2013 AMC 10A Problems/Problem 25: Difference between revisions
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==Solution== | ==Solution== | ||
If you draw a good diagram like the one below, it is easy to see that there are <math>\textbf{(A)}</math>, 49 points. | |||
<asy> | |||
size(14cm); | |||
pathpen = pink + 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] = magenta + 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> | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2013|ab=A|num-b=24|after=Last Problem}} | {{AMC10 box|year=2013|ab=A|num-b=24|after=Last Problem}} | ||
Revision as of 21:53, 7 February 2013
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
If you draw a good diagram like the one below, it is easy to see that there are 
, 49 points.
![[asy] size(14cm); pathpen = pink + 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] = magenta + 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/3/9/9/3994c52a4c47bb4b1bda13bae3d7836f035230fe.png)
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 | ||
