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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!
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.
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).
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.
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).
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:
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.
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.
45 degrees
ACE forms a right angled isosceles triangle
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
The missing radius is 11.
The points are nine of the twelve vertices of a regular dodecagon (12-sided polygon).
506
23√2 x 11√2
The probability of the centre of the circle being within the pentagon is 11/16.
The general formula for an n-sided shape is: 1 - n/(2^(n-1))
If you're interested, the derivation of the above formula is as follows:
The puzzle asked, for five points positioned at random, what is the probability of the pentagon formed by them containing the centre of the circle.
Let’s switch is round and ask what the probability is of missing the centre. For this to be the case, all of the points must be in the same half of the circle, but that half could be 0 to 180 degrees, or 180 to 360, or 63 to 243, or any of the countless other ways. So let us pick a point, one of the five random points, and (temporarily) call that our zero, our datum. What is the chance of the other 4 points being less than 180 degrees clockwise from the datum? Simply ½ x ½ x ½ x ½, or 1/16.
Each of the five points can be taken as the datum in turn. So is it just a matter of adding together those five probabilities (or, as they are equal, simply multiplying 1/16 by 5)? Well yes, since at most one datum can result in all the other points being less than 180 clockwise from that point, the five cases are mutually exclusive, and it is indeed just a matter of adding them together.
From here it is easy to see what the general equation is. For n points, the probability of not containing the centre is:
n/(2^(n-1))
Giving the sequence ¾, 4/8, 5/16, 6/32, 7/64, 8/128 etc (before simplification, to demonstrate that the numerator increases by one and the denominator doubles with each new term)
Then the probability of containing the centre is merely the complement of this, or 1 – n/(2^(n-1))
Resulting in the sequence ¼, ½, 11/16, 13/16, 57/64, 15/16 etc.
I'm delighted that the maths/puzzle writer Alex Bellos has used some of my puzzles in his Guardian column.
https://www.theguardian.com/science/2017/may/08/can-you-solve-it-have-a-punt-on-the-paddocks-puzzle
Thanks Alex!
Christian Festival = EASTER Food tester = TASTER
Annoy = BOTHER Parent = MOTHER
Childminder = SITTER Better adapted = FITTER
In each of these cases, the initial letter has ‘lurched’ two places to the right on a QWERTY keyboard.
So speak/complete = UTTER Animal = OTTER
The number I am thinking of has to contain at least five digits, as for any smaller number the total number of digits I write down cannot be exactly four times my number, as many of the numbers only contribute one to three digits each.
The total number of digits I write down up to 9,999 = 9 x 1 + 90 x 2 + 900 x 3 + 9,000 x 4 = 38,889
The total number of digits up to my number n, is therefore 38,889 + 5(n – 9,999) which must equal 4n.
Rearranging this equation gives 5n – 11,106 = 4n and so n = 11,106.