2013 AMC 10A Problems/Problem 25

Revision as of 13:29, 30 December 2020 by Asdff1234 (talk | contribs) (Solution 3 (Answer choices and reasoning))


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 (Answer choices and reasoning)

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 $5$ intersections counted multiple times (to get $\textbf{(B) }65$), leading us to the only reasonable answer, $\boxed{\textbf{(A) }49}$. -Lcz Note: You can easily prove this by looking at the simple case of the diagonals intersecting in the middle of the octagon. 4$"major" diagonals intersect here and only 1$ intersection point is counted so you can subtract 3$from 70.  Then look to the "middle" ring where certainly more than 2$ intersections are formed by the intersection of 3$ diagonals(a rough diagram is really enough).

Solution 4 (Drawing but easier)

Like solution one, we may draw. Except note that the octagon has eight regions, and each region has an equal number of points, so drawing only one of the eight regions and the intersection points suffices. One of the eight regions contains $8$ points (not including the octagon center). However each adjacent region share one side in common and that side contains $2$ intersection points, so in actuality there are $8 - 2 = 6$ points per region. We multiply this by $8$ to get $6\cdot 8 = 48$ and add the one center point to get $48 + 1 = \boxed{\textbf{(A) }49}$.


Video Solution by Richard Rusczyk


See Also

2013 AMC 10A (ProblemsAnswer KeyResources)
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