Difference between revisions of "2020 AMC 10B Problems/Problem 17"
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<math>\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15</math> | <math>\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | Consider the 10 people to be standing in a circle, where two people opposite each other form a diameter of the circle. | + | Consider the <math>10</math> people to be standing in a circle, where two people opposite each other form a diameter of the circle. |
Let us use casework on the number of pairs that form a diameter of the circle. | Let us use casework on the number of pairs that form a diameter of the circle. | ||
| Line 24: | Line 24: | ||
Case 3: <math>3</math> diameters | Case 3: <math>3</math> diameters | ||
| − | There are <math>5</math> possible diameters to draw. | + | There are <math>5</math> possible sets of <math>3</math> diameters to draw. |
| + | Notice we are technically choosing the number of ways to choose a pair of two diameters that are neighbors to each other. This means we can choose the first diameter in the pair, and have only two diameters to choose from for the second in the pair. This means we have <math>5*2=10</math> possibilities for choosing 5 neighboring diameters. However, notice that there are duplicates, so we divide the <math>10</math> possibilities by <math>2</math> to get <math>5</math>. | ||
Note that there cannot be a case with <math>4</math> diameters because then there would have to be <math>5</math> diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises. | Note that there cannot be a case with <math>4</math> diameters because then there would have to be <math>5</math> diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises. | ||
| Line 32: | Line 33: | ||
There is only <math>1</math> way to do this. | There is only <math>1</math> way to do this. | ||
| − | Thus, in total there are <math>2+5+5+1=\boxed{(C)13}</math> possible ways. | + | Thus, in total there are <math>2+5+5+1=\boxed{\textbf{(C) }13}</math> possible ways. |
| + | ~Pearl2008 (Minor Edits) | ||
| − | ==Video Solution== | + | ==Video Solution (HOW TO THINK CRITICALLY!!!)== |
| − | https://youtu.be/ | + | https://youtu.be/t1WQTZ8TF7k |
| − | |||
| − | ~ | + | ~Education, the Study of Everything |
| − | == Video Solution == | + | ==Video Solution by TheBeautyOfMath== |
| − | https://youtu.be/ | + | https://youtu.be/3BvJeZU3T-M?t=419 |
| − | + | https://youtu.be/0xgTR3UEqbQ | |
| − | ==Solution | + | == Video Solution by Sohil Rathi == |
| − | + | https://youtu.be/0W3VmFp55cM?t=2796 | |
| − | |||
==See Also== | ==See Also== | ||
Latest revision as of 15:41, 14 October 2024
- The following problem is from both the 2020 AMC 10B #17 and 2020 AMC 12B #15, so both problems redirect to this page.
Contents
Problem
There are 
people standing equally spaced around a circle. Each person knows exactly 
of the other 
people: the 
people standing next to her or him, as well as the person directly across the circle. How many ways are there for the people to split up into 
pairs so that the members of each pair know each other?
Solution 1
Consider the people to be standing in a circle, where two people opposite each other form a diameter of the circle.
Let us use casework on the number of pairs that form a diameter of the circle.
Case 1: 
diameters
There are ways: either 
pairs with , pairs with 
, and so on or pairs with , pairs with , etc.
Case 2: diameter
There are possible diameters to draw (everyone else pairs with the person next to them).
Note that there cannot be diameters since there would be one person on either side that will not have a pair adjacent to them. The only scenario forced is when the two people on either side would be paired up across a diameter. Thus, a contradiction will arise.
Case 3: diameters
There are possible sets of diameters to draw.
Notice we are technically choosing the number of ways to choose a pair of two diameters that are neighbors to each other. This means we can choose the first diameter in the pair, and have only two diameters to choose from for the second in the pair. This means we have 
possibilities for choosing 5 neighboring diameters. However, notice that there are duplicates, so we divide the possibilities by to get .
Note that there cannot be a case with diameters because then there would have to be diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises.
Case 4: diameters
There is only way to do this.
Thus, in total there are 
possible ways.
~Pearl2008 (Minor Edits)
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
Video Solution by TheBeautyOfMath
https://youtu.be/3BvJeZU3T-M?t=419
Video Solution by Sohil Rathi
https://youtu.be/0W3VmFp55cM?t=2796
See Also
| 2020 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 | ||
| 2020 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
