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? | ||
Line 7: | Line 7: | ||
[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
− | ==Solution 1 (Drawing | + | ==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 24: | 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 48: | 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 <math>4</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>. |
+ | |||
+ | ==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 | ||
+ | |||
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://artofproblemsolving.com/videos/amc/2013amc10a/359 | ||
==See Also== | ==See Also== |
Latest revision as of 13:35, 30 December 2020
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
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 |
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