[/quote]
,
be a triangle such that![\[ \left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 = \left( \dfrac{6s}{7r} \right)^2, \]](http://latex.artofproblemsolving.com/6/9/b/69b1964be5f92eb44a36b0b8604bf473fe27e210.png)
and 
denote its semiperimeter and its inradius, respectively. Prove that triangle 
whose side lengths are all positive integers with no common divisors and determine these integers.
.
and 
intersect 
and 
.
be the intersection point of these angle bisectors. Let 
and 
be the respective reflections of 
and 
.
.
,
of 
cards each from this deck. We would like to define a rule that declares one of them a winner. This rule should satisfy the following conditions:
cards: if the cards are laid down in increasing order face down and we are told which card belongs to which player, but not what numbers are written on them, we can still decide the winner.
and 
,
for all 
,
from the deck, 
then 
and 
for 
.
is increasing.
,
,
of 
such that ![\[\left|\sqrt{a_k}+\sqrt{b_k}+\sqrt{c_k}-2\sqrt{N}\right|<2023\]](http://latex.artofproblemsolving.com/9/8/0/98097fb721722b9103ef65d72322956877e262cb.png)
for every 
.
.
,
has at most 
be the sum of the digits of 
such that there are infinite positive integers 
.
be three diameters of the circumcircle of an acute triangle 
be an arbitrary point in the interior of 
,
be the orthogonal projection of 
,
be the point such that 
is the midpoint of 
,
be the point such that 
is the midpoint of 
,
be the point such that 
is the midpoint of 
.
is similar to triangle 
such that if 
and 
are relatively prime divisors of 
,
divides 
is an infinite sequence of integers satisfying the following two conditions:
divides 
for 
such that 
for all 
such that 
for all 
with complex coefficients for which all complex roots of the polynomials 
have absolute value 1.
Show that there exists a constant 
such that for all positive integer 
![\[\sum_{k\le n^{\varepsilon}}(n\text{ mod } k)>cn^{2\varepsilon}.\]](http://latex.artofproblemsolving.com/f/e/9/fe933e886e28c2ecf1fa880d0e650ac7a2998aaa.png)
Proposed by Cheng Jiang
such that 
be an isosceles trapezoid with 
.
and 
and 
,
is tangent to 
Y by cubres
Given a circle with chord AB. k and l are tangents to the circle at points A and B. C and E are in different half-planes with respect to AB and lie on k, and F and D are in different half-planes with respect to AB and lie on l. Furthermore, C and F are in the same half-plane with respect to AB and AC = BD; AE = BF. CD intersects the circle at P and R and EF intersects the circle at Q and S. P and Q are in the same half-plane with respect to AB and in different half-plane with R and S. Prove that PQRS is a parallelogram if and only if AB, CD, and EF intersect at one point.
This post has been edited 3 times. Last edited by iv999xyz, Apr 12, 2025, 10:26 AM
Reason: error
Reason: error
and 
are tangents to the circle at points 
a
a
a
a
a
a
.
i
a
a
i
an
.
i
a
