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.
and 
such that![\[
1^n + 2^n + 3^n + \cdots + n^n = x!
\]](http://latex.artofproblemsolving.com/6/4/b/64b5f5919fd3710cc15e96724918eae216c8e4f0.png)
crosses are placed. A set of 2025 cells we will call balanced if no two of them are in the same row or column. It is known that any balanced set has at least 
crosses. for which it is always possible to color crosses in two colors such that any balanced set has crosses of both colors.
is the foot of the altitude from 
of triangle 
. On the lines 
and 
points 
and 
are marked such that the circumcircles of triangles 
and 
are tangent, call this circles 
and 
respectively. Tangent lines to circles and at and intersect at 
.
.
Prove that 
Where 
such that 
is a prime or a perfect square of an integer.
be a function such that
. Determine 
.
be a scalene triangle such that 
is its incircle. is tangent to at 
. A point 
(
) is located on 
.
, 
, and 
be the incircles of the triangles 
, 
, and 
, respectively. and is also tangent to .
be a positive integer and let 
be a strictly increasing sequence of positive real numbers with sum equal to 2. Let be a subset of 
such that the value of![\[
\left|1-\sum_{i \in X} a_{i}\right|
\]](http://latex.artofproblemsolving.com/2/5/1/2515e074de1782f38453e1fce8ab7c2ec7cfe4c0.png)
is minimised. Prove that there exists a strictly increasing sequence of positive real numbers 
with sum equal to 2 such that![\[
\sum_{i \in X} b_{i}=1.
\]](http://latex.artofproblemsolving.com/3/7/4/37410820fc4c546f11d57212cfce53a7ec4941e3.png)
such that 
are collinear since they all lie on the perpendicular bisector of 
. If meets 
at 
, we have 
and 
are similar; also 
and 
are similar, so 
, so 
.
are perpendicular, and 
are perpendicular, triangles 
and 
are similar, so 
. Considering triangle 
, and the point 
on 
, by the converse of the external version of the Angle Bisector Theorem, 
is the external angle bisector of 
.
, 
, so 
, and similarly 
, so 
, so since bisects the external angle 
and 
bisects the internal angle , 
. Thus is tangent to the circumcircle of 
, and 
is similar to 
. Thus 
... (1).
is perpendicular to 
, bisects angle 
, so 
, and since , it follows that 
is similar to . Similarly, 
is similar to , and also . This gives us: 
... (2).
... (3)
, since 
and 
are similar, we have
.
, we have and 
are similar triangles, so 
, so 
, so 
, as required.
: about the midpoint of lies on the circumcircle. is the reflection about , lies on the circumcircle, and further, 
is a symmedian of triangle .
, so lies on the 
- Apollonius circle of .
the midpoint of 
, orthocenter of ,
.
symmetric point of wrt 
,. Since 
ie 
similarly 
so 
now diagonals of 
bisect each other so $ABA'Ç$ is a parallelogram. now 
but since 
we have 
are con-cyclic. so 
since 
ie 
and 
since 
and 
ie 
so 
since 
and 
ie 
but now 
and 
so 
therefore 
but since 
ie 
so 
be the orthic triangle. It is saying that 
is the intersection of 
with 
. Now, we cant solve this straight away from here, we will need a few other stuff. Indeed, note that if is the midpoint of that 
is the polar of , so it follows 
(given) so the 
and do symmetry for other. Then, by radical axis, 
so 
. Now, we have enough so we ignore and focus on what we need to prove. We need 
is cyclic. Let us prove 
is tangent to 
. Since 
is on the radical axis, it suffice to prove 
is tangent to 
, and by symmetry this means if 
then 
is tangent to . But, suppose the tangent at touches at 
. Then note 
is harmonic quad, so 
is tangent to and by symmetry 
so 
and we are done.
and 
be the circumcircles of 
and 
, respectively. Notice that 
is the perpendicular bisector of , so 
. Let 
. Let be on such that 
is tangent to the circumcircle of .
and 
, we know that the circumcircles of and 
are both tangent to , and is the radical axis, so that means 
, so is the midpoint of .. We have:![\[\angle APB = 180^{\circ} - \angle BAP - \angle PBA = 180^{\circ} - \angle ABC \\ \angle APC = 180^{\circ} - \angle PAC - \angle PCA = 180^{\circ} - \angle BCA\]](http://latex.artofproblemsolving.com/2/0/9/2098a6f2d4952f5a84afeb9058b949afca7c647e.png)
and 
. Therefore, 
.
, we have 
and 
, and multiplying these two together yields 
. Therefore, we have 
, so is tangent to the circumcircle of .
and 
, so 
is cyclic, so we know that 
passes through the circumcenter of 
. But 
, so is the isogonal conjugate of 
with respect to , so 
, and multiplying that equation by 
gives us the desired conclusion. 
be the circumcenters of triangles 
and 
altitude of .
, 
, from the above easily 
, so 
is angle bisector of 
; 
being perpendicular bisector of , 
are collinear, hence 
, so 
is external angle bisector of , i.e. 
, consequently 
is cyclic, of diameter 
and the altitude 
of 
is its isogonal w.r.t. 
, done.
the intersection of and . We have 
so is midpoint of . Now let 
be the reflection of in . 
so is on circumcircle of triangle .
so quadrilateral 
is harmonic hence tangents at and (on circumcircle of ) meet at one point,let's call it . 
so 
. Quadrilateral is cyclic so 
so 
.
,now let 
be the reflection of 
wrt .Now,we have to prove 
and since we have and 
,so it is enough to prove that triangles QPO' and QOA are congruent,but since we have QP=QA and QO=QO',it is enough to prove that 
,but this is true since is the circumcentre of 
and 
,so we are finished.
about to make , radical axis gives 
, 
so we need 
which is obvious as 
.
meet at 
Then since 
, we deduce that 
By writing the length ratios, it follows that ![\[MB^2 = MP \cdot MA.\]](http://latex.artofproblemsolving.com/f/a/8/fa8bad5e63322c148af4fea3909e5ff65dbc091f.png)
Similarly, we find that 
, and 
Hence, 
, so is the midpoint of 
In addition, note that ![\[\angle BPC = 180^{\circ} - \angle PBC - \angle PCB = 180^{\circ} - \angle PAB - \angle PAC - 180^{\circ} - A.\]](http://latex.artofproblemsolving.com/a/8/0/a803d9d05cc31223b0d2538879044b470ffb9775.png)
It follows that the reflection of about lies on 
, we have 
Similarly, we obtain 
Therefore, ![\[1 = \frac{PB}{AB} : \frac{PC}{AC} = \frac{P'B}{AB} : \frac{P'C}{AC}\]](http://latex.artofproblemsolving.com/3/8/8/38808884471aa0b30f63cebea0c8d84ce5a918fc.png)
which implies that the quadrilateral 
is harmonic. Then let the tangents to 
at and meet at 
Note that since is harmonic, we have 
Combining equal tangents with the fact that is the perpendicular bisector of 
and 
, it follows that ![\[Q'A = Q'P' = Q'P.\]](http://latex.artofproblemsolving.com/8/4/d/84db062b15fe26918a5e549b4f2f39b4e659159e.png)
Hence, 
It follows that 
, so the points 
are inscribed in a circle of diameter 
Then 
, which implies that 
, whence the desired result follows immediately. 
" this point is absolutely important.This point is on the median of ..that circumcircle tagents to .
as 
, 
as the midpoint of 
, and 
as the circle with diameter 
. Clearly 
is the perpendicular bisector of .
is the tangent to 
at 
and 
, we can immediately obtain that
or that is the midpoint of , and thus the center of . Furthermore, this implies that is the image of inversion of wrt , and thus is the radical axis of and the circle with radius 
at . Now, since is the radical axis of and , we must have that is the radical center of , , and , and so lies on radical axis of and , which is the desired tangent line.
and 
, so , , , and are concyclic, which implies that
Combining this with
yields that
and so 
. Q.E.D.
be the intersection point of lines and , it can be easily verified that 
and 
are similar, so 
, in the same way we can get 
, so 
, i.e. is the midpoint of . and as the circumcenters of 
and 
respectively. From the fact that 
and all lie on the perpendicular bisector of , then and are collinear. Also note that 
and 
are perpendicular to 
(by the fact that 
and 
). Next, we have 
is the external angle bisector of 
. Furthermore, we have 
and in the same way, 
. However, we know that 
, in the same way we get 
, so is the internal angle bisector of , hence 
, i.e. is tangent to .
is cyclic since 
. This implies that 
. But since , then 
, as desired.
and is the midpoint of is the 
point is on the A-appolonain circle 
inversion then 
is the perpindacular bisector of but 
is the intersection of tangents from 
to 
is tangent to 
is cyclic
be the reflection of over . Let 
be the intersection of the tangents to at 
and 
. Let be the intersection of lines and . Since is the -Humpty point of , is the midpoint of and 
. Thus 
if and only if 
. Thus we wish to prove is cyclic 
is tangent to .
, lies on . Let be the intersection between the tangents to at and . Proving thus finishes. Note that since lies on the -Apollonius circle of , 
. Thus is harmonic, and , , are collinear. Thus by La Hire's theorem, since lies on the polar of , must lie on the polar of with respect to 
lies on line . Further, note that 
, thus also lies on the perpendicular bisector of . Thus , as desired. 
is tangent to .
the midpoint of . Consider an inversion with center and radius 
. This swaps and , so 
is the radical axis of the degenerate circle at and the circle with diameter .
and the result follows.
is the -humpty point of 
so lies on the -median. Let 
. By the problem statement, we need 
would be cyclic, so 
, or is tangent to the circumcircle of . Thus, this prompts us to use Phantom Points, i.e. take some point s.t 
is tangent to the circumcircle of and try to prove that 
. From the properties of the Humpty point, we have 
and we also know that 
. Now we will show that this leads to the fact that 
is tangent to the circumcircle of 
. As a matter of fact, if we let the tangent from to the circumcircle of intersect at some point , then we have 
. Finally, by POP, we have 
, as claimed. 
is -humpty point of so lies on circle 
with ratio 
(familiar result). is the intersection of tangent at of and , so that's .
be midpoint of 
. We have 
is cyclic, which is exactly true. (Q.E.D)
be an acute scalene triangle with as its circumcenter. Point lies inside triangle with and . Point lies on line with 
. Prove that . is the A-humpty point, let the midpoint of , let the 2nd point in 
that lies on the A-symedian, let the midpoint of and 
the midpoint of 
. Its known that and are symetric w.r.t. hence lies on and is perpendicular to and 
meaning that is the intersection of the tangents from and to , hence 
is cyclic, now let 
meet 
at 
, by angle chasing:
Now the problem is equivalent to show that 
but now since 
we have that arcs 
and 
are equal in 
hence holds, thus we are done 
be the midpoint of 
Clearly 
tangent to 
so 
and is the exsimilicenter of these circles. Now from 
we get 
and hence 
. be the midpoint of , ray 
meet again at , 
, 
, 
, and the projection of onto be . is the -Humpty point, so lies on 
. Now, a well-known lemma regarding the -Apollonian Circle wrt yields 
. Thus, is actually the -Dumpty point, which means 
is cyclic with diameter 
. and 
are isogonal in 
, angle chasing yields 
. Thus, 
as desired. is cyclic, implying 
which is much simpler.
be the midpoint of ; it's known/not too hard to prove 
and 
. This means![\[
\frac{PB}{PC} = \frac{PB/PM}{PC/PM} = \frac{AB/BM}{AC/CM} = \frac{AB}{AC}.
\]](http://latex.artofproblemsolving.com/f/6/d/f6d64aca3fca2ec7052be6098fa8fa26f15b280c.png)
Now define 
. We claim that 
. To prove this, define the function 
via![\[
f(X) = PX^2 - \operatorname{pow}_{\odot(ABC)}(X);
\]](http://latex.artofproblemsolving.com/f/8/3/f83484b89bc0faca4c475248910bd4d4cc2c18eb.png)
remark that is an affine map. Observe 
, 
, and![\[
\frac{BQ_0}{Q_0C} = \frac{AB^2}{AC^2} = \frac{f(B)}{f(C)}.
\]](http://latex.artofproblemsolving.com/c/f/9/cf93a221fd5f7a1376dfb99d5c66f728d0afab99.png)
It follows that 
, i.e. .
, ergo is cyclic. The desired equality follows by an angle chase.
is the -Humpty point. Since and both lie on the -Apollonius circle, must be the center. It is well known this circle is orthogonal to , so 
. as the midpoint of . We then have 
cyclic, and hence![\[\angle MAQ = \angle MOQ \implies \angle PAQ = 90 - \angle OQB \implies \angle APQ = 2 \angle OQB.~\blacksquare\]](http://latex.artofproblemsolving.com/1/1/3/113c83dbfabfdb4e3028ab76c76d9d8ed2cdc1cd.png)
as . If we denote the intersection of tangents to the at and as , the midpoint of as , orthocenter as , and the intersection of and 
as . is the A-Humpty point of triangle so , , are collinear and 
will be perpendicular to . as the intersection of the tangent at wrt with . Then lies on which implies lies on the polar of wrt which implies that 
is tangent to . is isogonal to so that implies that 
and 
which leads to being the reflection of wrt which in turn implies that 
. Thus if we take midpoint of as , 
which makes 
.
so is cyclic implying 
finishing the proof.
is the -Humpty point. inversion, then goes to 
. The image of lies on the circle centred at 
passing through , and on , so it is the reflection of across the perpendicular bisector of . Call this point . Inverting back, we get that 
. be the midpoint of , then since 
, 
is concyclic. be the orthocentre, then
so we are done. 
is 
humpty point. Perform inversion and reflect over the angle bisector of 
. is a triangle 
where is the intersection of tangents to at and 
with 
. is the reflection of over . Prove that 
. be the midpoint of and be the intersection of symmedian with . Since 
and 
we see that 
are collinear. Also![\[(BC_{\infty},M;B,C)=-1=(A,S;B,C)=(DA,DS;DB,DC)=(BC_{\infty},DS\cap BC;B,C)\]](http://latex.artofproblemsolving.com/5/9/0/59050873f93ce77aa241f0b9095179abb920f805.png)
implies 
are collinear.![\[\frac{\measuredangle ATD}{2}+\measuredangle AA'D=\measuredangle ATM+\measuredangle TMA'=\measuredangle ASD=\measuredangle B-\measuredangle C\]](http://latex.artofproblemsolving.com/1/0/0/100625f79d2c6a9c42fd79cc27689377d7e7bd46.png)
As desired.
is Humpty Point. Let 
denote the intersection of internal and external angle bisectors of 
with respectively. Note that: is the center of the Apollonius circle, with diameter 
. Define couple of points:
. be the midpoint of .
(
).
b the midpoint of .
are collinear as 
( Apollonius circle is orthogonal to ). Also, note that 
are collinear with 
to be cyclic as 
. inversion at : maps to .
maps to .
.
.
and 
Prove that
where 
to 
.
at 
.
and 
,
is tangent to 
.
is tangent to the circumcircle of 
and as is 
.
since 
is tangent to the circumcircle of 
,
,
is cyclic, so 
since 
.
because of the fact that ![\[ \angle PB'T' = \angle PB'C' + \angle T'B'C' = \angle AC'B' + \angle PC'B' = \angle AC'P'
\]](http://latex.artofproblemsolving.com/c/4/6/c4650bf4fc7fad8963aaba08f60dba8da0de9780.png)
,
.
at 
and any point on 
'
,
.
,
is tangent to 
is tangent to the circumcircle of 
,
,
and thus 
.
.
,
since 
and 
.
,
.
is the circumcenter of 
,
is cyclic, so 
.
is the midpoint of 
since 
since 
is isosceles with symmetry axis 
.
.
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[/asy]](http://latex.artofproblemsolving.com/7/8/1/78137f02d66d35b402009dc4a9263a25bf770ae2.png)
be the orthic triangle of triangle 
the midpoint of segment 
be the midpoint of 
.
,
,
,
.
,
is tangent to 
.
,
and 
.
,
,
is cyclic. The rest is straightforward angle chasing. 
.
.
(
so 
and 
are similar. So 
and 
are similar too(why?) So we have 
and were done!!