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Geometry
gggzul   2
N 33 minutes ago by gggzul
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
2 replies
gggzul
2 hours ago
gggzul
33 minutes ago
thank you !
Piwbo   2
N 37 minutes ago by Piwbo
Given positive integers $a,b$ such that $a$ is even , $b$ is odd and $ab(a+b)^{2023}$ is divisible by $a^{2024}+b^{2024}$ .Prove that there exists a prime number $p$ such that $a^{2024}+b^{2024}$ is divisible by $p^{2025}$
2 replies
Piwbo
an hour ago
Piwbo
37 minutes ago
Inequality involving square root cube root and 8th root
bamboozled   2
N 39 minutes ago by bamboozled
If $a,b,c,d,e,f,g,h,k\in R^+$ and $a+b+c=d+e+f=g+h+k=8$, then find the maximum value of $\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$
2 replies
bamboozled
5 hours ago
bamboozled
39 minutes ago
find positive n so that exists prime p with p^n-(p-1)^n$ a power of 3
parmenides51   12
N an hour ago by n-k-p
Source: JBMO Shortlist 2017 NT5
Find all positive integers $n$ such that there exists a prime number $p$, such that $p^n-(p-1)^n$ is a power of $3$.

Note. A power of $3$ is a number of the form $3^a$ where $a$ is a positive integer.
12 replies
parmenides51
Jul 25, 2018
n-k-p
an hour ago
hard problem
Cobedangiu   5
N 2 hours ago by KhuongTrang
$a,b,c>0$ and $a+b+c=7$. CM:
$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+abc \ge ab+bc+ca-2$
5 replies
Cobedangiu
Yesterday at 4:24 PM
KhuongTrang
2 hours ago
Nordic 2025 P3
anirbanbz   9
N 2 hours ago by Tsikaloudakis
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
9 replies
anirbanbz
Mar 25, 2025
Tsikaloudakis
2 hours ago
Aime type Geo
ehuseyinyigit   1
N 2 hours ago by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
1 reply
ehuseyinyigit
Yesterday at 9:04 PM
ehuseyinyigit
2 hours ago
Arbitrary point on BC and its relation with orthocenter
falantrng   34
N 2 hours ago by Mamadi
Source: Balkan MO 2025 P2
In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus
34 replies
falantrng
Apr 27, 2025
Mamadi
2 hours ago
Which numbers are almost prime?
AshAuktober   5
N 3 hours ago by Jupiterballs
Source: 2024 Swiss MO/1
If $a$ and $b$ are positive integers, we say that $a$ almost divides $b$ if $a$ divides at least one of $b - 1$ and $b + 1$. We call a positive integer $n$ almost prime if the following holds: for any positive integers $a, b$ such that $n$ almost divides $ab$, we have that $n$ almost divides at least one of $a$ and $b$. Determine all almost prime numbers.
original link
5 replies
AshAuktober
Dec 16, 2024
Jupiterballs
3 hours ago
If $b^n|a^n-1$ then $a^b >\frac {3^n}{n}$ (China TST 2009)
Fang-jh   16
N 3 hours ago by Aiden-1089
Source: Chinese TST 2009 6th P1
Let $ a > b > 1, b$ is an odd number, let $ n$ be a positive integer. If $ b^n|a^n-1,$ then $ a^b > \frac {3^n}{n}.$
16 replies
Fang-jh
Apr 4, 2009
Aiden-1089
3 hours ago
Confusing inequality
giangtruong13   1
N 3 hours ago by Natrium
Source: An user
Let $a,b,c>0$ such that $a+b+c=3$. Find the minimum: $$P= \sum_{cyc} \frac{a}{b} + \sum_{cyc} \frac{1}{a^3+b^3+abc}$$
1 reply
giangtruong13
Yesterday at 8:04 AM
Natrium
3 hours ago
Problem 1 IMO 2005 (Day 1)
Valentin Vornicu   92
N Apr 3, 2025 by Baimukh
Six points are chosen on the sides of an equilateral triangle $ABC$: $A_1$, $A_2$ on $BC$, $B_1$, $B_2$ on $CA$ and $C_1$, $C_2$ on $AB$, such that they are the vertices of a convex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths.

Prove that the lines $A_1B_2$, $B_1C_2$ and $C_1A_2$ are concurrent.

Bogdan Enescu, Romania
92 replies
Valentin Vornicu
Jul 13, 2005
Baimukh
Apr 3, 2025
Problem 1 IMO 2005 (Day 1)
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Valentin Vornicu
7301 posts
#1 • 8 Y
Y by Davi-8191, anantmudgal09, Math-Ninja, Adventure10, mathematicsy, HWenslawski, megarnie, Mango247
Six points are chosen on the sides of an equilateral triangle $ABC$: $A_1$, $A_2$ on $BC$, $B_1$, $B_2$ on $CA$ and $C_1$, $C_2$ on $AB$, such that they are the vertices of a convex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths.

Prove that the lines $A_1B_2$, $B_1C_2$ and $C_1A_2$ are concurrent.

Bogdan Enescu, Romania
This post has been edited 1 time. Last edited by Valentin Vornicu, Oct 3, 2005, 1:59 AM
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schulmannerism
171 posts
#2 • 2 Y
Y by Adventure10, Mango247
Solution:
Click to reveal hidden text
EDIT: The argument I use to prove the first claim is wrong--it fails when the hexagon has acute angles. Many other solutions in this thread have similar flaws, see Darij's post later in the thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=44478&postorder=asc&start=26
This post has been edited 2 times. Last edited by schulmannerism, Apr 12, 2006, 11:21 PM
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Sly Si
777 posts
#3 • 2 Y
Y by Adventure10, Mango247
Please tell me if this hunch is on the right track:
Click to reveal hidden text
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al.M.V.
349 posts
#4 • 2 Y
Y by Adventure10, Mango247
Using schulmannerism's claim, one can also easily finish Ceva's Theorem using ratio of areas (half of product of two sides and the sine of the angle in between.) :)
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darij grinberg
6555 posts
#5 • 2 Y
Y by Adventure10, Mango247
Strange why most participants I have met used inequalities to solve this problem. My solution, at least, doesn't explicitely make use of them (of course, it uses the arrangement of the points, which can be considered as a kind of inequalities, but I guess there is no way to avoid using it...).

So here is, depending of how soon we will quit the internet cafe, my solution or just a rough outline of it:

Since the hexagon $A_1A_2B_1B_2C_1C_2$ has equal sidelengths, we have $C_2A_1=A_1A_2$; thus, the triangle $C_2A_1A_2$ is isosceles, so its base angle is $\measuredangle A_1C_2A_2=\frac{180^{\circ}-\measuredangle C_2A_1A_2}{2}=\frac{\measuredangle C_2A_1B}{2}$. Similarly, $\measuredangle C_1A_1C_2=\frac{\measuredangle A_1C_2B}{2}$. But by the sum of angles in triangle $C_2BA_1$, we have $\measuredangle C_2A_1B+\measuredangle A_1C_2B=180^{\circ}-\measuredangle C_2BA_1=180^{\circ}-\measuredangle ABC$. Finally, < ABC = 60°, since the triangle ABC is equilateral. Hence, if the lines $C_1A_1$ and $C_2A_2$ intersect at a point K, the exterior angle theorem in triangle $C_2KA_1$ yields

$\measuredangle A_2KA_2=\measuredangle A_1C_2K+\measuredangle KA_1C_2=\measuredangle A_1C_2A_2+\measuredangle C_1A_1C_2$
$=\frac{\measuredangle C_2A_1B}{2}+\frac{\measuredangle A_1C_2B}{2}=\frac{\measuredangle C_2A_1B+\measuredangle A_1C_2B}{2}$
$=\frac{180^{\circ}-\measuredangle ABC}{2}=\frac{180^{\circ}-60^{\circ}}{2}=60^{\circ}$.

So the lines $C_1A_1$ and $C_2A_2$ intersect at an angle of 60°. Similarly, the lines $A_1B_1$ and $A_2B_2$ intersect at an angle of 60°, and so do the lines $B_1C_1$ and $B_2C_2$. Hence, the triangle $A_2B_2C_2$ is the image of the triangle $A_1B_1C_1$ under a spiral similitude with rotation angle 60°. [On the exam, I avoided speaking of "angles between lines" without properly defining the orientation of these angles, and thus I rewrote the above argument in a non-transformational way.]

Now, let Z be the center of the spiral similitude mapping the triangle $A_1B_1C_1$ to the triangle $A_2B_2C_2$. Then, since the rotation angle of the similitude is 60°, the triangles $ZA_1A_2$, $ZB_1B_2$, $ZC_1C_2$ are all directly similar, and have the angles $\measuredangle A_1ZA_2=\measuredangle B_1ZB_2=\measuredangle C_1ZC_2=60^{\circ}$. Actually, since $A_1A_2=B_1B_2=C_1C_2$ (as the hexagon $A_1A_2B_1B_2C_1C_2$ has equal sidelengths), these triangles are even congruent. Thus, for instance, $ZA_1=ZB_1$. Together with $A_1A_2=B_1A_2$ (again since the hexagon $A_1A_2B_1B_2C_1C_2$ has equal sidelengths) and $ZA_2=ZA_2$, the triangles $ZA_1A_2$ and $ZB_1A_2$ are congruent, so that $\measuredangle B_1ZA_2=\measuredangle A_1ZA_2$. Since $\measuredangle A_1ZA_2=60^{\circ}$, we thus have $\measuredangle B_1ZA_2=60^{\circ}$. Hence,

$\measuredangle A_1ZA_2+\measuredangle B_1ZA_2+\measuredangle B_1ZB_2=60^{\circ}+60^{\circ}+60^{\circ}=180^{\circ}$.

And thus, the point Z lies on the line $A_1B_2$. Similarly, the point Z lies on the lines $B_1C_2$ and $C_1A_2$. And we are done.

Oh my goodness, I guess 75% of my above argumentation were redundant. But actually, the whole thing took me 25 minutes, and I found it too much (I expected a geometry problem 1 to be doable in 10 minutes, like the one last year, but this one was harder), so I did not try to simplify but rather wrote it up as quickly as possible to have the time for the other two, especially since problem 2 seemed to be absolutely hard :D . Anyway, a good first day, although not for me, since I could have done the third one...

Darij
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Anto
213 posts
#6 • 2 Y
Y by Adventure10 and 1 other user
The whole point of the problem is to prove that $A_1 B_1 C_1$ and $ A_2 B_2 C_2$ are equilateral.
To prove that let angle $C_1 B_1 B_2 = b$, $C_2 C_1 A_1 = c$ and $A_1 B_1 A_2 = a$.
Then we easily get : $ A_1 C_1 B_1 = 60 +b-c$, $ C_1 A_1 B_1 = 60 + c-a$ and $ A_1 B_1 C_1 = 60 + a-b$.
We can also prove in an easy way that : $ |a-b| < 30 $... and $ a<60$ ... respectivly.
Then by the cosine rule to the triangle $C_1 C_2 B_1$ we get that : $(C_1 B_1)^2 = (2x cos(b))^2 $ ...
By the sine rule to the trianlge $A_1 B_1 C_1$ and some calulation, we get the following system of equaltions :
(1) $ sin(a+b-c)\cdot cos(60 +b-a)=\frac{1}{2} \cdot sin(c) $
(2) $ sin(b+c-a)\cdot cos(60 +c-b)=\frac{1}{2} \cdot sin(a) $
(3) $ sin(a+c-b)\cdot cos(60 +a-c)=\frac{1}{2} \cdot sin(b) $
Now, if $a=b$ we easily get from (1) that $a=b=c$.
Let , wlog, that $a>b$. then from (1) we get that $sin(c) < sin(a+b-c)$ or $ 2c<a+b$ so $c<b$.
If $a>c$ then similarly from (3) $ 2b <a+c$ . Contradiction.
If $b>c>a$ then from (2) $sin(b+c-a) > sin(c)>sin(a)$ so $\frac{1}{2} > cos(60+c-b)$ so $60+c-b > 60$ or $c>b$ . Contradiction.
So $a=c$ but then from (3) we get $a=b=c$. Contradiction.
At last, we proved that $a=b=c$.
But then $A_1 B_1 C_1$ is equilateral and angle $C_1 C_2 A = B_1 A_2 C = A_1 C_2 B$. Simirly for the triangle $A_2 B_2 C_2$ and finally it is obvious that triangles $A C_1 B_2 = B A_1 C_2 = C A_2 B_1$.
The rest follows easily.
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Matija
53 posts
#7 • 2 Y
Y by Adventure10, Mango247
it seemed to me that this problem is harder than the 2nd, but only because I was looking for a more complicated solution :huh: :)

denote |A(1)A(2)|=d, |AB|=a, |A(2)C|=x, |B(2)A|=y, |C(2)B|=z.
let's assume x>y, then a-d-y>a-d-x, and from triangles BA(1)C(2) and CB(1)A(2) follows that z>x, then a-d-x>a-d-z and similarly from the 2 triangles BA(1)C(2) and AC(1)B(2) it follows that y>z.
So we have x>y>z>x, a contradiction. Similarly for x<y. So x=y and by symmetry x=y=z.
Hence A(1)B(1)C(1) and A(2)B(2)C(2) are equilateral and the lines from the problems are their heights (medians, bisectors.. ) and all meet in one point.
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shobber
3498 posts
#8 • 2 Y
Y by Adventure10, Mango247
Some say it is not an IMO problem? Well, I agree, since even I am able to solve it.

My solution to problem one:

Lemma(Poland 1965--1966)
If the diagnals AD, BE, CF half the area of the convex hexagon, then AD, BE and CF are concurrent.

Proof: Assume the do not meet at one point.
Let's say $O= AD \cap BE$, $P= CF \cap BE$, $Q= AD \cap CF$
Since AD, BE, CF half the area, so:
\[\triangle{AOB}=\triangle{DOE}, \triangle{EFP}=\triangle{BCP}, \triangle{CQD}=\triangle{AQF}\]
then:

$1=\frac{\triangle{AOB}}{\triangle{DOE}} \cdot \frac{\triangle{EFP}}{\triangle{BCP}} \cdot \frac{\triangle{CQD}}{\triangle{AQF}}= \frac{AO \cdot BO}{DO \cdot EO} \cdot \frac{EP \cdot FP}{BP \cdot CP} \cdot \frac{CQ \cdot DQ}{AQ \cdot FQ}$

$= \frac{AO}{AQ} \cdot \frac{BO}{BP} \cdot \frac{CQ}{CP} \cdot \frac{DQ}{DO} \cdot \frac{EP}{EO} \cdot \frac{FP}{FQ} $

$ \neq 1$

Hence they must be concurrent.

Now get back to this problem.

Since $C_1C_2 = B_1B_2= A_1A_2$, and $A=C=B=60^o$, so we shall have:
$R_{BA_1C_2}=R_{AC_1B_2}=R_{CA_2B_1}$
($R$ is the circumradius.)

Assume $A_2C > B_2A$. Since $BA_1+A_2C=CB_1+B_2A$, so $BA_1 < CB_1$.

Since $2R=\frac{BA_1}{\sin{BC_2A_1}}=\frac{CB_1}{\sin{B_1A_2C}}$

So $\angle{BC_2A_1} < \angle{CA_2B_1}$
So $ \angle{BA_1C_2} > \angle{CB_1A_2}$

Use the law of sines again, we get $BC_2 > CA_2$.
So $BC_2 > B_2A$. Hence $\angle{BA_1C_2} > \angle{AC_1B_2}$,
Hence $\angle{BC_2A_1} < \angle{AB_2C_1}$
Hence $BA_1 < AC_1$.
Because we already have $C_2B > CA_2$, so add them togather we have $AB-C_1C_2 > BC- A_1A_2$.

It is impossible.

By symmetry, we can know that $CA_2=BC_2=AB_2$, $A_1B=C_1A=B_1C$.

The rest is easy. We can prove that
$\triangle{AC_1B_2}=\triangle{CB_1A_2}=\triangle{BA_1C_2}$
$\triangle{A_1C_1C_2}=\triangle{B_1A_1A_2}=\triangle{C_1B_1B_2}$
$\cdots \cdots \cdots$
It is easy to prove that the diagnols half the area of the hexagon.
Use our lemma now...

Cheers!
How much point can this proof recieve?
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shobber
3498 posts
#9 • 2 Y
Y by Adventure10, Mango247
I shall also add a pic for the lemma.
P.S. How do I add two pictures in one post?
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xdj
1 post
#10 • 6 Y
Y by MathBoy23, k12byda5h, Adventure10, Mango247, NCbutAN, and 1 other user
This question is very easy.
Step 1: prove AC1=BA1=CB1.

Shift AC1 to A'C'', C2B to C''B', AB2 to A'B'', B1C to B''C'. Then it is easy to show that B'A'' and A''C' are also shifted BA1 and A2C. Since triangle A'B'C' and A''B''C'' are both equilateral triangles, A'C''=B'A''=C'B'' => AC1=BA1=CB1

Step 2: A2C1 is the middle line of C2B2 and B1A1 and so on. It is easy.

Step 3: triangle B2C2A2 is an equilateral triangle so the three lines meet.
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Virgil Nicula
7054 posts
#11 • 1 Y
Y by Adventure10
:) I note: $BA_1=a_1,CA_2=b_1;CB_1=a_2,AB_2=b_2;AC_1=a_3,BC_2=b_3.$
:roll: ...........$A_1A_2=B_1B_2=C_1C_2=x,a_1+a_2+a_3=l,$ $a_1+b_1=a_2+b_2=a_3+b_3=k.$
:blush: ****************************************************************************
$a_1^2+b_3^2-a_1b_3=a_2^2+b_1^2-a_2b_1=a_3^2+b_2^2-a_3b_2=x^2.$
$a_1^2+(k-a_3)^2-a_1(k-a_3)=a_2^2+(k-a_1)^2-a_2(k-a_1)...iff...$
$a_3^2-2ka_3+ka_1+a_1a_3=a_2^2-ka_2+a_1a_2...iff...$
$(a_3-a_2)(a_1+a_2+a_3)+k(a_1+a_2-2a_3)=0\ a.s.o.$
$l(a_3-a_1)=k(a_2+a_3-2a_1);\ l(a_1-a_2)=k(a_3+a_1-$ $2a_2);\ l(a_2-a_3)=k(a_1+a_2-2a_3).$ (*)
*****************************************************************************************
I pressupose that $a_1\ne a_2\ne a_3\ne a_1.$ Thus,

$\frac{a_2+a_3-2a_1}{a_3-a_1}=\frac{a_3+a_1-2a_2}{a_1-a_2}=\frac{a_1+a_2-2a_3}{a_2-a_3}=\frac{l}{k}...iff...$
$1-\frac{l}{k}=\frac{a_1-a_2}{a_3-a_1}=\frac{a_2-a_3}{a_1-a_2}=\frac{a_3-a_1}{a_2-a_3}=$ $\sqrt [3]{\frac{(a_1-a_2)(a_2-a_3)(a_3-a_1)}{(a_3-a_1)(a_1-a_2)(a_2-a_3)}}=1...iff...l=0$, what is absurd.

Thus, there are $i,j\in \{1,2,3\},\ i\ne j$, such that $a_i=a_j.$ From the relations (*) results $a_1=a_2=a_3,$ a.s.o. :first: :rotfl:

Thanks, Darij Grinberg, for correction of calculus !
This post has been edited 3 times. Last edited by Virgil Nicula, Jul 31, 2005, 6:48 AM
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vnmathboy
33 posts
#12 • 6 Y
Y by Tommy2000, Adventure10, Mango247, and 3 other users
I have beautiful solution:

I
is point inside convex hexagon
A_1A_2B_1B_2C_1C_2
such that
IA_1A_2
is regular triangle. It's easy to prove that
IC_1B_2
is regular triangle. Let the lines $A_1B_2$ and $A_2C_1$ meet at $M$. So
I, M, B_1
are collinear (because they lie on the perpendicular bisector of segment
B_2A_2
); and
I, M, C_2
are collinear (because they lie on the perpendicular bisector of segment
A_1C_1
). So
C_2, M, B_1
are conlinear or
A_1B_2, B_1C_2, C_1A_2
are concurrent.

[Moderator edit: Corrected some language mistakes and defined point M.]
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jarod
184 posts
#13 • 2 Y
Y by Adventure10, Mango247
I think only Darij 's proof and Anto 's proof are valuable .
Check your proofs again ! Something is vague .
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Piotr Achinger
6 posts
#14 • 4 Y
Y by Swistak, Adventure10, and 2 other users
check out this one:

as the three vectors A1A2, B1B2, C1C2 are similar (in the same scale) to BC, AC, AB respectively,
we have A1A2+B1B2+C1C2=0, so A2B1+B2C1+C2A1 is also a zero vector,
so we can build an equilateral triangle from vectors A2B1, B2C1, C2A1, which has its sides parallel
to the sides of the triangle made by the prolongations of A2B1, B2C1, C2A1,
so this triangle is also equilateral (denote its vertices as A',B',C').

as we have angle B1B'C1=60=B1AC1, B1B'AC1 is cyclic, so angle B'B1A=B'C1A, so the angles
B1 and C1 of the hexagon are equal.

analogically we can prove that the angles of the hexagon A1=B1=C1=alpha, A2=B2=C2=beta,
so A1B2, B1C2, C1A2 are the axes of symmetry of the hexagon....
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aodeath
69 posts
#15 • 2 Y
Y by Adventure10, Mango247
Steps of the proof:

1 - Construction of the hexagon.
2 - Show that
A_1B_1C_1
and
A_2B_2C_2
are equilateral triangles.
3 - Show that the point of concurrence is the circuncenter of the triangle
ABC
.

1 - It's easy to see that we can construct the hexagon described by the problem by the following manner:

Draw the equilateral triangle and its circuncircle. Choose other three points on the circuncircle and draw another equilateral triangle, different from
ABC
, named
\overline{A}\overline{B}\overline{C}
. The points where the triangles concur are exactly the vertices of the hexagon.

2 - By marking some anlgles and with the known fact of congruence, it's very easy to see that the triangles
A_1B_1C_1
and
A_2B_2C_2
are equilateral.

3 - Call
P
the circuncenter of the triangle
ABC
. Draw
AP,C_1P, B_1P
and
\angle{AC_1B_2}=\alpha
. We know that
\overline{C_1P}
bissects
\angle{B_2C_1C_2}
, and then
\angle{B_2C_1P}=\frac{\pi-\alpha}{2}
and using the fact that
C_1A\overline{A}B_1
are concyclic e conclude that
\angle{AB_1P}=\frac{\pi-\alpha}{2}
, since
\angle{AC_1P}+\angle{AB_1P}=\pi
then, this quadrilateral is cyclic to, which proves that
\overline{PC_1}=\overline{PB_1}
. By the same way, you prove the same for the other segment, which makes us conclude that
P
is the circuncenter of
A_1B_1C_1
. Analogously, you can prove the
P
is circuncenter of
A_2B_2C_2
. Since we know then that
P
belong to the mediatrice of the equilateral triangle, we conclude that the segments concur in
P


If you find any mistake, I'll be glad to be noticed =]

thanks =]
Z K Y
G
H
=
a