Difference between revisions of "2013 AMC 10A Problems/Problem 25"
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==Problem== | ==Problem== | ||
− | All 20 diagonals are drawn in a regular octagon. At how many distinct points in the interior | + | All <math>20</math> 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? | of the octagon (not on the boundary) do two or more diagonals intersect? | ||
<math> \textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128 </math> | <math> \textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128 </math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
− | ==Solution 1 ( | + | ==Solution 1 (Drawing)== |
− | If you draw a | + | If you draw a clear diagram like the one below, it is easy to see that there are <math>\boxed{\textbf{(A) }49}</math> points. |
<asy> | <asy> | ||
size(14cm); | size(14cm); | ||
− | pathpen = | + | pathpen = brown + 1.337; |
// Initialize octagon | // Initialize octagon | ||
pair[] A; | pair[] A; | ||
Line 23: | Line 24: | ||
pen[] colors; | pen[] colors; | ||
colors[1] = orange + 1.337; | colors[1] = orange + 1.337; | ||
− | colors[2] = | + | colors[2] = purple; |
colors[3] = green; | colors[3] = green; | ||
colors[4] = black; | colors[4] = black; | ||
Line 32: | Line 33: | ||
} | } | ||
} | } | ||
− | pathpen = | + | pathpen = blue + 2; |
// Draw all the intersections | // Draw all the intersections | ||
pointpen = red + 7; | pointpen = red + 7; | ||
Line 47: | Line 48: | ||
}</asy> | }</asy> | ||
− | ==Solution 2 ( | + | ==Solution 2 (Working Backwards)== |
− | Let the number of intersections be <math>x</math>. We know that <math>x\le \dbinom{8}{4} = 70</math>, as every 4 | + | 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>. |
+ | |||
+ | ==Solution 3 (Answer choices and reasoning)== | ||
+ | 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> intersections counted multiple times (to get <math>\textbf{(B) }65</math>), leading us to the only reasonable answer, <math>\boxed{\textbf{(A) }49}</math>. | ||
+ | -Lcz | ||
+ | |||
+ | Note: You can easily prove this by looking at the simple case of the diagonals intersecting in the middle of the octagon. <math>4</math> major diagonals intersect here and only <math>1</math> intersection point is counted so you can subtract <math>3</math> from <math>70</math>. Then look to the middle area of the octagon. In this area, if we label the major diagonal as the one where there are <math>3</math> 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 <math>8</math> intersection points of a major diagonal and <math>2</math> 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 <math>59</math>. 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 <math>8</math> points (not including the octagon center). However each adjacent region share one side in common and that side contains <math>2</math> intersection points, so in actuality there are <math>8 - 2 = 6</math> points per region. We multiply this by <math>8</math> to get <math>6\cdot 8 = 48</math> and add the one center point to get <math>48 + 1 = \boxed{\textbf{(A) }49}</math>. | ||
+ | |||
+ | ~skyscraper | ||
+ | |||
+ | ==Solution 5 (Case Work with Drawing)== | ||
+ | <asy> | ||
+ | size(8cm); | ||
+ | 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> | ||
+ | |||
+ | This problem is a counting problem of combinatoric geometry. There are 2 cases for the above diagram: | ||
+ | |||
+ | Case 1: Red Dots | ||
+ | |||
+ | The red dots are the intersection of 3 or more lines. It consists of 8 dots that make up an octagon and 1 dot in the center. Hence, there are <math>9</math> red dots. | ||
+ | |||
+ | Case 2: Blue Dots | ||
+ | |||
+ | The blue dots are the intersection of 2 lines. Each vertex of the octagon has 2 purple lines, 2 green lines, and 1 orange line coming out of it. There are 5 dots of intersection on a purple line, 6 dots on a green line, and 5 dots on an orange line. There are <math>2 \cdot 5+2 \cdot 6+5=27</math> dots that come out of 1 vertex, which includes 7 red dots already counted. So there are <math>27-7 = 20</math> blue dots coming out of 1 vertex. There are 8 vertices, but each blue dot is the intersection of 2 lines, corresponding to <math>2 \cdot 2 = 4</math> vertices. So there are <math>\frac{20 \cdot 8}{4} = 40</math> blue dots. | ||
+ | |||
+ | The number of intersection dots are the sum of the number of red and blue dots. Hence, the answer is <math>40 + 9 = \boxed{\textbf{(A) }49}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://artofproblemsolving.com/videos/amc/2013amc10a/359 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2013|ab=A|num-b=24|after=Last Problem}} | {{AMC10 box|year=2013|ab=A|num-b=24|after=Last Problem}} | ||
+ | {{MAA Notice}} |
Latest revision as of 08:21, 28 December 2021
Contents
Problem
All 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?
Solution 1 (Drawing)
If you draw a clear diagram like the one below, it is easy to see that there are points.
Solution 2 (Working Backwards)
Let the number of intersections be . We know that , as every 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 from this count, . Note that diagonals like , , and all intersect at the same point. There are of this type with three diagonals intersecting at the same point, so we need to subtract of the (one is kept as the actual intersection). In the end, we obtain .
Solution 3 (Answer choices and reasoning)
We know that the amount of intersection points is at most , as in solution . There's probably going to be more than intersections counted multiple times (to get ), leading us to the only reasonable answer, . -Lcz
Note: You can easily prove this by looking at the simple case of the diagonals intersecting in the middle of the octagon. major diagonals intersect here and only intersection point is counted so you can subtract from . Then look to the middle area of the octagon. In this area, if we label the major diagonal as the one where there are 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 intersection points of a major diagonal and 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 . 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 points (not including the octagon center). However each adjacent region share one side in common and that side contains intersection points, so in actuality there are points per region. We multiply this by to get and add the one center point to get .
~skyscraper
Solution 5 (Case Work with Drawing)
This problem is a counting problem of combinatoric geometry. There are 2 cases for the above diagram:
Case 1: Red Dots
The red dots are the intersection of 3 or more lines. It consists of 8 dots that make up an octagon and 1 dot in the center. Hence, there are red dots.
Case 2: Blue Dots
The blue dots are the intersection of 2 lines. Each vertex of the octagon has 2 purple lines, 2 green lines, and 1 orange line coming out of it. There are 5 dots of intersection on a purple line, 6 dots on a green line, and 5 dots on an orange line. There are dots that come out of 1 vertex, which includes 7 red dots already counted. So there are blue dots coming out of 1 vertex. There are 8 vertices, but each blue dot is the intersection of 2 lines, corresponding to vertices. So there are blue dots.
The number of intersection dots are the sum of the number of red and blue dots. Hence, the answer is .
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2013amc10a/359
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.