AoPS Academy
denote the set of positive integers. Find all functions 
satisfying the condition
for all 
be a scalene triangle, 
its medial triangle, and 
a point on the Thomson cubic (= 
) of . (Suppose that 
).
be the circumcevian triangle of wrt .
be the pedal triangle of wrt .
be the reflection in 
of 
. Define 
, 
cyclically.
be the reflection in 
of . Define 
, 
cyclically.
be the reflection in 
of . Define 
, 
cyclically.
, 
, 
and the orthocentroidal circle of are coaxal.
+1 w
lines are drawn on the plane such that no three of them are concurrent and no two are parallel. At least 
of the bounded regions formed are colored black. A triangle formed by three lines is called a \textit{good triangle} if it lies entirely within a black region. Prove that there are at least 
good triangles. (A good triangle is a bounded region with finite area.)
cities, including a capital city. Some pairs of cities are connected by two-way flights. Given a city 
, an ordered list of cities 
is called an antitour from if) appears in the list exactly once, and
, it is impossible to go from to 
by a sequence of exactly 
(not necessarily distinct) flights. for any city . Further, he can take a sequence of flights, starting from the capital and passing through each city exactly once. Find the least possible total number of antitours from the capital city.
, where 
. Verify by direct substitution that 
satisfies the quadratic equation 
and deduce that the value of is 
.
. Show that 
, show that the value of 
is 
.![$f \in C^1([0,1]), g \in C^2(f([0;1]))$](http://latex.artofproblemsolving.com/f/5/2/f52369171ef8571734950120c43aacac10154fcd.png)
.
?
, where 
are positive integers and 
is odd.
and 
.
with multiplication of matrixes operation is making an isomorphic-group structure with 
.
.
plane centered on the 
axis which are tangent to both the parabola 
and the line 
. Determine the lengths of the circles' diameters.
by 
and 
for all 
, where 
Show that 
is an integer for all 
.
, is defined in plane polar coordinates 
by the inequality
, where 
and 
.
from 
to 
and 
from 
to 
but in vain.
there exist nonnegative integers 
and integers 
such that 
be a triangle and let 
be a circle in its plane passing through and 
Suppose there exist circles 
such that for 
is externally tangent to 
and passes through 
and 
where 
for all . Prove that 
is determined by its center 
, since we know that it passes through 
. We also know that for all 
are collinear. We must prove that in these conditions, 
.
, the angles 
and 
are equal (where 
is the circumcenter of 
). The result would then follow, since 
would be an involution on the perpendicular bisector 
of 
, mapping a point on this line to its harmonic conjugate wrt and the intersection of the tangents in 
to the circumcircle of . Since 
, we must have , as desired.
be any point on 
, and put 
. Then, 
make equal angles with 
.
.
be a triangle and let 
be a circle in its plane passing through 
and 
Suppose there exist circles 
such that for 
is externally tangent to 
and passes through 
and 
where 
for all 
. Prove that 
and 
are externally tangent at 
, we see that is the midpoint of 
and 
.
and 
and those through and 
, though . . . maybe there is a way to multiply the coordinates by these factors such that it works out?
. Then, measures of the angles 
are easy to get, we find that 
, and we also need 
on the same side of 
(which follows from the fact that we need 
between 
and 
, blah.
by 
.As usual denote the side opposite to vertex 
by 
for 
.
,where is the centre of .
.
.
.Also because 
we note that 
passes through and .We thus obtain 
.
of for each . To this end, let 
, 
and 
.![[asy] size(8cm); defaultpen(fontsize(9pt)); pair A2 = (0,0); pair O1 = (-8,0); pair O2 = (5,0); pair A1 = abs(A2-O1)*dir(285)+O1; pair A3 = abs(A2-O2)*dir(245)+O2; pair O3 = extension(O2, A3, midpoint(A1--A3), midpoint(A1--A3)+dir(A1-A3)*dir(90));
filldraw(circle(O1,abs(A1-O1)), invisible, deepgreen); filldraw(circle(O2,abs(A2-O2)), invisible, deepgreen); filldraw(circle(O3,abs(A3-O3)), invisible, deepgreen); filldraw(A1--A2--A3--cycle, invisible, red);
pair J = (0,-3.2); draw(O1--O2--O3, deepcyan+dotted);
dot("$O_1$", O1, dir(90), deepcyan); dot("$O_2$", O2, dir(90), deepcyan); dot("$O_3$", O3, dir(-90), deepcyan); dot("$A_1$", A1, dir(A1-J), red); dot("$A_2$", A2, dir(A2-J), red); dot("$A_3$", A3, dir(A3-J), red);
label("$\alpha$", A1, 2.4*dir(J-A1), red); label("$\beta$", A2, 2*dir(J-A2), red); label("$\gamma$", A3, 2*dir(J-A3), red); [/asy]](http://latex.artofproblemsolving.com/f/c/4/fc494765ec438d8696fe8408d517d9a258a93ac6.png)
. Then we compute 
and repeating the same calculation another round gives ![\[ \measuredangle O_7 A_2 A_1 = k-(k-\theta) = \theta \]](http://latex.artofproblemsolving.com/0/f/a/0fa7599bb0b3a50b16fe7243944801c9e4a62bf5.png)
with 
. This implies 
, so 
.
be the center of for 
.![\[\measuredangle O_7A_1A_2=-\measuredangle A_2A_1A_3-\measuredangle A_3A_1O_6=-\measuredangle A_2A_1A_3-\measuredangle O_6A_3A_1=-\measuredangle A_2A_1A_3+\measuredangle A_1A_3A_2+\measuredangle A_2A_3O_5=\]](http://latex.artofproblemsolving.com/8/5/f/85f120aa56a3a988f80dddf9a0c33e58f857ef36.png)
![\[-\measuredangle A_2A_1A_3+\measuredangle A_1A_3A_2+\measuredangle O_5A_2A_3=-\measuredangle A_2A_1A_3+\measuredangle A_1A_3A_2-\measuredangle A_3A_2A_1-\measuredangle A_1A_2O_4.\]](http://latex.artofproblemsolving.com/e/6/3/e6375008fcddfadf9dd7fb940c8e2ca8c3c1cb39.png)
A similar angle chase finishes.
be the center of . Also let 
. Observe that is uniquely determined by 
, and is uniquely determined by 
, so if we can show those are equal we're done. Now, let 
. By basic angle chasing (since 
are collinear), we can find:
hence 
as desired. 
with center and ratio 
(it's not important). Define 
and 
. Note that 
is a line for 
because 
pass through . Also note that and 
are parallels for 
because and are tangents in . We want that 
, it's sufficient to prove that 
. Now we will use directed angles. Let 
:
and 
are parallels as desired.
be the center of . Define 
Define 
It suffices to show 
In fact, note that
Note that this process is the exact same for 
, and since function 
is cyclic with order 
, we are done.
, where is the center of . Furthermore, let 
be the angles of the triangle. Then, check that we have the sequence 
Upon applying another sequence of three transformations, this returns the identity 
, so the centers of and coincide, which suffices.
and for 
. Name these 
. Since 
is just 
reflected over the perpendicular bisecter of some side, and because the internal tangent is not dependent on the circle (it is in fact what makes the next circle), we just have to prove that 
.
, 
, and 
, , and 
is less than (otherwise 
does not exist). Thus we call this 
and we know that the angle formed by , , and is 
we can see that this is equal to the angle formed by , , and 
, so we know that the angle formed by , , is 
. Continuing this pattern for subsequent angles we find that the next angles are 
, 
, 
, and . Thus we find that 
and we are done. 
, let be the center of and 
. Then for 
, the tangency condition tells us that 
, and are collinear, so
Therefore, applying this identity in succession, we have
Thus, 
and 
. This is enough to imply that 
since we would otherwise have 
(using non-directed angles), which is impossible since 
. So, we are done.
denote the center of each circle . Let 
and 
denote respectively the perpendicular bisectors of segments 
and 
. Note that for all 
, the center 
of lies on the line 
. First, note that since is a center of a circle passing through points 
and 
, it must lie on 
. Further, since it is externally tangent to , which clearly passes through as well, also lies on 
.
and 
are collinear since then and thus as desired. To see why, simply note that![\[\measuredangle O_1A_1O_4 = \measuredangle O_4A_2O_1 = \measuredangle O_5A_2O_2 = \measuredangle O_2A_3O_5 = \measuredangle O_3A_3O_6 = \measuredangle O_6A_1O_3\]](http://latex.artofproblemsolving.com/7/5/2/7520c96bafde55ccfe30e6ef4b233f2722524045.png)
which implies that and are collinear finishing the proof.
be the angles of the triangle, and 
Then observe we obtain the sequence 
Doing this one more time yields 
the original angle so we are done. QED
and are collinear is simply the point of tangency of 
and 
, and similar result holds throughout the setup.
and intersect it with the perpendicular bisector of 
and so on.
map to 
.
and 
are both tangent and pass through 
and 
.
is a line through 
and similarly, 
passes through 
.
.
passes through both 
passes through 
.
and passes through 
is a line passing through 
is a line parallel to 
passes through 
is a circle tangent to 
is a line through 
.
and 
respectively and let 
and 
respectively. Since 
and 
are tangents to 
.
,
.![\[ \angle PA_2'A_3' = \angle PA_3'A_2'\implies \angle PA_3'A_2' = \angle QA_2'A_3'\implies QA_2'\parallel PA_3'\implies \omega_1'\parallel \omega_6'\implies \omega_1' = \omega_7'\]](http://latex.artofproblemsolving.com/4/b/8/4b89bde7aabab5eee9d9c2d6933e39d17a98f810.png)
and 
,
be 
.
are externally tangent at 
,
.
,
,
,
.
,
is the perpendicular bisector of the next side).
,
bisecting 
is possible)
as the centers of each respective circle. Let 
.
We can apply this transformation twice to get that,![\[\angle{A_{2}A_{1}O_{7}} = \alpha\implies \omega_{1} = \omega_{7}\]](http://latex.artofproblemsolving.com/1/3/a/13ae292ef9ffdc235200134f3e1d5e70660c53c7.png)