[/quote]
,
?
be a prime number and 
an integer polynomial of degree 
such that 
and 
is congruent to 
or 
modulo 
.
.
such that 
is an integer for all 
,
are all the positive divisors of 
.
be three positive reals such that 
.![\[ \frac { x^5-x^2 }{x^5+y^2+z^2} + \frac {y^5-y^2}{x^2+y^5+z^2} + \frac {z^5-z^2}{x^2+y^2+z^5} \geq 0 . \]](http://latex.artofproblemsolving.com/a/1/4/a14adf0f1e35df57c734c7df701d5a67da22dc3c.png)
where 
are lengths of parallel sides and 
are other two sides.
be a cyclic quadrilateral. Let 
,
be the feet of the perpendiculars from 
to the lines 
,
,
,
if and only if the bisectors of 
and 
are concurrent with 
.
be the feet of the altitudes from 
to 
to 
and 
as the reflections of 
be the circumcircle of 
.
with 
with 
can be inscribed within a circle.
be a triangle. A circle passing through 
intersect at 
,
and 
intersect at 
.
if and only if 
.
and 
be points on segment 
denote the circle passing through 
denote the circle passing through 
.
)
.
spheres centered at 
and 
and 
,
points 
outside of 
as the number of sets 
such that the midpoint of 
lies entirely inside of 
and 
.
.
has circumcenter 
.
and 
intersect at 
,
and 
intersect at 
and 
and 
intersect at 
.
are similar(with vertices in that order), than 
be the feet of the altitudes from 
respectively. Let 
be the projections of 
.
.
Y by GeoKing, Rounak_iitr, SatisfiedMagma
Let 
be an acute angled scalene triangle with circumcircle 
. Let 
and 
be the circumcenter and orthocenter of 
respectively. Let 
and 
be points on segments 
and , respectively, such that
Prove that 
and 
intersect on the circumcircle of 
.
Proposed by Antareep Nath
be an acute angled scalene triangle with circumcircle 
. Let 
and 
be the circumcenter and orthocenter of 
respectively. Let 
and 
be points on segments 
and , respectively, such that
Prove that 
and 
intersect on the circumcircle of 
.Proposed by Antareep Nath
This post has been edited 2 times. Last edited by taptya17, Dec 17, 2024, 6:08 AM
so that 
and 
are similar, we observe that ![\[\frac{BF}{CE} = \frac{BH}{HC}.\]](http://latex.artofproblemsolving.com/a/5/4/a540476b20d8c7ab9c063e3c6f2af249cb04aa87.png)
Since ![\[\frac{BQ}{QC} = \frac{BH}{HC}\]](http://latex.artofproblemsolving.com/a/3/4/a34da1a19aa4eca3c3a00424e3fb8c6cc7aa0394.png)
as well, which leads to 
cyclic from spiral-sim.
again at 
.
are reflections over 
.
,
.
.
and note that 
and thus 
we determine this isogonal conjugate to be 
,
which finishes aong with the previous paragraph. 
be the feet of the altitudes from 
to 
respectively. Note that 
q
.
are concyclic.
and 
.![\[\frac{EL.EA}{EB.EA}\overset{?}{=}\frac{FK.FA}{FC.FA}\iff \frac{HL.HK}{HB.HC}=\frac{EA.\frac{LH}{HC}}{EA.\frac{HB}{HK}}=\frac{EL}{EB}\overset{?}{=}\frac{FK}{FC}=\frac{FA.\frac{HK}{HB}}{FA.\frac{HC}{HL}}=\frac{HK.HL}{HB.HC}\]](http://latex.artofproblemsolving.com/7/9/9/79950946e13081bdd87b94dd65b3c8564b69da52.png)
Which proves the result
.![\[\overline{e}=\frac{1}{a}+\frac{1}{b}-\frac{e}{ab}, \ \ \frac{h-b}{h-e}=-\frac{\overline{h}-\overline{b}}{\overline{h}-\overline{e}}\implies \frac{1}{e-a-b-c}=\frac{1}{a+\frac{ec}{b}}\]](http://latex.artofproblemsolving.com/8/4/e/84eac33eb0e110fb5ee92cd5cd4994270cacabef.png)
Thus, 
.
.![\[e.\overline{e}=\frac{(2a+b+c)b}{b-c}.\frac{\overline{(2a+b+c)}.\frac{1}{b}}{c-b}=\frac{(2a+b+c)c}{c-b}.\frac{\overline{(2a+b+c)}.\frac{1}{c}}{b-c}=f.\overline{f}\]](http://latex.artofproblemsolving.com/3/e/7/3e7110f64541032f6d79d6234e6304a341915250.png)
So 
.
thus, 
.
and 
are concyclic and 
hence the perpendicular bisectors of 
intersect at another point than circumcenter of 
.
is an isosceles trapezoid and 
,
must hold. 
which proves the collinearity of 
as desired
.
and 
intersect on 
Indeed cool problem
,
and 
.
.
As 
is midpoint of 
,
.
we get 
is cyclic trapezium.
is point on 
.
where ![\[\Box AEHF \sim \Box CHBA'\]](http://latex.artofproblemsolving.com/4/c/8/4c8a1f5dba52eab47318ddc7215a994f39eae593.png)
then assuming 
we get 
Thus 
Hence 
lie on circle 
cyclic and to chase 
.
,
.
are tangents to 
.
and 
are similar and thus an angle chase tells us 
.
and 
then by symmetry 
must bisect 
.
.
be the feet from 
to 
respectively. By similar triangles we have 
,
.
.
is clearly a parallelogram, 
is the center of 
lies on the perpendicular bisector of 
cannot be 
.
intersect 
.
.
and 
This implies 
,
.
. 
, then we have 
.
we obtain 
collinear. Thus 
.
.
and 
meet 
which implies 
is an isosceles trapezoid.
.
are collinear.
so 
.
.
and 
is foot of 
foot of 
but this is because of 
and similarly defined altitude.
.![\[\frac{EE'}{EB}=\frac{EE'\times EB}{EB^2}=\frac{EH^2}{EB^2}=\sin^2{90^\circ-\angle A}=\cos^2{\angle A}\]](http://latex.artofproblemsolving.com/8/6/9/86924ac259c052f98ebe9f6762fc5c8fe088ce6a.png)
Now use ratio lemma on 
so![\[\frac{GH}{GH'}=\frac{QH}{QH'}\times\frac{\sin HQO}{\sin H'QO}\]](http://latex.artofproblemsolving.com/4/c/a/4ca815c645252e4918c13ff004f79732ce7c16ef.png)
now 
(because angle at centre theorem), and 
.
.
,![\[\cos{\angle A}=\frac{QH}{QH'}\]](http://latex.artofproblemsolving.com/d/5/9/d5972a06bf7c527bd87cb3f8f88c60b2add0db3c.png)
Now, doing more trig,![\[\frac{QH}{QH'}=\frac{AH \sin \angle QAH}{\sin\angle QAH}\times \frac{\sin \angle AQH'}{AH'}=\frac{AH}{AH'}\times \sin\angle AQH'\]](http://latex.artofproblemsolving.com/8/0/4/80470be4458bfee16177920731010053c938b5ee.png)
hence we just need ![\[\frac{AH}{AH'}=\frac{\sin \angle ACH}{\sin \angle ACH'}\]](http://latex.artofproblemsolving.com/3/e/3/3e339540f730c5167acd1469bd5f448641e8f7e3.png)
This is a direct consequence of ratio lemma on 
and the well known fact 
.
be orthic triangle of 
.![\[\frac{\text{Pow}(E,(AH))}{\text{Pow}(E,(ABC))}=\frac{\text{Pow}(F,(AH))}{\text{Pow}(F,(ABC))} \iff \frac{EZ}{EB}=\frac{FY}{FC}\]](http://latex.artofproblemsolving.com/e/e/2/ee2ba2980615681cfc987bb874b9f86067455915.png)
Which is true since 
because see that 
.
.
and so all we need to prove is that 
in 
,
and it is obvious that this must be ![\[\angle OFE=\angle HFC=90 ^{\circ}-\angle ACH=\angle BAC=\angle EAF\]](http://latex.artofproblemsolving.com/4/2/5/4255aa95a8cf5d47d1257dc3ef5b2f68907e9fac.png)
And so 
is tangent to 
and done.
,
,
and analogously 
are both tangents to the circumcircle of 
.
is the symmedian in 
.
.
where 
to 
.
and 
,
can be expressed in trigonometric ratios in 
,
which we can calculate analogously, thus the spiral similarity also sends 
invert then the whole problem turns into a very famous 
,
config which is very easy to prove.
