[/quote]
,
be real numbers satisfying 
and ![\[a_{i+1} - 2a_i + a_{i-1} = a^2_i\]](http://latex.artofproblemsolving.com/c/d/8/cd878e4ac230853182da75c54dcbf7be05b305bc.png)
for 
Prove that 
for 
![$ a,b\in [0,1] $](http://latex.artofproblemsolving.com/9/7/5/975edfc0cf5316d78283b313c6b5cfbdd7b92b6e.png)
. Prove that
Where 
Where 
,
letters 
can be placed in each cell of a table with 
columns, and has the following property: For any two distinct rows in the table, there exists at most one column, such that the entries of these two rows in such a column are the same letter.
be an continuous function defined in ![$\text{[0,2015]},f(0)=f(2015)$](http://latex.artofproblemsolving.com/c/9/9/c99f63e726b504aea8c1a65ba2926155a873effa.png)
pairs of 
such that 
such that ![\[f(f(x)) + xf(xy) = x + f(y)\]](http://latex.artofproblemsolving.com/a/6/f/a6f84dd53eec985b90bad3df6580eeaf29d73195.png)
for all positive real numbers 
and 
.
and 
. Prove that 
. Prove that
.
and 
,
be the point on line 
different from 
such that the line through 
is tangent to 
be the point on line 
such that the line through 
is tangent to 
intersect the circumcircle of triangle 
.
and 
be the midpoints of 
.
with![\[2x^3zf(z)+yf(y) \ge 3yz^2f(x)\]](http://latex.artofproblemsolving.com/a/b/f/abf10d7e679f6f04db556424a0a0eda26f371fd4.png)
for all 
.
be positive real numbers such that : 
. Prove that : 
.
points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so?
Y by
Let 
be the circumcenter of triangle 
. The altitudes from 
of triangle intersects the circumcircle of the triangle at 
respectively. 
meets 
at 
respectively. Prove that the circumcircles of triangles 
share two common points.
Proporsed by wassupevery1
be the circumcenter of triangle 
. The altitudes from 
of triangle intersects the circumcircle of the triangle at 
respectively. 
meets 
at 
respectively. Prove that the circumcircles of triangles 
share two common points.Proporsed by wassupevery1
This post has been edited 1 time. Last edited by Jalil_Huseynov, Dec 28, 2021, 12:33 PM
as the unit circle. Then 
.
is equivalent to 
,
,![\[z\bar z(b^2c^2-a^4)+za(a+b)(a+c)-\bar zabc(a+b)(a+c)-a(b+c)(a^2-bc)=0.\]](http://latex.artofproblemsolving.com/1/a/c/1ac8561c211e9898c2f94123f072865353e4cddf.png)
Analogous equations hold for 
.
-
from rows 
)![\[\det\begin{pmatrix}a(a+b)(a+c)&-abc(a+b)(a+c)\\b(b+a)(b+c)&-abc(b+a)(b+c)\end{pmatrix}=-abc(a+b)^2(b+c)(c+a)(a-b)\neq 0,\]](http://latex.artofproblemsolving.com/5/0/8/5087023c6fe08c83552d9fd9810650b44a99d84f.png)
![\[\det\begin{pmatrix}a(a+b)(a+c)&-abc(a+b)(a+c)&b^2c^2-a^4\\b(b+a)(b+c)&-abc(b+a)(b+c)&c^2a^2-b^4\\c(c+a)(c+b)&-abc(c+a)(c+b)&a^2b^2-c^4\end{pmatrix}\]](http://latex.artofproblemsolving.com/2/7/6/2766e045639b75cbd0af5e2b90883b784952f619.png)
![\[=-abc\det\begin{pmatrix}a(a+b)(a+c)&(a+b)(a+c)&b^2c^2-a^4\\(b-a)(b+a)(a+b+c)&(b-a)(b+a)&(a-b)(a+b)(a^2+b^2+c^2)\\(c-a)(c+a)(a+b+c)&(c-a)(c+a)&(a-c)(a+b)(a^2+b^2+c^2)\end{pmatrix}=0,\]](http://latex.artofproblemsolving.com/4/2/f/42f12865c2342b74a08c85aa75584ee9d2b3bdb4.png)
![\[\det\begin{pmatrix}a(a+b)(a+c)&-abc(a+b)(a+c)&a(b+c)(a^2-bc)\\b(b+a)(b+c)&-abc(b+a)(b+c)&b(c+a)(b^2-ca)\\c(c+a)(c+b)&-abc(c+a)(c+b)&c(a+b)(c^2-ab)\end{pmatrix}\]](http://latex.artofproblemsolving.com/1/a/4/1a45113a190319042d9bb214db7e9208cb03624a.png)
![\[=-abc\det\begin{pmatrix}a(a+b)(a+c)&(a+b)(a+c)&b^2c^2-a^4\\(b-a)(b+a)(a+b+c)&(b-a)(b+a)&(b-a)(b+a)(ab+bc+ca)\\(c-a)(c+a)(a+b+c)&(c-a)(c+a)&(c-a)(c+a)(ab+bc+ca)\end{pmatrix}=0,\]](http://latex.artofproblemsolving.com/2/f/0/2f0dda6c58e176fd3a932372c0441bb52cdc5f3e.png)
as desired.
with 
.
be the orthocenter of 
.
.
.![\[Pow(H,(BB_1EPQ))=HB.HB_1=HC.HC_1=Pow(H,(CC_1FPQ))\]](http://latex.artofproblemsolving.com/2/5/b/25bc338ce941d522fe15491d00a179e208cca6fe.png)
Thus, 
and 
which is 
are concyclic.![\[HA.HA_1=HB.HB_1=HP.HQ\]](http://latex.artofproblemsolving.com/d/5/7/d57704179b1889e955373b01a22d8b60e2278757.png)
Which gives the desired result
inversion and reflect over the angle bisector of 
.
with respect to 
at 
respectively. Parallel lines to 
through 
intersect 
respectively. 
is on 
.
and 
.
.
is the intersection of the altitude from 
are concyclic.
.![\[\measuredangle ESG=180-\measuredangle GCA=\measuredangle C=90-\measuredangle GAB\]](http://latex.artofproblemsolving.com/1/2/c/12cefada7c54ebbdc5d81900784eda1a18372d69.png)
Hence 
.
which implies 
.
we conclude that 
is an isosceles trapezoid. Thus, 
are concyclic
.
.
are concyclic.![\[\measuredangle ECB_1=90-\measuredangle A=\measuredangle EC_1H=\measuredangle EC_1B+\measuredangle BC_1H=\measuredangle EC_1B+\measuredangle DC_1B_1=\measuredangle EC_1B+\measuredangle C_1B_1D=\measuredangle(EC_1,B_1D)\]](http://latex.artofproblemsolving.com/6/e/f/6ef842bccc65a4cdc3cc7805d8f9361037b50199.png)
Hence 
are collinear. Similarily, 
are collinear.
.![\[\frac{BB_1}{BF}=\frac{BC}{BF}.\frac{AB_1}{CE}=\frac{BC}{CE}.\frac{AB_1}{BF}=\frac{CC_1}{CE}\]](http://latex.artofproblemsolving.com/b/1/8/b18ff94e60dcbea4d31a1d920787b0b6a66fbb32.png)
And 
subsequently 
.
which yields 
.
.![\[\measuredangle EMB_1=\measuredangle ECB_1=90-\measuredangle A=\measuredangle C_1BF=\measuredangle C_1NF\]](http://latex.artofproblemsolving.com/d/8/1/d81eb8f628468a4e4aafec7254587de7e126c419.png)
Thus, 
are concyclic. By using this,![\[TM.TF.TB_1.TK=(TK.TF)(TM.TB_1)=(TN.TC_1)(TE.TL)=(TN.TE).TC_1.TL=TM.TF.TB_1.TL\]](http://latex.artofproblemsolving.com/e/8/7/e8797d66f01e4ae16dc6c77942d72bc71554e109.png)
Hence 
we see that 
is an isosceles trapezoid whose diagonals intersect at ![\[Pow(D,(B_1EC))=DB_1.DL=DC_1.DK=Pow(D,(C_1FB))\]](http://latex.artofproblemsolving.com/5/a/1/5a1f90f28bb27efccb2c38545d8cd1ab2ff8c784.png)
Which completes our proof as desired
