AoPS Academy
of positive integers is called cardinal if contains the integer 
where denotes the number of distinct elements in . Let 
be a function from the set of positive integers to itself such that for any cardinal set , the set 
is also cardinal. Here denotes the set of all integers that can be expressed as 
where 
. Find all possible values of 
+1 w
amazing, if one can partition the set 
into nice subsets.
be an integer and 
is a sequence of real numbers for which 
and:
Prove that all elements of the sequence are positive.
, points 
, 
, and 
lie on sides 
, 
, and 
, respectively, such that each of the quadrilaterals 
, 
, and 
has an incircle. Prove that the inradius of is twice the inradius of 
.
be the triangle whose medial triangle is . We want to show and have the same inradius. We will instead show the following stronger claim, which finishes. and share the same incircle.
by Pitot, so 
. By the ``excircle'' version of the converse of Pitot, this implies the quadrilateral bounded by rays 
, 
, 
, and 
has an incircle. Call this 
, and the corresponding 
and 
analogs 
and 
. By Monge on , , and , we find that 
are collinear, contradiction. Hence two of 
are equal, which implies the desired claim, and finishes the proof. 
be inradius of 
, now expand every circle with radius 
. Let this map be . Now, consider the following 
and define 
similarly.
and midpoint of 
and midpoint of 
. Similarly define 
. Now, we have that 
has 
excircle of radius 
and 
has incircle of radius . Thus, excircle of 
is the incircle of .
has an incircle but we that the quadrilateral formed by 
has incircle the same as that as and the lines of the quadrilaterals are pairwise parallel. Thus, from 
, we must have that 
. Thus, 
.
has an incircle, we have from Brianchon that 
concur. Let this point be 
. Now, 
is a parallelogram, thus the midpoint of 
lies on 
, so is on 
. But then we get that 
concur!!, the incircle of 
went to the point circle . Thus, it must have had radius . Hence proved!
of 
. We will show that the incircles of 
and 
coincide.![[asy] size(7cm); defaultpen(fontsize(10pt)); pen pri=blue; pen pri2=lightblue; pen sec=lightred; pen tri=purple+pink; pen fil=invisible; pen fil2=invisible; pen sfil=invisible; pen tfil=invisible;
pair Dp = (11.23550,1.19023), Ep = (8.32,-4.02), Fp = (15.,-4.), D = (11.66,-4.01), EE = (13.11775,-1.40488), F = (9.77775,-1.41488), I = (11.43393,-2.18789), A = (11.80061,3.90725), B = (8.49173,-4.79838), C = (13.64143,-3.51694); filldraw(incircle(Dp,Ep,Fp),tfil,tri); filldraw(A--B--C--cycle,sfil,sec); filldraw(D--EE--F--cycle,fil2,pri2); filldraw(Dp--Ep--Fp--cycle,fil,pri);
dot("\(D'\)", Dp, N); dot("\(E'\)", Ep, SW); dot("\(F'\)", Fp, SE); dot("\(D\)", D, S); dot("\(E\)", EE, dir(30)); dot("\(F\)", F, dir(150)); dot("\(I\)", I, N); dot("\(A\)", A, N); dot("\(B\)", B, SW); dot("\(C\)", C, E); [/asy]](http://latex.artofproblemsolving.com/b/2/9/b29b60da941db55e4563622b5ddb7412d58bbed4.png)
, so there is a circle 
tangent to rays 
, 
, 
, 
. Define 
and 
analogously., , do not coincide. One can check that the pairwise exsimilicenters of , , are 
, 
, 
. By Monge's theorem, points 
, 
, 
are collinear, contradiction.
of 
(twice the inradius). We claim that and share a incircle, in which case we'd be done. Note that , and since 
and 
, we have 
. Thus, the quadrilateral formed by the rays 
has an incircle, call this 
(I don't know what else to call 
). We can do similar things with Pitot, and define 
equivalently. By Monge on those three circles, we see that 
are collinear, which is false, so two of the circles coincide, as desired. 
as that was the only way I could actually deal with the "twice inradius" statement.
and 
be the incircles of and , and let 
be the ratio of their inradii.
be the center of the positive homothety mapping the to . Notice by Monge's theorem on , , and the incircle of 
, we have that lies on 
. Similarly, it lies on 
and 
. mapping to maps , , and to 
, , 
. We have that and share an incircle. By Poncelet, this implies that 
all lie on a conic. Hence, it suffices to show the following problem: be a triangle, let be point in its interior, let 
meet at , and extend 
to with 
. Define 
similarly. Show that if 
are on a conic, then 
. lying on a conic is equivalent to 
, 
, and 
being collinear. Note that 
are on , 
, and 
,
, 
, 
are collinear. We have that for some fixed constants 
, by the hidden claim,![\[DBC\text{ collinear } \implies \frac{\alpha}{PD}+\frac{\beta}{PB}+\frac{\gamma}{PC}=0\]](http://latex.artofproblemsolving.com/8/8/2/882878c42f757ae75511c5cd1e108a65a28fc833.png)
![\[AEC\text{ collinear } \implies \frac{\alpha}{PA}+\frac{\beta}{PE}+\frac{\gamma}{PC}=0\]](http://latex.artofproblemsolving.com/8/7/f/87fa38b2f966e42cef7baff816c42cb6ae584388.png)
![\[ABF\text{ collinear } \implies \frac{\alpha}{PA}+\frac{\beta}{PB}+\frac{\gamma}{PF}=0\]](http://latex.artofproblemsolving.com/e/b/8/eb82e34d6f2f7c2fc11ddb0f5934b35e379c3cbe.png)
![\[XBF'\text{ collinear} \implies \frac{\alpha}{PX}+\frac{\beta}{PB}+\frac{\gamma}{\lambda PF}=0\]](http://latex.artofproblemsolving.com/6/0/a/60a2006c2afd336cac9e3d3ce87d0f403a78f3fe.png)
![\[D'YC\text{ collinear} \implies \frac{\alpha}{\lambda PD}+\frac{\beta}{PY}+\frac{\gamma}{PC}=0\]](http://latex.artofproblemsolving.com/1/8/1/18119b8097bc402e978b0102dd54c6eccbef55cd.png)
![\[AE'Z\text{ collinear} \implies \frac{\alpha}{PA}+\frac{\beta}{\lambda PE}+\frac{\gamma}{PZ}=0\]](http://latex.artofproblemsolving.com/7/d/8/7d8ea2e18c645bcd600ad8554cefb861d2a44920.png)
![\[XYZ\text{ collinear} \implies \frac{\alpha}{PX}+\frac{\beta}{PY}+\frac{\gamma}{PZ}=0\]](http://latex.artofproblemsolving.com/9/6/d/96d8a9d8fe59b011f8c447a9c82e62c233781f92.png)
Summing these up with coefficients 
gives![\[\left(1-\frac{2}{\lambda}\right) \left(\frac{\alpha}{PD}+\frac{\beta}{PE}+\frac{\gamma}{PF}\right)=0.\]](http://latex.artofproblemsolving.com/0/6/0/06074f13f739422824585cd629b8edaec693955b.png)
Since are not collinear, the second term is nonzero, so we must have , as desired.
denote the incircles of , , , , , respectively. Let 
be the centers of 
, 
, , , .
concur at the exsimilicenter of 
.
.
are concyclic. (Similarly, 
and 
are concyclic.)
. This follows from summing ![$\frac{PD}{AD} = \frac{[BPC]}{[BAC]}$](http://latex.artofproblemsolving.com/d/7/e/d7eac58d73498e8f9e137c0f2c16dc7c3f0e79f6.png)
, etc.
an applying the above fact on 
.
is the midpoint of 
.
. By Menelaus' theorem on 
and 
, we obtain
Similarly, we obtain
so summing gives 
as desired.
are collinear, and by Desargue's Theorem, 
are concurrent. be an incircle of .
which its incircle is and homothetic with and a homothetic center lies outside 
. and send to an incircle of , it will send 
and 
. Therefore, 
lies on , and similarly, 
lies on , 
lies on .
. We want to prove that is a midpoint of or 
. be a projective transformation that send to the center of . is incenter of 
. are symmetric with respect to . be symmetric with wrt . and are defined by intersections of lines tangent to and 
. move along , a segment 
r will not intersect 
or 
except when 
. Thus, is tangent to iff , but 
so 
as desired. 
before the transformation, is now a excentral triangle of .
, these point lie on image of a line at infinity.
. Pencil at 
, 
.
, these six points lie on a conic 
. Moreover, because 
and 
are symmetric wrt , a reflection of 
across lso lie on this conic. Denote this reflection by 
.
.
bisect 
. (since 
.)
as desired.
, , , and has an incircle.
are 1/4 perimeter of DEF ( are 1/2 homothety of excircles of DEF mapped with center D,E,F). Moreover, the tangent of incircle of ABC at each side are the symmetry of D,E,F wrt the midpoint of tangent of on each side. The incircle radius of ABC is then easily solved by the area relation.
be a triangle, let be point in its interior, let meet at , and extend to with . Define similarly. Show that if are on a conic, then .
be the anticomplementary triangle of . We finish by the following claim. and coincide.
, so using medial triangle lengths, we rewrite this as ![\[ AF - AE = D'E-D'F, \]](http://latex.artofproblemsolving.com/9/7/3/973d771994cdb7022d3557baf8b3d958b9ed2181.png)
so by Pitot again we have that there is an excircle tangent to 
. Analogously, we draw , . Suppose that , , are pairwise distinct. By Monge's theorem on , , , we have that , , and are collinear, a contradiction. Thus, at least two of these circles coincide, implying the claim.
, , and is the incenter of .
such that 
concur, and I used Pitot to prove that the incircle of 
were tangent, etc. This was all in an effort to start with 
be a fixed point and 
be fixed lines through it. Points 
are chosen on each of these lines. Then, the condition for 
being collinear is equivalent to some fixed homogenous linear function in 
,
,
being 
.
,
,
are the directed angles between 
and 
,
,![\[\frac{\sin(\theta_1)}{OP_1}+\frac{\sin(\theta_2)}{OP_2}+\frac{\sin(\theta_3)}{OP_3}=0.\]](http://latex.artofproblemsolving.com/c/c/0/cc03833e8754cf647ef9a251887fb53c6362ab65.png)
![[asy]
// Generated By AsyPadv2.0
import olympiad;
import markers;
import math;
import graph;
// change the unit size to fit your needs
unitsize(1.0cm);
// colored pens
pen c000000 = rgb("000000");
// dependency level 0
/* You can change the coordinates of these points of dependency level 0.
The drawing will retain the same relationships and qualities.
Please be aware that as a result of this some of the image may be clipped off. */
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pair P_3 = (1.34, 5.91); dot(P_3, c000000); label("$P_3$", P_3, dir(47.29061004263667));
// dependency level 1
// Do not change anything below, unless you are experienced in Asymptote.
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path lineO_P_2 = (O-100.0*unit(P_2-O))--(P_2+100.0*unit(P_2-O)); draw(lineO_P_2, c000000);
path lineP_1_O = (P_1-100.0*unit(O-P_1))--(O+100.0*unit(O-P_1)); draw(lineP_1_O, c000000);
path segP_2_P_3 = P_2--P_3; draw(segP_2_P_3, c000000);
path segP_2_P_1 = P_2--P_1; draw(segP_2_P_1, c000000);
// dependency level 2
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path parP_2_lineP_1_O = (P_2-100.0*dir(lineP_1_O))--(P_2+100.0*dir(lineP_1_O));
// dependency level 3
pair Y = intersectionpoint(lineP_1_O, parP_2_lineO_P_3); dot(Y, c000000); label("$Y$", Y, dir(252.64418515409656));
pair X = intersectionpoint(lineO_P_3, parP_2_lineP_1_O); dot(X, c000000); label("$X$", X, dir(204.6981593746918));
// dependency level 4
path segX_P_2 = X--P_2; draw(segX_P_2, c000000);
path segP_2_Y = P_2--Y; draw(segP_2_Y, c000000);
// clip the drawing view
clip((0.0, 0.0)--(0.0, 6.99)--(8.71, 6.99)--(8.71, 0.0)--cycle);
[/asy]](http://latex.artofproblemsolving.com/2/a/d/2ad079f49a3191a269f68ec8a453eaae1c39c4b6.png)
and 
.
,
.
and 
.
.
is the foot of perpendicular from 
it is well known that 
is an Isosceles trapezoid which implies the result
such that 
is the orthocenter of 
.
is concyclic. Suppose for the sake of contradiction that some 
works. Then any conic passing through 
,
,
,
,
,
must pass through six points which are alternatingly on and outside circle 
,
works, then any conic passing through 
,
,
be the incircles of 
,
,
.
be the centers of 
,
the centers of 
.
by properties of tangential quadrilaterals. We also know 
is the midpoint of 
.
at 
,
meet 
.
and 
meet at a point 
lie on a semicircle centered at 
meets 
at 
,
,
,
is a diameter, so the tangent at 
,
.
is the radical axis of 
,
,
passes through the Nagel point of 
are collinear) Thus, 
are collinear.
since 
is the radical axis of 
is the angle bisector of 
,
as well. But now we see that 
.
to 
from 
is 
is the midpoint of the 
,
.
and 
,
.
,
,