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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 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
Nice Collinearity
oVlad   10
N 15 minutes ago by Trenod
Source: KöMaL A. 831
In triangle $ABC$ let $F$ denote the midpoint of side $BC$. Let the circle passing through point $A$ and tangent to side $BC$ at point $F$ intersect sides $AB$ and $AC$ at points $M$ and $N$, respectively. Let the line segments $CM$ and $BN$ intersect in point $X$. Let $P$ be the second point of intersection of the circumcircles of triangles $BMX$ and $CNX$. Prove that points $A, F$ and $P$ are collinear.

Proposed by Imolay András, Budapest
10 replies
+1 w
oVlad
Oct 11, 2022
Trenod
15 minutes ago
cute geo
Royal_mhyasd   1
N 23 minutes ago by Royal_mhyasd
Source: own(?)
Let $\triangle ABC$ be an acute triangle and $I$ it's incenter. Let $A'$, $B'$ and $C'$ be the projections of $I$ onto $BC$, $AC$ and $AB$ respectively. $BC \cap B'C' = \{K\}$ and $Y$ is the projection of $A'$ onto $KI$. Let $M$ be the middle of the arc $BC$ not containing $A$ and $T$ the second intersection of $A'M$ and the circumcircle of $ABC$. If $N$ is the midpoint of $AI$, $TY \cap IA' = \{P\}$, $BN \cap PC' = \{D\}$ and $CN \cap PB' =\{E\}$, prove that $NEPD$ is cyclic.
PS i'm not sure if this problem is actually original so if it isn't someone please tell me so i can change the source (if that's possible)
1 reply
+1 w
Royal_mhyasd
3 hours ago
Royal_mhyasd
23 minutes ago
Hardest in ARO 2008
discredit   29
N 24 minutes ago by JARP091
Source: ARO 2008, Problem 11.8
In a chess tournament $ 2n+3$ players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least $ n$ next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
29 replies
discredit
Jun 11, 2008
JARP091
24 minutes ago
Macedonian Mathematical Olympiad 2019 problem 1
Lukaluce   5
N 28 minutes ago by AylyGayypow009
In an acute-angled triangle $ABC$, point $M$ is the midpoint of side $BC$ and the centers of the $M$- excircles of triangles $AMB$ and $AMC$ are $D$ and $E$, respectively. The circumcircle of triangle $ABD$ intersects line $BC$ at points $B$ and $F$. The circumcircle of triangle $ACE$ intersects line $BC$ at points $C$ and $G$. Prove that $BF\hspace{0.25mm} = \hspace{0.25mm} CG$ .
5 replies
Lukaluce
Apr 20, 2019
AylyGayypow009
28 minutes ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   9
N 29 minutes 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
9 replies
OgnjenTesic
May 22, 2025
JARP091
29 minutes ago
DE is tangent to a fixed circle whose radius is half the radius of (O)
parmenides51   1
N 29 minutes ago by TigerOnion
Source: 2017 Saudi Arabia JBMO Training Tests 2
Let $ABC$ be a triangle inscribed in circle $(O)$ such that points $B, C$ are fixed, while $A$ moves on major arc $BC$ of $(O)$. The tangents through $B$ and $C$ to $(O)$ intersect at $P$. The circle with diameter $OP$ intersects $AC$ and $AB$ at $D$ and $E$, respectively. Prove that $DE$ is tangent to a fixed circle whose radius is half the radius of $(O)$.
1 reply
parmenides51
May 28, 2020
TigerOnion
29 minutes ago
Orthocorrespondent of P on Euler line
Luis González   1
N 44 minutes ago by AuroralMoss
Let $O,G$ and $K$ be the circumcenter, centroid and symmedian point of $\triangle ABC,$ respectively. $P$ is an arbitrary point on Euler line $OG.$ Show that the orthocorrespondent of $P$ WRT $\triangle {ABC}$ falls on $GK.$
1 reply
Luis González
Feb 8, 2025
AuroralMoss
44 minutes ago
JBMO Shortlist 2023 N3
Orestis_Lignos   9
N an hour ago by Just1
Source: JBMO Shortlist 2023, N3
Let $A$ be a subset of $\{2,3, \ldots, 28 \}$ such that if $a \in A$, then the residue obtained when we divide $a^2$ by $29$ also belongs to $A$.

Find the minimum possible value of $|A|$.
9 replies
Orestis_Lignos
Jun 28, 2024
Just1
an hour ago
polygon's area doesn't add much when combined with its centric symmetry
mathematics2003   7
N an hour ago by sttsmet
Source: 2021ChinaTST test4 day2 P2
Find the smallest real $\alpha$, such that for any convex polygon $P$ with area $1$, there exist a point $M$ in the plane, such that the area of convex hull of $P\cup Q$ is at most $\alpha$, where $Q$ denotes the image of $P$ under central symmetry with respect to $M$.
7 replies
mathematics2003
Apr 14, 2021
sttsmet
an hour ago
a^2-bc square implies 2a+b+c composite
v_Enhance   41
N an hour ago by cursed_tangent1434
Source: ELMO 2009, Problem 1
Let $a,b,c$ be positive integers such that $a^2 - bc$ is a square. Prove that $2a + b + c$ is not prime.

Evan o'Dorney
41 replies
v_Enhance
Dec 31, 2012
cursed_tangent1434
an hour ago
IMO Shortlist 2008, Geometry problem 2
April   42
N an hour ago by s27_SaparbekovUmar
Source: IMO Shortlist 2008, Geometry problem 2, German TST 2, P1, 2009
Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE = \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$.

Proposed by Charles Leytem, Luxembourg
42 replies
April
Jul 9, 2009
s27_SaparbekovUmar
an hour ago
3^n + 61 is a square
VideoCake   25
N an hour ago by endless_abyss
Source: 2025 German MO, Round 4, Grade 11/12, P6
Determine all positive integers \(n\) such that \(3^n + 61\) is the square of an integer.
25 replies
VideoCake
Monday at 5:14 PM
endless_abyss
an hour ago
Problem 5
blug   3
N an hour ago by Jt.-.
Source: Czech-Polish-Slovak Junior Match 2025 Problem 5
For every integer $n\geq 1$ prove that
$$\frac{1}{n+1}-\frac{2}{n+2}+\frac{3}{n+3}-\frac{4}{n+4}+...+\frac{2n-1}{3n-1}>\frac{1}{3}.$$
3 replies
blug
May 19, 2025
Jt.-.
an hour ago
Inequality with xy+yz+zx=1
Kimchiks926   14
N 2 hours ago by math-olympiad-clown
Source: Baltic Way 2022, Problem 4
The positive real numbers $x,y,z$ satisfy $xy+yz+zx=1$. Prove that:
$$ 2(x^2+y^2+z^2)+\frac{4}{3}\bigg (\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\bigg) \ge 5 $$
14 replies
Kimchiks926
Nov 12, 2022
math-olympiad-clown
2 hours ago
perpendicularity involving ex and incenter
Erken   20
N May 7, 2025 by Baimukh
Source: Kazakhstan NO 2008 problem 2
Suppose that $ B_1$ is the midpoint of the arc $ AC$, containing $ B$, in the circumcircle of $ \triangle ABC$, and let $ I_b$ be the $ B$-excircle's center. Assume that the external angle bisector of $ \angle ABC$ intersects $ AC$ at $ B_2$. Prove that $ B_2I$ is perpendicular to $ B_1I_B$, where $ I$ is the incenter of $ \triangle ABC$.
20 replies
Erken
Dec 24, 2008
Baimukh
May 7, 2025
perpendicularity involving ex and incenter
G H J
Source: Kazakhstan NO 2008 problem 2
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Erken
1363 posts
#1 • 2 Y
Y by Adventure10, PikaPika999
Suppose that $ B_1$ is the midpoint of the arc $ AC$, containing $ B$, in the circumcircle of $ \triangle ABC$, and let $ I_b$ be the $ B$-excircle's center. Assume that the external angle bisector of $ \angle ABC$ intersects $ AC$ at $ B_2$. Prove that $ B_2I$ is perpendicular to $ B_1I_B$, where $ I$ is the incenter of $ \triangle ABC$.
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pohoatza
1145 posts
#2 • 6 Y
Y by jam10307, Titusir, Adventure10, Mango247, Cavas, PikaPika999
Let $ I_{a}$ ,$ I_{c}$ be the $ A$, $ C$-excenters, respectively. It is clear that $ B$, $ B_{1}$ and $ B_{2}$ are collinear; therefore, the perpendicularity of $ B_{2}I$ and $ B_{1}I_{b}$ is equivalent with the fact that $ I$ is the orthocenter of triangle $ I_{b}B_{1}B_{2}$. Thus, it is suffice to show that $ IB \cdot BI_{b} = BB_{1} \cdot BB_{2}$ (the power of $ I$ wrt. the circumcircle of $ I_{b}B_{1}B_{2}$). But, on the other hand, we know that $ I$ is the orthocenter of $ I_{a}I_{b}I_{c}$ and so $ IB \cdot BI_{b} = BI_{a} \cdot BI_{c}$. In this case, the problem reduces to proving that $ BB_{1} \cdot BB_{2} = BI_{a} \cdot BI_{c}$. But this is just a consequence of $ (B_{2}, I_{a}, B, I_{c}) = - 1$ and $ B_{1}I_{a} = B_{1}I_{c}$ (since the circumcircle of $ ABC$ is the nine-point center of $ I_{a}I_{b}I_{c}$).
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yetti
2643 posts
#3 • 10 Y
Y by futurestar, Bee-sal, NZQR, myh2910, starchan, CT17, Adventure10, Mango247, dxd29070501, PikaPika999
$ BI$ cuts the circumcircle $ (O)$ of $ \triangle ABC$ again at $ Y$ and $ (Y)$ is a circle with center $ Y$ and radius $ YA = YC = YI = YI_b.$ $ B_2I_b$ cuts $ (Y)$ again at $ Q.$ $ \overline{B_2Q} \cdot \overline{B_2I_b} = \overline{B_2A} \cdot \overline{B_2C} = \overline{B_2B} \cdot \overline{B_2B_1}$ $ \Longrightarrow$ $ BB_1I_bQ$ is cyclic and the angle $ \angle B_1QI_b = \angle B_1BI_b$ is right. Since $ II_b$ is a diameter of $ (Y),$ $ Q \in (Y)$ and $ B_1Q \perp I_bQ,$ $ B_1Q$ goes through $ I$ $ \Longrightarrow$ $ I$ is orthocenter of $ \triangle B_1B_2I_b.$
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math_13
48 posts
#4 • 2 Y
Y by Adventure10, PikaPika999
We use from vectors ($I_bB_1.B_2I=0$)
This post has been edited 1 time. Last edited by math_13, May 7, 2013, 12:12 PM
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BBAI
563 posts
#5 • 5 Y
Y by earthrise, futurestar, leru007, Adventure10, PikaPika999
We notice that if we prove $B_1I$ is perpendicular to $B_2I_b$ ,we are done.
Let $ \odot AIC \cap B_2I_b=L$.As $\odot AIC$ and $\odot ABC$ have $AC$ as the radical axis and as $B_1B_2,AC,B_2I_b $ are concurrent, then $B_1BLI_b$ is cyclic. So $ B_1L$ is $\perp$ to $B_2I_b$.So $ I$ sholud lie on $B_1L$ as $II_b$ is the diameter of $ \odot AIC$. Hence done.
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sunken rock
4401 posts
#6 • 3 Y
Y by Adventure10, Mango247, PikaPika999
Let $\{I, X\}\in B_2I\cap\odot (AIC)$; from power of $B_2$ w.r.t. $\odot(ABC),\odot(AIC)$ we get (already proven $B_1-B-B_2$ are collinear): $B_2B\cdot B_2B_1=B_2A\cdot B_2C=B_2I\cdot B_2X$, hence $BB_1XI$ is cyclic, i.e. $B_2X\cap B_1X$. As $II_b$ is a diameter of $\odot (AIC)$, we infer $I_bX\bot IX$, meaning $B_1-X-I_b$ are collinear, and we are done.

Best regards,
sunken rock
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highvalley
16 posts
#7 • 3 Y
Y by Adventure10, Mango247, PikaPika999
$ B_{1} $ is the midpoint of the arc $AC$ containing $B$, in the circumcircle of $\triangle ABC\cdot \cdot \cdot (1)$
$I_{b}$ is the $B$-excircle's center$\cdot \cdot \cdot (2) $
Angle bisector of $\angle ABC$ intersects $AC$ at $B_{2}\cdot \cdot \cdot (3)$
$ I  $ is the incenter of $\triangle ABC\cdot \cdot \cdot (4)$

By $(3)$ and $(4)$, $\angle IBB_{2}=\angle R\cdot \cdot \cdot (5)$
By $(1)$ and $(4)$,
\[\angle B_{1}BI\\=\angle IBC+\angle B_{1}BC
\\=\angle IBC+\angle B_{1}AC
\\=\angle IBC+(\angle R-\frac{1}{2}\angle AB_{1}C)
\\=\angle R+(\angle IBC-\frac{1}{2}\angle ABC)
\\=\angle R\cdot \cdot \cdot (6)\]
By $ (5) $ and $ (6) $, $  B $ and $B_{1},B_{2}$ are collinear$\cdot \cdot \cdot (7)$
By,$ (2) $ and $ (4) $, $\angle IAI_{b}=\angle ICI_{b}=\angle R\cdot \cdot \cdot (8)$
Let $ H $ be a point such that $ H $ is in $ B_{1}I $ and $ BH\bot HI_{b}. \cdot \cdot \cdot (9) $
By $ (6) $ and $(9)$, $B$ and $B_{1},I_{b},H$ are concyclic.$\cdot \cdot \cdot (10)$
By $ (8) $ and $(9)$, $A$ and $I,C,I_{b},H$ are concyclic.$\cdot \cdot \cdot (11)$
By $ (1) $ , $ A $ and $B,C,B_{1}$ are concyclic.$\cdot \cdot \cdot (12)$
By $ (10) $ and $(11),(12)$, $ I_{b} $ and $ H,BB_{1}\cap AC(=B_{2}) $ are collinear.$\cdot \cdot \cdot (13)$($\because$ $BB_{1}\cap AC\cap I_{b}H$ is a radical ceneter)
By $(9)$ and $(13) $, $B_{1}I\bot I_{b}B_{2}.\cdot \cdot \cdot (14)$
By $ (5) $ and $ (7) $, $ B_{1}B_{2}\bot BI_{b}.\cdot \cdot \cdot (15) $
By $ (14) $ and $ (15) $, I is orthocenter of $\triangle B_{1}B_{2}I_{b}$.
So $ B_{2}I\bot B_{1}I_{b} $.
$ (Q,E,D,) $
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yimingz89
222 posts
#8 • 3 Y
Y by Adventure10, Mango247, PikaPika999
Let $l$ be the line through $I$ perpendicular to $B_1I_B$. Define $P'=B_1I_B\cap l$. An easy angle chase shows $B_2,B,B_1$ are collinear on the external angle bisector of $B$ while $B_2,A,C$ are collinear by the definition. Now consider the circles $\Gamma_1=(ABC),\Gamma_2=(BB_1I),\Gamma_3=(AIC)$. Clearly $B_1\in\Gamma_1$ while $P'\in\Gamma_2$ since $\angle B_1P'I=\angle B_1BI=90^{\circ}$ and $P'\in\Gamma_3$ since $II_B$ is a diameter, where $I_B$ is the $B$-excenter, and $\angle IP'I_B=90^{\circ}$. It is easy to see that the Radical Axes of $\Gamma_1,\Gamma_2$ is $BB_1$, $\Gamma_2,\Gamma_3$ is $IP$, and $\Gamma_3,\Gamma_1$ is $AC$. By Radical Concurrence on $\Gamma_1,\Gamma_2,\Gamma_3$, these lines concur at $B_2$, which is enough to conclude that $B_2,I,P'$ are collinear, showing $P=P'$.
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Pluto1708
1107 posts
#9 • 3 Y
Y by Adventure10, Mango247, PikaPika999
Power of point!
This post has been edited 1 time. Last edited by Pluto1708, Sep 16, 2018, 7:25 AM
Reason: Sy
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WolfusA
1900 posts
#10 • 3 Y
Y by NZQR, Adventure10, PikaPika999
Complex numbers: vertices of triangle are $a^2,b^2,c^2$, and it's circumcircle is a unit circle. Then incenter is $-ab-bc-ca$, $B_2$ as intersection of lines $BB_1,AC$ has coordinates $\frac{(b^2+ac)ac-(a^2+c^2)b^2}{ac-b^2}$.
$\frac{B_2-I}{I_b-B_1}=\frac{(ab+bc+ca)(ac-b^2)+(b^2ac+a^2c^2-a^2b^2-b^2c^2)}{(ab+bc-2ac)(ac-b^2)}$
The conjugate of this number is $\frac{(a+b+c)(b^2-ac)+abc+b^3-bc^2-a^2b}{(c+a-2b)(b^2-ac)}$
Adding two last complex numbers we get $0$ (as you don't believe check here Click to reveal hidden text)
Hence $B_2I\perp IbB_1$
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AlastorMoody
2125 posts
#11 • 5 Y
Y by karitoshi, myh2910, Adventure10, Mango247, PikaPika999
Let $I_A,I_C$ be the $A$, $C-$ excenter, By Brokard's Theorem on Quadrilateral $I_CACI_A$ $\implies$ $I$ is the orthocenter of $\Delta B_2B_1I_B$
This post has been edited 1 time. Last edited by AlastorMoody, Feb 15, 2019, 8:07 AM
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Jupiter_is_BIG
867 posts
#12 • 2 Y
Y by Adventure10, PikaPika999
Erken wrote:
Suppose that $ B_1$ is the midpoint of the arc $ AC$, containing $ B$, in the circumcircle of $ \triangle ABC$, and let $ I_b$ be the $ B$-excircle's center. Assume that the external angle bisector of $ \angle ABC$ intersects $ AC$ at $ B_2$. Prove that $ B_2I$ is perpendicular to $ B_1I_B$, where $ I$ is the incenter of $ \triangle ABC$.

Was this question bonus or $I_b$ and $I_B$ the same?
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Kagebaka
3001 posts
#13 • 2 Y
Y by Adventure10, PikaPika999
It's well-known that under $\sqrt{ac}$ inversion, $\{I,I_B\},\{B_1,B_2\}$ swap, so we're done because then we must have$$BI\cdot BI_B = AB\cdot BC = BB_1\cdot BB_2,$$which means that $I$ is the orthocenter of $\triangle B_1B_2I_B.$ $\blacksquare$
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Dr_Vex
562 posts
#14 • 1 Y
Y by PikaPika999
We will prove that infact $I$ is the orthocenter of $\Delta B_{1}B_{2}I_{B}$.
Now let $B_{1}I\cap I_{B}B_{2}=F$. By PoP
$B_{2}F\cdot B_{2}I_{B} =B_{2}A\cdot B_{2}C=B_{2}B\cdot B_{2}B_{1}$. Hence
quadrilateral $B_{1}BI_{B}F$ is cyclic.
Now let $IE\perp B_{1}I_{B}$, it is also seen that there exists a circle $(I_{B}CEIAF)$. Hence, as $\angle B_{1}FI_{B}=\angle B_{1}BI_{B}=90^{\circ} \Rightarrow BIEB_{1}$ is cyclic too. As $BB_{1}\cap FI_{B}=B_{2}$ Its consequence leads to the fact that $B-I-E$
$\blacksquare$
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SenatorPauline
30 posts
#15 • 2 Y
Y by AlastorMoody, PikaPika999
Jupiter_is_BIG wrote:
Was this question bonus or $I_b$ and $I_B$ the same?
It was a bonus
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th1nq3r
146 posts
#16 • 1 Y
Y by PikaPika999
Notice that $\angle I_BBB_1 = 90$. (Indeed $\angle I_BBB_1 = \angle MBA + \angle B_1BA = \angle MB_1A + \angle B_1AC = 90$).

Denote by $P$ the intersection of line $B_2I_B$ with the circumcircle of $\triangle IAC$. It is immediate that \[B_2B \cdot B_2B_1 = B_2C \cdot B_2A = B_2P \cdot B_2I_B.\]Thus $B, B_1, P, I_B$ are concyclic. Now by the incenter/excenter lemma, we have that $II_B$ is the diameter of $(CAI_B)$. Using this, one obtains \[\angle I_BPI = 90 = \angle I_BBB_1 = \angle I_BPB_1.\]Therefore $P, I, B_1$ are collinear, and $I$ is the orthocenter of $B_2BI_B$, as desired. $\blacksquare$
This post has been edited 2 times. Last edited by th1nq3r, May 5, 2023, 1:35 PM
Reason: poor
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ricegang67
26 posts
#17 • 1 Y
Y by PikaPika999
We prove that $I$ is the orthocenter of $\triangle I_BB_1B_2$. In particular, it is equivalent to show that \[BI\cdot BI_B = BB_1\cdot BB_2.\]Let $D$ and $M$ be the intersections of line $BI$ with $AC$ and $(ABC)$. Since $(BD;II_B) = -1$, $BI\cdot BI_B = BD\cdot BM$. Then, observe that $MB\perp B_1B_2$ and $B_2D\perp B_1M$, so in fact $D$ is the orthocenter of $\triangle MB_1B_2$. Hence, \[BB_1\cdot BB_2 = BD\cdot BM = BI\cdot BI_B\]as desired.
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cursed_tangent1434
648 posts
#18 • 1 Y
Y by PikaPika999
Well, it is known that the external angle bisector of $\triangle ABC$ is simply $\overline{BB_1}$. Now, notice that,
\[\measuredangle IAI_b=\measuredangle ICI_b=90^\circ\]and thus, $I$,$A$,$C$ and $I_b$ are concyclic. Now, let $B_3=(IBB_1)\cap \overline{B_1I_b}$. It is easy to see that since $\measuredangle IB_3B_1=\measuredangle IBB_1 90^\circ$, $B_3$ also lies on $(IAC)$. Now, let $B_3'=\overline{B_2I} \cap (IBB_1)$. Then,
\[B_2I\cdot B_2B_3' = B_2B \cdot B_2B_1 = B_2A\cdot B_2C\]Thus, $B_3'$ must also lie on $(IAC)$ which implies that $B_3'=B_3$ and indeed, $B_2I \perp B_1I_b$ as required.
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aaravdodhia
2614 posts
#19 • 1 Y
Y by PikaPika999
[asy]
import olympiad; size(400);
draw(unitcircle);
pair B = dir(130), A = dir(200), C = dir(-20), M = midpoint(A--C), O = origin, B1 = intersectionpoint(O--(O+3*(O-M)),unitcircle);
pair I = incenter(A,B,C), L = intersectionpoint(B--(I+3*(I-B)),unitcircle), Ib = L + L-I;
pair exB = rotate(90,B) * I, B2 = extension(exB, B, A, C);
dot("$A$",A); dot("$B$",B); dot("$C$",C); dot("$I$",I); dot("$O$",O); dot("$M$",M); dot("$L$",L); dot("$I_B$",Ib); dot("$B_1$",B1); dot("$B_2$",B2);
draw(C--B2--B--C^^A--B--Ib^^Ib--B1--L^^B2--extension(B2,I,B1,Ib));
draw(B--B1);
// draw(incircle(A,B,C);
[/asy]

Let $D$ be the foot of angle bisector from $B$ to $AC$ and $M$ the midpoint of $AC$. Since $\angle B_2BI = 90 = \angle B_1BL$ (since $B1L$ is diameter), all $B$'s are collinear. Due to cyclic quad $BB_1MD$, $B_1B \cdot BB_2 = B_2B\cdot B_2B_1 - B_2B^2 = B_2D\cdot B_2M - B_2D^2 - B_2D^2 + BD^2$ (from right triangle $B_2BD$), equals $B_2D\cdot DM + BD^2 = BD\cdot DL + BD^2 = BD\cdot BL$ (from cyclic quad $B_2BML$). Also $\angle LAD = \angle LBA$ so from similar triangles $BL\cdot LD = LA^2 = IL^2 \implies BD\cdot BL = BL^2 - IL^2 = BI\cdot BI_B$. So in triangle $B_1B_2I_B$, we have $B_I\cdot BI_B = BB_1 \cdot BB_2$ so $I$ is the orthocenter and $B_2I \perp B_1I_B$.
This post has been edited 1 time. Last edited by aaravdodhia, Aug 30, 2024, 7:21 PM
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Primeniyazidayi
114 posts
#20 • 1 Y
Y by PikaPika999
Let the antipode of $B_1$ wrt $(ABC)$ be $M$ and let the intersection of $(I_BAIC)$ and $\overline{B_1I_B}$ be $X$.Because $I_B,A,I,X,C$ are concyclic by incenter/excenter lemma we have that $\angle IXI_B = 90$ and because $M$ is the antipode of $B_1$ we have that $\angle I_BBB_1 = \angle MBB_1 = 90$,so $B_1,B,I,X$ are concyclic.Then it succifies to show that $B_2,I,X$ are collinear which is trivial by the radical axis concurrence lemma on $(ABB_1CM),(B_1BIX),(AIXCI_B)$,which shows that $\overline{B_1B},\overline{XI},\overline{AC}$ are concurrent at $B_2$.
This post has been edited 2 times. Last edited by Primeniyazidayi, Apr 6, 2025, 6:40 PM
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Baimukh
11 posts
#21
Y by
Let $ID \bot B_2I_b$ and $ID \cup B_2I_b=D$. By the trident lemma, $B_2D\cdot B_2I_b=B_2C\cdot B_2A=B_2B\cdot B_2B_1\Longrightarrow \angle B_1BI_b=\angle B_1DI_b=90^\circ=\angle IDI_b\Longrightarrow B_1-I-D$ lie on the same line, and $I$ is the orthocenter of $\triangle B_1B_2I_b$
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