E36 Area of a Right Triangle

 A friend, Teddy O'Connell, sent me this geometry problem he found on the Web, which is not too hard and has a pretty neat solution.

Let ABC be a right triangle, with a right angle at C. Suppose that a circle has been inscribed in the triangle (the incircle) which is tangent to ABC at one point on each side including the point D on the hypotenuse AB, such that |AD| = 3 and |DB| = 5. Find (ABC), the area of triangle ABC. 

(drawing not to scale)

I will follow this posting with two comments. The first will contain a couple of hints, and the second (to be posted later) is my solution and a related question that I don't have an answer to.

A 19. Tiling an equilateral triangle

Consider an equilateral triangle. We wish to tile it with n congruent triangular tiles. We will call such a triangulation a CT-n tiling. Clearly there is a CT-n tiling if n = 2 (two 30-60-90 right triangles), n = 3 (three 30-120-30 isosceles triangles) or n = 4 (four equilateral triangles).

(1) Show that there is a CT-n for the following values of n: 1,2,3,4,6,8,9,12,16,18.

(2) Find 4 infinite sequences of n such that there is a CT-n for all values of n in the sequence. (The sequences may not overlap.)

Unsolved (by me) problems

(3) Find (with proof) at least one value of n for which there is no CT-n.

(4) Determine exactly which values of n yield a CT-n.

E 36. How many solutions?

(2020 Putnam Exam, problem A1)

How many positive integers N satisfy all of the following 3 conditions:

(i) N is divisible by 2020

(ii) N has at most 2020 decimal digits

(iii) The decimal digits of N are a string of consecutive ones followed by a string of consecutive zeros.

Note: This only requires pre-college level mathematics, but like most Putnam problems it is not easy.

A 18. Setting up breakout rooms

A Zoom session has N = 2^k people. There are n breakout sessions, each into two equal-sized breakout rooms (i.e., N/2 people in each) The facilitator wishes that every person in the session meets with every other person in a breakout room at least once. Show this can be done if n = k + 1.

Note: This is the simplest version. If you want more of a challenge, you could try the case where N is not a power of 2. Obviously if N is odd, the breakout rooms cannot be exactly equal in size, but there is always an efficient solution where the numbers of people in each breakout room are always within 2 of one another.

A 17. Square inscribed in a triangle

Given a triangle ABC with acute angles B and C, how do you construct a square PQRS with PQ in BC and vertices S and R in AB and AC, respectively?

I found this problem on Quora. It took me a while to solve it, but the construction turned out to be pretty simple and elegant.

A16. Treasure search

A treasure hunter is on an island which is in the shape of the unit disc, and they are located on the boundary at (1,0). The treasure hunter has a treasure-detector, which will light up if the detector is within 1/2 unit of any buried treasure. The treasure hunter can move along any path (say, continuous and piecewise infinitely differentiable) that starts at (1,0) and stays within the closed unit disk. The treasure hunter claims that they can discover whether or not there is treasure buried on the island (or in other words, that the union of all disks of radius 1/2 centered on points along the path covers the unit disk) by traveling along a path of length π. Prove or disprove the claim.

This is my revision of a problem sent to me by Apratim Roy.

A15. A Problem by Henry Dudeney

The houses on a long street have addresses 1, 2, 3, ... n. (Dudeney was British. Unlike the custom in most of the US, odd and even house numbers in Britain occur on the same side of the street. Assume all houses are on the same side of the street.) Call a house a half-way house if the sum of the numbers of the houses before it are equal to the sum of the numbers of the houses after it. Depending on n, there may or may not be a half-way house. For example, if n = 8, then there is a half-way house, namely 6, because 1 + 2 +3 + 4 + 5 = 7 + 8. You can check there is no half-way house if 1 < n < 8.

Find the value of n if you know that 50 < n < 500 and there is a half-way house.

You could do this easily by a brute-force search with a computer, so to make it interesting, no computer/calculator use is allowed.

I saw this problem presented online by the Mathologer. It connects with many fascinating parts of number theory, and there is an interesting connection with Ramanujan.

E35 - Three Triangles in a Regular Hexagon

Here is another problem from the Catriona Shearer collection. ABCDEF is a regular hexagon. Lines have been drawn to create three triangles, whose interiors have been shaded. What fraction of the hexagon is shaded?

The points labeled M_i are the midpoints of  their sides, and the unlabeled point marks the center of the hexagon.

E34. Tangent circles in a rectangle

This problem was called to my attention to Apratim Roy. It comes from a collection, on Twitter, of many dozens of interesting geometry problems, at the level of high school geometry, by Catriona Shearer, a math teacher in the UK, which she has posted on Twitter, @Cshearer41.

In the diagram below. you are given circles O and P, externally tangent at B. Circle O is tangent to two sides of rectangle DEFG, and circle P is tangent to three sides of the rectangle, as shown. Find the measure of angle ABC.

It is quite surprising that you don’t need to be given any measurements to answer the problem. (But no fair using that fact in your proof; that is, you can’t just prove a special case such as two circles of equal size.) I also like this problem because it is hard, but not too hard, and because there are several different nice ways to solve it.

A 14. Triangles in a regular hexagon

I came across a neat problem in Quora. When I found it, there were no solutions, and with crucial help from my student Apratim Roy, I was able to find a very elegant solution. Unfortunately, when I went back to Quora I was unable to find the original problem, so I cannot give credit to the proposer. If anyone wants, send me an email and I will send you my solution, or provide a hint.

Here is the problem: Let ABCDEF be a regular hexagon, and P a point inside the hexagon. Suppose Area(PAB) = 3, Area(PCD) = 5, and Area(PEF) = 8. Find Area(PBC).

E33. Tiling a checkerboard

This problem was shown to me by my student, Apratim Roy. Though it involves only elementary concepts, I found it rather difficult, and thought the solution was very surprising and elegant.

You are given a standard 8 x 8 checkerboard, with one square removed, and 21 3 x 1 tiles. In other words, there are exactly enough tiles to cover the modified board. Your task is to find a way to do this, without cutting any tile.

(a) Find out what square must be removed for the task to be possible. (4 possible answers).
(b) Describe the tiling. (Many possible answers).

You might guess that the removed square needs to be one of the four corners of the checkerboard, but you would be wrong.

A13. Points of tangency of an ellipse and a circle

Let E be the ellipse with equation x²/4 + y² = 1 and C(r) be the circle with center (1,0) and radius r. For which values of r do the curves E and C(r) have point(s) of tangency?

This is fairly routine, but still a bit challenging to find all solutions.

A12. Fifth powers final digit - generalized.

If numbers are expressed in base b, for which b is it true that n⁵ and n end in the same digit for all positive integers n?
This is an obvious generalization of Problem E32.

E32. Fifth powers final digit

The following rather neat problem occurs in Challenging Problems in Algebra by Alfred S. Posamentier and Charles T. Salkind. I think it is suitable for a bright high school student or, with some hints, even for average high school students.

Prove that n and n⁵ always end in the same digit (in ordinary base-10 representation).

Hippasus & The Discovery of Irrational Numbers

In November I gave a speculative talk to the New England Section of the Mathematical Association of America on the discovery of irrational numbers by the Pythagorean Hippasus through an examination of the mystic pentagram, the sacred symbol of the Pythagoreans. I have expanded it a bit, adding a method of recursively computing the golden ratio, phi. If you want to see how these topics are related, please check my paper on Scribd: https://www.scribd.com/doc/294006886/Irrational-Numbers-the-Mystic-Pentagram-and-Eigenvectors.

E32. Chimes

The following problem is given in Jacobs' Geometry: If it takes a clock 3 seconds to ring 3:00 (3 chimes), how long does it take the same clock to ring 6:00 (6 chimes)? The answer is not 6 seconds. The answer depends on making some assumptions, which I think are reasonable ones.

Embarrasing mistake

I just realized that my last post was far off the mark. There is a super-obvious example that shows that f(n) >= n/2, and so in fact f(n) = n/2. The example is the subset {n/2 + 1, n/2 + 2, ... , n} which contains n/2 elements and clearly has the non-divisible property. The problem has very little to do with prime numbers. The previously-published result was sharp.

A simple (?) number theory conjecture

Call a set S of positive integers non-divisible if, whenever a and b belong to S, it is not true that (a|b or b|a). For example, the set of prime numbers is non-divisible. Let S(n) be the set of the first n positive integers, and let f(n) be the cardinality of the largest non-divisible subset of S(n). Then clearly f(n) >= pi(n), where pi(n) denotes the number of prime numbers <= n. Furthermore, I know a pretty proof (published, but not by me) that shows f(n) <= n/2 (n even). That is, in any subset of n/2 + 1 positive integers all <= n, at least one integer divides another integer in the set. This proof involves the pigeonhole principle. So, pi(n) <= f(n) <= n/2. (n even) My conjecture is that, in fact, f(n) = pi(n) for all n > 1. Does anyone have a counterexample or a proof? It seems like it should be very simple.

E31. A packing polynomial

Funny how one comes across math problems. I was reading my Reed College alumni magazine and came across an article about Maddie Grant, class of '15, whose undergraduate thesis is apparently a substantial generalization of a 1923 result by Fueter and Polya. Feuter and Polya evidently discovered a polynomial function that maps the non-negative integers one-to-one and onto the pairs of non-negative integers. That is, they found a simple formula to express the inverse Cantor's "diagonal" mapping of pairs of non-negative integers to non-negative integers. Their function ‐ a so-called packing polynomial ‐ is given by

f(x,y) = (1/2)[(x + y)² + x + 3y],
where x and y are non-negative integers.

The proof that this function is one-to-one and onto the non-negative integers is actually pretty elementary if you look at it the right way. I will print a hint as a comment in a few days.

A11. Splitting a triangle and a tetrahedron

János Kurdics published this interesting problem in The Math Connection Linkedin group:
Given a triangle, find the shortest line segment that divides the triangle into two regions of equal area. A solution that does not involve calculus is preferred.
The solution that I know is very nice, and gives quite a workout in elementary trigonometry.

János says that he is working on three-dimensional analog (plane that divides a tetrahedron into two regions of equal volume and gives smallest possible cross-sectional area with the tetrahedron) but has only solved it for a regular tetrahedron.