# Four Points

By Stephen Meskin This problem concerns the number of ways of putting four distinct points in a (Euclidean) plane. But first, let me put the problem in context.

What about three distinct points? They determine a plane. But we are more interested in the three distances between the points. There are three possibilities:

1. their lengths could all be equal;
2. there could be two of the same length and one different; or
3. all three lengths could be different.

In the first case, we have all the equilateral triangles. While there are infinitely many of them, they are all similar to each other. However, in the second case—isosceles triangles—there are infinitely many dissimilar triangles, and the same is true in the third case.

The problem is to extend the above analysis from three points to four distinct points in the plane. With four distinct points, there are six distances between the various points. The problem is to determine and describe the similarity classes for configurations of four distinct points in the plane when there are only two different lengths among the six distances, because the other cases are fairly obvious. You will be replacing the question mark in the table with a finite number. Hints:

1. The answer is greater than 1 and less than infinite.
2. Placing the four points at the corners of a square is one configuration.
3. A tetrahedron is NOT a configuration because it is not in the plane.
4. A (non-square) rectangle is NOT a configuration because there are three different lengths between the points.
5. You should confirm that the configurations which you imagine and can even draw exist.
6. You should confirm that there are no more configurations than the ones you have proposed.

Solutions may be emailed to cont.puzzles@gmail.com.

In order to make the solver list, your solutions must be received by May 31, 2019.

Solution to Previous Puzzle:
The SuperContest

1. What is the probability of at least 1 contestant getting more winners than losers each of the 17 weeks of the contest? As each person makes five selections each week, there is a 50% chance the person gets more winners and losers each week. The probability a person does this 17 weeks in a row is 0.517=0.0008%. The probability a person has at least one losing week is therefore 1-0.517=99.9992%. Hence, the probability that everyone in the contest has at least one losing week is {1-(0.517)}500=99.62%.
2. If I pick more winners than losers for week one, what is my expected score? Notice that the number of winning picks is a binomial distribution, where the only possible answers are 3, 4 and 5. Therefore I have a 10 in 16 chance of getting 3 picks right, a 5 in 16 chance of getting 4 picks right, and a 1 in 16 chance of getting 5 picks correct. So (10/16)*3 + (5/16)*4 + (1/16)*5 = 3.44.
3. What is the median winning score of the SuperContest? A team’s score in the SuperContest is just a binomial distribution of 85 events, with each event having a 0.5 probability of occurring. Knowing this, we can come up with a probability of a team scoring exactly x number of points. We can convert this into a cumulative distribution for x by summing the probabilities of all the prior entries of x-1, x-2, x-3, etc. Once we have the cumulative distribution for one team, we can use the same logic as in question 1 to show that the probability of all 500 teams scoring below 56 points = 0.9989500 = 0.5775, and the probability of all 500 teams scoring below 55 points is 0.9977500=0.3233. So there is a 57.8% chance that no team will score 56 points, and a 32.3% chance that no team will score 55 points. Hence, the median winning score by definition is between 55 and 56, or very close to 56.
4. If I pick more winners than losers each week, what is the probability of me winning the contest? In question 3 we showed how to create a cumulative distribution of your opponent’s best score. Now we need to come up with a cumulative distribution of your own score, given the probabilities we calculated in question 2. Once you have this distribution, you can use the distribution computed in question 3 to come up with the probability of each individual score winning, and then weighing those by the probabilities of your individual score. Doing this leads to the answer that you have a 69.8% chance of winning outright, a 10.6% chance of tying for the win, and a 19.6% chance of losing the SuperContest.

P.S.: Kudos to all who caught the typo in the puzzle that 17 × 5 should equal 85, and not 65.

Solvers

Robert Bartholomew, Mike Crooks, Bob Conger, Bernie Erickson, Nick Feitcher / Josh Scholten / Michael Frushour, Bill Feldman, Yan Friedman, Rui Guo, Clive Keating, Douglas Levi, David Lovit, David Promislow, Tomasz Serbinowski, John Snyder, Al Spooner, Zig Swistunowicz, Daniel Wade.

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