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forced vertices in graphs
Davdav1232   0
an hour ago
Source: Israel TST 7 2025 p2
Let \( G \) be a graph colored using \( k \) colors. We say that a vertex is forced if it has neighbors in all the other \( k - 1 \) colors.

Prove that for any \( 2024 \)-regular graph \( G \), there exists a coloring using \( 2025 \) colors such that at least \( 1013 \) of the colors have a forced vertex of that color.

Note: The graph coloring must be valid, this means no \( 2 \) vertices of the same color may be adjacent.
0 replies
Davdav1232
an hour ago
0 replies
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Can this sequence be bounded?
darij grinberg   70
N an hour ago by ezpotd
Source: German pre-TST 2005, problem 4, ISL 2004, algebra problem 2
Let $a_0$, $a_1$, $a_2$, ... be an infinite sequence of real numbers satisfying the equation $a_n=\left|a_{n+1}-a_{n+2}\right|$ for all $n\geq 0$, where $a_0$ and $a_1$ are two different positive reals.

Can this sequence $a_0$, $a_1$, $a_2$, ... be bounded?

Proposed by Mihai Bălună, Romania
70 replies
1 viewing
darij grinberg
Jan 19, 2005
ezpotd
an hour ago
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weird conditions in geo
Davdav1232   0
an hour ago
Source: Israel TST 7 2025 p1
Let \( \triangle ABC \) be an isosceles triangle with \( AB = AC \). Let \( D \) be a point on \( AC \). Let \( L \) be a point inside the triangle such that \( \angle CLD = 90^\circ \) and
\[ CL \cdot BD = BL \cdot CD. \]Prove that the circumcenter of triangle \( \triangle BDL \) lies on line \( AB \).
0 replies
Davdav1232
an hour ago
0 replies
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find angle
TBazar   4
N an hour ago by vanstraelen
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
4 replies
TBazar
Today at 6:57 AM
vanstraelen
an hour ago
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Polys with int coefficients
adihaya   4
N an hour ago by sangsidhya
Source: 2012 INMO (India National Olympiad), Problem #3
Define a sequence $<f_0 (x), f_1 (x), f_2 (x), \dots>$ of functions by $$f_0 (x) = 1$$$$f_1(x)=x$$$$(f_n(x))^2 - 1 = f_{n+1}(x) f_{n-1}(x)$$for $n \ge 1$. Prove that each $f_n (x)$ is a polynomial with integer coefficients.
4 replies
adihaya
Mar 30, 2016
sangsidhya
an hour ago
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Italian WinterCamps test07 Problem4
mattilgale   89
N 2 hours ago by cj13609517288
Source: ISL 2006, G3, VAIMO 2007/5
Let $ ABCDE$ be a convex pentagon such that
\[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE. \]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$.

Proposed by Zuming Feng, USA
89 replies
mattilgale
Jan 29, 2007
cj13609517288
2 hours ago
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Simple triangle geometry [a fixed point]
darij grinberg   49
N 2 hours ago by cj13609517288
Source: German TST 2004, IMO ShortList 2003, geometry problem 2
Three distinct points $A$, $B$, and $C$ are fixed on a line in this order. Let $\Gamma$ be a circle passing through $A$ and $C$ whose center does not lie on the line $AC$. Denote by $P$ the intersection of the tangents to $\Gamma$ at $A$ and $C$. Suppose $\Gamma$ meets the segment $PB$ at $Q$. Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ does not depend on the choice of $\Gamma$.
49 replies
darij grinberg
May 18, 2004
cj13609517288
2 hours ago
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Kosovo MO 2010 Problem 5
Com10atorics   19
N 2 hours ago by CM1910
Source: Kosovo MO 2010 Problem 5
Let $x,y$ be positive real numbers such that $x+y=1$. Prove that
$\left(1+\frac {1}{x}\right)\left(1+\frac {1}{y}\right)\geq 9$.
19 replies
Com10atorics
Jun 7, 2021
CM1910
2 hours ago
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Hard combi
EeEApO   1
N 2 hours ago by EeEApO
In a quiz competition, there are a total of $100 $questions, each with $4$ answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?

Now, suppose my spouse and I move into a new home. Every year, we have twins. Starting at the age of $16$, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
1 reply
EeEApO
3 hours ago
EeEApO
2 hours ago
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Problem on symmetric polynomial
ayan_mathematics_king   5
N 2 hours ago by bjump
If $a^3+b^3+c^3=(a+b+c)^3$, prove that $a^5+b^5+c^5=(a+b+c)^5$ where $a,b,c \in \mathbb{R}$
5 replies
ayan_mathematics_king
Jul 28, 2019
bjump
2 hours ago
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Weird ninja points collinearity
americancheeseburger4281   1
N Apr 27, 2025 by MathLuis
Source: Someone I know
For some triangle, define its Ninja Point as the point on its circumcircle such that its Steiner line coincides with the Euler line of the triangle. For an triangle $ABC$, define:
[list]
[*]$O$ as its circumcentre, $H$ as its orthocentre and $N_9$ as its nine-point centre.
[*]$M_a$, $M_b$ and $M_c$ to be the midpoint of the smaller arcs.
[*]$G$ as the isogonal conjugate of the Nagel point (i.e. the exsimillicenter of the incircle and circumcircle)
[*]$S$ as the ninja point of $\Delta M_aM_bM_c$
[*]$K$ as the ninja point of the contact triangle
[/list]
Prove that:
$(a)$ Points $K$, $N_9$ and $I$ are collinear, that is $K$ is the Feuerbach point.
$(b)$ Points $H$, $G$ and $S$ are collinear
1 reply
americancheeseburger4281
Apr 26, 2025
MathLuis
Apr 27, 2025
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Weird ninja points collinearity
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americancheeseburger4281
31 posts
americancheeseburger4281
#1 h  Apr 26, 2025, 10:12 PM
Y by
For some triangle, define its Ninja Point as the point on its circumcircle such that its Steiner line coincides with the Euler line of the triangle. For an triangle $ABC$, define:
  • $O$ as its circumcentre, $H$ as its orthocentre and $N_9$ as its nine-point centre.
  • $M_a$, $M_b$ and $M_c$ to be the midpoint of the smaller arcs.
  • $G$ as the isogonal conjugate of the Nagel point (i.e. the exsimillicenter of the incircle and circumcircle)
  • $S$ as the ninja point of $\Delta M_aM_bM_c$
  • $K$ as the ninja point of the contact triangle
Prove that:
$(a)$ Points $K$, $N_9$ and $I$ are collinear, that is $K$ is the Feuerbach point.
$(b)$ Points $H$, $G$ and $S$ are collinear
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MathLuis
1524 posts
MathLuis
#2 h  Apr 27, 2025, 3:00 AM • 1 Y
Y by Yiyj1
Denote $\mathcal H$ as the feuerbach hyperbola, then redefine $S$ as the antigonal conjugate of $H$ in $\triangle ABC$, clearly its isogonal conjugate in $\triangle ABC$ is $\infty_{OI}$ which therefore shows that (notice $OI$ is by homothety euler line of $\triangle M_AM_BM_C, \triangle DEF$ where incircle of $\triangle ABC$ touches $BC,CA,AB$ at $D,E,F$ respectively) in fact $S$ is our desired ninja point.
Now redefine $K$ as the center of $\mathcal H$ then let $L,J$ orthocenters of $\triangle AIB, \triangle AIC$ respectively, then they also lie on $\mathcal H$ but also let $ID \cap EF=M$ and $N$ midpoint of $BC$, either from ratio lemma or homothety we can see that $A,M,N$ are colinear but also recall that the poles of $L,J$ w.r.t. $(DEF)$ are the $B,C$-midbases of $\triangle ABC$ respectively therefore $L,D,J$ are colinear and lie on the polar of $N$ w.r.t. $(DEF)$ so by homothety and letting $LJ \cap EF=V$ we have to notice from La'Hire twice that $M$ lies on the polar of $V$ w.r.t. $(DEF)$ and therefore $-1=(E, F; M, V)$, we will use this later.
Now recall from Iran Lemma twice that $AL, DE, BI$ meet at a point $P$ and $AJ, DF, CI$ meet at a point $Q$ and however $P,Q$ lie on $(AI)$ and the $A$-midbase on $\triangle ABC$ which is sufficient to tell that $I$ is the orthocenter of $\triangle PDQ$, now let $IN \cap LJ=R$ then $\angle IRD=90$ but we will prove $R$ is the $D$-queue point of $\triangle PDQ$ because $-1=(E, F; M, V) \overset{D}{=} (DP, DQ; DI, DR)$ which is enough by the radax that gives an harmonic showing it lies on line $DR$ however by the angle condition mentioned we force $R$ to be our desired point.
And this is so useful because now notice on $\triangle ALJ$ that since $A,L,J \in \mathcal H$ we have that by invariant of pedal circles through the center of $\mathcal H$ that $QKPRD$ is cyclic but also reflecting $K$ over $DE,DF$ let them be $K_1,K_2$ then you also reflect this recently found circle to see that $K_2QID, K_1PID$ are cyclic and thus note $\measuredangle K_2ID=\measuredangle K_2QD=\measuredangle DQK=\measuredangle DPK=\measuredangle K_1PD=\measuredangle K_1ID$ which shows that $K_1,I,K_2$ are colinear so we have that $K$ is indeed anti-steiner of $IO$ as desired.
And now we are basically done because $S$ is just $H$ reflected over $K$ so the colinearity $H,G,K,S$ is trivial by considering homothety at $G$ sending $\triangle DEF$ to $\triangle M_AM_BM_C$ thus we are done :cool:.
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