be the real solution of the system of equations 
,
,
.
can be written in the form 
,
and 
are relatively prime positive integers. Find 
.
cube is given, and a 
grid of square unit cells is drawn on each of its six faces. A beam is a 
rectangular prism. Several beams are placed inside the cube subject to the following conditions:
faces of each beam coincide with unit cells lying on opposite faces of the cube. (Hence, there are 
possible positions for a beam.)
faces of each beam touch either a face of the cube or the interior of the face of another beam.
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
and 
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
