[/quote]
,
such that : 
-
and 
such that 
.
written on the first page and 
empty pages. Suppose in each of the next 
be the largest possible number such that for any 
,
such that for all 
,![\[n^{0.99n}\leqslant F(n)\leqslant n^{1.01n}.\]](http://latex.artofproblemsolving.com/5/7/e/57e2b7e3d97564665d8de28a3afd74bdcf74ba5b.png)
of positive integers for which there exist positive integers 
and 
holds for all integers 
(Note that 
denotes the greatest common divisor of integers 
and 
)
.
on one side and a 
on the other. Harry has 
coins showing 
the process starting with the configuration 
would be 
,
,
be the number of operations before Harry stops. For example, 
and 
.
possible initial configurations 
be the foot of the altitude from the vertex 
in the acute-angled 
.
meets line 
at 
,
board where we identify the squares with pairs 
where 
and 
denote the row and column number of that square, respectively.
and places a pawn at the bottom left corner (i.e. on 
)
or 
if 
and 
if 
Prove that
be real numbers satisfying 
Prove that 
and determine all cases of equality.
-
Where 
inscribed in a circle with center 
intersect 
at 
.
be the Lemoine points of triangle 
respectively. Prove that 
are collinear.
denote the set of positive integers. Find all functions 
such that ![\[\left\lfloor \frac{f(mn)}{n} \right\rfloor=f(m)\]](http://latex.artofproblemsolving.com/2/3/1/231c0bb0911db66d1a3d2354da7e5dff3004c47f.png)
for all positive integers 
.
Y by A-Thought-Of-God, Rounak_iitr, ItsBesi
In acute triangle 
, 
is the midpoint of minor arc 
, 
is the circumcenter of 
and 
is its diameter. The line parallel to 
through meets segment 
at 
, and 
extended at 
. Lines 
and 
meet at 
, lines 
and 
meet at 
. Prove that 
.
, 
is the midpoint of minor arc 
, 
is the circumcenter of 
and 
is its diameter. The line parallel to 
through meets segment 
at 
, and 
extended at 
. Lines 
and 
meet at 
, lines 
and 
meet at 
. Prove that 
.This post has been edited 2 times. Last edited by mofumofu, Mar 11, 2021, 9:53 AM
Reason: typo
Reason: typo
.
is its reflection point WRT 
is a diameter of 
inter sects 
,
respectively. Suppose 
,
.
.
,
.
we get 
,
.
we have 
,
.
,
.
it's obvious to see 
has the same expression. So 
. It is glad to see that there are finally some nice problems appearing on the CMO![[asy]
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[/asy]](http://latex.artofproblemsolving.com/b/c/6/bc6e6a26b97baf9e879d6d36d59481477224e3f4.png)
about 
.
.
.
by symmetry
Therefore,
as desired.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/7/d/d/7dd70cd75cea51b555d3980761211a7c188bfd1d.png)
and 
,
,
.
,
.
.
.
and 
,
corresponds to 
under this symmetry).
and 
interact with 
,
and 
are cyclic and hence 
as well as 
.
,
i.e. we need to find out how 
and 
are collinear from where our claim immediately follows.
resp. 
.
![\[\angle PCB = \angle KCB = \angle KAB = \angle OAB = \angle OBA = \angle OBD,\]](http://latex.artofproblemsolving.com/8/8/c/88c231dc0a01737d8267b691802c8c538665ba2c.png)
and ![\[\angle PBC = \angle MBC = \angle MCB = \angle MAB = \angle ODB.\]](http://latex.artofproblemsolving.com/9/e/0/9e018f6a7efff8336388165bf9420a1baf5b1f14.png)
So, we have 
Similarly, 
.![\[\frac{BC}{DB} = \frac{PC}{OB} = \frac{PC}{OC},\]](http://latex.artofproblemsolving.com/c/9/1/c9149c0b3341e7192dffa813e65233aae1fcf8da.png)
here we get 
,
.
,
also, 
Whence, ![\[\angle OQC + \angle ODC = 90^{\circ} - \angle A + \angle BEO + \angle ODC\quad(*).\]](http://latex.artofproblemsolving.com/4/a/f/4af5ae9d682bea16fcea4a8e0259cc8b846d37d8.png)
Next, ![\[\angle OPB + \angle OPC = \angle BPC \overset{(1)}{=} \angle DOB = 180^{\circ} - \angle OBD - \angle BAM = 90^{\circ}+\angle C - \frac{A}{2}\]](http://latex.artofproblemsolving.com/7/7/5/775a3dcbd171f4b9b58b05281c8ab9f440bfe438.png)
yields
Adding 
to both sides of the above equation, and comparing it with 
gets us done. 
is isosceles with 
;
is isosceles, with 
,
Further, if we let 
as 
,
,
.
are segments from the same line and 
,![\[ \angle ODB = \angle MAB = \dfrac{\angle A}{2} = \angle CAM = \angle CEO. \]](http://latex.artofproblemsolving.com/e/8/d/e8d39f47be8406f925150d1a73ed19951018a564.png)
Also, 
concylic and 
internal angle bisector implies![\[ \angle MCB = \angle MBC = \angle MAC = \dfrac{\angle A}{2}. \]](http://latex.artofproblemsolving.com/2/f/5/2f5739fe002a2d15740ddbd6979f6365a7c268d4.png)
For simplicity, call 
.
in a clockwise angles to accomodate 
)![\[ \angle DBO = \angle PCB = 90 - \angle C, \angle ECO = \angle QBC = 90 - \angle B. \]](http://latex.artofproblemsolving.com/5/9/c/59cc6758299df2c78a50d236a58001da75904287.png)
From here, we can see that 
,
counterpart.
with each other are rare, let's transform this to a almost-too-good-to-be-true spiral! Namely, draw 
is an isosceles trapezoid with bases 
.
can be substituted with 
!![\[ \angle OP'C + \angle OEB = \angle OQ'B + \angle ODC \Longleftrightarrow \angle OP'C - \angle ODC = \angle OQ'B - \angle OEB. \]](http://latex.artofproblemsolving.com/e/6/6/e66377128aa4f32abe6dfafd55b6d01faa5eca55.png)
and so forth, but I'd argue that grouping 
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\draw [fill=uuuuuu] (-10.614559541876705,8.505293629731504) circle (2pt);
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\end{tikzpicture}");[/asy]](http://latex.artofproblemsolving.com/f/2/e/f2eb6d7db035e6b713f046b8a65a64d07528eb30.png)
![\[ B,O,C; \quad D,P' \]](http://latex.artofproblemsolving.com/2/6/a/26a31b9fd3758a270c1e817b6e41845cb9aa4ff9.png)
We Claim that given 
,
and![\[ \angle ODB = \angle P'CB = \alpha \]](http://latex.artofproblemsolving.com/7/6/6/7661f91e9b28e077adaf6e4e5cf1ee0ba83985af.png)
(with 
'
now), then![\[ \angle OP'C - \angle ODC = 90-A. \]](http://latex.artofproblemsolving.com/2/e/b/2eb59d1c2824baf2da78f616a045f76b00703fa7.png)
by 
and resize it so that![\[ \text{The image of} \ \triangle DOB = \triangle CP'B. \]](http://latex.artofproblemsolving.com/4/b/a/4ba73678c3a043bf17d952a73a20cd537f70e31c.png)
Name the image of 
(this name intends to say that 
to be the point on 
so that![\[ (COB) \cap \overline{CP'} = \{C,F_1\} \]](http://latex.artofproblemsolving.com/5/5/c/55c1fca8950468ddbadd6f415f19cd9f44cb4d2c.png)
and 
also on 
.![\[ \angle CF_1O = \angle F_1F_2B = \angle F_1BF_2 = 90-A. \]](http://latex.artofproblemsolving.com/9/8/3/983eea9a7f50add358feb2af5a732dc2f8c89286.png)
![\[ \angle D_RCP' = \angle P'OF_1. \]](http://latex.artofproblemsolving.com/1/0/4/104f5592f115627a9da4fe085b0d90495cf2871d.png)
,![\[ \triangle COB \sim \triangle D_RP'B \sim \triangle BF_1F_2 \]](http://latex.artofproblemsolving.com/5/3/e/53e4b32504ad69a92f084fc2ab449ae63e4d1ebc.png)
but I presume that this has already been included in the second set of similar triangles; since this is a![\[ (2A) - (90-A) - (90-A) \]](http://latex.artofproblemsolving.com/8/8/2/882db3a635a801e463ac94d0e1ce73b63db3dea7.png)
triangle.
,
cyclic. In turn, this will cause![\[ \angle CF_2D_R = \angle P'F_2D_R = \angle P'BD_R = 90-A = \angle P'F_1O. \]](http://latex.artofproblemsolving.com/0/b/e/0be46ce76090d88f8d84a5f1a11cbcb58e63aa0d.png)
Now we prove that![\[ \dfrac{OF_1}{P'F_1} = \dfrac{CF_2}{D_RF_2}, \ \text{or} \ \dfrac{OF_1}{CF_2} = \dfrac{PF_1}{D_RF_2}. \]](http://latex.artofproblemsolving.com/b/d/6/bd6254612e2d47d7295e7ce048812a183b7d51bc.png)
We can count the LHS manually: repeated sine rule on the circle 
yields![\[ \text{LHS} = \dfrac{CF_1+F_1B}{F_1O} = \dfrac{\sin{(\alpha)}+\sin{(3\alpha)}}{\sin{(90-\alpha)}} = \dfrac{2 \sin{(2\alpha)} \cos{(\alpha)}}{\cos{(\alpha)}} = 2 \sin{A} = \dfrac{BF_2}{BF_1}. \]](http://latex.artofproblemsolving.com/2/f/f/2fffce8b8631296096162aca5bf8191fda147fc8.png)
Finally, we end this by proving![\[ \triangle P'F_1B \sim \triangle D_RF_2B \]](http://latex.artofproblemsolving.com/4/c/4/4c4d07ce49531fc5f015a86a6e7bb09298343130.png)
which directly establishes this Section's Claim.
by cyclicity and 
.
(with respect to 
(with respect to 
and 
are pretty hard to exactly compute; and I knew that 
and 
are both fixed lengths (therefore producing a fixed ratio no matter how 
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![\[ \angle B + \angle C = 180 - A \]](http://latex.artofproblemsolving.com/b/b/9/bb98f260df6d86e0424a57b3808f66125a19d864.png)
but I couldn't be certain that 
would be constant after perturbations. In contrast, having a pair of similar triangles gives better odds for some sort of invariant relation between the angle substractions.
and the projection of 
.
and 
.
towards some angle centered at ![\[ \angle P'OF_1 = \angle D_RO (\text{pink point}) \]](http://latex.artofproblemsolving.com/8/f/c/8fcde821650f1e66a30b57e207dbd3c3a0d6ce0e.png)
but I could not find any good identities involving said point.
,
was supposed to be cyclic, following![\[ (D_RP'BF_2) \ \text{cyclic, and} \ OF_1 \parallel BF_2. \]](http://latex.artofproblemsolving.com/d/d/1/dd1119fd40f6978576788a1b0c7ab3fc1a9c423a.png)
![\[ P'= F_1. \]](http://latex.artofproblemsolving.com/6/b/7/6b720d967c28c75a2bf3bc1ed9d31e67a14a0858.png)
![\[\begin{tabular}{p{12cm}}
$\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\
\end{tabular} \\ \]](http://latex.artofproblemsolving.com/8/4/2/842413e566043c96b7a8f2300401b08af9d8ba48.png)
Final Comments (for my sol): Strangely enough, the situation in this problem is very similar to the APMO problem: where virtually all the angles around 
and 
angles in between are unknown.
wasn't due to symmetry reasons: but with experience of solving the APMO Problem.
be a cyclic convex quadrilateral and 
be its circumcircle. Let 
.
be the center of the circle tangent to sides 
,
opposite 
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as the incenter of 
;
the excenter of those triangles.![\[ MI = MJ = MB = MC = MI' = MJ' \]](http://latex.artofproblemsolving.com/d/0/b/d0b56b53f2c405682d65010f3c1b21523d2b42cd.png)
the six-point circle is cyclic with center 
and 
collinear.
opposite 
is the incenter!)
as the excenter of 
the intersection of 
and 
,
as the intersection of 
and 
.![\[ I_EM \perp T_1T_2, LM \perp K_1K_2. \]](http://latex.artofproblemsolving.com/a/c/6/ac634902fa70271bab3d4775d8dffcb721881886.png)
So, to prove that 
collinear, it's enough to prove that 
,![\[ \dfrac{\sin{(\angle I'T_1T_2)}}{\sin{(\angle T_2T_1C)}} = \dfrac{\sin{(\angle IK_1K_2)}}{\sin{(\angle K_2K_1B)}} \]](http://latex.artofproblemsolving.com/b/a/4/ba4b69cc9acb12676cc04447a97f9f356bf32ea1.png)
as we know that 
.
and its 
counterpart, then deduce that![\[ \dfrac{I'T_2}{T_2C} = \dfrac{I'J'}{BC} = \dfrac{IJ}{BC} = \dfrac{IK_2}{K_2B} \]](http://latex.artofproblemsolving.com/f/6/8/f68732efd01d9cd0c9defcce2d5ed3dcff6b78db.png)
to complete the proof.
How about not doing that parallel stuff and sine ratio completely?
and run with the money.
collinear, yet doing the exact same thing but rearranged yields direct results)![[asy]usepackage("tikz");label("\begin{tikzpicture}[scale=0.85]
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,![\[ \overline{CN} \cap \overline{AB} = X, \overline{BN} \cap \overline{AC} = Y. \]](http://latex.artofproblemsolving.com/c/2/1/c2183a414565abce649954243682a38e702ccc7b.png)
Pascal gives us 
and 
collinear.
.
.![\[ \angle CXD = 90+\dfrac{A}{2}-B, \angle COD = 360-\angle COB - \angle BOD = 270-\dfrac{3}{2}A-C. \]](http://latex.artofproblemsolving.com/2/0/d/20d7a043e2a5987255852480298dbe09951c784f.png)
So, 
,
.![\[ \angle OXC + \angle OQC = \angle OYB + \angle OPB. \]](http://latex.artofproblemsolving.com/2/5/0/2504167044c19be5e7e035d1591dc198f961d331.png)
![\[ \angle XOQ = \angle YOP, \angle XCQ = \angle NCM = \angle NBM = \angle YBP; \]](http://latex.artofproblemsolving.com/1/c/f/1cf49815bcea66429018dec7afe5005192ec60c5.png)
the first one by collinearity and the second one by symmetry. So,![\[ OXC + OQC = 360 - XOQ - XCQ = 360 - YOP - YBP = OYB + OPB. \]](http://latex.artofproblemsolving.com/9/9/7/9971ed528d2325b434221f3f858b4512e3be3283.png)
We are done.
.
so 
Also note that 
so 
So 
and 
.
.
.
is an orthocentric system.So, 
.
as desired. 
and 
.
,
,
implying the desired similarity. ![\[\angle OPB - \angle ODC = \angle BPC - \angle OPC - \angle ODC = \angle BOD - \angle DCB - \angle ODC = \angle OBC\]](http://latex.artofproblemsolving.com/e/1/0/e10a2aa1722772282b9e6f922d080514d8d8c26d.png)
and analogously, 
,
and 
,
![\[\angle OPB - \angle OQC = \angle ODC - \angle OEB.\]](http://latex.artofproblemsolving.com/5/d/7/5d763a5f7f646f1bab537c20a036f45ee7528b65.png)
![\[\angle BCP = \angle BAK = \angle ABO = \angle DBO,\]](http://latex.artofproblemsolving.com/c/0/b/c0bf252f1950578b8a3ccf138d384bd8421c9709.png)
![\[\angle CBP = \angle CAM = \angle BAM = \angle BDO.\]](http://latex.artofproblemsolving.com/3/e/f/3ef9796b11d6d41aa3716049f1e3607513d059bd.png)
,![\[\frac{BO}{BD} = \frac{CO}{BD} = \frac{CP}{CB} \implies \frac{CO}{CP} = \frac{BD}{BC}.\]](http://latex.artofproblemsolving.com/e/f/3/ef3f8c0b648462e1ded6376f6fd6de4532615294.png)
![\[\angle CBD = \angle ABC = \angle AKC = \angle OCP.\]](http://latex.artofproblemsolving.com/e/2/c/e2ccb4e15e9f5e3785fde4032714809c929c1222.png)
.
.
and 
.
and 
.
and 
and 
,
yielding 
in a similar manner and it remains pure angle chasing: if 
and 
,
,
and 
,
.
