Solution of the Week #520 - Irregular Octagon

We can do this the easy way or the hard way.

The hard way involves setting up and solving a series of equations involving a,b,s and r in the above figure. And then noting that the area of the octagon is (2r)^2-2ab.

The easy way is to note that by drawing lines from each vertex to the centre, the octagon can be dissected into eight triangles, each of base s and height r. So the area is simply 4rs, which given s=17 and r=19 gives an area of 1292.

I recently used the fact that there are two ways to calculate the area to provide a proof of Pythagoras Theorem.

Solution of the Week #519 - Word Pairs

COPPER, CHOPPER, HOP, HOPE, FATTEN, FLATTEN, SPIT, SPLIT, RUSE, ROUSE

The extra letters spell HELLO.

Solution of the Week #518 - Four Rectangles

In the general case of two rectangles overlaid such that they have one corner in common and each has a corner coinciding with the side of the other rectangle, it will produce a pair of similar triangles as below. From this we can tell that C/A=B/D. Cross-multiplying shows that AB=CD, in other words the two rectangles have the same area. So in the original figure, all four rectangles have the same area. We can’t know the exact length and width of the red and blue rectangles, but since the green and magenta rectangles also share an edge, that gives us enough information to solve the puzzle. If we call the height of the magenta rectangle ‘x’, then the height of the green rectangle is x-14.

36*(x-14)=27x

9x=36*14

x=56

Solution of the Week #517 - Friday April 18th

If the days of the week are arranged in alphabetical order, Friday is first and Wednesday is last. For the months of the year April is first and September is last, and for the dates when written out in words, eighteenth is first and twenty third is last.

Solution of the Week #516 - Two Pairs of Circles

The answer is that the radius of the incircle is 7. I only really know that because I already knew that when I constructed the puzzle. I have an extremely messy and complicated way to find it only given the information in the question, which I won’t bore you with, but if you found it in a neat way please do let me know.

Edit: see comments on the puzzle page, neat solutions to be found there.

Solution of the Week #515 - Three Circles in a Trapezoid

Following the previous puzzles and the relationships between lengths of a right trapezium with an inscribed circle, if we were to follow that through to equations I believe we would end up with quartic equations awkward to work with.

More practically in this instance we can just use trial and error, and very quickly find that each trapezoid as we descend the figure is 50% bigger than the one above. So the horizontal lines measure 40, 60, 90 and 135. The sloping lines will then be 51, 78 and 117. And the vertical lines will be 48, 72 and 108, giving the overall height of 228.

Solution of the Week #514 - Two More Circles in Trapezoids

If we call the length of the middle horizontal x, we can use the formula for the radius of the incircles as before to determine the height of each trapezoid in terms of x.

Since we are told the angle on the right is a right angle, we can find two similar triangles above and below it (in fact they are congruent, but we don’t need to assume that for our solution).

So (x-5)(x-12)=10x/(x+5) * 24x/(x=12)

(x+5)(x-5)(x+12)(x-12)=240x^2

(x^2-25)(x^2-144)=240x^2

Let’s make a substitution and call x^2 y

(y-25)(y-144)=240y

y^2-409y+3600=0

y=9 or 400

x therefore is +/-3 or +/-20

In the diagram, it is clearly bigger than 12, so must be equal to 20.

Plugging this value into the height expressions above gives the overall height as 8+15=23.

Solution of the Week #513 - Two Circles in a Trapezoid

The trivial answer is if A=B=C=D, wherein A/B = 1.

To find the other solution, first consider that the top part of the shape is similar, in the mathematical sense, to the bottom part. Let’s say that the top horizontal has length 1, then the middle horizontal has some unknown factor x, and the bottom horizontal will therefore be x^2.

Next let’s look at the properties of a right trapezoid with a circle inscribed within it.

Splitting the lengths at the tangent points we call the radius of the circle R and the other two lengths S and T, forming a right angled triangle to combine them.

Using Pythagoras, (T+S)^2-(T-S)^2=(2R)^2, therefore ST=R^2.

Now let’s define S as U-R and T as V-R.

(U-R)(V-R)=R^2

UV-R(U+V)+R^2=R^2

UV=R(U+V)

R=UV/(U+V)

Therefore we can find the radius of a circle within a right trapezoid as simply the product of the top and bottom sides divided by their sum. The height of the trapezoid will be twice that.

And for completeness, the sloping side will be the sum of the top and bottom sides, less the height of the trapezium.

Let’s now look at the top trapezoid in the puzzle, where we have defined the top edge to be 1 and the bottom edge to be x.

The left side will be 2x/(1+x). That denominator is awkward, so let’s scale everything up by (1+x). We can do that because ultimately it is a ratio between sides that we are seeking.

Now remember everything in the bottom trapezoid is the same as the top one, but scaled up by x. We can therefore find expressions for A, B, C and D in terms of x:

A=2x^2

B=x+1

C=2x

D=x^3+x

We are told that A+C=B+D

2x^2+2x=x^3+2x+1

x^3-2x^2+1=0

We know that one solution (the trivial solution) is that x=1, which leaves the quadratic x^2-x-1=0. This has one positive solution: (1+sqrt(5))/2, or the golden ratio.

Finally, plugging this into the expressions for A and B gives a ratio for A/B of exactly 2.

So the solution is A/B = 2.

If we rescale back to letting B=1, A=2, C will be sqrt(5)-1 and D is sqrt(5). The diagonal line will have a gradient of -2. And of course the middle horizontal will be equal to the golden ratio.

Solution of the Week #513 - Two Circles in a trapezoid

The trivial answer is if A=B=C=D, wherein A/B = 1.

I couldn’t find an elegant way to get to the other solution, so I won’t attempt to present one. I’ll just tell you that the solution is A/B = 2.

If we let B=1, A=2, C will be sqrt(5)-1 and D is sqrt(5). The diagonal line will have a gradient of -2. The middle horizontal will be equal to the golden ratio.

Solution of the Week #512 - 4x4x4 Cube

The argument for the 3x3x3 version hinges on the fact that the central cube will have 6 newly cut faces, and therefore will require 6 slices to form, regardless of rearrangement. For the 4x4x4 cube, there are 8 inner cubes, each of which need 6 slices directly acting on them. This can be done as long as the first cut in each direction is the middle one, and then the two halves can be lined up to simultaneously perform the other cuts in that direction. So perhaps counter-intuitively, a 4x4x4 cube also only needs 6 slices.

Solution of the Week #511 - Seven Circles

They will have a radius of 2. The base of the triangle will be 40, but the other sides cannot be determined.

17 circles of radius 1 will fit along the baseline.

 In general I have found that if one circle has radius A and two circles have radius B, then n circles will have radius:

 r_n = AB/((n-1)A-(n-2)B)

 In our case A=5, B=4. When n=7, r_7 = 20/(30-20) = 2.

 Reverse engineering r_n = 1:

 1 = 20/(5(n-1)-4(n-2)) = 20/(n-5+8)

20 = n-5+8

n = 17

Solution of the Week #510 - Pair of Circles 2

As before, the largest possible pair of circles will be tangent to the longest side of the triangle: the 15 side.

Using Heron’s formula we can find that the area of the triangle is 84.

From this we can work out that the inradius of the entire triangle will be 4.

The version with two circles will effectively be a scaled down version of the full triangle with one incircle, split apart and a 2r wide strip placed between.

The ratio of the inradius to the base length is the same for this split-apart triangle as it is for the full triangle.

(15-2r)/r = 15/4

60-8r = 15r

60 = 23r

r = 60/23

Solution of the Week #509 - Pair of Circles

By Pythagoras, a 21, 28, 35 triangle has a right angle opposite the 35 side.

If we want the two circles to be as large as possible, they need to both be tangent to the long side of the triangle, each tangent to one of the other sides of the triangle, and tangent to one another. This is shown in the above diagram.

The small triangle in the middle has sides parallel to each of the sides of the overall triangle, and is therefore similar to it. We know the long side is twice the radius, therefore we can calculate the other sides in terms of r. The rest of the horizontal line can be calculated to ensure the overall length is the given 21. Likewise for the vertical side. Because of tangent lines from a point, these lengths can be copied to the hypotenuse of the triangle. We have the resulting equation for the length of the hypotenuse:

35 = 21-6r/5-r + 2r + 28-8r/5-r

35 = 49-14r/5

14r/5 = 14

r=5

And so therefore the radius of the circles is 5 units.

Solution of the Week #507 - Triangle in a Quarter Circle

Since two opposite vertices of the above quadrilateral are right angles, the quadrilateral is cyclic and we are able to make use of Ptolemy’s theorem, which states that the product of the diagonals is equal to the sum of products of opposite sides.

We know the two diagonals, and two of the sides. Let’s call the other two sides a and b. The area we seek is ab/2.

By Pythagoras: a^2+b^2 = (10*sqrt(2))^2 = 200

By Ptolemy: 10a+10b=140*sqrt(2)

a+b=14*sqrt(2)

Let’s square both sides:

(a+b)^2=392

a^2+b^2+2ab=392

But we already know the value of a^2+b^2

200+2ab=392

ab=96

Area = ab/2 = 48 square units.

Solution of the Week #506 - Hexagon and Circle

If you’re familiar with the intersecting chords theorem, we will make use of it here. However the chords we will use actually intersect outside of the circle. Our point is the lower left vertex.

Looking at the diagonal line across the hexagon, its total length is 4x (since a regular hexagon is simply six equilateral triangles), therefore the length to the far side of the circle is 3x, and we are given the distance to the near side of the circle: 1.

The other line going through the lower left vertex is the base of the hexagon, which is tangent to the circle. In this case the distance to the nearside and the distance to the far side are both x.

By the intersecting chords theorem, 1*3x = x*x

3x = x^2

Therefore x=3

To find what the radius of the circle is, draw a line straight up from the tangent point to form another pair of intersecting chords with the main diagonal. The main diagonal within the circle will be split into lengths of 5 and 3, and the vertical line will have lengths of sqrt(27) (half the height of the hexagon) and (2r – sqrt(27)). This eventually works out as the radius r = 7/sqrt(3) = 4.041…

Solution of the Week #505 - Linked Values

The angles between the sides will not change as a and b change, since it is always an isosceles triangle with sides in the ratio 2,2,1. I have dropped an altitude to the midpoint of the base, giving us a right-angled triangle, whose smallest angle is x.

sin(x) = (ab/4)/ab = 1/4.

By Pythagoras the length of the altitude is ab*sqrt(15)/4.

cos(x) = ab*(sqrt(15)/4)/ab = sqrt(15)/4.

To find cos(2x) we can use the double angle identity:

cos(2x)=(cos(x))^2-(sin(x))^2 = 14/16 = 7/8.

Now focusing on the upper right triangle that has angle 2x, hypotenuse a(b-1) and adjacent side b(a-1).

cos = adj/hyp

7/8 = b(a-1)/(a(b-1))

7a(b-1) = 8b(a-1)

7ab-7a = 8ab-8b

8b-ab = 7a

b(8-a) = 7a

b = 7a/(8-a)

7a+ab = 8b

a(7+b) = 8b

a = 8b/(7+b)

Let’s see what happens when we let b increase. As b becomes massive the 7 in the denominator will become insignificant, and a will get closer and closer to 8, but never quite reach it, so an open upper bound for a is 8.

ab must be greater than a for the line across the triangle to be within the triangle, therefore an open lower bound for b is 1. This corresponds to an open lower bound for a of 1 also.

So the integer values that a could take are 2,3,4,5,6 or 7. Calculating the corresponding values of b, it is only an integer when a is 4, 6 or 7

The three integer solutions are a=4,b=7; a=6,b=21; a=7,b=49.

Solution of the Week #504 - Sum and Product

By Pythagoras the long side of the triangle is 87. Since a+b is 100 we can change b to 100-a. The small triangle at the top shares an angle with the large triangle, and since both are also right-angled the two triangles are similar.

87/63 = (63-a)/(a-13)

87(a-13) = 63(63-a)

87a-1131 = 3969-63a

150a = 5100

a = 34

b = 100-a = 66

a*b = 34*66 = 2244

Solution of the Week #503 - Semicircle in a Triangle

If we split the triangle into two triangles by drawing a line from the apex to the centre of the semicircle, we can see that the ‘height’ of each triangle is equal to the radius. So the total area of the triangle is 87r/2 + 75r/2 = 81r

 We can also calculate the area of the triangle using Heron’s formula, which states that the area is equal to the square root of the product s(s-a)(s-b)(s-c), where a, b and c are the sides of the triangle and s is the semi-perimeter.

 s=(87+75+108)/2 = 135

s(s-a)(s-b)(s-c) = 135*48*60*27 = 10497600

Area = sqrt(10497600) = 3240

 Now we have two different expressions for the triangle’s area we can equate them:

81r = 3240

And then find the radius

r = 3240/81 = 40. 

Solution of the Week #502 - Almost a Square

Here are the solutions. I have placed within each triangle what the primitive form of that Pythagorean triangle is.