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

(Solution 2 (elimination))
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[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
  
==Solution 1 (drawing)==
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==Solution 1 (Drawing)==
  
If you draw a good diagram like the one below, it is easy to see that there are <math>\boxed{\textbf{(A) }49}</math>, points.
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If you draw a good diagram like the one below, it is easy to see that there are <math>\boxed{\textbf{(A) }49}</math> points.
  
 
<asy>
 
<asy>

Revision as of 14:56, 27 January 2015

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 1 (Drawing)

If you draw a good diagram like the one below, it is easy to see that there are $\boxed{\textbf{(A) }49}$ 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 = 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]

Solution 2 (elimination)

Let the number of intersections be $x$. We know that $x\le \dbinom{8}{4} = 70$, as every 4 points forms a quadrilateral with intersecting diagonals. However, four diagonals intersect in the center, so we need to subtract $\dbinom{4}{2} -1 = 5$ from this count. $70-5 = 65$. Note that diagonals like AD, CG, and BE all intersect at the same point. There are $8$ of this type with three diagonals intersecting at the same point, so we need to subtract $2$ of the $\dbinom{3}{2}$ (one is kept as the actual intersection). In the end, we obtain $65 - 16 = \boxed{\textbf{(A) }49}$

See Also

2013 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
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All AMC 10 Problems and Solutions

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