AoPS Academy
![$f : \mathbb{Q}[\sqrt{2}] \to \mathbb{Q}[\sqrt{2}]$](http://latex.artofproblemsolving.com/c/a/1/ca172b912bf5bf7178cbbd5745fbc0a6a8b27503.png)
such that for all ![$x, y \in \mathbb{Q}[\sqrt{2}]$](http://latex.artofproblemsolving.com/9/f/e/9fede0b2fd22836372b8afc488ccd8a14db5d0c2.png)
,
where ![$\mathbb{Q}[\sqrt{2}] = \{ a + b\sqrt{2} \mid a, b \in \mathbb{Q} \}$](http://latex.artofproblemsolving.com/f/a/c/fac8bf68761534a139d570bf66b1aa5dfb59e8db.png)
.
such that Bob can always find the two coins in at most guesses?
such that the following two conditions hold:
and 
satisfying 
, We have 
for all real numbers 
.
and 
, there exist real numbers and , such that and 
.
help in one of my classrooms, and one of the Halpers posted. Is this a glitch or a bug(see screenshot)?
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]
size(9cm);
defaultpen(fontsize(11pt));
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 .
