Geometry Forum - Problem of the Week
Solutions - Shaking Hands, Sept. 19-23, 1994
Annie says:We had a good response to this week's problem. There were two ways that folks went about solving the problem. First was to make a chart and look for patterns. The other was answers such as Al and Matt's from Shaler Area High School: the answer for part a of the problem of the week would be to take the number of people at the party and multiply that number by the number of people minus one. This would give you the number of total handshakes. Then divide this number by two to find the number of hands shook with no one shaking the same person's hand twice.
Jon Star's Class
Advanced 7th Grade Math Class, Albuquerque Academy, Albuquerque, New MexicoThis chart shows the number of handshakes for certain number of people:
Handshake Chart
People 1 2 3 4 5 6 7 8 H.S. (a) 0 1 3 6 10 15 21 28 H.S. (b,i) 0 0 1 3 6 10 15 21 H.S. (b,ii) 0 0 0 2 5 9 14 20
Let's say there are P people at the party. (a) For P people, the number of handshakes would be the sum of the numbers from 1 to (P-1). An easy way to figure out this sum is: # handshakes = (1/2)*P*(P-1) (b) This answer depends on whether the table is a circle or a long row-type table. (i) For row-type tables: Having 5 people at a row table is like having 4 people not at a table, the tables shows. So instead of using P in the formula, use P-1. So: # handshakes = (1/2)*(P-1)*(P-2) OR: you can subtract (P-1) from each number in part (a). So: # handshakes = (1/2)*P*(P-1) - (P-1) (ii) For circular tables: The number is one less than for row-type tables. So the formula would be: # handshakes = (1/2)*(P-1)*(P-2) - 1 OR # handshakes = (1/2)*P*(P-1) - (P-1) - 1 This second formula can be re-written as: # handshakes = (1/2)*P*(P-1) - P
Andrew Liu
A. To figure out how many handshakes you could use the formula [N(N-1)] / 2. The reason for this would be N would represent the total amount of people there. (N-1) would be all the people one person would shake. Over 2 would be because you shake hands once... like person A shakes hands with person B. Person B doesn't shake back, therefore it counts as one for both.B. This formula is merely a variation of the formula in question. [N(N-3)] / N. If the answer is negative, make 0. You can't shake with a negative number of people. The change to N-3 is because you are not shaking hands with yourself and also the two people sitting on your right and left. All the other remains the same.
Anna Ustazewska
Anna used the same word argument as Andrew Liu's explanation above, yet she derived it by drawing a network with each person as a node... later polygons with each handshake represented by a diagonal of the polygon.Michele Gibney
Michele found solutions to several easy numbers and plotted them on a graphing calculator. She used the regression function to derive a curve. For an unknown number of people at the party... "If they brought a graphing calculator ( a very high tech one ) or a lot of paper, a ruler, pencils, erasers, calculators or a math expert, things might be a lot easier."Cecilia Merediz
a) If each person shakes hands with everyone else at
the party, how many handshakes will there be?
People (n) Handshakes
2 1
3 3
4 6
5 10
6 15
7 21
8 28
9 36
10 43
The number of handshakes is equal to the number of diagonals plus
the number of sides, n, of a polygon.
b) What if everyone is seated at a table and shakes hands with
everyone except the two people sitting on either side of them?
People (n) Handshakes
4 2
5 5
6 9
7 14
8 20
The number of handshakes is equal to the number of diagonals of a
polygon with n sides.
Braulio hasn't said how many people are coming and probably
doesn't know for sure, but several people are trying to figure
out the answers to the contests before the party anyway. Can they
do this?
They can't figure out how many people will be there.
What could they bring to the party in order to come up with the
answers quickly?
They could bring a table, like the one above, to come up with the
answers quickly.
Karen Lannutti
The guests would be able to use a formula to figure out the number of handshakes in both a and b. By plugging in the value for the first unknown, x, or the number of people, they could determine the second unknown, y, or the number of handshakes. Let x = # of guests Let y = # of handshakes a) y= x(x-1)/2 ; y= (x^2-x)/2 Since everyone is going to be shaking hands with everyone else, I first multiplied the number of guests by itself. I then remembered that, obviously, you can't shake hands with yourself, so I subtracted one from one of the x's. I then put this equation over two because it takes two people for one handshake. b) y= x(x-3)/2 ; y= x^2-3/2 The same idea applies here. This time, however, I subtracted 3 from one of the x's because you can't shake hands with yourself, the person on the right and the person on the left. Actually, I didn't approach this problem algebraically. I plugged in different numbers of guests and determined the number of handshakes for each. From my examples, I developed my formulas.
Sarah McLaughlin
Yes, it can be done. The person could bring a calculator to make it easier. a) n=number of people n(n-1); You have to subtract 1 because you can't shake hands with yourself; n people interacting with n-1 people. It takes 2 people to make a handshake, so, n(n-1)/2 = (n^2-n)/2 handshakes. Example: let n=3 (3^2-3)/2= 3 handshakes b) Let n= number of people n(n-3) - subtract 3 because you can't shake with yourself or the 2 people next to you. n(n-3)/2 - divide by 2 because it takes 2 people to make a handshake. Example: Let n=4; (16-12)/2 = 2 handshakes
Meg Tenny's Group
There are 420 people. How many handshakes are there in all, if each person shakes hands with all 419 other people? This is the problem we were given in class. We were instructed to solve it using inductive reasoning, which is, you have to admit, a whole lot easier than drawing 420 people and all the people that they shook hands with. The only problem is that you have to decipher the pattern first, before you can do anything else. Most people would try to solve the problem by using some simpler methods, (which by the way, don't work.) I think my group tried about everything, since the activity was equal to ten points. For example, we tried 420 x 420, and 419 x 419 and even, 419 x 420. None of these were correct. Finally, we were left with charts to try to figure out the pattern. We started with: # of people 0 1 2 3 4 5 # of handshakes 0 0 1 3 6 10 We determined how many handshakes there were between 1, 2, 3, and 4 people. By the time we got to 5 people, we had a pretty good idea of our pattern. The next step was to figure out the problem, how many handshakes were there between 420 people. To do this, first you must discover the true pattern, according to n, the number on the top portion of the chart. Our group realized the pattern between the numbers in the chart, by using inductive reasoning. # of people 420 n# of handshakes [420(420-1)] / 2 n(n-1) / 2 We saw that n, (any number on the first line of the chart) multiplied to the number one less than n, (n-1) and divided by two, was the answer. For example, take 4 people. [4(4-1)] divided by 2. 12/2=6. 6 is the number located below 4 in the first chart above. So, if you substitute 420 for n, you have 420 (420-1) divided by 2. The final answer, was 87,990. 87,990 handshakes took place between 420 people.
Erik Ostrowski
Sorry for being late for the problem of the week but our class
just started it. I have figured out the solution for part a of
the September 19-23 problem. The solution of the problem is if
there are n people, that is equal to the number of people minus
one, then multiplied by the number of people, then divided by
two.
n = (n-1)n
------
2
# of people # of handshakes
2 1
3 3
4 6
5 10
6 15
7 21
8 28
n use above formula
Annie Chan
a) Let x be the unknown number of people: x(x-1)
--------
2
First, I start drawing diagrams of 2 people shaking hands, then 3
people, 4, and 5. I found a pattern. If we have 2 people,
there's going to be one hand shake. If 3, the 1st person shake
his hand twice and the 2nd person shake his hand one. If 4, 1st
person shake his hand 3 times, 2nd person 2 times, 3rd person 1
time. If 5, 1st person shakes 4 times, 2nd person 3 times, 3rd
person 2 times, 4th person 1 time. In every situation the last
person don't get his turn to shake had with others because
everyone have already had a chance first to shake with him.
2 people: 2-1
3 people: (3-1)+(3-2)
4 people: (4-1)+(4-2)+(4-3)
5 people: (5-1)+(5-2)+(5-3)+(5-4)
x people: x(x-1)-x(x-1)
_______
2
=x(x-1)
______
2
b) Think of this question as find the number of diagonals in a n
sides polygon.
sides of polygon | number of diagonals
_________________________________________
4 | 1
5 | 2
6 | 3
7 | 4
n | n-3
n-3 is the number of people you shake hands with. So n(n-3) is
basically the number of total hand shakes. But you have to
divide it by 2 to get the correct answer because if person A
shake hands with person B, person B doesn't come again to shake
hands with person A. So the answer I come up with is n(n-3)
______
2
Al Trezza and Matt Walker
The answer for part a of the problem of the week would be to take the number of people at the party and multiply that number by the number of people minus one. This would give you the number of total handshakes. Then divide this number by two to find the number of hands shook with no one shaking the same person's hand twice.Louise, Ryan, and Ben
This is an answer from Louise (age 9), Ryan (age-10), and Ben (age-11). All are learners at Colorado Creative Education's Learning Center. We made a table that looks like this: # people: 2 3 4 5 6 7 8 9 10 11 12 # shakes: 1 3 6 10 15 21 28 36 45 55 66 We saw a pattern of adding the two vertical numbers together and getting the next amount of shakes. This table would help you. If one more person came the answer would be +13, one more would make +14 ... If they sat at a table and didn't shake two people's hands the difference would be : 10 people minus two = -17; 9 people minus 2 =-15; 8 people minus 2 = -13 ...
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