Difference between revisions of "2020 AMC 10B Problems/Problem 17"
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Case 3: <math>3</math> diagonals | Case 3: <math>3</math> diagonals | ||
| + | There are <math>5% possible diagonals to draw. | ||
Note that there cannot be a case with 4 diagonals because then there would have to be 5 diagonals for the two remaining people, thus a contradiction. | Note that there cannot be a case with 4 diagonals because then there would have to be 5 diagonals for the two remaining people, thus a contradiction. | ||
| − | Case 4: <math>5< | + | Case 4: </math>5<math> diagonals |
| − | There is <math>1< | + | There is </math>1<math> way to do this. |
| − | Thus, in total there are <math>2+5+5+1=\boxed{13} | + | Thus, in total there are </math>2+5+5+1=\boxed{13}$ possible ways. |
==Video Solution== | ==Video Solution== | ||
Revision as of 10:29, 26 March 2020
- 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
Let us use casework on the number of diagonals.
Case 1: 
diagonals
There are ways: either 
pairs with , pairs with 
, and so on or pairs with , pairs with , etc.
Case 2: diagonal
There are possible diagonals to draw (everyone else pairs with the person next to them.
Note that there cannot be 2 diagonals.
Case 3: diagonals
There are $5% possible diagonals to draw.
Note that there cannot be a case with 4 diagonals because then there would have to be 5 diagonals for the two remaining people, thus a contradiction.
Case 4:$ (Error compiling LaTeX. Unknown error_msg)5
1$way to do this.
Thus, in total there are$ (Error compiling LaTeX. Unknown error_msg)2+5+5+1=\boxed{13}$ possible ways.
Video Solution
https://youtu.be/3BvJeZU3T-M (for AMC 10) https://youtu.be/0xgTR3UEqbQ (for AMC 12)
~IceMatrix
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. 
