2013 AMC 10A Problems/Problem 25
Contents
Problem
All 20 diagonals are drawn in a regular octagon. At how many distinct points in the interior of the octagon (not on the boundary) do two or more diagonals intersect?
Solution 1 (Drawing: The REALLY Dumb YET "it still works" Way)
If you draw a good diagram like the one below, it is easy to see that there are 
points.
![[asy] size(14cm); pathpen = white + 1.337; // Initialize octagon pair[] A; for (int i=0; i<8; ++i) { A[i] = dir(45*i); } D(CR( (0,0), 1)); // Draw diagonals // choose pen colors pen[] colors; colors[1] = orange + 1.337; colors[2] = blue; colors[3] = green; colors[4] = black; for (int d=1; d<=4; ++d) { pathpen = colors[d]; for (int j=0; j<8; ++j) { D(A[j]--A[(j+d) % 8]); } } pathpen = blue + 2; // Draw all the intersections pointpen = red + 7; for (int x1=0; x1<8; ++x1) { for (int x2=x1+1; x2<8; ++x2) { for (int x3=x2+1; x3<8; ++x3) { for (int x4=x3+1; x4<8; ++x4) { D(IP(A[x1]--A[x2], A[x3]--A[x4])); D(IP(A[x1]--A[x3], A[x4]--A[x2])); D(IP(A[x1]--A[x4], A[x2]--A[x3])); } } } }[/asy]](http://latex.artofproblemsolving.com/b/4/b/b4b9ebad19769c57ae5bdb090ea24aabbbbc4907.png)
Solution 2 (Elimination)
Let the number of intersections be 
. We know that 
, as every 
points forms a quadrilateral with intersecting diagonals. However, four diagonals intersect in the center, so we need to subtract 
from this count. 
. Note that diagonals like 
, 
, and 
all intersect at the same point. There are 
of this type with three diagonals intersecting at the same point, so we need to subtract 
of the 
(one is kept as the actual intersection). In the end, we obtain 
.
See Also
| 2013 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
