AoPS Academy
of the numbers 
such that
is divisible by 
for all 
.
be an acute-angled triangle with circumcenter 
. Let 
be a circle with center on the altitude from 
in , passing through vertex and points 
and 
on sides 
and 
. Assume that![\[BP\cdot CQ = AP\cdot AQ.\]](http://latex.artofproblemsolving.com/1/c/2/1c22443c125d448bf3f48d2729ae4f2eb375a405.png)
Prove that is tangent to the circumcircle of triangle 
.
is a right-angled triangle and the sides of the polygon (dark region) are parallel or coincide with one of the legs of the triangle. Calculate 
so that the area of the polygon is equal to the area of the triangle.
was bent so that vertex 
was taken in the point 
on side 
. The fold is 
, with 
over and 
over 
. It is known that 
, 
and 
.
.. and 
have sides 
cm and 
cm respectively. Calculate of the side of the square 
.
and 
marked in the figure below? be a right triangle at . Consider M and 
points on the hypotenuse 
such that 
. The points 
and 
are such that 
and 
. Calculate the area of quadrilateral 
, knowing that triangle has an area of 
cm
. below, is the intersection of the angle bisectors and H is the intersection of the altitudes. it is known that 
. Calculate .
. is right-angled at . Let be the center of the inscribed circle in and be the midpoint of the side . If 
, what is the angle 
?
meter to the East, 
meters to the North, to the West, 
to the South, 
to the East, 
to the North,..., 
meters to the North,
for the West and 
for the South. Calculate, in meters, the nearest integer value distance between the traveler's start and end positions., let 
be a point on the side such that 
, 
and 
. Knowing that the circle circumscribed to triangle 
has a radius equal to that of the circle circumscribed to triangle 
, calculate, the area of triangle . be an acute triangle with 
. Let be the foot of the altitude relative to the side and the midpoint of side . If 
, calculate, the area of triangle . and are squares of side 
cm. Knowing that is the midpoint of and 
is the midpoint of 
, calculate the highlighted area in cm. be the point on the side of the triangle such that 
and 
the point on the side such that 
is perpendicular to . Knowing that 
cm and that the centroid of the triangle belongs to the segment , calculate 
.
grid, one can find 
points such that no four of them are vertices of a square with sides parallel to lines of the grid. Imagine yourself as Erdos (!) and guess what is the best exponent instead of 
!
grid, one can find points such that no four of them are vertices of a square with sides parallel to lines of the grid. Imagine yourself as Erdos (!) and guess what is the best exponent instead of !
or 
points per row, but asymptotically this is irrelevant; I'll assume there are points per row.
points from the grid uniformly and at random, where 
is a positive constant to be determined later.![\[\frac{\frac{(n - 1)n(2n - 1)}{6}}{\frac{(n^2)(n^2 - 1)(n^2 - 2)(n^2 - 3)}{24}} \le Cn^{-5}\]](http://latex.artofproblemsolving.com/6/d/f/6df6f010753baa49ea21e442264eb27576338675.png)
for some positive constant 
points, the expected number of squares with sides parallel to the grid is ![\[Cn^{-5}\binom{cn^{\frac{5}{3}}}{4} \le \frac{c^4Cn^{\frac{5}{3}}}{24} \]](http://latex.artofproblemsolving.com/e/c/1/ec1c90ecc98e5bd78395dd10da8b0f580f6da558.png)
In particular, there exists a configuration in which there are at most this many squares. Removing all vertices of every such square, we are left with ![\[cn^{\frac{5}{3}} - \frac{c^4Cn^{\frac{5}{3}}}{6} = (c - \frac{c^4C}{6})n^{\frac{5}{3}}\]](http://latex.artofproblemsolving.com/9/5/6/956e9b69f4fe41d6401e1761fc668df50d100001.png)
points. For some positive sufficiently small, the constant multiplier is positive, and we are done.
is the best lower bound.
be a real number to be decided later. Choose each of the points randomly and independently with probability 
. After doing so, remove one vertex of each square present.
as originally we would expect there to be 
points and each square(of which there are 
would be present with probability 
) so LOE gives the stated value. Now its not so hard to see that if we choose 
, then the expected value gives us our construction.
