Solution of the Week #470 - Word List

They can all be followed by a pair of consecutive letters of the alphabet to form a six-letter word:

SCARAB     THOUGH     NAPALM     SOLEMN     TECHNO     DOLLOP     DIVERS     DIVEST     GALAXY

They are placed in order according to the letters that have been added.

Of the four words in the second list, one can also be followed by a pair of consecutive letters: PARA -> PARADE, and therefore belongs between SCAR and THOU in the list.

The other three could be followed by a pair of consecutive letters in reverse order: BECKON, CLEFTS, PASTED.

Solution of the Week #469 - Temperature Teaser

Both scales are linear, so we can (mentally) draw straight lines between the data points given in the question, but we currently have two lines to think about, the Fahrenheit and Celsius. However since we are only interested in the point where they are positive and negative by the exact same amount, we can think of a single line which is the average of the two (or the sum of the two would also work), and find the point where it equals 0.

So we have 16 for the first data point and -40 for the second. If we mentally split that into seven ‘chunks’ of 8 degrees each, we can see that we want to be 2 chunks away from the point where Celsius is 0, where 7 chunks would get all the way to -40. So Celsius will be 2/7 * -40, or -80/7, which is -11 3/7 degrees. Predictably, at that point, Fahrenheit will be 11 3/7 degrees.

Solution of the Week #468 - Quarter Circles

Calling the radius of the top left circle ‘r’, and the missing length we seek ‘x’, we can work our way around the rectangle, and express each of the radii in terms of r and x. We can form two right angles triangles and form their Pythagorean equations as below:

(1) x^2+(3-r)^2=(3+x-3r)^2

(2) x^2+r^2=(6-x-3r)^2

Taking the difference of these two, the x^2 terms cancel:

(3-r)^2-r^2=(3+x-3r)^2-(6-x-3r)^2

Factoring each side as a difference of squares:

(3-2r)(3)=(3+x-3r-6+x+3r)(3+x-3r+6-x-3r)

(9-6r)=(-3+2x)(9-6r)

If (9-6r)=0, r=3/2, but that would mean the top right circle would disappear entirely, so that’s not possible.

Since (9-6r) is not equal to 0, we can divide both sides by it:

1=(-3+2x)

4=2x

x=2

 And that’s the answer!

Solution of the Week #467 - Sequence

One part is obvious: that the initial digit increases with each term, so the missing term will start with a 6. The other part is that the product of the digits of each number is the same number but with the initial digit omitted, eg 4x3x8x4=384. Specifically each term is the smallest it could be (excluding the use of 0). So the missing term will be 612.

11, 236, 315, 4384, 5175, 612, 7735, 8128, 9135

Solution of the Week #466 - Perimeter

The above figure represents one of the 1x1 squares, and how it interacts with the edge of the 3x3 square represented by the line through BD. Because of the fact that the inner point of the square F lies on a corner of the middle square of the figure we can say that FC is equal to 1 in length. ABCF is a kite, and therefore AB=BC. Likewise CDEF is a kite, and so CD=DE. The overall change to the perimeter from this square is that we lose BD, but gain AB, DE, AF and FE. But BD is just AB+DE so they cancel out, and AF=FE=1, so overall we gain 2. Since there are four such squares, we gain 8 compared to the original 3x3 square which has a perimeter of 12. And so therefore the overall perimeter = 20.

Solution of the Week #465 - Circle Array

There’s likely a more elegant way of doing this, but this was my approach. I’ve rotated the figure and put in place a coordinate system with the origin at the centre of the lower left circle. Without loss of generality I’m using unit circles. By Pythagoras the top right circle is at (10,0) as part of a 3,4,5 triangle. Since the distance between adjacent circles is 2, and we know the triangles formed with the horizontal are in the ratio 3,4,5, we can easily find the coordinates of the circles next to the origin circle, and by extension, any circle we wish. The red line is the critical line that goes from (0,1) to (10,1). The first two circles it passes through are centred 0.6 off the line. The third circle is indeed tangent as appears from the figure. The other two circles it passes through are centred 0.2 off the line. Using Pythagoras we can find the lengths of the chords to be 1.6 and 0.8*sqrt(6) respectively. The overall proportion then is 4/25(2+sqrt(6)) or just under 0.72.

Solution of the Week #464 - Mystery Word Game

After each guess, the player is told whether each individual letter in that position in the word is earlier or later in the alphabet than the guessed letter.

After three guesses she knew the first letter was C-H, the second A-H, the third E-R, the fourth A-D and the final letter U-Z. DELAY obviously fits this, along with some other possibilities such as FANCY, which I hid in plain sight in the question.

I based this puzzle on the game ‘Wordle Peaks’ which I play every day. You can find it at: https://vegeta897.github.io/word-peaks/

Solution of the Week #463 - Pentagon into Kites

For the solution below I have started with a large rhombus, then three large kites from left to right, leaving a scalene triangle. I have then constructed the incircle of that triangle and drawn lines from the incentre to meet the three sides of the triangle at right angles, forming the final three kites. Upon inspection, all seven kites are different in both size and shape.

Solution of the Week #462 - Rectangle into Kites

Start by drawing a semicircle whose diameter is on the short side of the rectangle. Draw a line tangent to this semicircle from one of the far vertices, dividing the rectangle into a triangle and a quadrilateral. Draw the incircle of the triangle. Draw lines from the centre of the incircle to each of the three points where the incircle touches the triangle. Draw a line from the centre of the semicircle to the point where the tangent line touches the semicircle. Et voila, a square and four different kites.

There is a special case when the ratio of the sides of the rectangle is (1+sqrt(2))/2, when the upper and left kites are identical. We avoid this problem by drawing the semicircle on the long side instead in that case. Another special case is when the rectangle is a square, when the lower left and mid left kites are identical. This cannot be avoided by using the other side of the square, which is why the rectangle was specified in the question as non-square.

By the way, it’s a nice little fact that the largest kite, the right one in the diagram, will always be exactly half the area of the overall rectangle.

Solution of the Week #461 - Tangents

Looking at the general case above, each tangent line a,b,c,d is by definition at right angles to the radius of the circle, r, so we can define a,b,c,d in terms of e,f,g,h,r using Pythagoras:

a^2=e^2-r^2

b^2=f^2-r^2

c^2=g^2-r^2

d^2=h^2-r^2

We can then also define e,f,g,h in terms of the perpendicular distances from the centre of the circle to the edges of the rectangle j,k,m,n:

e^2=j^2+n^2

f^2=j^2+k^2

g^2=k^2+m^2

h^2=m^2+n^2

Next I’m going to use the above to find an expression for a^2+c^2:

a^2+c^2=j^2+n^2+k^2+m^2-2r^2

And also for b^2+c^2:

b^2+d^2=j^2+k^2+m^2+n^2-2r^2

Both these are equal to the same thing, so a^2+c^2=b^2+d^2

Returning to the numbers in the puzzle:

10^2+15^2=6^2+?^2

?^2=289

?=17

Solution of the Week #460 - Neighbours

In 1990, Leoni 29, Joshua 11.

In 2024, Leoni 63, Joshua 45.

Solution of the Week #459 - Eight Unit Squares

We can place two pairs of similar triangles within the shape as shown above. The red triangles are clearly in the ratio of 3,4,5. It’s less obvious what the ratio of the sides of the blue triangles is, until you construct a grid as below which demonstrates that if you want to add an angle to the small angle of a 3,4 right triangle to total 45 degrees, that new triangle will have legs in the ratio of 7 to 1. (This generalises to a triangle with legs a,b pairing with a triangle with legs a+b,|a-b|).

So c/b=7, c/a=4/3 and a+b=1. From this we can work out that a=21/25, b=4/25 and c=28/25. By Pythagoras, d is therefore 4/5*sqrt(2). By similar triangles, e/5=c/d, so e=7/2*sqrt(2).

Since the triangle is right-angled and isosceles, its area is ((d+e)^2)/2, which equals exactly 18.49.

Solution of the Week #458 - Phone Charger

You would need 6 chargers to form a ring as per the diagram below:

Solution of the Week #457 - Sloping Square

The first thing we need to is to establish the length and width of the rectangle, let’s call them x and y respectively. We can make use of the fact that the small triangle above the 196 square is similar to the large triangle to the right of it. Since 14^2=196, the side length of the square is 14, and so:

(x-14)/14 = 14/(y-14)

(x-14)(y-14) = 196

xy-14(x+y)+196 = 196

But we already know that xy=882, as that is given in the question,

882 = 14(x+y)

x+y = 63

Knowing the sum and product of x and y we could solve as a quadratic, but I’m going to use the similar method of middle and difference. m is the middle of x and y, and is therefore 31.5, and d is the amount by which x is greater than m and y is less than m.

(m+d)(m-d)=882

m^2-d^2=882

d^2=m^2-882

d=10.5

And so therefore x and y are 42 and 21 respectively.

Now looking at the other half of the rectangle, the triangles surrounding the sloping square are also similar to the two in the lower half, and will have legs in the ration of 42 to 21, or 2 to 1.

Looking along the diagonal, if we call the side length of the sloping square ‘s’, we find that 2s + s + s/2 makes up the diagonal, which has an overall length of sqrt(42^2+21^2) = 21*sqrt(5).

So 3s/2 = 21*sqrt(5)

s = 6*sqrt(5)

The area of the sloping square is s^2, which is 180.

Solution of the Week #456 - Trigonometry Conundrum

Draw a line between the top of the left triangle and the top of the right triangle.

The length (d) of this line is √(a2+b2), and the angle (α) at the top of the left triangle is arctan(b/a).

The angle between this new line and the vertical is (α+θ).

Sliding this line down by the height of the right triangle forms a triangle with the left and lower lines, without changing the angle or the length d.

Therefore arcsin(c/d) = (α+θ), or θ = arcsin(c/d)- α

Or written out in full:

θ = arcsin(c/√(a2+b2))- arctan(b/a)

Solution of the Week #455 - Triangle

Let’s call the a+3 length ‘c’ and the base ‘B’.

The area of the triangle is half base times height, and that can be done in two different ways, so ac = 4B.

Using Pythagoras, c^2 + a^2 = B^2

Let’s expand (c-a)^2:

(c-a)^2 = c^2 + a^2 – 2ac

But we know that (c-a) is equal to 3, and we know the values of c^2+a^2 and ac in terms of B, so:

9 = B^2 – 8B

Solving this quadratic we get B is 9 or -1. As it’s a length it must be positive, so the base B is equal to 9.

If you’re interested, the value of a is 3(sqrt(17)-1)/2, which is about 4.68.

Solution of the Week #454 - Heptagon Circle Maximisation

So to solve the heptagon puzzle, we would hope to reduce it to this case: a bite made of two equal lines, which we can then solve by letting the vertex lie on the circle centre. We want to deform the four edges of the heptagon to two equal lengths, whilst maintaining the two endpoints where it meets the circle, and also maintaining the area of the heptagon.

If I draw an line between two non-adjacent vertices, then draw a parallel line through the vertex between them, and extend the next edge of the heptagon to meet that parallel line, and then move the middle vertex to that new intersection, then the area is preserved, but the number of edges has been reduced. I can do the same thing again to reduce the cut to two edges:

Finally to make those two edges equal, I draw a line between the two fixed vertices, drop a perpendicular bisector, and find the intersection between that and the parallel line through the remining internal vertex.

Finally, centring the circle on this point relative to the heptagon and rescaling the figure so that the arc length is equal to 1, we do indeed find this to give the maximum area, which is approximately equal to 0.0942943...

Solution of the Week #453 - The Long Game

You will find that you have to take the second day off, since you can’t play the same letter as day one, but you also can’t play another card for the first time. At the very end of the game, when you only have one card to play, you must again take a day off, since it will either be following the 6th playing of the same letter, or the 7th playing of a different letter. However these are the only days off you will need to take, and so you can play all 98 cards in 100 days.

For example (underscore denotes a day off, spaces are inserted for readability):

A_AB ABC ABCD ABCDE ABCDEF ABCDEFG BCDEFGH CDEFGHI DEFGHIJ EFGHIJK FGHIJKL GHIJKLM HIJKLMN IJKLMN JKLMN KLMN LMN MN_N

Solution of the Week #452 - Find the Radius

Firstly notice that the two smaller triangles are identical, and that the height of each of them is the same as the height of the rectangle. If we call the radius of the blue circle ‘r’, then a small triangle is 4r tall. Let’s work out what the base is:

Using Pythagoras: (ar+3r)^2 = (ar+r)^2 + (4r)^2

Divide throughout by r^2 and multiply the brackets:

a^2+6a+9 = a^2+2a+1+16

4a = 8, a=2, therefore the base of the small triangle, and so also of the rectangle, is 3r.

Now let’s look at a small section within the rectangle:

Using Pythagoras again on the right-angle triangle in the middle of this figure:

r^2+18r+81 = r^2 + (9/4)r^2-27r+81

18r = (9/4)r^2-27r, since we know r is not equal to 0 we can divide throughout by r:

45 = 9r/4, r = 20.

So the radius of the blue circles is 20.

Solution of the Week #451 - Answer Sequence

ELF – FILE – LIFER – RIFFLE – FIREFLY (or the reverse order)