Can You Turn a Right Triangle Into an Isosceles Triangle?

Starting with a 3-4-5 right triangle, make an isosceles triangle by appending a second triangle to one of its sides. How many distinct ways can you do this?

Welcome to Fiddler on the Proof! The Fiddler is the spiritual successor to FiveThirtyEight’s The Riddler column, which ran for eight years under the stewardship of myself and Ollie Roeder.

Each week, I present mathematical puzzles intended to both challenge and delight you. Beyond these, I also hope to share occasional writings about the broader mathematical and puzzle communities.

Puzzles come out Friday mornings (8 a.m. Eastern time). Most can be solved with careful thought, pencil and paper, and the aid of a calculator. Many include “extra credit,” where the analysis gets particularly hairy or where you might turn to a computer for assistance.

I’ll also give a shoutout to 🎻 one lucky winner 🎻 of the previous week’s puzzle, chosen randomly from among those who submit their solution before 11:59 p.m. the Monday after that puzzle was released. I’ll do my best to read through all the submissions and give additional shoutouts to creative approaches or awesome visualizations, the latter of which could receive 🎬 Best Picture Awards 🎬.

This Week’s Fiddler

From Dean Ballard and his daughter comes a puzzle they recalled from “the Greater Internet Hivemind”:

Beginning with a 3-4-5 right triangle, it’s possible to append another triangle to one of its sides, thereby making an isosceles triangle. For example, here is how you can make a 5-5-8 isosceles triangle:

Including the one given above, how many distinct ways can you append a triangle to a 3-4-5 right triangle to make an isosceles triangle?

This Week’s Extra Credit

Now suppose you have a right triangle with legs of length a and b and a hypotenuse of length c. And suppose further that there are N distinct ways to append a triangle to this a-b-c right triangle to make an isosceles triangle.

What are all the possible values of N? (Note that any appended triangle may not be degenerate, meaning it must have a positive area. Also, some of the resulting isosceles triangles may be congruent to each other, but they should be counted as distinct if the appended triangles are attached to different sides, or have different positions or orientations.)

Making the Rounds

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing another quick nugget from Daniel Litt, which I’ll paraphrase here, spoiler-free:

You have 10,000 coins. Among these, 9,999 are fair, while the remaining coin is rigged so that it always lands on heads. You choose one of the 10,000 coins at random and flip it 10 times. To your surprise, it comes up heads all 10 times. Is this coin more likely to be a fair coin or the rigged coin?

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Davide Mancusi Pa 🎻 from Paris, France. I received 57 timely submissions, of which 54 were correct—good for an impressive 95 percent solve rate.

Suppose you had two real numbers, like 3.14 and 2.718. If you rounded these two numbers and added their rounded values together, you got 3 + 3, or 6. Alternatively, if you added the original two numbers and then rounded the sum, you still got 6.

But rounding then adding didn’t always give you the same result as adding then rounding. For example, if the two numbers were 2.4 and 3.4, rounding then adding gave you 5 (i.e., 2 + 3), whereas adding then rounding gave you 6 (i.e., 2.4 + 3.4 = 5.8, which rounded to 6).

How likely was it that rounding then adding gave you the same result as adding then rounding?

To be more precise, suppose you randomly, uniformly, and independently picked two real numbers between 0 and 1. What was the probability that rounding the two numbers and then adding gave you the same result as adding the two numbers and then rounding?

Solver Peter Breyfogle wrote some code that simulated a whopping 10,000,000 pairs of numbers between 0 and 1, and checked for how many of these pairs rounding then adding gave the same result as adding then rounding. The answer was suspiciously close to 75 percent of the pairs. Was the answer in fact 75 percent, or was there something else going on here?

Most solvers, like David Gedye, took a geometric approach. Here was David’s sketch:

The larger squares in the image are unit squares, with 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. The variables x and y represented the two randomly chosen numbers, which meant round(x) + round(y) was the result from rounding then adding, while round(x+y) was the result of adding then rounding. Either expression could have been 0, 1, or 2, as indicated by the corresponding regions in the diagram. So round(x) + round(y) was 0 a quarter of the time, 1 a half of the time, and 2 another quarter of the time. Meanwhile, round(x+y) was 0 an eighth of the time, 1 three-quarters of the time, and 2 another eighth of the time.

The question posed by the puzzle was essentially this: For what fraction of points in the unit square did round(x) + round(y) equal round(x+y)?

Whenever round(x+y) was 0, round(x) + round(y) was also 0. That was the bottom left corner of the unit square. Similarly, whenever round(x+y) was 2, round(x) + round(y) was also 2. That was the top right corner of the unit square. At the same time, whenever round(x) + round(y) was 1, round(x+y) was also 1. That was the top left and bottom right quarters of the unit square. In the end, the region for which round(x) + round(y) equaled round(x+y) was the region labeled “0” in the final square above. Indeed, three-quarters of the unit square had the same result when adding then rounded versus rounding then adding, which meant the answer was precisely 75 percent.

For some additional fun, check out Sophia Wood’s codepen, which runs countless simulations and plots the regions of the unit square where adding then rounding equals rounding then adding.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Michael Montuori 🎻 from Morris Plains, New Jersey. I received 29 timely submissions, of which 19 were correct—good for a 68 percent solve rate (a good deal higher than I was expecting for this beast of a puzzle!).

Instead of picking two numbers from the interval between 0 and 1, suppose you randomly and independently picked N numbers.

What was the probability that rounding each of the N numbers and then adding gave you the same result as rounding the sum of the N numbers?