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

Revision as of 10:41, 26 September 2021

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 (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 area of the octagon. In this area, if we label the major diagonal as the one where there are $3$ points between the two points forming the diagonal, and the semi-minor diagonal the diagonal where there is one less point between the two diagonal forming points, there are $8$ intersection points of a major diagonal and $2$ semi-minor diagonals. This means that these eight points would be, not double-counted -which the calculation by Lcz accounts for- but triple-counted. Thus, taking away one for each of these points is more than enough to see the value of the answer has to be less than or equal to $59$ . Choice A is the only answer that works.

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 $48 + 1 = \boxed{\textbf{(A) }49}$ .

~skyscraper

Solution 5 (Case work)

$[asy] size(10cm); pathpen = black; // draw the circle pair[] A; for (int i=0; i<8; ++i) { A[i] = dir(45*i); } D(CR((0,0), 1)); // draw the octagon and diagonals // choose pen colors pen[] colors; colors[1] = yellow; colors[2] = purple; colors[3] = green; colors[4] = orange; for (int d=1; d<=4; ++d) { pathpen = colors[d]; for (int j=0; j<8; ++j) { D(A[j]--A[(j+d) % 8]); } } // draw the 3 or more line intersections pointpen = red + 5; D(IP(A[0]--A[4], A[2]--A[6])); // center of the circle for (int x1=0; x1<8; ++x1) { for (int x2=0; x2<8; ++x2) { int y1 = (x1 + 4)%8; int y2 = (x2 + 3)%8; if (x1 != x2 && y1 != y2 && x1 != y1 && x1 != y2 && x2 != y1 && x2 != y2) D(IP(A[x1]--A[y1], A[x2]--A[y2])); } } // draw the 2 line intersections pointpen = blue + 4; for (int x1 = 0; x1 < 8; ++x1) { int x2 = (x1 + 1)%8; D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+2)%8])); D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+3)%8])); D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+4)%8])); D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+5)%8])); D(IP(A[x1]--A[(x1+3)%8], A[x2]--A[(x2+5)%8])); } [/asy]$

There are 2 cases for the diagram above.

Case 1: Red Dots The red dots are the intersection of 3 or more lines. It is consisted of 8 dots that make up an octagon, and 1 dot in the center. Hence, there are $9$ dots.

Case 2: Blue Dots

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2013amc10a/359

@@ Line 111: / Line 111: @@
 There are 2 cases for the diagram above.
 Case 1: Red Dots
-The red dots are the intersection of 3 or more lines. It is consisted of 8 dots that make up an octagon, and 1 dot in the middle. Hence, there are <math>9</math> dots.
+The red dots are the intersection of 3 or more lines. It is consisted of 8 dots that make up an octagon, and 1 dot in the center. Hence, there are <math>9</math> dots.
+Case 2: Blue Dots
 ==Video Solution by Richard Rusczyk==

2013 AMC 10A (Problems • Answer Key • Resources)
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