AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
denote the infinite parallel ruler with the parallel edges being at distance 
from each other. The following construction steps are allowed using ruler 
:
parallel to a given line 
at distance (there are two such lines, both of which can be constructed using this step);
and 
with 
two parallel lines at distance such that one of them passes through , and the other one passes through (if 
, there exists two such pairs of parallel lines, and both can be constructed using this step).
be a given positive integer. Prove that based on the three points and there exists 
such that for any 
it is possible to construct points using only on one of the excircles of the triangle.
and 
and 
is non-decelerating if 
holds for all positive integers 
. We say that a strictly increasing sequence 
is convergence-inducing, if the following statement is true for all real sequences 
: if subsequence 
is convergent and tends to 
for all positive integers 
, then sequence is also convergent and tends to . Prove that a non-decelerating sequence is convergence-inducing if and only if sequence 
, 
, 
is bounded from above.
light bulbs are arranged around a circle, and are consecutively numbered with
. Each bulb can be in one of two states: either it is on or off. In the initial configuration, days we change the current on/off configuration as
, on the -th day we start from the -th bulb and moving in clockwise direction-th day, coincides with the initial one.
let 
and 
denote the feet of the altitudes from 
and 
, respectively. Let line 
intersect circumcircle at points 
. Similarly, let line 
intersect circumcircle at points 
. Prove that the radical axis of circles 
and 
passes through the orthocenter of triangle
, 
, and![\[
a_{2k+1} = 2 + 2a_k, \quad a_{2k+2} = 2 + a_k + a_{k+1},
\]](http://latex.artofproblemsolving.com/4/e/c/4ec892df19b4ec97af9bb1e5d1954e6c26be3172.png)
for all integers 
. Determine all positive integers 
such that![\[
\frac{a_n}{n}
\]](http://latex.artofproblemsolving.com/3/d/1/3d1d012e7c41e1e09abd14626a79b1fb1cbab34d.png)
is an integer.
, find positive integer solutions 
to the equation![\[
\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} + \frac{1}{x_1 x_2 \cdots x_n} = 1
\]](http://latex.artofproblemsolving.com/5/c/4/5c40a3f0eaa58e438c8dcaad395d7ab9e8f4f439.png)
be the incenter of triangle 
. Let 
, 
, and 
be the orthocenters of triangles 
, 
, and 
, respectively. Prove that the lines through , , and , parallel to 
, 
, and 
, respectively, are concurrent.
, a single parity swapping is to choose 
terms in this sequence, say 
and 
, such that 
is odd, then switch their placement, while the other terms stay in place. This creates a new sequence.
to 
, using only single parity swapping.
be a fixed integer. A game is played by players sitting in a circle. Initially, each player draws three cards from a shuffled deck of 
cards numbered 
. Then, on each turn, every player simultaneously passes the smallest-numbered card in their hand one place clockwise and the largest-numbered card in their hand one place counterclockwise, while keeping the middle card.
denote the configuration after 
turns (so 
is the initial configuration). Show that is eventually periodic with period , and find the smallest integer for which, regardless of the initial configuration, 
.
be a positive integer. The integers 
are written on a board. In a move, Alice can pick two integers written on the board 
such that 
is an even number, erase both 
and 
from the board and write the number 
on the board instead. Find all for which Alice can make a sequence of moves so that she ends up with only one number remaining on the board.
, Alice changes 
to 
and can't make any further moves.
with vertices may have, so that after deleting an arbitrary cycle, is not connected anymore.
be a triangle with centroid . Points 
and 
are chosen on rays 
and 
, respectively, such that![\[ \angle ABS=\angle ACR=180^\circ-\angle BGC.\]](http://latex.artofproblemsolving.com/0/a/d/0ad01e29194358ea0d3e10cb74b0769d1bd3a656.png)
Prove that 
.
hopefully this is correct, pretty similar to v_enhances solution so it should be
, 
be the midpoints of 
, 
respectively.
and 
.; the other equality can be proved in a similar manner.
; hence![\[\triangle MGC \sim RMC \implies \frac{MC}{MG} = \frac{MR}{MC} \implies MC^2 = MG \cdot MR; \]](http://latex.artofproblemsolving.com/b/d/5/bd5b95949260675e257d0df31cbe5727dc6e752e.png)
but by definition 
hence![\[MA^2 = MG \cdot MR \implies \triangle MAG \sim \triangle MRA\]](http://latex.artofproblemsolving.com/f/3/3/f33bdb2617362bd0443991395b3b256eb183d09d.png)
and the angle equality follows. 
![\[ \measuredangle RAS+\measuredangle BAC= \measuredangle RAC + \measuredangle BAS = \measuredangle MGA + \measuredangle AGN = \measuredangle MGN = \measuredangle BGC\]](http://latex.artofproblemsolving.com/e/8/2/e82150b4845ef492edf2054af6b1a27c375c3c6e.png)
as desired. 
are counter-clockwise.
. Scale such that 
. Let 
.
on the other side of as , 
. Thus the line 
is, for real 
, ![\[\frac{z-b}{a-b}=x\frac cb \implies z = b+x(a-b)\frac cb = b+x(-2b-c)\frac cb = b-2cx-\frac{c^2}{b}x.\]](http://latex.artofproblemsolving.com/5/5/f/55fca268426694142e121a5db0635d0dd7d22cae.png)
are collinear so 
. Thus ![\[\frac bc - 2x - \frac cbx = \frac{1/b}{k/c}-2x-\frac{k/c}{1/b}x,\]](http://latex.artofproblemsolving.com/5/b/8/5b8206981948aee2a1ba63117cf8107f974c9ed2.png)
giving 
. Then 
.
the line ![\[\frac{z-c}{a-c}=x\frac bc \implies z=c+x\frac bc (-b-2c) = c-2bx-\frac{b^2}{c}x.\]](http://latex.artofproblemsolving.com/0/b/c/0bca9bcb75be8abd943c75a5e8e57f7cb293f5aa.png)
Now 
, so ![\[\frac cb - 2x - \frac bc x = \frac{k/c}{1/b}-2x-\frac{1/b}{k/c}x,\]](http://latex.artofproblemsolving.com/d/a/a/daa3b317f83debb924ef2d721b75bb183ff283e0.png)
giving 
. Then 
.
. But ![\[\frac{s-a}{r-a} \frac{c-a}{b-a} \frac{b}{c} = \frac{\frac{1}{bk}(2b^2k+2bc+c^2+cbk)(-2c-b)(b)}{\frac{1}{c}(2c^2+2bck+b^2k+bc)(-2b-c)(c)} = \frac{1}{k} \cdot \frac{(c+bk)(c+2b)(b+2c)}{(c+bk)(b+2c)(c+2b)}=\frac{1}{k} \in \mathbb R.\]](http://latex.artofproblemsolving.com/5/3/0/53009a69937c4cde77d9d7425bb8b45c94730580.png)
be the midpoints of 
. Then,
hence 
. Similarly, 
and so
to be the midpoints of and respectively. Note that, 
is tangent to 
and 
is tangent to 
. Now notice that lies on the radical axis of 
and , which implies 
is tangent to . Using a similar argument, we get 
is tangent to 
. To finish, note that
![[asy]
unitsize(3cm);
pair A = dir(95); pair B = dir(190); pair C = dir(350); pair M=(A+C)/2; pair N=(A+B)/2; pair G=(A+B+C)/3; pair P=IP(Line(G, M, 10), circumcircle(A, C, G), 1); pair R = 2M-P; pair Q=IP(Line(G, N, 10), circumcircle(A, B, G), 1); pair S = 2N-Q;
draw(A--B--C--cycle,heavygreen); draw(circumcircle(A,C,G),red); draw(B--P, heavygreen); draw(C--N,heavygreen); draw(A--P,red); draw(C--P,red); draw(A--G,heavygreen); draw(A--R--S--cycle,red);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$G$",G,NE); dot("$N$",N,dir(N)); dot("$M$",M,dir(M)); dot("$R'$",P,dir(P)); dot("$R$",R,dir(R)); dot("$S$", S, dir(S));
[/asy]](http://latex.artofproblemsolving.com/3/7/8/378aa34c2d383f0210803cfe74d7732fce7d5d0a.png)
be the midpoint of 
and be the midpoint of 
. Let 
denote the reflection of over , so that 
. Then 
is a parallelogram, and we can angle chase to get
so quadrilateral 
is cyclic. Thus, we have
Analogously, we have 
. Thus, to finish, note that
as desired.
and denote the midpoints of and respectively.
, and since point lies on ray , we infer that the line is tangent to circle 
.
and lies on ray , therefore the line is tangent to circle 
.
.
.
but we also know that 
. Thus, 
. lies on segment and point lies outside of the segment , we have :
which is what we wanted to prove. lies outside of the segment and point lies on segment , we have:
which again is the result.
be and let the midpoint of be . Notice by the condition we have![\[\angle MBS = \angle BGM \implies \triangle BMG \sim \triangle SMB.\]](http://latex.artofproblemsolving.com/e/a/b/eab0d0a84cd14ea685d9582c1b2343454f048269.png)
![\[\frac{MG}{MB} = \frac{MB}{MS} \implies MA^2=MB^2=MG \cdot MS.\]](http://latex.artofproblemsolving.com/f/c/0/fc05f125ba186417238a116a284080dbc7c6b970.png)
![\[\frac{MA}{MS} = \frac{MG}{MA} \implies \triangle MAS \sim \triangle GAM\]](http://latex.artofproblemsolving.com/e/b/a/ebabe32581e2328428c4003b4b52af7053e10dfd.png)
![\[\implies \angle MAG = \angle MSA.\]](http://latex.artofproblemsolving.com/b/d/c/bdc441d0b061ada93e5dcbc214a1bad44464c88a.png)
. Thus, we have
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
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/* draw figures */
draw((5,12)--(-4.169023255813958,-0.12167441860465383), linewidth(1));
draw((-4.169023255813958,-0.12167441860465383)--(14.853302325581398,-0.25674418604651433), linewidth(1));
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draw((10.549253914691349,1.5903140779807183)--(-4.169023255813958,-0.12167441860465383), linewidth(1));
draw((0.41548837209302114,5.939162790697673)--(14.853302325581398,-0.25674418604651433), linewidth(1));
draw((5,12)--(-1.2177286743345868,1.1331786769882735), linewidth(1));
draw((5,12)--(10.549253914691349,1.5903140779807183), linewidth(1));
draw((5,12)--(5.2280930232558145,3.8738604651162776), linewidth(1));
/* dots and labels */
dot((5,12),dotstyle);
label("$A$", (5.083255813953487,12.214697674418606), NE * labelscalefactor);
dot((-4.169023255813958,-0.12167441860465383),dotstyle);
label("$B$", (-4.4789767441860505,0.10344186046511367), NE * labelscalefactor);
dot((14.853302325581398,-0.25674418604651433),dotstyle);
label("$C$", (14.893348837209305,-0.03162790697674684), NE * labelscalefactor);
dot((9.926651162790698,5.871627906976743),linewidth(4pt) + dotstyle);
label("$N$", (10.013302325581394,6.046511627906976), NE * labelscalefactor);
dot((5.2280930232558145,3.8738604651162776),linewidth(4pt) + dotstyle);
label("$G$", (5.228372093023254,4.042976744186045), NE * labelscalefactor);
dot((10.549253914691349,1.5903140779807183),linewidth(4pt) + dotstyle);
label("$S$", (10.643627906976745,1.769302325581393), NE * labelscalefactor);
dot((-1.2177286743345868,1.1331786769882735),linewidth(4pt) + dotstyle);
label("$R$", (-1.6299534883720963,1.219069767441858), NE * labelscalefactor);
dot((0.41548837209302114,5.939162790697673),linewidth(4pt) + dotstyle);
label("$M$", (0.0133953488372065,6.1140465116279055), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/7/0/c/70cec9632ef995608ca392b998d6aeacc89e5991.png)
about the midpoints of 
to 
respectively, denote the midpoint of 
as 
.
are parallelograms, 
is a cyclic quadrilateral. Similarly, we have 
is a cyclic quadrilateral. Thus, we have 
. As 
, the problem is done.
is tangent to and 
is tangent to .
is tangent to 
.
.
.
be the midpoints of 
respectively. Notice that 
so we have that 
Therefore, 
and similarly 
Adding these up yields the desired result. QED
and denote the reflections of and across the midpoints 
and 
of sides and respectively.
immediately implies that is tangent to side at . Similarly note that it also follows that is tangent to side at . Then, we note that,![\[M_cA \cdot M_cB = M_cB^2 = M_cG\cdot M_cS = M_cG \cdot M_cS'\]](http://latex.artofproblemsolving.com/f/5/3/f53918167a0d7f80dc2c2fdf1d1f96ae64e692c1.png)
which implies that lies on 
. Similarly we have that lies on 
. Now, let 
. Note that 
and 
are parallelograms by construction. Thus, 
and 
. This implies that 
.![\[\measuredangle GBP = \measuredangle GAS' = \measuredangle GAB + \measuredangle BAS' = \measuredangle GAB + \measuredangle BGC\]](http://latex.artofproblemsolving.com/6/d/2/6d281b11730c7752f80a155d096e81ca3ae5fb44.png)
Similarly we have,![\[\measuredangle PCG = \measuredangle CAG + \measuredangle BGC\]](http://latex.artofproblemsolving.com/9/4/9/949d907c700fa6b07089a7284e2adc11bbb9ab66.png)
Thus,![\[\measuredangle SAR + \measuredangle CAB = \measuredangle BPC + \measuredangle CAB = \measuredangle PBG + \measuredangle GCP + \measuredangle BGC + \measuredangle CAB = \measuredangle CGB \]](http://latex.artofproblemsolving.com/7/e/0/7e0d4954d52be5314eb093a9d36a04e8032fb111.png)
which proves the desired result.
and 
be the reflections of over midpoints of side and respectively. Then clearly, quadrilaterals 
and 
are parallelograms.
and 
are concyclic.
which proves that quadrilateral is concyclic. Similarly, we can prove that quadrilateral is concyclic, which finishes the claim.
which finishes the problem.
be the midpoints of 
Let 
be reflected over 
Notice that 
is a parallelogram. Next, I cliam that 
is cyclic. Observe that 
Hence, we have that
Also notice that 
Hence, 
Similarly, 
Thus, 
![[asy]
size(12cm);
pair A = dir(97); pair B = dir(190); pair C = dir(350); pair M = midpoint(A--C); pair N = midpoint(A--B); pair G = extension(B, M, C, N); draw(A--G, blue); pair Y = A*dir((C-G)/(B-G))**2;
pair X = A*dir((B-G)/(C-G))**2;
pair S = extension(B, Y, C, G); pair R = extension(C, X, B, G); filldraw(A--B--C--cycle, invisible, blue); draw(B--M, blue); draw(C--N, blue); draw(R--A--S, lightred); draw(C--R, deepgreen); draw(B--S, deepgreen);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$G$", G, dir(280)); dot("$S$", S, dir(270)); dot("$R$", R, dir(150));
/* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ !size(12cm); A = dir 97 B = dir 190 C = dir 350 M = midpoint A--C N = midpoint A--B G 280 = extension B M C N A--G blue !pair Y = A*dir((C-G)/(B-G))**2; !pair X = A*dir((B-G)/(C-G))**2; S 270 = extension B Y C G R 150 = extension C X B G A--B--C--cycle / 0.1 lightblue / blue B--M blue C--N blue R--A--S lightred C--R deepgreen B--S deepgreen */
[/asy]](http://latex.artofproblemsolving.com/1/a/0/1a029236070ae0349760663c57d9b556ca77bf3b.png)
,![\[ MA^2 = MC^2 = MG \cdot MR \implies \measuredangle RAC = \measuredangle MGA. \]](http://latex.artofproblemsolving.com/4/4/8/4487257c6d8cab40bd82884c5c7809994324918c.png)
Analogously, ![\[ \measuredangle BAS = \measuredangle AGN. \]](http://latex.artofproblemsolving.com/2/f/7/2f71cb89acaf6d2ae03538f8bbff02d9fabe3dbf.png)
Hence, ![\[ \measuredangle RAS + \measuredangle BAC = \measuredangle RAC + \measuredangle BAS = \measuredangle MGA + \measuredangle AGN = \measuredangle MGN = \measuredangle BGC. \]](http://latex.artofproblemsolving.com/0/f/0/0f07028ae973115b7c13900c40ce853a8fca3892.png)
Now we find the coordinate of 
,
Since 
(as directed angles), we have
Now
and so we have![$$\frac{(a-c)(a+c-2b)}{(r-c)(a+b-2c)} = \frac{b(a-c)(a+c-2b)(2ac-ab-bc)}{(2ab-ac-bc)\left[r(2ac-ab-bc) - (a^2c-a^2b+ab^2-abc+ac^2-b^2c)\right]}$$](http://latex.artofproblemsolving.com/4/1/d/41da16a8708d298bc3103a1076c10be8121d4651.png)
Canceling and clearing denominators gives
Solving for 
The numerator expands out to
which factors as
Thus we write
Similarly,
So we have the vectors
So
which is real. 
and hence 
But 
so 
.
and in particular 
Similarly, we find that 
.
as desired
we have 
.
so 
so
and by symmetry 
.
and 
now rest is trivial
and then complex bash :skull:
,
respectively
a
are parallelogram
are cyclic
be the midpoints of 
and 
,so 
is tangent to 
and 
.
,
denote the midpoints of 
Which means
Since 
,
.
So
So
and 
.
,
.
.
,
.
and 
.
and we are done.
.
and
Suppose 
is the midpoint of 
.
.
as desired.
and 
.
and 
.
implying 
.
.