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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
Random concyclicity in a square config
Maths_VC   3
N 12 minutes ago by Assassino9931
Source: Serbia JBMO TST 2025, Problem 1
Let $M$ be a random point on the smaller arc $AB$ of the circumcircle of square $ABCD$, and let $N$ be the intersection point of segments $AC$ and $DM$. The feet of the tangents from point $D$ to the circumcircle of the triangle $OMN$ are $P$ and $Q$ , where $O$ is the center of the square. Prove that points $A$, $C$, $P$ and $Q$ lie on a single circle.
3 replies
Maths_VC
Tuesday at 7:38 PM
Assassino9931
12 minutes ago
Basic ideas in junior diophantine equations
Maths_VC   2
N 14 minutes ago by Assassino9931
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
2 replies
1 viewing
Maths_VC
Tuesday at 7:54 PM
Assassino9931
14 minutes ago
Prime number theory
giangtruong13   2
N 14 minutes ago by RagvaloD
Find all prime numbers $p,q$ such that: $p^2-pq-q^3=1$
2 replies
+1 w
giangtruong13
44 minutes ago
RagvaloD
14 minutes ago
An algorithm for discovering prime numbers?
Lukaluce   2
N 17 minutes ago by Assassino9931
Source: 2025 Junior Macedonian Mathematical Olympiad P3
Is there an infinite sequence of prime numbers $p_1, p_2, ..., p_n, ...,$ such that for every $i \in \mathbb{N}, p_{i + 1} \in \{2p_i - 1, 2p_i + 1\}$ is satisfied? Explain the answer.
2 replies
Lukaluce
May 18, 2025
Assassino9931
17 minutes ago
Problem 2
delegat   147
N 32 minutes ago by math-olympiad-clown
Source: 0
Let $n\ge 3$ be an integer, and let $a_2,a_3,\ldots ,a_n$ be positive real numbers such that $a_{2}a_{3}\cdots a_{n}=1$. Prove that
\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]

Proposed by Angelo Di Pasquale, Australia
147 replies
delegat
Jul 10, 2012
math-olympiad-clown
32 minutes ago
Coloring points of a square, finding a monochromatic hexagon
goodar2006   6
N 41 minutes ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part 2-P1
Prove that for each coloring of the points inside or on the boundary of a square with $1391$ colors, there exists a monochromatic regular hexagon.
6 replies
goodar2006
Sep 15, 2012
quantam13
41 minutes ago
Van der Warden Theorem!
goodar2006   7
N an hour ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part 2-P2
Suppose $W(k,2)$ is the smallest number such that if $n\ge W(k,2)$, for each coloring of the set $\{1,2,...,n\}$ with two colors there exists a monochromatic arithmetic progression of length $k$. Prove that


$W(k,2)=\Omega (2^{\frac{k}{2}})$.
7 replies
goodar2006
Sep 15, 2012
quantam13
an hour ago
Maxi-inequality
giangtruong13   0
an hour ago
Let $a,b,c >0$ and $a+b+c=2abc$. Find max: $$P= \sum_{cyc} \frac{a+2}{\sqrt{6(a^2+2)}}$$
0 replies
giangtruong13
an hour ago
0 replies
Isosceles triangles among a group of points
goodar2006   2
N an hour ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part1-P2
Consider a set of $n$ points in plane. Prove that the number of isosceles triangles having their vertices among these $n$ points is $\mathcal O (n^{\frac{7}{3}})$. Find a configuration of $n$ points in plane such that the number of equilateral triangles with vertices among these $n$ points is $\Omega (n^2)$.
2 replies
goodar2006
Jul 27, 2012
quantam13
an hour ago
APMO Number Theory
somebodyyouusedtoknow   12
N an hour ago by math-olympiad-clown
Source: APMO 2023 Problem 2
Find all integers $n$ satisfying $n \geq 2$ and $\dfrac{\sigma(n)}{p(n)-1} = n$, in which $\sigma(n)$ denotes the sum of all positive divisors of $n$, and $p(n)$ denotes the largest prime divisor of $n$.
12 replies
somebodyyouusedtoknow
Jul 5, 2023
math-olympiad-clown
an hour ago
My Unsolved Problem
ZeltaQN2008   0
an hour ago
Source: IDK
Let \( P(x) = x^{2024} + a_{2023}x^{2023} + \cdots + a_1x + a_0 \) be a polynomial with real coefficients.

(a) Suppose that \( 2023a_{2023}^2 - 4048a_{2022} < 0 \). Prove that the polynomial \( P(x) \) cannot have 2024 real roots.

(b) Suppose that \( a_0 = 1 \) and \( 2023(a_1^2 + a_2^2 + \cdots + a_{2023}^2) \leq 4 \). Prove that \( P(x) \geq 0 \) for all real numbers \( x \).
0 replies
ZeltaQN2008
an hour ago
0 replies
Points of a grid
goodar2006   2
N an hour ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part1-P4
Prove that from an $n\times n$ grid, one can find $\Omega (n^{\frac{5}{3}})$ points such that no four of them are vertices of a square with sides parallel to lines of the grid. Imagine yourself as Erdos (!) and guess what is the best exponent instead of $\frac{5}{3}$!
2 replies
goodar2006
Jul 27, 2012
quantam13
an hour ago
Classical NT FE
Kimchiks926   6
N 2 hours ago by math-olympiad-clown
Source: Baltic Way 2022, Problem 16
Let $\mathbb{Z^+}$ denote the set of positive integers. Find all functions $f:\mathbb{Z^+} \to \mathbb{Z^+}$ satisfying the condition
$$ f(a) + f(b) \mid (a + b)^2$$for all $a,b \in \mathbb{Z^+}$
6 replies
Kimchiks926
Nov 12, 2022
math-olympiad-clown
2 hours ago
Hagge circle, Thomson cubic, coaxal
kosmonauten3114   0
2 hours ago
Source: My own (maybe well-known)
Let $\triangle{ABC}$ be a scalene triangle, $\triangle{M_AM_BM_C}$ its medial triangle, and $P$ a point on the Thomson cubic (= $\text{K002}$) of $\triangle{ABC}$. (Suppose that $P \notin \odot(ABC)$ ).
Let $\triangle{A'B'C'}$ be the circumcevian triangle of $P$ wrt $\triangle{ABC}$.
Let $\triangle{P_AP_BP_C}$ be the pedal triangle of $P$ wrt $\triangle{ABC}$.
Let $A_1$ be the reflection in $BC$ of $A'$. Define $B_1$, $C_1$ cyclically.
Let $A_2$ be the reflection in $M_A$ of $A'$. Define $B_2$, $C_2$ cyclically.
Let $A_3$ be the reflection in $P_A$ of $A'$. Define $B_3$, $C_3$ cyclically.

Prove that $\odot(A_1B_1C_1)$, $\odot(A_2B_2C_2)$, $\odot(A_3B_3C_3)$ and the orthocentroidal circle of $\triangle{ABC}$ are coaxal.
0 replies
kosmonauten3114
2 hours ago
0 replies
Symmedian line
April   93
N May 18, 2025 by aidenkim119
Source: All Russian Olympiad - Problem 9.2, 10.2
Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$. The line $ BD$ intersects the circumcircle $ \Omega$ of triangle $ ABC$ at $ B$ and $ E$. Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$.
93 replies
April
May 10, 2009
aidenkim119
May 18, 2025
Symmedian line
G H J
Source: All Russian Olympiad - Problem 9.2, 10.2
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April
1270 posts
#1 • 8 Y
Y by Amir Hossein, dave_mathsguru, 799786, itslumi, Adventure10, Mango247, fuzzball2023, and 1 other user
Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$. The line $ BD$ intersects the circumcircle $ \Omega$ of triangle $ ABC$ at $ B$ and $ E$. Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$.
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mathVNpro
469 posts
#2 • 6 Y
Y by Olemissmath, Psyduck909, Adventure10, Mango247, fuzzball2023, ehuseyinyigit
Let $ M_b$ be midpoint of $ AC$.
We have $ \angle FAM_b = \angle FBC$, $ \angle AM_bF\equiv \angle DM_bF = \angle DEF = \angle BEF = \angle BCF$
$ \Rightarrow \triangle FAM_b\sim \triangle FBC$, $ (i)$. Thus, through the spiral similarity with center $ F$, denote this $ f$ then:
$ f: \triangle FAM_b\mapsto \triangle FBC$
$ \Rightarrow f: A\mapsto B$, $ M_b\mapsto C$
$ \Rightarrow f: A\mapsto M_b$, $ B\mapsto C$
$ \Rightarrow f: \triangle FAB\mapsto \triangle FM_bC$, $ (1)$
But it is easy to notice that: $ \triangle BAF\sim \triangle BM_bC$ (a.a), $ (2)$
$ (1),(2)\Rightarrow \triangle CM_bB\sim \triangle FM_bC$
$ \Rightarrow \frac {CB}{CF} = \frac {CM_b}{FM_b} = \frac {M_bB}{M_bC}$, $ (*)$.
With the same argument, we get $ \triangle BAM_b\sim \triangle BFC$
$ \Rightarrow \triangle BAM_b\sim \triangle AFM_b$ (thanks to $ (i)$).
$ \Rightarrow \frac {BA}{AF} = \frac {M_bB}{M_bA}$, $ (**)$
But $ M_bA = M_bC \Rightarrow \frac {BA}{BC} = \frac {FA}{FC}$ (Combine $ (*)\&(**)$)
$ \Longrightarrow ABCF$ is a harmonic quadrilateral. Therefore, $ BF$ passes through the Pole of $ AC$ wrt $ (O)$, where $ O$ is the circumcenter of $ \triangle ABC$. Then we will get the result of the problem.
Our proof is completed then :lol:
This post has been edited 1 time. Last edited by mathVNpro, May 11, 2009, 4:56 AM
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silouan
3952 posts
#3 • 6 Y
Y by ultralako, Adventure10, Mango247, fuzzball2023, and 2 other users
Let $ M$ the midpoint of $ AC$ , $ O$ the circumcenter of $ \Omega$ and let the tangent to $ \Omega$ at $ B$ meets $ AC$ at $ R$ .
Then $ BOMR$ is cyclic (let $ Q$ its circumcenter) . Then $ R,Q,O$ are collinear . So if the circumcircle of $ BOMR$ meets $ \Omega$ at $ F'$ , then $ BF'\bot RO$ . So by polars $ RF'$ is tangent to $ \Omega$ . By well known theorem we have that $ BF'$ is the $ B$ symmedian. Finally if $ BE$ meets the circumcircle of $ BOMR$ at $ S$ , then $ SM=SF$ . But also because $ RF'=RB$ we have that $ D$ is the incenter of $ F'BM$ so $ SD=SM=SF$ and because $ DM\bot ME$ we also have $ SD=SM=SF=SE$ . This shows that $ MDF'E$ is cyclic and so $ F\equiv F'$ and we are done .
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yetti
2643 posts
#4 • 5 Y
Y by Adventure10, Mango247, MS_asdfgzxcvb, and 2 other users
External bisector of $ \angle B$ cuts $ CA$ at $ Y.$ $ I, I_b$ are incenter and B-excenter of $ \triangle ABC.$ $ (E)$ is a circle with center $ E$ and radius $ EA = EC = EI = EI_b.$ The cross ratio $ (B, D, I, I_b)$ is harmonic and $ E$ the midpoint of $ II_b$ $ \Longrightarrow$ $ \overline{BD} \cdot \overline{BE} = \overline{BI}\cdot \overline{BI_b}$ $ \Longrightarrow$ $ BY \perp BDE$ is radical axis of $ \omega, (E).$ Pairwise radical axes $ EF, CA, BY$ of $ \Omega, \omega, (E)$ meet at their radical center $ Y.$ Since $ DF \perp EFY$ and $ BY \perp BDE,$ $ BDFY$ is cyclic. Its circumcircle is the B-Apollonius circle of $ \triangle ABC$ $ \Longrightarrow$ $ BF$ is the B-symmedian.
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Virgil Nicula
7054 posts
#5 • 7 Y
Y by Amir Hossein, Adventure10, Mango247, fuzzball2023, MS_asdfgzxcvb, and 2 other users
PP. Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$ . The line $ BD$ intersects the circumcircle $ \Omega$ of
$\triangle ABC$ at $ B$ and $ E$ . Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$ .


Lemma. Let $ ABC$ be a triangle with the circumcircle $ w$ . Denote the following points : the middlepoint $ M$ of the side $ [BC]$ ;

the point $ D\in (BC)$ for which $ \angle DAB\equiv\angle DAC$ ; the point $ E$ for which $ \{A,E\} = AD\cap w$ ; the $ A$-symmedian $ AL$ ,

where $ L\in (BC)$ ; the point $ S$ for which $ \{A,S\} = AL\cap w$ . Then $ SD\perp SE$ , i.e. the quadrilateral $ MDSE$ is cyclically.


Proof. Denote $ XX$ - the tangent in $ X\in w$ to $ w$ . Denote $ T\in BB\cap CC$ . Observe that $ \overline {MET}\perp BC$ and $ \angle ASB\equiv\angle ACB$ .

From the first well-known property, $ T\in \overline {ALS}$ and the line $ \overline{ALST}$ is the $ A$-symmedian in $ \triangle ABC$ $ \Longrightarrow$ $ \left\|\begin{array}{c} \angle BAS\equiv\angle MAC \\
\ \angle SAE\equiv\angle DAM\end{array}\right\|$ .

From the second well-known property, the division $ \{A,S;L,T\}$ is harmonically. Since $ ML\perp MT$ results $ \angle DMA\equiv\angle DMS$ .

Show easily that $ \triangle BAS\sim\triangle MAC$ (a.a.) . Thus, $ \frac {AB}{AS} = \frac {AM}{AC}$ , i.e. $ AS\cdot AM = AB\cdot AC$ .

From the third well-known property $ AD\cdot AE = AB\cdot AC$, obtain $ \boxed {AS\cdot AM = AD\cdot AE = AB\cdot AC}$ , i.e. $ \frac {AS}{AE} = \frac {AD}{AM}$ .

Since $ \angle SAE\equiv\angle DAM$ obtain $ \triangle SAE\sim\triangle DAM$ and in consequence, $ \angle DMA\equiv\angle SEA$ . Since $ \angle DMA\equiv\angle DMS$

and $ \angle SEA\equiv\angle SED$ obtain $ \angle DMS\equiv\angle SED$ , i.e. the quadrilateral $ MDSE$ is cyclically. In conclusion, $ SD\perp SE$ .

Remark. Here are another interesting metrical relations : $ \left\|\begin{array}{ccc} ABS\sim CMS & \implies & \boxed {SB\cdot SC = SA\cdot SM} \\
 \\
AMC\sim CMS & \implies & \boxed {MA\cdot MS = MB^2}\end{array}\right\|$ .

The proposed problem is an immediate consequence of the above lemma.
See PP13 from here
This post has been edited 4 times. Last edited by Virgil Nicula, Aug 11, 2014, 1:54 AM
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jayme
9801 posts
#6 • 3 Y
Y by AlastorMoody, Adventure10, fuzzball2023
Dear Mathlinkers,
1. Let A' be the midpoint of BC, X, Y the second meetpoint of AA', EA' with the circumcircle of ABC and Te the tangent to this circumcircle at E.
2. Te // DA'
3. The circle with diameter ED passes through A'
4. By a converse of Reim's theorem applied to this circle and the circumcircle, FD goes through Y;
5. By a converse of Pascal's theorem applied to EAXFYTe, FX //DA' (BC)
and we are done...
Sincerely
Jean-Louis
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Agr_94_Math
881 posts
#7 • 3 Y
Y by Adventure10, Mango247, ehuseyinyigit
Very nice application of Reim's Theorem jayme.Though a very obvious result, it has great application in many synthetic geometry results.
Thanks.
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mathVNpro
469 posts
#8 • 2 Y
Y by Adventure10, Mango247
jayme wrote:
Dear Mathlinkers,
1. Let A' be the midpoint of BC, X, Y the second meetpoint of AA', EA' with the circumcircle of ABC and Te the tangent to this circumcircle at E.
2. Te // DA'
3. The circle with diameter ED passes through A'
4. By a converse of Reim's theorem applied to this circle and the circumcircle, FD goes through Y;
5. By a converse of Pascal's theorem applied to EAXFYTe, FX //DA' (BC)
and we are done...
Sincerely
Jean-Louis
Do you have any document about Reim's theorem, Jayme? I am so excited about this. :blush:
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jayme
9801 posts
#9 • 3 Y
Y by Adventure10, Mango247, fuzzball2023
See for example my website
http://perso.orange.fr/jl.ayme then: A propos
You will see all the possibility.
Sincerely
Jean-Louis
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ma 29
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#10 • 16 Y
Y by MexicOMM, Wizard_32, Durjoy1729, HolyMath, Siddharth03, Greenleaf5002, myh2910, snakeaid, centslordm, Jalil_Huseynov, Adventure10, Mango247, GustavoPudim, fuzzball2023, and 2 other users
Dear friends.
I have a short solution for the nice problem,here :) .
//cdn.artofproblemsolving.com/images/4a3755e6df266a6313f50f6a8a6c7c8845715061.jpg

Denote by $ A'$ the midpoint of segment $ AC$.

$ EO$ meets $ (O)$ again at $ P$

By ${ \angle {DFE} = 90^0}$ ,we have $ F,D,P$ are collinear.

From $ BPEF$ is cyclic and $ BDA'P$ is cylic ,we have :

$ \angle {EBF} = \angle {EPF} = \angle{DBA'}$.

Therefor,$ BF$is the symmedian line of the triangle $ BAC$.
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armpist
527 posts
#11 • 3 Y
Y by Adventure10, Mango247, and 1 other user
All are very nice solutions indeed, I must admit.



T.Y.
M.T.
This post has been edited 1 time. Last edited by armpist, May 11, 2009, 8:29 PM
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Virgil Nicula
7054 posts
#12 • 2 Y
Y by Adventure10, Mango247
After the Ma29's shortest proof, this problem seems very simple ...
In the another proofs appeared many nice properties of this configuration.
Thanks to all for their interest and proofs.
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ma 29
40 posts
#13 • 2 Y
Y by Adventure10, Mango247
Hello my friends.
I think all proof above are very nice. :)
Here is another solution to this problem:
//cdn.artofproblemsolving.com/images/c51085c9ea4036fa17fa17def72321254472bf4e.jpg

Denote $ V = EF \cap AC$.

We have :$ \bar {VD} .\bar{VA'} = \bar{VF}.\bar{VE} = \bar{VA}.\bar{VC}$ (1)

But $ A'$ is the midpoint of segment $ AC$.(2)

From (1) ,(2) and Maclaurin's theorem , we obtain the division $ {VDAC}$ is harmonically,so $ V,B,P$ are collinear.

Denote $ Q = BC\cap {PA}$ and denote by $ H$ the intersection of two tangent to $ (O)$ at $ A,C$ ,respectively.

We have $ Q,D,H$ lie on the polar of $ V$ wrt $ (O)$,i.e $ Q,D,H$ are collinear.

By Pascal's theorem applied to six point $ A,A,P,C,B,F$ ,we obtain $ B,F,H$ are collinear.

Therefore ,$ BF$ is the symedian line of the triangle $ ABC$.
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bravado
19 posts
#14 • 2 Y
Y by Adventure10, Mango247
It can also be viewed as an easy application of : http://www.mathlinks.ro/viewtopic.php?p=1082899#1082899

Indeed, from the above problem it follows that the line $ FM$ (where $ M$ is the midpoint of $ AC$) meets the circumcircle of $ \triangle{ABC}$ at a point $ T$ such that $ BT||AC$. By symmetry,
\[ \angle{MBC} = \angle{MTA} = \angle{ABF}.
\]
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Virgil Nicula
7054 posts
#15 • 2 Y
Y by Adventure10, Mango247
We can use my found property at the proposed problem from here.
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