Solution of the Week #587 - Greatest Quadrilateral

I didn’t want to put a spoiler in the question, but the area of the greatest triangle was 1344 square miles, and this is explained fully in the previous solution. The quadrilateral needs to have an area therefore of 2016 square miles (1 1/2 times 1344).

The method of achieving the greatest area with four runners is actually much more straightforward than for a triangle. The shortest and longest distances head out in diametrically opposite directions, while the other two head out in perpendicular directions. 

So if the distances were, in increasing order, a b c d, the area of the quadrilateral will be (a+d)*(b+c)/2.

We know three distances: 25,33,39. The only slight issue is that we don’t know if the fourth distance will be the largest, the smallest, or somewhere in between. If we assume it lies between 25 and 39, we can say that 25 and 39 are the minimum and maximum distances, therefore:

(25+39)(33+x)/2=2016

64(33+x)=4032

64(33+x)=4032

(33+x)=63

x=30

And since this does indeed fall between 25 and 39, it is the correct answer.

(If we had assumed x was the maximum it would work out as x=31, which is not the maximum, and is we had assumed x was the minimum it would be 30 15/29, which is not the minimum).

The fourth runner covers 30 miles.

Solution of the Week #586 - Greatest Triangle

If we call the starting point H, then for a given BH and CH, the area of the triangle is maximised when AH is perpendicular to BC. The same is true for the other sides, so BH is perpendicular to AC and CH is perpendicular to AB. This means that H is the orthocentre of the triangle (the point where the three altitudes cross).

Now, for any triangle, the distance between a vertex and the orthocentre is the diameter of the circumcircle multiplied by the cosine of the angle at that vertex, so

25=DcosA

33=DcosB

39=DcosC

This means that the cosines are in the ratio 25:33:39, and because they are vertices of a triangle, the angles sum to 180 degrees.

Let’s use the following identity that is true for angles of a triangle:

(cosA)^2 + (cosB)^2 + (cosC)^2 + 2*cosA*cosB*cosC = 1

Using cosA=25/D etc:

(25^2+33^2+39^2)/D^2 + 2*25*33*39/D^3 = 1

3235D+64350=D^3

Which has the obvious integer root of D=65.

Now we know the circum-diameter is 65 we can calculate the cosines and consequently the side lengths of the triangle:

cosA = 25/65, by Pythagoras the opposite side is 60

cosB = 33/65, opposite side is 56

cosC = 39/65, opposite side is therefore 52.

Now we know the sides of the triangle we can use Heron’s formula to calculate the area.

Area is sqrt(s(s-a)(s-b)(s-c)), where s is the semiperimeter, in our case 84

Area = sqrt(84*24*28*32) = 1344 square miles

Solution of the Week #584 - Simple Triangle Puzzle

The triangle we start with is famously right-angled and so has an area of 3x4/2 = 6.

Rotate the triangle so that the hypotenuse is now the base and the right angle is the apex. Since the area is still 6, the height in this orientation must be such that 5xh/2=6, so the height is 2.4 units.

Bring that apex directly downwards towards the base of length 5. It will now be 0.4 units above the base.

The new area is therefore 5x0.4/2, which is equal to 1.

Solution of the Week #583 - Multiply or Add

It’s a trick question, the two expressions are exactly equivalent!

I constructed this myself and then went looking for other examples, but struggled to find any others.

The way I stumbled across this was to try to find a simpler expression for the product of pairwise sums of three numbers:

(a+b)(a+c)(b+c)

I reasoned that each of those sums could be thought of as the sum of all three numbers less one of them. If I call the sum of all three ‘s’ it becomes:

(s-c)(s-b)(s-a)

Multiplying this out I get:

s^3 – (a+b+c)s^2 + (ab+ac+bc)s – abc

However since (a+b+c) is equal to s, the first two terms cancel each other and we find that:

(a+b)(a+c)(b+c)=(ab+ac+bc)(a+b+c)-abc

Adding abc to each side and we get an equation demonstrating the equivalence of the two expressions we started with, which have to beautiful property of interchanging the multiplications and additions.

(a+b)(a+c)(b+c)+abc=(ab+ac+bc)(a+b+c)

Solution of the Week #581 - Inevitable Divisibility

The product will always be divisible by 34560.

For divisibility by each number we need to spread our six numbers as evenly as possible across the modulo classes, so for divisibility by 2, the best we can do is have 3 odd numbers and 3 even numbers. This will result in 6 of the 15 differences being even, contributing a factor of 2^6 = 64 to the final answer.

For 3 we can have 2 of the six divisible by 3, 2 that leave a remainder of 1 and 2 that leave a remainder of 2. This contributes 3^3 = 27 to the final answer.

As 4 is a power of a prime, we also need to consider it, but bear in mind that the factors of 2 have already been included. As a minimum, 2 of the 15 difference will be divisible by 4, contributing 2^2 = 4 to the final answer.

Finally there must be at least 1 pair that differ by 5, contributing a factor of 5 to the answer.

So 64 x 27 x 4 x 5 = 34560.

In the general case, for n numbers, the answer will be the product of all the factorials up to (n-1)! In this case 1 x 2 x 6 x 24 x 120 = 34560.

Solution of the Week #579 - Trip Counter

The solution is as simple as dividing the initial 55664.9 mileage by 9 and rounding down to the next 0.1 of a mile. That will give the trip mileage of 6184.9, at which point the overall mileage will be 61849.8 but it will display 61849. This happened yesterday (August 2nd) as I drove down the A42 near Ashby. Very satisfying it was too!

Solution of the Week #578 - Prime Bridge

It is tempting to think that the number 2 should be the designated driver for the whole endeavour:

2,3,5

2

2,7,11

2

2,13,17

2

2,19,23

which would take a total of 62 minutes. However we can get it under an hour:

2,3,5

2

7,11,13

3

17,19,23

5

2,3,5

which takes just 56 minutes.

But even this isn’t optimal. If we combine the two methods we can shave another minute off:

2,3,7

2

2,11,13

2

17,19,23

3

2,3,5

Giving a total of 55 minutes.

Solution of the Week #575 - 166669

We only need to look at prime number, since if a number isn’t divisible by p, it won’t be divisible by any multiple of p.

We’ll start at the bottom and work our way up. Since the number always ends with 9, it will never be divisible by 2.

For divisibility by 3, the sum of the digits would need to be divisible by 3, but since the digits 6 and 9 are divisible by 3, the leading 1 means the number as a whole will never be.

Since the number ends with 9 it will never be divisible by 5.

Next let’s look at 7. If we multiply a number of the form 16..9 by 3 we will always get a number of the form 50..7. if we then subtract 7 we will have a number that is clearly just the product of 2s and 5s. Since multiplying by 3 and subtracting 7 would have preserved any divisibility by 7, this proves that 16..9 is never a multiple of 7.

For divisibility by 11, the sums of alternate digits need to either be equal, or differ by a multiple of 11, for example 1342 is divisible by 11 because the first and third digits add to the same number as the second and fourth digits.

Let’s consider when we have an odd number of digits. Then 1+k+9 would need to equal (or differ by 11n) 6+k. k in each case is some indeterminate number of 6s, but would cancel out in this case. 10 – 6 is 4, not a multiple of 11. But if the number of digits is even we have 1+k and k+9. The difference here is 8, again not a multiple of 11. So 166…9 can never be a multiple of 11.

Finally 169 is 13 x 13, and so 13 is the smallest number that divides a number from this sequence.

Solution of the Week #573 - 127 Cubes

The simplest way is to start by dissecting it by 4x4x4 = 64. If you then dissected one of these 64 cubes into 2x2x2=8 cubes you have increased the number of cubes by 7. You can repeatedly do this, either to more of the original 64 cubes, or even some of the new smaller cubes. Either way, dissecting 9 cubes into 8 will increased the number of cubes from 64 to 127.

In fact you can dissect a cube into almost any number of smaller cubes. There is just a short list of small numbers that cannot be achieved, the highest being 47.

Solution of the Week #572 - Quadrilteral

The most efficient way of enclosing the circles is first to arrange them so that each circle is tangent to the other two, draw an isosceles triangle around them, and then draw a fourth side, tangent to larger circle, but parallel to the shorter side of the isosceles triangle.

(The other candidate shape is placing the circles in a row and building a longer slimmer trapezoid around them, but for a quick sanity check, that shape would have to be greater in area than 196^2 + 144^2 + 144^2, which is 79888, and as we shall see the shape below is smaller).

 Using coordinate geometry I ascertained the dimensions of this trapezoid. I won’t bore you with the messy details, especially as there’s doubtless a far more elegant way to do it. In any case the base measures 336, the top 147, the sloping sides 337.5, the height 324 and therefore the area is 78246 square units.

Solution of the Week #571 - Crossing Triangles

Take the right-way-up large equilateral triangle. If we work out what the area of it is we can just subtract 3, 12 and 27 to find the hexagon area.

The relationship between area A and side length s of an equilateral triangle is A = sqrt(3)/4 * s^2

The reverse of this is that s = sqrt(4A/sqrt(3))

So the side length of the 12, 48 and 27 triangles are respectively approximately 5.264, 10.529 and 7.896, summing to about 23.689.

Then the area of the right-way-up large triangle comes out as exactly 243.

The hexagon area is therefore exactly 201

Solution of the Week #570 - Cube

There must be a better way to solve this but this is what I did:

 If we call the side length x, and the coordinates of the point in space relative to the middle vertex, (a,b,c) we can form four equations:

 a^2 + b^2 + c^2 = 583^2

(a+x)^2 + b^2 + c^2 = 637^2

a^2 + (b+x)^2 + c^2 = 713^2

a^2 + b^2 + (c+x)^2 = 727^2

Subtracting the first equation from each of the other three in turn gives:

2ax + x^2 = 65880

2bx + x^2 = 168480

2cx + x^2 = 188640

Rearrange each to isolate a,b,c

a = (65880 - x^2)/2x = 32940/x - x/2

b = (168480 - x^2)/2x = 84240/x - x/2

c = (188640 - x^2)/2x = 94320/x - x/2

Let’s square each of those expressions

a^2 = 32940^2/x^2 – 32940 + x^2/4

b^2 = 84240^2/x^2 – 84240 + x^2/4

c^2 = 94320^2/x^2 – 94320 + x^2/4

Adding them together and equating the sum to 583^2:

(32940^2+84240^2+94320^2)/x^2 – 211500 + 3(x^2)/4 = 583^2

Multiply every term by 4x^2 to eliminate the x^2 term in the denominator:

Solve the quadratic in x^2 to x^2 = 32400 or 2108356/3, so x is either 180 or 838.322929…, however in this second case the point would be inside the cube.

The cube’s side length is therefore 180