A new December and a new bunch of puzzles from mscroggs.co.uk.

Official answer: “By adding a point inside a triangle, you can turn one triangle into three triangles. By adding a point outside all the current triangles, you can add either one or two more triangles. As we are after the maximum number of triangles, we will choose to add two triangles for each point we add. Four points make 3 triangles. Adding 286 more points will add 572 more triangles, giving a total of 575.”

I think, but I can’t quite prove, the construction of the most triangles possible goes like this:

1. Begin with 2 points and the line between them.
2. Add 2 points close to this and construct the 3 triangles possible.
3. Choose one of the outside lines of the figure.
4. Go back to #2 and repeat.

This means: 2 points, 0 triangles; 4 points, 3 triangles; 6 points, 6 triangles; … ; 2(n+1) points, 3n triangles.

290 = 2(n+1)

145 = n+1

144 = n

432 = 3n

The answer is 432.

This week the question is: Can You Squeeze the Sheets?

Two large planar sheets have parallel semicircular cylindrical ridges with radius 1. Neighboring ridges are separated by a distance L ≥ 2. The sheets are placed so that the ridges extrude toward each other, and so that the sheets cannot shift relative to each other in the horizontal direction, as shown in the cross-section below:

Which value of L (again, that’s the spacing between ridges) maximizes the empty space between the sheets?

To be clear, you are maximizing the volume of empty space per unit area of one flat sheet. In the cross-section shown above, that’s equivalent to maximizing the area of empty space per unit length of one flat sheet.

And for extra credit:

Instead of cylindrical ridges, now suppose the sheets have any number (greater than zero) of hemispherical deformations with radius 1 that extrude toward each other. This time, the sheets need not be the same as each other.

As before, the distance between the centers of any two deformations on the same sheet must be at least 2. What is the minimum empty space, again expressed as volume per unit area of one flat sheet?

Highlight to reveal (possibly incorrect) solution:

Video. Figure 1, 2, 3 and 4.

Construct (in the cross section of the 2 sheets) a triangle. 1 side is horizontal, 1 side vertical, and 1 side connects the centres of 2 neighboring ridges-semicircles.

The horizontal side is L/2. The hypotenuse is exactly 2. Therefore the vertical side is √(4 – L²/4). Let’s call that h for height.

Looking at a rectangle containing 2 of these triangles, its area (again, a cross section of the real 3d situation) is hL/2. Some of this area is taken up by circle parts; exactly π/2. The rest is empty: hL/2 – π/2. We want to normalize this, to correspond to a unit step in the horizontal direction. This gives (hL/2 – π/2)/(L/2). This function has a maximum for about L = 2.6581.

And for extra credit:

Figure 1, 2, 3 and 4.

Option a: This is the same solution as the optimal stacking of balls. In my figures there’s a triangle between the centres of the 3 blue hemispheres. The base of the triangle is 2, and the height is √3. Adding the centre of the red hemisphere, we get a tetrahedron. This has height 2√2/√3.

Looking at the volume with the 2 green triangles as the base, we have a complete red hemisphere (the bits chopped off have friends included elsewhere) and a complete blue hemisphere (2 x 1/3 and 2 x 1/6). The space taken up is π4/3, appr. 4.19. The complete space is 2*√3*2*√2/√3 = 4*√2, appr. 5.66. The empty space is appr. 1.45. The base of the volume is 2*√3, appr. 3.46. The empty volume pr. unit area is 1.45/3.46, appr. 0.42.

Calculations and animation.

Option b: Let L be the distance between 2 centres of hemispheres on a sheet. Vary L and find a minimum. Oh! That minimum occurs when L = 2, so we get the same solution, appr. 0.42.

Option c: Defies the my imagination.

A new December and a new bunch of puzzles from mscroggs.co.uk.

The circle all around is 360°.

The angle to the minute hand, corresponding to 22 minutes is 360 * 22/60 = 132.

The angle to the hour hand, going clockwise, corresponding to 8 hours and 22 minutes is 360 * (8 + 22/60) / 12 = 251.

The angle between the hands is 251 – 132 = 119.

This December I participated in Advent of Code.

Days 1-6 happened without a lot of drama (only 1 question on the forum). This phase also mostly saw me coding in a sandbox (exception: day 3, part 2). Day 7 I permanently changed to my PC. The tasks were rather easy for me. I didn’t really try to come up with a fast answer, in part because the sandbox + my tablet was a slow way to code. My examples below manipulate the test data, but ran on the actual data as well.

A new December and a new bunch of puzzles from mscroggs.co.uk.

n = 99999…999

= 10⁵⁵ – 1

n³ = (10⁵⁵ – 1)³

= (10⁵⁵)³ + 3*(10⁵⁵)²*(-1) + 3*(10⁵⁵)*(-1)² + (-1)³

= 10¹⁶⁵ – 3*10¹¹⁰ + 3*10⁵⁵ – 1

 10000...0000...0000...000
-         300...0000...000
+                300...000
-                        1
==========================
  9999...9700...0299...999

So now we just have to count the 9s and 7s and the 2.

The 1 lives in position 1. The 3 above it has a 3 in position 56. The next 3 is in position 111. Finally the 1 above it is in position 166.

The result of adding and subtracting is 9s from position 1 to 55, a 2 at position 56, then some 0s, a 7 at position 111 and finally 9s from position 112 to 165. 55*9+2+7+54*9 = 990. Which happens to be 55*18.

Just to test my logic here: if n had been 9999 (4 digits), n³ would be 999,700,029,999. 4*9+2+7+3*9 = 72 = 4*18. Checks out.

A new December and a new bunch of puzzles from mscroggs.co.uk.

n + n+1 + n+2 + … n+10 = 2024

Let m = n+5

m-5 + m-4 + … m + … m+5 = 2024

11m = 2024

m = 2024/11 = 184

n = m-5 = 179

A new December and a new bunch of puzzles from mscroggs.co.uk.

I think the example is also a hint.

A number with 1 factor would be 1. And that’s not 13.

A number with 2 factors would be a prime. Number p, factors 1 and p. The geometric mean would be (1*p)^1/2. We can’t get 13 to fit there.

A number with 3 factors has a special property. There’s a square here somewhere. The number, p², has factors 1, p and p². This accounts for p² = 1 * p² = p * p, the 2 different ways to write p². Like, 9 = 1*9 = 3*3. The geometric mean works out to be p, isn’t that neat? So my guess is that we’re looking for 13² = 169.

A new December and a new bunch of puzzles from mscroggs.co.uk.

My first attempt is further down. But I thought of a better way.

These are equivalent:

• Writing n as a sum of m odd, positive integers (5 = 1+3+1).
• Writing n+m as a sum of m even, positive integers (8 = 2+4+2).
• Writing (n+m)/2 as a sum of m positive integers (4 = 1+2+1).

And how many ways can I write the last sum? Stars and bars. Imagine writing 4 as 1 1 1 1. We want to convert this to 3 positive integers. We do this by adding 2 bars in between, like 1 | 1 1 | 1. (3-1 = 2.) The bars could be added in 3 different positions, between 2 1s. (4-1 = 3.) Only 1 bar in each position. We can do this in (3 c 2) different ways. This is 3*2/2*1 = 3. Anyway, 1 | 1 1 | 1 corresponds to 1 | 1+1 | 1 = 1 | 2 | 1, corresponds to the 4 = 1+2+1.

Going back, writing (n+m)/2 as a sum of m positive integers can be done in ((n+m)/2-1 c m-1) different ways.

Anyway.

• Write 14 as m odd, positive integers.
• Write 14+m as m even, positive integers.
• Write (14+m)/2 as m positive integers.

To write 14 as a sum of m odd, positive integers, m must be at least 2 and at most 14. Also m is even, because the sum of an odd number of odd integers would be odd. Let’s look at these.

And when we add up the right column, we get 377.

First attempt:

There’s probably a smarter way to do this. But I needed an intermediary step, before I could answer the primary question. First: How many ways can I write a sum of odd, positive integers, if I don’t care about the order?

(1) I find this result for 2 by adding a 1 to the result for 1. 1 = 1*1. 2 = 1*1+1 = 2*1.

(2) I find this result for 3 by first adding a 1 to the result for 2 (getting the first new result), and then adding a 3 to the result for 0 (getting the second new result).

I only actually need the last cell. Let’s look a little closer at it. E.g. 3*1+1*11 is 1+1+1+11 (4 elements), when we don’t care about the order. Let’s look at how many different ways each sum can be written (how many permutations, like 1+1+1+11 = 1+1+11+1 = 1+11+1+1 = 11+1+1+1, 4 different ways).

(1) Among the 12 positions, I choose 1 for the 3.

(2) Among the 10 positions, I choose 2 for a 3. I can do this in “10 choose 2” ways.

(3) Among the 8 positions, I choose 1 for the 3. I can do this in 8 different ways. Among the remaining 7 positions, I choose 1 for the 5. I can do this in 7 different ways.

(4) Combining methods 2 and 3.

Finally we have to add up the right column. The result is 373. I did some of this manually, I must have missed a step somewhere. Oh yeah, I think I missed 3*3+5, good for 4 further permutations.

A new December and a new bunch of puzzles from mscroggs.co.uk.

14 is even, so 2 is a factor. The other factor is 7, a prime, so it can’t be constructed by multiplying lower numbers.

100 d100: One factor is 2, to get an even product the lowest possible way. Then the other factor of the product is a prime p, in this case 101, to make the product impossible. (Anything lower could be reached with 1⁹⁸ * 2 * p.) The product is 202.

A new December and a new bunch of puzzles from mscroggs.co.uk.

The only way to reach the sum 16 with 5 different positive integers is 1+2+3+4+6.

The lowest possible sum with 5 different positive integers is 1+2+3+4+5 = 15. In order to reach 16, one of these integers has to be 1 higher. If the integers are still different after this operation, the only candidate is 5.

Alternate proof: Play Killer Sudoku.

The product is 1*2*3*4*6 = 144.