An integer set [i][b]T[/b][/i] is orensan if there exist integers[b] a<b<c[/b], where [b]a [/b]and [b]c[/b] are part of [i][b]T[/b][/i], but [b]b[/b] is not part of [b][i]T[/i][/b]. Count the number of subsets [b][i]T[/i][/b] of {1,2,...,2019} which are orensan.

Let $a$,$b$,$c$ be the roots of the equation $x^{3} - 2018x +2018 = 0$. Let $q$ be the smallest positive integer for which there exists an integer $p, \, 0 < p \leq q$, such that $$\frac {a^{p+q} + b^{p+q} + c^{p+q}} {p+q} = \left(\frac {a^{p} + b^{p} + c^{p}} {p}\right)\left(\frac {a^{q} + b^{q} + c^{q}} {q}\right).$$ Find $p^{2} + q^{2}$.

A fudgeflake is a planar fractal figure with a $120^{\circ}$ rotational symmetry such that three identical fudgeflakes in the same orientation fit together without gaps to form a larger fudgeflake with its orientation $30^{\circ}$ clockwise of the smaller fudgeflakes' orientation, as shown below. If the distance between the centers of the original three fudgeflakes is $1$, what is the area of one of those three fudgeflakes? Justify your answer. [asy] defaultpen(linewidth(.7)); string s = "00d08c8520612022202288272220065886,00e0708768822888788866683,01006c8765,01206b88227606,01208c858768588616678868160027,017068870228728868822872272220600278886,0190988886872272882228166060,02209486868506,02304b282022852282022828888828228166060066002000668666,02304d8882858688886166702,023050,0230637222282272220786,0240918681,02505f,0260908527222027285886852728522820766,02905b81,03904422888888766686882288888607,03908c58588685876668688228882078688228822220258587685886166702,049053852282000027888666,0490fa852061112222282282000702202220065868,04a0ae868822888888866602768688228888860728228822820228586888766702,04a0de2821666868822888500602,04b0ac5812022882227200070,04c04a8220228822272012882876882288227606,04c0da288282220228886882288227600660020060668,04c0f2,04e0fa88868588616668688228882078688228822220786,05e0ba878605,05e0d287688872006,05e102786720,05f0b88220228822816606066860,06002786882288886888666868886166600668666868887,0600748207,0600ce88616,0600ff86816,0610258222282220228887882288228202285868886668688228858688228822827066,062023223,0620cd88227600,0620fe8527222027686,064074555555555555555555555555555555555552882288888876668688866606060276866686882288878886668668522888868527282822816660222228588688861668678886166600668666868886660602782228,0650c985877,0680b0865282888882822202288878886668678282858768522882822200602788860660606,06a0fa88886070220,06c0b48830,0700fe8527222206702,0790fb867888666868822888788228722722720128288768886668688866600668666027686882288886888606786882288888607,0870b9822202288222022888886058,0880428582288868886668688228828768822882282166606060602,088070816,08903e7870,08a06c877,08b02a82858868886067222006,08b03b867202285272220786,08c06b8528220676,09003572,0900745555555555555555555555555555555555555555555555555555555555555555555555555558207,0920668858888285272220786,09606c,09812e88228166,09a128868827,09b12682202288222766060061,09c0648867,09c094868672020228160,09d04b86787228828868886668688866616006686605820228,09d0e38207,09e0908282822202288888828282220228886888681666868522888888285228202282886685886066886888606606666678868160027,09f0478765,0a10468822220276866606,0a411e828886882288227606060602,0a90428888681288227600,0aa046,0aa091,0b111f8282886788766866852288586882288228216660606,0b2122,0b30438861,0b5042858868886668688228882078688228822760600660,0b805f28868686868686850633,0be09d,0c3127868681020222022882276066660,0c40de86788228728868886668688866606066860678688228886888681666868527285886872272882228166602,0c504b88688810600220222027686,0c60da86816,0c70a8,0c80d98527222027686,0c90b08886888666868822888886058228822820228287688868166768886816606006786667021,0cb046827222025888685272850,0cc11f8702288888605822882281660602,0d00d588886070220,0d60d98527222070070,0d811a86868850607,0d90b8786783,0da0268768822882876888666868886661600668666868886816611,0da107272858868886166786888616660066870202,0db0b586816,0dc02527201288227600,0dd0b48527222027686,0e301e8861,0e501d88888605822882222027686,0e50b0888860702,0eb0b48527222206702,0f401d88868886166702,0f40b1867888666868822888788228822202786,0f50298822882888860652288227600,0fc02d,1010b98765,10206f82222282220228888882828222006588606600660020066660660066868,10207288228888685866686888681636160706768166686882288878886706786882288868822885228166,10303e28216668688228887685886166702,1030b8220258527227,10502b87,10503a288282220065886066006,10602958882078688866683,10904e8527285886882288227600,10a0b127282887685272858122816,1160ad86885,11706327683,1170ab8216,11904f877,11905f87022872812220605886,119097882886888666868886661600658,11b04e858868886166783"; string[] k = split(s,","); for(string str:k) { string a = substr(str,0,3),b=substr(str,3,3),c=substr(str,6); real x=hex(a),y=hex(b); for (int i=0;i<length(c);++i) { int next = hex(substr(c,i,1)); real yI=(int)((next-next%3)/3),xI=next%3; --xI; --yI; draw((x,-y)--(x+xI,-(y+yI))); x+=xI; y+=yI; } }[/asy]

Suppose $P(x)$ is a non-constant polynomial with real coefficients, and even degree. Bob writes the polynomial $P(x)$ on a board. At every step, if the polynomial on the board is $f(x)$, he can replace it with 1. $f(x)+c$ for a real number $c$, or 2. the polynomial $P(f(x))$. Can he always find a finite sequence of steps so the final polynomial on the board has exactly $2020$ real roots? What about $2021$? [i]~Sutanay Bhattacharya[/i]

Is it is possible to choose 24 distinct points in the space such that no three of them lie on the same line and choose 2019 distinct planes in a way that each plane passes through at least 3 of the chosen points and each triple belongs to one of the chosen planes?

Rohit is counting the minimum number of lines $m$, that can be drawn so that the number of distinct points of intersection exceeds $2020$. Find $m$. (A)$63$ (B)$64$ (C)$65$ (D)$66$

Find the maximum possible number of kings on a $12\times 12$ chess table so that each king attacks exactly one of the other kings (a king attacks only the squares that have a common point with the square he sits on).

A triangle $ABC$ is given in the plane with $AB = \sqrt{7},$ $BC = \sqrt{13}$ and $CA = \sqrt{19},$ circles are drawn with centers at $A,B$ and $C$ and radii $\frac{1}{3},$ $\frac{2}{3}$ and $1,$ respectively. Prove that there are points $A',B',C'$ on these three circles respectively such that triangle $ABC$ is congruent to triangle $A'B'C'.$

Two circunmferences $ \Gamma_1$ $ \Gamma_2$ intersect at $ A$ and $ B$ $ r_1$ is the tangent from $ A$ to $ \Gamma_1$ and $ r_2$ is the tangent from $ B$ to $ \Gamma_2$ $ r_1 \cap r_2\equal{}C$ $ T\equal{} r_1 \cap \Gamma_2$ ($ T \neq A$) We consider a point $ X$ in $ \Gamma_1$ which is distinct from $ A$ and $ B$. $ XA \cap \Gamma_2 \equal{}Y$ ($ Y \neq A$) $ YB \cap XC\equal{}Z$ Prove that $ TZ \parallel XY$

Let $P$ be an arbitrary point inside triangle $ABC$, denote by $A_{1}$ (different from $P$) the second intersection of line $AP$ with the circumcircle of triangle $PBC$ and define $B_{1},C_{1}$ similarly. Prove that $\left(1 \plus{} 2\cdot\frac {PA}{PA_{1}}\right)\left(1 \plus{} 2\cdot\frac {PB}{PB_{1}}\right)\left(1 \plus{} 2\cdot\frac {PC}{PC_{1}}\right)\geq 8$.

Let $a,b,c$ be real numbers $a + b +c = 0$. Show that [list=a] [*] $\displaystyle \frac{a^2 + b^2 + c^2}{2} \cdot \frac{a^3 + b^3 + c^3}{3} = \frac{a^5 + b^5 + c^5}{5}$. [*] $\displaystyle \frac{a^2 + b^2 + c^2}{2} \cdot \frac{a^5 + b^5 + c^5}{5} = \frac{a^7 + b^7 + c^7}{7}$. [/list] [I]Folklore[/I]

For a certain value of $x$, the sum of the digits of $10^x - 100$ is equal to $45$. What is $x$? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$

Call a pair of integers $a$ and $b$ square makers , if $ab+1$ is a perfect square. Determine for which $n$ is it possible to divide the set $\{1,2, \dots , 2n\}$ into $n$ pairs of square makers.

If $x$ and $y$ are nonnegative real numbers such that $x+y=1$, determine minimal and maximal value of $$A=x\sqrt{1+y}+y\sqrt{1+x}$$

Let $a,b$ be two constants within the open interval $(0,\frac{1}{2})$. Find all continous functions $f(x)$ such that \[f(f(x))=af(x)+bx\] holds for all real $x$. This is much harder than the problems we had in the 1st TST...

Given a triangle $ABC$ with $\angle ACB = 120^{\circ}.$ A point $L$ is marked in the side $AB$ such that $CL$ bisects $\angle ACB.$ Points $N$ and $K$ are chosen in the sides $AC$ and $BC $ such that $CK+CN=CL.$ Prove that the triangle $KLN$ is equilateral.