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.
, the point 
is the midpoint of 
and 
is a point on the side such that 
. The cotangents of the angles 
, 
, and 
are positive integers 
.
is also an integer and equals 
.
, some of which consist of Fibonacci numbers.
+1 w
such that for all positive real numbers 
and 
:
and 
be two isogonal conjugate with respect to triangle . Let 
and 
be two points lying on the circle 
such that 
and 
are perpendicular and parallel to bisector of , respectively. Prove that 
and 
bisect two arcs containing 
and not containing , respectively, of 
.
be a triangle such that 
and 
. The tangent at to the circumcircle 
of triangle meets the line at 
. Let 
be the reflection of in the line , let 
be the foot of the perpendicular from to 
, and let 
be the midpoint of the segment 
. Let the line 
intersect the circle again at 
.
is tangent to the circumcircle of triangle 
.![\[\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1\]](http://latex.artofproblemsolving.com/f/a/6/fa6c46c12e7e6bf0b04f6f9a910df6c2b452fe6a.png)
for all real numbers , , 
, each different from 
, and satisfying 
., , , each different from , and satisfying .
there exists an -digit number divisible by 
all of whose digits are odd.
be a strictly increasing sequence of positive integers such that for all positive integers 
![\[d(2025^{a_n}-1)|a_{n+1}.\]](http://latex.artofproblemsolving.com/8/f/d/8fda531266d39c7f11c58d3c1077d2cab3357bea.png)
Show that for any positive real number 
there is a positive integers 
such that 
for all 
.
denotes the number of positive divisors of the positive integer 
.
such that for all real numbers 
and 
, ![\[f(f(xy)+y)=(x+1)f(y).\]](http://latex.artofproblemsolving.com/d/1/f/d1f369109456464bc8b0102763211802770d983d.png)
be positive real numbers such that 
. Show that![\[ \frac{x^2-yz}{x^2+x}+\frac{y^2-zx}{y^2+y}+\frac{z^2-xy}{z^2+z}\leq 0. \]](http://latex.artofproblemsolving.com/e/a/1/ea1f56cc594862673040411c3ad322cb00cfe904.png)
formed by using the digits 
once each such that the number 
is divisible by 
for 
.
be an odd prime and 
is a bijection and 
is an integer such that 
Show that 
is a multiple of .
be a set of 
positive integers with the property that the product of any two distinct members of will not exceed 1994. Show that there are numbers 
and 
in the set such that the gcd of and is greater than 1.
and : 
, knowing that the median 
makes an angle of 
with side 
and an angle of 
with side 
.
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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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 .
