Difference between revisions of "2013 AMC 10A Problems/Problem 25"

Line 7: Line 7:
  
 
==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?

$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$

Solution

If you draw a good diagram like the one below, it is easy to see that there are $\textbf{(A)}$, 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

2013 AMC 10A (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS