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.
be a fixed integer. Find the least constant 
such the inequality![\[\sum_{i<j} x_{i}x_{j} \left(x^{2}_{i}+x^{2}_{j} \right) \leq C
\left(\sum_{i}x_{i} \right)^4\]](http://latex.artofproblemsolving.com/d/d/e/dde2d1dd0c412d8ba640f1d4c044e2fa2dcccc82.png)
(the sum on the left consists of 
summands). For this constant , characterize the instances of equality.
, let 
be the least prime that does not divide 
, and define 
to be the product of all primes less than . In particular, 
For having 
, define 
. Consider the sequence 
defined by 
and ![\[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \]](http://latex.artofproblemsolving.com/8/2/5/825be3bbef6a29de69aead03392ca7215093f320.png)
for 
. Find all 
such that 
.
such that 
. Find the minimum value of:
be a cyclic quadrilateral. Let 
be the midpoint of the arc 
of its circumcircle which does not contain or 
. Let the lines 
and 
meet at 
and the lines 
and 
meet at 
. Prove that the lines 
and are parallel.
be an acute triangle with circumcenter 
and orthocenter 
. Let 
denote the circumcircle of triangle , and 
the midpoint of 
. The tangents to at 
and , and the line through perpendicular to line 
, determine a triangle whose circumcircle we denote by 
. Define 
and 
similarly., and are concurrent on line .
such that for all positive integers 
and ,![\[f(m+nf(m))=f(n)^m+2024! \cdot m.\]](http://latex.artofproblemsolving.com/8/b/9/8b90a9326c0fa448a5a574809c184f43513f768f.png)
Jaedon Whyte
, and an odd prime 
not dividing any of them, and an integer 
such that if one considers the infinite sequence 
and looks at the highest power of that divides each of them, these powers are not all zero, and are all at most . Prove that there exists some 
(which may depend on 
) such that whenever divides an element of this sequence, the maximum power of that divides that element is exactly 
.
, 
, 
be an infinite sequence of positive integers. Suppose that there is an integer 
such that, for each 
, the number
is an integer. Prove that there is a positive integer such that 
for all 
.
be a right triangle with 
. Point lies on the line 
such that is between and . Let 
be the midpoint of 
and let be the seconf intersection point of the circumcircle of 
and the circumcircle of 
. Prove that as varies, the line 
passes through a fixed point.
and 
intersect each other at 
and 
. The common tangent to two circles nearer to touch and at 
and 
respectively. Let 
and 
be the reflection of and respectively with respect to . The circumcircle of the triangle 
intersect circles and respectively at points 
and 
(both distinct from ). Show that the line 
is the second tangent to and .
to 
face-down on a table. On each move, the Detective can point to two cards and inquire if the numbers on them are consecutive. The Magician replies truthfully. After a finite number of moves, the Detective points to two cards. She wins if the numbers on these two cards are consecutive, and loses otherwise.
questions.
= 2∠AMP.
is the midpoint of the arc 
, containing 
, in the circumcircle of 
, and let 
be the -excircle's center. Assume that the external angle bisector of 
intersects at 
. Prove that 
is perpendicular to 
, where 
is the incenter of .
with 
and the circle 
of radius centered at 
Let be the midpoint of 
The line 
intersects a second time at 
Let be a point on such that 
Let be the intersection of 
and Prove 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
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 
.
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)