\} \to \{1, 2, \ldots, n\}. $$
Proposed by Giorgi Kekenadze, Georgia
5 replies
miiirz30
Today at 11:07 AM
OGMATH
2 hours ago
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Inequality with one variable rational functions
liliput   14
N 2 hours ago by IEatProblemsForBreakfast
Source: 2022 Junior Macedonian Mathematical Olympiad P2
Let $a$, $b$ and $c$ be positive real numbers such that $a+b+c=3$. Prove the inequality
$$\frac{a^3}{a^2+1}+\frac{b^3}{b^2+1}+\frac{c^3}{c^2+1} \geq \frac{3}{2}.$$
Proposed by Anastasija Trajanova
14 replies
liliput
Jun 7, 2022
IEatProblemsForBreakfast
2 hours ago
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$7^{7^n}+1$ is the product of at least $2n + 3$ primes
N.T.TUAN   46
N 2 hours ago by reni_wee
Source: USAMO 2007
Prove that for every nonnegative integer $n$, the number $7^{7^{n}}+1$ is the product of at least $2n+3$ (not necessarily distinct) primes.
46 replies
N.T.TUAN
Apr 26, 2007
reni_wee
2 hours ago
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n x n square and strawberries
pohoatza   19
N 2 hours ago by atdaotlohbh
Source: IMO Shortlist 2006, Combinatorics 4, AIMO 2007, TST 4, P2
A cake has the form of an $ n$ x $ n$ square composed of $ n^{2}$ unit squares. Strawberries lie on some of the unit squares so that each row or column contains exactly one strawberry; call this arrangement $\mathcal{A}$.

Let $\mathcal{B}$ be another such arrangement. Suppose that every grid rectangle with one vertex at the top left corner of the cake contains no fewer strawberries of arrangement $\mathcal{B}$ than of arrangement $\mathcal{A}$. Prove that arrangement $\mathcal{B}$ can be obtained from $ \mathcal{A}$ by performing a number of switches, defined as follows:

A switch consists in selecting a grid rectangle with only two strawberries, situated at its top right corner and bottom left corner, and moving these two strawberries to the other two corners of that rectangle.
19 replies
pohoatza
Jun 28, 2007
atdaotlohbh
2 hours ago
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Consecutive squares are floors
ICE_CNME_4   7
N 2 hours ago by ICE_CNME_4

Determine how many positive integers \( n \) have the property that both
\[ \left\lfloor \sqrt{2n - 1} \right\rfloor \quad \text{and} \quad \left\lfloor \sqrt{3n + 2} \right\rfloor \]are consecutive perfect squares.
7 replies
ICE_CNME_4
Today at 1:50 PM
ICE_CNME_4
2 hours ago
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Hard geo finale with the cursed line
hakN   12
N 3 hours ago by ihategeo_1969
Source: 2024 Turkey TST P9
In a scalene triangle $ABC,$ $I$ is the incenter and $O$ is the circumcenter. The line $IO$ intersects the lines $BC,CA,AB$ at points $D,E,F$ respectively. Let $A_1$ be the intersection of $BE$ and $CF$. The points $B_1$ and $C_1$ are defined similarly. The incircle of $ABC$ is tangent to sides $BC,CA,AB$ at points $X,Y,Z$ respectively. Let the lines $XA_1, YB_1$ and $ZC_1$ intersect $IO$ at points $A_2,B_2,C_2$ respectively. Prove that the circles with diameters $AA_2,BB_2$ and $CC_2$ have a common point.
12 replies
hakN
Mar 18, 2024
ihategeo_1969
3 hours ago
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Concurrent perpendicular bisectors from three circles in a triangle
darij grinberg   5
N Jul 31, 2022 by darij grinberg
Source: German TST 2022, probably on exam 6-7, again proposed by me
Given a triangle $ABC$ and three circles $x$, $y$ and $z$ such that $A \in y \cap z$, $B \in z \cap x$ and $C \in x \cap y$.

The circle $x$ intersects the line $AC$ at the points $X_b$ and $C$, and intersects the line $AB$ at the points $X_c$ and $B$.
The circle $y$ intersects the line $BA$ at the points $Y_c$ and $A$, and intersects the line $BC$ at the points $Y_a$ and $C$.
The circle $z$ intersects the line $CB$ at the points $Z_a$ and $B$, and intersects the line $CA$ at the points $Z_b$ and $A$.
(Yes, these definitions have the symmetries you would expect.)

Prove that the perpendicular bisectors of the segments $Y_a Z_a$, $Z_b X_b$ and $X_c Y_c$ concur.
5 replies
darij grinberg
Jul 19, 2022
darij grinberg
Jul 31, 2022
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Concurrent perpendicular bisectors from three circles in a triangle
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Source: German TST 2022, probably on exam 6-7, again proposed by me
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darij grinberg
6555 posts
darij grinberg
#1 h  Jul 19, 2022, 4:24 PM • 2 Y
Y by cadaeibf, asdf334
Given a triangle $ABC$ and three circles $x$, $y$ and $z$ such that $A \in y \cap z$, $B \in z \cap x$ and $C \in x \cap y$.

The circle $x$ intersects the line $AC$ at the points $X_b$ and $C$, and intersects the line $AB$ at the points $X_c$ and $B$.
The circle $y$ intersects the line $BA$ at the points $Y_c$ and $A$, and intersects the line $BC$ at the points $Y_a$ and $C$.
The circle $z$ intersects the line $CB$ at the points $Z_a$ and $B$, and intersects the line $CA$ at the points $Z_b$ and $A$.
(Yes, these definitions have the symmetries you would expect.)

Prove that the perpendicular bisectors of the segments $Y_a Z_a$, $Z_b X_b$ and $X_c Y_c$ concur.
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MathSaiyan
77 posts
MathSaiyan
#2 h  Jul 19, 2022, 6:24 PM • 1 Y
Y by Mango247
Nice one!
solution
This post has been edited 2 times. Last edited by MathSaiyan, Jul 19, 2022, 6:26 PM
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cadaeibf
701 posts
cadaeibf
#3 h  Jul 19, 2022, 9:44 PM • 1 Y
Y by Mango247
Let $X=Y_aY_c\cap Z_aZ_b$ and cyclically define $Y$ and $Z$. By the cyclicities, we have $$\measuredangle Y_aZ_aX=\measuredangle BZ_aZ_b=\measuredangle BAZ_b=\measuredangle Y_cAC=\measuredangle Y_cY_aC=\measuredangle XY_aZ_a$$Therefore, $XZ_aY_a$ is isosceles, meaning that the axis of $Z_aY_a$ is also the internal bisector of $\angle Z_aXY_a$, which is also the internal angle bisector of $\angle YXZ$. Therefore, the three axes must concur at the inventer of $XYZ$.
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Tintarn
9045 posts
Tintarn
#4 h  Jul 20, 2022, 8:01 AM
Y by
It was Exam 6, Problem 1, by the way. :)
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darij grinberg
6555 posts
darij grinberg
#5 h  Jul 31, 2022, 12:29 AM
Y by
MathSaiyan wrote:
$BM_A-CM_A = 2(Z_AC-Y_AB)$

How are you getting this? (The first solution I discovered was very close to yours, but the formula for $BM_A - CM_A$ is a bit longer, and the formula for $BM_A^2 - CM_A^2$ does not have the factors of $2$ that yours has.)
cadaeibf wrote:
Let $X=Y_aY_c\cap Z_aZ_b$ and cyclically define $Y$ and $Z$. By the cyclicities, we have $$\measuredangle Y_aZ_aX=\measuredangle BZ_aZ_b=\measuredangle BAZ_b=\measuredangle Y_cAC=\measuredangle Y_cY_aC=\measuredangle XY_aZ_a$$Therefore, $XZ_aY_a$ is isosceles, meaning that the axis of $Z_aY_a$ is also the internal bisector of $\angle Z_aXY_a$, which is also the internal angle bisector of $\angle YXZ$. Therefore, the three axes must concur at the inventer of $XYZ$.

Be careful -- the bisector isn't always internal. This is the shortest solution if one assumes that the bisectors come out "right" (i.e., either all three are internal, or exactly two are external), but it takes a bit of angle chasing to confirm this rigorously (the proper statement to prove is that $\measuredangle\left(p, Z_a Z_b\right) + \measuredangle\left(q, X_b X_c\right) + \measuredangle\left(r, Y_c Y_a\right) = 90^\circ$, where $p, q, r$ are the perpendicular bisectors of $Y_a Z_a, Z_b X_b, X_c Y_c$, respectively).

Note: Your solution is the one Georg Schröter found as well.
Tintarn wrote:
It was Exam 6, Problem 1, by the way. :)

Thanks -- Georg never told me what exam he used this in.
This post has been edited 2 times. Last edited by darij grinberg, Jul 31, 2022, 12:31 AM
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darij grinberg
6555 posts
darij grinberg
#6 h  Jul 31, 2022, 12:48 AM
Y by
And here is my favorite solution (since it shows something extra):

Let $X, Y, Z$ be the centers of the circles $x, y, z$.
Let $X', Y', Z'$ be the reflections of the points $X, Y, Z$ through the midpoints of the segments $YZ, ZX, XY$.
Thus, triangle $X'Y'Z'$ is the antimedial triangle of triangle $XYZ$. Hence, the homothety with center in the centroid of $\triangle XYZ$ and ratio $-2$ sends $X, Y, Z$ to $X', Y', Z'$. Let $h$ denote this homothety. Thus, $X' = h\left(X\right)$.

Let $O$ the circumcenter of $\triangle ABC$. We shall prove that the point $h\left(O\right)$ belongs to the perpendicular bisectors of $Y_a Z_a, Z_b X_b, X_c Y_c$. This will clearly solve the exercise (and yield an Euler-line-like collinearity to boot).

We let $M_{UV}$ denote the midpoint of any given segment $UV$. For any point $P$, we let $\overline{P}$ denote the projection of $P$ onto the line $BC$. It is well-known (and follows from the theorem about the midline in a trapezoid) that $\overline{M_{UV}} = M_{\overline{U}\overline{V}}$ for any segment $UV$. We shall refer to this equality as the midpoint-projection identity.

The quadrilateral $XYX'Z$ is a parallelogram (by the construction of $X'$). Hence, the midpoints of its diagonals $XX'$ and $YZ$ coincide. In other words, $M_{XX'} = M_{YZ}$. Hence, $\overline{M_{XX'}} = \overline{M_{YZ}}$. However, the midpoint-projection identity yields $\overline{M_{XX'}} = M_{\overline{X}\overline{X'}}$ and $\overline{M_{YZ}} = M_{\overline{Y}\overline{Z}}$. Thus, $M_{\overline{X}\overline{X'}} = \overline{M_{XX'}} = \overline{M_{YZ}} = M_{\overline{Y}\overline{Z}}$.

However, $Y$ is the center of the circle $y$, and $\overline{Y}$ is the projection of this center onto the chord $C Y_a$ of this circle. Therefore, $\overline{Y}$ is the midpoint of this chord $C Y_a$. In other words, $\overline{Y} = M_{C Y_a}$. Similarly, $\overline{Z} = M_{B Z_a}$. Thus, $M_{\overline{Y}\overline{Z}} = M_{M_{C Y_a} M_{B Z_a}} = M_{M_{BC} M_{Y_a Z_a}}$. (Here, we have used the fact that $M_{PQ} M_{UV} = M_{UP} M_{QV}$ for any four points $P, Q, U, V$. This fact is clear from vector algebra or from the Varignon parallelogram, although the latter parallelogram is degenerate in the case where we are using this fact.)

So we now know that $M_{\overline{X}\overline{X'}} = M_{M_{BC} M_{Y_a Z_a}}$. On the other hand, $X$ is the center of the circle $x$, and $\overline{X}$ is the projection of this center onto the chord $BC$ of this circle. Therefore, $\overline{X}$ is the midpoint of this chord $BC$. In other words, $\overline{X} = M_{BC}$. In view of this, we can rewrite $M_{\overline{X}\overline{X'}} = M_{M_{BC} M_{Y_a Z_a}}$ as $M_{M_{BC} \overline{X'}} = M_{M_{BC} M_{Y_a Z_a}}$. From this, we obtain $\overline{X'} = M_{Y_a Z_a}$ (because if three points $P, U, V$ satisfy $M_{PU} = M_{PV}$, then $U = V$).

The perpendicular bisector of the segment $Y_a Z_a$ passes through the midpoint $M_{Y_a Z_a}$ (by definition), i.e., through the point $\overline{X'}$ (since $\overline{X'} = M_{Y_a Z_a}$). Furthermore, this bisector is orthogonal to the line $BC$ (since $Y_a$ and $Z_a$ lie on $BC$). Hence, this bisector is the perpendicular to $BC$ through $\overline{X'}$. But this latter perpendicular clearly passes through $X'$ (since the construction of $\overline{X'}$ entails $X'\overline{X'} \perp BC$). As an upshot, we see that the perpendicular bisector of the segment $Y_a Z_a$ passes through $X' = h\left(X\right)$.

Now, let $\mathbf{m}\left(UV\right)$ denote the perpendicular bisector of any segment $UV$. We claim that $\mathbf{m}\left(Y_a Z_a\right)$ is the image of $\mathbf{m}\left(BC\right)$ under the homothety $h$. Indeed, the perpendicular bisectors $\mathbf{m}\left(BC\right)$ and $\mathbf{m}\left(Y_a Z_a\right)$ are both orthogonal to $BC$, and thus are parallel to one another. Moreover, the former bisector passes through $X$ (since $X$ is the center of the circle $x$, while $BC$ is a chord of this circle), while the latter passes through $h\left(X\right)$ (as we saw in the previous paragraph). Hence, the homothety $h$ sends the former bisector to the latter (since a homothety sends any line to a parallel line). Since the former bisector contains $O$ (because $O$ is the circumcenter of $\triangle ABC$), we thus conclude that the latter bisector contains $h\left(O\right)$.

In other words, the point $h\left(O\right)$ belongs to the perpendicular bisector of $Y_a Z_a$. Similarly, it belongs to the perpendicular bisectors of $Z_b X_b$ and $X_c Y_c$ as well. This proves our claim and solves the exercise.
This post has been edited 2 times. Last edited by darij grinberg, Jul 31, 2022, 1:42 AM
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