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.
to 
(including ). He writes some of the numbers on the numerator, and writes 
(multiplication) between each number. Then he writes the rest of the numbers in the denominator and also writes between each number. There is at least one number both in numerator and denominator. The teacher ensures that the fraction is equal to the smallest possible integer possible.
. Prove that ![\[\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}\le \sqrt{2}(a+b+c)\]](http://latex.artofproblemsolving.com/6/e/d/6ed654c1d3616a8274ff0838e7bb13dfa0a11b8a.png)
light bulbs are operated by some switches. Each switch works on a subset of the light bulbs. When we use a switch, all the light bulbs in the subset change their state: bulbs that were on turn off, and bulbs that were off turn on. We know that every light bulb is operated by at least one of the switches. Initially, all lamps were off. Find the biggest number 
for which we can surely turn on at least light bulbs.
denote the infinite parallel ruler with the parallel edges being at distance 
from each other. The following construction steps are allowed using ruler 
:
parallel to a given line 
at distance (there are two such lines, both of which can be constructed using this step);
and 
with 
two parallel lines at distance such that one of them passes through , and the other one passes through (if 
, there exists two such pairs of parallel lines, and both can be constructed using this step).
be a given positive integer. Prove that based on the three points and there exists 
such that for any 
it is possible to construct points using only on one of the excircles of the triangle.
and 
and 
+2 w
is non-decelerating if 
holds for all positive integers . We say that a strictly increasing sequence 
is convergence-inducing, if the following statement is true for all real sequences 
: if subsequence 
is convergent and tends to 
for all positive integers 
, then sequence is also convergent and tends to . Prove that a non-decelerating sequence is convergence-inducing if and only if sequence 
, 
, 
is bounded from above.
light bulbs are arranged around a circle, and are consecutively numbered with
. Each bulb can be in one of two states: either it is on or off. In the initial configuration, days we change the current on/off configuration as
, on the -th day we start from the -th bulb and moving in clockwise direction-th day, coincides with the initial one.
let 
and 
denote the feet of the altitudes from 
and 
, respectively. Let line 
intersect circumcircle at points 
. Similarly, let line 
intersect circumcircle at points 
. Prove that the radical axis of circles 
and 
passes through the orthocenter of triangle
, 
, and![\[
a_{2k+1} = 2 + 2a_k, \quad a_{2k+2} = 2 + a_k + a_{k+1},
\]](http://latex.artofproblemsolving.com/4/e/c/4ec892df19b4ec97af9bb1e5d1954e6c26be3172.png)
for all integers 
. Determine all positive integers 
such that![\[
\frac{a_n}{n}
\]](http://latex.artofproblemsolving.com/3/d/1/3d1d012e7c41e1e09abd14626a79b1fb1cbab34d.png)
is an integer.
, find positive integer solutions 
to the equation![\[
\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} + \frac{1}{x_1 x_2 \cdots x_n} = 1
\]](http://latex.artofproblemsolving.com/5/c/4/5c40a3f0eaa58e438c8dcaad395d7ab9e8f4f439.png)
be the incenter of triangle 
. Let 
, 
, and 
be the orthocenters of triangles 
, 
, and 
, respectively. Prove that the lines through , , and , parallel to 
, 
, and 
, respectively, are concurrent.
, a single parity swapping is to choose 
terms in this sequence, say 
and 
, such that 
is odd, then switch their placement, while the other terms stay in place. This creates a new sequence.
to 
, using only single parity swapping.
be a fixed integer. A game is played by players sitting in a circle. Initially, each player draws three cards from a shuffled deck of 
cards numbered 
. Then, on each turn, every player simultaneously passes the smallest-numbered card in their hand one place clockwise and the largest-numbered card in their hand one place counterclockwise, while keeping the middle card.
denote the configuration after 
turns (so 
is the initial configuration). Show that is eventually periodic with period , and find the smallest integer for which, regardless of the initial configuration, 
.
be a positive integer. The integers 
are written on a board. In a move, Alice can pick two integers written on the board 
such that 
is an even number, erase both 
and 
from the board and write the number 
on the board instead. Find all for which Alice can make a sequence of moves so that she ends up with only one number remaining on the board.
, Alice changes 
to 
and can't make any further moves.
, the rays 
meet at 
, and the rays 
meet at 
. 
is the projection of 
on 
. Prove that there is a circle inscribed in if and only if the incircles of triangles 
are visible from under the same angle.
and 
be the centers of incircles 
and 
of triangles 
and 
respectively. Denote their inradii by 
and 
respectively. Let lines and 
meet at 
(on the projective plane). The incircles are seen from under the same angle 
bisects angle 
is an Apollonius circle of triangle 
is harmonic the external common tangents to and meet at 
Now, if a circle 
is inscribed into 
then and meet at simply by the Three Circles Theorem applied to 
Conversely, if and meet at 
then let 
be the excircle of triangle opposite to 
Applying the Three Circles Theorem to 
yields that the external common tangents to and meet on 
so they have to meet at and thus 
is tangent to as well.
.
, meaning that bisects 
. Furthermore, since 
and the external and internal bisectors of an angle are perpendicular, it is seen that 
is the exterior bisector of . Thus, 
and 
is harmonic.
be center of circle which inscribed in quadrilateral 
and line 
, we can show without using harmonic quadruples.
be the incircles of 
, and 
the incenters. Let 
. Consider the dilation about 
(with negative factor) which takes 
into 
. Then 
, and define 
as the images of 
. Let 
.
subtend equal angles from , which means 
, or 
lies equidistant from parallel lines 
. Thus is the midpoint of 
. If 
is the point at infinity of , then 
are harmonic. Projecting from 
onto 
yields that 
are harmonic as well (
), where 
is the angle bisector in 
. This means that is the excenter of opposite 
. Similarly it is the excenter of 
opposite , and we are done. To prove the converse simply reverse the steps. are harmonic, but can be easily shown by projecting the points from onto the line through parallel to 
.
to be the incenter of 
and 
to be the incenter of 
. Let be the intersection of 
(external bisector of 
). Clearly, the ratio of the inradii of and is 
to 
, so the condition is equivalent to 
. This is equivalent to 
being the angle bisector of 
, which due to 
is equivalent to 
.
has an inscribed circle, call it . Now let 
. Since 
is the foot of the interior angle bisector of 
, it can be seen that 
. Since 
, 
, and 
all pass through , observe that![\[ (K, M; D, F) \stackrel{I}{\doublebarwedge} (P, Q; R, F) = -1 \]](http://latex.artofproblemsolving.com/e/4/2/e42c14eadcecfd2bddce05d5e9412edcf372366b.png)
as desired. and 
intersect at 
, then let 
. Clearly 
by a perspecitivity from , so 
, and the angle bisectors of 
, 
, and 
concur at a point . Observe that this implies that is the incenter; indeed, ![\[ \delta(I, AB) = \delta(I, AD) = \delta(I, DC) = \delta(I, BC) \]](http://latex.artofproblemsolving.com/f/5/9/f5992e7cac945f4a9b6e94375fa948deb6d68506.png)
as desired.
under the same angle. Let be the incenter of triangle 
and 
be the incenter of triangle 
. Clearly, 
is the external bisector of angle 
. Let be the intersection of the lines and 
. It is clear that has equal distance from both pairs of opposite sides of the quadrilateral . Therefore, it is sufficient to show that lies on the internal bisector of 
. Let 
and let 
. Since the angle of visibility is the same, by the sine rule in corresponding triangle, on the half of the visibility angle yields that 
. Since 
we conclude that 
. Projecting from we get that 
. Since is the external bisector of angle we get that 
and so 
bisects angle . It is clear that all these steps can be reversed and the only if implication follows as well. The result clearly follows.
![[asy]
unitsize(50);
pen n_purple = rgb(0.7,0.4,1),
n_blue = rgb(0,0.6,1),
n_green = rgb(0,0.4,0),
n_orange = rgb(1,0.4,0.1),
n_red = rgb(1,0.2,0.4);
pair pole(pair a, pair b) {return extension(a, rotate(90, a) * (0, 0), b, rotate(90, b) * (0, 0));}
pair S1, S2, S3, S4, A, B, C, D, P, Q, H, I, I1, I2;
S1 = dir(173);
S2 = dir( 53);
S3 = dir(313);
S4 = dir(253);
A = pole(S4, S1);
B = pole(S1, S2);
C = pole(S2, S3);
D = pole(S3, S4);
P = extension(B, A, C, D);
Q = extension(B, C, A, D);
H = foot(D, P, Q);
I1 = incenter(A, D, P);
I2 = incenter(C, D, Q);
I = extension(P, I1, Q, I2);
draw(unitcircle, n_blue);
draw(A--B--C--D--cycle);
draw(P--A^^P--D^^Q--C^^Q--D, gray(0.6));
draw(P--Q, n_green);
draw(D--H, n_green + dashed);
draw(P--I^^Q--I, gray(0.6));
draw(I1--I2, gray(0.6));
draw(H--I1^^H--I2, n_green);
draw(incircle(A, D, P)^^incircle(C, D, Q), n_purple);
dot(A^^B^^C^^D^^P^^Q^^H^^I1^^I2^^I);
label("$A$", A, dir(160));
label("$B$", B, dir(110));
label("$C$", C, dir(40));
label("$D$", D, dir(300));
label("$P$", P, dir(230));
label("$Q$", Q, dir(320));
label("$H$", H, dir(270));
label("$I$", I, dir(90));
label("$I_1$", I1, dir(140));
label("$I_2$", I2, dir(60));
[/asy]](http://latex.artofproblemsolving.com/6/1/d/61d9fa386de990ac15334c9c09021b1f15a5cf2b.png)
be the incircles of , and let 
be their centers. Let 
be their radii, and lines 
and 
intersect at .
of quadrilateral exists. Then by Monge on 
we deduce that the exsimilicenter of is on line . This exsimilicenter also must lie on line , hence it is the point . is the insimilicenter of , 
. Thus 
; as 
we obtain 
bisects 
. Finally by Angle Bisector Theorem we deduce 
; thus 
and are visible from under the same angle. are visible from under the same angle. Then , so 
, thus is the internal bisector of .
we obtain 
, thus 
. Since is the insimilicenter of , must be the exsimilicenter, so the exsimilicenter lies on 
. be the -excircle of . Then the exsimilicenter of 
is , so by Monge on we deduce that the exsimilicenter of 
is on . Since line 
is an external common tangent, this exsimilicenter must be point . Thus, is tangent to line 
, so is an incircle for quadrilateral .
(call them 
from now on.) be the exsimilicenter of 
and let 
be the projection of onto 
.Note that by Monge-de Alambert we have that 
.We have:
.Also form similarity we have 
analogously for 
we have that 
and as 
,taking in account that 
is coaxial with we're done.
?
and 
respectively. Denote by 
are collinear since 
and 
both externally bisect 
.
.
.
.
it follows that 
.
.
.
we conclude that the exsimilicenter of 
lies on 
.
is tangent to both 
is tangent to 
it must be tangent to 
by definition, it follows that 
and