[/quote]
,
,
,
and 
,
is the intersection point of the diagonals 
and 
.
and radius 
intersects 
and 
.
is tangent to the circles with diameters 
and 
.
satisfying the equation![\[
(f(x+y))^2= f(x^2) + f(2xf(y) + y^2), \quad \forall x, y \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/2/5/e/25eed911f303ed7cc4e03765a2f943ad4641c451.png)
be a triangle inscribed in a circle, where the tangents to the circle at points 
and 
intersect at the point 
.
be a point on the arc 
and 
.
and 
intersect at point 
.
be the reflection of 
and 
intersect at point 
,
.
.
be non-negative real numbers such that 
Prove that
such that 
and 
for any real numbers 
and 
.
is written on the blackboard. Every time, Carmela can erase the number 
on the black board and replace it with a new number 
,
is a perfect square. Prove or disprove that all positive integers 
can be written exactly once on the blackboard.
and 
as ![\[\sqrt{2}=1.a_1a_2a_3\dots\quad\quad\sqrt{3}=1.b_1b_2b_3\dots\]](http://latex.artofproblemsolving.com/4/d/5/4d5049d5c3b26757a25c5dd34462c5462228b83d.png)
where each 
or 
is a digit between 0 and 9. Prove that there exist at least 1000 values of 
between 
and 
such that 
.
of positive integers with 
is called good if we can colour each integer red, blue, green or purple, in such a way that
consecutive integers at least one is coloured red;
of each 
consecutive integers at least one is coloured blue;
of each 
consecutive integers at least one is coloured green;
of each 
consecutive integers at least one is coloured purple.
.
,
points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least 
scalene triangles with vertices of that color.
such that :
.
such that 
,
and 
and 
,
be the intersection of the altitude through 
be the intersection of the altitude through 
.
be the intersection of the median through 
.
either the sum of the first 
.
Y by Davi-8191, HWenslawski, ImSh95, Adventure10, Mango247, Rounak_iitr
There is given a convex quadrilateral 
. Prove that there exists a point 
inside the quadrilateral such that
![\[
\angle PAB + \angle PDC = \angle PBC + \angle PAD = \angle PCD + \angle PBA = \angle PDA + \angle PCB = 90^{\circ}
\]](//latex.artofproblemsolving.com/1/b/e/1be300b009f7eb1f1af5c69af25a13a768a6e9b4.png)
if and only if the diagonals 
and 
are perpendicular.
Proposed by Dusan Djukic, Serbia
. Prove that there exists a point 
inside the quadrilateral such that![\[
\angle PAB + \angle PDC = \angle PBC + \angle PAD = \angle PCD + \angle PBA = \angle PDA + \angle PCB = 90^{\circ}
\]](http://latex.artofproblemsolving.com/1/b/e/1be300b009f7eb1f1af5c69af25a13a768a6e9b4.png)
if and only if the diagonals 
and 
are perpendicular.Proposed by Dusan Djukic, Serbia
This post has been edited 1 time. Last edited by djmathman, Jan 26, 2020, 8:29 AM
Reason: centered equation
Reason: centered equation
and let 
be the diagonal intersection. Let 
be intersection, other than 
of circumcircles 
with diameters 
Let 
of circumcircles 
Then
and 
are similar 
are similar 
and 
to 
at 
pairwise intersect at 
Cyclic quadrilaterals 
are similar 
is cyclic, let 
be its circumcircle. Let 
be poles of 
WRT 
Then 
and from similar right 
with center 
and radius 
Similarly, 
with center 
and radius 
and 
is radical center of 
and 
is radical axis of 
Since 
are both perpendicular to 
the circumcenter 
are pedals of 
Let 
be pedals of 
and let 
be pedals of 
From collinearity of 
and from
intersect at 
and let 
be circles with centers 
and equal radii 
Since polar of 
of 
are coaxal and 
Using sine theorem for 
with 
and 
bisects 
and circle 
bisects angle formed by circles 
takes circles 
such that 
bisects 
Circle 
centered at the concurrency point 
of 
Since 
Cyclic quadrilateral 
with diagonal lines 
inscribed in 
By symmetry, 
are congruent 
are similar. On the other hand, 
is pedal triangle of 
WRT 
is inversion image of 
are also similar (well known and posted before). As a result, 
are similar. In exactly the same way, 
are also similar. Let 
be the corresponding similarity transformations. Due to 
these similarities map 
to the same point 
Since 
it follows that 
is diagonal intersection of 
,
,
.
if and only if 
means the isogonal conjugate of 
)
the projections of 
and by 
the intersections of 
with 
respectively. Then
are all concyclic.
and we shall see that this point satisfies the desired property.
otherwise some of the angles in the 
sums would be obtuse.
of circumcircles 
Let 
be intersection, other than 
WLOG, let 
follow on the corresponding circumcircles in this order. (Otherwise, relabel 
)
and 
The case 
(when 
)
of the circles 
follow on the corresponding circumcircles in this order. (Proof for the other order 
is the same.)
(when 
)
of the circles 
This means that 
be the Miquel point for the complete quadrilateral formed by the lines 
.
.
and 
are orthogonal by the problem condition, thus the image of the circle 
is a circle of diameter 
.
,
is the intersection of the perpendicular 
and 
.
the angle bisectors of 
and 
coincide 
.
:
,
.
,
,
.
and 
of all these angles are positive.
.![\[AD^2+BC^2=a^2+b^2+c^2+d^2-2(ad-bc)\cos{\angle APD}\]](http://latex.artofproblemsolving.com/9/4/4/9442d590d77aed60801a4b4c7f324228684a8583.png)
and ![\[AB^2+CD^2=a^2+b^2+c^2+d^2-2(ab-cd)\cos{\angle APB}\]](http://latex.artofproblemsolving.com/9/4/f/94fdbf33c1fdfc2e337b5c209b2db163272989f3.png)
. So 
.
.![\[\frac{a}{d}\cdot \sin{\theta_2} = \sin{\theta_1}\]](http://latex.artofproblemsolving.com/0/d/4/0d42175db2ae6787a1ac0ddc7a64ba92d6a2c511.png)
![\[\frac{b}{c}\cdot {\cos{\theta_2}}=\cos{\theta_1}\]](http://latex.artofproblemsolving.com/f/7/d/f7d85703bd667477e5f06a67a2d4107706b8a2fb.png)
. Squaring the 2 above equations, and summing, we get that ![\[\frac{a^2}{d^2}\cdot\sin^2{\theta_2}+\frac{b^2}{c^2}\cdot \left(1-\sin^2{\theta_2}\right) = 1 \implies \sin{\theta_2} = d \cdot \sqrt{\frac{c^2-b^2}{a^2c^2-b^2d^2}} \implies \cos{\theta_2} = c \cdot \sqrt{\frac{a^2-d^2}{a^2c^2-b^2d^2}}\]](http://latex.artofproblemsolving.com/4/7/f/47fb62861fd593cabd999684818cd3c88dec1fc2.png)
(the sine and cosine are both positive). Now we easily get that ![\[\sin{\theta_1} = a \cdot \sqrt{\frac{c^2-b^2}{a^2c^2-b^2d^2}} \hspace{1cm} \cos{\theta_1} = b \cdot \sqrt{\frac{a^2-d^2}{a^2c^2-b^2d^2}}\]](http://latex.artofproblemsolving.com/1/0/d/10da792679d3f6566bc29e15a573a099fd94e5d7.png)
.![\[\cos{\angle APD} = -\cos({\theta_1+\theta_2}) = -(\cos{\theta_1}\cos{\theta_2}-\sin{\theta_1}\sin{\theta_2}) \\ = -\frac{a^2bc-bcd^2}{a^2c^2-b^2d^2}+\frac{ac^2d-ab^2d}{a^2c^2-b^2d^2} \\= -\frac{(ac+bd)(ab-cd)}{(ac+bd)(ac-bd)}\\ = -\frac{ab-cd}{ac-bd} \\ \implies (ad-bc)\cos{\angle APD} = -\frac{(ab-cd)(ad-bc)}{ac-bd}\]](http://latex.artofproblemsolving.com/f/2/9/f29ea1e0f1267e25fcfb783c42658f0542cb2084.png)
. Since this is symmetric, 
.
,
existence of 
.
,
,
.
denote the angle between 2 lines (positive or negative). Finally, let 
be 
respectively.
concur. The other triplet follows similarly. We use Ceva's theorem (sine form) for this. Note that 
.![\[\frac{\sin(\theta_1-\theta_2)}{\sin{\theta_2}}\cdot\frac{\sin(\frac{\pi}{2} - \theta_3)}{\sin(\frac{\pi}{2}-\theta_1)}\cdot{\frac{\sin{\theta_4}}{\sin(\theta_3-\theta_4)} = \frac{\sin{\theta_1}\cos{\theta_2}-\sin{\theta_2}\cos{\theta_1}}{\sin{\theta_2}}\cdot\frac{\cos{\theta_3}}{\cos{\theta_1}}\cdot{\frac{\sin{\theta_4}}{\sin{\theta_3}\cos{\theta_4}-\sin{\theta_4}\cos{\theta_3}}}}\]](http://latex.artofproblemsolving.com/f/7/d/f7d6bd1e1b14e79b2d4767a563c61de4b43ce568.png)
![\[\\ = \frac{\frac{OB}{AB}\cdot\frac{OA}{AD}-\frac{OD}{AD}\cdot\frac{OA}{AB}}{\frac{OD}{AD}}\cdot\frac{\frac{OC}{BC}}{\frac{OA}{AB}}\cdot{\frac{\frac{OD}{DC}}{\frac{OB}{BC}\cdot\frac{OC}{CD}-\frac{OD}{DC}\cdot\frac{OC}{BC}} = 1}\]](http://latex.artofproblemsolving.com/c/2/4/c2410b5e8bea73314b80698a6f52125af3720237.png)
(after elementary calculation). So the result follows.
wrt 
)
.
(
)
Now from this we have 
giving 
similarly we have that 
and it's easy to check that 
we consider 
which are isogonal conjugates of 
.
form a rectangle now note that if 
cut 
at 
we have that 
are homothethic(because 
meaning that 
are collinear). But using the if part(on 
we know that 
but now since 
we have that the homothethy that sends 
which means 
meet at point 
,
and 
meet at points 
respectively. Let the reflections of 
be 
respectively. By some old USAMO problem, whose link I shall post later on, we know that 
be the centre of this circle. It is evident that 
.
be the centre, then it is the isogonal conjugate of 
which means that 
,
,
,
,
,
,
,
.
and this means 
.
,
.
and clearly 
works.
so 
there exists an isogonal conjugate 
which obviously satisfies the conditions.
and any point 
inside the square, the quadrilateral induced by the lines through 
and perpendicular to the segment through 
is orthodiagonal. This is easy to see by Cartesian coordinate bash. To prove that the orthodiagonal condition implies the angle condition, consider some orthodiagonal quadrilateral 
,
,
eventually concur. The technique is actually somewhat simple: find two distinct positions and fix an 
has a smaller 
,
If 
Reflect 
adjacent sides. 
Thus 
and 
,
.
by a property of isogonal conjugation in triangles, and thus 
and 
are isogonal in both this triangle and 
Similarly 
and thus 
and 
are isogonal in 
are isogonal conjugates in the triangle with sides lines 
is equidistant to the feet from 
Similarly it is equidistant to the feet from 
be the feet from 
respectively. We see ![\[\angle EFG=\angle EFP+\angle PFG=\angle EAP+\angle PBG=\angle PBC+\angle PAD=90^\circ,\]](http://latex.artofproblemsolving.com/f/7/a/f7a5441d5ac89c7dd32fbc4f31aaeae02b2b68f1.png)
and similarly 
In particular 
is cyclic. Thus by our lemma 
We can check that the angle conditions imply 
so 
then 
,
be 
,
,
,
are concyclic. Then, we have
but since 
,
is also cyclic. Now, we claim that this means 
.
.
,
,
as desired.
°
°
and 
are collinear and 
