On Friday I posted this puzzle….

Yesterday I saw a column in the middle of town. It was exactly 200 feet high and 16 feet 8 inches in circumference. Someone had wrapped a spiral garland around the column exactly five times. What was the length of the garland?

If you have not tried to solve it, have a go now. For everyone else, the answer is after the break….

Imagine that the column was made from a large sheet of paper that has been rolled into a tube. Next, imagine cutting that sheet of paper into five smaller column, each of which had the same height. Finally, imagine flattening out just one of these mini-columns. The garland would run as a diagonal across the piece of paper. So, the height of the mini column would be 40 ft (a fifth of 200), and the length would be 16ft 8 inches, making the diagonal 43ft 4inches. As a result, the entire garland would be five times this, or 216ft 8 inches.

Did you solve it? Any other solutions?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle** (UK here and USA here) and on the **iBookstore** (UK here in the USA here).

### Like this:

Like Loading...

*Related*

Yes. No.

That seems a strange way to look at it from my eyes: I just treated the column as a giant roll of paper, which I then unrolled along with the garland. Since it was wrapped five times around, the garland becomes the hypotenuse of a triangle 200 ft high and (16′ 8″)*5 wide. Convert to inches, apply Pythagoras, and you’re done.

Yup, that’s the way I did it too.

I figure its easier to picture the ‘unwrapping’ if you consider just one turn of the garland?

This was my way of visualising it also.

That’s exactly what I did. Although I got the units a bit mixed up, but when converted back into feet the answer came out the same. Stupid Imperial system…

but that way the answer comes 200 ft 8 in which is incorrect

Got it.

The way to arrive at it is to realize that upon unrolling the column/garland five times, you’ve got a right-angle triangle with legs of 200 ft and 5 x the circumference of 16 2/3 ft.

The mildly interesting result is that the length of the garland (the hypotenuse) is 200 ft + 1 x 16 2/3 ft, which (in units of 16 2/3 ft), corresponds to a Pythagorean triangle with sides of 12, 5, and 13 units.

Strictly speaking, the garland would actually have to be a bit longer because its finite thickness would hold it away from the column surface a bit, thereby making the

effectivecircumference slightly larger.Note that the answer turns out to be equal to the height plus circumference. This is just a co-incidence though, due to the particular ratios of the original numbers. Most other staring values will give you a garland length which does NOT equal the sum of circumference and height.

Looking at the number Richard gave, if you take the ratios of those measurements of the triangle, (by dividing each measurement by 3 feet 4 inches), then you end up with side ratios of (5, 12, 13). This is a pythagorean triad, and additionally, the longest 2 sides differ in value by 1.

So, by finding other pythagorean triads that differ in their longest sides by a value of 1, you can calculate some other initial values which will give a similar result for the garland length (that is, the garland length equals the sum of circumference and height).

Using the triad 3,4,5 as a start, you must multiply the 4 by 50 to get a column height of 200 metres. The circumference is therefore 50 metres, and you have to wrap 3 times. Work that out properly, and you can see the garland would have to be 250m long.

Using the triad 9, 40, 41, you must multiple the 40 by 5 metres to get a column height of 200 metres. The circumference is therefore 5 metres, and you have to wrap 9 times. Work that out, you get a garland length of 205 metres.

Actually, you can start with other right triangles, that aren’t pythagorean triads, and still end up with the garland length equaling the sum of the circumference and height, but any other right triangles will contain non-integers, which will make the question and answer look ugly…

nice

Actually after a few more calcs, I found that it doesn’t get THAT messy without pythagorean triads. Turns out if you choose a number of wrappings, let’s call that “w”. We’ll assume our column is always 200m high.

Then, the circumference which will give a nice answer (where garland length equals sum of circumference and height) is given by the following:

circ = 400 / (w^2 – 1)

For instance, with 7 wrappings, you get a circumference of 8.333333…

Continuing that train of thought… A similarly simple way to find if such a wrapping is possible:

1 + 2 * (Column Height) / (Column Circumference)

is a perfect square.

The number of wrappings will just be the square root. To take Richard’s example…

note: 16 feet 8 inches converts to 50/3 feet

1 + 2 * (200) / (50/3)

= 1 + 2 * 12

= 25

So the number of wrapping is root(25) = 5

Another example is a column of height 35 with circumference 2:

1 + 2 * (35)/(2)

= 1 + 35

= 36

the number of wrappings for this particular case is 6.

The interesting thing to note is the column height and circumference need not be even rational numbers, so long as their ratio is rational.

Hmm. I imagined turning the column horizontal and rolling it 5 revolutions along the ground, unwrapping the garland on the way to get the garland laid out as a straight line that I could solve with pythagoras. Richard’s explanation seems baffling to me, though this could just be the way my brain works.

Alternatively, it’s fairly easy to just set this up as a line integral.

Pray tell.

And here’s me thinking the answer was 42. Last time I refer to Douglas Adams, lol.

The pythagoras 5,12,13 triangle and the use of feet and inches (12s) give the elegant solution of 200 feet and 200 inches.

I forgot to check the puzzles last Friday, so, I give it a got today, before reading the solution. Took maybe half minute being the worst part how to convert inches into feet (Now, that’s a difficult puzzle to me :-)).

I got the same result, using a slight different path: I used the Pythagoras Theorem on the length of the column and 5 times the circumference. In retrospective, I think Richard’s way of getting the solution to be more easy to explain and understand.

That’s the answer I got, but it still logically doesn’t make sense to me. In essence, the final length is the same as the column height, plus one circumference.

To my mind, the garland has gone around the column 5 times, not just once, so it doesn’t _seem_ right, although the maths proves that it is…

I know what you mean. It’s one of those crazy things in maths where you can get the right answer and check it a thousand times, but it still doesn’t feel right.

But then I guess that’s why we have maths, the mind can be easily fooled, but if you calculate correctly you will always get the right answer! 🙂

I think that’s the whole point of Pythagoras.

a² + b² = c² which leads to

a + b > c which can be translated as

The shortest route between two points is a straight line, not zig-zag.

I was thinking 3D on friday and got several stupid answers including one that was shorter than the column. Once I’d rolled it out on saturday it was easy. What a long thin column.

I did it considering the line was a helix (i.e. a spiral stair case)

L = (H^2 + C^2)^(0.5)

L = length

H= height of each complete rotation (200/5 in this case as you can see 5 rotations on the garland)

c= circumference

Lots of examples online if you google it.

Yes, I found it the same way. But I didn’t count the rotations on the column in the picture because I believe the picture is for illustrative purposes as it does seem to have two garlands per column, so one garland would have two and a half wraps. But the problem does state that the garland is wrapped around the column exactly five times, so it works out as descirbed above.

Also after computing the above forumla you will get the length of one rotation (43.334 feet). You need to multiply the answer by 5 to get the total length of 216.67 feet or 216′-8″.

For me, the fact that it came down to a 5, 12, 13 pythagorean triple was one of the two interesting points about this puzzle (the unrolling to a flat surface with a straight diagonal being the other, obviously). I’m disappointed Richard doesn’t find it worth mentioning. After all, someone had set up the numbers carefully to get this property.

Did any of you notice that given the illustration, there either a) is actually two garlands wrapped 2,5 times round OR b) the one garland goes horizontal behind the column.

In case a) I would calculate using the aforementione way of “unwrapping the column” and having the garland as the hypotenuse.

In case b) I would be using the method Richard gave with the execption that the length of the garland would be calculated with hypotenuse of 1/4 circumference + 1/2 of the circumference.

If I didn’t a brainfart here somewhere.

Passed on this one. It looked too complicated for a math-deficient soul to attempt. So I was right about that anyway.

Why do you need to separate it into five smaller columns? I just used one big sheet of paper to get the answer.

After converting the 16 feet 8 inches into 16.66666666 feet and doing the math that way I realized it was easier to work in inches. 200 for the circumference and 2400 for the height.

Okay, I didn’t think of this, I just thought that the column is round and Pi is needed to calculate circumference…but the question of how many actual complete rounds the garland really makes? Hmm…okay, nope, Pi is not applicaple. The circles are elongated since the garland reaches upwards.

Pythagoras again. 🙂 Just don’t make me calculate anything. I always confuse the numbers even though I understood how to make the calculation. And the feets are just something very alien to me. Why can’t you just have centimeters and meters?!

I solved it the same way as Richard, by splitting it into 5 parts, but the answer is the same. (Although I first I thought I had it wrong because my answer was 2,600 inches … then I did a head slap … doh! … and did the conversion to feet.)

However, this puzzle is completely unrealistic because there is no way the stringers of this garland would have or perhaps even could have (without damaging the column) made it perfectly straight, so it would have drooped and the real answer would have been significantly larger.

Why was it done in an antiquated imperial system? It puts those of us on a different continent who use a more logical system at a disadvantage.

His answer is so concise – it took me forever to justify my answer to my boyfriend. I struggled to find a word to describe one “spiral” on the column – since it’s not really a “spiral” . . is it?

I settled on the word “rotation”. #NerdOut!

[…] Answer to the Friday Puzzle…. On Friday I posted this puzzle…. […]

Hooray! I got it. Basically, I used the method described by Richard.

While of course the Pythagorean theorem is the best way to do it, it is possible to calculate the answer using parametric curves and calculus—not a bad Calc II exercise.

Let C be the circumference, H be the height, and W the number of times it is wrapped. Then the curve (a helix) can be represented as x=(C/2pi)cos(2pi*t), y=(C/2p)sin(2pi*t), z=Ht/W (for 0≤t≤W). The arc length formula is int_0^W sqrt((dx/dt)^2+(dy/dt)^2+(dz/dt)^2)dt. This looks messy, but if you work it out, it is actually the integral of a constant. You end up with a length of sqrt(W^2C^2+H^2)—which gives the correct answer if you use the values from the problem.

Well I must have had my contrarian specs on because my first thought was to calculate the answer as described above, then I thought hang on a minute, the illustrations may be deliberately deceiving. There are 2 other ways to read the wording. One is that the length of the garland reaches around the column exactly 5 times, but is wound around only, not in a spiral from bottom to top (or top to bottom). So the answer is 5 X the circumference of 16ft 8ins. The second interpretation is that the garland winds around the column 5 times but may be longer – ie, that it is at least as long as 5 X the circumference, but potentially could be longer. In that case, the answer is that the length of the garland is unknown.

I worked it out the way Richard did. My fourteen year old worked it out in pretty much the same way, but instead of using Pythagoras he went the long way round and used trigonometric functions.

Not exactly tricky – before maths standards went down in schools it would have been a question for 11 year olds.

However, I can’t help but think that the column in question is rather tall and thin. At 200 feet high it’s 30 foot taller than Nelson’s Column and less than half the average diameter. Indeed if you take just the column part, ignoring the statue and the plinth, then Nelson’s Column is only about 150 foot. Indeed Nelson’s Column was reduced in height by 30 metres from the original design due to worries about its stability.

So would be interested to know how this 200 foot column, with it’s very narrow diameter is constructed as I might not want to be near it on a windy day.

I don’t even understand the explanation!!

“Note that the answer turns out to be equal to the height plus circumference. ”

Which gives us another puzzle – given that the garland is wrapped around the pillar n times, what ratio of height to diameter gives you a total length equal to height + circumference?

From comments by Gib and Kelly (above)…

n wrappings:

(n^2 – 1 ) / 2 = (Column Height) / (Column Circumference)

or equivalently

PI * (n^2 – 1) / 2 = (Column Height) / (Column Diameter)

A little easier to read…

n wrappings:

(n^2 – 1) / 2 = Height / Circumference

or equivalently

PI * (n^2 – 1) / 2 = Height / Diameter