I don't have full solutionsAlphaTauri wrote:Winneratlife: If you have full solutions, by all means post them. I really want to see how to do 20 and 25.
Only my solutions.So here goes:
You had to recognize that a chord of length r equaled an arc angle of 60 degrees (think about it, if you draw the 2 radii to the end of either chord, you have an equilateral triangle with side length r)
Now, draw one of the chords. The other chord will intersect it if either of it's 2 points touch the arc "cut off" by the first chord. That arc length is [math]\frac{1}6[/math], and since either of the 2 points could be in that arc, our solution is [math]2 * \frac{1}6 = \frac{1}3[/math] or [math]\boxed{D}[/math]For this problem, first you had to understand it. In essence, if we have a square, the point is "n-ray partitional" if it can divide the square into n triangles with identical areas. So, for a point to be 100-ray partitional, you had to be able to draw 100 rays from it that cut the square into 100 parts. Now comes the tricky part.
What you had to realize was that in order for all the parts to be triangles, 4 of the rays had to go to the 4 corners; otherwise you ended up with quadrilaterals in the corners. After drawing those 4 rays, you can think about the opposite pairs of triangles. Each pair had half the area (In a square with side length [math]b[/math] and with the two triangles having heights of [math]h_{1}[/math] and [math]h_{2}[/math], the sum of the triangles' areas is [math]\frac{1}2*h_{1}*b+\frac{1}2*h_{2}*b[/math]. Recognizing that [math]h_{1}+h_{2}=b[/math], we distribute out [math]\frac{1}2*b[/math] and find the sum of the triangles' areas' to be [math]\frac{1}2*b^2[/math], or half the area of the square.
Now, in order for each triangle to contain an integral number of triangles, it must have a height that is a multiple of [math]\frac{1}{50}*b[/math]. This is because the area of the triangle would be [math]\frac{1}2*bh[/math], and it must also be equal to [math]\frac{1}{100}*b^2*n[/math], where [math]n[/math] is some integer (it must hold some integral number of triangles with area 1/100th of the squares). Therefore, there are 49 such heights, and 49 such widths (nothing is special in particular about the height, the same rules apply to width), so there are [math]49^2=2401[/math] such points.
Now we must find the points that are both 60-ray partitional and 100-ray partitional. This happens to be all of the points that are 20-ray partitional, or have triangle heights of [math]\frac{n}{10}[/math] (if you draw it you'll see it). There are 9 such heights, so we must subtract [math]9^2=81[/math] such points from [math]2401[/math], yielding [math]2320[/math] or [math]\boxed{C}[/math]
I'm a freakin' member...
Even though I personally found this year harder, mainly because I blew combatorics that day, and 25 was the most epic jump in problem difficulty I've ever seen.
I only got 6 right on the 12B version yesterday. well, i'm a fish, no need to worry about it.