[/quote]
,
,
.
points and some chords between them is given.
In every step we can take two chords 
with a common point other than 
and erase exactly one of 
let 
be the minimum of chords after some steps. Find the maximum of 
In every step we can take two chords 
be a positive integer and let 
and 
be positive integers congruent to 1 modulo 4. Prove that there exists a positive integer 
such that at least one of the numbers 
and 
is divisible by 
be the lengths of the sides of a triangle. Prove that![\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]](http://latex.artofproblemsolving.com/7/4/b/74b7aa5621bc52977d9acd448d74293dc27a8e00.png)
is a positive integer. Prove that
and 
be arbitrary points on the sides 
and 
of triangle 
,
meets for the second time the circumcircle of 
at point 
.
meets line 
at point 
,
meets line 
at point 
.
and 
pass through the same point.
.
and 
intersect the altitude from 
to 
and 
respectively. 
is the circumcenter of triangle 
and 
is the reflection of 
is the second intersection of circumcircles of triangles 
and 
.
are collinear.
and 
Find the value of 
and 
such that 
and 
is a prime number
such that 
is the cube of a positive integer. Find 
is greater than 0 if 
.
and since 
?
Y by
Rodrigo has a very large sheet of graph paper. First he draws a line segment connecting point 
to point 
and colors the 
cells whose interiors intersect the segment, as shown below. Next Rodrigo draws a line segment connecting point 
to point 
. How many cells will he color this time?
![[asy]
filldraw((0,4)--(1,4)--(1,3)--(0,3)--cycle, gray(.75), gray(.5)+linewidth(1));
filldraw((0,3)--(1,3)--(1,2)--(0,2)--cycle, gray(.75), gray(.5)+linewidth(1));
filldraw((1,2)--(2,2)--(2,1)--(1,1)--cycle, gray(.75), gray(.5)+linewidth(1));
filldraw((1,1)--(2,1)--(2,0)--(1,0)--cycle, gray(.75), gray(.5)+linewidth(1));
draw((-1,5)--(-1,-1),gray(.9));
draw((0,5)--(0,-1),gray(.9));
draw((1,5)--(1,-1),gray(.9));
draw((2,5)--(2,-1),gray(.9));
draw((3,5)--(3,-1),gray(.9));
draw((4,5)--(4,-1),gray(.9));
draw((5,5)--(5,-1),gray(.9));
draw((-1,5)--(5, 5),gray(.9));
draw((-1,4)--(5,4),gray(.9));
draw((-1,3)--(5,3),gray(.9));
draw((-1,2)--(5,2),gray(.9));
draw((-1,1)--(5,1),gray(.9));
draw((-1,0)--(5,0),gray(.9));
draw((-1,-1)--(5,-1),gray(.9));
dot((0,4));
label("$(0,4)$",(0,4),NW);
dot((2,0));
label("$(2,0)$",(2,0),SE);
draw((0,4)--(2,0));
draw((-1,0) -- (5,0), arrow=Arrow);
draw((0,-1) -- (0,5), arrow=Arrow);
[/asy]](//latex.artofproblemsolving.com/b/5/6/b56819d6759e8e4881a403198207737a774512e9.png)
to point 
and colors the 
cells whose interiors intersect the segment, as shown below. Next Rodrigo draws a line segment connecting point 
to point 
. How many cells will he color this time?![[asy]
filldraw((0,4)--(1,4)--(1,3)--(0,3)--cycle, gray(.75), gray(.5)+linewidth(1));
filldraw((0,3)--(1,3)--(1,2)--(0,2)--cycle, gray(.75), gray(.5)+linewidth(1));
filldraw((1,2)--(2,2)--(2,1)--(1,1)--cycle, gray(.75), gray(.5)+linewidth(1));
filldraw((1,1)--(2,1)--(2,0)--(1,0)--cycle, gray(.75), gray(.5)+linewidth(1));
draw((-1,5)--(-1,-1),gray(.9));
draw((0,5)--(0,-1),gray(.9));
draw((1,5)--(1,-1),gray(.9));
draw((2,5)--(2,-1),gray(.9));
draw((3,5)--(3,-1),gray(.9));
draw((4,5)--(4,-1),gray(.9));
draw((5,5)--(5,-1),gray(.9));
draw((-1,5)--(5, 5),gray(.9));
draw((-1,4)--(5,4),gray(.9));
draw((-1,3)--(5,3),gray(.9));
draw((-1,2)--(5,2),gray(.9));
draw((-1,1)--(5,1),gray(.9));
draw((-1,0)--(5,0),gray(.9));
draw((-1,-1)--(5,-1),gray(.9));
dot((0,4));
label("$(0,4)$",(0,4),NW);
dot((2,0));
label("$(2,0)$",(2,0),SE);
draw((0,4)--(2,0));
draw((-1,0) -- (5,0), arrow=Arrow);
draw((0,-1) -- (0,5), arrow=Arrow);
[/asy]](http://latex.artofproblemsolving.com/b/5/6/b56819d6759e8e4881a403198207737a774512e9.png)
This post has been edited 1 time. Last edited by fruitmonster97, Oct 11, 2024, 4:43 PM
be relatively prime positive integers. When a 
rectangle is split up into 
unit squares, exactly 
unit squares are crossed by the diagonal of this rectangle.
The diagonal has endpoints at 
,
Now, suppose the diagonal goes through the corner point 
,
and 
.
,
implying that 
can be reduced, contradicting the fact that 
horizontal lines and 
vertical lines, so there are 
total lines crossed by the diagonal. This doesn't include the square in the bottom left corner, crossed initially. Therefore, there are 
unit squares crossed by the diagonal and our claim is proven.
unit squares are crossed by the diagonal of this rectangle.
,
,
and 
,
and 
.
rectangles, strung together at the diagonal. An example for 
is shown below, where two 
rectangles are strung together:![[asy]
unitsize(1cm);
draw((0,0)--(6,4),linewidth(1));
currentpen = linewidth(.5);
for (real i = 0; i <= 6; ++i) {
draw((i, 0)--(i, 4));
}
for (real i = 0; i < 5; ++i) {
draw((0, i)--(6, i));
}
currentpen = linewidth(1.5);
for (real i = 0; i <= 3; ++i) {
draw((i, 0)--(i, 2));
}
for (real i = 0; i <= 2; ++i) {
draw((0, i)--(3, i));
}
for (real i = 3; i <= 6; ++i) {
draw((i, 2)--(i, 4));
}
for (real i = 3; i <= 5; ++i) {
draw((3, i-1)--(6, i-1));
}
[/asy]](http://latex.artofproblemsolving.com/a/5/7/a57601f184ed9bd3706a9f27f99680a76aa4278c.png)
diagonals crossed in each rectangle. There are 
rectangles, so the number of crossed diagonals in total is
,
to the point 
when 
are relatively prime positive integers intersects 
unit squares. Translating, the number of unit squares intersected from 
to 
is the same as the number of unit squares intersected from 
.
times the number of squares intersected from 
,
.
into the equation 
to see what we get.
which is an integer, so we can stop here.
squares.
then multiply by 7/12 by intuition which is C
