AoPS Academy
be vertices of regular pentagon inscribed in a circle whose radius is 
and center is at 
. Find all possible values of 
are integers whose greatest common divisor is 1. Let 
be a set of integers with the following properties:
, 
.
(not necessarily distinct), 
.
, if 
, then 
. must be equal to the set of all integers.
.
, there exist indices 
such that 
is divisible by .
, find the minimum possible value of 
.
that satisfies![\[
a^3+b^3=1957
\]](http://latex.artofproblemsolving.com/5/d/8/5d8d2a40734ec6b0a9149f9a4ae3a5828d5055f3.png)
![\[
(a+b)(a+1)(b+1)=2014
\]](http://latex.artofproblemsolving.com/3/0/0/300f2d16e669a1d8d9c54b357a6eaf7a8eb5d826.png)
Hence, find the value of 
be the roots of the equation 
. Hence, find the value of 
if![\[
Z=\frac{\alpha+\beta+\gamma+\delta}{\alpha(\beta+\gamma+\delta)+\beta(\gamma+\delta)}+\alpha\beta\gamma\delta[\alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta)]
\]](http://latex.artofproblemsolving.com/4/2/5/425d9100db9e90264b50fc7c4b062ddb53c64068.png)
denote the greatest integer less than or equal to 
. Hence find the value of 
, if![\[
M = \left\lfloor \frac{1^2}{3} \right\rfloor + \left\lfloor \frac{2^2}{3} \right\rfloor + \left\lfloor \frac{3^2}{3} \right\rfloor + \dots + \left\lfloor \frac{39^2}{3} \right\rfloor + \left\lfloor \frac{40^2}{3} \right\rfloor
\]](http://latex.artofproblemsolving.com/1/d/4/1d4576c5f14b3f43a99f273eb2e8133388945135.png)
if within that number exists a digit 
that appears exactly 
times in that number, hence find the number of that consist of 4 digits (Numbers may contain a 0)
be real numbers, such that![\[
a^2+b^2+c^2+abc=5
\]](http://latex.artofproblemsolving.com/7/c/8/7c8044347b7e25cb21dbef42c65ec351cefd4699.png)
![\[
a+b+c=3
\]](http://latex.artofproblemsolving.com/6/4/a/64a3034ae02dd28aa6220ee5c37a22d608074173.png)
Prove that atleast one of the numbers is equal to 
.
. A circle O with center C has a radius of 2. Let P be a point on the circle O.
?
?
,there are 
equal to 
, and 
equal to 
. Find the sum of all the products 
, where 
.
be triangle with point 
and 
on 
and 
are on the same side of 
be midpoint of segment and 
be midpoint of segment 
be intersection of 
with 
and 
be triangle with point and on and are on the same side of be midpoint of segment and be midpoint of segment be intersection of with and 
be triangle with point and on and are on the same side of be midpoint of segment and be midpoint of segment be intersection of with and 
true for D E F symmetrically ?
![[asy]
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pair A = (-0.25,2);
pair B = (-1,0);
pair C = (2,0);
path C1 = circle(B,1);
path C2 = circle(C,1);
pair E1 = intersectionpoint(C1, A--B);
pair F = intersectionpoint(C2, A--C);
pair M = (E1+F)/2;
pair N1 = (B+C)/2;
pair P = 2N1-A;
pair Q = 2M-A;
pair R = extension(Q,F,B,P);
pair S1 = extension(E1,Q,C,P);
pair T = extension(B,F,C,E1);
pair D = extension(T,P,B,C);
pair U = extension(F,Q,C,E1);
draw(A--B--C--cycle, black);
draw(E1--Q--F, black);
draw(B--P--C, black);
draw(R--Q--S1, black+dashed);
draw(E1--F, black);
draw(B--F,blue);
draw(C--E1,blue);
draw(T--P, blue+dashed);
draw(M--N1, red+dashed);
dot(A);
dot(B);
dot(C);
dot(E1);
dot(F);
dot(P);
dot(Q);
dot(R);
dot(S1);
dot(M,red);
dot(N1,red);
dot(T,blue);
dot(D);
label("$A$", A, N, black);
label("$B$", B, SW, black);
label("$C$", C, E, black);
label("$E$", E1, NW, black);
label("$F$", F, NE, black);
label("$Q$", Q, 1.3W+0.5S, black);
label("$P$", P, S, black);
label("$R$", R, SW, black);
label("$S$", S1, SE, black);
label("$N$", N1, SW, black);
label("$M$", M, N, black);
label("$G$", T, S+0.5W, black);
label("$D$", D, NE, black);
[/asy]](http://latex.artofproblemsolving.com/4/a/5/4a5c8083b91859c80df2379908b173908762b3c8.png)
and 
denote the reflection of 
across and , respectively.
, and 
, and 
. It implies that 
, and 
are parallelograms., we have 
. This indicates that 
is a rhombus. Furthermore, the line 
must bisect 
.
:
so, lies on the angle bisector of ; that is 
. also lies on , and it could be proceed using Menelaus' theorem on 
and the line 
:
The converse of Menelaus' theorem also holds for 
, so 
. is the homothetic center sending 
. Therefore, 
.
as desired.
and . linearly along 
, then moves linearly along 
and moves linearly. Assume WLOG 
, then let the value of when 
be . Then when 
and we get that lies on the line through and the midpoint of 
. A homothety of scale factor from takes this line through parallel to the -angle bisector. A simple bary (or MMP) argument shows that moves on a line (since it's degree 
). When 
we get 
which does lie on the desired line. When 
we get 
which does lie on the desired line. 
, 
, and 
. Then, 
and 
, so 
has direction 
. In addition, 
, so 
also has direction 
. Therefore, and are parallel.
or ) that is parallel to the angle bisector of 
. by the angle bisector property.
or ) that is parallel to the angle bisector of . by the angle bisector property.
is the reflection of wart to then lies on the angle bisector of , for which it suffices to prove that the distances from to 
and are equal. Since 
it suffices to prove that ![$[G'CF]=[G'EB]$](http://latex.artofproblemsolving.com/5/3/a/53a4c302f9c1b01a0f80107fa69645346f39d588.png)
. Now, observe that![$$[G'CF]=[G'CB]=[G'EB],$$](http://latex.artofproblemsolving.com/8/9/5/8954d64f559da05554039b7c37f44bf721f45902.png)
because 
and 
. Done.
be the arc midpoint of 
and intersect and at 
and 
respectively. will be the center of spiral similarity that sends 
and 
and 
then let 
is the center of spiral similarity that sends 
and a fixed point 
is cyclic
is a fixed point
such that 
is a parallelogram the problem reduces to-
be a point on the angle bisector of and let 
and 
be lines from 
and 
such that and is parallel to and . Let 
and 
. Prove .
Prove that
,
.
.
