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.