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.