AoPS Academy
: 
such that:
for all x,y E 
, 
, 
, 
be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, 
. Prove that 
for all 
.
with rational coordinates lie on a plane. It turned out that the distance between every pair of points is an integer. Prove that there exist points 
with integer coordinates such that 
for every pair 
be a trapezoid inscribed in circle 
, 
. Let 
is the symmetric point of 
across 
, 
intersects at 
. Let 
is midpoint of and 
is orthocenter of triangle 
. On 
take a point 
so that 
. Prove that: 
coincide
Product of all even divisors of 
If 
dice are rolled, Then number of ways of getting sum at least 
is
foot radius. Bruno completes 
laps every hour, while Brutus completes 
laps every hour. If they start at the same point but run in opposite directions, how far along the track’s circumference (in feet) from the starting point are they when they meet for the sixth time? Note: Do not count the moment they start running as a meeting point.
, 
, 
, 
such that ![\[\left(a_1+\frac1{a_1}\right)\left(a_2+\frac1{a_2}\right)\cdots\left(a_n+\frac1{a_n}\right)=2020.\]](http://latex.artofproblemsolving.com/3/6/2/362fccfa6353ffcbc4d29f24ccbfce73269080bf.png)
Compute 
. be an isosceles triangle such that 
. Let be a dot on the 
side.
at point intersects the circumcircle of 
at 
. Prove that CD
AB
have angles 
and 
such that 
and 
. Moreover, suppose that the product of the side lengths of the triangle is equal to its area. Let 
denote the circumcircle of . Let 
intersect 
at , 
intersect 
at 
, and 
intersect 
at . If the area of 
can be written as 
for relatively prime integers 
and 
and squarefree 
, find the sum of all prime factors of , counting multiplicities (so the sum of prime factors of 
is 
), given that has 
divisors.
.
such that 
is a positive integral power of a prime positive integer."
be mid-points of 
respectively 
and since, 
and 
are isosceles 
is a kite,
and we also know, 
is isosceles
is altitude of 
,such, 
.........Let 
Therefore, 
, 
, 
, 
. Thus Midpoint of 
and midpoint of 
. Using distance formula we get 
.
such that 
are collinear. Let be the midpoint of 
, thus it is also the midpoint of 
. Now 
and 
. By pythagorean theorem, 
. And using MPT, the required answer is .
be the altitude through 
.
and 
be the midpoint of 
)
and 
,
and 
.
,
.
using 
and 
.
,
and 
on 
,
.
by the Midpoint Theorem.
,
.
is a right triangle.