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.
be a triangle, and let 
and 
be points on segment 
symmetric with respect to the midpoint of . Let 
denote the circle passing through and tangent to line 
at 
. Similarly, let 
denote the circle passing through and tangent to line 
at 
. Let the circles and intersect again at point 
(
). Prove that 
.
be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
. If is an integer, then 
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
, and that, for every such pair 
, the "next" such pair can be calculated as
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
be an acute triangle with 
the feet of the altitudes lying on 
respectively. One of the intersection points of the line 
and the circumcircle is 
The lines 
and 
meet at point 
Prove that 
be an acute triangle with orthocenter 
and circumcenter 
. Let 
and 
be points on the line segments and respectively such that 
is a parallelogram. Prove that 
.
cities, 
airline companies in a country. Between any two cities, there is exactly one -way flight connecting them which is operated by one of the two companies. A female mathematician plans a travel route, so that it starts and ends at the same city, passes through at least two other cities, and each city in the route is visited once. She finds out that wherever she starts and whatever route she chooses, she must take flights of both companies. Find the maximum value of .
be inscribed in the circumcircle 
and circumscribed about the incircle 
, with 
. The incircle touches the sides , 
, and at 
, , and , respectively. A line through 
, perpendicular to 
, intersects , , and at 
, 
, and 
, respectively. The line meets at 
(distinct from ). The circumcircle of triangle 
intersects at 
(distinct from ). Let 
be the midpoint of the arc 
of . The line cuts segments and 
at 
and 
, respectively, and the tangents to the circle 
at and intersect at 
. Prove that 
.
known by both players, Arnaldo picks a number 
but doesn't tell it to anyone. Bernaldo thens pick an arbitrary positive integer 
(not necessarily in ). Then Arnaldo tells the number of divisors of 
. Show that Bernaldo can choose in a way that he can find out the number 
chosen by Arnaldo.
be a set of 2025 positive real numbers. For a subset 
, define 
as the median of 
when all elements of are arranged in increasing order, with the convention that 
. Define![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set of 2025 positive real numbers, the following inequality holds:![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in .
in 
colors, there exists an arithmetic progression of size 
. How can we derive Van der Waerden's theorem for 
from this?
be circumcenter of a non-isosceles triangle and be a point in the interior of 
. Let 
be foots of perpendicular lines from to 
. Suppose that 
is cyclic and is the circumcenter of , 
. Prove that 
bisects
interesting if is composite and for every divisor 
of at least one of 
and 
is also a divisor of with 
be a convex pentagon such that 
and 
Let be the midpoint of 
and let be the circumcenter of triangle 
Given that 
prove that 
be a point on 
such that 
Then 
and hence 
are cyclic. Let be a point on 
such that 
Since 
and are parallel to each other, we conclude that 
is a parallelogram. Then 
are collinear. Since 
and 
we conclude that 
and 
are perpendicular. Let 
is also a parallelogram and therefore 
Let 
and 
Then since 
we get 
Hence 
and 
and therefore 
We shall show that 
and 
we get
and since 
we get So, we are done.
be the reflections of 
with respect to . Then 
becomes the perpendicular bisector of 
meaning , so that 
lies on 
. Note that is the midpoint of 
and 
so 
is a parallelogram, implying 
. Then 
so that 
are collinear. Let 
be such that 
. Then given the condition 
this implies 
and clearly 
is isosceles. Now by a simple angle chase, 
, i.e. 
is cyclic. Then 
. But 
. But 
hence 
. Thus, 
is cyclic. Then 
. Now 
. So 
.
in the line , just draw the perpendicular from to the line and then draw in the circle with the foot of this perpendicular as the centre, with on the circumference. Then 's antipode is .
. and be the reflections of and with respect to , respectively. We find that 
, hence is isosceles. Hence 
, so is the angular bisector of 
and 
. is cyclic, because that would yield 
, as we have to prove.
, we get 
and hence 
is cyclic. be the intersection of the lines en 
. Then 
, so 
is also a cyclic quadrilateral. And so and all lie on one circle. Furthermore, since 
is the reflection of with respect to , that quadrilateral is also cyclic.
and so is cyclic and we are done., the condition 
becomes equivalent with lying on a certain circle. So in this way you've got rid of all the ugly conditions, and therefore you get a much easier angle chase 
a point on 
such that 
.
and 
a point such that 
.
a quadrilatere, this one is a parallelogram if and only if his diagonals meet each other in her midpoints.
and 
, and 
is the midpoint of 
, we have 
is a parallelogram and is the midpoint of 
. Otherwise, since we must prove that : 
because the triangle 
is iscoscele. And so we are left to prove that points 
are concyclic.
and is the midpoint of 
and then the triangle 
is iscoscele, hence points 
are concyclic. Even since is the common midpoint of the 
we get that 
are parallelograms. Hence : 
. Moreover, since 
and
and 
hence 
which give us : 
.
which means,
i.e : 
.
, are concyclic and hence we get the desired result which is :
.
with 
.Suppose 
and so 
.That implies 
. Also from 
we've 
. So solving we get 
. Now from the condition 
we get 
and basically so, 
. Now 
.Also as 
so, 
. Now take 
. From the condition 
directly we get 
. Now from the condition directly we've 
. Now just eliminating 
from above two expressions we get 
. Here we're asked to show 
. So we need to show 
.Suppose 
. If 
then 
, which is indeed true, so 
, that's what all we needed to show,so we're done.
be the midpoint of 
.Then 
and 
,so 
is cyclic.Thus 
since 
.Let be the midpoint of .Then it is well known that 
so is the circumcenter of 
.Consequently from straightforward angle chasing we get that 
are the internal angle bisectors of 
and 
respectively.Now applying sine rule in 
and 
we get
.Dividing the expressions and using the fact that 
we get 
.Thus 
,as desired.
about to get 
respectively.
implies that 
are concyclic. Now we have by the length conditions, 
are collinear and 
.
are concyclic and since 
by Fact 5 we have 
. This completes the proof. 
about to get respectively. implies that are concyclic. Now we have by the length conditions, are collinear and . are concyclic and since by Fact 5 we have . This completes the proof. 
about to get respectively.
?)
on 
such that 
and 
. Then observe that 
so 
are perpendicular bisectors of 
respectively. Hence bisects and 
respectively.
. From 
, we get 
. Moreover, if 
, then notice that![\begin{align*}
\angle EAQ &= 180^{\circ}-\angle ABC \\
&= 180^{\circ}-\angle CDE \\
&= \angle DCP \\
&= \angle DBP \\[4pt]
\angle AEP &= 180^{\circ}-\angle BCD = \angle BPD
\end{align*}](http://latex.artofproblemsolving.com/a/6/f/a6fd783cb04b2bcde5e38130cb8b0170f57f3e70.png)
So 
which means 
or 
. Now 
and 
or 
are tangents to 
. Let be the midpoint of 
. Then by symmedian lemma, 
. Finally, angle chase
so we are done.
![[asy]
size(8cm);
pair A,B,D,P,E,M,C,O,X,T;
A=dir(215);
B=dir(305);
P=dir(120);
E=0.2*P+0.8*A;
M=midpoint(P--B);
C=2*M-E;
D=dir(100);
T=2*M-D;
O=circumcenter(B,C,D);
X=foot(O,D,M);
T=2*X-D;
filldraw(E--P--D--cycle,gray);
filldraw(C--B--T--cycle,gray);
draw(A--B--C--D--P--cycle,linewidth(1.2));
draw(unitcircle);
draw(B--P);
draw(E--C);
draw(B--D);
draw(D--T);
draw(E--D);
draw(T--C);
draw(T--B);
draw(circumcircle(B,T,C),dashed);
dot("$A$",A,dir(200));
dot("$B$",B,dir(-30));
dot("$C$",C,dir(0));
dot("$D$",D,dir(80));
dot("$E$",E,dir(180));
dot("$P$",P,dir(120));
dot("$M$",M,2*dir(240));
dot("$T$",T,dir(270));
[/asy]](http://latex.artofproblemsolving.com/4/e/7/4e7dc83b81dadcf4c2f9820a9c6965173d297e7d.png)
on 
such that 
.Now by the problem stipulation we have that 
is parallelogram.So lies on 
.Because 
we have 
and so 
.Whence it suffices to show 
or 
is cyclic. be the reflection of about .By the problem condition we know 
is cyclic.Whence 
.Now a simple claim:
is the reflection of 
about 
and whence 
.Now compute: 
because .Whence 
is cyclic and we are done 
be the reflection of wrt .
, we get that 
are collinear and thus, 
. be the midpoint of 
, be the midpoint of . Thus, as 
, we have 
cyclic quadrilateral. Hence,
and as 
and hence we conclude that 
is a cyclic quadrilateral. Thus, 
we are done.
across to . Let be a point on such that 
. By the given length condition, 
as well.
, 
are collinear.
. So, since is midpoint of hypotenuse in 
, 
. Since , this means 
be midpoints of and . Since we're given 
, this means 
is cyclic. are midpoints, there is a homothety centered at with ratio that takes 
to 
and so the reflection of across , call it , lies on 
, reflecting stuff across , we have that 
and so is cyclic.
and so we are done. 
and be the reflections of and over 
Notice lies on 
as 
is a parallelogram and lies on as 
Also, the length condition implies 
is cyclic.
and 
are the reflections of 
and 
over 
respectively; hence, is cyclic. Since 
is a parallelogram, 
and be the reflections of and across . We have 
, so . Thus, is a cyclic trapezoid, and 
is also a cyclic trapezoid which is the reflection across .
so collinear. ![\[\angle BDB'=\angle CDE=\angle ABC=180^\circ-\angle BAE=180^\circ-\angle BAB'\]](http://latex.artofproblemsolving.com/3/5/a/35ac324cd0c7acd8909fd11d4712e42639874ad5.png)
so is cyclic. Note that 
so ![\[\angle ADB=\angle AB'B=90^\circ-\frac12 \angle B'AB=\frac12 \angle BDB'=\frac12 \angle CDE\]](http://latex.artofproblemsolving.com/a/6/7/a67e2530bbd8e1eea6bcb4caee79ed5c63f4e915.png)
as desired.
(such that triangle 
is isosceles) and then proving the two claims
collinear, where is the midpoint of 
, and
collinear, where is the reflection of over ?
cyclic.) Unfortunately, I wasn't able to make much headway proving either claim, but they intuitively do not feel hard to prove. (Thus I suspect that they're actually wrong -- which might make one of the funniest coincidences in geometry.)
such that 
such that 
.
at 
intersect 
.
,
meet this circle again at 
intersect 
at the distinct points 
.
and the perpendicular bisector of 
.
.
i.e. 
.
.
be the midpoint of 
,
is cyclic.
,
,
.
is cyclic.
.
