AoPS Academy
be an acute triangle. Let 
be the reflection of point 
over the line 
. Let 
and 
be the circumcenter and the orthocenter of triangle , respectively, and let 
be the midpoint of segment 
. Let 
and 
be the points where the reflection of line 
with respect to line 
intersects the circumcircle of triangle , where point lies on the arc not containing . If 
is a point on the line 
such that 
, prove that 
.
be a positive integer and let 
be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of around the circle such that the product of any two neighbors is of the form 
for some positive integer 
.
be real numbers satisfying ![\[u_{1}+u_{2}+\cdots+u_{2019}=0 \quad \text { and } \quad u_{1}^{2}+u_{2}^{2}+\cdots+u_{2019}^{2}=1.\]](http://latex.artofproblemsolving.com/2/d/0/2d09d5e894fdd13ec466aa664f3f10da5ca8d134.png)
Let 
and 
. Prove that![\[
a b \leqslant-\frac{1}{2019}.
\]](http://latex.artofproblemsolving.com/8/7/0/870973eb68893864ee8f8494efcae4c34c11d2d8.png)
such that 
is divisible by 
and 
. Prove that is a composite number.
be positive integers. Maximize 
if 
.
, where 
, then maximum value of 
.
where 
is a natural number.What is the remainder when is divided by 
and 
are the roots of the cubic equation 
satisfies ![$$\log_{\sqrt[3]3}(\log_3 x)+\log_3(\log_{27}x)+\log_{27}(\log_{\sqrt[3]3}x)=1.$$](http://latex.artofproblemsolving.com/4/f/9/4f9bef4f56017af4af3893afdfc10be5d986a462.png)
Compute 
.
be a convex quadrilateral such that diagonals 
and 
intersect at right angles, and let 
be their intersection. Prove that the reflections of across 
are concyclic.
brings the reflections back to the feet of perpendiculars simplifying the solution
,
,
and 
at the points 
,
,
and 
, respectively. And let's name the points of the reflections ,
,
and 
across the sides ,, and , respectively. Since the quadrilaterals 
and 
have parallel sides, they are similar. So, we must to show that the quadrilateral is cyclic: from the cyclic quadrilaterals 
,
,
and 
we have 
, 
,
and 
, respectively. And since 
and 
. From equal angles we get 
, as desired.
brings the reflections back to the feet of perpendiculars simplifying the solution
? Can anyone explain reason of ratio being ?
over 
respectively be 
respectively.
be the circles centered at 
with radius 
.
and 
.
centered at with an arbitary radius, (let the circle be 
) we see that 
Radical Axis of and 
. Similarly for others. Let 
respectively .
. Hence, 
. Similarly for others.
then 
is a rectangle which is actually a cyclic quadrilateral. Now again inverting back we get that 
is a cyclic quadrilateral. Hence, proved. 
.
with scale factor , which brings the reflection of across each side of 
to the projection from onto that side. Because dilations preserve circles, it suffices to prove that these four projections are concyclic, so let the projection from onto 
, and 
be 
, and 
, respectively.
, we have that 
is cyclic and thus 
, and by right angles, 
. Because 
, we have that 
is cyclic as well, so thus 
, and therefore 
. Now we do basically the same angle chase on the "other side": we know 
, so 
, and by right angles, 
. Because 
, we have that 
is cyclic as well, so thus 
, and therefore 
. Hence, 
, so 
, so thus 
is cyclic, and we are done.
be foot of altitudes from to sides. obviously 
is cyclic. with scaling with center of and 
we are done.
denote the foot of the perpendicular from 
respectively. Let the reflection of over 
be 
. 
are 
merely dilations of each other, so by homothety, it suffices to prove that 
is cyclic. Noting that 
are all cyclic,
Hence, 
, which implies that is cyclic, as desired.
be a convex quadrilateral such that diagonals and intersect at right angles, and let be their intersection. Prove that the reflections of across are concyclic.
Prove that
and 
Prove that
should be equal to 
Prove that
Where 
Prove that
Let 
Prove that
and let 
intersect 
.
is cyclic,and 
,
,
,
,
,
,
because 
,
.
be the foot of perpendicular from 
,
are concyclic.![\[\angle EPA = \tfrac12\pi = \pi - \tfrac12\pi = \pi - \angle ESA\]](http://latex.artofproblemsolving.com/5/8/1/581a91b1f43afd2e670376c4ce6ba283d8d92b95.png)
Thus, 
are concyclic.
be the reflections of 
,
,
.
has isogonal conjugate 
)
are reflections of 
i
.
)
around 
r
c
centered at 
since 
is cyclic, 
,
since 
is cyclic. So, we have 
.
.
and 
,
,
,
,
and 
,
respectively and 
,
,
and 
,
,
and 
with 
= 
and 
= 
.
and 
are cyclic and
is right angled at ![\[\angle ESP = \angle EAP = \angle EAB = 90^{\circ} - \angle ABE = 90^{\circ} - \angle PBE = 90^{\circ} - \angle PQE\]](http://latex.artofproblemsolving.com/e/f/8/ef82eadf0721f7a9270dae779015ac8fadad4d86.png)
.
Hence, quadrilateral 
at 
.
across 
be 
respectively, and let 
intersect 
respectively. Using midlines, we find 
,
,
,
.
,
,
,
,
with ratio 
.
is cyclic, we need 
,
.
are cyclic, we have 
due to the perpendicular diagonals, thus we are done.
.
Summing implies
Therefore,
Hence, 
,
be the midpoints of 
,
,
,
,
lies on 
,
.
.
,
,
,
are cyclic.
However, 
,
. Therefore, 
as 
.
.
.
are also cyclic.
.
,
,
,
,
,
which are the feet of the perpendiculars from ![\[\angle EIF = \angle EAF = 90^{\circ} - \angle ABE = 90^{\circ} - \angle FBE = 90^{\circ} - \angle FGE\]](http://latex.artofproblemsolving.com/a/f/b/afb2432543a5a44ddaacc9691dfec2ef27260541.png)
![\[\angle EIH = \angle EDH = 90^{\circ} - \angle ECD = 90^{\circ} - \angle ECH = 90^{\circ} - \angle EGH\]](http://latex.artofproblemsolving.com/6/1/f/61fe26475e1cac3fa4cff58f973bef20db36ad4b.png)
Therefore, ![\[\angle FIH = \angle FIE + \angle EID = 90^{\circ}- \angle FGE + 90^{\circ} - \angle EGH = 180^{\circ} - \angle FGH \]](http://latex.artofproblemsolving.com/b/6/8/b68579ae0c28249035e60c08a1653f126f4d4676.png)
so quadrilateral 
and 
be the reflection points. Take a homothety 
Let 
denote the midpoints of sides 
be the midpoint of 
.
,
From 
so 
(since points E, P, A, and S are concyclic, as are points P, B, Q, and E). Similarly, 
We want to prove that 
or 
Since 
and 
,
.
.
![[asy]
import olympiad;
size(250);
pair A = (-3,0), B = (0,3), C = (3,0), D = (0,-2);
pair E = intersectionpoint(A--C, B--D);
pair E_AB = 2*foot(E, A, B) - E;
pair E_BC = 2*foot(E, B, C) - E;
pair E_CD = 2*foot(E, C, D) - E;
pair E_DA = 2*foot(E, D, A) - E;
pair O = circumcenter(E_AB, E_CD, E_DA);
real r = abs(O - E_AB);
pair K = (E + E_AB)/2;
pair L = (E + E_BC)/2;
pair M = (E + E_CD)/2;
pair N = (E + E_DA)/2;
draw(A--B--C--D--cycle, linewidth(1.2) + red);
draw(A--C, dashed + blue);
draw(B--D, dashed + green);
draw(E--E_AB, dotted + orange);
draw(E--E_BC, dotted + purple);
draw(E--E_CD, dotted + cyan);
draw(E--E_DA, dotted + magenta);
draw(circle(O, r), linewidth(1.2) + heavygreen);
dot("$A$", A, SW, red);
dot("$B$", B, NW, blue);
dot("$C$", C, SE, green);
dot("$D$", D, S, orange);
dot("$E$", E, NE, purple);
dot("$F$", E_AB, W, cyan);
dot("$G$", E_BC, NE, magenta);
dot("$H$", E_CD, SE, yellow);
dot("$I$", E_DA, S, brown);
dot("$O$", O, NE, heavygreen);
dot("$K$", K, NW, red);
dot("$L$", L, NE, blue);
dot("$M$", M, SE, green);
dot("$N$", N, SW, orange);
[/asy]](http://latex.artofproblemsolving.com/d/f/e/dfe32174dfcdf803339fdefd8469adae72976474.png)
and
Then from 
we have 
,
and 
are also cyclic. Also, since 
is the midpoint of 
and 
is the midpoint of 
,
and similarly for
Since 
to 
;
is cyclic or
which is true since 
and 
are right triangles, so we are done.
be the feet of 
,
are all cylic because of the right angles. Now with some angle chase we get that:
and 
,