Solution of the Week #254 - Three Squares in a Hexagon

To make life easier we can divide each of the areas by two, (as long as we remember to multiply the final answer by two to compensate). The reason this makes things easier is that the areas will then be square numbers, and therefore the lengths of the edges of the squares will be whole numbers, namely 10, 9 and 17.

Consider the area of the triangle in the very centre. Knowing the lengths we can find the area using Heron’s formula:

A^2 = s(s-a)(s-b)(s-c), where s is the semiperimeter (a+b+c)/2

The area works out to be 36.

Now consider the triangle between the top two squares. The area sine rule states that the area of a triangle is half the product of two adjacent sides, multiplies by the sine of the angle in between. At the moment we don’t know the sine of angle a. In fact I have no intention of calculating it!

I’m going to use the fact that the sine of an angle (a) is the same as the sine of 180-a. This is because the sine graph between 0 and 180 forms a symmetrical curve centred on 90. Any two angles adding up to 180 (or to put it another way, averaging 90) will have the same sine.

Now if we look at angles a and b, they are a full 360 less two right angles (the corners of the squares), and so therefore will add up to 180. Sin a = sin b. Angle b is also between two lengths of 10 and 9. We can therefore see that the area of the top triangle is identical to the area of the middle triangle, which we already know to be 36.

We can use the exact same reasoning to show that the other two triangles are also 36.

Therefore the area of the whole hexagon is 100+81+289+(4x36), and not forgetting to double the answer to get the final answer of 1228.

Solution of the Week #252 - Hexagon Angle

To start with we can assume each side of the hexagon to be 1 unit without affecting the size of the angle. Now work out the lengths of the lines AG and AH using Pythagoras (having first worked out the horizontal and vertical components of each). It transpires that both AG and AH are equal to sqrt(7)/2 (~1.323). This alone is not of much use in working out the angle in question, however if we now calculate the length of the dashed line GH we discover that this too has a length of sqrt(7)/2. Therefore AGH is an equilateral triangle, and the angle we seek is 60 degrees.

If you’re interested about how I created this puzzle, it is a special case of an interesting theorem I stumbled across: if you have three non-overlapping chords of length x, inside a circle of radius x, and then you connect the ends of those chords to form three new chords (which now don’t necessarily have to be x long), then the midpoints of those new chord form an equilateral triangle.

In my puzzle, the x length chords are AB, CD and FA. The fact that the circumscribed circle has radius x is implicit in the fact that those chords are part of a regular hexagon. Therefore the midpoints A, G and H form an equilateral triangle.

Solution of the Week #251 - Trispace

a=39

Below is the figure fully dimensioned.

Quite interestingly, in all cases the total of the three trapezoid perimeters will equal 5a+b. Since in this case the total was 198, and a=13b, this provides a useful shortcut to the fact that b=3 and therefore a=39 (although of course deriving this fact is more work in the short term).

Despite the fact that you don’t need to calculate the lengths of each of the trapezoids, there is only one figure which satisfies the conditions.

Solution of the Week #250 - Squarespace

There are just the right degrees of freedom to uniquely define this, however the equations are not easy so I was kind and made each length in the figure an integer, so that people could chance upon the solution by trying a few integer combinations. The area of the square space is 100.

Solution of the Week #249 - Tetrahedron Packing

It’s tempting to think that the answer is 125, as the space is the volume of 125 unit tetrahedrons, however tetrahedrons cannot pack without leaving gaps.

To the best of my knowledge the most you can fit in is 65:

On the bottom level you can fit 25 pointing upwards and 6 pointing downwards; on the next level up you can fit 16 pointing upwards and 3 pointing downwards; on the middle level you can fit 9 pointing upwards and 1 pointing downwards; on the next level you can fit 4, and a further 1 at the very top, giving a total of 65.

Solution of the Week #248 - Quartet

Call the four numbers we are looking for, in ascending order, a, b, c and d. We know they are strictly ascending because if any were the same, some of the product pairs would be the same too.

ab must be the lowest product 6.

The second lowest product has to be ac = 8.

cd must be the highest product 32.

The second highest product has to be bd = 24.

However for the other two products we have a choice. In the original quartet of 2,3,4 and 8, ad = 16 and bc = 12. If we now assign those products the other way round we should end up with a different quartet of numbers. So:

ab=6

ac=8

ad=12

bc=16

bd=24

cd=32

If we multiply ab by ac and then divide by bc we will have a^2 = 3. Therefore a = sqrt(3) = 1.732

We can easily work out the others now:

b = 6/sqrt(3) = (2)sqrt(3) = 3.464

c = 8/sqrt(3) = (8/3)sqrt(3) = 4.619

d = 12/sqrt(3) = (4)sqrt(3) = 6.928

The sum of this quartet is 16.743, so is indeed less than the 17 total of the original quartet.

Solution of the Week #247 - Diophantine Parallelogram

By alternate angles ADC is also equal to the two equal corners at point A. By isosceles triangle, ABD is also the same angle.

Therefore triangles ADB and ACD are similar. AD:AB = CD:AD.

Let CD = x^2 and AB = y^2. It follows that AD =xy.

We are told y^2 – x^2 = 13, therefore (y+x)(y-x) = 13.

The only integer solutions to this is y=7, x=6, therefore AD = 42.

(Pedantic note: x and y don’t necessarily need to be integers for x^2, xy, y^2 and (y^2)-(x^2) all to be integers, eg sqrt(2) and sqrt(8) would satisfy. Basically either x and y are integers, or y/x is an integer, (or both). However since they represent the sides of a triangle, y/x must be between 1 and 2, so y=7, x=6 is the only solution).

Solution of the Week #246 - Ghost Crossword

Method of solving: Numbers 1, 12 and 15 appear in both lists, so because of rules 1 and 2, they must either be on the top-left-bottom-right diagonal, or in pairs in corresponding position either side of it. Rule 6 means that 1 must be on the top row, so is either in top corner or in corresponding position in top row that 12 is in left-hand column. 1 cannot be in second column, as there are too many numbers for 12 to be in second row. If 1 is in third column or beyond, then either a 2x2 black square appears in the corner, or another number must lie on TLBR diagonal. Neither of these are allowed, so 1 must be in top left corner. 

Now 4 across must also be on the top row to mirror 1 across (as 1 across is not allowed to fill the entire row). 4 must either be in 6th column or 7th column, depending on whether 1 across and 4 across are three or four ‘letters’ long. As 4 is across only, and not down, any attempt to make it three ‘letters’ long leads to a contradiction, therefore it is in the 6th column, and the square immediately to the left and below it are black. When you place a white or a black square, you can use rule 2 to transfer it around the grid, and very quickly the grid takes shape. 

Solution of the Week #245 - Three Digits

ABC = 239.

239 and 293 are prime.

329 = 7 x 47.

923 = 13 x 71.

392 = 2^3 x 7^2. Its proper divisors are: 1, 2, 4, 7, 8, 14, 28, 49, 56, 98 and 196. Their sum is 463, which is 71 greater than 392.

932 = 2^2 x 233. Its proper divisors are: 1, 2, 4, 233 and 466. Their sum is 706, which is 226 less than 932.

Solution of the Week #244 - Three Way Split

The following is simple proof that the 1 2/3 solution can be improved upon:

However, we can do even better using circular arcs, such that any junctions with the edge of the square are at right angles and any junctions within the square are at 120 degrees:

As far as I can ascertain, this is now the shortest overall cut length but if you can do better I would be interested in knowing about it!

Solution of the Week #243 - Shaded Circles

The ring option is fairly easy to calculate: each of the nine circles will have an area of pi/4, therefore the shaded are is 9pi/4 = 7.069…

The crescent area is trickier.

The radius of the largest shaded circle is 1, the next is 2/3, the next 1/3, then 2/11… Naming the largest circle the zeroth, in general the radius of the nth circle is:

2/(n^2 + 2). The area of each circle is pi.r^2 or:

pi*(2/(n^2+2))^2. Usefully this also works for negative values of n, so will cover both arms of the crescent.

At this point I consulted the Wolfram Alpha website to see if the infinite sum of this boils down to anything simple:

sum_(n=-∞)^∞ π (2/(n^2 + 2))^2 = 1/2 π^2 (sqrt(2) coth(sqrt(2) π) + 2 π csch^2(sqrt(2) π))

Either using Wolfram Alpha or Excel we can estimate the infinite sum as ≈ 6.998…

Therefore the ring option gives a slightly larger shaded area.

Solution of the Week #242 - Where are they from?

Each person+place contains the five vowels exactly once each. Caroline is only missing the letter U so she is from Hull (or anywhere else with a ‘u’ and no other vowels).

Solution of the Week #241 - Irregular Polyhedron

Using Euler’s formula for vertices, edges and faces of a polyhedron:

V+F-E = 2

If we let the number of pentagonal face be n, then the number of faces is (n+2), the number of edges is (14+5n)/2 (since each edge belongs to two faces), and the number of vertices is (14+5n)/3 (since each vertex belongs to three faces).

(14+5n)/3 + (n+2) – (14+5n)/2 = 2

Which boils down to n = 14, so there are 14 pentagonal faces.

Below is such a polyhedron. All of the edges and all of the vertices are visible, just one heptagonal face is on the far side.

Solution of the Week #240 - Base 4 Code

1-103  113-11  1-13-11  113-11  12-21-32-10  113-11  1-102-11  32-33-110  32-11-1-102-30-121  1-103  113-21-103-11  1-103  113-11  20-33-100-11-10  110-33  2-11-3-33-31-11.

“As we age we find we are not nearly as wise as we hoped to become.”

(I deliberately wrote a sentence that would result in ambiguous words, as ‘we’ and ‘age’ both encode to 11311).

Solution of the Week #239 - Nearly Equal Corners

x is 714. One possible set of dimensions are as below:

My expectation being that the solver would try some numbers and arrive at this or another integer solution. However it is solvable using algebra as below (excuse my handwriting):

Alternatively, there is a far quicker way to arrive at an approximate solution by looking at a similar but easier problem:

Solution of the Week #238 - Circle Box

A score of 100(pi) is possible: (36+25+16+9+4+10).

If you’re interested, I also did this for a 100x100 grid, which results in a score of 44246(pi), shown below.

Solution of the Week #225 - Six Integers - submitted by Simon Richards

The first challenge is to spot the neat trick.

1 The first equation

X^2 = A^2 + 4B + 1

Consider (A + 1)^2 will equal A^2 + 2A + 1

Hence if we solve for X being 1 + A then  4B +1 = 2A + 1 Hence A = 2B (This is the neat trick)

2 We now need to discount X being A + n where n is greater than 1.

If n = 2 then  A^2 + 4A + 4

the 2B +1 = 4A + 4  hence 2B = 4A + 3 this can be discounted as B would be greater than A.

3 The second  equation using the same logic as above B = 2C

4 Hence A = 2B = 4C

5 Then start with the third formula find the value of Z^2 which has an integer square root.

6 Then do the same with the other equations.

7 The answer is C = 24, B=48 and A=96 giving Z=31, Y=49 and X-97

8 NOTE there is almost but not quite another solution C, B, A = 4, 8 ,16 give Z, Y, X = 8, 8, 17.

Solution of the Week #211 - Angle

45 degrees

ACE forms a right angled isosceles triangle