AoPS Academy
. Find 
.
be a graph colored using 
colors. We say that a vertex is forced if it has neighbors in all the other 
colors.
-regular graph that contains no triangles or quadrilaterals, there exists a coloring using 
colors such that at least 
of the colors have a forced vertex of that color.
vertices of the same color may be adjacent.
be a triangle with 
. The tangent to the circumcircle of 
at 
intersects 
at 
. The angle bisector of 
intersect at 
. Suppose that the perpendicular bisector of 
intersect 
at 
, respectively. Show that 
defined by 
and![\[
u_{n+1} = \frac{1}{2}u_n^2 + 4 \quad \text{for all } n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/1/a/5/1a5c8b3599d063a6eb1ad59ac9540a937b5b8027.png)
a) Prove that there exist infinitely many positive integers 
such that 
.![\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]](http://latex.artofproblemsolving.com/f/f/1/ff174e7431cbfbc17c650d109651241286756a1a.png)
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);
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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]
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[/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 
then gives
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 
.
be the midpoints of 
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 
.
.