Tolstoy is considered one of the greatest novelists of all time. But Tolstoy was also fond of one math puzzle, according to Perelmen’s Algebra Can Be Fun, which I have slightly re-phrased.

As punishment, a team had to mow the grass on two fields. One field was twice the size of the other. The team worked for 1/2 a day on the larger field. Then the team split into 2 equal groups. The first group stayed in the larger field and finished mowing by evening. The second group mowed the smaller field, but by evening there remained a portion to do. This remaining portion was completed the next day by 1 person in a single day’s work.

How many people were there in the team?

As usual, watch the video for a solution.

Tolstoy’s Math Puzzle

Or keep reading.

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Answer To

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

In such puzzles, the rate each person works is assumed to be constant.

As punishment, a team had to mow the grass on two fields. One field was twice the size of the other. The team worked for 1/2 a day on the larger field. Then the team split into 2 equal groups. The first group stayed in the larger field and finished mowing by evening. The second group mowed the smaller field, but by evening there remained a portion to do. This remaining portion was completed the next day by 1 person in a single day’s work.

Method 1: Algebra

Define variables:

n = number of people in the team
r = rate 1 person works in 1 day
t = time in days

Then we have:

(number people)(rate)(time) = work done

The team worked for 1/2 a day on the larger field translates to:

nr/2 = work

Then 1/2 team worked for 1/2 a day on the larger field, which is:

(n/2)r/2 = work
(nr)/4 = work

This work is sufficient to complete the large field.

nr/2 + nr/4 = work for large field

Also 1/2 team worked for 1/2 a day on the smaller field, which is:

(n/2)r/2 = work
(nr)/4 = work

Finally 1 person worked 1 day to finish the smaller field:

1(r)1 = r

This is the work for the smaller field.

nr/4 + r = work for smaller field

The larger field is 2 times the size, so 2 times the work of the large field, so we have:

nr/2 + nr/4 = 2(nr/4 + r)
3nr/4 = 2r(n/4 + 1)
3n/4 = 2(n/4 + 1)
3n/4 = 2n/4 + 2
n/4 = 2
n = 8

Method 2: Area Model

Let the larger field have an area of 1. Then the smaller field has an area of 1/2.

The entire team works 1/2 day on the large field, then 1/2 of the team works 1/2 day. The entire team working 1/2 day is equivalent to two 1/2 teams working 1/2 day. So this work is equivalent to three 1/2 teams working 1/2 day, meaning 1/2 team 1/2 day is 1/3 of the large field.

So in the small field, 1/2 team working 1/2 day completes an area of 1/3. This means the remaining portion is 1/2 – 1/3 = 1/6. A single person working 1 day can complete this.

How much work can a team do in 1 day? That is equal to four times the amount that 1/2 team does in 1/2 day, which is 4 times 1/3, or 4/3 = 8/6. Thus the team does 8 times as much work as a single person, and there must be 8 people in the team.

Reference

Algebra Can Be Fun by Yakov Perelman
https://archive.org/details/perelman-algebra-can-be-fun-mir-1979/page/46/mode/2up?q=tolstoy

Lev Tolstoy
https://en.wikipedia.org/wiki/Leo_Tolstoy

These puzzles are fun, but in order to be solvable you have to make assumptions about angles and lengths from the diagram. This is problematic because in real life many diagrams are not drawn to scale, so you do not want to be in the habit of making such assumptions.

It is an interesting mental challenge to convert a puzzle into a more precise mathematical statement. I did some research and found this work was already done, as the problem appeared in the 2009 AMC 10B (one of the qualifying tests for the U.S. mathematical Olympiad team).

The AMC problem is stated like this:

A keystone arch, shown below, is composed of 9 congruent isosceles trapezoids fitted together along the non-parallel sides. The bottom sides of the two end trapezoids are horizontal. Let x be the angle measure in degrees of the larger interior angle of the trapezoid. What is x ?

The mathematical statement is more precise, however, it is also more verbose. So it is a challenge of puzzle makers to present problems precisely, but without too much jargon which would alienate the casual puzzle solver.

I thought it was interesting to juxtapose the two presentations of this problem. With that said, let’s work out the answer!

As usual, watch the video for a solution.

Angle In Arch Puzzle

Or keep reading.

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Answer To Angle In Arch Puzzle

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

There are many ways to solve this problem. I will present 4 methods that I found interesting.

Method 1: triangles

Extend the non-parallel sides of each trapezoid.

Most people just assume all the line segments meet at a common point, but one should justify this is the case.

Consider two adjacent trapezoids, say the leftmost and the next one. The trapezoids are congruent, so the angles and lengths of the triangle formed by the extended line segments are congruent, so the adjacent diagram is a mirror image about the common line segment. Thus the line segments meet at the same point. Continue this reasoning to each adjacent trapezoid, so all trapezoids meet at the same common point. By symmetry this point will be the center of the horizontal line segment.

We have now formed 9 congruent angles y that form a straight line, so 9y = 180° and y = 20°.

We have also formed 9 congruent isosceles triangles with a vertex angle of y and non-vertex angles of z. So we have 2z + y = 180° and z = 80°.

Finally x + z = 180°, so x = 100°.

Method 2: decagon

Form a decagon with the horizontal line and the 9 periphery sides of the trapezoids.

The sum of the interior angles can be computed by the well known formula with n = 10 to get:

180°(n – 2)
= 180°(10 – 2)
= 1440°

The larger interior angle in each trapezoid is x. Let y be the smaller interior angle in each trapezoid. In a quadrilateral the sum of interior angles is 306°, so 2x + 2y = 360° and y = 180° – x.

The 10 sided polygon has 2 interior angles of y, and the remaining 8 are formed by 2y, so we have a total sum of 2y + 8(2y) = 18y. (We could equivalently see there are 9 trapezoids and each contributes 2y to the interior angle sum, to make 9(2y) = 18y.) Thus we have:

18y = 1440°
y = 80°

Finally we have:

y = 180° – x
100 = 180° – x
x = 100°

Method 3: octadecagon

We can similarly reflect the arch to form a regular 18-sided polygon.

Using the interior angle sum formula we get:

180°(n – 2)
= 180°(18 – 2)
= 2880°

We can also sum the interior angles of the 18-sided polygon in terms of the angles y of the trapezoid:

18(2y = 2880°
y = 80°

Finally we have:

y = 180° – x
100 = 180° – x
x = 100°

Method 4: turning angle

We can also employ exterior angles to solve this puzzle.

Consider the angle z a trapezoid has to be turned to be in the orientation of the adjacent trapezoid. A pen that is turned 9 such times will be rotated upside down 180°.

Thus we have:

9z = 180°
z = 80°

We form a straight line between two angles of z and one of y, so we have:

2z + y = 180°
2(20°) + y = 180°
y = 80°

Finally as before we have:

y = 180° – x
100 = 180° – x
x = 100°

What a fun puzzle!

Special thanks this month to:

Daniel Lewis
Lee Redden
Kyle

Thanks to all supporters on Patreon and YouTube!

References

Puzzle on X
https://x.com/sonukg4india/status/2038832197256585556

2009 AMC 10B Problems/Problem 24 via AoPS
https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_24

So where is the mistake?

As usual, watch the video for a solution.

Proof that 2 does not exist. Where is the mistake?

Or keep reading.

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Answer To Proof that 2 does not exist. Where is the mistake?

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

The mistake in reasoning happens when saying the probability of a given prime is even is zero.

It is not possible to assign a uniform distribution to a countably infinite set. The reason is if you set the probability to be 0 for each event, then the total probability is 0 (less than 1). If you set the probability to be greater than 0, then summing over the infinite set will become unbounded (greater than 1). So it’s impossible to set a uniform distribution to a countably infinite set.

So yes, the number 2 does exist after all.

Reference

X
https://x.com/Math_files/status/2039328376247484763