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

Solution of the Week #568 - Dice

Each sequence will continue until I roll either a 1 or a 6, so will end with a probability of 1/3, therefore making the average length 3. The fact that the sequences that end in a 1 instead of a 6 are discarded is irrelevant for the average sequence length.

Solution of the Week #567 - Basketball Tournament

The most a team could play is 12, if they played all of the other six teams twice each.

The fewest possible is 6, if they played each of the other teams once. So on the face of it it looks as if the seven different possible numbers of games are 6,7,8,9,10,11,12.

However if a team played each other team once there can’t be team that played every team twice, as that would include the first team, leading to a contradiction.

So there are two scenarios regarding number of games played: either 6,7,8,9,10,11 or 7,8,9,10,11,12. If you were expecting seven different possibilities, reread the question and find that only the other six team’s numbers of games have to be different, and Nuneaton Predators number of games can be a repeat of one of those. That is the key to finding the number of games played by the Predators.

Let’s just consider the 7,8,9,10,11,12 scenario. One team, let’s call it A, must play every team twice, to become a team with 12 games. Another team, say B, plays A twice and every other team once, totalling 7 games. Team C plays A twice and B once, and every other team twice to be the team with 11 games. Team D plays teams A and C twice each, and team B and every other team only once, to total 8. Team E plays B and D once each and every other team twice, making 10. Teams F and G have played A, C and E twice, and B and D once, making 8 so far. If they play each other once, they will both be on 9 games, but if they play each other twice they will both be on 10 games, along with team E. In that second situation, no team has played exactly 9 times, so that can’t be the correct situation. Those two teams play 9 games each, and one them must be Nuneaton Predators.

If you do the same thing considering the 6,7,8,9,10,11 scenario, you will again find that the repeated number of games is 9, and so one of those teams must be Nuneaton Predators.

Nuneaton Predators played nine times.

Solution of the Week #566 - Central Area

No matter the proportions of the rectangle, it can be stretched by some factor in the horizontal and by the reciprocal of that same factor in the vertical, preserving the are of each of the regions. That being the case we can assume the rectangle is a square.

The upper right region will have sides of sqrt(2). The other two area 1 regions will have sides of ‘s’ and ‘s’-sqrt(2). We can therefore show that the side length ‘s’ must be equal to (sqrt(2)+sqrt(10))/2, or sqrt(2) times the golden ration.

The overall area of the square will then be sqrt(5)+3, and so the central area is sqrt(5).

Solution of the Week #564 - Comparea

If we assume the square is a unit square, then the triangular region has area 1/2.

The radius of the large circle is sqrt(2)/2

The radius of the smaller circle is 2-sqrt(2).

The area of the crescent region is therefore: (4*sqrt(2)-11/2)*pi, or about 0.493.

Therefore the triangle area is very slightly larger than the crescent area.

Solution of the Week #563 - Powers

a+b = s

a*b = p

What is a^4 + b^4 in terms of s and p?

 

Let’s start by squaring (a+b):

s^2 = a^2 + 2ab + b^2

Let’s isolate (a^2+b^2)

(a^2+b^2) = s^2 – 2p

Let’s square both sides:

a^4 + 2a^2b^2 + b^4 = s^4 – 4ps^2 + 4p^2

Again let’s isolate (a^4+b^4)

a^4 + b^4 = s^4 – 4ps^2 + 4p^2 – 2p^2

 

a^4 + b^4 = s^4 – 4ps^2 + 2p^2

 

And of course this works, and is real, even if the choice of s and p mean that a and b must be complex. For instance if the sum and product were 1 and 2 respectively, the sum of the fourth powers would be, using the formula, 1. But if you calculate a and b you would find messy expressions involving i and the square root of 7.

Solution of the Week #562 - Zig Zag Sequence

We saw previously how a related sequence 1,2,2,3,3,3,4,4,4,4 etc could be expressed as a square root rounded to the nearest integer.

a(n)= round(sqrt(2*n))

Next I’m going to square each term in that sequence:

1,4,4,9,9,9,16,16,16,16,etc.

b(n)= (a(n))^2

Our target sequence 1,3,2,6,5,4,10,9,8,7,etc. Can be expressed in terms of n and b(n):

c(n)= b(n)+1-n

For instance, the fourth term will be 9+1-4=6.

So the full expression will be:

c(n) = (round(sqrt(2*n)))^2 + 1 – n

 

If you were curious about what, say, the millionth term would be:

C(1000000) = (1414)^2 + 1 – 1000000 = 999397

Solution of the Week #560 - Concentric Circles

You could of course, assuming the actual radii are not important, let the smallest circle vanish to a point, at which point it becomes clear that the radii of the other two circles which must be 65 and 33, become the hypotenuse and one leg of a right triangle, with x as the other leg. x is therefore 56.

If you want to be more rigorous, let the radii be a<b<c.

33^2 = b^2 – a^2

65^2 = c^2 – a^2

x^2 = c^2 – b^2

65^2 – 33^2 = c^2 – a^2 – b^2 +2^2 = c^2 – b^2

x^2 = 65^2 – 33^2

x = 56

Solution of the Week #559 - Minimal Combinations

(2k)! must have at least three more factors of 5 within it that twice k! (the same goes for factors of 2 as well, but that will be achieved much earlier).

If k=63, 63 has 14 factors of 5 (floor(63/5) + floor(63/25)).

2*63 = 126, 126 has 31 factors of 5 (floor(126/55) + floor(126/25) + floor (126/125)).

For any smaller k (2k)!/(2*k!) is less than 3.

Solution of the Week #558 - Isosceles Trapezoid

The area of the lower region is 1/2*54*d

The area of the upper region is 1/2*30*e

Since those areas are equal we know that e/d = 54/30 = 9/5

Since the triangles with hypotenuse a and 13 are similar we know that a/13 is also 9/5.

a is therefore 23.4.

If the lower area is x, then the total area below the dashed line is 9x/5.

Since the total area must be 3x, the area above the dashed line must be 6x/5. It follows from this that b/30 must be 6/5.

b is therefore 36.

Since 54 = 36 + 2c, c=9.

To find e we use Pythagoras on the ace triangle. e = 21.6.

The total area of the trapezoid is therefore 21.6*(54+36)/2 = 972

Solution of the Week #556 - Triangle and Semicircle

If we call the radius ‘r’ and the height from the centre of the circle to the diagonal line ‘a’, then by Pythagoras we have:

49^2 = r^2 + a^2

Using intersecting chords theorem, 49*31 = (r+a)(r-a)

49*31 = r^2 – a^2

 If we take the first equation and subtract the second we get:

 49(49-31) = 2a^2

a^2 = 49*9

a = 21

 The height of the triangle is twice this distance ‘a’, and is therefore 42. 

Solution of the Week #555 - Regular Nonagon

The area of the third region is 4.

I’m going to show that this is the case a little back-to-front, by presupposing that the overall area of the nonagon is 9, and then proving that that is in fact the case.

First of all I hope it’s evident that there is exactly one size of nonagon for which, if you form a triangle of area 2 using one side as the base and a vertex somewhere opposite, that the region to the left has area 3. That being the case, if we find a size of nonagon that works, that is THE solution.

So next I’m going to claim that the overall area is 9 (and therefore the missing are is 4) and show that that does indeed work.

Drawing lines from each of the vertices to a point in the exact centre will divide the nonagon into 9 equal regions of area 1, therefore the combined area of the a regions is 2, of the b regions is 3, and of the c regions is 4.

The red lines are all parallel – the left and right ones are parallel because of the inherent symmetry of a regular nonagon, and I have defined the middle red line to be parallel to the others and passing through the centre of the nonagon.

In the second figure we can see that if we moved the point that all of the vertices of the nonagon were connected to (the hub if you will), but ensured it stayed on this middle red line, that the combined area of the b regions would remain 3. This is because the left shaded region is unaffected by the change, and the other part of the b region is a triangle with a fixed base and a fixed altitude.

The exact same rationale applies to the c region, which will remain 4 regardless of where on the middle red line the hub is.

Finally the a region will remain area 2, because the overall area of 9 is unaffected, and the a region includes everything not in either the b or c regions. This being the case we can let the hub move all the way to the top edge of the nonagon, meaning that the a region becomes a single triangle, and the b and c regions remain 3 and 4 respectively.

Solution of the Week #554 - Fiftieth Birthday

For the person to have their 50th birthday on the same weekday as they were born, they will have to have been alive through 13 leap days. (50 extra days, one for each year, plus 13 leap days = 63: 9 weeks).

Ordinarily this will happen if they were born in the two year period immediately preceding a leap day. If they were born in the two year period immediately after a leap day they will only have lived through 12 leap days. However the situation is complicated by the fact that century years are only leap years if the number of the century is divisible by 4, so 1700, 1800 and 1900 were not leap years. So for people born in the latter half of the 1600s, 1700s or 1800s, it’s impossible to live through 13 leap days in their first 50 years. This means that the probability of the person celebrating their 50th birthday on the same weekday as they were born is roughly 5/16.

This deals with every case except for if the person was born on the 29th of February, which is difficult to account for as it is unclear exactly when they would celebrate their birthday in a non-leap year (which their 50th would certainly be).

 

Solution of the Week #553 - Unknown Primes

The only solution is: (2+1/3)(2+1/5)(2+1/7) = 11 (or some ordering of a,b,c = 3,5,7).

7/3 * 11/5 * 15/7 = 11

 

(2+1/a)(2+1/b)(2+1/c) = d

((2a+1)/a)((2b+1)/b)((2c+1)/c) = d

(2a+1)(2b+1)(2c+1) = abcd

All of the three terms on the left are odd, so a b c and d must all be odd primes.

Each term (2+1/p) must therefore be between 2 and 2 1/3, and so d must be between 8 and 12.7, and there is only one prime number in that range: 11, so d=11

One of the (2p+1) terms must be a multiple of 11. Without loss of generality we can say it is (2a+1). a must therefore be of the form 5+11n. 5 is prime, so let’s explore that possibility:

Let’s say that a is 5 and d is 11.

This leaves (2b+1)(2c+1)=5bc. Let’s say (2b+1) is divisible by 5.

b must be of the form 2+5n. 2 is prime, however we know we are looking for an odd prime. Let’s try 7. That leaves:

15(2c+1)=35c

30c+15=35c

5c=15

c=3, which is prime, so everything works.

I’ll leave it to the reader to satisfy themselves that this is the only solution.

Solution of the Week #552 - Five Answers

ab + c = 12

bc + a = 12

ca + b = 12

The fact that we are told there are five answers is a clue as to what they might look like. If there was a solution with a b and c all distinct, that would form 6 answers alone, so can’t be possible. If there is a solution where two of the numbers are equal and the third is different, that would account for three answers. Assuming that is the case the remaining two answers must be when all three values are the same.

The solutions are that either two of the numbers are 1 and the other is 11 (this accounts for three of the five answers), or all three are 3, or all three are -4.

Solution of the Week #551 - Seven Bricks

If you divide each brick into four as shown, each of those quarters will have the same proportions as the original brick. We can say without losing generality, that the height of each of these new bricks is 1. If we call the length of each of the mini bricks ‘x’, this value will be the proportion we are seeking.

If we make a Pythagorean triangle in two different ways as shown, their hypotenuses will both be equal to the circle radius R.

R^2 = (2x)^2+3^2 = (3x)^2 + 1^2

4x^2+9 = 9x^2+1

5x^2 = 8

x^2 = 8/5

x = sqrt(8/5) = ~1.265

so the proportion of each brick’s length to its height is ~1.265