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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.

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[*]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
combi/nt
blug   0
6 minutes ago
Prove that every positive integer $n$ can be written in the form
$$n=2^{a_1}3^{b_1}+2^{a_2}3^{b_2}+..., $$where $a_i, b_j$ are non negative integers, such that
$$2^x3^y\nmid 2^z3^t$$for every $x, y, z, t$.
0 replies
blug
6 minutes ago
0 replies
Interesting inequalities
sqing   2
N 13 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0 , a(b+c)=k.$ Prove that
$$\frac{1}{a+1}+\frac{2}{b+1}+\frac{1}{c+1}\geq  \frac{4\sqrt{k}-6}{ k-2}$$Where $5\leq  k\in N^+.$
Let $ a,b,c\geq 0 , a(b+c)=9.$ Prove that
$$\frac{1}{a+1}+\frac{2}{b+1}+\frac{1}{c+1}\geq \frac{6}{7}$$
2 replies
1 viewing
sqing
2 hours ago
sqing
13 minutes ago
Find smallest value of (x^2 + y^2 + z^2)/(xyz)
orl   7
N 22 minutes ago by Bryan0224
Source: CWMO 2001, Problem 4
Let $ x, y, z$ be real numbers such that $ x + y + z \geq xyz$. Find the smallest possible value of $ \frac {x^2 + y^2 + z^2}{xyz}$.
7 replies
orl
Dec 27, 2008
Bryan0224
22 minutes ago
easy substitutions for a functional in reals
Circumcircle   9
N 44 minutes ago by Bardia7003
Source: Kosovo Math Olympiad 2025, Grade 11, Problem 2
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ with the property that for every real numbers $x$ and $y$ it holds that
$$f(x+yf(x+y))=f(x)+f(xy)+y^2.$$
9 replies
Circumcircle
Nov 16, 2024
Bardia7003
44 minutes ago
writing words around circle, two letters
jasperE3   1
N an hour ago by pi_quadrat_sechstel
Source: VJIMC 2000 2.2
If we write the sequence $\text{AAABABBB}$ along the perimeter of a circle, then every word of the length $3$ consisting of letters $A$ and $B$ (i.e. $\text{AAA}$, $\text{AAB}$, $\text{ABA}$, $\text{BAB}$, $\text{ABB}$, $\text{BBB}$, $\text{BBA}$, $\text{BAA}$) occurs exactly once on the perimeter. Decide whether it is possible to write a sequence of letters from a $k$-element alphabet along the perimeter of a circle in such a way that every word of the length $l$ (i.e. an ordered $l$-tuple of letters) occurs exactly once on the perimeter.
1 reply
jasperE3
Jul 27, 2021
pi_quadrat_sechstel
an hour ago
Interesting inequality
imnotgoodatmathsorry   0
an hour ago
Let $x,y,z > \frac{1}{2}$ and $x+y+z=3$.Prove that:
$\sqrt{x^3+y^3+3xy-1}+\sqrt{y^3+z^3+3yz-1}+\sqrt{z^3+x^3+3zx-1}+\frac{1}{4}(x+5)(y+5)(z+5) \le 60$
0 replies
imnotgoodatmathsorry
an hour ago
0 replies
Arithmetic Sequence of Products
GrantStar   19
N an hour ago by OronSH
Source: IMO Shortlist 2023 N4
Let $a_1, \dots, a_n, b_1, \dots, b_n$ be $2n$ positive integers such that the $n+1$ products
\[a_1 a_2 a_3 \cdots a_n, b_1 a_2 a_3 \cdots a_n, b_1 b_2 a_3 \cdots a_n, \dots, b_1 b_2 b_3 \cdots b_n\]form a strictly increasing arithmetic progression in that order. Determine the smallest possible integer that could be the common difference of such an arithmetic progression.
19 replies
GrantStar
Jul 17, 2024
OronSH
an hour ago
Inequality Involving Complex Numbers with Modulus Less Than 1
tom-nowy   0
an hour ago
Let $x,y,z$ be complex numbers such that $|x|<1, |y|<1,$ and $|z|<1$.
Prove that $$ |x+y+z|^2 +3>|xy+yz+zx|^2+3|xyz|^2 .$$
0 replies
tom-nowy
an hour ago
0 replies
Inequality
nguyentlauv   2
N an hour ago by nguyentlauv
Source: Own
Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3$ and $k\ge 0$, prove that
$$\frac{\sqrt{a+1}}{b+c+k}+\frac{\sqrt{b+1}}{c+a+k}+\frac{\sqrt{c+1}}{a+b+k} \geq \frac{3\sqrt{2}}{k+2}.$$
2 replies
nguyentlauv
May 6, 2025
nguyentlauv
an hour ago
japan 2021 mo
parkjungmin   0
an hour ago

The square box question

Is there anyone who can release it
0 replies
parkjungmin
an hour ago
0 replies
easy sequence
Seungjun_Lee   17
N an hour ago by GreekIdiot
Source: KMO 2023 P1
A sequence of positive reals $\{ a_n \}$ is defined below. $$a_0 = 1, a_1 = 3, a_{n+2} = \frac{a_{n+1}^2+2}{a_n}$$Show that for all nonnegative integer $n$, $a_n$ is a positive integer.
17 replies
Seungjun_Lee
Nov 4, 2023
GreekIdiot
an hour ago
Japan MO Finals 2023
parkjungmin   0
an hour ago
It's hard. Help me
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parkjungmin
an hour ago
0 replies
I Brazilian TST 2007 - Problem 4
e.lopes   77
N an hour ago by alexanderhamilton124
Source: 2007 Brazil TST, Russia TST, and AIMO; also SL 2006 N5
Find all integer solutions of the equation \[\frac {x^{7} - 1}{x - 1} = y^{5} - 1.\]
77 replies
e.lopes
Mar 11, 2007
alexanderhamilton124
an hour ago
Japan MO Finals 2024
parkjungmin   0
an hour ago
Source: Please tell me the question
Please tell me the question
0 replies
parkjungmin
an hour ago
0 replies
Simple Geometry
rkm0959   8
N Sep 3, 2022 by HamstPan38825
Source: 2017 FKMO Day 1 Problem 1
A acute triangle $\triangle ABC$ has circumcenter $O$. The circumcircle of $OAB$, called $O_1$, and the circumcircle of $OAC$, called $O_2$, meets $BC$ again at $D ( \not=B )$ and $E ( \not= C )$ respectively. The perpendicular bisector of $BC$ hits $AC$ again at $F$. Prove that the circumcenter of $\triangle ADE$ lies on $AC$ if and only if the centers of $O_1, O_2$ and $F$ are colinear.
8 replies
rkm0959
Mar 25, 2017
HamstPan38825
Sep 3, 2022
Simple Geometry
G H J
G H BBookmark kLocked kLocked NReply
Source: 2017 FKMO Day 1 Problem 1
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rkm0959
1721 posts
#1 • 5 Y
Y by Davi-8191, tenplusten, Adventure10, Mango247, Rounak_iitr
A acute triangle $\triangle ABC$ has circumcenter $O$. The circumcircle of $OAB$, called $O_1$, and the circumcircle of $OAC$, called $O_2$, meets $BC$ again at $D ( \not=B )$ and $E ( \not= C )$ respectively. The perpendicular bisector of $BC$ hits $AC$ again at $F$. Prove that the circumcenter of $\triangle ADE$ lies on $AC$ if and only if the centers of $O_1, O_2$ and $F$ are colinear.
Z K Y
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ThE-dArK-lOrD
4071 posts
#2 • 1 Y
Y by Adventure10
The following generalized is true.
rkm0959 wrote:
A acute triangle $\triangle ABC$ has circumcenter $O$. The circumcircle of $OAB$, called $O_1$, and the circumcircle of $OAC$, called $O_2$, meets $BC$ again at $D ( \not=B )$ and $E ( \not= C )$ respectively. Let $T$ be the circumcenter of $\triangle ADE$ and let the perpendicular bisector of $BC$ intersect $O_1O_2$ at $K$. Prove that $A,K,T$ collinear.
My solution uses some painful and lengthy angle chasing.
This post has been edited 1 time. Last edited by ThE-dArK-lOrD, Mar 25, 2017, 12:35 PM
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babu2001
402 posts
#3 • 2 Y
Y by Adventure10, Mango247
Let the angles of $\triangle ABC $ be $A,B,C $. We shall show that both conditions are equivalent to $2B=90^{\circ}+C $.
Suppose centres of $O_1,O_2$ and $F $ are collinear. Since the radical axis of $O_1,O_2$ is $AO $ we conclude that the perpendicular bisectors of $AO,BC $ intersect on $AC $ at $F $. Considering $F $ as the intersection of perpendicular bisector of $AO$ with $AC $, we have $\angle OFC=2\angle OAF=2 (90^{\circ}-B) $. Considering $F $ as the intersection of perpendicular bisector of $BC $ with $AC $, we get $\angle OFC=90^{\circ}-C $. Thus, $2 (90^{\circ}-B)=90^{\circ}-C\implies 2B=90^{\circ}+C $, as required.
Suppose the circumcentre of $\triangle ADE $ lie on $AC $ then $D$ lies on segment $BC $ while $E$ lies on $BC $ extended so $B>A>C $. Now $\angle ADE = A-C+B $ by simple angke chase while $\angle EAC=B-A $. Hence if the circumcentre lies on $AC $, then $\angle ADE+\angle EAC=90^{\circ}\implies  (A-C+B)+(B-A)=90^{\circ}\implies 2B=90^{\circ}+C $, as required.
Hence we are done.
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lminsl
544 posts
#4 • 1 Y
Y by Adventure10
babu2001 wrote:
Let the angles of $\triangle ABC $ be $A,B,C $. We shall show that both conditions are equivalent to $2B=90^{\circ}+C $.
Suppose centres of $O_1,O_2$ and $F $ are collinear. Since the radical axis of $O_1,O_2$ is $AO $ we conclude that the perpendicular bisectors of $AO,BC $ intersect on $AC $ at $F $. Considering $F $ as the intersection of perpendicular bisector of $AO$ with $AC $, we have $\angle OFC=2\angle OAF=2 (90^{\circ}-B) $. Considering $F $ as the intersection of perpendicular bisector of $BC $ with $AC $, we get $\angle OFC=90^{\circ}-C $. Thus, $2 (90^{\circ}-B)=90^{\circ}-C\implies 2B=90^{\circ}+C $, as required.
Suppose the circumcentre of $\triangle ADE $ lie on $AC $ then $D$ lies on segment $BC $ while $E$ lies on $BC $ extended so $B>A>C $. Now $\angle ADE = A-C+B $ by simple angke chase while $\angle EAC=B-A $. Hence if the circumcentre lies on $AC $, then $\angle ADE+\angle EAC=90^{\circ}\implies  (A-C+B)+(B-A)=90^{\circ}\implies 2B=90^{\circ}+C $, as required.
Hence we are done.

Exactly same with me :)
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rkm0959
1721 posts
#5 • 3 Y
Y by vsathiam, Adventure10, Mango247
You can actually just prove both implications, but yes, it is easily equivalent to $2B=90+C$. Same as mine.
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jlammy
1099 posts
#6 • 2 Y
Y by Adventure10, Mango247
We prove ThE-dArK-lOrD's generalisation.
ThE-dArK-lOrD wrote:
A acute triangle $\triangle ABC$ has circumcenter $O$. The circumcircle of $OAB$, called $O_1$, and the circumcircle of $OAC$, called $O_2$, meets $BC$ again at $D ( \not=B )$ and $E ( \not= C )$ respectively. Let $T$ be the circumcenter of $\triangle ADE$ and let the perpendicular bisector of $BC$ intersect $O_1O_2$ at $K$. Prove that $A,K,T$ collinear.
Claim. $O_1,O_2$ are the midpoints of arcs $\widehat{AD},\widehat{AE}$.

Proof. Since $AO=CO$, $EO$ bisects $\angle DEA$. But $O_1A=O_1D$, so $O_1$ must be the arc midpoint; similarly, $O_2$ is an arc midpoint. $\square$

Now we use complex numbers with $(ADE)$ as the unitcircle. Let $a^2,d^2,e^2$ be the complex numbers of $A,D,E$. Then $o_1=-ad$, $o_2=-ae$ and $o=-(ab+bc+ca)$. Redefine $K=AT\cap O_1O_2$; then by the chord intersection formula, $k=\frac{-a^3(b+c)}{a^2+bc}$. Now $$\frac{k-o}{b^2-c^2}=\frac{bc(a+b)(a+c)}{(b+c)(b-c)\left(a^2+bc\right)}=-\frac{\frac{1}{bc}\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{a}+\frac{1}{c}\right)}{\left(\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{a^2}+\frac{1}{bc}\right)}=-\overline{\left(\frac{k-o}{b^2-c^2}\right)},$$so $OK\perp BC$, and the result follows.
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Vrangr
1600 posts
#7 • 1 Y
Y by Adventure10
ThE-dArK-lOrD wrote:
A acute triangle $\triangle ABC$ has circumcenter $O$. The circumcircle of $OAB$, called $O_1$, and the circumcircle of $OAC$, called $O_2$, meets $BC$ again at $D ( \not=B )$ and $E ( \not= C )$ respectively. Let $T$ be the circumcenter of $\triangle ADE$ and let the perpendicular bisector of $BC$ intersect $O_1O_2$ at $K$. Prove that $A,K,T$ collinear.
I'll prove ThE-dArK-lOrD's generalization:
Claim 1: $O$ is the incenter of $\triangle ADE$.
Proof. $AO = BO$, so $O$ is the midpoint of arc $\widehat{AB}$ of $\odot(AOB)$. Thus, $DO$ bisects $\angle ADE$. Similarly $EO$ bisects $\angle ADE$. $\square$

Claim 2: $O_1, O_2$ are the midpoints of arcs $\widehat{AD}$ and $\widehat{AE}$ of $\odot(ADE)$, respectively.
Proof.
\[\measuredangle AOO_1 = 90^{\circ} + \measuredangle OAB = \measuredangle ACB = \measuredangle AOE\]So, $EOO_1$ are collinear. Along with this, $O_1$ lies on the perpendicular bisector of $AD$, so, $O_1$ is the midpoint of arc $\widehat{AD}$. $\square$

[asy]
unitsize(2.5 cm);
import geometry; import olympiad;
pair A = dir(100), B = dir(200), C = -1/B, O = (0, 0);
circle c_1 = circle((point) A,(point)  B,(point)  O), c_2 = circle((point) A, (point) C,(point)  O);
pair D = intersectionpoints(line(B, C), c_1)[1], E = intersectionpoints(line(B, C), c_2)[0];
pair O_1 = circumcenter(A, B, O), O_2 = circumcenter(A, C, O);
pair K = extension(O_1, O_2, (B+C)/2, O);
pair T = circumcenter(A, D, E);

draw(A--B--C--cycle); dot(A^^B^^C); dot(O^^O_1^^O_2); draw(c_1^^c_2, linewidth(0.4)+dashed); dot(D^^E);
draw(D--A--E--cycle, linewidth(1));
dot(K); draw(circumcircle(A, D, E), dashed+linewidth(0.7));
dot(T); draw(O_1--O_2);
draw(A--K--T, linewidth(1.1));
draw(K--(B+C)/2);

label("$A$", A, A);
label("$B$", B, B);
label("$C$", C, C);
label("$D$", D, dir(240));
label("$E$", E, dir(-60));
label("$O$", O, dir(0));
label("$O_1$", O_1, dir(150));
label("$O_2$", O_2, dir(60));
label("$K$", K, dir(150));
label("$T$", T, dir(-60));
[/asy]
So the problem now transforms into:
Quote:
$ABC$ is a triangle with incenter $I$ and circumcenter $O$. $Y, Z$ are the intersection points of $\odot(ABC)$ with $BI$ and $CI$, respectively. Let $K$ be the intersection point of $AO$ with $YZ$.

Prove that $KI \perp BC$.
From here, it's very easy to complex bash it. I'll provide a synthetic solution.
Let $X$ be the intersection of $AI$ and $\odot(ABC)$.
Note that $YZ$ is the perpendicular bisector $AI$. So, $KA = KI$.
Also, clearly, $OA = OX$. Therefore, we have
\[\measuredangle KIA = \measuredangle IAK = \measuredangle XAO = \measuredangle AXO\]Therefore, $IK \parallel OX$. Since, $X$ is the midpoint of arc $\widehat{BC}$, $OX \perp BC$.

Thus, $IK \perp BC$.

[asy]
unitsize(2.5 cm);
import geometry;
pair A = dir(110), B = dir(210), C = -1/B, O = (0, 0), I = incenter(A, B, C);
pair X = circumcenter(I, B, C), Y = circumcenter(I, C, A), Z = circumcenter(I, A, B);
pair K = extension(A, O, Y, Z), D = projection(B, C)*I;

draw(unitcircle); draw(A--B--C--cycle); draw(X--Y--Z--cycle);
dot(A^^B^^C); dot(X^^Y^^Z); dot(O^^I); dot(K);
draw(O--A--X--O); draw(K--D, linewidth(1.2));
markangle(I, A, K, n=2, radius=0.7 cm); markangle(K, I, A, n=2, radius=0.7 cm); markangle(O, X, A, n=2, radius=0.7 cm);
markrightangle(K, D, B); markrightangle(O, (B+C)/2, C);
label("$A$", A, A);
label("$B$", B, B);
label("$C$", C, C);
label("$X$", X, X);
label("$Y$", Y, Y);
label("$Z$", Z, Z);
label("$O$", O, dir(0));
label("$I$", I, -dir(0));
label("$K$", K, dir(-30));
[/asy]
This post has been edited 1 time. Last edited by Vrangr, Sep 11, 2018, 8:35 AM
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mathaddiction
308 posts
#8 • 2 Y
Y by Mango247, Rounak_iitr
[asy]
size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -20.13839182578881, xmax = 24.678453195216232, ymin = -20.864724143538304, ymax = 16.934536603715053;  /* image dimensions */pen zzttff = rgb(0.6,0.2,1); pen qqwuqq = rgb(0,0.39215686274509803,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); draw(arc((3.9048088393944984,3.417712349159264),1.196179137571309,-151.68988080175083,-114.63464561747348)--(3.9048088393944984,3.417712349159264)--cycle, linewidth(2) + qqwuqq); draw(arc((3.9048088393944984,3.417712349159264),1.196179137571309,-88.16226320961216,-51.10702802533482)--(3.9048088393944984,3.417712349159264)--cycle, linewidth(2) + qqwuqq);  /* draw figures */draw(circle((4.1440446669087585,-4.038470941701886), 7.460020311506091), linewidth(0.8) + zzttff); draw(circle((-3.4620030175639,-0.5505855990750161), 8.367634405361127), linewidth(0.8) + qqwuqq); draw((3.9048088393944984,3.417712349159264)--(-1.662239610071297,-8.722375930381483), linewidth(0.8) + fuqqzz); draw((11.24983850229547,-1.766996306786579)--(-1.662239610071297,-8.722375930381483), linewidth(0.8)); draw((11.24983850229547,-1.766996306786579)--(3.9048088393944984,3.417712349159264), linewidth(0.8)); draw(circle((6.839384262002882,-0.22005980400870417), 4.673876244547078), linewidth(0.8) + qqwuqq); draw(circle((1.7195675443454639,-1.347653441134675), 5.24251758536787), linewidth(0.8) + linetype("4 4") + blue); draw((-3.4620030175639,-0.5505855990750161)--(6.839384262002882,-0.22005980400870417), linewidth(0.8)); draw((4.1440446669087585,-4.038470941701886)--(3.9048088393944984,3.417712349159264), linewidth(0.8) + fuqqzz); draw((-3.4620030175639,-0.5505855990750161)--(3.9048088393944984,3.417712349159264), linewidth(0.8)); draw((6.839384262002882,-0.22005980400870417)--(3.9048088393944984,3.417712349159264), linewidth(0.8)); draw((-3.4620030175639,-0.5505855990750161)--(5.704572246887101,-4.754077982147205), linewidth(0.8)); draw((6.839384262002882,-0.22005980400870417)--(2.3715829157948893,-6.549467219835816), linewidth(0.8));  /* dots and labels */dot((4.1440446669087585,-4.038470941701886),dotstyle); label("$O$", (2.987738167256496,-4.3175794071352005), NE * labelscalefactor); dot((3.9048088393944984,3.417712349159264),dotstyle); label("$A$", (3.9845541152325867,4.733509400487703), NE * labelscalefactor); dot((-3.4620030175639,-0.5505855990750161),dotstyle); label("$O_1$", (-5.0665346923903165,-0.21069770147370676), NE * labelscalefactor); dot((-1.662239610071297,-8.722375930381483),linewidth(4pt) + dotstyle); label("$C$", (-2.35519531389535,-9.979493991639396), NE * labelscalefactor); dot((2.16791200013861,-0.369946614549803),linewidth(4pt) + dotstyle); label("$F$", (1.0339789092233584,0.1481560397976859), NE * labelscalefactor); dot((11.24983850229547,-1.766996306786579),linewidth(4pt) + dotstyle); label("$B$", (12.676789181584098,-1.7657305803164083), NE * labelscalefactor); dot((6.839384262002882,-0.22005980400870417),linewidth(4pt) + dotstyle); label("$O_{2}$", (7.493346252108426,-1.0878957356926666), NE * labelscalefactor); dot((2.3715829157948893,-6.549467219835816),linewidth(4pt) + dotstyle); label("$E$", (2.1105401330375364,-7.906116819849127), NE * labelscalefactor); dot((5.704572246887101,-4.754077982147205),linewidth(4pt) + dotstyle); label("$D$", (5.579459631994332,-6.191593389330251), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
[/asy]
CLAIM. $A,O_1,O_2,E,D$ are concyclic.
Proof.
Notice that by angle chasing we easily have $EA=EB$ and $DA=DC$. Hence $E,O,O_2$ and $D,O,O_1$ are collinear.
Hence $\angle O_2O_1D=\frac{1}{2}\angle OO_1A=\angle ACO=\angle CAO=\angle OED=\angle O_2ED$. Meanwhile $\angle O_2O_1D=\angle ACO=90^{\circ}-\angle ABD=\angle O_2AD$ as desired. $\blacksquare$
$\noindent\rule{15.5cm}{0.1pt}$
Now
\begin{align*}
&\text{The circumcenter of }\triangle AED\hspace{1pt} \text{lies on} AC\\
\iff &\angle O_1AC=\angle O_2AO\\
\iff &90^{\circ}-\angle AEB=90^{\circ}-\angle ABO\\
\iff &2B=C+90^{\circ}\\
\iff &\angle AFO=\angle AOC\\
\iff &\triangle FAO\sim \triangle OAC\\
\iff &FA=FO\\
\iff &F \text{lies on the perpendicular bisector of} AO=O_1O_2
\end{align*}as desired.
This post has been edited 1 time. Last edited by mathaddiction, Oct 2, 2020, 6:09 AM
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HamstPan38825
8861 posts
#9
Y by
We show that both conditions are equivalent to $\angle C + 90^\circ = 2\angle B$.

First Restriction. Suppose $F$ lies on $\overline{O_1O_2}$. Then $\overline{O_1F}$ is a perpendicular bisector of $\overline{AO}$, so $AO=OF$. But $$\angle AFO = 180^\circ- (90^\circ-\angle C) = \angle C + 90^\circ$$and $\angle OAC = 90^\circ - B$, so $$\angle C + 90^\circ + 180^\circ - 2\angle B = 180^\circ \iff \angle C + 90^\circ = 2\angle B.$$
Second Restriction. Suppose the circumcenter $O_3$ of $\triangle ADE$ lies on $\overline{AC}$. Then $$\angle O_3AD = \angle A - \angle BAD = \angle A - 180^\circ + \angle B + 2\angle C = \angle C.$$Because $\angle AO_3D = 2\angle AED = 360^\circ - 4\angle B$, $$360^\circ - 4\angle B + 2\angle C = 180^\circ \iff \angle C + 90^\circ + 2\angle B.$$Thus both statements are equivalent to the same statement, and they must be equivalent.
This post has been edited 2 times. Last edited by HamstPan38825, Sep 3, 2022, 2:47 AM
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