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The locus of P with supplementary angles condition
WakeUp   3
N 41 minutes ago by Nari_Tom
Source: Baltic Way 2001
Given a rhombus $ABCD$, find the locus of the points $P$ lying inside the rhombus and satisfying $\angle APD+\angle BPC=180^{\circ}$.
3 replies
WakeUp
Nov 17, 2010
Nari_Tom
41 minutes ago
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Inspired by JK1603JK
sqing   6
N an hour ago by sqing
Source: Own
Let $ a,b,c\geq 0 $ and $ab+bc+ca=1.$ Prove that$$\frac{abc-2}{abc-1}\ge \frac{4(a^2b+b^2c+c^2a)}{a^3b+b^3c+c^3a+1} $$
6 replies
sqing
Today at 3:31 AM
sqing
an hour ago
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Geometry problem
kjhgyuio   1
N 2 hours ago by Mathzeus1024
Source: smo
In trapezium ABCD,AD is parallel to BC and points E and F are midpoints of AB and DC respectively. If
Area of AEFD/Area of EBCF =√3 + 1/3-√3 and the area of triangle ABD is √3 .find the area of trapezium ABCD
1 reply
kjhgyuio
Apr 1, 2025
Mathzeus1024
2 hours ago
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D1018 : Can you do that ?
Dattier   1
N 2 hours ago by Dattier
Source: les dattes à Dattier
We can find $A,B,C$, such that $\gcd(A,B)=\gcd(C,A)=\gcd(A,2)=1$ and $$\forall n \in \mathbb N^*, (C^n \times B \mod A) \mod 2=0 $$.

For example :

$C=20$
$A=47650065401584409637777147310342834508082136874940478469495402430677786194142956609253842997905945723173497630499054266092849839$

$B=238877301561986449355077953728734922992395532218802882582141073061059783672634737309722816649187007910722185635031285098751698$

Can you find $A,B,C$ such that $A>3$ is prime, $C,B \in (\mathbb Z/A\mathbb Z)^*$ with $o(C)=(A-1)/2$ and $$\forall n \in \mathbb N^*, (C^n \times B \mod A) \mod 2=0 $$?
1 reply
Dattier
Mar 24, 2025
Dattier
2 hours ago
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2025 Caucasus MO Juniors P3
BR1F1SZ   1
N 2 hours ago by FarrukhKhayitboyev
Source: Caucasus MO
Let $K$ be a positive integer. Egor has $100$ cards with the number “$2$” written on them, and $100$ cards with the number “$3$” written on them. Egor wants to paint each card red or blue so that no subset of cards of the same color has the sum of the numbers equal to $K$. Find the greatest $K$ such that Egor will not be able to paint the cards in such a way.
1 reply
BR1F1SZ
Mar 26, 2025
FarrukhKhayitboyev
2 hours ago
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1 area = 2025 points
giangtruong13   0
2 hours ago
In a plane give a set $H$ that has 8097 distinct points with area of a triangle that has 3 points belong to $H$ all $ \leq 1$. Prove that there exists a triangle $G$ that has the area $\leq 1 $ contains at least 2025 points that belong to $H$( each of that 2025 points can be inside the triangle or lie on the edge of triangle $G$)X
0 replies
giangtruong13
2 hours ago
0 replies
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Burak0609
Burak0609   0
2 hours ago
So if $2 \nmid n\implies$ $2d_2+d_4+d_5=d_7$ is even it's contradiction. I mean $2 \mid n and d_2=2$.
If $3\mid n \implies d_3=3$ and $(d_6+d_7)^2=n+1,3d_6d_7=n \implies d_6^2-d_6d_7+d_7^2=1$,we can see the only solution is$d_6=d_7=1$ and it is contradiction.
If $4 \mid n d_3=4$ and $(d_6+d_7)^2-4d_6d_7=1 \implies d_7=d_6+1$. So $n=4d_6(d_6+1)$.İt means $8 \mid n$.
If $d_6=8 n=4.8.9=288$ but $3 \nmid n$.İt is contradiction.
If $d_5=8$ we have 2 option. Firstly $d_4=5 \implies 2d_2+d_4+d_5=17=d_7 d_6=16$ but $10 \mid n$ is contradiction. Secondly $d_4=7 \implies d_7=2.2+7+8=19 and d_6=18$ but $3 \nmid n$ is contradiction. I mean $d_4=8 \implies d_7=d_5+12, n=4(d_5+11)(d_5+12) and d_5 \mid n=4(d_5+11)(d_5+12)$. So $d_5 \mid 4.11.12 \implies d_5 \mid 16.11$. If $d_5=16 d_6=27$ but $3 \nmid n$ is contradiction. I mean $d_5=11,d_6=22,d_7=23$. The only solution is $n=2024$.
0 replies
Burak0609
2 hours ago
0 replies
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Good Partitions
va2010   25
N 3 hours ago by lelouchvigeo
Source: 2015 ISL C3
For a finite set $A$ of positive integers, a partition of $A$ into two disjoint nonempty subsets $A_1$ and $A_2$ is $\textit{good}$ if the least common multiple of the elements in $A_1$ is equal to the greatest common divisor of the elements in $A_2$. Determine the minimum value of $n$ such that there exists a set of $n$ positive integers with exactly $2015$ good partitions.
25 replies
va2010
Jul 7, 2016
lelouchvigeo
3 hours ago
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An inequality on triangles sides
nAalniaOMliO   7
N 3 hours ago by navier3072
Source: Belarusian National Olympiad 2025
Numbers $a,b,c$ are lengths of sides of some triangle. Prove the inequality$$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \geq \frac{a+b}{2c}+\frac{b+c}{2a}+\frac{c+a}{2b}$$
7 replies
nAalniaOMliO
Mar 28, 2025
navier3072
3 hours ago
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D is incenter
Layaliya   3
N 3 hours ago by rong2020
Source: From my friend in Indonesia
Given an acute triangle \( ABC \) where \( AB > AC \). Point \( O \) is the circumcenter of triangle \( ABC \), and \( P \) is the projection of point \( A \) onto line \( BC \). The midpoints of \( BC \), \( CA \), and \( AB \) are \( D \), \( E \), and \( F \), respectively. The line \( AO \) intersects \( DE \) and \( DF \) at points \( Q \) and \( R \), respectively. Prove that \( D \) is the incenter of triangle \( PQR \).
3 replies
Layaliya
Yesterday at 11:03 AM
rong2020
3 hours ago
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Constructing sequences
SMOJ   6
N 4 hours ago by lightsynth123
Source: 2018 Singapore Mathematical Olympiad Senior Q5
Starting with any $n$-tuple $R_0$, $n\ge 1$, of symbols from $A,B,C$, we define a sequence $R_0, R_1, R_2,\ldots,$ according to the following rule: If $R_j= (x_1,x_2,\ldots,x_n)$, then $R_{j+1}= (y_1,y_2,\ldots,y_n)$, where $y_i=x_i$ if $x_i=x_{i+1}$ (taking $x_{n+1}=x_1$) and $y_i$ is the symbol other than $x_i, x_{i+1}$ if $x_i\neq x_{i+1}$. Find all positive integers $n>1$ for which there exists some integer $m>0$ such that $R_m=R_0$.
6 replies
SMOJ
Mar 31, 2020
lightsynth123
4 hours ago
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The incircle problem
danil_e   1
N Dec 1, 2023 by ancamagelqueme
Given triangle \(ABC\) inscribed in circle \((O)\) and circumscribed about circle \((I)\). A circle passing through \(B\) and \(C\) is tangent to \((I)\) at \(N_a\) and intersects \(AB\) and \(AC\) at \(A_c\) and \(A_b\) respectively. Similarly, define \(B_c, B_a, N_b\) and \(C_b, C_a, N_c\) correspondingly. Let \(XYZ\) be the triangle formed by the radical axis of circles \((N_aBC)\), \((N_bCA)\), \((N_cBA)\) (as shown in the figure); \(MNP\) is the triangle formed by the intersection of lines \(A_cC_a\), \(B_cC_b\), \(A_bB_a\) (as shown in the figure).

a) Prove that: Triangle \(XYZ\) and triangle \(MNP\) are in perspective axially.
b) Let \(S\) be the point of concurrence of \(XM\), \(YN\), \(ZP\). Prove that: \(S\), \(I\), \(O\) are collinear.
1 reply
danil_e
Dec 1, 2023
ancamagelqueme
Dec 1, 2023
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The incircle problem
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danil_e
25 posts
danil_e
#1 h  Dec 1, 2023, 4:23 PM
Y by
Given triangle \(ABC\) inscribed in circle \((O)\) and circumscribed about circle \((I)\). A circle passing through \(B\) and \(C\) is tangent to \((I)\) at \(N_a\) and intersects \(AB\) and \(AC\) at \(A_c\) and \(A_b\) respectively. Similarly, define \(B_c, B_a, N_b\) and \(C_b, C_a, N_c\) correspondingly. Let \(XYZ\) be the triangle formed by the radical axis of circles \((N_aBC)\), \((N_bCA)\), \((N_cBA)\) (as shown in the figure); \(MNP\) is the triangle formed by the intersection of lines \(A_cC_a\), \(B_cC_b\), \(A_bB_a\) (as shown in the figure).

a) Prove that: Triangle \(XYZ\) and triangle \(MNP\) are in perspective axially.
b) Let \(S\) be the point of concurrence of \(XM\), \(YN\), \(ZP\). Prove that: \(S\), \(I\), \(O\) are collinear.
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ancamagelqueme
104 posts
ancamagelqueme
#2 h  Dec 1, 2023, 11:44 PM
Y by
The point $S$ is NOT aligned with $I=X_1$ and $O=X_3$. If it is on the line $X_{57}X_{2346}$.

The barycentric coordinates of $S$, with respect to $ABC$, are given by the triangle center function

f(a,b,c)= a (a+b-c) (a-b+c)(a(b+c)-(b-c)^2)((b-c)^6 (b+c)^2 (b^2-3 b c+c^2)-2 (b-c)^4 (4 b^5-b^4 c-11 b^3 c^2-11 b^2 c^3-b c^4+4 c^5) a+(b-c)^2 (27 b^6+4 b^5 c-39 b^4 c^2-112 b^3 c^3-39 b^2 c^4+4 b c^5+27 c^6) a^2-2 (b-c)^2 (24 b^5+55 b^4 c+97 b^3 c^2+97 b^2 c^3+55 b c^4+24 c^5) a^3+2 (21 b^6+56 b^5 c+141 b^4 c^2+140 b^3 c^3+141 b^2 c^4+56 b c^5+21 c^6) a^4-2 b c (49 b^3+167 b^2 c+167 b c^2+49 c^3) a^5-2 (21 b^4+7 b^3 c-34 b^2 c^2+7 b c^3+21 c^4) a^6+2 (24 b^3+35 b^2 c+35 b c^2+24 c^3) a^7+(-27 b^2-41 b c-27 c^2) a^8+8 (b+c) a^9-a^10)
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