Click here to download a .pdf version of this newsletter.

Return to Main Page

It's a Puzzlement
Four Points, Two Distances

How many ways are there to arrange four points in a plane so that there are exactly two possible distances between pairs of points? Arrangements that are rotations, expansions, or contractions of each other are considered the same. Here’s one example: the four corners of a square define two distances—the side lengths are one distance and the diagonals are another. Four points on a straight line, with a uniform distance between adjacent points, does not work—that arrangement defines three distances.

I am told that most people who try to solve this puzzle miss at least one arrangement, so be careful!

Martin Gardner

The puzzle was as follows:

My wife and I recently attended a party at which there were four other married couples. Various handshakes took place. No one shook hands with himself (or herself) or with his (or her) spouse, and no one shook hands with the same person more than once. After all the handshakes were over, I asked each person, including my wife, how many hands he (or she) had shaken. To my surprise each gave a different answer. How many hands did my wife shake?

Robert Thomas’ solution is as follows:

You have 9 people with 9 answers, all different. So the outcomes are {0, 1, 2, 3, 4, 5, 6, 7, 8} since 8 is the maximum number of handshakes.

First Couple: Someone has 8 shakes, their spouse must have zero because everyone else has 1 already.

Second Couple: Both start with 1. Someone has to have 7, so they shake every remaining hand but their spouse’s, giving everyone but their spouse 2 and thus their spouse is 1 and done.

Next couple: Both start with 2. Someone has to have 6. One of them shakes the next two couples' hands with 2 in the bank, their spouse has 2 already and is done since the remaining people have 3.

Next Couple: Both start with 3. Someone has to have 5. One shakes the next couples’ hands, the spouse has 3 already. For the last couple each has 4.

Last couple: Each has 4.

Since everyone gave a different answer, the “last couple” must be the speaker and his wife and each have 4.

Orin Linden reminded me that he had suggested this problem when I first ran it in August 1987.

David Uhland generalized the problem to n couples and showed that the spouse shook hands with n – 1 people.

Amy Angell, Charles Bernatchez, John Captain, Bob Conger, Sol Feinberg, Walter Fransen, John Jansen, Rob Kahn, Frank Karlinski, Stuart Klugman, Vant Lammers, John Nauss, Dat Tien Nguyen, David Oakden, Rachel Tritz Rutledge, Zac Rutledge, Eric Savage, Matt Schutz, Joan Skurnick, Bryan Starke, Rob Thomas, Kimberley Ward, Chad Wilson, Ian Winograd, and Walter Wright also solved the puzzle.   

Click here to write a Letter to the Editors

Copyright © 2018 Casualty Actuarial Society. All Rights Reserved.