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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)!

Be sure to mark your calendars for the following upcoming events:
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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
German-Style System of Equations
Primeniyazidayi   1
N 2 minutes ago by Primeniyazidayi
Source: German MO 2025 11/12 Day 1 P1
Solve the system of equations in $\mathbb{R}$

\begin{align*}
\frac{a}{c} &= b-\sqrt{b}+c \\
\sqrt{\frac{a}{c}} &= \sqrt{b}+1 \\
\sqrt[4]{\frac{a}{c}} &=\sqrt[3]{b}-1
\end{align*}
1 reply
Primeniyazidayi
11 minutes ago
Primeniyazidayi
2 minutes ago
gcd nt from switzerland
AshAuktober   5
N 11 minutes ago by Siddharthmaybe
Source: Swiss 2025 Second Round
Let $a, b$ be positive integers. Prove that the expression
\[\frac{\gcd(a+b,ab)}{\gcd(a,b)}\]is always a positive integer, and determine all possible values it can take.
5 replies
AshAuktober
an hour ago
Siddharthmaybe
11 minutes ago
Shortlist 2017/G1
fastlikearabbit   92
N 30 minutes ago by Ilikeminecraft
Source: Shortlist 2017
Let $ABCDE$ be a convex pentagon such that $AB=BC=CD$, $\angle{EAB}=\angle{BCD}$, and $\angle{EDC}=\angle{CBA}$. Prove that the perpendicular line from $E$ to $BC$ and the line segments $AC$ and $BD$ are concurrent.
92 replies
fastlikearabbit
Jul 10, 2018
Ilikeminecraft
30 minutes ago
set construction nt
top1vien   2
N an hour ago by top1vien
Is there a set of 2025 positive integers $S$ that satisfies: for all different $a,b,c,d\in S$, we have $\gcd(ab+1000,cd+1000)=1$?
2 replies
top1vien
Yesterday at 10:04 AM
top1vien
an hour ago
strange geometry problem
Zavyk09   0
an hour ago
Source: own
Let $ABC$ be a triangle with circumcenter $O$ and internal bisector $AD$. Let $AD$ cuts $(O)$ again at $M$ and $MO$ cuts $(O)$ again at $N$. Point $L$ lie on $AD$ such that $(AD, LM) = -1$. The line pass through $L$ and perpendicular to $AD$ intersects $NC, NB$ at $P, Q$ respectively. Let circumcircle of $\triangle NPQ$ cuts $(O)$ at $G \ne N$. Prove that $\angle AGD = 90^{\circ}$.
0 replies
Zavyk09
an hour ago
0 replies
A sharp one with 3 var (3)
mihaig   3
N an hour ago by JARP091
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$a^2+b^2+c^2+5abc\geq8.$$
3 replies
mihaig
Yesterday at 5:17 PM
JARP091
an hour ago
Dophantine equation
MENELAUSS   2
N an hour ago by Assassino9931
Solve for $x;y \in \mathbb{Z}$ the following equation :
$$3^x-8^y =2xy+1 $$
2 replies
MENELAUSS
Yesterday at 11:35 PM
Assassino9931
an hour ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   11
N an hour ago by JARP091
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
11 replies
OgnjenTesic
May 22, 2025
JARP091
an hour ago
Shortest number theory you might've seen in your life
AlperenINAN   9
N an hour ago by Assassino9931
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.
9 replies
AlperenINAN
May 11, 2025
Assassino9931
an hour ago
Inequality about number of spanning trees of graph
CBMaster   0
2 hours ago
Let \( k(G) \) be the number of spanning trees in a graph \( G \), where \( G \) may have multiple edges and loops.

For two edges \( e \) and \( f \) of \( G \), let \( G/e \), \( G/f \), and \( G/\{e,f\} \) denote the graphs obtained by contracting the edges \( e \), \( f \), and both \( e \) and \( f \) in $G$, respectively.

Find a combinatorial proof of the following inequality:
\[
k(G/\{e,f\}) \cdot k(G) \leq k(G/e) \cdot k(G/f)
\]
0 replies
CBMaster
2 hours ago
0 replies
1,2,...,2011 around circle such that 8 of 25 successive multiples of 5 and/or 7
parmenides51   1
N 2 hours ago by ririgggg
Source: 2011 Belarus TST 2.1
Is it possible to arrange the numbers $1,2,...,2011$ over the circle in some order so that among any $25$ successive numbers at least $8$ numbers are multiplies of $5$ or $7$ (or both $5$ and $7$) ?

I. Gorodnin
1 reply
parmenides51
Nov 8, 2020
ririgggg
2 hours ago
Sipnayan JHS 2021 F-9
PikaVee   1
N 2 hours ago by PikaVee
Matt and Sai are playing a game of darts together. Matt has a slightly more accurate aim than Sai. In
fact, Matt can hit the bullseye 80% of the time while Sai can only hit it 60% of the time. They take turns
in playing and the first player is determined by a flip of a fair coin. If the probability that Sai scores the
first bullseye is given by $ \frac {a}{b} $ where a and b are relatively prime integers, what is b − a?
1 reply
PikaVee
2 hours ago
PikaVee
2 hours ago
Standart looking FE
Kimchiks926   13
N 2 hours ago by math-olympiad-clown
Source: Baltic Way 2022, Problem 5
Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(0)+1=f(1)$ and for any real numbers $x$ and $y$,
$$ f(xy-x)+f(x+f(y))=yf(x)+3 $$
13 replies
1 viewing
Kimchiks926
Nov 12, 2022
math-olympiad-clown
2 hours ago
A sharp one with 3 var (2)
mihaig   4
N 2 hours ago by mihaig
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$a+b+c+\sqrt{abc}\geq4.$$
4 replies
mihaig
May 26, 2025
mihaig
2 hours ago
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.
Z K Y
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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 :)
Z K Y
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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
8868 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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