[/quote]
,
,
are chosen such that when going around the triangle, the points occur in the order 
.![\[
AD_1 = AD_2 = BE_1 = BE_2 = CF_1 = CF_2.
\]](http://latex.artofproblemsolving.com/4/c/f/4cf9e7bc009736836bcd10c1d8ba46bd3a91cd7b.png)
Prove that the perimeters of the triangles formed by the lines 
and 
are equal.
Prove that :
Where 
such that 
Prove that 
Where 
Prove that
Where 
be a triangle such that 
.
,
at points 
,
,
be its centre. Let 
be a point on the side 
and 
intersect for the second time at 
.
be the foot of the perpendicular from 
.
,
beyond 
,
beyond 
.
distinct points in a plane such the circumradius of any 
of them is the same.
,
.
and 
at 
,
and 
,
intersects the circles 
and 
again at 
and 
,
and circumcentre 
.
and 
lie on 
and 
.
meet 
meet 
and 
have an intersection lie on line 
.
with three fixed points 
and 
.
on 
be the point on the segment 
such that 
. Let 
and 
.
.
,
.
is tangent to 
.
be the midpoints of segments 
intersects 
.
,
meet at point 
and 
and 
is equal to the distance between the orthocenters of triangles 
and 
.
be the touchpoints of the incircle with 
,
is the Euler line of 
.
and medial triangle 
.
.
is the midpoint of segment 
,
is isoceles.
Y by HA5x5, yshk, rstenetbg, deplasmanyollari, Funcshun840
Let 
be a positive integer. Given is a subset 
of 
with 
elements. Prove that there exist three elements 
from such that 
.
Proposed by Dominik Burek and Tomasz Ciesla, Poland
be a positive integer. Given is a subset 
of 
with 
elements. Prove that there exist three elements 
from such that 
.Proposed by Dominik Burek and Tomasz Ciesla, Poland
This post has been edited 3 times. Last edited by VicKmath7, Dec 22, 2022, 4:09 PM
,
.
for all 
,
.
.
.
(with the idea in mind that we do not win more if we do not pick the largest element of 
and 
to be consecutive in the above sequence (in any other case the difference 
shall be larger). If we prove that 
for at least one 
,
for all 
yields 
and so 
,
.
,
,
and 
with 
.
such that 
such that there are no 
and 
are chosen at the moment. Take 
,
,![$\left[\frac{5^n}3\right]+1$](http://latex.artofproblemsolving.com/8/8/8/8882ebdd0721f62d56f6f5da3b64e411d592c7c5.png)
cannot be chosen, so the smallest number can be chosen is greater than 
.![$\left[\frac{5^n+\frac23\cdot 5^n}3\right]+1$](http://latex.artofproblemsolving.com/d/e/e/dee088c497e0d00c3330ee9e1e43625a50e1473b.png)
cannot be chosen, so the smallest number can be chosen this time is greater than 
.
times.
and 
for all 
.
is greater than 
.
.
,
,
.
.
does the inductive step.
time, is greater than 
,
the maximal element of 
inclusive in above. Therefore, there is no way we can choose numbers such that there are no 
with 
.
for all 
.
by induction on 
,
.
Therefore, 
since 
.
,
,
.
such that 
for all 
with 
If the elements of 
then 
for all 
Induct on 
the base case is true because 
Take 
by our assumption, we have ![\[s_{4n+2}+2s_{k-1}\ge 3s_k.\]](http://latex.artofproblemsolving.com/6/b/8/6b8499bcc81210f319a40f1c7b70dbacf0dc4bb1.png)
Therefore,
as desired. 
This is impossible, as 
is an integer less than 
(of course, let 
)
,
,
,
by multiplying them. Then 
,
.
,
satisfies the condition. Assume not. Let 
.
.
.
and 
.
which is an obvious contradiction so we are done.
.
of indices 
such that 
.
,
also works.
,
.![\[2(x_{j-1} - x_{4n+2}) \le 3(x_{j} - x_{4n+2}) \implies x_{4n+2}-x_j \le \frac{2}{3} (x_{4n+2}-x_{j-1}) \]](http://latex.artofproblemsolving.com/6/0/2/60205e37ef5f2f5ac2d980b0ba34a022bc1241f4.png)
In particular, this means ![\[x_{4n+2} - x_{4n+1} \le \left ( \frac{2}{3} \right )^{4n} (x_{4n+2} - x_1) \le \left ( \frac{2}{3} \right )^{4n} (5^n - 0) = \left ( \frac{80}{81} \right )^n < 1,\]](http://latex.artofproblemsolving.com/2/d/7/2d74879839a8a4afe1d02a92da44922e0436be04.png)
a contradiction.
.
.
;
for all 
.
.![\[a_{4n+1}-a_{4n}\ge \frac{d}{2}\]](http://latex.artofproblemsolving.com/9/1/c/91c38c781ac5d1a2e078392a871baec3a15db236.png)
![\[a_{4n}-a_{4n-1}\ge \frac{3d}{4}\]](http://latex.artofproblemsolving.com/f/3/c/f3cb7d93f9bda97fe48524f6b2ac058a7d5f496c.png)
![\[\vdots\]](http://latex.artofproblemsolving.com/7/6/4/76403adfdd35fb846675aaeaac92daec680a4fba.png)
Finally we obtain 
which is a contradiction. We are done. 
,
if 
Assume FTSOC that 
for all 
Thus, we get that 
-
If 
is 
is 
-
is 
-
-
-
is 
-
We claim that it is forced to have 
To do this, we show that 
This is true, since 
Thus, we reach a contradiction, so we're done.
and let 
Suppose FTSOC that 
for all 
which implies 
for all 
We claim by induction that 
for 
Indeed, 
and 
implies 
completing the induction. Notice 
since 
and 
Hence, 
so 
a contradiction. 
for all 
in 
Then 
where 
we have for all 
and considering the triple 
that 
and 
and 
etc. We can easily see that in general we have by induction 
by induction, as 
where the last inequality holds for all 
as 
follows from 
This is a contradiction. 
,
,![\[a_{4n + 2} + 2a_{k} \leq 3a_{k + 1} \implies \frac{a_{4n + 2} - a_{k + 1}}{a_{4n + 2} - a_k} \leq \frac{2}{3}.\]](http://latex.artofproblemsolving.com/e/9/c/e9c8828317f5173dd9cabcdbf6aaa769593e3dc2.png)
But then ![\[\frac{a_{4n + 2} - a_{4n + 1}}{a_{4n + 2} - a_1} = \prod_{k = 1}^{4n} \frac{a_{4n + 2} - a_{k + 1}}{a_{4n + 2} - a_k} \leq \left(\frac{2}{3}\right)^{4n} < \frac{1}{5^n},\]](http://latex.artofproblemsolving.com/b/2/a/b2a1ae8de89f318cd952ba8aeec373bf82af05c3.png)
contradiction (as we must have 
and 
)
Now, let our set of positive integers be 
in increasing order. Now, applying the fact that 
to 
gives us that
Now, by an easy induction, we find that
Now,
so 
But, these numbers are distinct, so we have a contradiction.
for 
such that 
.
.
so 
.
,
.
,
,
such that 
.
.
and dividing out common terms on both sides 
.
.
,
.
with 
.
for 
.
such that 
and 
.
.
,
.
.
.
for 
.
.
,
which is consistent with our conjecture.
we have that for all 
,
.
.
.
.
.
,
for all 
.
,
.![$i\in[1,4n]$](http://latex.artofproblemsolving.com/e/e/e/eee855628e884d69410582d1dc980e9fff88efbd.png)
,
But since 
and 
,
which is a contradiction, as desired. 
.
for all 
,
,
,
,
,
,
Let 
.
to
So, 
and as 
and 
we finally have
which is a contradiction.
and 
.
,
.
implying that 
,
and assume FTSOC 
.
and exploit the confinment of 
But using the miraculous fact that 
we get an obvious contradiction.
to be the elements of 
.
,
.
.
and thus, contradiction.
.
can be selected without needing to change 
.
for 
must have each term at least a factor of 
greater than the previous term. Since all elements of 
and therefore 
.
producing a contradiction, since this implies the difference between some two elements of 
where 
.
such that 
.
vary it is clear that 
is a necessary and sufficient condition.
These imply
Which clearly is a contradiction.
,
and 
for some ![\[c < 3a_{k+1} - 2a_k \implies a_{k+1} > \frac{2}{3} a_k + \frac{c}{3} \qquad \text{for all } 1 \leq k \leq 4n \]](http://latex.artofproblemsolving.com/5/1/c/51cbd66458b776fb2b0b8cda5e88c1fd3beafea8.png)
Now this implies that, since 
,
.![\[ a_{k+1} - c > \frac{2}{3} (a_k - c) \]](http://latex.artofproblemsolving.com/8/5/d/85d0e0a7daa67b18284ce137e6e91fcda26fb250.png)
to get 
.![\[ c > a_{4n+1} > \left( 1 - \left(\frac{2}{3}\right)^{4n} \right) c \]](http://latex.artofproblemsolving.com/a/3/2/a32e6bcf7112e93b95b0d46b552442566954054e.png)
Now I will show that![\[ \left( 1 - \left(\frac{2}{3}\right)^{4n} \right) c > c-1 \]](http://latex.artofproblemsolving.com/6/e/9/6e9a2c9a7e161ad70c2a73e14d51f544492bb5d0.png)
.
.![\[ 1 > \left(\frac{16}{81}\right)^n c,\]](http://latex.artofproblemsolving.com/9/1/5/9150ccb93cf17309c757c5ad011ae95d8c3431d2.png)
which is true since 
.![\[ c > a_{4n+1} > c-1 \]](http://latex.artofproblemsolving.com/f/c/8/fc8e5e5a4577bd8d1e0e57f678eac58b2f363c93.png)
which leaves 
that doesn't satisfy the conditions but exists entirely within the bound 
.
,
,
.
,
,
(where 
is essentially the distance between the last term and term 
is at most 
is at least 
,
