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

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Let the number of intersections be <math>x</math>.  We know that <math>x\le \dbinom{8}{4} = 70</math>, as every <math>4</math> 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 <math>\dbinom{4}{2} -1 = 5</math> from this count,  <math>70-5 = 65</math>.  Note that diagonals like <math>\overline{AD}</math>, <math>\overline{CG}</math>, and <math>\overline{BE}</math> all intersect at the same point.  There are <math>8</math> of this type with three diagonals intersecting at the same point, so we need to subtract <math>2</math> of the <math>\dbinom{3}{2}</math> (one is kept as the actual intersection). In the end, we obtain <math>65 - 16 = \boxed{\textbf{(A) }49}</math>.
 
Let the number of intersections be <math>x</math>.  We know that <math>x\le \dbinom{8}{4} = 70</math>, as every <math>4</math> 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 <math>\dbinom{4}{2} -1 = 5</math> from this count,  <math>70-5 = 65</math>.  Note that diagonals like <math>\overline{AD}</math>, <math>\overline{CG}</math>, and <math>\overline{BE}</math> all intersect at the same point.  There are <math>8</math> of this type with three diagonals intersecting at the same point, so we need to subtract <math>2</math> of the <math>\dbinom{3}{2}</math> (one is kept as the actual intersection). In the end, we obtain <math>65 - 16 = \boxed{\textbf{(A) }49}</math>.
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==Solution 3 (Using the answer choices)==
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We know that the amount of intersection points is at most <math>\dbinom{8}{4} = 70</math>, as in solution <math>2</math>. There's probably going to be more than <math>5</math> (to get <math>(B) 65</math>), leading us to the only reasonable answer, <math>\boxed{\textbf{(A) }49}</math>.
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-Lcz
  
 
==See Also==
 
==See Also==

Revision as of 12:14, 7 January 2020

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 (ProblemsAnswer KeyResources)
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Problem 24
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