Difference between revisions of "2013 AMC 10A Problems/Problem 25"
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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 22:53, 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.
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 |