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.
) for all their time and support. We received immense help from TestX01 for the proof of claim 2.19 and it's associated lemma.
by 
, 
for every positive integer 
. Prove that 
leave distinct remainders when divided by 
.
such that 
Prove that 
. Find all 
such that for all 
be real numbers such that 
find the largest value of 
be a triangle with 
, and let 
be the foot of the altitude from 
. A point 
is chosen inside the triangle 
so that 
bisects 
. Let 
be the intersection point of the lines 
and . Let 
be the semicircle with diameter that meets the segment 
at an interior point. A line through is tangent to at 
. Prove that the lines 
and meet on .
![\[ x^n + 1 = y^{n+1}, \]](http://latex.artofproblemsolving.com/4/a/8/4a80f616d2a091187cdd1c57a5b6e1f09cdc9396.png)
where is a positive integer not smaller then 2, has no positive integer solutions in 
and 
for which and 
are relatively prime.
, the sides 
and 
are parallel and the angles 
and 
are acute. Show that it is possible to divide the triangle into 4 disjoint triangle 
and the triangle 
into 4 disjoint triangles 
such that the triangles 
and 
are congruent for all 
.
and 
have the same radii. over 
to 
; then as 
, quadrilateral 
is cyclic, proving the claim. 
.
to intersect 
at a point 
, and let 
be the circumcenter of . As this circle and 
have the same radius, can be considered as the reflection of across the line . Reflect over to a point 
; then 
and by the logic in the previous sentence 
. Now we can prove 
is an isosceles trapezoid, which is just a lot of messy work (it's ; I can't seem to find a shorter way to do this darn), which immediately implies 
is a parallelogram.
and since 
passes through , 
as well. Hence 
. 
is acute.Since 
,we obtain that 
are concyclic.Note that 
,so 
.
,and hence 
,then 
are concyclic,as desired.
, and let 
other than the parallelogram point. So 
(because the other intersection of 
is diametrically opposite 
) so 
is on 
.
are cyclic by perpendiculars. Also 
is cyclic (the nine-point circle of .) By radical axis on these three we get 
concurrent at say . Now the next part is a bit strange: Note 
and 
, so is the orthocenter of 
. Since 
are collinear and 
, then 
is the foot from 
to 
, so 
are collinear. Finally, the desired result follows from the converse of the radical axis theorem on 
.
in , call it , is just the reflection of through .(For a proof of this fact, go to @post #5's blog, which has a lot of good olympiad stuffs 
) Notice that 
means that 
is on the circle with diameter 
so we are done.
Prove that 
be real numbers.Prove that
,
,
,
.
,
.
or 
,![\[AM^2 = bc\cos{A}+x^2 = bc\sin{(90-A)}+x^2 = AF \cdot c + x^2\]](http://latex.artofproblemsolving.com/1/6/5/165225b4b8436ced226bdd5e0a730864736bf4fd.png)
From similar triangles, 
.
,
.![\[\angle BCK = \angle KCM = \angle CAM = \angle EAK = \angle EHK = 180 - \angle BHK\]](http://latex.artofproblemsolving.com/6/1/d/61de1be9d98a0332e5b5ad9f66cfc599e88f8df6.png)
The result follows.
.
is concylic.
is right, the center of this circle must be 
,
.
.
must be the perpendicular bisector of 
.
(don't know how to prove this; could someone help me?), we have that 
is a parallelogram.
.
is perpendicular to 
.
,
we have 
,
we have 
(okay I know this is SSA similarity which is not valid, but it's clear that 
and 
are either both obtuse or both acute). Hence 
,
.
where 
is the circumradius of 
![[asy]
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pair M = midpoint(B--C);
pair K1 = intersectionpoint(E--F, A--M);
pair K = intersectionpoint(M--K1, circumcircle(A,E,F));
pair Ap = rotate(180, M)*A;
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dot(E); label("$E$", E, dir(0));
dot(F); label("$F$", F, dir(180));
dot(H); label("$H$", H, dir(-90));
dot(M); label("$M$", M, dir(M));
dot(K); label("$K(K')$", K, dir(-K));
dot(Ap); label("$A'$", Ap, dir(Ap));
[/asy]](http://latex.artofproblemsolving.com/9/c/f/9cffc8ddb64d47a7b8c356717aafe59faa0f63dd.png)
be the antipode of 
.
,
,
is a parallelogram, with 
collinear. By Reim's theorem on 
and 
are collinear . Since 
,
,
is cyclic.
).
meet on 
so 
.
is cyclic is a pain though, so we deal with the weird orthocenter thing.
![[asy]
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pair O = circumcenter(A,B,C);
pair M = midpoint(B--C);
pair K1 = intersectionpoint(E--F, A--M);
pair K = intersectionpoint(M--K1, circumcircle(A,E,F));
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draw(A--D, rgb(0.4, 0.6, 0.9));
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dot(B); label("$B$", B, dir(B));
dot(C); label("$C$", C, dir(C));
dot(D); label("$D$", D, S);
dot(E); label("$E$", E, dir(0));
dot(F); label("$F$", F, dir(180));
dot(H); label("$H$", H, dir(-90));
dot(M); label("$M$", M, S);
dot(K); label("$K$", K, dir(-K));
draw(A--B,rgb(0.4, 0.6, 0.9));
draw(A--C,rgb(0.4, 0.6, 0.9));
draw(circumcircle(A,B,C),green);
draw(E--(-6,0),rgb(0.4, 0.6, 0.9));
draw(B--(-6,0),rgb(0.4, 0.6, 0.9));
dot((-6,0));
label("$K'$",(-6,0),S);
dot((-1.28,3.93));
draw(A--(-6,0),magenta+dashed);
draw(circumcircle(H,D,M),magenta+dotted);
label("$M'$",(-1.28,3.93),NW);
[/asy]](http://latex.artofproblemsolving.com/2/4/e/24eaa3405e9fe29dabad8ea5b2f1d05c58ca7710.png)
and 
are collinear. Let 
or that 
are concyclic.
.
are collinear. This inversion sends 
be the inverse of 
,
so 
be the inverse of 
and 
.
are collinear which follows from applying radical axis theorem for 
.
