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 #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 #449 - Swap Grid

The reason the number grid takes 20 swaps is that it can be split into five cycles of 5 (for instance, 1,8,15,17,24 is a cycle). A cycle of n cells can be solved in n-1 swaps, so each of the cycles takes 4 swaps to solve.

In the letter grid, you want to find smaller cycles if possible. It turns out you can find four cycles of 2 letters each (where a single swap would solve both letters), three cycles of 3 letters, and two cycles of 4 letters, as coloured below. This will take 16 swaps in total.

To advance the solution, swaps should only be made within those minimum cycles (in the number grid, since 1 and 15 were in the same cycle, swapping them advanced the solution, without solving either letter in the swap, although it did change a 5-cycle into a 2-cycle and a 3-cycle).

In the letter grid, if any swaps were made within the 2-cycles or 3-cycles, letter would be solved, so we need to look only at the 4-cycles. Swapping one pair within each of the two 4-cycles as shown will turn each 4-cycle into two 2-cycles, advancing the solution without solving any letters, so the answer to the bonus question is 2.

 

Solution of the Week #445 - Functional Equation

As we’re hoping to evaluate f(4) we might try letting x = 4 initially:

2*f(-3) – f(4) = 540

But now we need to know the value of f(-3), so:

2*f(1/2) - f(-3)= -405

We plough on:

2*f(6/5) - f(1/2) = 135/2

2*f(5/3) - f(6/5) = 162

2*f(9/4) - f(5/3) = 225

And just as we start to lose hope:

2*f(4) - f(9/4) = 1215/4

 

So now we have 6 equations in 6 unknowns, let’s multiply the second equation by 2, the next by 4, the next by 8 etc:

 

2*f(-3) - f(4) = 540

4*f(1/2) - 2*f(-3) = -810

8*f(6/5) - 4*f(1/2) = 270

16*f(5/3) - 8*f(6/5) = 1296

32*f(9/4) - 16*f(5/3) = 3600

64*f(4) - 32*f(9/4) = 9720

 

Adding all of these together cancels out all of the f() terms except for f(4):

 

63*f(4) = 14616

Therefore:

f(4) = 232