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

Solution of the Week #180 - Letter Wall - submitted by Philip Morris Jones

There are 26 letters and you draw 2 at random so the first letter is one of 26 and the second one of 25 so there are 26*25 = 650 ways to pick two letters.

The letter A is adjacent to B and H so that is 2 ways to pick adjacent letters with A as the first letter.

B is next to 4 letters so there are 4 ways to pick adjacent letters with B as the first, and AB is different to BA.

If you go through the wall you get the following ways per brick

2 4 4 4 4 4 2

5 6 6 6 6 5

3 6 6 6 6 6 3

3 4 4 4 4 3

and that sums to 116 so you get 116/650 which simplifies to 58/325

Solution of the Week #178 - Tangent Curve

The missing radius is 11.

Solution of the Week #176 - Nine Point Circle

The points are nine of the twelve vertices of a regular dodecagon (12-sided polygon).

Solution of the Week #174 - Rectangular Area

506

23√2 x 11√2