Sequence and prime factors

by USJL, Mar 26, 2025, 6:29 AM

Let $a_0,a_1,\ldots$ be a sequence of positive integers with $a_0=1$ , $a_1=2$ and
$a_n = a_{n-1}^{a_{n-1}a_{n-2}}-1$ for all $n\geq 2$ . Show that if $p$ is a prime less than $2^k$ for some positive integer $k$ , then there exists $n\leq k+1$ such that $p\mid a_n$ .

number theory

Taiwan

Jury Meeting Lasting for Twenty Years

by USJL, Mar 26, 2025, 6:27 AM

2025 IMO leaders are discussing $100$ problems in a meeting. For each $i = 1, 2,\ldots , 100$ , each leader will talk about the $i$ -th problem for $i$ -th minutes. The chair can assign one leader to talk about a problem of his choice, but he would have to wait for the leader to complete the entire talk of that problem before assigning the next leader and problem. The next leader can be the same leader. The next problem can be a different problem. Each leader’s longest idle time is the longest consecutive time that he is not talking.
Find the minimum positive integer $T$ so that the chair can ensure that the longest idle time for any leader does not exceed $T$ .

Proposed by usjl

combinatorics

Taiwan

Find the value

by sqing, Mar 26, 2025, 5:22 AM

Let $a,b,c$ be real numbers such that $abc\neq 0,2a-b+c= 0$ and $a-2b-c=0.$ Find the value of $\frac{a^2+b^2+c^2}{ab+bc+ca}.$
Let $a,b,c$ be real numbers such that $abc\neq 0,a+2b+3c= 0$ and $2a+3b+4c=0.$ Find the value of $\frac{ab+bc+ca}{a^2+b^2+c^2}.$

algebra

value

integral points

by jhz, Mar 26, 2025, 1:14 AM

Prove: there exist integer $x_1,x_2,\cdots x_{10},y_1,y_2,\cdots y_{10}$ satisfying the following conditions:
$(1)$ $|x_i|,|y_i|\le 10^{10}$ for all $1\le i \le 10$
$(2)$ Define the set $S = \left\{ \left( \sum_{i=1}^{10} a_i x_i, \sum_{i=1}^{10} a_i y_i \right) : a_1, a_2, \cdots, a_{10} \in \{0, 1\} \right\},$ then $|S| = 1024$ ，and any rectangular strip of width 1 covers at most two points of S.

This post has been edited 1 time. Last edited by jhz, 6 hours ago

combinatorics

equal angles

by jhz, Mar 26, 2025, 12:56 AM

In convex quadrilateral $ABCD, AB \perp AD, AD = DC$ . Let $E$ be a point on side $BC$ , and $F$ be a point on the extension of $DE$ such that $\angle ABF = \angle DEC>90^{\circ}$ . Let $O$ be the circumcenter of $\triangle CDE$ , and $P$ be a point on the side extension of $FO$ satisfying $FB =FP$ . Line BP intersects AC at point Q. Prove that $\angle AQB =\angle DPF.$

geometry

Additive Combinatorics!

by EthanWYX2009, Mar 25, 2025, 12:49 AM

Let $X$ be a finite set of real numbers, $d$ be a real number, and $\lambda_1, \lambda_2, \cdots, \lambda_{2025}$ be 2025 non-zero real numbers. Define
$A = \left\{ (x_1, x_2, \cdots, x_{2025}) : x_1, x_2, \cdots, x_{2025} \in X \text{ and } \sum_{i=1}^{2025} \lambda_i x_i = d \right\},$ $B = \left\{ (x_1, x_2, \cdots, x_{2024}) : x_1, x_2, \cdots, x_{2024} \in X \text{ and } \sum_{i=1}^{2024} (-1)^i x_i = 0 \right\},$ $C = \left\{ (x_1, x_2, \cdots, x_{2026}) : x_1, x_2, \cdots, x_{2026} \in X \text{ and } \sum_{i=1}^{2026} (-1)^i x_i = 0 \right\}.$ Show that $|A|^2 \leq |B| \cdot |C|$ .

combinatorics

Additive combinatorics

2 degree polynomial

by PrimeSol, Mar 24, 2025, 6:13 AM

Let $P_{1}(x)= x^2 +b_{1}x +c_{1}, ... , P_{n}(x)=x^2+ b_{n}x+c_{n}$ , $P_{i}(x)\in \mathbb{R}[x], \forall i=\overline{1,n}.$ $\forall i,j ,1 \leq i<j \leq n : P_{i}(x) \ne P_{j}(x)$ .
$\forall i,j, 1\leq i<j \leq n : Q_{i,j}(x)= P_{i}(x) + P_{j}(x)$ polynomial with only one root.
$max(n)=?$

This post has been edited 8 times. Last edited by PrimeSol, Mar 24, 2025, 6:34 AM

algebra

polynomial

D1010 : How it is possible ?

by Dattier, Mar 10, 2025, 10:49 AM

Is it true that $\forall n \in \mathbb N^*, (24^n \times B \mod A) \mod 2 = 0$ ?

A=1728400904217815186787639216753921417860004366580219212750904
024377969478249664644267971025952530803647043121025959018172048
336953969062151534282052863307398281681465366665810775710867856
720572225880311472925624694183944650261079955759251769111321319
421445397848518597584590900951222557860592579005088853698315463
815905425095325508106272375728975

B=2275643401548081847207782760491442295266487354750527085289354
965376765188468052271190172787064418854789322484305145310707614
546573398182642923893780527037224143380886260467760991228567577
953725945090125797351518670892779468968705801340068681556238850
340398780828104506916965606659768601942798676554332768254089685
307970609932846902

This post has been edited 6 times. Last edited by Dattier, Mar 16, 2025, 10:10 AM

number theory

Computer solutions

7 triangles in a square

by gghx, Oct 12, 2024, 11:29 AM

Seven triangles of area $7$ lie in a square of area $27$ . Prove that among the $7$ triangles there are $2$ that intersect in a region of area not less than $1$ .

combinatorics

area

Mesmerising Isogonals

by MathPassionForever, Sep 27, 2019, 2:59 PM

Finally what you expect from me: a GEOMETRY blog post!!
So I was having a conversation with Naruto.D.Luffy and we came up with this as a result of a problem's generalisation.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -21.70393503651915, xmax = 16.96733784512575, ymin = -19.527249887479634, ymax = 8.639647115834748; /* image dimensions */ pen qqffff = rgb(0,1,1); pen ffxfqq = rgb(1,0.4980392156862745,0); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen ffqqff = rgb(1,0,1); draw((-8.204062972546984,5.370109264277287)--(-12.676912150369088,-11.22009495891742)--(11.029188492088062,-10.732147775882282)--cycle, linewidth(1.2) + qqffff); /* draw figures */ draw((-8.204062972546984,5.370109264277287)--(-12.676912150369088,-11.22009495891742), linewidth(1.2) + qqffff); draw((-12.676912150369088,-11.22009495891742)--(11.029188492088062,-10.732147775882282), linewidth(1.2) + qqffff); draw((11.029188492088062,-10.732147775882282)--(-8.204062972546984,5.370109264277287), linewidth(1.2) + qqffff); draw((xmin, -7.573770491803278*xmin-56.76558079009495)--(xmax, -7.573770491803278*xmax-56.76558079009495), linewidth(0.8)); /* line */ draw((xmin, -3.1740105245712344*xmin-20.669672954832006)--(xmax, -3.1740105245712344*xmax-20.669672954832006), linewidth(0.8) + green); /* line */ draw((xmin, -1.9247973507800558*xmin-10.421049410913895)--(xmax, -1.9247973507800558*xmax-10.421049410913895), linewidth(0.8)); /* line */ draw((xmin, -48.58333333333346*xmin-393.21061681863137)--(xmax, -48.58333333333346*xmax-393.21061681863137), linewidth(0.8) + ffxfqq); /* line */ draw((xmin, 0.13203463203463203*xmin-9.546303527808082)--(xmax, 0.13203463203463203*xmax-9.546303527808082), linewidth(0.8) + dotted + wrwrwr); /* line */ draw((xmin, 0.5195352121586896*xmin-4.633992715358346)--(xmax, 0.5195352121586896*xmax-4.633992715358346), linewidth(0.8) + dotted + wrwrwr); /* line */ draw((xmin, 0.5195352121586896*xmin-16.46219955905743)--(xmax, 0.5195352121586896*xmax-16.46219955905743), linewidth(0.8) + dotted + wrwrwr); /* line */ draw((xmin, 0.13203463203463203*xmin-12.188382620075727)--(xmax, 0.13203463203463203*xmax-12.188382620075727), linewidth(0.8) + dotted + wrwrwr); /* line */ draw(circle((-0.8238618291405168,-10.976121367399855), 5.340092852868624), linewidth(0.8) + ffqqff); draw((xmin, 1.194444444444445*xmin + 3.92177233180122)--(xmax, 1.194444444444445*xmax + 3.92177233180122), linewidth(0.8) + linetype("4 4") + wrwrwr); /* line */ draw((xmin, -0.2696078431372549*xmin-7.758592054976186)--(xmax, -0.2696078431372549*xmax-7.758592054976186), linewidth(0.8) + linetype("4 4") + wrwrwr); /* line */ draw(circle((-3.644468734917405,-4.7288267331171525), 3.29918604211756), linewidth(0.8) + blue); draw(circle((-4.669738921552697,-13.377517176814951), 5.680356398464501), linewidth(0.8) + red); draw((-6.9212894030477266,-4.345323344061346)--(-2.3675406461868462,-5.86401344726935), linewidth(0.4) + wrwrwr); draw((-4.598175673547878,-1.5704930560476382)--(-6.709460220477019,-5.949668956318167), linewidth(0.4) + wrwrwr); draw((xmin, 0.5138512998065294*xmin-1.5080719407995526)--(xmax, 0.5138512998065294*xmax-1.5080719407995526), linewidth(0.8) + linetype("2 2") + red); /* line */ draw((xmin, -2.2148760330578505*xmin-12.800873187314375)--(xmax, -2.2148760330578505*xmax-12.800873187314375), linewidth(0.8) + linetype("2 2") + red); /* line */ /* dots and labels */ dot((-8.204062972546984,5.370109264277287),linewidth(4pt) + dotstyle); label("$A$", (-8.04141391153527,5.695407386300712), NE * labelscalefactor); dot((-12.676912150369088,-11.22009495891742),linewidth(4pt) + dotstyle); label("$B$", (-12.514263089357375,-10.894796836893994), NW * labelscalefactor); dot((11.029188492088062,-10.732147775882282),linewidth(4pt) + dotstyle); label("$C$", (11.191837553099775,-10.406849653858856), NE * labelscalefactor); dot((-5.784885220120563,-12.952187811476929),linewidth(3pt) + dotstyle); label("$P$", (-5.642340261612507,-12.724598773275764), NE * labelscalefactor); dot((-2.3675406461868462,-5.86401344726935),linewidth(3pt) + dotstyle); label("$Q$", (-2.186047715113608,-5.608702354013328), NE * labelscalefactor); dot((2.471492725556317,-15.178172061536873),linewidth(3pt) + dotstyle); label("$R$", (2.6527618499848495,-14.920361096933886), E * labelscalefactor); dot((-6.1277538820978,-10.355379256829654),linewidth(3pt) + dotstyle); label("$S$", (-5.967638383635932,-10.122213797088358), NE * labelscalefactor); dot((-0.8238618291405126,-10.976121367399852),linewidth(3pt) + dotstyle); label("$M$", (-0.6408816355023359,-10.732147775882282), SW*2); dot((-7.8646224521743635,-11.121042683825934),linewidth(3pt) + dotstyle); label("$I$", (-7.7161157895118455,-10.894796836893994), NE * labelscalefactor); dot((-7.978106032039922,-5.60763209535758),linewidth(3pt) + dotstyle); label("$H$", (-7.797440320017702,-5.3647287624957585), W*2); dot((-1.6085513735097214,-7.32491398858876),linewidth(3pt) + dotstyle); label("$R'$", (-1.4541269405609003,-7.072543903118744), E * labelscalefactor); dot((-4.598175673547878,-1.5704930560476382),linewidth(3pt) + dotstyle); label("$Q'$", (-4.42247230402466,-1.339164502455867), N * labelscalefactor); dot((-6.9212894030477266,-4.345323344061346),linewidth(3pt) + dotstyle); label("$S'$", (-6.740221423441568,-4.104198539654985), W); dot((-6.709460220477019,-5.949668956318167),linewidth(3pt) + dotstyle); label("$P'$", (-6.536910097176928,-5.690026884519185), NE * labelscalefactor); dot((0.8592501594336901,-12.0749318414492),linewidth(3pt) + dotstyle); label("$W$", (1.026271239867721,-11.830028937711344), NE * labelscalefactor); dot((-4.979841699456322,-19.049402672901163),linewidth(3pt) + dotstyle); label("$Z$", (-4.829094956553942,-18.823938561214995), NE * labelscalefactor); dot((-0.4252879624658302,-9.602456267441015),linewidth(3pt) + dotstyle); label("$Y$", (-0.27492124822598196,-9.349630757282723), NE * labelscalefactor); dot((-6.441321998897964,-7.980486306638234),linewidth(3pt) + dotstyle); label("$X$", (-6.292936505659358,-7.723140147165594), W * labelscalefactor); dot((-6.0724568386395825,-4.628411780353551),linewidth(3pt) + dotstyle); label("$T$", (-5.926976118383004,-4.388834396425482), NE * labelscalefactor); dot((-4.138486506330636,-3.634638611309332),linewidth(3pt) + dotstyle); label("$H_{A}$", (-3.9751873862424496,-3.372277765102277), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
Problem: Perpendiculars $BP,CS$ are dropped on line $l$ through $A$ and perpendiculars $BQ,CR$ are dropped on lime $m$ , which is the isogonal conjugate of $l$ with respect to $\angle A$ such that $P,S$ lie on $l$ , $Q,R$ lie on $m$ . $M$ is the midpoint of $BC$ and $I$ is the foot of $A$ -altitude on $BC$ . Then prove that:
(i) $PQRS$ is cyclic with circumcenter $M$ .
(ii) $MIPR$ and $MIQS$ are cyclic.

Proof:

And this is the question from where we got this:

Elnino2k wrote:

Given the circle $(ABC)$ and $BC$ fixed, $A$ moves on $(ABC)$ . The circles with diameter $AB$ and $AC$ intersect at $K$ . Let $E,F$ be the intersection of $AD$ with $(AB),(AC)$ respectively where $D$ is midpoint of $BC$ . Suppose that $(KDF)$ intersects $(AB)$ at $N$ and $(KDE)$ intersects $(AC)$ at $M$ .
Prove that $MN$ passes through a fixed point as $A$ moves.

Solution:

Edit:
Supercali and BOBTHEGR8 helped me study this configuration even more.

Let $P',Q',R',S'$ be intersections of $BH,CH$ with $l,m$ as shown in the diagram. Then:
1) $P'Q'R'S'$ and $PQRS$ have similicenter $A$ .
2) Let $T=P'Q' \cap R'S'$ . Then $H,T,H_A$ are collinear, where the last point is the $A$ -Humpty point.

This post has been edited 25 times. Last edited by MathPassionForever, Apr 1, 2020, 6:43 PM

n-variable inequality

by ABCDE, Jul 7, 2016, 7:34 PM

Suppose that a sequence $a_1,a_2,\ldots$ of positive real numbers satisfies $a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}$ for every positive integer $k$ . Prove that $a_1+a_2+\ldots+a_n\geq n$ for every $n\geq2$ .

Submit

Cum'on post smth man
by Commander_Anta78, Dec 12, 2021, 8:55 AM
:spam: :spam: :spam: :spam: :spam: :spam: :spam: :spam::spam: :spam: :spam: :spam: :spam: :spam::spam: :spam: :spam: :spam:
by Project_Donkey_into_M4, Nov 8, 2021, 7:58 AM
Owners offline blogs dead,time for :spam: :spam: :spam: :spam: :spam: :spam: :spam: :spam: :spam: :spam: :spam: :spam: :spam::spam: :spam: :spam: :spam: :spam: :spam::spam: :spam: :spam: :spam: :spam: :spam:
by HoRI_DA_GRe8, Nov 8, 2021, 7:54 AM
？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？、？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？
by whatagreatday7, Dec 11, 2020, 2:22 PM
@3 below, that's an april fools prank ig?
by Synthetic_Potato, Apr 20, 2020, 9:03 AM
Kya yaar kuch bhi yaar
by MathPassionForever, Apr 1, 2020, 6:50 PM
Nice blog yaar geo pro
by Wizard_32, Apr 1, 2020, 6:02 PM
Lol okay, guess I will restart soon.
by MathPassionForever, Apr 1, 2020, 12:51 PM
No celebratory posts?
by Hexagrammum16, Mar 4, 2020, 6:02 AM
Nice blog, especially the Artzt Parabola post!
by PhysicsMonster_01, Jan 5, 2020, 1:09 PM
Uhh, looks like I'm a bit too stuck b/w JEE+Schoolwork+Oly prep. Perhaps I'll blog a bit after Mains. Sorry for no posts for now. And The algebraic to geometric ineq, oh man that'll take a LOT of time.
by MathPassionForever, Dec 4, 2019, 8:07 PM
New Post ??
by gamerrk1004, Nov 24, 2019, 6:43 AM
When are you releasing the article on Algebraic to Geometric inequalities?
by Math-wiz, Nov 5, 2019, 8:43 AM
HIGH LAVIL ... Not for me
by gamerrk1004, Oct 15, 2019, 12:40 PM
I found the properties actually!
I'd like to see some more geo posts
by Physicsknight, Oct 9, 2019, 7:57 PM
Chill, couldn't find that huge a compilation of properties anywhere
by MathPassionForever, Sep 28, 2019, 6:09 PM
Yayyy!!! Geo Posts!!!
I earlier thought that we made a groundbreaking discovery of the Dumpty Parabola, but bery sad to know that it was already known, namely Artzt Parabola
by AlastorMoody, Sep 27, 2019, 7:52 PM
Shocked!!!!
by Math-wiz, Sep 10, 2019, 6:18 AM
HURRAAYYYY!! #MPF Hai Lavil!! Next Post geo?!!
by AlastorMoody, Sep 9, 2019, 6:03 PM
yeeee this is nicer
by Hexagrammum16, Sep 5, 2019, 8:30 AM
Nice blog!
by Mathotsav, Sep 4, 2019, 10:08 AM
Okay, maybe after yet another inequality post.
by MathPassionForever, Sep 4, 2019, 8:03 AM
Get back to geo please
by Hexagrammum16, Sep 4, 2019, 6:36 AM
Shout!!!
Expecting some very interesting stuff here
by Naruto.D.Luffy, Sep 3, 2019, 6:25 PM
$\frac{1}{\cos{C}}$ blog..
by Mr.Chagol, Aug 28, 2019, 9:44 AM
Let me try this shout thing too.(Nothing much to say as of now,..well, I hope to see some interesting stuff here).
by Mathotsav, Aug 28, 2019, 7:15 AM
With that kind of support, you can expect nonzero blog posts soon

Also, those who know what I shall be posting, don't spoil for the others
by MathPassionForever, Aug 23, 2019, 5:07 PM
Even before you get to write I'm shouting
by Hexagrammum16, Aug 23, 2019, 4:47 PM
No posts yet I'll give a shout
by Pluto1708, Aug 23, 2019, 3:37 PM