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.
of positive integers such that ![\[t^{2}+1=s(s+1).\]](http://latex.artofproblemsolving.com/f/6/6/f665bfe4fd6cec17dbc5b90d42d201274ff7c564.png)
is the orthocentre of 
. 
, 
, 
are on the circumcircle of such that 
. 
, 
, 
are the semetrical points of , , with respect to 
, 
, 
. Show that 
lie on the same circle.
be three diameters of the circumcircle of an acute triangle 
. Let 
be an arbitrary point in the interior of 
, and let 
be the orthogonal projection of on 
, respectively. Let 
be the point such that is the midpoint of 
, let 
be the point such that is the midpoint of 
, and similarly let 
be the point such that is the midpoint of 
. Prove that triangle 
is similar to triangle .
is call 
if it satisfies the following properties:
and 
for all indices 
.
.
for which: Every sequence, there always exist two terms whose diffence is not less than 
. (where is given positive integer)
be a triangle with incentre 
and circumcentre 
. Let be the touchpoints of the incircle with , , respectively. Prove that 
is the Euler line of 
.
of integers that has the following property:
(possibly 
) then 
for every integer .
of nonzero integers such that the only admissible set containing both 
and is the set of all integers. satisfies 
, and its incenter is . is the foot from to 
，and is a point on segment such that 
. Line 
meets 
at point 
, and the orthocenter of 
is .
is tangent to 
.
be a non-cyclic convex quadrilateral. The feet of perpendiculars from to 
are 
respectively, where 
lie on segments 
and 
lies on 
extended. The feet of perpendiculars from to 
are 
respectively, where 
lie on segments 
and lies on 
extended. Let the orthocenter of 
be . Prove that the common chord of circumcircles of 
and 
bisects 
.
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x^2 + y) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/c/7/c/c7cd7c9a4d4372afc23e4503bb1c05fc7d07e788.png)
be an acute-angled triangle with 
and let be the foot of the-angle bisector on . The reflections of lines and in line meet and at points and respectively. A line through meets and at 
and respectively such that and 
while lies strictly between 
and . Prove that the circumcircles of
and 
are tangent to each other.
of natural numbers such that
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.
such that 
then find minimal value of 
such 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 
,
,
,
respectively 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 
,
,
,
,
,
.
.
,
,
,
are cyclic.
However, 
,
. Therefore, 
as 
.
.
.
are also cyclic.
.
,
,
,
,![\[\angle EIF = \angle EAF = 90^{\circ} - \angle ABE = 90^{\circ} - \angle FBE = 90^{\circ} - \angle FGE\]](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 
,