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Difference between revisions of "2020 AMC 10B Problems/Problem 17"

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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.
  
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Case 4: <math>5</math> diagonals
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Case 4: <math>5</math> diameters
  
 
There is only <math>1</math> way to do this.
 
There is only <math>1</math> way to do this.

Revision as of 00:06, 17 August 2020

The following problem is from both the 2020 AMC 10B #17 and 2020 AMC 12B #15, so both problems redirect to this page.

Problem

There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know each other?

$\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15$

Solution

Consider the 10 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 diameters.

Case 1: $0$ diameters

There are $2$ ways: either $1$ pairs with $2$, $3$ pairs with $4$, and so on or $10$ pairs with $1$, $2$ pairs with $3$, etc.

Case 2: $1$ diameter

There are $5$ possible diameters to draw (everyone else pairs with the person next to them).

Note that there cannot be $2$ 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: $3$ diameters

There are $5$ possible diameters to draw.

Note that there cannot be a case with $4$ diameters because then there would have to be $5$ diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises.

Case 4: $5$ diameters

There is only $1$ way to do this.

Thus, in total there are $2+5+5+1=\boxed{13}$ possible ways.

Video Solution

https://youtu.be/3BvJeZU3T-M?t=419 (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

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