You most certainly have remembered the area of an equilateral triangle correctly.

And you can also do subtraction quite well too

You most certainly have remembered the area of an equilateral triangle correctly.

And you can also do subtraction quite well too

Hi Don,

I think one way to confirm there are only those two solutions is to start with the rectangle of sides a and b so the perimeter = area condition gives you

2a+2b=ab or rewriting 1/a + 1/b = 1/2 and rearranging for a gives

a = 2b/(b-2).

Now instead of starting small and looking for ever larger rectangles let’s start at (infinitely) large b and see where the first one we find is as we reduce b. For very large b the right hand side of the eq. above has a limit of a = 2. As we decrease b from this large value a will increase very slowly and the next integer value it hits is a=3 (at b=6). Looking at it this way you can be sure there are no larger rectangles which satisfy the area = perimeter criterion.

Now maybe a rigorous mathematician who knows number theory will tell us that b=infinity in the space of integers and a=2 is a solution but I’m not losing sleep worrying about that.

Don, do I have this right, (1) you add a new coin on each turn. (2) Can you jump coins for a second time once they are on the board? (3) Can you jump over the centre yellow circle eg A - yellow - F

cheers

Ian

(1) Yes ! You start with a pile of 9 coins. You put the first coin on any of the red circles A to J. You move the coin in a straight line, jumping over one red circle and leaving that first coin on the next red circle. eg Coin on A. Moves to D by jumping over B. Red circle D is now occupied and cannot be used a starting point. You take Coin No 2 and put it on any unoccupied red circle. You move that coin likewise in a straight line. eg 2nd Coin on C. Moves to F by jumping over D (even though D is occupied). Coin No 2 on C could have moved to J by jumping over B (even though B is unoccupied). Either way, you now have two red circles occupied and no longer available as the starting point for Coin No 3.

(2) No !

(3) No !

I hope that the above is clear and more importantly, actually helps !! If not, just shout ! (or better still, ask again and eventually i’ll get the question right )

ok thanks. I’m enjoying the golf from Pebble beach right now so I’ll have a think about it later.

PS I saw on another thread that you are a golfer too so if you are ever in Scotland and fancy a game @ North Berwick West links give me a shout.

No rush Ian. these teasers are for relaxing into !!

Mrs D and myself **enjoy** playing golf but we aren’t any **good** at playing golf !

Our Canadian son-in-law used to be the Head Pro at Preditor Ridge and then Vernon Golf & Country Club, both in the Okagagan Valley BC, but despite 20 years of coaching, he wasn’t able to improve our game. He now helps run the family business, and is having more success. Meanwhile he relegated us to the OAP’s Hillview course run by his (best ?) friend.

But next time we plan to make it past Newcastle, I let you know. many thanks.

ok, I think one ( of many) solutions to the star problem is :

J over H occupies G

C over B occupies J

F over D occupies C

I over H occupies F

A over B occupies D

H over J occupies A

E over F occupies H

B over D occupies E

B over J occupies I

leaving B empty.

Found by a combination of trial and error, keeping the coins spread out as one goes along and figuring out what the last move or two must look like.

Best way I can describe a typical solution is as follows

Circle around the star in the same direction, always selecting your previous starting point as your next end point.

EG A to D; H to A; E to H etc

I am disappointed !

I posted my solution to the “Tethered Goat” problem a few days ago (see posts 76 to 85 above). And nobody has pulled me up over the (deliberate ?) mistake on post …

…well, **if anybody can identify it, all is forgiven**

Fortunately, it didn’t affect the algebraic calculation at all. I only spotted it today whilst browsing through this thread again (sad really, I know!). The mistake doesn’t appear in my PowerPoint presentation, but somehow exists in my subsequently saved JPEG version.

Apologies to all those who have been baffled about this mistake and especially those who have lost countless night sleep over it !

I was going to give it a go, but I wasn’t sure if the grass was all the same hieght and density…

The mistake is in Post No 82 (use the slider to select it)

First box

Line 2

*Triangle ABC is isosceles AB = BC = R*

should read

*Triangle ABC is isosceles AB = AC = r = 1*

Fortunately it was an unnecessary line in the development of the geometry

But apologies to anybody who found it confusing !

There is a cistern of volume 48. It has two inlet valves and one outlet valve.

The first inlet alone, fully opened, can fill the cistern in 12 hours. The second inlet, again by itself and fully open, can fill the cistern in 6 hours. The outlet can empty the cistern in 8 hours when fully open.

If the cistern is emptied and all three valves are fully opened, will the cistern ever fill up, and if so, how long will it take to fill the cistern ?

Valve #1 has a throughput of 4 volume units per hour, #2 8 and #3 6. So, let x be hours:

4x+8x-6x=48 => x= 8 hours.

An area of 100 sq cubits (eg 10 x 10 cubits) is equal to the area of two smaller squares together. The side of one of these lesser squares is ½ + ¼ the side of the other lesser square. What are their sides ?

Spot-on Dozey.

Was this “Trial & Error”, a “good guess” a “methodical mathematical/algebraic calculation” or “I could just **see** it !”

FWIW, mine was a bit of trial and error, followed by algebra to confirm.

But more importantly, well done !