Easy Combinatorics

by JetFire008, Mar 16, 2025, 4:32 PM

How many positive integers not exceeding $2001$ are multiples of $3$ or $4$ but not $5$?
AMC
AMC 12
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2025 ROSS Program

by scls140511, Mar 16, 2025, 2:36 AM

Since the application has ended, are we now free to discuss the problems and stats? How do you think this year's problems are?
Ross Mathematics Program
Ross
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Stanford Math Tournament (SMT) Online 2025

by stanford-math-tournament, Mar 9, 2025, 6:45 PM

Register for Stanford Math Tournament (SMT) Online 2025

:surf: Stanford Math Tournament (SMT) Online is happening on April 13, 2025! :surf:
https://i.ibb.co/qYCw0RxB/Merch-Back-DRAFT.png

Register and learn more here:
https://www.stanfordmathtournament.com/competitions/smt-2025-online

When? The contest will take place April 13, 2025. The pre-contest puzzle hunt will take place on April 12, 2025 (optional, but highly encouraged!).

What? The competition features a Power, Team, Guts, General, and Subject (choose two of Algebra, Calculus, Discrete, Geometry) rounds.

Who? You!!!!! Students in high school or below, from anywhere in the world. Register in a team of 6-8 or as an individual.

Where? Online - compete from anywhere!

Check out our Instagram: https://www.instagram.com/stanfordmathtournament/

Register and learn more here:
https://www.stanfordmathtournament.com/competitions/smt-2025-online

https://i.ibb.co/fY93MfMb/Year-Logo-DRAFT.jpg

:surf: :surf: :surf: :surf: :surf:
Stanford Math Tournament
Stanford
Math Competitions
High school math
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[Registration Open] Mustang Math Tournament 2025

by MustangMathTournament, Mar 8, 2025, 10:26 PM

Mustang Math is excited to announce that registration for our annual tournament, MMT 2025, is open! This year, we are bringing our tournament to 9 in-person locations, as well as online!

Locations include: Colorado, Norcal, Socal, Georgia, Illinois, Massachusetts, New Jersey, Nevada, Washington, and online. For registration and more information, check out https://mustangmath.com/competitions/mmt-2025.

MMT 2025 is a math tournament run by a group of 150+ mathematically experienced high school and college students who are dedicated to providing a high-quality and enjoyable contest for middle school students. Our tournament centers around teamwork and collaboration, incentivizing students to work with their teams not only to navigate the challenging and interesting problems of the tournament but also to develop strategies to master the unique rounds. This includes a logic puzzle round, a strategy-filled hexes round, a race-like gallop round, and our trademark ‘Mystery Mare’ round!

Awards:
  • Medals for the top teams
  • Shirts, pins, stickers and certificates for all participants
  • Additional awards provided by our wonderful sponsors!

We are also holding a free MMT prep seminar from 3/15-3/16 to help students prepare for the upcoming tournament. Join the Google Classroom! https://classroom.google.com/c/NzQ5NDUyNDY2NjM1?cjc=7sogth4
This post has been edited 4 times. Last edited by MustangMathTournament, Mar 14, 2025, 2:01 AM
MMT
Mustang Math
Mustang Math Tournament
Competition
middle school math contests
Math Competitions
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Did this get posted yet

by pog, Oct 11, 2024, 2:09 PM

What is the ones digit of \[222{,}222-22{,}222-2{,}222-222-22-2?\]
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }6\qquad\textbf{(E) }8$
2024 AMC 8
AMC
AMC 8
turning into flamewar
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average FE

by KevinYang2.71, Mar 21, 2024, 4:00 AM

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy
\[ f(x^2-y)+2yf(x)=f(f(x))+f(y) \]for all $x,y\in\mathbb{R}$.

Proposed by Carl Schildkraut
This post has been edited 1 time. Last edited by KevinYang2.71, Mar 21, 2024, 3:13 PM
USA(J)MO
USAJMO
function
functional equation
PointWise Trap
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d_k-eja Vu

by ihatemath123, Mar 20, 2024, 4:02 AM

Find all integers $n \geq 3$ such that the following property holds: if we list the divisors of $n!$ in increasing order as $1 = d_1 < d_2 < \dots < d_k = n!$, then we have
\[ d_2 - d_1 \leq d_3 - d_2 \leq \dots \leq d_k - d_{k-1}. \]
Proposed by Luke Robitaille.
This post has been edited 1 time. Last edited by ihatemath123, Mar 20, 2024, 4:57 AM
USAMO
number theory
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apparently circles have two intersections :'(

by itised, Jun 21, 2020, 11:00 PM

Let $\omega$ be the incircle of a fixed equilateral triangle $ABC$. Let $\ell$ be a variable line that is tangent to $\omega$ and meets the interior of segments $BC$ and $CA$ at points $P$ and $Q$, respectively. A point $R$ is chosen such that $PR = PA$ and $QR = QB$. Find all possible locations of the point $R$, over all choices of $\ell$.

Proposed by Titu Andreescu and Waldemar Pompe
This post has been edited 2 times. Last edited by djmathman, Jun 22, 2020, 5:29 AM
USA(J)MO
geometry
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Too Bad I'm Lactose Intolerant

by hwl0304, Apr 18, 2018, 10:59 PM

Let \(a,b,c\) be positive real numbers such that \(a+b+c=4\sqrt[3]{abc}\). Prove that \[2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.\]
This post has been edited 1 time. Last edited by hwl0304, Apr 18, 2018, 11:15 PM
2018 USAMO Problem 1
2018 USAJMO Problem 2
Inequality
homogenous
Hi
WHAT DOES THE TITLE MEAN
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New Results for Orthocenters

by XmL, Dec 22, 2015, 7:30 PM

Given $ABC$ with isogonal conjugates $P,Q$ and orthocenter $H$.

Property1: Reflect $P$ over $BC$ to get $P_A$, Prove that the lines through $H,Q$ perpendicular to $AP,HP_A$ resp. are concurrent with $BC$.

Proof: Denote the two lines defined in the statement $l,j$ resp. Reflect $H$ over $BC$ to get $H_A$ which lies on the circumcircle of $ABC$, with center $O$. Let $AQ\cap (O)=D\ne A$. It is not hard to show that $l,H_AD$ are symmetric over $BC$. If $j\cap (BQC)=E\ne Q$, then it suffices to show $QE,BC,DH_A$ concur, which is equivalent to $H_A,D,Q,E$ are concyclic by the radical axis theorem

Let $F$ be the isogonal conjugate of $P$ wrt $BH_AC$. By angle relations of conjugates, $F\in (QBC)$. Define $E'=H_AF\cap (QBC)\ne F$, thus $\angle (E'Q,E'H_A)=\angle (CQ,CF)=\angle (CA,CH_A)=\angle (DA,DH_A)$ because $\angle (CH_A,CF)=\angle (CP,CB)=\angle (CA,CQ)$. Hence $H_AEQD$ is cyclic, and it suffices to show $E\equiv E'\iff QE'\perp HP_A$.

Define $k$ the reflection of $\overline {OH_A}$ over $AH_A$. Since $(H_AH,H_AO), (H_AE',H_AP)$ are isogonals wrt $\angle BH_AC$, therefore $\angle (H_AO,H_AE')=\angle (H_AP,H_A)=\angle (H_AH,k)$. Hence $\angle (H_AE,k)=\angle (H_AO,H_AH)$ $\implies \angle (E'Q,E'H_A)+ \angle (E'H_A,k)=90^{\circ}$, i.e $k\perp QE'$. Because $k\parallel HP_A$, it has been shown that $QE'\perp HP_A \Box$.

Property2: If $X=HP_A\cap AP$, then $X$ lies on the radical axis of $(H,HP), (Q,QP_A)$. The second circle is also known as the circumcircle of the reflection triangle of $P$ wrt $ABC$.

Proof: It suffices to show that the intersections of $AP$ with $(H,HP)$ and $HP_A$ with $(Q,QP_A)$ are concyclic $\iff $ the perpendicular bisectors of the two intersection segments is concurrent with the perpendicular bisector of $PP_A$, i.e $BC$. This is equivalent to Property1, thus Property2 is proven. $\Box$

Property3: Define $X=HP_A\cap AP$, $Y=HP_B\cap BP$, $Z=HP_C\cap CP$. Then $X,Y,Z$ are collinear and $HQ\perp \overline {XYZ}$.

This is immediate from property2, since $X,Y,Z$ lie on the radical axis of $(H,HP), (Q,QP_A)$. $\Box$

Theorem1: $R_{PQ}^2-HP^2-HQ^2=~pot~(H,(O))$, where $R_{PQ}$ denote the circumradius of the reflection triangle of $P$ wrt $ABC$.

Proof: Let the radical axis of $(H,HP), (Q,QP_A)$ intersect $QH$ at $K$. Then $KH^2-HP^2=KQ^2-QP_A^2=KQ^2-R_{PQ}^2$ $\implies R_{PQ}^2-HP^2-HQ^2=KQ^2-KH^2-HQ^2=QH[KQ+KH-HQ]=2QH\cdot HK$. If $AH\cap BC=D$, then it suffices to show that $QH\cdot HK=\frac {~pot~(H,(O))}{2}=AH\cdot HD$.

Let the lines through $H,Q$ perpendicular to $AP, HP_A$ resp. meet at $L$; from property1, $L\in BC$. Define $M=AP\cap HL, N=HP_A\cap QL$, $X=AP\cap HP_A$, then $XK\perp QH$ by property3. Hence the multiple cyclic quadrilaterals give $QH\cdot HK=HN\cdot HX=HM\cdot HL=AH\cdot HD$. $\Box$
This post has been edited 1 time. Last edited by XmL, Jul 28, 2017, 7:08 AM
Reason: edited title
geometry

Consequences of Tangent Circles Configurations

by XmL, Dec 5, 2015, 6:30 AM

Proposition1(Sawayama): Circle $\omega$ is internally tangent to circle $\Omega$ at $P$. $AC,BD$ are two chords of $\Omega$ that are tangent to $\omega$ at $Y,X$ respectively. Prove that the incenters of $ABC,BDC$ lie on $XY$.

Lemma1: If $\frac {BX}{BP}$ is constant for $B\in \Omega$, and $BX$ is tangent to $\omega$ at $X$.

Let $PB\cap \omega=B'$. $BX^2=BP'\cdot BP\implies \frac {BX}{BP}^2=\frac {BP'}{BP}=1-\frac {R(\omega)}{R(\Omega)}$ where $R(X)$ is the radius of circle $X$, and this proves our lemma.

Correlary1: $PX$ bisects $\angle BPD$ by the converse of angle bisector theorem: $\frac {BX}{BP}=\frac {DX}{DP}$.

Proof of our proposition: Let $I_1$ denote the incenter of $ABC$, $BI_1\cap \Omega=E\ne B$. By Correlary1 $E,Y,P$ are collinear since $E$ is the midpoint of arc $AC$. By similarity, $\frac {EA}{EP}=\frac {AY}{AP}$, which is the constant described in Lemma1, therefore $EA^2=EY\cdot EP$, the power of $E$ wrt $\omega$. Since $EI_1=EA$, thus$\frac {EI_1}{EP}=\frac {DX}{DP}$ and $\triangle EPI_1\sim\triangle DPX$ $\implies BXP=\angle BI_1P$. Since $\triangle EYI_1\sim \triangle EI_1P$, therefore $\angle I_1YP=\angle BI_1P=\angle BXP=\angle XYP$. The last angle equality implies $Y,I_1,X$ are collinear and we are done because the incenter of $BDC$ lies on $XY$ by symmetry.

Some properties I want to highlight and are trivial from the proof above:
Let $I_2$ denote the incenter of $BCD$.
1. $\odot (E,EA)$ and $\omega$ are orthogonal
2. $XBPI_1, YCPI_2$ are cyclic quadrilaterals.
3. $\triangle PBI_1 \sim \triangle PXY \sim \triangle PI_2C$
4. $PI_1,PD$ are isogonal wrt $\angle XPY$.
5. $B,C,I_2,I_1$ are concyclic.


Property6: $K$ is the midpoint of arc $BC$ that doesn't contain $A$. Prove that $I_1,I_2,K,P$ are concyclic.

Proof: By property3 $PI_1,PI_2$ are isogonal wrt $\angle BPC$. Since $PK$ externally bisects $\angle BPC$, it also externally bisects $\angle I_1PI_2$. As $KI_1=KI_2=KB$ by properties of incenters, therefore $PKI_2I_1$ is cyclic.

Correlary2: $PK,BC,XY$ are concurrent by radical axis theorem.

Property7: $M$ is the midpoint of arc $BC$ that does not contain $A$. Prove that $PM\cap CD,X,Y$ are collinear. This is immediate by pascal's theorem.

Proposition 2: $\omega _1, \omega _2$ are internally tangent to $\Omega$ at $P_1,P_2$. $AC,BD$ are chords of $\Omega$ that are also internal common tangents of $\omega _1, \omega _2$ like so in the diagram. Prove that the external tangent of these two circles closer to $DC$ than $AB$ is parallel to $DC$.

Proof:

Let $M$ be the midpoint of arc $DC$ that doesn't contain $A$. Define $H_1,Z_1=MP_1\cap \omega _1, DC$ and $H_2,Z_2$ similarly. Thus $Z_1Z_2$ is antiparallel to $P_1P_2$ wrt $\angle P_1MP_2$. It suffices to show that $H_1H_2||Z_1Z_2$, which implies $H_1H_2\perp (OM||O_1H_1||O_2H_2)\implies H_1H_2$ is the external tangent. We will show that $\frac {Z_1H_1}{Z_1M}=\frac {Z_2H_2}{Z_2M}$.

$AC$ is tangent to $\omega _1, \omega _2$ at $Y_1,X_2$, $DB$ is tangent to the same circles at $X_1,Y_2$. By property7 we know $X_1,Y_1,Z_1$ are collinear. Thus $Z_1H_1\cdot Z_1P_1=Z_1Y_1\cdot Z_1X_1$. In addition, $Z_1M\cdot Z_1P_1=CZ_1\cdot DZ_1$. Dividing these two gives us $\frac {Z_1H_1}{Z_1M}=\frac {Z_1Y_1}{CZ_1}\cdot \frac {Z_1X_1}{DZ_1}$. Since $X_1Y_1||X_2Y_2$, the analogous expression for $\frac {Z_2H_2}{Z_2M}$ equals $\frac {Z_1H_1}{Z_1M}$ and we are done.
This post has been edited 2 times. Last edited by XmL, Dec 5, 2015, 6:36 AM
Reason: .
geometry

Segment has Length Equal to Circumradius

by djmathman, Apr 30, 2014, 9:30 PM

Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\angle BAC$. Let $X$ be the circumcenter of triangle $APB$ and $Y$ the orthocenter of triangle $APC$. Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$.
This post has been edited 1 time. Last edited by djmathman, Apr 30, 2014, 9:56 PM
geometry
circumcircle
vector
geometric transformation
parallelogram
trigonometry
complex bash
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Still an immature proof writer :)

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XmL
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  • Good blog! I hope to see more posts here!

    by WizardMath, Feb 3, 2018, 6:57 PM

  • Thanks Harry Potter :D

    by XmL, Jul 30, 2014, 10:47 PM

  • Hey dude. Nice blog. I like your work.

    by gamjawon, Jul 30, 2014, 7:43 AM

  • Thanks...although most of the posts here won't be about geometry...rightnow I'm trying to improve my skills in other areas such as number theory and combo.

    by XmL, Jun 5, 2014, 1:11 AM

  • Hi Geometer,,nice blog..:D

    by vanu1996, Jun 4, 2014, 1:27 PM

  • Check your private massage :)

    by zabihpourmahdi, Jan 11, 2013, 5:34 PM

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