A lot of tangent circle
by ItzsleepyXD, Mar 23, 2025, 7:53 AM
Let 
be a triangle with circumcircle 
and circumcenter 
. Let 
and 
represent the 
-excircle and -excenter, respectively. Denote by 
the circle tangent to 
, 
, and on the arc not containing , and similarly for 
. Let the tangency points of 
with line be 
, respectively. Let 
be the intersection point of 
and . Define 
as the point on segment 
such that 
. Suppose that 
intersects again at 
. Let 
be the touch point of the -mixtilinear incircle and , and let 
be the antipode of with respect to . Let 
be the intersection of 
and 
.
Show that the line
is the radical axis of and .
be a triangle with circumcircle 
and circumcenter 
. Let 
and 
represent the 
-excircle and -excenter, respectively. Denote by 
the circle tangent to 
, 
, and on the arc not containing , and similarly for 
. Let the tangency points of 
with line be 
, respectively. Let 
be the intersection point of 
and . Define 
as the point on segment 
such that 
. Suppose that 
intersects again at 
. Let 
be the touch point of the -mixtilinear incircle and , and let 
be the antipode of with respect to . Let 
be the intersection of 
and 
.Show that the line
is the radical axis of and .
and 
be points on circle 
,
,
tangent to 
be any point on arc 
,
,
,
to circles 
.
,
,
,
be points on circle 
equal parts. Construct 
tangent to 
,
,
to circles 
of 
(where 
)
and 
Prove that
Where 
denote the set of positive integers
such that for all 
it holds that 
Note: consider 
and 
all distinct points on opposite sides; for Pascal, no two opposite sides parallel)
,
on 
,
is above, 
is below 
,
is a plane, etc.
,
,
,
,
are coplanar, so 
and 
intersect at a point on the line equal to the intersection of planes 
and 
,
and 
,
,
,
(possibly degenerate, but then the result's obvious) intersects plane 
,
,
,
be positive rational numbers and let 
be integers greater than 1. If ![$\sum_{i=1}^{n}\sqrt[k_i]{a_i}\in\mathbb{Q}$](http://latex.artofproblemsolving.com/6/d/3/6d3268640d63424770f2abb2e6c42657d054ace0.png)
,![$\sqrt[k_i]{a_i}\in\mathbb{Q}$](http://latex.artofproblemsolving.com/1/e/9/1e98b3af4db054d32f00d529032658c15875251a.png)
for all 
.
and go by contradiction. For motivation we look at the 
case. WLOG 
are ``minimal'' in the sense that 
for 
.
for all 
is the minimal polynomial of 
over the rationals, so 
.
;
and we can easily get a contradiction by considering the ![$[x^{k_1-1}]$](http://latex.artofproblemsolving.com/a/1/3/a131a3600dec6f0f8fc7d7d771d8e0224b7b78cc.png)
coefficients.![\[x^{k_1}-a_1\mid \prod_{\omega_i^{k_i}=1,2\le i\le n}(S-x-\sum_{i=1}^{n}\omega_i a_i^{1/k_i})\in\mathbb{Q}[x],\]](http://latex.artofproblemsolving.com/1/c/e/1ce1cb394f58859b98d50fe5b410b813d4c9a5a1.png)
t
and (any) nontrivial 
t
,
.![\[ 0=\Re(S-S) = \Re(a_1^{1/k_1}(1-\omega_1))+\sum_{i=2}^{n}\Re((1-\omega_i)a_i^{1/k_i}) > \sum_{i=2}^{n}\Re((1-\omega_i)a_i^{1/k_i}) \ge 0,\]](http://latex.artofproblemsolving.com/2/3/9/2399ccddd04a1f97962cffb620d26fa61541fa66.png)
are all positive, but then 
would be in 
.
square board contains either 
or 
.
there are two; we restrict our attention to 
.
.
of each entry so we're working in 
instead; we're choosing some of the 
entires to be 1. Viewing this as the equation 
(where boundary cases are 0 for convenience), we can set up a matrix equation 
where 
iff entries 
(there are 
total) are neighboring, and 
has a 1 for each entry 
that 
(whence 
must be the all-zero vector and there's exactly one solution)---note that for 
(as we're working 
that contributes 1 to the sum (i.e. doesn't vanish), draw an arrow from 
for each 
board into disjoint directed cycles of size 1 or 
for some 
(arrows connect neighboring elements---these directed cycles correspond to permutation cycles). These cycles are nontrivially ``reversible'' iff 
,
)
,
,
tiles. Reflecting over the horizontal axis we're left with horizontally symmetric tilings; now reflecting over the vertical axis we're left with tilings that are both horizontally and vertically symmetric.
it's easy to check that 
where 
,
,
is Fibonacci, as all dominoes intersecting one of two axes must lie within the axes (by symmetry). By induction we get 
.
)
boards using the same recursive linear algebra ideas, but when both of 
are even it seems hard to reduce the problem.
be a connected graph disconnected by the removal of (all of the edges of) any odd cycle. Prove that 
be two positive integers. A graph 
is 
-
such that 
and 
are 
-
-
to a vertex colored 
in 
,
,
and 
,
for 
and 
and define 
by the first and second coordinates, respectively---the induced subgraphs 
can easily be split in the desired form.) 
and 
are both bipartite.
without inducing an odd cycle. Now assume for the sake of contradiction that 
)
,
.
and 
(indices modulo 
)
are connected in 
for all 
.
is connected in 
to at least one vertex of 
,
denote the time it takes for car 
when in and out of town, respectively; let 
(not both zero) be the distances from flag 
to flag 
along in and out of town routes. Then car 
to get to flag 
for some 
for all 
pairwise distinct cars (WLOG cars 
for all 
and 
.
iff![\[(x_k,y_k)\in S_{i,j}=\{(x,y)\mid L_{i,j}(x,y)=(a_i-a_j)x+(b_i-b_j)y+(t_i-t_j)>0\},\]](http://latex.artofproblemsolving.com/f/9/f/f9f56603a6e3049e2438f5de19aa67b1403e9a91.png)
and 
iff 
.
in the Cartesian plane for 
and 
)
(where 
is the number of lines) distinct regions (well-known, e.g. 
and 
are in the same region iff the orderings at flags 
are equal, so we're done by pigeonhole.
(in reduced form) is infinitely often not a prime power.
and primes from 
to 
for 
,
,
as well. Subtracting dudes, get a contradiction.
July 2014
June 2014