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

## 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 clear diagram like the one below, it is easy to see that there are $\boxed{\textbf{(A) }49}$ points.

$[asy] size(14cm); pathpen = brown + 1.337; // Initialize octagon pair[] A; for (int i=0; i<8; ++i) { A[i] = dir(45*i); } D(CR( (0,0), 1)); // Draw diagonals // choose pen colors pen[] colors; colors[1] = orange + 1.337; colors[2] = purple; colors[3] = green; colors[4] = black; for (int d=1; d<=4; ++d) { pathpen = colors[d]; for (int j=0; j<8; ++j) { D(A[j]--A[(j+d) % 8]); } } pathpen = blue + 2; // Draw all the intersections pointpen = red + 7; for (int x1=0; x1<8; ++x1) { for (int x2=x1+1; x2<8; ++x2) { for (int x3=x2+1; x3<8; ++x3) { for (int x4=x3+1; x4<8; ++x4) { D(IP(A[x1]--A[x2], A[x3]--A[x4])); D(IP(A[x1]--A[x3], A[x4]--A[x2])); D(IP(A[x1]--A[x4], A[x2]--A[x3])); } } } }[/asy]$

## Solution 2 (Working Backwards)

Let the number of intersections be $x$. We know that $x\le \dbinom{8}{4} = 70$, as every $4$ 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 $\dbinom{4}{2} -1 = 5$ from this count, $70-5 = 65$. Note that diagonals like $\overline{AD}$, $\overline{CG}$, and $\overline{BE}$ 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 $65 - 16 = \boxed{\textbf{(A) }49}$.

## Solution 3 (Using the answer choices)

We know that the amount of intersection points is at most $\dbinom{8}{4} = 70$, as in solution $2$. There's probably going to be more than $5$ (to get $(B) 65$), leading us to the only reasonable answer, $\boxed{\textbf{(A) }49}$. -Lcz

## See Also

 2013 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast 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

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