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 #508 - Jigsaw
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.
Solution of the Week #501 - Rectangle and Quarter Circle
The first thing we can do is to work out the radius of the quarter circle. We can form a right triangle on the figure that has hypotenuse r and legs 6 and (r-2).
Using Pythagoras this works out as r=10.
If we make a copy of the figure, rotate it 90 degrees and place it next to the original figure we can see what the length and width of the rectangle are in terms of the radius.
Since we already know r=10, we find that the rectangle measures 14 x 4 units.
Solution of the Week #500 - Rectangle
If we call the length of the rectangle ‘x’ and the height of the rectangle ‘y’, then the value we are seeking is 2x+2y-xy.
If we just look at one half of the figure, replacing the circle with some radii and showing tangent lines as equal we get:
We now have a right-angled triangle with sides x, y and (x+y-2).
x^2+y^2 = (x+y-2)^2
x^2+y^2 = x^2+2xy-4x+y^2-4y+4
Subtract x^2+y^2 from both sides:
0 = 2xy-4x-4y+4
Subtract 2xy-4x-4y from both sides:
4x+4y-2xy = 4
Divide all terms by 2
2x+2y-xy = 2
The value of 2x+2y-xy is what we are seeking, and so the answer is 2. There isn’t enough information to determine x and y, but the difference between the perimeter and the area will be constant.
Solution of the Week #499 - Octagon to Square
The above shows how the dissection would be achieved. If we call the side length of the octagon s and the radius of the small circles r, then we can say that an octagon with side s, subtract an octagon with side 2r must have the same area as a square with side 2s.
The formula for the area of a regular octagon is (2+2sqrt2)s^2. The area of the inner octagon is 4(2+2sqrt2)r^2, and the area of the full square is 4s^2.
(2+2sqrt2)s^2 - 4(2+2sqrt2)r^2 = 4s^2
Let’s use this to find a relationship between s and r.
r = s/2(sqrt(3-2sqrt2))
If we let the square be a unit square, s is 1/2 and therefore r is (sqrt(3-2sqrt2))/4. The area of each of the circular holes is pi*r^2 = pi*(3-2sqrt2)/16.
The shaded part of the square is 1 less the area of five holes,
1-5*pi*(3-2sqrt2)/16
which is approximately equal to 0.83156…
So the square is about 83.156% shaded.
Solution of the Week #498 - Three Week Odyssey
Say I walked x miles on the first day. In the first week I will have walked 7x+21 miles. In the second week, 7x+70 miles. In the third week, 7x+119 miles.
These distances need to form a right angles triangle, so we can use Pythagoras Theorem:
(7x+21)^2 + (7x+70)^2 = (7x+119)^2
49x^2+294x+441 + 49x^2+980x+4900 = 49x^2+1666x+14161
49x^2-392x-8820=0
x^2-8x-180=0
(x+10)(x-18)=0
So the distance travelled on the first day was either -10 miles or 18 miles. Since the distance needs to be positive the answer is 18 miles. In total I walked 147+196+245 = 588 miles.
A shortcut to the solution would be to note that the difference between week 2 and week 1 is the same as the difference between week 3 and week 2: 49. So we are looking for a scaled up version of a primitive Pythagorean triple in arithmetic progression. Since 3,4,5 is the only such triangle we need only scale this up by a factor of 49 to find our distances.
Solution of the Week #497 - Six Regions
Here are the coordinates of all of the junctions. The marked point has coordinates of (588,1064).
Solution of the Week #496 - Four Towns
Since Gunton is due north of Kipton and Lawton is due east, Gunton-Kipton-Lawton form a right angle. Since we don’t know the speed we can’t yet directly equate the given times and the given distances. Let’s say we are travelling at ‘s’ miles per minute. We can create the following diagram:
We can use the law of cosines to give an expression for cos(a) and cos(90-a) in terms of s:
Cos(a) = (15^2+(78s)^2–(102s-15)^2)/(2*15*78s)
Cos(a) = (17-24s)/13
Cos(90-a) = (16^2+(78s)^2–(82s-16)^2)/(2*16*78s)
Cos(90-a) = (41-10s)/39
But, cos(90-a) is also expressible as sin(a), and (sin(a))^2+(cos(a))^2=1
(51-72s)^2+(41-10s)^2 = 39^2
5284s^2-8164s+2761=0
s can be 1/2 or 2761/2642
If we use this second figure and plot the resulting distances, Torton would be way to the north-west, not generally north-east as stated, so the speed is 1/2 mile/minute or 30mph. The distance between Kipton and Torton is therefore 39 miles.
Solution of the Week #495 - Smallest Triangle
If we call the base b, and the other sides a and c, such that a>c (a can’t be equal to c as the mirror line would be parallel with the base). We want a and c to be as close as possible, so that the angle of the mirror line is as shallow as possible. Since they are integers the smallest difference is 1, so let’s say a-1=c. With that being the case the length that we want to be 45 will be bc (in general it would be bc/(a-c)). To minimise the perimeter we want b to be roughly double c. If we had said we wanted the dashed line to be 50 instead of 45, we could achieve that exactly with b=10, c=5 and a=6.
Here the closest is b=9, c=5, a=6, with a perimeter of 20, and area of 10*sqrt(2).
Solution of the Week #494 - Five Dice
The fact that none of the rearrangements are even means that each individual roll is odd.
The fact that we are told that b>d and we are expected to figure out the entire sequence must mean that there are four rolls of one number and one roll of a second number.
The gives few enough possibilities for trial and error. We need to try each of five rearrangements of 13333, 15555, 31111, 35555, 51111, 53333 for divisibility by 7.
The only one that is never divisible by 7 is 35555, therefore the sequence of dice rolls was 55535.
For completeness, the sequences that are never divisible by 7 are:
11111 and its multiples 22222 33333 44444 55555 66666
11112 and its multiples 22224 33336
The result of subtracting those three from 77777: 66665 55553 44441
Every other sequence of dice rolls can be arranged to be a multiple of 7.
Solution of the Week #493 - Grid Fill 3
As we are told that there are 5 black squares that limits the possible arrangements. Since each of the four edges of the grid must contain an even number of black squares, they must appear in an odd number of corners. That leaves just seven distinct arrangements of black squares: four with one black corner and three with three black corners. One of the seven arrangements of black squares leads to a symmetrical solution. Five more arrangements don’t lead to any solutions.
Here is the asymmetrical solution:
And if you’re interested, here are the other 18 solutions I found:
