2013 AMC 10A Problems/Problem 25
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)
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 = yellow + 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/0/1/c/01c199264a970271ceb63ff73b9ba5cec2a4378b.png)
Solution 2 (elimination)
Let the number of intersections be 
. We know that 
, as every 4 points forms a quadrilateral with intersecting diagonals. However, four diagonals intersect in the center, so we need to subtract 
from this count. 
. One might be tempted to choose 65 at this point, but one then sees that diagonals like AD, CG, and BE all intersect at the same point. There are 
of this type with three diagonals intersecting at the same point. We need to subtract 
of the 
(one is kept as the actual intersection), so we get 
See Also
| 2013 AMC 10A (Problems • Answer Key • Resources) | ||
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