AoPS Academy
and 
for 
. Let 
. Prove that there exist infinitely many natural numbers 
for which 
is divisible by 
. Does there exist for which 
is divisible by ?
satisfy,
Where {
} is a permutation of . Prove that 
be a fixed square and 
a variable point on segment 
. The square 
is constructed such that 
is on segment 
and 
is on segment 
. Let 
be the intersection point of lines 
and 
. Find the locus of as varies along segment .
be a function taking in positive integers and outputting nonnegative integers, defined as follows:
is the number of positive integers 
with 
such that the equation 
has an integer solution 
.
such that
and 
Adit Aggarwal, India.
, 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