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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 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
USAMO 1985 #4
Mrdavid445   6
N 3 minutes ago by anudeep
There are $n$ people at a party. Prove that there are two people such that, of the remaining $n-2$ people, there are at least $\left\lfloor\frac{n}{2}\right\rfloor-1$ of them, each of whom either knows both or else knows neither of the two. Assume that knowing is a symmetric relation, and that $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$.
6 replies
Mrdavid445
Jul 26, 2011
anudeep
3 minutes ago
Polygon formed by the edges of an infinite chessboard
AlperenINAN   0
10 minutes ago
Source: Turkey TST 2025 P5
Let $P$ be a polygon formed by the edges of an infinite chessboard, which does not intersect itself. Let the numbers $a_1,a_2,a_3$ represent the number of unit squares that have exactly $1,2\text{ or } 3$ edges on the boundary of $P$ respectively. Find the largest real number $k$ such that the inequality $a_1+a_2>ka_3$ holds for each polygon constructed with these conditions.
0 replies
AlperenINAN
10 minutes ago
0 replies
Nice FE as the First Day Finale
swynca   0
13 minutes ago
Source: 2025 Turkey TST P3
Find all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x,y \in \mathbb{R}-\{0\}$,
$$ f(x) \neq 0 \text{ and } \frac{f(x)}{f(y)} + \frac{f(y)}{f(x)} - f \left( \frac{x}{y}-\frac{y}{x} \right) =2 $$
0 replies
swynca
13 minutes ago
0 replies
Arithmetic Sequence about Number of Primes
swynca   0
21 minutes ago
Source: 2025 Turkey TST P2
For all positive integers $n$, the function $\gamma: \mathbb{Z}^+ \to \mathbb{Z}_{\geq 0}$ is defined as, $\gamma(1) = 0$ and for all $n > 1$, if the prime factorization of $n$ is $n = p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k},$ then $\gamma(n) = \alpha_1 + \alpha_2 + \dots + \alpha_k$. We have an arithmetic sequence $X = \{x_i\}_{i=1}^{\infty}$. If for a positive integer $a > 1$, the sequence $\{ \gamma(a^{x_i} -1) \}$ is also an arithmetic sequence, show that the sequence $X$ has to be constant.
0 replies
swynca
21 minutes ago
0 replies
interesting number theory problem
Zavyk09   0
24 minutes ago
Source: forgotten
Let $a$, $b$ are coprime positive integers. Find all $(x, y) \in \mathbb{N}^*\times\mathbb{N}^*$ such that:
$$a^n + b^n \vert x^n + y^n, \forall n \in \mathbb{N}^*$$
0 replies
Zavyk09
24 minutes ago
0 replies
unnecessary wrapped FE on Q
iStud   1
N 25 minutes ago by jasperE3
Source: Monthly Contest KTOM March 2025 P3 Essay
Find all functions $f:\mathbb{Q}\to\mathbb{Q}$ such that
\[f(f(f(\frac{x+y}{2}))+x+y)=f(x)+f(y)+f(\frac{x+y}{2})\]for all rational numbers $x,y$.

Hint
1 reply
iStud
3 hours ago
jasperE3
25 minutes ago
Minimal Grouping in a Complete Graph
swynca   0
27 minutes ago
Source: 2025 Turkey TST P1
In a complete graph with $2025$ vertices, each edge has one of the colors $r_1$, $r_2$, or $r_3$. For each $i = 1,2,3$, if the $2025$ vertices can be divided into $a_i$ groups such that any two vertices connected by an edge of color $r_i$ are in different groups, find the minimum possible value of $a_1 + a_2 + a_3$.
0 replies
1 viewing
swynca
27 minutes ago
0 replies
Like Father Like Son... (or Like Grandson?)
AlperenINAN   0
27 minutes ago
Source: Turkey TST 2025 P4
Let $a,b,c$ be given pairwise coprime positive integers where $a>bc$. Let $m<n$ be positive integers. We call $m$ to be a grandson of $n$ if and only if, for all possible piles of stones whose total mass adds up to $n$ and consist of stones with masses $a,b,c$, it's possible to take some of the stones out from this pile in a way that in the end, we can obtain a new pile of stones with total mass of $m$. Find the greatest possible number that doesn't have any grandsons.
0 replies
1 viewing
AlperenINAN
27 minutes ago
0 replies
limit of function at 5pi
RenheMiResembleRice   0
30 minutes ago
Find f(x)
0 replies
RenheMiResembleRice
30 minutes ago
0 replies
Slovenia 2019 TST2 P2
pj294   5
N 37 minutes ago by jasperE3
Source: 2019 Slovenia 2nd TST Problem 2
Determine all non-negative real numbers $a$, for which $f(a)=0$ for all functions $f: \mathbb{R}_{\ge 0}\to \mathbb{R}_{\ge 0} $, who satisfy the equation $f(f(x) + f(y)) = yf(1 + yf(x))$ for all non-negative real numbers $x$ and $y$.
5 replies
pj294
Feb 14, 2019
jasperE3
37 minutes ago
"Mistakes were made" -Luke Rbotaille
a1267ab   9
N an hour ago by asbodke
Source: USA TST 2025
Let $a_1, a_2, \dots$ and $b_1, b_2, \dots$ be sequences of real numbers for which $a_1 > b_1$ and
\begin{align*}
    a_{n+1} &= a_n^2 - 2b_n\\
    b_{n+1} &= b_n^2 - 2a_n
\end{align*}for all positive integers $n$. Prove that $a_1, a_2, \dots$ is eventually increasing (that is, there exists a positive integer $N$ for which $a_k < a_{k+1}$ for all $k > N$).

Holden Mui
9 replies
a1267ab
Dec 14, 2024
asbodke
an hour ago
Yet another circle!
Rushil   7
N an hour ago by FireMaths
Source: INMO 1999 Problem 4
Let $\Gamma$ and $\Gamma'$ be two concentric circles. Let $ABC$ and $A'B'C'$ be any two equilateral triangles inscribed in $\Gamma$ and $\Gamma'$ respectively. If $P$ and $P'$ are any two points on $\Gamma$ and $\Gamma'$ respectively, show that \[ P'A^2 + P'B^2 + P'C^2 = A'P^2 + B'P^2 + C'P^2. \]
7 replies
Rushil
Oct 7, 2005
FireMaths
an hour ago
Binary Operator from AMC 10
pinetree1   36
N an hour ago by Ilikeminecraft
Source: USA TSTST 2019 Problem 1
Find all binary operations $\diamondsuit: \mathbb R_{>0}\times \mathbb R_{>0}\to \mathbb R_{>0}$ (meaning $\diamondsuit$ takes pairs of positive real numbers to positive real numbers) such that for any real numbers $a, b, c > 0$,
[list]
[*] the equation $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ holds; and
[*] if $a\ge 1$ then $a\,\diamondsuit\, a\ge 1$.
[/list]
Evan Chen
36 replies
pinetree1
Jun 25, 2019
Ilikeminecraft
an hour ago
Interesting inequality
sqing   4
N 2 hours ago by lbh_qys
Source: Own
Let $ a,b\geq 2  . $ Prove that
$$(a^2-1)(b^2-1) -6ab\geq-15$$$$(a^2-1)(b^2-1)  -7ab\geq  -\frac{58}{3}$$$$(a^3-1)(b^3-1)  -\frac{21}{4}a^2b^2\geq -35$$$$(a^3-1)(b^3-1)  -6a^2b^2\geq-\frac{2391}{49}$$
4 replies
sqing
2 hours ago
lbh_qys
2 hours ago
Right angles on incircle
DynamoBlaze   36
N Jan 19, 2024 by SilverBlaze_SY
Source: RMO 2018 P6
Let $ABC$ be an acute-angled triangle with $AB<AC$. Let $I$ be the incentre of triangle $ABC$, and let $D,E,F$ be the points where the incircle touches the sides $BC,CA,AB,$ respectively. Let $BI,CI$ meet the line $EF$ at $Y,X$ respectively. Further assume that both $X$ and $Y$ are outside the triangle $ABC$. Prove that
$\text{(i)}$ $B,C,Y,X$ are concyclic.
$\text{(ii)}$ $I$ is also the incentre of triangle $DYX$.
36 replies
DynamoBlaze
Oct 7, 2018
SilverBlaze_SY
Jan 19, 2024
Right angles on incircle
G H J
G H BBookmark kLocked kLocked NReply
Source: RMO 2018 P6
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DynamoBlaze
170 posts
#1 • 4 Y
Y by FreakLM10, MatBoy-123, Adventure10, Rounak_iitr
Let $ABC$ be an acute-angled triangle with $AB<AC$. Let $I$ be the incentre of triangle $ABC$, and let $D,E,F$ be the points where the incircle touches the sides $BC,CA,AB,$ respectively. Let $BI,CI$ meet the line $EF$ at $Y,X$ respectively. Further assume that both $X$ and $Y$ are outside the triangle $ABC$. Prove that
$\text{(i)}$ $B,C,Y,X$ are concyclic.
$\text{(ii)}$ $I$ is also the incentre of triangle $DYX$.
Z K Y
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Mate007
69 posts
#2 • 2 Y
Y by Adventure10, Rounak_iitr
First part is easy as it has only angle chasing.
And showing angle in the same segment.
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stroller
894 posts
#3 • 1 Y
Y by Adventure10
Second part
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jayme
9767 posts
#4 • 3 Y
Y by danepale, Adventure10, Mango247
Dear Mathlinkers,
for the second part

http://www.oei.es/oim/revistaoim/numero52/DuranSolucElem_a_probelem.pdf

Sincerely
Jean-Louis
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AlastorMoody
2125 posts
#5 • 2 Y
Y by Adventure10, Mango247
This was the easiest P6 ever on RMO
Z K Y
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PPKHardy
20 posts
#6 • 2 Y
Y by Adventure10, Mango247
I agree but my first part just got half of it wrong.
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TheDarkPrince
3042 posts
#7 • 3 Y
Y by Delta0001, Adventure10, Mango247
Giveaway :( This was stupidly way too easy for RMO.
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PPKHardy
20 posts
#8 • 2 Y
Y by Adventure10, Mango247
Any one #P4
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e_plus_pi
756 posts
#9 • 1 Y
Y by Adventure10
My Solution
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Vrangr
1600 posts
#10 • 1 Y
Y by Adventure10
Just like last year, this year's second geometry problem was just a restatement of a lemma/problem in EGMO. *sigh*
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omriya200
317 posts
#11 • 2 Y
Y by Adventure10, Mango247
it is one of the easiest question in the paper.
This post has been edited 1 time. Last edited by omriya200, Oct 7, 2018, 3:44 PM
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MathematicalPhysicist
179 posts
#12 • 2 Y
Y by Adventure10, Mango247
Vrangr wrote:
Just like last year, this year's second geometry problem was just a restatement of a lemma/problem in EGMO. *sigh*

I think this is so because EGMO is such a complete book; it's hard to create a geometry problem that is not doable by methods taught in EGMO.
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wu2481632
4233 posts
#13 • 6 Y
Y by AlastorMoody, RudraRockstar, Adventure10, Mango247, Mango247, Mango247
I'm sure it is possible without lifting a problem directly from EGMO.
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Arthur.
133 posts
#14 • 2 Y
Y by AlastorMoody, Adventure10
MathematicalPhysicist wrote:
Vrangr wrote:
Just like last year, this year's second geometry problem was just a restatement of a lemma/problem in EGMO. *sigh*

I think this is so because EGMO is such a complete book; it's hard to create a geometry problem that is not doable by methods taught in EGMO.

I'd make a large distinction between 'problems solvable by techniques in EGMO' and 'restatements of EGMO problems'.
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Fouad-Almouine
72 posts
#15 • 2 Y
Y by Adventure10, Mango247
In fact it is an old well known lemma :D
It states that there is a spiral similarity at $B$ that maps $I$ to $F$ and $C$ to $Y$, with $\widehat{CYI}=90$...
$\widehat{BYX}=\widehat{BYF}=\widehat{BCI} $
Hence $BXYC$ and $IDCY$, the conclusion follows and we are done.$\blacksquare$
This post has been edited 1 time. Last edited by Fouad-Almouine, Oct 7, 2018, 10:19 PM
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biomathematics
2564 posts
#16 • 2 Y
Y by Adventure10, Mango247
This is just too well-known :P
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math_pi_rate
1218 posts
#17 • 2 Y
Y by Adventure10, Mango247
Posting my solution from contest just for completeness sake (and also cause I have been using this link in the answers of some other questions too :P): We have $$\angle YEC=\angle AEF=90^{\circ}-\frac{A}{2}=180^{\circ}-\left(90^{\circ}+\frac{A}{2} \right)=180^{\circ}-\angle BIC=\angle YIC$$which means that $CIEY$ is cyclic. Similarly, we get that $BIFX$ is also cyclic. Thus, $$\angle CYI=\angle CEI=90^{\circ}=\angle BFI=\angle BXI$$This means that $X$ and $Y$ lie on the circle with $BC$ as diameter, proving Part (i).

As for Part (ii), notice that, as $D$ also lies on $\odot (CIE)$, and cause $ID=IE$, i.e. $I$ is the midpoint of minor arc $DE$ in $\odot (CDIEY)$, we get that $YI$ bisects $\angle DYE$, i.e. $\angle DYX$. Similarly, $XI$ bisects $\angle DXY$, giving that $I$ is the incenter of $\triangle DXY$. $\blacksquare$

REMARK: Part (ii) can also be proved in another manner. Let $H$ be the orthocenter of $\triangle BIC$. Then $I$ is the orthocenter of $\triangle BHC$, and $DXY$ is its orthic triangle, and so by a well known property, we get that $I$ is the incenter of $\triangle DXY$.
This post has been edited 2 times. Last edited by math_pi_rate, Nov 28, 2018, 1:23 PM
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Wizard_32
1566 posts
#18 • 3 Y
Y by AlastorMoody, Adventure10, Mango247
The part 2 can also be proved using congruency, which I am astonished no one in the posts above has pointed out. Probably because congruency is not used outside school :P
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TheDarkPrince
3042 posts
#19 • 2 Y
Y by Adventure10, Mango247
Wizard_32 wrote:
The part 2 can also be proved using congruency, which I am astonished no one in the posts above has pointed out. Probably because congruency is not used outside school :P

Or probably no one wanted to waste time on this problem ;)
This post has been edited 1 time. Last edited by TheDarkPrince, Oct 19, 2018, 10:33 AM
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AlastorMoody
2125 posts
#20 • 2 Y
Y by Adventure10, Rounak_iitr
https://artofproblemsolving.com/community/c6h1712858p11053650

Part 1 of the problem was on FBH regional 2012 for grade 9
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AlastorMoody
2125 posts
#21 • 3 Y
Y by Adventure10, Mango247, Rounak_iitr
Seems really easy right now:
part(i)
part(ii)
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yayups
1614 posts
#22 • 5 Y
Y by AlastorMoody, Wizard_32, Hexagrammum16, Adventure10, Mango247
Wizard_32 wrote:
The part 2 can also be proved using congruency, which I am astonished no one in the posts above has pointed out. Probably because congruency is not used outside school :P

I can't even remember the last time I used congruent triangles :p
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AlastorMoody
2125 posts
#23 • 5 Y
Y by Wizard_32, Pluto1708, amar_04, Adventure10, Mango247
INMO 1989 P6
This post has been edited 1 time. Last edited by AlastorMoody, May 8, 2019, 4:12 AM
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AlastorMoody
2125 posts
#24 • 2 Y
Y by amar_04, Adventure10
Nordic Math Contest 2015 P1


@below Let's hope for a well-known Lemma this year too...I am getting very scared :( :(
This post has been edited 1 time. Last edited by AlastorMoody, Sep 12, 2019, 5:41 PM
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Delta0001
1422 posts
#25 • 3 Y
Y by AlastorMoody, Adventure10, Mango247
I am sorry for unnecessary bump, but is this problem a joke?
Vrangr wrote:
Just like last year, this year's second geometry problem was just a restatement of a lemma/problem in EGMO. *sigh*

Also, could someone please tell which lemma is being referred to?
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Kamran011
678 posts
#26 • 1 Y
Y by Adventure10
Delta0001 wrote:
I am sorry for unnecessary bump, but is this problem a joke?
Vrangr wrote:
Just like last year, this year's second geometry problem was just a restatement of a lemma/problem in EGMO. *sigh*

Also, could someone please tell which lemma is being referred to?

I don't think that it's a Lemma, a non-tricky angle chasing :)
Solution
This post has been edited 2 times. Last edited by Kamran011, Sep 12, 2019, 6:36 PM
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Delta0001
1422 posts
#27 • 2 Y
Y by Adventure10, Mango247
@Above, it's easy and I do know the solution, but I wanted to see if another problem from Chapter-4 appeared in the RMO
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PhysicsMonster_01
1445 posts
#28 • 3 Y
Y by AlastorMoody, Adventure10, Mango247
For storage
This post has been edited 1 time. Last edited by PhysicsMonster_01, Oct 20, 2019, 4:53 AM
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Math-wiz
6107 posts
#30 • 5 Y
Y by amar_04, AlastorMoody, Bumblebee60, lilavati_2005, Rounak_iitr
Finally solved it, phew! Angle-chased diagram attached for the sake of completeness, as well as reference. A bit too detailed solution compared to other solutions..

Let $\angle A=2\alpha, \angle B=2\beta, \angle C=2\gamma$. I guess the rest of angle chase is clear from figure.
(i) $\angle XFB=\angle XIB\implies BIFX$ is cyclic$\implies \angle BFI=\angle BXI\implies\angle BXI=90^{\circ}$.
Similarly, $\angle YEC=\angle YIC\implies YEIC$ is cyclic$\implies \angle IEC=\angle IYC\implies \angle IYC=90^{\circ}$
So, $\angle BXC=\angle BYC\implies BXYC$ is cyclic, as desired.
(ii)As proved in (i), $BXYC$ ic cyclic $\implies \angle XCB=\angle XYB\implies \angle XYB=\gamma$. Similarly, $\angle YBC=\angle YXC\implies \angle YXC=\beta$.
Also, as proved in (i), $\angle BXC=90^{\circ}=\angle BDI\implies BDIX$ is cyclic $\implies\angle DXI=\angle DBI=\beta$.
Similarly, $\angle BYC=90^{\circ}=\angle CDI\implies CDIY$ is cyclic$\implies\angle IYD=\angle ICD=\gamma$.
So, $IX$ bisects $\angle DXY$ and $IY$ bisects $\angle DYX$. Thus $I$ is also the incenter of $\triangle DXY$, as desired $\blacksquare$.
Attachments:
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Stormersyle
2785 posts
#31
Y by
Trivial by angle chasing: Redefine $K=EF\cap BI$. Note that $\angle{EKI}=\angle{FKI}=180-\angle{B}/2-(90+\angle{A}/2)=\angle{C}/2$, meaning $EKCI$ is cyclic. Thus, since $\angle{IEC}=90$, we have $\angle{IKC}=90$ as well. Now let $M$ be the midpoint of $BC$ and $N$ be the midpoint of $AC$. Note that since $\triangle{KBC}$ is right, we have $KM=BM=CM$, so $\angle{KMB}=180-2\cdot (\angle{B}/2)=180-\angle{B}$, meaning $K, M, N$ are collinear.
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jelena_ivanchic
151 posts
#32 • 3 Y
Y by Jupiter_is_BIG, Illuzion, Mango247
By Iran lemma, $\angle BYC=\angle CXB=90$ and so, $BCXY$ is cyclic. Also, as $DIEYC, DIFXB$ are cyclic with $IF=ID=IE,$ by fact 5 we get $I$ is the incenter of $DXY$. $\blacksquare$
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Rajdeep111
8 posts
#33 • 1 Y
Y by jelena_ivanchic
Ig a different solution than others, here we go

[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(15cm); 
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 = -13.764483846876923, xmax = 5.176303477974182, ymin = -1.5715650474047884, ymax = 9.629055475231542;  /* image dimensions */
pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen ffqqff = rgb(1,0,1); 
 /* draw figures */
draw((-6.207733319596109,6.131918499823255)--(-8.523569086328425,2.153812191539418), linewidth(0.8)); 
draw((-8.523569086328425,2.153812191539418)--(-1.5963622702582592,1.8575894042284895), linewidth(0.8)); 
draw((-6.207733319596109,6.131918499823255)--(-1.5963622702582592,1.8575894042284895), linewidth(0.8)); 
draw(circle((-5.8337845690589685,3.624772291402155), 1.5845331914338108), linewidth(0.8) + wvvxds); 
draw((-7.602495703688477,4.362399043957692)--(-3.3158178513850256,5.001770541894226), linewidth(0.8)); 
draw((-3.3158178513850256,5.001770541894226)--(-8.523569086328425,2.153812191539418), linewidth(0.8)); 
draw((-7.602495703688477,4.362399043957692)--(-1.5963622702582592,1.8575894042284895), linewidth(0.8)); 
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draw((-2.611734153564187,5.386812176199431)--(-5.17857829741542,-0.7680700896066753), linewidth(0.8)); 
draw((-5.17857829741542,-0.7680700896066753)--(-8.097069626094108,4.5686571203275985), linewidth(0.8)); 
draw((-6.207733319596109,6.131918499823255)--(-5.17857829741542,-0.7680700896066753), linewidth(0.8)); 
draw((-7.602495703688477,4.362399043957692)--(-5.901480869656607,2.041685861152432), linewidth(0.8)); 
draw((-3.3158178513850256,5.001770541894226)--(-5.901480869656607,2.041685861152432), linewidth(0.8)); 
draw((-7.602495703688477,4.362399043957692)--(-8.523569086328425,2.153812191539418), linewidth(0.8)); 
draw((-3.3158178513850256,5.001770541894226)--(-1.5963622702582592,1.8575894042284895), linewidth(0.8)); 
draw(circle((-5.0599656782933415,2.0057007978839554), 3.4667687481403786), linewidth(0.8) + linetype("4 4") + ffqqff); 
 /* dots and labels */
dot((-6.207733319596109,6.131918499823255),linewidth(4pt) + dotstyle); 
label("$A$", (-6.1588223129470405,6.229740513121389), NE * labelscalefactor); 
dot((-8.523569086328425,2.153812191539418),linewidth(4pt) + dotstyle); 
label("$B$", (-8.469867377115445,2.2557212228847003), NE * labelscalefactor); 
dot((-1.5963622702582592,1.8575894042284895),linewidth(4pt) + dotstyle); 
label("$C$", (-1.5489599362724982,1.9500274313280317), NE * labelscalefactor); 
dot((-5.8337845690589685,3.624772291402155),linewidth(4pt) + dotstyle); 
label("$I$", (-5.779762011416773,3.7230514223567086), NE * labelscalefactor); 
dot((-5.901480869656607,2.041685861152432),linewidth(4pt) + dotstyle); 
label("$D$", (-5.853128521390373,2.1334437062620326), NE * labelscalefactor); 
dot((-4.756624001282189,4.786869768373093),linewidth(4pt) + dotstyle); 
label("$E$", (-4.7037198651373044,4.884687830272049), NE * labelscalefactor); 
dot((-7.203178462372859,4.421958467718602),linewidth(4pt) + dotstyle); 
label("$F$", (-7.149270197590643,4.517855280404047), NE * labelscalefactor); 
dot((-3.3158178513850256,5.001770541894226),linewidth(4pt) + dotstyle); 
label("$X$", (-3.260845168989835,5.10478736019285), NE * labelscalefactor); 
dot((-7.602495703688477,4.362399043957692),linewidth(4pt) + dotstyle); 
label("$Y$", (-7.552786002445444,4.456716522092712), NE * labelscalefactor); 
dot((-8.097069626094108,4.5686571203275985),linewidth(4pt) + dotstyle); 
label("$Y_1$", (-8.054123820598377,4.664588300351247), NE * labelscalefactor); 
dot((-2.611734153564187,5.386812176199431),linewidth(4pt) + dotstyle); 
label("$X_1$", (-2.5638633242406335,5.483847661723119), NE * labelscalefactor); 
dot((-5.17857829741542,-0.7680700896066753),linewidth(4pt) + dotstyle); 
label("$Z_1$", (-5.131691173316638,-0.6667114243970499), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */[/asy]

Notice that $XY\parallel X_{1}Y_{1}$ as $\Delta X_{1}Y_{1}Z_{1}$ has its orthocenter as $I\Rightarrow AI\perp X_{1}Y_{1}$ and $AI\perp \overline{YFEX}$. Then $\exists$ a homothety sending $XY\leftrightarrow X_{1}Y_{1}$, hence $(BY_{1}X_{1}C)\leftrightarrow (BYXC)$ and we complete our first part.

Now for the second part, by Iran lemma or EGMO lemma 1.45 we know that $\angle CXB = 90^{\circ}.  \angle XYC = \angle XBC = \angle DYC $( because $\square YIDB$ is cyclic). Hence $CY$ bisects $\angle DYX$
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Commander_Anta78
58 posts
#35
Y by
Solution
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lifeismathematics
1188 posts
#36
Y by
[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(40cm); 
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 = -10.222, xmax = 13.736, ymin = -4.304, ymax = 10.216;  /* image dimensions */
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); 

draw((-4.216,8.17)--(-8.044,2.032)--(9.49,0.118)--cycle, linewidth(2) + rvwvcq); 
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draw((-4.216,8.17)--(-8.044,2.032), linewidth(2) + rvwvcq); 
draw((-8.044,2.032)--(9.49,0.118), linewidth(2) + rvwvcq); 
draw((9.49,0.118)--(-4.216,8.17), linewidth(2) + rvwvcq); 
draw((xmin, -0.3313292275453181*xmin + 3.2623143694050687)--(xmax, -0.3313292275453181*xmax + 3.2623143694050687), linewidth(2)); /* line */
draw((xmin, 0.48579311234662764*xmin + 5.9397197957162735)--(xmax, 0.48579311234662764*xmax + 5.9397197957162735), linewidth(2)); /* line */
draw(circle((-3.2766273734056255,4.3479567859894), 2.819610910551786), linewidth(2)); 
draw((-5.669096776867536,5.840034478471021)--(-1.8483924116576476,6.7790780460139635), linewidth(2)); 
draw((-5.669096776867536,5.840034478471021)--(-9.342,5.134), linewidth(2)); 
draw((-1.8483924116576476,6.7790780460139635)--(7.862,9.6), linewidth(2)); 
draw((-8.044,2.032)--(-7.004925504045343,5.58325092567291), linewidth(2)); 
draw((9.49,0.118)--(7.047687174717885,9.363437683167886), linewidth(2)); 
draw((-4.216,8.17)--(-3.276627373405626,4.3479567859894), linewidth(2)); 
draw((-3.276627373405626,4.3479567859894)--(-3.5825967298465926,1.5449961298577837), linewidth(2)); 
draw((-3.276627373405626,4.3479567859894)--(-5.669096776867536,5.840034478471021), linewidth(2)); 
draw((-3.276627373405626,4.3479567859894)--(-1.8483924116576476,6.7790780460139635), linewidth(2)); 
 /* dots and labels */
dot((-4.216,8.17),dotstyle); 
label("$A$", (-4.128,8.39), NE * labelscalefactor); 
dot((-8.044,2.032),dotstyle); 
label("$B$", (-7.956,2.252), NE * labelscalefactor); 
dot((9.49,0.118),dotstyle); 
label("$C$", (9.578,0.338), NE * labelscalefactor); 
dot((-3.276627373405626,4.3479567859894),dotstyle); 
label("$I$", (-3.182,4.562), NE * labelscalefactor); 
dot((-3.5825967298465926,1.5449961298577837),dotstyle); 
label("$D$", (-3.49,1.768), NE * labelscalefactor); 
dot((-5.669096776867536,5.840034478471021),dotstyle); 
label("$F$", (-5.58,6.058), NE * labelscalefactor); 
dot((-1.8483924116576476,6.7790780460139635),dotstyle); 
label("$E$", (-1.752,7.004), NE * labelscalefactor); 
dot((-9.342,5.134),dotstyle); 
label("$H$", (-9.254,5.354), NE * labelscalefactor); 
dot((7.862,9.6),dotstyle); 
label("$I_{1}$", (7.95,9.534), NE * labelscalefactor); 
dot((-7.004925504045343,5.58325092567291),dotstyle); 
label("$X$", (-6.922,5.794), NE * labelscalefactor); 
dot((7.047687174717885,9.363437683167886),dotstyle); 
label("$Y$", (7.136,9.578), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

I) denote $\angle{AIF}=A, \angle{IAE}=A , \angle{IBD}=\angle{IBE}=B , \angle{ICD}=\angle{ICE}=C$

now we notice that $\angle{BIC}=180^{\circ}-(B+C) \implies \angle{BIX}=B+C$

now in $\triangle{AIB}=180^{\circ}-(B+A)$ , so we get $\angle{IXB}=90^{\circ}-A$ , also we have $\angle{AFE}=90^{\circ}-A=\angle{BFX}$ , hence $X,F,I,B$ are concyclic points, hence $\angle{BXI}=90^{\circ}$

similarly, we can prove that points $E,I,C,Y$ are concyclic points , hence $\angle{IYC}=90^{\circ}$ hence we have points $B,X,Y,C$ concyclic $\blacksquare$

II) we notice that $\angle{XBI}=A$ and we also quadrilateral $XBDI$ to be cyclic quadrilateral hence we have $\angle{XDI}=A$ and similarly $\angle{XDY}=A$ , by using the fact that quadrilateral $XBYC$ is also cyclic , it is not hard to observe that $\angle{IXD}=\angle{IXY}$ hence we have $I$ as the incenter of triangle $DXY$ $\blacksquare$
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Om245
163 posts
#37
Y by
Amazing problem , we have to prove that BXFID is cyclic with BI diameter and CDIEY is cyclic with CI diameter,
if we think about this configuration as its also come in 1989 INMO as P6
DI is radical axis and BC is radical axis of BCXY
so D is radical center........
This post has been edited 1 time. Last edited by Om245, Aug 22, 2023, 9:26 AM
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Jishnu4414l
154 posts
#38
Y by
Part 1: This is just Iran Lemma. $\angle BXC=\angle CYB=90^{\circ}$.
Part 2: Note that points $I,D,C,X,E$ are concyclic, So $\angle ICD=\angle IXD$ and $\angle ICE=\angle IXE$
$\implies$ $XI$ is the angle bisector of $\angle YXD$. Similarly chase angles for the other side and we are done.
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SilverBlaze_SY
66 posts
#39 • 1 Y
Y by Rounak_iitr
Let $P= AB \cap CX$, $Q= AC \cap BY$. As $I$ is the incentre of $\Delta ABC$, $\angle ACI= \angle BCI = c$ (say). $\angle CBI = \angle ABI=b$ (say).
$\therefore \angle A=2a$, $\angle B=2b$, $\angle C=2c$.
Then, by angle chasing in $\Delta BQC$, we get that $\angle BQC= 90^{\circ}+a-c$. As $AE$ and $AF$ are tangents to the incircle of $\Delta ABC$, we get that $\angle AFE= \angle AEF= 90^{\circ}-a$. $\angle AEF = \angle YEC$ (vertically opposite). Thus, by angle chasing in $\Delta YEQ$, we get that $\angle EYB= \angle EYQ= c \Rightarrow BCYX$ is cyclic. Part (i) proved.

Then, we have that $\angle CBI= \angle CXY=b$, as $BCYX$ is cyclic.
Also, $\angle EYI = \angle ECI= c \Rightarrow EICY$ is cyclic $\Rightarrow \angle IEC = \angle IYC = 90^{\circ}$. Then, $\angle IYC + \angle IDC = 180^{\circ} \Rightarrow IDCY$ is cyclic $\Rightarrow \angle DCI = \angle DYI=c$. Thus, $YI$ bisects $\angle XYD$. Similarly, $BDIX$ is cyclic and $XI$ bisects $\angle DXY$. Then $I$ is the incentre of $\Delta DYX$. Proved :)
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