Summer is a great time to explore cool problems to keep your skills sharp!  Schedule a class today!

G
Topic
First Poster
Last Poster
k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Euler Line Madness
raxu   75
N 41 minutes ago by lakshya2009
Source: TSTST 2015 Problem 2
Let ABC be a scalene triangle. Let $K_a$, $L_a$ and $M_a$ be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of $AK_aL_a$ intersects $AM_a$ a second time at point $X_a$ different from A. Define $X_b$ and $X_c$ analogously. Prove that the circumcenter of $X_aX_bX_c$ lies on the Euler line of ABC.
(The Euler line of ABC is the line passing through the circumcenter, centroid, and orthocenter of ABC.)

Proposed by Ivan Borsenco
75 replies
raxu
Jun 26, 2015
lakshya2009
41 minutes ago
Own made functional equation
Primeniyazidayi   8
N an hour ago by MathsII-enjoy
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
8 replies
Primeniyazidayi
May 26, 2025
MathsII-enjoy
an hour ago
IMO ShortList 2002, geometry problem 7
orl   110
N an hour ago by SimplisticFormulas
Source: IMO ShortList 2002, geometry problem 7
The incircle $ \Omega$ of the acute-angled triangle $ ABC$ is tangent to its side $ BC$ at a point $ K$. Let $ AD$ be an altitude of triangle $ ABC$, and let $ M$ be the midpoint of the segment $ AD$. If $ N$ is the common point of the circle $ \Omega$ and the line $ KM$ (distinct from $ K$), then prove that the incircle $ \Omega$ and the circumcircle of triangle $ BCN$ are tangent to each other at the point $ N$.
110 replies
orl
Sep 28, 2004
SimplisticFormulas
an hour ago
Cute NT Problem
M11100111001Y1R   6
N an hour ago by X.Allaberdiyev
Source: Iran TST 2025 Test 4 Problem 1
A number \( n \) is called lucky if it has at least two distinct prime divisors and can be written in the form:
\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]where \( p_1, \dots, p_k \) are distinct prime numbers that divide \( n \). (Note: it is possible that \( n \) has other prime divisors not among \( p_1, \dots, p_k \).) Prove that for every prime number \( p \), there exists a lucky number \( n \) such that \( p \mid n \).
6 replies
M11100111001Y1R
May 27, 2025
X.Allaberdiyev
an hour ago
China MO 2021 P6
NTssu   23
N an hour ago by bin_sherlo
Source: CMO 2021 P6
Find $f: \mathbb{Z}_+ \rightarrow \mathbb{Z}_+$, such that for any $x,y \in \mathbb{Z}_+$, $$f(f(x)+y)\mid x+f(y).$$
23 replies
1 viewing
NTssu
Nov 25, 2020
bin_sherlo
an hour ago
Prove that the circumcentres of the triangles are collinear
orl   19
N 2 hours ago by Ilikeminecraft
Source: IMO Shortlist 1997, Q9
Let $ A_1A_2A_3$ be a non-isosceles triangle with incenter $ I.$ Let $ C_i,$ $ i = 1, 2, 3,$ be the smaller circle through $ I$ tangent to $ A_iA_{i+1}$ and $ A_iA_{i+2}$ (the addition of indices being mod 3). Let $ B_i, i = 1, 2, 3,$ be the second point of intersection of $ C_{i+1}$ and $ C_{i+2}.$ Prove that the circumcentres of the triangles $ A_1 B_1I,A_2B_2I,A_3B_3I$ are collinear.
19 replies
orl
Aug 10, 2008
Ilikeminecraft
2 hours ago
c^a + a = 2^b
Havu   9
N 2 hours ago by Havu
Find $a, b, c\in\mathbb{Z}^+$ such that $a,b,c$ coprime, $a + b = 2c$ and $c^a + a = 2^b$.
9 replies
Havu
May 10, 2025
Havu
2 hours ago
An algorithm for discovering prime numbers?
Lukaluce   4
N 2 hours ago by alexanderhamilton124
Source: 2025 Junior Macedonian Mathematical Olympiad P3
Is there an infinite sequence of prime numbers $p_1, p_2, ..., p_n, ...,$ such that for every $i \in \mathbb{N}, p_{i + 1} \in \{2p_i - 1, 2p_i + 1\}$ is satisfied? Explain the answer.
4 replies
Lukaluce
May 18, 2025
alexanderhamilton124
2 hours ago
Orthocentroidal circle, orthotransversal, concurrent lines
kosmonauten3114   0
2 hours ago
Source: My own
Let $\triangle{ABC}$ be a scalene oblique triangle, and $P$ a point on the orthocentroidal circle of $\triangle{ABC}$ ($P \notin \text{X(4)}$).
Prove that the orthotransversal of $P$, trilinear polar of the polar conjugate ($\text{X(48)}$-isoconjugate) of $P$, Droz-Farny axis of $P$ are concurrent.

The definition of the Droz-Farny axis of $P$ with respect to $\triangle{ABC}$ is as follows:
For a point $P \neq \text{X(4)}$, there exists a pair of orthogonal lines $\ell_1$, $\ell_2$ through $P$ such that the midpoints of the 3 segments cut off by $\ell_1$, $\ell_2$ from the sidelines of $\triangle{ABC}$ are collinear. The line through these 3 midpoints is the Droz-Farny axis of $P$ wrt $\triangle{ABC}$.
0 replies
kosmonauten3114
2 hours ago
0 replies
inequality
Hoapham235   0
2 hours ago
Let $ 0 \leq a, b, c \leq 1$. Find the maximum of \[P=\dfrac{a}{\sqrt{2bc+1}}+\dfrac{b}{\sqrt{2ca+1}}+\dfrac{c}{\sqrt{2ab+1}}.\]
0 replies
Hoapham235
2 hours ago
0 replies
3^n + 61 is a square
VideoCake   28
N 3 hours ago by Jupiterballs
Source: 2025 German MO, Round 4, Grade 11/12, P6
Determine all positive integers \(n\) such that \(3^n + 61\) is the square of an integer.
28 replies
VideoCake
May 26, 2025
Jupiterballs
3 hours ago
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   5
N 3 hours ago by AshAuktober
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
5 replies
BR1F1SZ
May 5, 2025
AshAuktober
3 hours ago
An easy number theory problem
TUAN2k8   0
4 hours ago
Source: Own
Find all positive integers $n$ such that there exist positive integers $a$ and $b$ with $a \neq b$ satifying the condition that,
$1) \frac{a^n}{b} + \frac{b^n}{a}$ is an integer.
$2) \frac{a^n}{b} + \frac{b^n}{a} | a^{10}+b^{10}$.
0 replies
TUAN2k8
4 hours ago
0 replies
Polynomial having infinitely many prime divisors
goodar2006   12
N 4 hours ago by quantam13
Source: Iran 3rd round 2011-Number Theory exam-P1
$P(x)$ is a nonzero polynomial with integer coefficients. Prove that there exists infinitely many prime numbers $q$ such that for some natural number $n$, $q|2^n+P(n)$.

Proposed by Mohammad Gharakhani
12 replies
goodar2006
Sep 19, 2012
quantam13
4 hours ago
2021 ELMO Problem 6
reaganchoi   26
N Jun 29, 2024 by meql
In $\triangle ABC$, points $D$, $E$, and $F$ lie on sides $BC$, $CA$, and $AB$, respectively, such that each of the quadrilaterals $AFDE$, $BDEF$, and $CEFD$ has an incircle. Prove that the inradius of $\triangle ABC$ is twice the inradius of $\triangle DEF$.
26 replies
reaganchoi
Jun 24, 2021
meql
Jun 29, 2024
2021 ELMO Problem 6
G H J
G H BBookmark kLocked kLocked NReply
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
reaganchoi
5289 posts
#1 • 2 Y
Y by centslordm, megarnie
In $\triangle ABC$, points $D$, $E$, and $F$ lie on sides $BC$, $CA$, and $AB$, respectively, such that each of the quadrilaterals $AFDE$, $BDEF$, and $CEFD$ has an incircle. Prove that the inradius of $\triangle ABC$ is twice the inradius of $\triangle DEF$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pad
1671 posts
#2 • 46 Y
Y by mormonmath, a_n, CANBANKAN, GorgonMathDota, Imayormaynotknowcalculus, Rg230403, Inconsistent, L567, Aimingformygoal, IndoMathXdZ, PRMOisTheHardestExam, srijonrick, p_square, Muaaz.SY, khina, NJOY, edfearay123, centslordm, ApraTrip, HamstPan38825, electrovector, Lcz, mijail, 606234, Supercali, Pluto1708, franchester, brainiacmaniac31, Aryan-23, Taco12, Offset, AforApple, JVAJVA, diegoca1, Kagebaka, GeoKing, mathleticguyyy, Kobayashi, OlympusHero, Math00954, megarnie, YesToDay, CyclicISLscelesTrapezoid, Sedro, Aneves, L13832
Let $\triangle D'E'F'$ be the triangle whose medial triangle is $\triangle DEF$. We want to show $\triangle D'E'F'$ and $\triangle ABC$ have the same inradius. We will instead show the following stronger claim, which finishes.

Key Claim: In fact, $\triangle ABC$ and $\triangle D'E'F'$ share the same incircle.

Proof: We know $AF+DE=AE+DF$ by Pitot, so $AF+D'F=AE+D'E$. By the ``excircle'' version of the converse of Pitot, this implies the quadrilateral bounded by rays $FB$, $D'F$, $D'E$, and $EC$ has an incircle. Call this $\omega_A$, and the corresponding $B-$ and $C-$ analogs $\omega_B$ and $\omega_C$. By Monge on $\omega_A$, $\omega_B$, and $\omega_C$, we find that $D,E,F$ are collinear, contradiction. Hence two of $\{\omega_A,\omega_B,\omega_C\}$ are equal, which implies the desired claim, and finishes the proof. $\blacksquare$

Remarks
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Rg230403
222 posts
#3 • 16 Y
Y by L567, lneis1, PRMOisTheHardestExam, centslordm, Hyperbolic_, Pluto1708, Aritra12, 606234, CANBANKAN, Aryan-23, Kobayashi, OlympusHero, megarnie, Mango247, Mango247, Mango247
@above, wow!! What an awesome solution!! I'll still post mine since its very different.

I did dilatation/expansion!! Interested people can read about it from Handout by Nikolai Beluhov or in short

Now, direct DEF, ABC positively. Now, let $r$ be inradius of $\Delta DEF$, now expand every circle with radius $-2r$. Let this map be $f$. Now, consider the following $D'=f(DE^+)\cap f(FD^+)$ and define $E', F'$ similarly.

Also let $f(BC^+)\cap f(DE^+)=D_Y, f(BC^+)\cap f(DF^+)=D_Z$ and midpoint of $D'D_Y=L_Y$ and midpoint of $D'D_Z=L_Z$. Similarly define $E_X, E_Z, F_X, F_Y,M_X,M_Z,N_X,N_Y$. Now, we have that $D'D_YD_YZ$ has $D'-$ excircle of radius $2r$ and $D'E'F'$ has incircle of radius $r$. Thus, excircle of $D'L_YL_Z$ is the incircle of $D'E'F'$.

Also, we have from problem statement that the quadrilateral formed by $f(DE^+), f(BC^+), f(AC^+), f(EF^+)$ has an incircle but we that the quadrilateral formed by $f(DE^+), f(EF^+), L_YL_Z, N_XN_Y$ has incircle the same as that as $D'E'F'$ and the lines of the quadrilaterals are pairwise parallel. Thus, from $E'$, we must have that $L_YN_Y\parallel D_YF_Y$. Thus, $L_YN_Y\parallel AC$.

Now, since $L_YL_ZN_XN_YM_ZM_X$ has an incircle, we have from Brianchon that $L_YN_Y, L_ZM_Z, M_XN_X$ concur. Let this point be $T$. Now, $D'L_YTL_Z$ is a parallelogram, thus the midpoint of $D'T$ lies on $L_YL_Z$, so $T$ is on $D_YD_Z$. But then we get that $D_XD_Z, E_XE_Z, F_XF_Y$ concur!!

But then under $f$, the incircle of $ABC$ went to the point circle $T$. Thus, it must have had radius $2r$. Hence proved!
This post has been edited 1 time. Last edited by Rg230403, Jun 24, 2021, 8:01 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
TheUltimate123
1740 posts
#4 • 4 Y
Y by p_square, centslordm, tree_3, kiemsibongtoi
Consider the anticomplementary triangle \(D'E'F'\) of \(\triangle DEF\). We will show that the incircles of \(\triangle ABC\) and \(\triangle D'E'F'\) coincide.

[asy]         size(7cm); defaultpen(fontsize(10pt));         pen pri=blue;         pen pri2=lightblue;         pen sec=lightred;         pen tri=purple+pink;         pen fil=invisible;         pen fil2=invisible;         pen sfil=invisible;         pen tfil=invisible;

pair Dp = (11.23550,1.19023), Ep = (8.32,-4.02), Fp = (15.,-4.), D = (11.66,-4.01), EE = (13.11775,-1.40488), F = (9.77775,-1.41488), I = (11.43393,-2.18789), A = (11.80061,3.90725), B = (8.49173,-4.79838), C = (13.64143,-3.51694);         filldraw(incircle(Dp,Ep,Fp),tfil,tri);         filldraw(A--B--C--cycle,sfil,sec);         filldraw(D--EE--F--cycle,fil2,pri2);         filldraw(Dp--Ep--Fp--cycle,fil,pri);

dot("\(D'\)", Dp, N);         dot("\(E'\)", Ep, SW);         dot("\(F'\)", Fp, SE);         dot("\(D\)", D, S);         dot("\(E\)", EE, dir(30));         dot("\(F\)", F, dir(150));         dot("\(I\)", I, N);         dot("\(A\)", A, N);         dot("\(B\)", B, SW);         dot("\(C\)", C, E);     [/asy]

First note that \(AF-AE=DF-DE=D'E-D'F\), so there is a circle \(\omega_A\) tangent to rays \(AE\), \(AF\), \(D'E\), \(D'F\). Define \(\omega_B\) and \(\omega_C\) analogously.

Assume for contradiction the circles \(\omega_A\), \(\omega_B\), \(\omega_C\) do not coincide. One can check that the pairwise exsimilicenters of \(\omega_A\), \(\omega_B\), \(\omega_C\) are \(\overline{BC}\cap\overline{E'F'}=D\), \(\overline{CA}\cap\overline{F'D'}=E\), \(\overline{AB}\cap\overline{D'E'}=F\). By Monge's theorem, points \(D\), \(E\), \(F\) are collinear, contradiction.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
starchan
1610 posts
#5 • 1 Y
Y by centslordm
pad wrote:
By the ``excircle'' version of the converse of Pitot, this implies the quadrilateral bounded by rays $FB$, $D'F$, $D'E$, and $EC$ has an incircle.

Could someone explain what that version states?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
franzliszt
23531 posts
#6 • 2 Y
Y by centslordm, Negar
Darn those are nice solutions. :o

How much would this get?
Attachments:
2021 ELMO P6.pdf (231kb)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
starchan
1610 posts
#7 • 2 Y
Y by centslordm, L567
Loved the half talking you were doing in the solution. Is it advisable to write solutions this way?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
franzliszt
23531 posts
#8 • 1 Y
Y by centslordm
What is a half taking?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
606234
80 posts
#9 • 1 Y
Y by CANBANKAN
Solution (Forgot to submit oops, same as above ones though): Take the anticomplementary triangle $D'E'F'$ of $DEF$ (twice the inradius). We claim that $ABC$ and $D'E'F'$ share a incircle, in which case we'd be done. Note that $AF+DE=AE+DF$, and since $D'F = DE$ and $DF=D'E$, we have $AF+D'F = AE+D'E$. Thus, the quadrilateral formed by the rays $AE, AF, D'E, D'F$ has an incircle, call this $w_a$ (I don't know what else to call :blush: ). We can do similar things with Pitot, and define $w_b, w_c$ equivalently. By Monge on those three circles, we see that $D, E, F$ are collinear, which is false, so two of the circles coincide, as desired. $\square$

Remark: I added the anticomplementary of $DEF$ as that was the only way I could actually deal with the "twice inradius" statement.
This post has been edited 1 time. Last edited by 606234, Jul 23, 2021, 4:48 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mira74
1010 posts
#10 • 9 Y
Y by khina, franchester, Imayormaynotknowcalculus, tree_3, SK_pi3145, Kagebaka, Inconsistent, centslordm, CyclicISLscelesTrapezoid
This solution is kinda weird. Use directed lengths throughout.

Let $\omega_1$ and $\omega_2$ be the incircles of $DEF$ and $ABC$, and let $\lambda$ be the ratio of their inradii.

Let $P$ be the center of the positive homothety mapping the $\omega_1$ to $\omega_2$. Notice by Monge's theorem on $\omega_1$, $\omega_2$, and the incircle of $AEDF$, we have that $P$ lies on $AD$. Similarly, it lies on $BE$ and $CF$.

Now, suppose the homothety centered at $P$ mapping $\omega_1$ to $\omega_2$ maps $D$, $E$, and $F$ to $D'$, $E'$, $F'$. We have that $D'E'F'$ and $ABC$ share an incircle. By Poncelet, this implies that $ABCD'E'F'$ all lie on a conic. Hence, it suffices to show the following problem:
new problem wrote:
Let $ABC$ be a triangle, let $P$ be point in its interior, let $AP$ meet $BC$ at $D$, and extend $PD$ to $D'$ with $PD'=\lambda PD$. Define $E,E',F,F'$ similarly. Show that if $A,B,C,D',E',F'$ are on a conic, then $\lambda=2$.

We establish a claim (hidden b/c it's sorta known).
claim

By Pascal, $A,B,C,D',E',F'$ lying on a conic is equivalent to $X=AD' \cap BF'$, $Y=BE' \cap CD'$, and $Z=CF' \cap AE'$ being collinear. Note that $X,Y,Z$ are on $PD$, $PE$, and $PF$,

Now, suppose $X$, $Y$, $Z$ are collinear. We have that for some fixed constants $\alpha,\beta,\gamma$, by the hidden claim,
\[DBC\text{ collinear } \implies \frac{\alpha}{PD}+\frac{\beta}{PB}+\frac{\gamma}{PC}=0\]\[AEC\text{ collinear } \implies \frac{\alpha}{PA}+\frac{\beta}{PE}+\frac{\gamma}{PC}=0\]\[ABF\text{ collinear } \implies \frac{\alpha}{PA}+\frac{\beta}{PB}+\frac{\gamma}{PF}=0\]\[XBF'\text{ collinear} \implies \frac{\alpha}{PX}+\frac{\beta}{PB}+\frac{\gamma}{\lambda PF}=0\]\[D'YC\text{ collinear} \implies \frac{\alpha}{\lambda PD}+\frac{\beta}{PY}+\frac{\gamma}{PC}=0\]\[AE'Z\text{ collinear} \implies \frac{\alpha}{PA}+\frac{\beta}{\lambda PE}+\frac{\gamma}{PZ}=0\]\[XYZ\text{ collinear} \implies \frac{\alpha}{PX}+\frac{\beta}{PY}+\frac{\gamma}{PZ}=0\]Summing these up with coefficients $1,1,1,-2,-2,-2,2$ gives
\[\left(1-\frac{2}{\lambda}\right) \left(\frac{\alpha}{PD}+\frac{\beta}{PE}+\frac{\gamma}{PF}\right)=0.\]Since $DEF$ are not collinear, the second term is nonzero, so we must have $\lambda=2$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathleticguyyy
3217 posts
#11 • 3 Y
Y by franchester, mango5, centslordm
Darn those are nice solutions. :o

How much would this get?

Incomplete bashes are generally worth 0 points.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Fouad-Almouine
72 posts
#12 • 3 Y
Y by Mango247, Mango247, Mango247
Can someone complete this :(
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mira74
1010 posts
#13 • 1 Y
Y by centslordm
@above i think you forgot to use a sharing link, so we can't see it :(
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Fouad-Almouine
72 posts
#14
Y by
mira74 wrote:
@above i think you forgot to use a sharing link, so we can't see it :(

Oh, my bad. It should work now ^^
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
edfearay123
92 posts
#15 • 1 Y
Y by diegoca1
Solution with metrics and trig bash, by @diegoca1. Part 1
Attachments:
problem6 1-2.pdf (335kb)
problem6 3-4.pdf (453kb)
problem6 5-6.pdf (391kb)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
edfearay123
92 posts
#16 • 1 Y
Y by diegoca1
Solution with metrics and trig bash, by @diegoca1. Part 2
Attachments:
problem6 11-12.pdf (335kb)
problem6 7-8.pdf (344kb)
problem6 9-10.pdf (289kb)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
edfearay123
92 posts
#17 • 1 Y
Y by diegoca1
Solution with metrics and trig bash, by @diegoca1. Part 3
Attachments:
problem6 13.pdf (150kb)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MarkBcc168
1595 posts
#18 • 3 Y
Y by wwn13, pad, Aneves
Here is a convoluted solution using Menelaus' Theorem

Let $\Omega, \omega, \omega_A, \omega_B, \omega_C$ denote the incircles of $\triangle ABC$, $\triangle DEF$, $AFDE$, $BDEF$, $CEFD$, respectively. Let $I,J,I_A, I_B, I_C$ be the centers of $\Omega$, $\omega$, $\omega_A$, $\omega_B$, $\omega_C$.

Claim: $AD, BE, CF$ concur at the exsimilicenter $P$ of $\omega, \Omega$.

Proof: Follows from Monge's theorem applied on $\Omega, \omega, \omega_A$. $\blacksquare$
Claim: $D, I_B, I_C, J$ are concyclic. (Similarly, $E,I_A,I_C,J$ and $F,I_A,I_B,J$ are concyclic.)

Proof: Angle chasing:
$$\angle I_BDI_C = 180^{\circ} - \frac{\angle BDE+\angle CDF}{2} = 90^{\circ} + \frac{\angle EDF}{2}. \blacksquare$$
Before the next step, we would like to recall the well-known fact that $\frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF}=1$. This follows from summing $\frac{PD}{AD} = \frac{[BPC]}{[BAC]}$, etc.

Claim: We also have the relation
$$\frac{JI_A}{DI_A}  + \frac{JI_B}{EI_B} + \frac{JI_C}{FI_C}  = 1$$
Proof: Follows from inversion around $J$ an applying the above fact on $\triangle I_A^*I_B^*I_C^*$. $\blacksquare$
Claim: $J$ is the midpoint of $PI$.

Proof: Let $t = \tfrac{PI}{JI}$. By Menelaus' theorem on $\triangle DJP$ and $\overline{AI_AI}$, we obtain
$$\frac{PI}{IJ}\cdot \frac{JI_A}{I_AD}\cdot\frac{DA}{AP} = 1 \implies t\cdot \frac{JI_A}{I_AD} = \frac{AP}{AD}.$$Similarly, we obtain
\begin{align*}
t\cdot \frac{JI_B}{I_BE} &= \frac{BP}{BE} \\
t\cdot \frac{JI_C}{I_CF} &= \frac{CP}{CF}, \\
\end{align*}so summing gives $t=2$ as desired. $\blacksquare$
The first and the last claim combined imply the problem.
This post has been edited 1 time. Last edited by MarkBcc168, Jun 25, 2021, 8:28 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
iman007
270 posts
#19 • 2 Y
Y by tarannom, MZLBE
There is another way to solve this problem here i just give an sketch
Click to reveal hidden text
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MZLBE
8 posts
#20
Y by
iman007 wrote:
There is another way to solve this problem here i just give an sketch
Click to reveal hidden text

Nice. But $P'IPG$ is a parallelogram, in fact.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
iman007
270 posts
#21
Y by
MZLBE wrote:
iman007 wrote:
There is another way to solve this problem here i just give an sketch
Click to reveal hidden text

Nice. But $P'IPG$ is a parallelogram, in fact.
no It's not look
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
psi241
49 posts
#22 • 2 Y
Y by SK_pi3145, MarkBcc168
You have seen a lot of normal solutions. Now, let's see this weird one, and I'm also not sure if it is totally correct.

By Monge's Theorem with three incircles, $BC\cap EF,CA\cap FD, AB\cap DE$ are collinear, and by Desargue's Theorem, $AD,BE,CF$ are concurrent.

Let $\omega$ be an incircle of $\triangle DEF$.
Construct $\triangle A'B'C'$ which its incircle is $\omega$ and homothetic with $\triangle ABC$ and a homothetic center lies outside $AA'$.

If we homothety with center at $D$ and send $\omega$ to an incircle of $AFDE$, it will send $A'B'\to AB$ and $A'C'\to AC$. Therefore, $A'$ lies on $AD$, and similarly, $B'$ lies on $BE$, $C'$ lies on $CF$.

Let $X=B'C'\cap AD$. We want to prove that $X$ is a midpoint of $PD$ or $(P,D;X,\infty_{AD})$.

The fun begins here. Let $T$ be a projective transformation that send $P$ to the center of $\omega$.
For convenience, a name of point after this will referred to a point after the transformation as we will not convert it back.
Clearly, $P$ is incenter of $\triangle DEF,\triangle A'B'C'$.

Claim: $\triangle DEF,\triangle A'B'C'$ are symmetric with respect to $P$.
Let $\triangle D'E'F'$ be symmetric with $\triangle DEF$ wrt $P$.
$B'$ and $C'$ are defined by intersections of lines tangent to $\omega$ and $PE,PF$.
We see that as $A'$ move along $AP$, a segment $B'C'$ r $BC$ will not intersect $E'F'$ or $EF$ except when $A'\in\{D',D\}$. Thus, $B'C'$ is tangent to $\omega$ iff $A'\in\{D',D\}$, but $A'\neq D$ so $A'=D'$ as desired. $\square$

Since $(A,P;EF\cap AD,D)=-1$ before the transformation, $\triangle ABC$ is now a excentral triangle of $\triangle DEF$.
Let $K=BC\cap B'C',L=CA\cap C'A',M=AB\cap A'B'$, these point lie on image of a line at infinity.
We want to show that $(D,P;X,KL\cap AD)=-1$. Pencil at $K$, $K(D,P;X,KL\cap AD)=(C,P;C',KL\cap CP)$.

Now another unexpected thing come. By Reverse Pascal's theorem on $LA'DKB'E$, these six points lie on a conic $\mathcal{C}$. Moreover, because $A'B'$ and $DE$ are symmetric wrt $P$, a reflection of $L$ across $P$ lso lie on this conic. Denote this reflection by $L'$.
Now $L(C,P;C',KL\cap CP)=L(E,L';A',K)=B'(E,L;A',K)$.
We know that $B'E,B'L'$ bisect $\angle KB'A'$. (since $B'L'//EL$.)
Therefore, the last cross-ratio is equal to $-1$ as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Flip99
82 posts
#23 • 1 Y
Y by Aneves
starchan wrote:
pad wrote:
By the ``excircle'' version of the converse of Pitot, this implies the quadrilateral bounded by rays $FB$, $D'F$, $D'E$, and $EC$ has an incircle.

Could someone explain what that version states?

Can anyone explain it to me too, please?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
duanby
76 posts
#24
Y by
Consider the structure based on triangle DEF, it is easy to see(calculating the tangent length) that the condition is equivalent to tangent length from D,E,F to $\omega_{a,b,c}$ are 1/4 perimeter of DEF ($\omega_{a,b,c}$ are 1/2 homothety of excircles of DEF mapped with center D,E,F). Moreover, the tangent of incircle of ABC at each side are the symmetry of D,E,F wrt the midpoint of tangent of $\omega_{a,b,c}$ on each side. The incircle radius of ABC is then easily solved by the area relation.
This post has been edited 1 time. Last edited by duanby, Dec 4, 2022, 1:52 PM
Reason: typo
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
BlizzardWizard
108 posts
#25
Y by
mira74 wrote:
new problem wrote:
Let $ABC$ be a triangle, let $P$ be point in its interior, let $AP$ meet $BC$ at $D$, and extend $PD$ to $D'$ with $PD'=\lambda PD$. Define $E,E',F,F'$ similarly. Show that if $A,B,C,D',E',F'$ are on a conic, then $\lambda=2$.

Alternate finish
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Leo.Euler
577 posts
#28
Y by
Let $\triangle D'E'F'$ be the anticomplementary triangle of $\triangle DEF$. We finish by the following claim.

Claim: The incircles of $\triangle ABC$ and $\triangle D'E'F'$ coincide.
Proof. By Pitot, $AF+DE=AE+FD$, so using medial triangle lengths, we rewrite this as \[ AF - AE = D'E-D'F, \]so by Pitot again we have that there is an excircle $\omega_A$ tangent to $AC, AB, D'F', D'E'$. Analogously, we draw $\omega_B$, $\omega_C$. Suppose that $\omega_A$, $\omega_B$, $\omega_C$ are pairwise distinct. By Monge's theorem on $\omega_A$, $\omega_B$, $\omega_C$, we have that $D$, $E$, and $F$ are collinear, a contradiction. Thus, at least two of these circles coincide, implying the claim.
:yoda:

Remark: The excircle construction by Pitot very similar to that in 2022 China TST/4/2 (which also used Monge, with one of the three circles being the constructed one), although in the China TST problem the finish was more interesting; the claim that Monge induced resulted in the existence of an ellipse, and DIT was applied on the ellipse.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
meql
20 posts
#29
Y by
I was thinking about this diagram recently and reremembered the following (hard) fact.

Problem: Show that the radical center of the incircles of $AFDE$, $BDEF$, and $CEFD$ is the incenter of $DEF$.

My own solution is fairly long, so I would love to see if there are any shorter solutions.

Solution
This post has been edited 1 time. Last edited by meql, Jun 29, 2024, 2:55 AM
Z K Y
N Quick Reply
G
H
=
a