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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)!

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[*]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]
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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
Parallel lines..
ts0_9   9
N 4 minutes ago by OutKast
Source: Kazakhstan National Olympiad 2014 P3 D1 10 grade
The triangle $ABC$ is inscribed in a circle $w_1$. Inscribed in a triangle circle touchs the sides $BC$ in a point $N$. $w_2$ — the circle inscribed in a segment $BAC$ circle of $w_1$, and passing through a point $N$. Let points $O$ and $J$ — the centers of circles $w_2$ and an extra inscribed circle (touching side $BC$) respectively. Prove, that lines $AO$ and $JN$ are parallel.
9 replies
ts0_9
Mar 26, 2014
OutKast
4 minutes ago
KMN and PQR are tangent at a fixed point
hal9v4ik   4
N 5 minutes ago by OutKast
Let $ABCD$ be cyclic quadrilateral. Let $AC$ and $BD$ intersect at $R$, and let $AB$ and $CD$ intersect at $K$. Let $M$ and $N$ are points on $AB$ and $CD$ such that $\frac{AM}{MB}=\frac{CN}{ND}$. Let $P$ and $Q$ be the intersections of $MN$ with the diagonals of $ABCD$. Prove that circumcircles of triangles $KMN$ and $PQR$ are tangent at a fixed point.
4 replies
hal9v4ik
Mar 19, 2013
OutKast
5 minutes ago
one cyclic formed by two cyclic
CrazyInMath   40
N 14 minutes ago by HamstPan38825
Source: EGMO 2025/3
Let $ABC$ be an acute triangle. Points $B, D, E$, and $C$ lie on a line in this order and satisfy $BD = DE = EC$. Let $M$ and $N$ be the midpoints of $AD$ and $AE$, respectively. Suppose triangle $ADE$ is acute, and let $H$ be its orthocentre. Points $P$ and $Q$ lie on lines $BM$ and $CN$, respectively, such that $D, H, M,$ and $P$ are concyclic and pairwise different, and $E, H, N,$ and $Q$ are concyclic and pairwise different. Prove that $P, Q, N,$ and $M$ are concyclic.
40 replies
CrazyInMath
Apr 13, 2025
HamstPan38825
14 minutes ago
geometry problem
Medjl   5
N 21 minutes ago by LeYohan
Source: Netherlands TST for IMO 2017 day 3 problem 1
A circle $\omega$ with diameter $AK$ is given. The point $M$ lies in the interior of the circle, but not on $AK$. The line $AM$ intersects $\omega$ in $A$ and $Q$. The tangent to $\omega$ at $Q$ intersects the line through $M$ perpendicular to $AK$, at $P$. The point $L$ lies on $\omega$, and is such that $PL$ is tangent to $\omega$ and $L\neq Q$.
Show that $K, L$, and $M$ are collinear.
5 replies
Medjl
Feb 1, 2018
LeYohan
21 minutes ago
Connected, not n-colourable graph
mavropnevma   7
N 31 minutes ago by OutKast
Source: Tuymaada 2013, Day 1, Problem 4 Juniors and 3 Seniors
The vertices of a connected graph cannot be coloured with less than $n+1$ colours (so that adjacent vertices have different colours).
Prove that $\dfrac{n(n-1)}{2}$ edges can be removed from the graph so that it remains connected.

V. Dolnikov

EDIT. It is confirmed by the official solution that the graph is tacitly assumed to be finite.
7 replies
mavropnevma
Jul 20, 2013
OutKast
31 minutes ago
Homothety with incenter and circumcenters
Ikeronalio   8
N 34 minutes ago by LeYohan
Source: Korea National Olympiad 2009 Problem 1
Let $I, O$ be the incenter and the circumcenter of triangle $ABC$, and $D,E,F$ be the circumcenters of triangle $ BIC, CIA, AIB$. Let $ P, Q, R$ be the midpoints of segments $ DI, EI, FI $. Prove that the circumcenter of triangle $PQR $, $M$, is the midpoint of segment $IO$.
8 replies
Ikeronalio
Sep 9, 2012
LeYohan
34 minutes ago
2-var inequality
sqing   11
N an hour ago by ytChen
Source: Own
Let $ a,b>0 , a^2+b^2-ab\leq 1 . $ Prove that
$$a^3+b^3 -\frac{a^4}{b+1}  -\frac{b^4}{a+1} \leq 1 $$
11 replies
sqing
May 27, 2025
ytChen
an hour ago
Sums of products of entries in a matrix
Stear14   0
an hour ago
(a) $\ $Each entry of an $\ 8\times 8\ $ matrix equals either $\ 1\ $ or $\ 2.\ $ Let $\ A\ $ denote the sum of eight products of entries in each row. Also, let $\ B\ $ denote the sum of eight products of entries in each column. Find the maximum possible value of $\ A-B.\ $ In other words, find
$$ {\rm max}\ \left[ \sum_{i=1}^8\ \prod_{j=1}^8\ a_{ij} - 
\sum_{j=1}^8\ \prod_{i=1}^8\ a_{ij} \right]
$$
(b) $\ $Same question, but for a $\ 2025\times 2025\ $ matrix.
0 replies
Stear14
an hour ago
0 replies
a father and his son are skating around a circular skating rink
parmenides51   2
N an hour ago by thespacebar1729
Source: Tournament Of Towns Spring 1999 Junior 0 Level p1
A father and his son are skating around a circular skating rink. From time to time, the father overtakes the son. After the son starts skating in the opposite direction, they begin to meet five times more often. What is the ratio of the skating speeds of the father and the son?

(Tairova)
2 replies
parmenides51
May 7, 2020
thespacebar1729
an hour ago
Sums of n mod k
EthanWYX2009   1
N an hour ago by Martin.s
Source: 2025 May 谜之竞赛-3
Given $0<\varepsilon <1.$ Show that there exists a constant $c>0,$ such that for all positive integer $n,$
\[\sum_{k\le n^{\varepsilon}}(n\text{ mod } k)>cn^{2\varepsilon}.\]Proposed by Cheng Jiang
1 reply
EthanWYX2009
May 26, 2025
Martin.s
an hour ago
Easy P4 combi game with nt flavour
Maths_VC   1
N 4 hours ago by p.lazarov06
Source: Serbia JBMO TST 2025, Problem 4
Two players, Alice and Bob, play the following game, taking turns. In the beginning, the number $1$ is written on the board. A move consists of adding either $1$, $2$ or $3$ to the number written on the board, but only if the chosen number is coprime with the current number (for example, if the current number is $10$, then in a move a player can't choose the number $2$, but he can choose either $1$ or $3$). The player who first writes a perfect square on the board loses. Prove that one of the players has a winning strategy and determine who wins in the game.
1 reply
Maths_VC
May 27, 2025
p.lazarov06
4 hours ago
Central sequences
EeEeRUT   14
N 4 hours ago by HamstPan38825
Source: EGMO 2025 P2
An infinite increasing sequence $a_1 < a_2 < a_3 < \cdots$ of positive integers is called central if for every positive integer $n$ , the arithmetic mean of the first $a_n$ terms of the sequence is equal to $a_n$.

Show that there exists an infinite sequence $b_1, b_2, b_3, \dots$ of positive integers such that for every central sequence $a_1, a_2, a_3, \dots, $ there are infinitely many positive integers $n$ with $a_n = b_n$.
14 replies
EeEeRUT
Apr 16, 2025
HamstPan38825
4 hours ago
Elementary Problems Compilation
Saucepan_man02   32
N 6 hours ago by atdaotlohbh
Could anyone send some elementary problems, which have tricky and short elegant methods to solve?

For example like this one:
Solve over reals: $$a^2 + b^2 + c^2 + d^2  -ab-bc-cd-d +2/5=0$$
32 replies
Saucepan_man02
May 26, 2025
atdaotlohbh
6 hours ago
Random Points = Problem
kingu   5
N 6 hours ago by happypi31415
Source: Chinese Geometry Handout
Let $ABC$ be a triangle. Let $\omega$ be a circle passing through $B$ intersecting $AB$ at $D$ and $BC$ at $F$. Let $G$ be the intersection of $AF$ and $\omega$. Further, let $M$ and $N$ be the intersections of $FD$ and $DG$ with the tangent to $(ABC)$ at $A$. Now, let $L$ be the second intersection of $MC$ and $(ABC)$. Then, prove that $M$ , $L$ , $D$ , $E$ and $N$ are concyclic.
5 replies
kingu
Apr 27, 2024
happypi31415
6 hours ago
Prove concyclic and tangency
syk0526   41
N Apr 1, 2025 by endless_abyss
Source: Japan Olympiad Finals 2014, #4
Let $ \Gamma $ be the circumcircle of triangle $ABC$, and let $l$ be the tangent line of $\Gamma $ passing $A$. Let $ D, E $ be the points each on side $AB, AC$ such that $ BD : DA= AE : EC $. Line $ DE $ meets $\Gamma $ at points $ F, G $. The line parallel to $AC$ passing $ D $ meets $l$ at $H$, the line parallel to $AB$ passing $E$ meets $l$ at $I$. Prove that there exists a circle passing four points $ F, G, H, I $ and tangent to line $ BC$.
41 replies
syk0526
May 17, 2014
endless_abyss
Apr 1, 2025
Prove concyclic and tangency
G H J
G H BBookmark kLocked kLocked NReply
Source: Japan Olympiad Finals 2014, #4
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syk0526
202 posts
#1 • 7 Y
Y by itslumi, KrazyGuy, mathleticguyyy, Adventure10, Mango247, ItsBesi, Rounak_iitr
Let $ \Gamma $ be the circumcircle of triangle $ABC$, and let $l$ be the tangent line of $\Gamma $ passing $A$. Let $ D, E $ be the points each on side $AB, AC$ such that $ BD : DA= AE : EC $. Line $ DE $ meets $\Gamma $ at points $ F, G $. The line parallel to $AC$ passing $ D $ meets $l$ at $H$, the line parallel to $AB$ passing $E$ meets $l$ at $I$. Prove that there exists a circle passing four points $ F, G, H, I $ and tangent to line $ BC$.
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nima1376
111 posts
#2 • 4 Y
Y by mathematiculperson, Mathematicsislovely, Adventure10, Mango247
let parallel line to $AC$ passing $D$ meet $BC$ at $X$.
$\frac{BD}{AD}=\frac{BX}{CX}=\frac{AE}{EC}\Rightarrow XE\left | \right |AB,$
$E,X,I$ in same line.
$\widehat{HAB}=\widehat{ACB}=\widehat{HXB}\Rightarrow$ $AXBH$ is cycle$\Rightarrow$ $HD.DX=AD.BD=DF.DG\Rightarrow$ $HFXG$ is cycle.similar we find $FXGI$is cycle.$\Rightarrow$ $HFXGI$ is cycle($o$).
$\widehat{AIX}=\widehat{ACB}=\widehat{HXB}$ $\Rightarrow$ $o$ is tangent to $BC$ at $X$.
done
Z K Y
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saturzo
55 posts
#3 • 3 Y
Y by mathematiculperson, Adventure10, Mango247
Let $HD \cap BC = P$. So, $\frac{BP}{PC}=\frac{BD}{DA}=\frac{AE}{EC} \implies EP \parallel AB \implies P \in EI$.
Now, $\measuredangle CPI = \measuredangle CBA = \measuredangle IAC \implies EI.EP = EA.EC = EF.EG \implies I \in \odot PFG$. Similarly, $H \in \odot PFG$.
$\therefore FGHIP$ cyclic. Now, $\measuredangle CPI = \measuredangle PBA = \measuredangle PHI \implies CP, i.e., BC$ touches $\odot FGHI$ at $P$.
Z K Y
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sayantanchakraborty
505 posts
#4 • 2 Y
Y by mathematiculperson, Adventure10
Let $DX$ be the line parallel to $AC$ with $X$ on $BC$.Then $\frac{AE}{EC}=\frac{DB}{AD}=\frac{BX}{XC} \implies EX \parallel AB$.Thus the parallels meet at the same point on $BC$.Now we see that $\angle{HXB}=C$ and since $HI$ is the tangent at $A$,$\angle{HAB}=C$.Thus points $A,H,B,X$ ar concyclic.So $FD \cdot DG=AD \cdot DB=DH \cdot DX$ or points $F,H,G,X$ concyclic.Similarly points $G,I,F,X$ are concyclic.Thus $FHIGX$ is a cyclic pentagon.Now note that $\angle{HIX}=\angle{HAB}=C=\angle{HXB}$ or $\odot{HFXGI}$ is tangent to $BC$ at $X$,as desired.
Z K Y
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62861
3564 posts
#5 • 2 Y
Y by aopsuser305, Adventure10
not really adding anything new, but here it is.

Let $X$ be the point on $\overline{BC}$ satisfying $\tfrac{BD}{DA} = \tfrac{AE}{EC} = \tfrac{BX}{XC}$. Then $\overline{XD} \parallel \overline{AC}$ so $X$ lies on line $HD$; likewise $X$ lies on line $IE$.

Now from
\[\measuredangle HXB
= \measuredangle ACB
= \measuredangle HAB\]we deduce $HAXB$ is cyclic, so from $DH \cdot DX = DA \cdot DB = DF \cdot DG$ we obtain that $HFXG$ is cyclic. Likewise $IFXG$ is cyclic.

To finish, we need to show that $(HIX)$ is tangent to $\overline{BC}$: this follows as
\[\measuredangle HIX
= \measuredangle HAB
= \measuredangle HXB.\]
Z K Y
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Kimchiks926
256 posts
#6 • 1 Y
Y by mkomisarova
Nice problem. The key step is to prove that $HD$ and $EI$ intersects on $BC$.
Let $X$ be intersection of $HD$ and $EI$. It is obvious that $ADXE$ is parallelogram. Therefore $\frac{BD}{DA}=\frac{AE}{EC}=\frac{BD}{XE}=\frac{DX}{EC}$. We conclude that $\triangle BDX \sim \triangle XEC$. As a result $\angle DXB = \angle ECX$. Note that $\angle EXC=180 -\angle A -\angle DXB$ and $\angle A = \angle DXE$. Therefore $\angle BXC = 180$ and it means that $X$ lies on $BC$.
The rest is easy. Note that $\angle HAB=\angle ACB= \angle HXB$. This implies that $HAXB$ is cyclic. As a result $AD \cdot DB=HD \cdot DX$, but $AD \cdot DB=FD \cdot DG$, therefore $HD \cdot DX=FD \cdot DG$. As a result $HFXG$ is cyclic. Similarly we get that $IGXF$ is cyclic. We conclude that $HFXGI$ is cyclic.
We claim that the circumcircle of $HFXGI$ is tangent to $BC$ at $X$. This is obvious, since $\angle BXH=\angle ACB=\angle HIX $.
This post has been edited 1 time. Last edited by Kimchiks926, Apr 4, 2020, 8:07 PM
Reason: typo
Z K Y
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v_Enhance
6882 posts
#7 • 7 Y
Y by itslumi, srijonrick, Mathematicsislovely, v4913, HamstPan38825, Mango247, MS_asdfgzxcvb
Solution from OTIS office hours:

We start with the following critical definition: Define $X$ on $BC$ such that \[ \frac{BD}{DA} = \frac{BX}{XC} = \frac{AE}{EC}. \]Then $ADXE$ is obviously a parallelogram, so that $X = \overline{HD} \cap \overline{IE}$.
[asy] pair A = dir(110); pair B = dir(210); pair C = dir(330); filldraw(unitcircle, invisible, blue); draw(A--B--C--cycle, blue);
real r = 0.7; pair D = r*B+(1-r)*A; pair E = r*A+(1-r)*C; pair X = r*B+(1-r)*C;
pair H = extension(X, D, A, dir(90)*A+A); pair I = extension(X, E, A, dir(90)*A+A);
draw(H--I--X--cycle, red); filldraw(circumcircle(H, I, X), invisible, red);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(210)); dot("$E$", E, dir(45)); dot("$X$", X, dir(270)); dot("$H$", H, dir(H)); dot("$I$", I, dir(I));
/* TSQ Source:
A = dir 110 B = dir 210 C = dir 330 unitcircle 0.1 lightcyan / blue A--B--C--cycle blue
! real r = 0.7; D = r*B+(1-r)*A R210 E = r*A+(1-r)*C R45 X = r*B+(1-r)*C R270
H = extension X D A dir(90)*A+A I = extension X E A dir(90)*A+A
H--I--X--cycle red circumcircle H I X 0.1 lightred / red
*/ [/asy]

Claim: The circumcircle of $\triangle HIX$ is tangent to $X$ at $BC$.
Proof. Because $\measuredangle HIX = \measuredangle HAB = \measuredangle ACB = \measuredangle HXB$. $\blacksquare$

Claim: Points $D$ and $E$ lie on the radical axis of $(HIX)$ and $(ABC)$.
Proof. We saw already $\measuredangle HAB = \measuredangle ACB = \measuredangle HXB$ so quadrilateral $HAXB$ is cyclic. Then $D$ is the radical center of $(HIX)$, $(ABC)$, $(HAXB)$. $\blacksquare$

Thus $F$ and $G$ are the two intersections of the circles, as needed.
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niyu
830 posts
#8 • 1 Y
Y by Mango247
Let $M, N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$, let $E' \neq E$ be on $\overline{AC}$ such that $NE = NE'$, and let $T = \overline{HD} \cap \overline{EI}$. The ratio condition implies that $\overline{DE'} \parallel \overline{BC}$; in particular, this implies that $\overline{DE}$ is bisected by $\overline{MN}$. As $ADTE$ is a parallelogram, it follows that $\overline{AT}$ is bisected by $\overline{MN}$ as well, implying that $T$ lies on $\overline{BC}$.

Now, we have
\begin{align*}
	\angle HAD = \angle HAB = \angle BCA = \angle BTH,
\end{align*}implying that $HATB$ is cyclic. By Power of a Point, we now have
\begin{align*}
	DH \cdot DT = DA \cdot DB = DF \cdot DG,
\end{align*}implying that $FTGH$ is cyclic. Similarly, $FTGI$ is cyclic, so $FGHIT$ is cyclic. To finish, we have
\begin{align*}
	\angle HIT = \angle HAD = \angle BTH,
\end{align*}implying that $\overline{BC}$ is tangent to $(FGHIT)$. We are done! $\Box$
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mathlogician
1051 posts
#9
Y by
Let lines $\overline{HD}$ and $\overline{BC}$ meet at a point $T$. Now note that $TC/TB = DA/DB = EC/EA$, so $T$ lies on $\overline{IE}$. Note that $\angle HAB = \angle ACB = \angle DTB = \angle HTB$, so $AHBT$ is cyclic. Similarly, $AICT$ is cyclic. Now $HD \cdot TD = AD \cdot BD = FD \cdot GD$, which implies that $HFTG$ is cyclic. By symmetry $IGTF$ is also cyclic. Now $\angle HTB = \angle HAB = \angle HIT$, so $\overline{BTC}$ is tangent to $(HIGTF)$, as desired.
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jj_ca888
2726 posts
#10
Y by
Heavy 2020 APMO 1 vibes.

Let $X_1$ be the point on $BC$ such that $DX_1 \parallel AC$ and $X_2$ be the point on $BC$ such that $EX_1 \parallel AB$. We see that\[\frac{BX_1}{CX_1} = \frac{BD}{DA} = \frac{AE}{EC} = \frac{BX_2}{CX_2}\]hence $X_1 = X_2$ are the same point. Let this point be $X$, on $BC$ such that $ADXE$ is a parallelogram. Since\[\angle HAB = \angle ACB = \angle HXB \text{    and    } \angle IAC = \angle ABC = \angle IXC\]we see that $HAXB$ and $IAXC$ are cyclic. Now through PoP we get that\[FD \cdot GD = BD \cdot AD = HD \cdot XD\]\[FE \cdot GE = AE \cdot CE = IE \cdot XE\]thus $HFXG$ and $IGXF$ are cyclic and thus $HFXGI$ are all cyclic. Indeed, since $\angle BXH = \angle BCA = \angle XIH$, this circle is tangent to $BC$, as desired. $\blacksquare$
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IndoMathXdZ
694 posts
#11
Y by
Wonderful problem!

Claim. $DH \cap EI \in BC$.
Proof. To prove this, we use phantom point. Suppose that $X \in BC$ such that $DX \parallel AC$ and $E' \in AC$ such that $XE' \parallel AB$. We will prove that $E \equiv E'$. Now, notice that we have $\triangle BDX \sim \triangle XE'C$ and $AEX'D$ being a parallelogram. Therefore, we must have
\[ \frac{BD}{DX} = \frac{XE'}{E'C} \Rightarrow \frac{BD}{AE'} = \frac{AD}{E'C} \Rightarrow \frac{BD}{AD} = \frac{AE'}{E'C} \]However, $E \in AC$ such that $\frac{AE}{EC} = \frac{BD}{AD} = \frac{AE'}{E'C}$, which forces $E = E'$.
Now, notice that $(HIX)$ tangent to $BC$ at $X$ since $\measuredangle HIX = \measuredangle HXB$. It suffices to prove that $F, G \in (HIX)$.
Claim. $AICX$ and $AXBH$ is cyclic.
Proof. Simple angle chasing yields $\measuredangle AIX = \measuredangle HAB = \measuredangle ACB \equiv \measuredangle ACX$. Similarly, we have $\measuredangle AHX = \measuredangle IAC = \measuredangle ABC \equiv \measuredangle ABX$.

Claim. $DE$ is the radical axis of $(HXI)$ and $(ABC)$.
Proof. Radical axis theorem on $(HIX), (ABC), (AHBX)$ gives us $D = AB \cap HX$ lies on radical axis of $(HXI)$ and $(ABC)$. Similarly, radical axis theorem on $(HIX), (ABC), (AICX)$ gives us $E = AC \cap IX$ lies on radical axis of $(HIX)$ and $(ABC)$. Therefore, $DE$ is the radical axis of $(HXI)$ and $(ABC)$.
Now, notice that $(HIX)$ must intersect $(ABC)$ at two points, both of which lies in its radical axis. However, we know that $DE \cap \Gamma = F,G$, and therefore, $F,G \in (HIX)$ as well.
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Spacesam
596 posts
#12
Y by
First interpret the length conditions as $P$ lying on $\overline{BC}$, which can be done with some simple ratio manipulations to get $\frac{BD}{PD} = \frac{PE}{CE}$, which is $\triangle BDP \sim \triangle PEC$.

We continue with angle chase: $$\measuredangle HAB = \measuredangle ACB = \measuredangle HPB,$$implying $HABP$ cyclic. Similarly, $AICP$ is cyclic. Then, $$\angle HIP = \angle AIP = \angle ACP = \angle HPB,$$thus $(HIP)$ is tangent to $\overline{BC}$ at $P$. Next, redefine $F'$ and $G'$ to be $(PHI) \cap (ABC)$. However, from $HABP$ cyclic, we get $HD \cdot PD = AD \cdot BD$, aka $D$ lies on the radical axis of $(PHI)$ and $(ABC)$, and similarly for $E$. Thus $\{F', G'\} = \{F, G\}$ and we are done.
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rcorreaa
238 posts
#13
Y by
Let $P= EI \cap BC$. Since $EP \parallel AB \implies \frac{CP}{BP}= \frac{CE}{EA}= \frac{AD}{BD} \implies PD \parallel AC \implies P$, $D$ and $H$ are collinear.

Now, observe that since $AI$ is tangent to $\Gamma$ and $AC \parallel PH$, $\angle ABC= \angle CAI= \angle PHA \implies AHBP$ is cyclic $\implies DH.DP=DA.DB=DF.DG \implies HFPG$ is cyclic. Similarly, $IGPF$ is cyclic $\implies HFGI$ is cyclic. Moreover, $\angle IPC= \angle ABC= \angle PHI \implies$ since $P \in (HFGI) \implies BC$ is tangent to $HFGI$ at $P$, as desired.

$\blacksquare$
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amuthup
779 posts
#14 • 3 Y
Y by Mango247, Mango247, Mango247
Let the line through $D$ parallel to $\overline{AC}$ intersect $\overline{BC}$ at $X.$ The given ratio condition implies that $\overline{EX}\parallel\overline{AB}.$

$\textbf{Key Claim: }$ $B,H,A,X$ are concyclic.

$\emph{Proof: }$ Since $\overline{HA}$ is tangent to $\Gamma,$ we have $\angle BAH = \angle BCA = \angle BXH.$ Thus, quadrilateral $BHAX$ is cyclic, as desired. $\blacksquare$

Now, by power of a point, we have $$\text{Pow}_{(XHI)}D=DH\cdot DX=DA\cdot DB=\text{Pow}_{\Gamma}D.$$Therefore, $D$ lies on the radical axis of $(XHI)$ and $\Gamma.$ We can similarly show that $E$ lies on this radical axis, so pentagon $FGHIX$ is cyclic.

Finally, note that $\angle IXC = \angle ABX = \angle AHX=\angle IXH.$ Hence, $(FGHIX)$ is tangent to $\overline{BC}$ at $X,$ as needed.
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bora_olmez
277 posts
#15
Y by
Very cool problem - same as the solutions above, yet, I may or may not have written out the PoP rather than quoting Radical Axes.
Solution
This post has been edited 1 time. Last edited by bora_olmez, Aug 16, 2021, 1:43 PM
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hakN
429 posts
#16
Y by
Because of the given ratio condition, $HD$ and $EI$ meet on $BC$, call it $J$.

By angle chase, $HAJB$ and $AICJ$ are cyclic.

By PoP through $D$, $DF \cdot DG = DA \cdot DB = DH  \cdot DJ$, so $HFJG$ is cyclic.

By PoP through $E$, $EG \cdot EF = EA \cdot EC = EI \cdot EJ$, so $IGJF$ is cyclic.

So, $HFJGI$ is cyclic, and $\angle DJB = \angle ACB = \angle AIJ$, hence $(HFGJI)$ is tangent to $BC$ at $J$. We are done.
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554183
484 posts
#17
Y by
[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 = -13.795602650594365, xmax = 13.801198427350894, ymin = -7.525540163790313, ymax = 7.536835997072933;  /* image dimensions */
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label("$E$", (-0.3131960170943919,0.7692998548668937), NE * labelscalefactor); 
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Claim : $HD \cap IE=J \in BC$
Proof. Let $HD \cap BC=J_1, IE \cap BC = J_2$. We have
$$\frac{BJ_1}{J_1C} =\frac{BD}{DA}$$and $$\frac{CJ_2}{J_2B}=\frac{CE}{EA}$$Therefore $$\frac{BJ_1}{J_1C} \times \frac{CJ_2}{J_2B}=\frac{BD}{DA} \times \frac{AD}{DB}=1 \implies \frac{BJ_1}{J_1C}=\frac{BJ_2}{J_2C}$$Finally, $J_1 \equiv J_2$, as desired.

Back to the main problem, Note that $AHBJ, AJCI$ is cyclic. Now, by PoP, see that
$$DF \times DG =DA \times DB= DH \times DJ$$Therefore $FGJH$ is cyclic. Applying PoP is a very similar way, we see that $GIFJ$ is cyclic. This implies that $FGJHI$ is cyclic, proving the first part.
For the second part, note that
\begin{align*} 
\angle{HJB} &= \angle{ACB} \\
&=\angle{JIA}
\end{align*}As desired $\blacksquare$
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L567
1184 posts
#18 • 1 Y
Y by starchan
Let $X$ be on $BC$ such that $DX || AC$, by the length condition, we have $EX || AB$ too, so $ADXE$ is a parallelogram and $H,D,X$ and $I,E,X$ are collinear.

Note that $(HIX)$ is tangent to $BC$ since $\angle HIX = \angle AIE = 180 - \angle IAE - \angle IEA = 180 - \angle ABC - \angle BAC = \angle ACB = \angle HXB$

Now, note that $AHXB$ and $AICX$ are cyclic and so by PoP, we get $AD.DB = XD.DH$ and $AE.EC = EX.EI$ which means $D,E$ lie on the radical axis of $(HIX)$ and $(ABC)$, so $F,G \in (HIX)$ too

So, there does indeed exist a circle passing through $F,G,H,I$, and tangent to $BC$ at $X$, as desired. $\blacksquare$
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mathleticguyyy
3217 posts
#19
Y by
Illegally cute

Denote the given ratio as $r$, and let $X$ be the point on $BC$ such that $\frac{BX}{XC}=r$. We claim that $X$ is the desired tangency point. Note that triangles $BXD,XCE$ are similar, which gives that $X=IE\cap HD$.

Claim: $(AIX)$ is tangent to $BC$.

Proof: Because $AC\parallel HX$, we have $\angle EXC=\angle B=\angle IAC=\angle IHX$.

Now, by since $\triangle XEC\sim\triangle AEI$ we have $EA\cdot EC=EI\cdot EX$ and similar for $D$, hence $\overline{DE}$ is the radical axis of $(ABC),(AIX)$ and it follows that $H,G$ are precisely $(AIX)\cap (ABC)$.
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Math-48
44 posts
#20
Y by
Let $J=IE\cap BC$ so we have :
$BD : DA=AE : EC=BJ : CJ$
and so $H,D,J$ are collinear
note that $\angle{HAB}=\angle{ACB}=\angle{HJB}$
$\implies AHBJ$ is cyclic
$\implies DH.DJ=DA.DB=DF.DG$
$\implies FHGJ$ is cyclic
and similarly $FIGJ$ is cyclic
and so $FHIGJ$ is cyclic and we have :
$\angle{IHJ}=\angle{ABC}=\angle{IJC}$
So $(FHIGJ)$ is tangent to $BC$ .
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AwesomeYRY
579 posts
#21
Y by
Let $X$ be the point on $BC$ such that $\frac{BX}{XC} = \frac{BD}{DA}=\frac{AE}{EC}$. Then, clearly, $X,D,H$ and $X,E,I$ are collinear.

Now, note that $\angle HAB = \angle ACB = \angle HXB$, so $(AXBH)$ is cyclic. Thus,
\[Pow_{(XHI)}(D) = DX\cdot DH = DB\cdot DA = Pow_{(ABC)}(D)\]Similarly, $Pow_{(XHI)}(E) = Pow_{(ABC)}(E)$. Thus, $DE$ is the radical axis of $(XHI)$ and $(ABC)$, so $F,G = DE\cap (ABC)$ also lie on $(XHI)$ and we are done. $\blacksquare$.
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MathsLion
113 posts
#22 • 1 Y
Y by Johnweak
I am really surprised how no one realized that condition $ BD: DA= AE: EC $ is completely redundant. The reason for this is probably that the ratio condition strongly motivates you to introduce the key point from all previous solutions. In fact, the problem is true for arbitrary points $D, E$. Unfortunately, this makes this problem much less cute.
Let $FG \cap HI =J$. By power of a point it suffices to prove $JI\cdot JH=JG\cdot JF$ or $JI\cdot JH=JA^2$. This is true because from $AE||DH$ we have $\frac{JA}{JH}=\frac{JE}{JD}$ and from $IE||AD$ we have $\frac{JE}{JD}=\frac{JI}{JA}$. Now $\frac{JA}{JH}=\frac{JI}{JA}$ yields desired conclusion.
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David_Kim_0202
384 posts
#23
Y by
In this picture, if we grab point $A$ that is the intersection of $IF$ and $HE$.
Let's show that K is the tangent point of circle $FIHG$ and $BC$. First because of the condition of the problem $ BD : DA= AE : EC $ and if we use Chebas theorem at triangle $ABC$ we get $IF$ and $EH$ meet at $K$ that is on the line $BC$. And if we look at parallel lines and the angles we can earn $IABK$ and $AHCK$ are all concyclic. Also if you use the power at point $E, F$ you will get $IFKG, HFKG$ are concyclic. So overall we get $IFKGH$ are concyclic. And because triangle $IKH$ is tangent to $BC$(trivial) we get circle $IFKGH$ is tangent to $BC$ at point $K$.
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signifance
140 posts
#24
Y by
I hope this works, it's written up sloppily and I'm not sure if i ever used circular reasoning/forgot to say something. Note that most results are analogous if omitted.


Define J as HD\cap BC; we claim that this is the desired tangency point. Indeed, note that \frac{CE}{CA}=\frac{AD}{AB}=\frac{CJ}{CB}\Rightarrow EJ\parallel AB, so JEI is a line. Then HAB=ACB=DJB and analogous results gives HD\cdot DJ=BD\cdot DA=AE\cdot EC=IE\cdot EJ, so DE is the radax of (HIJ) and (ABC), whence if F and G lie on (ABC) they also lie on (HIJ); used in tandem with the fact that HJB=ACJ=HIJ, BC is indeed tangent to (FHIGJ) at J, as desired.
This post has been edited 1 time. Last edited by signifance, Sep 22, 2023, 8:56 PM
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IAmTheHazard
5005 posts
#25
Y by
It is clear (by phantom points) that the length condition is simply equivalent to the parallel to $\overline{AC}$ through $D$ and the parallel to $\overline{AB}$ through $E$ intersecting on $\overline{BC}$ at some point $X$. I claim that the desired tangency point is (unsurprisingly) just $X$.

Observe that $\measuredangle HAB=\measuredangle ACB=\measuredangle DXB$, hence $HABX$ and $IACX$ are cyclic. Furthermore, since $\measuredangle HAB=\measuredangle HIX$, it follows that $(HIX)$ is tangent to $\overline{BC}$.

Now, since $DH\cdot DX=DA\cdot DB$ and $EI\cdot EX=EA\cdot EC$, $D$ and $E$ lie on the radical axis of $(HIX)$ and $(ABC)$, hence $F$ and $G$ are the intersections of the circles, so we're done. $\blacksquare$
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math_comb01
662 posts
#26
Y by
Very easy problem[asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(13cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
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draw((1.5065826571655647,-2.116548967617623)--(4.133906450997055,-0.47265540409717827), linewidth(0.4)); 
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draw((-1.7197432167156428,0.5222543975122591)--(7.74,-3.45), linewidth(0.4)); 
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label("$C$", (7.830014159002851,-3.4054448922611367), NE * labelscalefactor); 
label("$D$", (3.119853549920445,-3.3651871092775263), NE * labelscalefactor); 
label("$E$", (-1.6305648421455705,0.5599467316244714), NE * labelscalefactor); 
label("$F$", (6.360605080101075,5.270107340706868), NE * labelscalefactor); 
label("$G$", (-0.02025352280115828,-0.7685601068346664), NE * labelscalefactor); 
label("$H$", (4.206813690477923,-0.42636895147397935), NE * labelscalefactor); 
label("$I$", (-1.1072136633586365,-0.8490756728018868), NE * labelscalefactor); 
label("$J$", (7.769627484527436,-0.14456447058870772), NE * labelscalefactor); 
label("$K$", (-8.11206790250683,-3.2444137603266956), NE * labelscalefactor); 
label("$O$", (3.4419158137893278,-1.8555202473921426), NE * labelscalefactor); 
label("$Q_A$", (1.5900577965432539,-2.076938053801999), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */[/asy] Labels in this soloution are different but it is clear from diagram.
Claim 1: $(AEBD),(AFCD)$ are cyclic
Proof
CLaim 2: $EIJD,FIJD$ is cyclic
Proof
Claim 3: $(EJIFD)$ is tangent at $D$ to $BC$.
Proof
This post has been edited 1 time. Last edited by math_comb01, Nov 18, 2023, 8:35 AM
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naonaoaz
334 posts
#27
Y by
Let $K = \overline{HD} \cap \overline{IE}$.
Claim: $B,K,C$ are collinear.
Proof: Let $\overline{HD}$ hit $\overline{BC}$ at $D'$ and define $E'$ similarly. It suffices to prove $D' = E'$. Since $\overline{IE} \parallel \overline{AB}$,
\[\frac{CE'}{CB} = \frac{CE}{CA} = \frac{1}{1+\frac{AE}{CE}} = \frac{1}{1+\frac{BD}{DA}} = \frac{DA}{AB}\]However, since $\overline{HD} \parallel \overline{AC}$, $\frac{DA}{AB} = \frac{CD'}{CB}$. Combining these two gives $D' = E'$. $\square$.

Claim: $HFKG$ and $IGKF$ are cyclic.
Proof: By symmetry, it suffices to prove $HFKG$ cyclic. Notice that
\[\angle HAB = \angle ACB = \angle HKB\]so $HAKB$ cyclic. Therefore,
\[HD \cdot DK = AD \cdot DB = FD \cdot DG\]which concludes. $\square$

The above claim directly implies that $FGHI$ is cyclic. Also, $(FGHI)$ must be tangent to $\overline{BC}$ at $K$ since
\[\angle IHK = \angle AHK = \angle ABC = \angle IKC\]
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OronSH
1748 posts
#28 • 2 Y
Y by megarnie, Spectator
In triangle $APQ,$ let $B,C$ be on sides $AP,AQ$ respectively with $PB/BA=AC/CQ.$ Let $BC$ intersect $(APQ)$ at $M,N.$ Let $X$ be on the same side of $AC$ as $B$ such that $XA\parallel BC$ and $\angle AXP=\angle BAC.$ Similarly, let $Y$ be on the same side of $AB$ as $C$ such that $YA\parallel BC,\angle AYQ=\angle BAC.$ Let $D$ be the $A$ dumpty point of $\triangle APQ.$

Claim. $AXPD,AYQD$ are cyclic.
Proof. By properties of the dumpty point we have $\angle PXA=\angle PAQ=\angle PAD+\angle DAQ=\angle PAD+\angle APD=180-\angle PDA.$

Claim. $ABDC$ is cyclic.
Proof. The spiral similarity at $D$ taking $PA$ to $AQ$ takes $B$ to $C$ by the length ratio. Thus $180-\angle ABD=\angle PBD=\angle ACD.$

Claim. $D,B,X$ and $D,C,Y$ are collinear.
Proof. The spiral similarity gives $\angle PDB=\angle ADC=\angle ABC=\angle XAP=\angle XDP,$ so $D,B,X$ are collinear. The other side is similar.

Claim. $MDNYX$ is cyclic.
Proof. By PoP $BD\cdot BX=BA\cdot BP=BM\cdot BN$ so $X$ lies on $(MDN).$ Similarly $Y$ lies on $(MDN)$ as well.

Claim. $(ABC)$ is tangent to $(XMNY).$
Proof. Since $BC\parallel XY$ there is a homothety at $D$ taking $BC$ to $XY,$ which takes $(BDC)=(ABC)$ to $(XDY)=(XMNY).$

To finish, simply invert the diagram about $A.$

inversion :love: :love:
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Spectator
657 posts
#29 • 1 Y
Y by megarnie
Let $X$ be $HD\cap BC$ and $X'$ be $EI\cap BC$. Note that $\angle{HAD} = \angle{C} = \angle{DXB} \implies HABX$ is cyclic. Also, $\angle{DBX} = \angle{A}$ implying that $\triangle DBX \sim \triangle DHA\sim \triangle ABC$. We can see from symmetry that
\[\triangle{DBX}\sim\triangle{EAI}\sim\triangle{EXC}\sim\triangle{DHA}\sim\triangle{ABC}\]Notice that
\[BD\cdot EC = AD\cdot AE \]and that
\[\frac{AE}{HD} = \frac{IE}{AD} \implies BD\cdot EC = HD\cdot IE \implies \triangle{HDB}\sim \triangle{CEI}\]Let $\angle{BDH} = \angle{IGE} = \alpha$. Notice how, $\angle{AXC} = \angle{B}+\alpha = \angle{AX'C}$ which means that $X = X'$. We can see that $ADXE$ is a parallelogram because $EX \parallel AD$ and $AE\parallel DX$ and that $\triangle{ABC} \sim \triangle{XHI}$.

Consider the circumcircle of $\triangle{HIX}$. Because $\angle{HIX} = \angle{C}$, $(HIX)$ is tangent to $BC$. Notice that $POW_{(ABC)}(D) = AD\cdot BD = HD\cdot DX = POW_{(HIX)}(D)$. The same can be said for $E$. Thus, $DE$ is the radical axis of $(HIX)$ and $(ABC)$ implying that $FG$ lies on $(HIX)$ ending the proof.
This post has been edited 3 times. Last edited by Spectator, Feb 5, 2024, 2:10 AM
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dolphinday
1329 posts
#30
Y by
Construct point $X$ on $AB$ so that $AX$, $BE$ and $CD$ all concur. Then note that $BD : DA = AE : EC = CX : BX$, by Ceva. From this, we also find that $AEXD$ is a parallelogram, by angle chasing and similarity. From angle chasing, we have $\angle HIX = \angle HAD = \angle C = \angle DXB \implies BC$ is a tangent to $(IXH)$. Note that $(AHBX)$ is cyclic, so the radical axises of $(HIX)$, $(AHBX)$ and $(ABC)$ concur at $D$. This implies that $F, G = (HIX) \cap (ABC)$ which finishes.
This post has been edited 1 time. Last edited by dolphinday, Feb 12, 2024, 8:44 PM
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HamstPan38825
8872 posts
#31
Y by
Surprisingly standard. Let $J$ be the point on $\overline{BC}$ such that $ADJE$ is a parallelogram, which exists by the ratio condition. Then, as $\angle AHJ =\angle ABJ$, $AHBJ$ is concyclic, hence $$DH \cdot DJ = DA \cdot DB = DF \cdot DG,$$i.e. $HFGJ$ is concyclic. Similarly, $IFGJ$ is concyclic, so $FHIGJ$ is concyclic and tangent to $\overline{BC}$ at $J$ as $\angle HJB = \angle C = \angle HIJ$.
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alsk
28 posts
#32
Y by
More radax usage.

Let $P = HD \cap IE$. By the length ratios given in the problem, we must have $P \in BC$.

Now, notice that $\angle HIP = \angle HAB = \angle C = \angle HPB$, so $(HIP)$ is tangent to $BC$ at $P$.

Then, Notice that we have $H, A, P, B$ cyclic by $\angle HAB = \angle HPB$, so we have $\text{Pow}_{(HIP)}(D) = DH \cdot DP = dB \cdot DA = \text{Pow}_{(ABC)}(D)$, so $D$ lies on the radical axis of $(HIP), (ABC)$. Using a similar argument for $E$, we can deduce that $DE$ is the radical axis of $(HIP), (ABC)$. Since $(ABC)$ intersects $DE$ at $F, G$, it follows that both $F, G$ lie on $(HIP)$.

Thus, we have that $(HIFG)$ is tangent to $BC$ at $P$, done.
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Markas
150 posts
#33
Y by
Let $\angle BAC = \alpha$ and $\angle ABC = \beta$. Let P be a point on BC, where $\frac{BD}{DA} = \frac{BP}{PC}$ $\Rightarrow$ from Tales $DP \parallel AC$ $\Rightarrow$ H, D, P lie on one line, also from $\frac{BD}{DA} = \frac{AE}{EC}$ it follows that $\frac{AE}{EC} = \frac{BP}{PC}$ $\Rightarrow$ $EP \parallel AB$ $\Rightarrow$ I, E, P lie on one line.

We will now want to prove that HFPGI is cyclic and that BC is tangent to this circle in point P.

Since AI is tangent to the circumcircle of $\triangle ABC$, we have that $\angle IAC = \angle ABC = \beta$. Since $HP \parallel AC$, we have that $\angle IAC = \angle IHP = \beta$, also since $EP \parallel AB$, we have $\angle EPC = \angle ABC = \beta$. Now we have that $\angle PBA = \angle PHA = \beta$ $\Rightarrow$ PBHA is cyclic $\Rightarrow$ PD.DH = AD.DB, but since AFBG is cyclic, we have that AD.DB = FD.DG $\Rightarrow$ PD.DH = FD.DG $\Rightarrow$ FPGH is cyclic.

Also $\angle CPI = \angle CAI = \beta$ $\Rightarrow$ CPAI is cyclic $\Rightarrow$ PE.EI = AE.EC, but since AFCG is cyclic, we have that AE.EC = FE.EG $\Rightarrow$ PE.EI = FE.EG $\Rightarrow$ FPGI is cyclic.

We have that FPGH is cyclic and that FPGI is cyclic $\Rightarrow$ F, P, G, H, I lie on one circle $\Rightarrow$ the only thing which is left to show is that BC is tangent to this circle in P, which is equivalent to showing that BC is tangent to the circumcircle of $\triangle HPI$ at point P. Since $\angle CPI = \angle PHI = \beta$, than this is true $\Rightarrow$ we proved that there exists a circle passing through points F, G, H, I, and tangent to line BC $\Rightarrow$ we are ready.
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shendrew7
799 posts
#34
Y by
The length condition forces $K := DH \cap EI$ to lie on $BC$. Notice that $(HIK)$ is tangent to $BC$ at $K$, as
\[\measuredangle BKH = \measuredangle BCA = \measuredangle BAH = \measuredangle HIK.\]
To show that $F$, $G$ lie on this circle, first note that $BHAK$ and $CIAK$ are cyclic since
\[\measuredangle ABK = \measuredangle IAC = \measuredangle AHK, \quad \measuredangle KCA = \measuredangle BAH = \measuredangle KIA.\]
As a result, Power of a Point tells us line $DE$ is the radical axis of $(ABC)$, $(HIK)$, as desired. $\blacksquare$
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L13832
268 posts
#35 • 1 Y
Y by alexanderhamilton124
INMO Mock and boredom :play_ball:
Storage
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Eka01
204 posts
#36
Y by
sketch
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Saucepan_man02
1361 posts
#37
Y by
Phantom Points :yup:

Define $X$ to be a point on $BC$ such that: $\frac{CX}{XB} = \frac{CE}{EA}, \frac{BX}{XC} = \frac{BD}{DA}$ (which exists due to given condition).
Note that: $DX \parallel AC, EX \parallel AB$ and $(H, D, X), (X, E, I)$ are collinear. Therefore: $\triangle ABC \sim \triangle XDE$. Thus: $DE \parallel HI$.
Notice that: $$\angle EXC = \angle ABC = \angle RDX = \angle IHX$$which implies $(HIX)$ is tangent to $BC$ at $X$. Since $AHBX$ is cyclic, it implies $FG$ is the radical axis of $(HIX), (ABC)$. For any two intersecting circles, the radical axis passes through the two intersection points. Thus, we must have $F, G \in (HIX)$, and we are done.
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ehuseyinyigit
840 posts
#38
Y by
Let the line passing through $D$ parallel to $AC$ meet $BC$ at $X$. Since $\dfrac{BD}{DA}=\dfrac{BX}{XC}=\dfrac{AE}{EC}$ we have that $EX\parallel AB$. Thus, we have $\overline{H-D-X}$ and $\overline{E-I-X}$. On the other hand
$$\angle XIA=\pi-\angle BAC-\angle CEI=\pi-\angle BAC-\angle CBA=\angle BCA$$implying the points $A$, $X$, $C$ and $I$ are concyclic. Similarly $X\in (ABH)$. Thus a quick PoP gives
$$HD\cdot DX=AD\cdot DB=FD\cdot DG$$So points $H$, $X$, $F$ and $G$ are concyclic. Similarly
$$XE\cdot EI=AE\cdot EC=FE\cdot EG$$implies $I$, $X$, $F$ and $G$ are concyclic which finishes as the points $F$, $G$, $H$, $I$ and $X$ all lie on a common circle as desired. Tangency is easy to obtain.
This post has been edited 2 times. Last edited by ehuseyinyigit, Dec 15, 2024, 8:34 PM
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cherry265
16 posts
#39
Y by
Sol sketch: show that $HD, BC, IE$ concur at a point $X$ and angle chase gives $HAXB$ and $AICX$ cyclic. Conjecture $X$ is tangency point and from here it’s just details with power of a point and angle chase.
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ItsBesi
147 posts
#40
Y by
Here is my solution
Attachments:
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Ilikeminecraft
674 posts
#41
Y by
construct parallelogram $ADRE.$
Clearly, $R\in BC.$
Furthermore, $\angle HIR = \angle HAB = \angle ACB = \angle HRB,$ so $(HRI)$ is tangent to $BC$ AND $HARB$ cyclic
Thus, $HD\cdot HR = AD\cdot AB = FD\cdot DG$
Thus, $F$ lies $(HRI).$
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endless_abyss
69 posts
#42
Y by
2015/4 Canada is similar! :)
https://aops.com/community/p4756480
Although in this case the construction of the parallelogram is much more natural.
Claim - $H D$, $E I$, $B C$ concur
Let $X$ be the intersection of $H D$ and $I E$,
we have $D B X$ similar to $E X C$ through the ratio conditions
and
$\angle D X E = a$,
$\angle D X B = c$,
$\angle E X C = b$,

sums up to 180, so $B - X - C$ are collinear.
Claim - $H X B A$ and $A C X I$ are concyclic
We use alternate segment theorem,
$\angle B A H = \angle B C A = \angle H X B$
and similarly, for $A C X I$
Claim - $H F X E I$ lie on a circle, and $X$ is the point of tangency
We prove that $H F X G$ and $I E X F$ are concyclic using Power of Point.

$(A D) (B D) = (H D) (D X) = (F D) (D G) $
$(A E) (E C) = (G E) (E F) = (X E) (E I) $

Next, simply note that $X$ is the tangency point because of alternate segment theorem -
$\angle I X C = \angle I H X = \angle A B C$

$\square$
Spamming PoP is so satisfying :P
:starwars:
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