A, B, C and D are 3, 4, 5 and 2 respectively, since 34^5 = 45435424
Solution of the Week #282 - Heptagon Revisited
Using the fact that an angle in a semi-circle is always a right angle, we can see why the given circle passes through the midpoints shown. Similarly we can see why the smaller circle in the second figure also passes through midpoints of the heptagon. Now we only need to find the diameter of that circle. I only give you an overview of the solution for you to fill in the gaps if you desire.
We can find angle ACB easily enough, and since triangle ABC is isosceles, we can find the length of AC.
Triangle ABC is similar to triangle ACD, therefore AB:AC = AC:CD, so we know length CD.
By examining the angles triangle CDE is isosceles, therefore CE = CD.
Since the heptagon is regular, we know the angles within isosceles triangle CEF, so we can know the length of CF.
Moving to the next figure, GH = CE, and IJ = CF.
GHIJ and IJKL are similar trapezoids (sides and diagonals parallel between the two figures) therefore GH:IJ = IJ:KL.
Therefore if we know AB = 6296, we also know that KL = 1247 (to the nearest integer, 1247.000015… to a greater degree of accuracy).
If you’re interested, the exact ratio between AB and KL is:
(1+2sin(270/7))^2
Solution of the Week #281 - Unit Fractions
Since both 8 and 188 divide evenly into 43992, the original sum of three fractions can be written as (5499+234+1)/43992. This further cancels to 61/468. So we have:
61/468 = 1/a + 1/b multiply each term by 468ab to eliminate denominators:
61ab = 468b + 468a multiply each term by 61, and move everything over to the left:
61x61ab – 61x468a – 61x468b = 0 add 468x468 to both sides:
61x61ab – 61x468a – 61x468b + 468x468 = 468x468
So far we just seem to be needlessly complicating it, but in fact we’ve been leading up to the next step, which is to factorise the left hand side:
(61a-468)x(61b-468) = 468x468
Now here’s the really clever bit. Since a and b have to be whole numbers, the remainder when divisible by 61 of (61a-468) or (61b-468) will remain the same regardless of the value of a and b. Their ‘modulo 61’ value will each be 20. The point of doing this is that however we factorise the right hand side, the two parts that correspond to (61a-468) and (61b-468) must also have a value of 20, modulo 61.
468x468 = 219024 = 2x2x2x2x3x3x3x3x13x13, so there are plenty of options for pairs of numbers that multiply to make 219024:
1x219024, 2x109512, 3x73008, 4x54756, 6x36504, 8x27378, 9x24336, 12x18252, 13x16848, 16x13689, 18x12168, 24x9126, 26x8424, 27x8112, 36x6084, 39x5616, 48x4563, 52x4212, 54x4056, 72x3042, 78x2808, 81x2704, 104x2106, 108x2028, 117x1872, 144x1521, 156x1404, 162x1352, 169x1296, 208x1053, 216x1014, 234x936, 312x702, 324x676, 338x648, 351x624, 432x507, 468x468,
However we are only interested in numbers that are equal to 20 mod 61. 81 is such a number, and its complement 2704 is the only other.
So we can let 61a-468=81 and 61b-468=2704 (or vice versa). These solve simply to a=9 and b=52 (or vice versa), and so these are the only solutions to the puzzle.
To conclude:
1/8 + 1/188 + 1/43992 = 1/9 + 1/52
Solution of the Week #280 - Sum of Cubes
There are infinitely many solutions but the smallest is that a, b, c and d are (in some order) -11,-2,7 and 10.
Their sum is 4, and the sum of their cubes (-1331, -8, 343, 1000) is also 4.
I arrived at this answer partly by narrowing the scope of where I was looking, and thereafter trial and inspection. I remembered that ALL cube numbers are either 0,1 or 8 modulo 9 (in other words they all cube numbers are either a multiple of 9, or one away either side). You can verify this astonishing fact by just trying the first nine cube numbers and observing that this is in fact true for those, and because we are working in module 9, it is therefore true for all cube numbers.
For our four cube numbers to add to 4, they must all be of the +1 variety. This happens whenever the number to be cubed is 1 more than a multiple of 3, ie: 1, 4, 7, 10, etc and -2, -5, -8, -11, etc. So we only need to look at the cubes of these numbers. Obviously the trivial 1 appears in this list, as do each of the numbers in my answer: -11, -2, 7 and 10.
Solution of the Week #269 - Heptagon Figure
The angles marked in this figure are in units of 180/7°. So the internal angles in a heptagon will add up to 35, the internal angles in a pentagon will add up to 21, in a quadrilateral will sum to 14 , and in a triangle or on a line will sum to 7.
By definition all of the angles in the regular heptagon will be 5. From pentagon NBCDE, angle ENB is 1 (same for angle GPD).
Since H and K are midpoints of the heptagon, FGHK is a trapezoid, therefore HK is parallel to GF, and so angle HKE is 5 (same for JLD). The supplementary angles MKN and MLP are therefore 2.
From the inherent symmetry of the figure, and the angles in the quadrilateral NPDE, angles ENP and DPN are each 2. Therefore from triangle NMK, angle NMK is 3 (same for PML). From the pentagon MKEDL, angle KML is 1.
Finally, since NMK is isosceles, NM = MK (same for MP). So if MK is 7, NP = 14.
Solution of the Week #268 - Dice Game
It should come as little surprise that winning the first round gives you an advantage as the game continues. You are already more likely to win the second round, and if you do your advantage is further increased.
It is however by no means a foregone conclusion. Examining all of the possibilities the probability of the winner of the first round going on to win works out at about 63.5%, with the chance of the initial underdog seeing a reversal of fortune as the remaining 36.5%.
The exact values, should you be interested, are as follows:
Initial winner winning: 137009/215622
Initial loser winning: 78613/215622
The most likely specific sequence of results by a long way is for the initial winner to simply win every round. This occurs with a probability of 3315/10648 or just over 31% and so alone accounts for almost half of the initial winner’s possible routes to victory.
Solution of the Week #267 - Tetrahedron Game
As it rolls around the grid, the tetrahedron will always present the same face to particular triangle of the grid. In fact the grid can be subdivided into four classes of triangles, each of which can only be landed on by one particular face of the tetrahedron. That being the case, a selection of painted triangles is solvable if and only if it contains one each of the four classes of triangles. Grid 3 is the only such grid.
Interestingly, so long as the grid is solvable, it doesn’t matter where you initially place the tetrahedron, and you can also decide to end up on any particular triangle of the grid (even if it isn’t initially painted).
Solutions
As you know if you follow this site, for a few years I didn’t publish the solutions to my Puzzles of the Week, but starting this year I have published a solution each Monday for the puzzle of the previous Friday. This has proved popular and I’ve found in writing them that the solution is often more interesting than the puzzle itself. I’m very pleased that I made the decision to start including solutions, but I only wish I’d started earlier.
It is my intention to eventually fill in the gaps and provide solutions for all the previous puzzle, Squarespace allows me to back-date them so that they will slot in between the puzzles.
With currently 232 solutions missing, this is not a quick task. If there are any old puzzles you would like me to provide a solution for, just let me know and I’ll be very happy to bring them to the top of the to do list.
Solution of the Week #266 - Base Four Code
Anyone who stops learning is old, whether twenty or eighty. Anyone who keeps learning stays young. The greatest thing you can do is keep your mind young.
This has been variously attributed to Henry Ford and Mark Twain.
Solution of the Week #265 - Not Enough Information Surely?!
Solution of the Week #264 - What is the Angle?
Here are a couple of different approaches:
Firstly I’m going to make a copy of triangle ABC, rotated 90 degrees anti-clockwise around point A to form triangle AB’C’.
Since ACEC’ is a 35 by 35 square, we can easily find out what DE and B’E are: 21 and 20 respectively. This forms a Pythagorean triple with the hypotenuse, B’D, equal to 29.
However since BD is also (14+15=) 29, AB equals AB’ and AD is shared by both, triangles ABD and AB’D are similar and angle x is equal to the angle we seek.
But since angle BAB’ = 90 degrees (as B’ was formed by rotating B by 90 degrees), the angle we seek must be half of this, namely 45 degrees.
An alternative approach is to scale down the two parts of the puzzle triangle as below. This can be done because it doesn’t change the angle we are seeking to find.
Similar to before we can rotate the left hand triangle 90 degrees anti-clockwise. Now to complete a rectangle we just need a copy of the 5x2 triangle, which means the triangular space in the middle is a right angled isosceles triangle, so x is 45 degrees. Since x + ? = 90 degrees, ? is also 45 degrees.
Solution of the Week #263 - 24-gon
That was where my original solution ended, however following feedback after the puzzle went live on Friday, it isn’t necessarily clear that three equally spaced segments have a constant combined sum, so I have added proof of that. Since the edge length of the 24-gon is the same we only need to consider the perpendicular distance to the internal point. Since this three segments are equally spaced, if we extended the bases we would form an equilateral triangle, making it equivalent to the situation illustrated below.
Solution of the Week #262 - Hexagon Area
If you extend two of the lines until they meet, as shown, and then also draw a line from this new point to the point inside the hexagon you will form two new triangles with the same base and height as two of the given triangles. The sum of these, and therefore the area of the quadrilateral, will be 12. But this 12 is also the sum of the other given triangle and the new equilateral triangle formed by extending the lines. This equilateral triangle therefore has an area of 7. This is also one sixth of the area of the original hexagon. The ultimate answer is therefore 42.
Solution of the Week #261 - Su-Dot-U
Solution of the Week #260 - Semi-grams
ALERT + RATIO = RETAL + IATOR = RETALIATOR
METRO + SENSE = REMOT + ENESS = REMOTENESS
NOTED + ROAST = DETON + ATORS = DETONATORS
TIMES + GIANT = ESTIM + ATING = ESTIMATING
UPSET + SOUND = STUPE + NDOUS = STUPENDOUS
Alternatively you could swap the second halves of DETONATORS and ESTIMATING
Solution of the Week #259 - Triangle Area (Tough)
Imagine a couple of (differently) scaled down versions of the triangle, one where the corner square becomes area 1, and one where the sloping square becomes area 1. From using similar triangles and chasing the angle ‘a’ around the figure we can see that the areas of some of the other parts of the triangle will be the same across the two figures. We can call these triangular areas x, y and z.
(I should probably point out at this point, while I’m introducing lots of new variables, that I’ve no intention of ever calculating what a, x or y evaluate to. z on the other hand…).
Each of the triangles with area x, y or z has one right angle, one angle ‘a’, and one side of length 1. You can express x y and z in terms of trig functions of a:
x = tan(a)/2 ( = sin(a)/2cos(a) )
y = cot(a)/2 ( = cos(a)/2sin(a) )
z = sin(a)cos(a)/2
If you multiply x and y together you get simply xy = 1/4
If you add the reciprocals of x and y:
1/x + 1/y = 2cos(a)/sin(a) + 2sin(a)/cos(a) = 2((cos(a))^2+(sin(a))^2)/ sin(a)cos(a)
But since (cos(a))^2+(sin(a))^2 = 1 by definition:
1/x + 1/y = 1/z
Putting these two facts together we get (x+y)/(1/4)=1/z
4(x+y)=1/z
x+y = 1/4z
Now scaling the two triangles back up so that the square areas are 363 and 336 respectively, and equating the areas of the two triangles we find that:
Area = 363(1+x+y) = 336(1+x+y+z)
Substituting x+y = 1/4z
Area = 363(1+1/4z) = 336(1+1/4z+z)
I’m now going to multiply each term by z and group terms on one side:
336zˆ2 – 27z – 27/4 = 0
Using the quadratic formula we see that:
z = (27 +/- sqrt(27^2 + 27x336))/(2x336)
z = (27+99)/672 (since z is an area and must be positive)
z = 3/16
Plugging that back into one of the area formulae:
Area = 363(1+1/4z) = 363+484 = 847, which is the final answer.
Solution of the Week #258 - Su-Dot-U
First I used the dots around the edge to place the 1s in the top and bottom rows and first and last columns, allowing me to place some lines and find some places where there are definitely not lines.
Since there are no dots above or below the 1 in the first column, but there is a downward dot in the square above the 1, that square must contain a 3. This means the lower 5 squares of column 1 must be part of the same shape, along with one square from the second column. It must be the one in the second row as if there was a line separating it from column 1, then the dot tells us that square contains a 1, but it can’t as we already have a 1 in the second column. This completes the left hand shape and the dots now tell us exactly which numbers are where.
Similarly the third square on the top row must be a 3, one square from the second row must complete that shape. Since the squares in the third and fourth columns would lead to a second 1 in that shape, it must be the square in the fifth column. Using the dots and the numbers that have already been placed, the top shape can also now be fully filled in.
The two remaining 1s can now be placed. The 1 in the fifth column is interesting because the lack of dots tells us it is open on three sides. This effectively gives us the entire shape. We can place the 5 in this shape too, to complete the top row.
The dotted square in the fourth row cannot be a 1 as 1s are already present in that column and row. It also can’t be greater than 2 as there is a line in place 2 below it. It must therefore be a 2. The square two above it can only now be a 4. That and a 3 at the end of the second row complete that row.
The final two dotted squares are confirmed to contain a 2 and a 3 respectively. Through normal Sudoku rules the remaining numbers can be placed in the top right shape. We can therefore complete the last two columns.
Since there are two 2s in close proximity in the bottom right corner they must be in different shapes. The one on the bottom row must have lines above and to the right. The bottom right shape must extend a further square above the 3 on the bottom row, but cannot extend further in that direction as it would disrupt the 2 in the third column. We can therefore complete the lines of this shape, and now there is only one way to extend the middle shape without splitting the bottom shape into two pieces. All lines have now been placed.
The 3 and 5 in the third row can be placed since if it were the other way around there would be a dot in the 3 square. The 6 then completes this shape. This disambiguates the 5 and 6 in column four which means completing the second column is now straightforward.
Solution of the Week #257 - Two Squares
There are a few ways of solving this but the most intuitive is probably to dissect the figure into identical triangles and then simply count them. There are 18 in the tilted square and 16 in the lower square, so therefore their areas are in a ratio of 9:8.
Solution of the Week #256 - Su-Dot-U
I start by using the dots at the edge of the grid to place the 1s in the first two rows and the bottom row, and the extra lines that that leads to. This fully determines the shape of the top left region, so I am also able to place the 2 in the top row. The 2 can also be placed in the second row, and the lower line from this fully determines the shape of the top right region. I can now use the dots to place the 2 in the third row, and the lower line from that. The remaining two 1s can be placed, along with the extra lines they tell us. It is now straightforward to place all of the other lines. This fixes the 2, 3 and 4 of the lower right shape (and so therefore also the 5). The final 2 can be placed in the bottom corner. The dot tells us the position of the 3 in the backwards L region, and the lack of dot tells us where the 3 goes in the S region. The last two 3s can now also be placed. From here on in, there is always a row, column or shape that only needs one more number to be complete.
Solution of the Week #255 - Three Squares in a Triangle
I don’t know if having a related puzzle last week is a help or a hindrance, as my approach to solving, for what it’s worth, is not at all similar to last week.
I notice that the given areas are each the sum of two squares (5=4+1, 8=4+4, 17=16+1). In general, if the area is a^2+b^2, then it’s at least possible that each side of the square is a units in one direction, and b units in a perpendicular direction.
On a hunch I hopefully set out lattice points of a coordinate system, and sure enough, each of the corners of each of the squares lies exactly on a lattice point. It is then straightforward to work out what the coordinates of the triangle corners are and then calculate the area of the triangle is 64.
