Difference between revisions of "2013 AMC 10A Problems/Problem 25"
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Let the number of intersections be <math>x</math>. We know that <math>x\le \dbinom{8}{4} = 70</math>, as every <math>4</math> vertices on the octagon forms a quadrilateral with intersecting diagonals which is an intersection point. However, four diagonals intersect in the center, so we need to subtract <math>\dbinom{4}{2} -1 = 5</math> from this count, <math>70-5 = 65</math>. Note that diagonals like <math>\overline{AD}</math>, <math>\overline{CG}</math>, and <math>\overline{BE}</math> all intersect at the same point. There are <math>8</math> of this type with three diagonals intersecting at the same point, so we need to subtract <math>2</math> of the <math>\dbinom{3}{2}</math> (one is kept as the actual intersection). In the end, we obtain <math>65 - 16 = \boxed{\textbf{(A) }49}</math>. | Let the number of intersections be <math>x</math>. We know that <math>x\le \dbinom{8}{4} = 70</math>, as every <math>4</math> vertices on the octagon forms a quadrilateral with intersecting diagonals which is an intersection point. However, four diagonals intersect in the center, so we need to subtract <math>\dbinom{4}{2} -1 = 5</math> from this count, <math>70-5 = 65</math>. Note that diagonals like <math>\overline{AD}</math>, <math>\overline{CG}</math>, and <math>\overline{BE}</math> all intersect at the same point. There are <math>8</math> of this type with three diagonals intersecting at the same point, so we need to subtract <math>2</math> of the <math>\dbinom{3}{2}</math> (one is kept as the actual intersection). In the end, we obtain <math>65 - 16 = \boxed{\textbf{(A) }49}</math>. | ||
Revision as of 19:00, 22 December 2019
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 clear diagram like the one below, it is easy to see that there are 
points.
![[asy] size(14cm); pathpen = brown + 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] = purple; 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/3/a/8/3a8a2f2837978820f77e3bbaf68d676be0114ff7.png)
Solution 2 (Working Backwards)
Let the number of intersections be 
. We know that 
, as every 
vertices on the octagon forms a quadrilateral with intersecting diagonals which is an intersection point. 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. 
