## November 09, 2009

### Target Practice

Robin Hood and Friar Tuck were engaging in some target practice. The target was a series of concentric rings, lying between successive circles with radii 1, 2, 3, 4, 5. (The innermost circle counts as a ring.)

Friar Tuck and Robin both fired a number of arrows.

“Yours are all closer to the centre than mine,” said Tuck ruefully.

“That’s why I’m the leader of this outlaw band,” Robin pointed out.

“But let’s look on the bright side,” Tuck replied. “The Total area of the rings that I hit is the same as the total area of the rings you hit. So that makes us equally accurate, right?”

Naturally, Robin pointed out the fallacy...but:

Which rings did the two archers hit? (A ring may be hit more than once, but it only counts once towards the area.)

For a bonus point: what is the smallest number of rings for which this question as two or more different answers?

For a further bonus point: if each archer’s rings are adjacent – no gaps where a ring that has not been hit lies between two that have – what is the smallest number of rings for which this question has two or more different answers?

### 7 comments by 1 or more people

1. #### Sebastien Dutrieu

it doesnt matter what rings they hit, as long as they both hit the same ring with all their arrows, and that Robin’s were all closer to the bull than Tuck’s, whilst still staying in the same ring.

is that right?!?

09 Nov 2009, 11:29

2. #### Andrew Ingram

rings 1,2,3 (area = 1 + 3 + 5 = 9) for Robin, ring 5 (area = 9) for Tuck
rings 1,2,4 (area = 1 + 3 + 7 = 11) for Robin, ring 6 (area = 11) for Tuck

So, smallest number of rings is 6 for two different answers

if the rings have to be adjacent:

rings 2,3,4 (area = 3 + 5 + 7 = 15) for Robin, ring 8 (area = 15) for Tuck

so 8 rings

Unless I completely lost my maths-fu of course…

09 Nov 2009, 11:34

3. #### David Buckley

We can see easily that Andrew’s reasoning for the first one is correct since Hood must his an even number more rings than Tuck as the area of every ring is 1 modulo 2. In particular he must shoot at least 3, so the 1,2,3;5 example is minimal, and the 1,2,4;6 example is the next smallest. With only 5 rings there can be no other examples, so the answer 6 is correct.

However, the second answer is not. Clearly 1,2,3;5 is the minimal example again, but we can beat 2,3,4;8 perhaps counter-intuitively by increasing Hood’s total and using 2,3,4,5;6,7. There is no smaller example where Hood hits 4 rings as this would give 1,2,3,4;4,5 which has Hood and Tuck hitting the same ring. So the last part has answer 7, not 8.

It seems likely that in all examples which have minimal target size for a given number of Hood rings, Tuck has exactly 2 rings fewer than Hood. This is not the case, however. If 5 consecutive rings are hit, we find that the sum is 25+10i where i is the number of inner rings missed. Tuck must then hit 3 rings (if his first is j, his total is 6j+3, so there’s a solution only when i=2 mod 3, the first example of which is 3,4,5,6,7;7,8,9; the next example of this is 6,7,8,9,10;12,13,14. On the other hand, 1,2,3,4,5;13 is a valid ring selection, so in the case of 5 Hood rings, the minimal example has only 1 Tuck ring. And Tuck shouldn’t be let near a bow.

Extra bonus credit as I looked at the problem late?! What about when all the rings are adjacent like with 2,3,4,5;6,7?

This seems to come down to a quadratic Diophantine equation; if Hood skips i, hits n and Tuck hits m we find (i+n)2-i2=(m+n+i)2-(i+n)2, so 2(n+i)2=(m+n+i)2+i^2. In particular, both i and m must be odd (or we find powers of 2 don’t add up). The equation can probably be solved in a similar way to the way one generates Pythagorean Triples. The next example with i=1 seems to be 46 rings; with i=3 we find 4,...,15;16,...,21 provides the first example. (I cheated these solutions using Dario Alpern’s generic two variable quadratic diophantine equation solver at http://www.alpertron.com.ar/QUAD.HTM – for a given i, the first example seems to be i+1,...,5i;5i+1,...,7i)

So, forcing /all/ of the rings to be adjacent, we seem to find that 21 rings are sufficient for there to be 2 answers.

10 Nov 2009, 02:24

4. #### David Buckley

A bit more playing reveals I’m wrong on the second answer; there’s a few more examples with i=7, including 8,...,13;14,...,17. Up to i=11 there’s no further complications, though, so 17 sounds like a safe bet for the real answer.

That broken equation above should of course read (i+n)2 – i2 = (m+n+i)2 – (i+n)2, so 2(n+i)2 = (m+n+i)2 + i2

10 Nov 2009, 02:32

5. #### Eleanor Lovell

Sebastien – nice try but it wasn’t a trick question. As Andrew and David said, – Tuck hit the outer ring (5), while Robin hit the three inner rings (1, 2 and 3).

Prof Ian Stewart’s explanation is as follows:

The areas of successive circles are $\pi$r$^2$, where r = 1, 2, 3, 4, 5, respectively:

$\pi$, 4$\pi$, 9$\pi$, 16$\pi$, 25$\pi$

The areas of the rings are the differences between these:

$\pi$, 3$\pi$, 5$\pi$, 7$\pi$, 9$\pi$

These are $\pi$ multiplied by consecutive odd integers. Robin’s integers are less than or equal to Tuck’s since Robin’s arrows are closer to the centre. The two sets of odd integers must have the same sum. The only possibility is 1 + 3 + 5 = 9

Bonus point: Six rings. The sixth ring has area 11$\pi$, so 1 + 3 + 7 = 11 is a second solution.

Further bonus point: Eight rings. Robin’s odd numbers must be consecutive, and so must Tucks. The next two rings have areas 13$\pi$, 15$\pi$, and 3 + 5 + 7 = 15 is a second solution with consecutive odd numbers.

10 Nov 2009, 15:30

6. #### David Buckley

Did Ian get the further bonus point wrong in the book, or is that your reasoning, Eleanor?

As above: If Robin hits 2, 3, 4 and 5, his total is 3+5+7+9 = 24. If Tuck his 6 and 7, his total is 11+13=24. This uses only 7 rings.

12 Nov 2009, 01:16

7. #### Ian Stewart

Dave is right. My book needs a correction! It’s always the little things that can’t possibly go wrong that catch you out. (Well, they catch me out, anyway.) Well spotted, Dave! I’ll tell the publisher and next time we reprint, we’ll put it right.

25 Nov 2009, 09:41

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