Solution of the Week #537 - Non-Prime Sequence

You might notice that each pair of adjacent terms has a GCD greater than 1, for instance 21 and 119 are both divisible by 7, and 697 and 4059 are both divisible by 41. In fact on further inspection, each term appears to be the product of adjacent terms of a second sequence: 1,3,7,17,41,99,etc. It looks as if the definition of this second sequence (after the initial two terms) is that each term is twice the previous term, plus the term before that. If that is confirmed to be the case, the only way a term in the product sequence can be prime is if one of the two factors is 1 (and the other is itself prime). This is only the case for the term ‘3’, and so therefore this is the only prime number that will appear in the entire infinite sequence.

 

But how do we prove that the given sequence is the products of adjacent terms of the second sequence?

That is equivalent to proving the claim that if a=2b+c, b=2c+d and c=2d+e then it follows that ab=5bc+5cd-de.

 

Let’s start with the product relation but substitute a=2b+c

(2b+c)b=5bc+5cd-de

2bb+bc=5bc+5cd-de

2bb=4bc+5cd-de

Next replace any instances of b with 2c+d

2(2c+d)(2c+d)=4(2c+d)c+5cd-de

8cc+8cd+2dd=8cc+4cd+5cd-de

2dd=cd-de

Finally replace c with 2d+e

2dd=(2d+e)d-de

2dd=2dd+de-de

Since both sides are evidently equal, that proves that the three relations are equivalent to the product relation.

 

Solution of the Week #536 - Find the Radius

First let’s figure out the radius of the semicircles. Lines from the ends of the chord to the centre of the circle will bisect the angles of the equilateral triangles, forming 30 degrees angles. From this we can work out that the perpendicular distance from the centre of the unit circle to the chord is 1/2.

We can now set up a Pythagorean triangle with the radius R of the semicircles as an unknown.

(1-R)^2 = R^2 + 1/4

1 - 2R + R^2 = R^2 + 1/4

1 - 2R = 1/4

3/4 = 2R

R = 3/8

 If the distance from the centre to the chord is 1/2, then so too is the distance from the chord to the top of the circle. Again we can set up a Pythagorean triangle, this time with the radius of the small circle as the unknown:

(3/8+r)^2 = (3/8)^2 + (1/2-r)^2

(3/8)^2 + 3r/4 + r^2 = (3/8)^2 + 1/4 – r +r^2

3r/4 = 1/4 - r

7r/4 = 1/4

r = 1/7

Solution of the Week #535 - Three Sets of Words

If the letters of the alphabet were put in rows of three then each set of words can be made from only the letters in a given column.

 

A    B    C

D    E    F

G    H    I

J    K    L

M    N    O

P    Q    R

S    T    U

V    W    X

Y    Z

 

Trying the same thing with two columns is a problem since all of the vowels A, E, I, O and U, and even the occasional vowels Y and W will all appear in the left hand column, and the right hand column would be reduced to onomatopoeic non-words like BZZZ and PFFT.

Solution of the Week #534 - Triomino Packing

The first thing to do is to establish the size of the square. If we assume that the L-triomino is made up of three unit squares, the diameter of the circle will be the hypotenuse of a triangle with base 5 units and height 3 units.

This diameter, and therefore also the side length of the square, is the square root of 34.

If we place an L-triomino in each corner, then tilt the square 45 degrees, how large a square can we fit in the remaining space?

The side of the whole square is sqrt(34), so the diagonal is sqrt(68). From this we need to take off twice the diagonal of a 1.5 x 1.5 square to find the side length of the inner square.

These combine to be sqrt(18). So sqrt(68)-sqrt(18) = 4.0036…

Just to be safe we need to confirm that the diagonal of the inner square doesn’t exceed the side length of the larger square, or else the corners will extend beyond the original square. This isn’t the case so we are safe to continue.

How many L-triominoes can we fit in this 4x4 square? If we place one in each corner we have space for a fifth in the middle. Added to the four we already placed means that overall we can fit 9 L-triominoes in the square of area 34 square units.

Of course this doesn’t prove that 10 or 11 isn’t possible (more than 11 is impossible as the area of the triominoes would be greater than that of the square), but such a proof is a little more involved.

Solution of the Week #532 - Relative Areas

If we assume that the red squares measure 1x1, then the red region has an area of 10 square units.

The next step is to establish the dimensions of the blue square.

The top left and bottom left triangles in the above figure are similar, so (2+x)/1 = 2/x. Cross-multiplying and solving the resulting quadratic we find that x = sqrt3-1, or about 0.732.

Using Pythagoras the side length of the blue square is therefore sqrt(8+2sqrt3), and the area of the full square is 8+2sqrt3.

From that we have to subtract the two triangles A and B. A has height 1 and base x. B has base 1+x and is similar to A. From this we find that A has an area of (sqrt3-1)/2 and B has an area of 3(sqrt3-1)/2.

This means that the visible blue region is also 10 square units.

The horizontal component of the lower right side of the blue square is the same as the vertical component of the lower left side, so 2 units, so the overall rectangle has sides 5 and 6. Therefore the green region also has area of 10 square units.

 The areas of the red, blue and green regions are all identical.

Solution of the Week #531 - Mountain Range

If you scale the entire figure by x in the horizontal direction and 1/x in the vertical dimension all of the areas will be preserved. By setting x to the fourth root of 3 (approximately 1.316) the equilateral triangles are all transformed into right-angled isosceles triangles with height half of their bases, so that a triangle with area n^2 will have height n and base 2n. Now it is simple to find the area of the trapezoid of base 19, and then subtract the triangles and half triangles to find the required area.

The trapezoid has area 19x7/2, from which we must subtract 16/2, 9, 1, 4 and 9/2.

The answer is therefore 40 square units.

Solution of the Week #530 - Rhombus Stack

Opposite sides of a rhombus are parallel and the same length, therefore the length and orientation of the lower left side of the square are identical to the right hand slope of the triangle with base 18. Similarly the lower right side of the square has the same length and orientation as the left slope of the 38 base triangle. We can now dispense with the rest of the diagram and only look at those two triangles, armed with the knowledge that the angle between them is 90 degrees. This also means that the length of the other two triangle bases was irrelevant, so long as they produced valid triangles.

Dropping perpendicular bisectors from each we get two congruent right angled triangle with legs 9 and 19.

Since the area we seek is simply x^2, we can use Pythagoras to find that the area of the original square is 9^2+19^2, which is equal to 442.

Solution of the Week #527 - Allesley Park part 3

A Linean Cycle is not possible for a cube.

If we mark the doubly traversed edges in red, there are two distnict ways to colour the graph, and since at each vertex we arrive and depart, and since the red edges are responsible for net zero arrivals-departures, so must the black edges. Without loss of generality we can add directionality to the black edges such that they form a loop. A valid path must go red – black – red – black etc, since not doing so would result in a red edge being traversed consecutively. With one of the arrangements of red edges, the black edges form two smaller loops and so we must try them both going the same way, or going different ways.

In the first configuration, if we start with AB we go ABFGCDHE(AB…) completing a loop before all edges have been traversed.

Likewise in the second: ABFEHGCD(AB…), and the third: ABFE(AB…)

I’ll leave it for the reader to make a Linean cycle for the tetrahedron, dodecahedron or icosahedron.

Solution of the Week #526 - Allesley Park part 2

It is possible to run four of the repeated segments bidirectionally, and only incur penalty points on two paths. The network of paths is topologically equivalent to the edges of a truncated tetrahedron, and the same-direction repeated edges will be opposite edges of one of the hexagonal faces. That being the case the ‘cheapest’ edges to take the penalty points for are BL and HI, incurring only 5 penalty points in total.

The way to achieve this is A-C-B-L-J-K-E-D-C-A-B-L-K-J-H-I-G-F-D-E-F-G-H-I-A (or its reverse, or beginning at any other point in the cycle).  

Solution of the Week #525 - Allesley Park part 1

You might be aware that to complete a route and return to where you started you need to visit each junction point an even number of times. However, each of the 12 junction points on the map has an odd number of paths coming from them, so we need to repeat some of the paths to change all of the odds nodes to even. This can be done in several ways but the one that has the least extra mileage is to repeat AC, BL, JK, HI, FG and DE.

So the best we can do is to run all the paths and return to where we started in 2.24 miles.

Solution of the Week #522 - Geometric Sequence

Unlike before, we cannot simply take the mean of each sequence and make them equal. The geometric means of the sequences would be equal, but we aren’t given the product of each sequence so that doesn’t help. As before we don’t know how many terms in the sequence, but we do know it must be of the form 6m+1.

The formula for the sum of a (finite) geometric sequence is:

S = a*(r^n-1)/(r-1) (r not equal to 1)

We are told our initial term is 1, so the a can be omitted from the formula.

For the main sequence the common ratio is simply r, and the number of terms n is 6m+1:

127 = (r^(6m+1)-1)/(r-1)

127(r-1) = r^6m*r – 1; let’s call r^6m = k

127r-126 = kr

k = 127-126/r

 

For our second sequence there are 3m+1 terms and the common ratio is r^2:

85 = (r^2(3m+1)-1)/(r^2-1)

85(r^2-1) = r^(6m+2)-1

85r^2-84 = kr^2

Let’s plug in the expression for k we found above:

85r^2-84 = (127-126/r)r^2

85r^2-84 = 127r^2-126r

 

42r^2 – 126r + 84 = 0

 

Solve the quadratic to find r = 1 or 2.

The formula for the sum of a geometric sequence specifies that r should not be exactly 1, but we’ll try it anyway if only to eliminate it as a possibility. The full sequence would just be 127 instances of the number 1. The second sequence would have 64 instances of 1 and therefore total 64, but this is not the case, so r=1 is not a valid solution.

If r = 2, the sequence goes 1,2,4,8 etc. The first 7 terms total 127. But if we want to be strictly systematic we know that k=127-126/2, so k=64. r^6m = k, so 2^6m = 64. m is therefore equal to 1. n, the number of terms in the sequence, is 6m+1, and therefore equal to 7.

Now to check the second sequence, 1+4+16+64=85, which is correct.

Finally to answer the question of the sum of the third sequence:

1+8+64 = 73.

Solution of the Week #521 - Arithmetic Sequence

We don’t know how many terms in the whole sequence but if all three sequences start and end at the same point, there must be both an odd number of terms and one more than a multiple of three terms. So there must be 6m+1 terms.

S is the sum of all 6m+1 terms.

88 is the sum of 3m+1 terms.

60 is the sum of 2m+1 terms.

All three sequences will have the same average since they are arithmetic sequences beginning and ending on the same terms, therefore:

S/(6m+1)=88/(3m+1)=60/(2m+1)

Taking just the last two terms and cross-multiplying:

88(2m+1)=60(3m+1)

176m+88=180m+60

4m=28

m=7

There are therefore 6*7+1=43 terms in the whole sequence.

Now just taking the first two terms and using the newly discovered value for m:

S*22=88*43

S=172

In all three sequences the average of the sequence will be 4. Therefore the middle (22nd)term of the full sequence will be 4.

If the first term is 0, and it needs 21 steps to get from there to the 22nd term, the common difference will be 4/21.