Solution of the Week #399 - What's Wrong

That OAB and BDC are equal presupposes that AC and BD are parallel. This is an unwarranted assumption.

Instead we should work out what the angle AOB is. If OAB is indeed 32 degrees, then to complete the 180 degrees of the triangle OAB, angle AOB should be 58 degrees.

Alternatively, if we assume CDB is indeed 32 degrees, then from the central angle theorem, AOB is equal to twice 32: 64 degrees.

So both marked angle being 32 degrees leads to a contradiction.

If OAB is 32 degrees, BDC is 29 degrees, whereas if BDC is 32 degrees, OAB is 26 degrees. The only time they can be equal is if they are both equal to 30 degrees.

Solution of the Week #398 - Find the Radius

If we draw a line from the origin, directly through the centre of the circle and to the point where the arc and the circle meet, that line is clearly equal to R, but it is also equal to (sqrt(2)+1)*r, where r is the radius of the circle. So we have R = (sqrt(2)+1)*r, and if we square both sides (you’ll see why in just a moment) we get:

R^2 = (2*sqrt(2)+3)*r^2

We can also form a right-angle triangle using the origin and the line of length 1, to get:

R^2 = (2r)^2 + 1 = 4r^2 + 1

So we have two alternative equations for R^2, which we can equate:

(2*sqrt(2)+3)*r^2 = 4r^2 + 1

Making r^2 the subject gives:

r^2 = 1/(2*sqrt(2)–1)

Which we can make the denominator rational by multiplying top and bottom by 2*sqrt(2)+1:

r^2 = (2*sqrt(2)+1)/7

We can find R^2 by using either of the relations we worked out previously:

R^2 = 4(2*sqrt(2)+1)/7 + 1

R^2 = (11+8*sqrt(2))/7

Finally R= sqrt((11+8*sqrt(2))/7) ~ 1.7854

 

Now for the 3-point solution, let R^2 = x

x = (11+8*sqrt(2))/7

but we would like it to look like the quadratic formula (-b +/- sqrt(b^2-4ac))/2a , so the first thing we need is an even number in the denominator, and we also need to bring the 8 under the square root:

x = (22+sqrt(512))/14

So, a = 7, b = -22, and b^2-4ac = 512, 484 - 28c = 512, therefore c = -1

So the polynomial in x is:

7x^2 – 22x - 1

And back in terms of R is:

7R^4 - 22R^2 - 1

This polynomial has four roots, however two are complex, and one is negative, and so the only positive real root is the value for R that we previously found.

 

 

Solution of the Week #397 - Starting Squares

16974441796

1 = 1^2

16 = 4^2

169 = 13^2

169744 = 412^2

16974441796 = 130286^2

 

1,4 or 9 would satisfy the first part

16 or 49 would satisfy the next part

169 is the only number that satisfies the next part

To proceed we need a square number that lies between 169000 and 169999. By taking the square root of both those numbers, we find the only whole number between them is 412, therefore 412^2 = 169744 is the only number which satisfies the next part.

Similarly, we take the square root of 16974400000 and 16974499999 and find only one whole number between them: 130286. Therefore the answer is 130286^2 = 16974441796.

Solution of the Week #395 - Vaguely Defined Pentagon

The first thing to determine is where on the shape are the three right angles. There are only two possible cases: the two non-right angles are adjacent, or they are not.

Let’s consider the case where they are adjacent, as above. Clearly t>r and s>p. The only way this is possible is if q is the second shortest length. q is the hypotenuse of a triangle whose other sides are (s-p) and (t-r). One of these will be equal to 2 and the other 3, and so q equals sqrt(13). This is not a whole number, and so therefore this case is eliminated.

In this second case, the dashed line is simultaneously the hypotenuse of q and r, and of t and (s-p). Checking through the cases where t is the shortest, second shortest, or third shortest side, there are no solutions. I’ll demonstrate this by assuming that it is the third shortest. As I’ve shown s>p, we’ll consider q=n, p=n+1, t=n+2, s=n+3, r=n+4.

The square of the dashed line = n^2 + (n+4)^2 = (n+2)^2 + 2^2

Expanding each squared bracket gives:

2n^2+8n+16 = n^2+4n+8

Clearly the left hand side is always greater than the right, for any positive whole number n.

Next let’s consider when t is the longest length:

p=n, q=n+1, r=n+2, s=n+3, t=n+4.

The square of the dashed line = (n+1)^2 + (n+2)^2 = 3^2 + (n+4)^2

n^2+2n+1 + n^2+4n+4 = 9 + n^2+8n+16

n^2 – 2n – 20 = 0

Using the quadratic formula, and eliminating the case when n is negative, we get n = 1+sqrt(21). Clearly not a whole number.

Finally let’s look at the last remaining case, where t is the second longest side. r=n, q=n+1, p=n+2, t=n+3, s=n+4.

The square of the dashed line = n^2 + (n+1)^2 = (n+3)^2 +2^2

2n^2+2n+1 = n^2+6n+13

n^2 – 4n – 12 = 0

n = (4 + sqrt(64))/2 = 6

So we now know that the side lengths are 6,7,8,9,10 starting at r and going anti-clockwise. Finally, to find the area, we work out the area of the trapezium to the left of the dashed line and the triangle to the right: 81+21 = 102. And that’s the answer!

 

Solution of the Week #394 - Coin Paradox

Statements 1, 2 and 3 are true.

Statement 1: HTH and HTT will each happen 1/8 of the time.

Statement 2: at some point HT will come up on consecutive throws. The next throw will be H or T with equal probability and so HTH and HTT will have an equal chance of coming up.

Statement 3: the average number of throws to see HT is 4, and as we have seen, the game will finish on the next throw regardless of what it is, so the average number of flips ending in either HTH or HTT is 5.

Statement 4: this seems like it should be true, but it isn’t. On average it will take me 10 throws to see HTH but my friend only 8 throws to see HTT. This is because if my friend reaches HT, she will either finish on the next throw, or have an H to start off a new attempt at HTT. If I’m on HT I will either finish on the next throw or be back at square one.

 

Interestingly, if we make the target sequences differ earlier, for instance HHH vs HTT, then even statements 2 and 3 will be false. HTT would win 3 times out of 5, and the average number of flips when HHH wins would be 5 2/5, whereas the average number of flips when HTT wins would be 5 11/15.

 

Solution of the Week #393 - Two Point Oh Eight

Let’s begin by seeing what happens with four digits, five digits, etc.

With four digits there are two numbers that work: 5252 and 5772.

Since in each case abcd/dcba is equivalent to 52/25, abcd must be a multiple of 52 and dcba must be a multiple of 25. The same multiple in fact.

Let’s take a look at exactly what that multiple is for the first few:

52/25 = 1

572/275 = 11

5252/2525 = 101

5772/2775 = 111

52052/25025 = 1001

57772/27775 = 1111

You might notice that the length of each of these factors is always one less than the number of digits of the final number, because multiplying by 52 or 25 increases the digits by 1. You might also notice that each number is palindromic, because the denominator needs to be the reverse of the numerator. And that it only uses digits 1 and 0, because if you had any digits greater than 1, there would be carrying, which would disrupt the digits of the numerator and the denominator in different ways.

For the 25-digit numbers of the question, we need to find all palindromic 24-digit numbers that are made up of 1s and 0s.

The first digit needs to be a 1 (otherwise the number would not be 25 digits long). The next 11 digits can be either 0 or 1, and the remaining 12 digits are defined by the fact that it needs to be a palindrome.

So we have 11 binary decisions to make. 2^11 = 2048. Therefore there are 2048 25-digit numbers which, when divided by their reversal, equal 2.08. You’ll forgive me for not listing them all, but I’ll pick one at random by flipping a coin:

100010110010010011010001:

5200525720520520572520052/2500252750250250275250025 = 2.08

 

 

Solution of the Week #392 - Shaded Areas

If we let the dimensions of the rectangle be 2x and 2/x, then whatever the value of x, its area will be 4.

This means that the radii of the three semicircles will be:

x, 1/x and (x+1/x)

The area of a semicircle is (pi*r^2)/2, and we want the area of the larger semicircle, less the area of the smaller two. For simplicity we can take out the pi/2:

Area = pi/2 * ((x+1/x)^2 – x^2 – 1/x^2)

Area = pi/2 * (x^2 + 2(x/x) + 1/x^2 – x^2 – 1/x^2)

x gets cancelled out

Area = pi/2 * (2)

Area = pi

And that’s the answer!

Solution of the Week #391 - Five Towns 2

The original puzzle was solved using Pythagorean triples. It’s impossible to find a Pythagorean triple where none of the numbers is divisible by 3.

Instead we can use the cosine rule, which is a kind of generalised version of Pythagoras theorem.

 

a^2 + b^2 - 2ab*cosC = c^2, where C is the angle opposite the side c. (When C is 90, cosC is zero and that whole term disappears).

 

Clearly if a b and c are integers, 2ab*cosC needs to be too. cosC can be anything from 1 to -1. We don’t care if C is rational (in whatever units: degrees, radians, etc), only that cosC is rational.

We need two adjacent central angles to add to 180 degrees. It turns out, conveniently for our purposes, that if C and C’ add to 180 degrees, then cosC and cosC’ add to zero.

 

So now we just need to find a convenient rational value for cosC, for example 1/13, then find a triple a,b,c that satisfies

a^2 + b^2 - 2ab/13 = c^2

none of which is a multiple of 3, for example 13,20,23

and other triples that satisfy

a^2 + b^2 + 2ab/13 = c’^2

with again, no multiples of 3, for example 10,13,17 and 13,40,43

Making a double size copy of 13,20,23 and also doubling 10,13,17 gives us four triangles of alternating central angles, which fit together as shown:

I thank Philip Morris Jones for this solution.

There is one more solution with a total mileage of less than 400 miles. It is based on a value of cosC of 1/5, and triples of 4,5,7 and 7,10,11, with roads leading from the centre with lengths of 50,40,28 and 35, and around the outside of 70,44,49 and 55, giving a total mileage of 371.

Solution of the Week #389 - Strings

There are two overall cases: the 3 digit in the sum could be made from a 1 in the first number and a 2 in the second number, or vice versa.

In the first case, that means the first four digits of the first number must also be 1s, and so the first four digits of the second number must all be 3s. We have no choice. However for the ‘444’ at the end of the sum, we have some choice. The first of those 4s must be 2+2 (as otherwise the first number wouldn’t have any 2s). The last digit must be made of 3+1. However the middle of those 4s could be either, giving us two possibilities.

If instead the 3 in the sum is made of 2 in the first number plus 1 in the second number, now the second half of the sum is completely determined and we have some choice towards the start of the number, three possibilities in fact.

So altogether the five possibilities are:

 

11111223+

33332221

 

11111233+

33332211

 

11122333+

33321111

 

11222333+

33221111

 

12222333+

32221111

 

Solution of the Week #388 - Ten Pin Bowling

As the letter frames are generated randomly from a scrabble set, I can play along myself. I personally achieved a modest score of 114. I began well with a decent spare on frame 1 but frame 2 was a bit of a horror show. On both frames 4 and 7 I made judicious use of the blank to get spares. On frame 5 I only scored 5+4 but there is a scrabble-legal ten letter word available for a strike: DETOXICANT. I didn’t get that without help, so can’t fairly claim it for my own score. However I did score over 100 so I can be happy with my performance.

Solution of the Week #387 - Five Circles

If you start with the smallest triangle and ensure it is a Pythagorean triple, then the radii of all of the circles is bound to be rational, and so can always be scaled up by an appropriate factor to obtain whole numbers. Beginning with a larger triangle and working backwards will not necessarily ensure this.

The best solution is found by starting with a 5-12-13 Pythagorean triple. The radii of the tangent circles for this are found by the semi-perimeter less each of the triangle edges, so 10-3-2.

Using Pythagoras on the second triangle, (10+3),(3+x),(x+10), we find that x needs to be 39/7. We scale up all the numbers we have by a factor of 7 to maintain whole numbers.

On the third and final triangle we now have (70+39),(39+x),(x+70), we find that x is 4251/31. Scaling everything up by a factor of 31, the five radii are:

2170, 434, 651, 1209 and 4251, for a total of 8715.

We might have used a different triangle to begin the construction, but we would have found that the procedure fails when the ratio between the radii of the zero-th circle and the first circle (2170 and 434 in our figure) is 4 or less. So for instance the 3-4-5 triangle would not have worked.

Solution of the Week #386 - Optimal Cone

The formulas for the surface area and volume of an equal right cone are:

Area = pi*r^2*(sqrt(5)+1)

Volume = 2/3*pi*r^3

We want the difference to be a maximum:

A-V = pi*r^2*(sqrt(5)+1)-2/3*pi*r^3

Let’s differentiate:

pi*2r*(sqrt(5)+1)-2/3*pi*3r^2

2*pi*r*(sqrt(5)+1-r)

The (Area-Volume) will be at a maximum when the above expression is equal to 0, or more specifically, when the term in brackets is equal to 0, which is obviously when r = sqrt(5)+1.

To answer the original question, the surface area will be pi*(sqrt(5)+1)^3 ~ 106.464.

Solution of the Week #385 - Odd Prime Cycle

Believe it or not, there are only two possible starting numbers that eventually returns to themselves, 5 and 13:

 

As we saw in the example:

5+1 is divisible by 3

3+2 is divisible by 5

 

13+1 is divisible by 7

7+2 is divisible by 3

3+3 is divisible by 3

3+4 is divisible by 7

7+5 is divisible by 3

3+6 is divisible by 3

3+7 is divisible by 5

5+8 is divisible by 13

 

Any other starting number will at some point reach a power of 2, which has no odd prime factors, without first getting back to the starting number.

For example, the longest sequence (for at some points you have a choice of odd primes) beginning with 101 is: 101 – 17 – 19 – 11 - 5 – 5 – 11 – 3 – 11 – 5 – 3 – 7 – 19 – (32).

Solution of the Week #384 - Mystery Numbers

Let’s first assume that all a,b,c,d are at least 2.

If they were all exactly 2, the triple sum would be 4x2x2x2=32, and the pair sum would be 6x2x2=24, so the triple sum is greater. If we increase any of the numbers, say we change a to 2+e, the triple sum would be increase to 32 + 12e , but the pair sum would only increase to 24 + 6e. In short, the pair sum can never catch up with the triple sum. So our assumption that all numbers were at least 2 is false. Without loss of generality we can let a=1. The equation then becomes:

 

bc + bd + cd + bcd = b + c + d + bc + bd + cd

 

And so bcd = b + c + d

 

Looking for a trio of positive integers whose sum is the same as their product, the only candidate is 1,2,3.

So if our a,b,c,d are (in some order) 1,1,2,3, the triple sum and the pair sum are both equal to 17.

 

Solution of the Week #383 - Smallest Square

Intuitively it makes sense that the basic shape of inner region to be a square with rounded corners. I’ll leave it to the reader to try to prove this.

We need to know what the radius of the rounded corners needs to be.

We can do this by considering a circle, letting the radius vary, and finding when the value of (perimeter – area) is at a maximum.

Perimeter = 2*pi*r

Area = pi*r*r

Perimeter – Area = r*(2-r)*pi

As pi is a constant, we just need the value of r where r*(2-r) is at a maximum. Clearly this is achieved when r=(2-r), so when r=1.

So now we need to find a square with radius=1 rounded corners, whose area is equal to its perimeter. The best way is to split the shape into regions:

The four quarter circles combine to form an entire circle, and then we have an inner square and four rectangles.

Area = pi + 4x + x^2

Perimeter = 2*pi + 4x

Making these equal tells us that x^2 = pi, so x = sqrt(pi).

The side length of the bounding square is therefore sqrt(pi)+2 = 3.772…