such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and 
.
and 
be two sequences of integers such that 
and 
for all 
.
divides 
for an infinity of integers 
such that 
for all 
are coprime. Consider the sequence 
given by 
.
such that 
is even and 
is an integer. Prove that the numerator in its simplest form of 
is an odd integer greater than 
.
be an integer and let 
be 
distinct subsets of 
.
such that the n subsets 
are also disjoint.
s
.
Y by Adventure10, Mango247
For a triangle 
we take the point 
on the side 
such that 
, the point 
on the segment 
such that 
and, if possible, the point 
on the ray (
such that 
. We denote by 
the set of all triangles 
for which
. Prove that all triangles from are similar and find the measure of their smallest angle.
we take the point 
on the side 
such that 
, the point 
on the segment 
such that 
and, if possible, the point 
on the ray (
such that 
. We denote by 
the set of all triangles 
for which
. Prove that all triangles from are similar and find the measure of their smallest angle.Attachments:
This post has been edited 1 time. Last edited by orl, Nov 14, 2004, 10:20 PM
![[asy]unitsize(3cm);
string foo(pair X, pair Y, pair Z) {
return string(degrees(X-Y)-degrees(Z-Y));
};
string n[] = {"B","C","A","X","Y","Z","X'"};
real r = 110.4962;
pair A[] = {
(0,0),(3,0),3*dir(r)
};
A[3] = (5*A[0]+4*A[2])/9;
A[4] = (2*A[3]+A[1])/3;
A[5] = IP(L(A[1],A[2]),L(A[3],A[3]+dir(180-r)*(A[4]-A[3])));
for(int i = 0;i<A.length;++i) {
MP(n[i],D(A[i]));
};
MP("\angle XYZ: "+foo(A[3],A[4],A[5])+", \angle BAC:"+foo(A[0],A[1],A[2]),(-2,0));[/asy]](http://latex.artofproblemsolving.com/e/c/b/ecbd0b6da6e879cd813059c053d5fa9ce6ea8219.png)
)
past 
,
as the problem states.
on line 
and 
on line 
(so that 
are in that order) with figure 
similar to 
.
and 
.
such that 
,
,
,
,
,
,
,
.
denote the length of the altitude 
from 
,
.![\[\frac{h}{CC'}=\sin\angle{A'C'B}=\sin\angle{ACX}=\frac{2[ACX]}{AC\cdot CX}=\frac{AX\cdot h}{AC\cdot CX}=\frac{4uh}{3bv},\]](http://latex.artofproblemsolving.com/8/8/b/88b59895d67384ae45cafb713670f569d9688600.png)
so![\[CC'=\frac{3bv}{4u}.\]](http://latex.artofproblemsolving.com/2/7/9/27907684b07389f2273fe55b8c5eba2b345e801a.png)
Since 
,
,
.
Plugging (2) into (1) yields![\[h^2+(h+2vt)^2 = \frac{9b^2v^2}{16u^2}=\frac{9v^2}{16u^2}(h^2+(5u+vt-h)^2).\]](http://latex.artofproblemsolving.com/4/0/7/407de37cec1ae164f6cf6a49cd564bc34970fbac.png)
Equating 
with (3), we find![\[h=\frac{4u^2-4tuv+9v^2}{8u},\]](http://latex.artofproblemsolving.com/c/4/1/c416399aece71e92efbdb09776b289f1ad563321.png)
and putting this in (3) we now find![\[16u^4+32tu^3v+(80t^2-288)u^2v^2-144tuv^3+81v^4=0,\]](http://latex.artofproblemsolving.com/4/7/3/473a9fa7ab06eca249abc45e9b735db878057048.png)
or![\[(80u^2v^2)t^2+(32u^3v-144uv^3)t+(16u^4-288u^2v^2+81v^4)=0.\]](http://latex.artofproblemsolving.com/1/7/6/17674f6bf9da3b154c328a165a0635dd3e33a465.png)
The discriminant of this quadratic in 
is![\[\Delta_t=-64u^2v^2(64u^4-1296u^2v^2+81v^4),\]](http://latex.artofproblemsolving.com/6/b/e/6be1001b42a68dccd6cc134538c431aa0162403c.png)
so for infinitely many pairs of fixed 
,
must have sides in a constant ratio, i.e.![\[\frac{CY'}{BY'}=\frac{h\sqrt{2}}{vt}=\frac{4u^2-4tuv+9v^2}{4uvt\sqrt{2}}\]](http://latex.artofproblemsolving.com/4/d/b/4dbc3eebd6bf973bca5dd1fd953fe8c62536f197.png)
must be constant for all solutions 
,![\[(u,v,t)=\left(1,1,\frac{14+\sqrt{1151}}{20}\right)\]](http://latex.artofproblemsolving.com/3/c/f/3cf67d7252ecd9833961dc0f0db3d21842c755a7.png)
and![\[(u,v,t)=\left(2,1,\frac{2+\sqrt{4079}}{40}\right).\]](http://latex.artofproblemsolving.com/f/3/3/f338fdeb6859f048a320dabdec77eefeea52c2b6.png)
,
,
is the unique solution and the official diagram has 
(even if they meant directed angles, it's just completely silly and misleading).
.
