Brain Teasers ? or 50 Years On........... ?

Well Don I wonder what the UK State Pension age will be in 50 years time (assuming of course that the state pension is still in existence ?)

But being happily retired myself, I thought I would give your latest teaser a go. I stated with the obvious trigonometry approach for the inscribed circle. The solution soon got ugly, so I went in search for the elegant method.

So with a lot of help from Mr Google I got the solutions as:

Radius of inscribed (inner) circle: 6/23 

Radius of circumscribed circle (outer): 6

Thanks for once again sharing a very interesting problem !

Regards,

Peter

sophiebear0_0 posted:

Well Don I wonder what the UK State Pension age will be in 50 years time (assuming of course that the state pension is still in existence ?)

But being happily retired myself, I thought I would give your latest teaser a go. I stated with the obvious trigonometry approach for the inscribed circle. The solution soon got ugly, so I went in search for the elegant method.

So with a lot of help from Mr Google I got the solutions as:

Radius of inscribed (inner) circle: 6/23 

Radius of circumscribed circle (outer): 6

Thanks for once again sharing a very interesting problem !

Regards,

Peter

 

 

Hi Peter,

Yes, the "elegant" solution is  very neat and incredibly easy to use. Glad you enjoyed it !

OTOH, raw trig or geometry needs a bit of care and careful reasoning, but it does deliver the results.......eventually.

I'll leave it for a day or so to see if anyone else wants to try the "hard" trig or find the elegant formula. Meanwhile, well done !

Earlier this week, I set a puzzle involving four circles. Sophiebear (Peter) solved it and said I would post the answer after giving others time to try it for themselves. Here it is.

• A maximum of four circles can be made to 'just' touch one another
• Curvature (c) is the reciprocal of radius (r), hence c=1/r
  Curvature is positive when circles touch externally
  Curvature is negative when circles touch internally
  Four circles A,B,C and D have curvatures a,b,c and d respectively
• 2(a² + b² + c² + d²) = (a + b + c + d)²

If the radii corresponding to a,b,c are 1,2,3 then the radius corresponding to d will be either 6/23 or 6. (as Peter correctly answered!)

Take care with the arithmetic......eg b² = 1/4.....it does work. Let me know if you have difficulty!

Cheers

Don

The other day myself and Mrs D walked along part of the Kennet & Avon Canal between Bedwyn and Honey Street. We watched a few narrow boats navigate the locks and passed the Grafton steam pumping station which on special days still pumps water into the upper pound, ie the stretch of canal from Burbage to Wooton - the rest of the time this pumping is done by an insignificant electric pump.

Anyway, watching these narrow boats reminded me that an ex-university friend now works for the Hydraulics Research Station at Wallingford (now HR Wallingford). A couple of years ago he was in someway involved in the development of the Panama Canal. The capacity of the canal locks is being increased and there are significant penalties if the locks don’t perform as specified eg filling/emptying times.

This combination of events got me thinking as we wandered along in the sunshine.

• Does the time that a lock takes to fill, take the same amount of time regardless of the displaced tonnage of ships in the lock ?

• Does the time a lock takes to fill, depend on whether it is fed by over-topping a given sluice, or by opening a sluice at the bottom of the lock wall or lock gate – assuming each sluice has the same cross-section ?

BTW, I don't have any definitive answers to the two lock questions that I posed above.

I was just "musing" as I walked along the canal.

Please feel free to "muse aloud"

The Mid-Week Lotto numbers today.....obviously the usual probability - 1 in 45 million, but

to see 1,2,3,4,20,21 and the Bonus 22............well...........

about as likely as Leicester City winning the Premier League

Regarding the locks, my thought experiment went like this: 

1) Imagine a boat sailing into the open lock. A certain amount of water will leave the lock, equal to the displacement of the boat. The water level remains the same as the level outside the open lock. A hundred more boats, or a thousand could do the same thing. Provided that the lock is deep enough and large enough, they all remain floating and the water level remains the same. 

2) Now close the lock. The water level in the lock is the same as if no boats had entered it. The water level is equal to the water level outside the entrance gate of the lock. There is a lesser volume of water in a lock full of boats than in a lock bereft of boats, but the level is the same. 

3) All of the filling action serves to raise the already-floating boats to the level outside the exit gate. 

My answer to the first question is that there is no difference.

I agree. It's in the water to start with, and still there once it gets to the top. The amount of extra water needed to raise the levels is the same regardless of the size of the boat. 

Pas to the second question, perhaps opening the sluice at the bottom will be quicker, as there is greater water pressure. But I'm a biologist rather than a physicist, so am very likely wrong. 

Agree.

As for the first question, if you close the lock and then add boats (or lemmings jump in), the water level rises and it takes less time to fill the lock. But that's not how locks work. I'm not sure how lemmings work.

Apologies, I had forgotten that the "Lock" question was still "open" (if you will forgive the pun!)

I agree that JR Hardee is spot on with his assessment.

I have prepared a couple of sketches to illustrate what is happening and will try to post in the next day or so.

Ok, not the best impression, I admit. But hopefully it backs up JR Hardee's explanation.

Cheers

Don

Yup. The key to the visualization is that the lock gate is open, so water seeks its own level as boats enter.

Last winter we were in Canada and in between skiing, trekking and sleigh rides, we enjoyed snowball fights with the kids !

We soon got into the habit of stacking snowballs in a pyramid ready to repel the next attack. I made a tetrahedral pyramid with 10 snowballs and my grandson did the same. So we had 20 snowballs ready to hand. However, this was only just enough to fend off the next attack by Mrs D and our granddaughter, so we decided to double our stockpile.

We found that two tetrahedral pyramids, each of ten snowballs, could be re-stacked as a single tetrahedral pyramid of twenty snowballs. We built two of them and that worked ok to repel the subsequent two or three attacks.

My grandson then suggested that because I could make snowballs faster than him, I should make a “big” stack and he a “little” stack which we could then combine into a single “massive” stack using all the snowballs from the “big” and “little” stacks – with no shortages or remainders……………

We lost the next attack because we took too long to figure out how many snowballs were needed for the “little” and “big” stacks etc.

We also lost the next attack, because we were too busy making all the snowballs needed for the two stacks………….never mind combining them into a single stack……

So, what is the minimum number of snowballs we needed, so as to make one large tetrahedral pyramid from two unequal tetrahedral pyramids. ?

ie “little stack” X snowballs + “Big stack” Y snowballs = “Massive stack” Z snowballs

PS, you can  assume that all the snowballs are perfect spheres, each of the same size.

Each of the three different symbols has a different value associated with it. Adding up the value of the symbols in each row and column gives you the value for that row or column. What is the value of each symbol and what is the missing value of the column ?

Hi Don,

the missing value of the second column is 102. It contains the same symbols as the first row. The sum of all rows and colums is also the same this way (382).

The sun has a value of 17, the drop is 21 and the snowflake is 32.

By looking at the first two rows, you can see that the drop has a higher value than the sun. These two are nearly the same, only a single symbol is different. A drop is replaced by a sun in the second row and the sum drops from 102 to 98. This means that the value of these two symbols has difference of 4.

Using the same approach on the first row and first column, I get a higher value for the snowflake, compared to the drop. This difference is 11.

The sun has the lowest value, the drop is next at sun+4 with the snowflake at sun+15 (=sun+drop+11).

Looking at the second row I used the formula 4*sun + 30 = 98. Subtracting 30 at both sides gives 4*sun=68. 68 divided by 4 gives a value of 17 for the sun.

Spot-on Mulberry.

And a very nice explanation !

Pig of a problem…………?

Earlier in this thread, a goat was eating half the grass in a circular field whilst tethered to a post on the circumference. It was a pig of a problem.

So I thought I’d try a pig tied to post………..  

The farmer has a triangular field, actually an equilateral triangular field of side 100m. Tethered to a post at one of the corners of the field is a pig. The farmer wants the pig to be able to eat half the grass in the field. How long should the tethering rope be ?

(You can assume the distance between the end of the rope and the mouth of the pig is negligible ! ditto for the necessary knots etc.)

I thought a simple diagram might help.

The Pig is able to eat the grass within the blue arc, tethered by the red rope.

64.304m I think.

Used trig to calculate height of triangle (100m/cos30) and multiplied by half the base i.e. 50m to get the area of the triangle. As half of this was equal to one sixth of the area of a circle of radius L (i.e. πL²) since all angles of the triangle are 60°, it is possible to calculate L² by multiplying the area of the triangle by 3/π and taking the square root to find L.

Hi sj

You are on the right track. But I get a different answer..............

Hm, my calculation is as follows:

Height of triangle = 50m * √3 = 86.6m

Area of triangle = 50m * 86.6m = 4,330 sq m

Area under 60º Arc with radius L = 4,330/2 = 2,165 sq m

This produces an area of a circle with radius L of  6 * 2,165 = 12,990 sq m

So π * L² = 12,990 sq m

L² = 12,990/π = 4,134.845 sq m

L = √4,134.845 m

L = 64.303 m

I also get 64.3 metres ?

Well, this is embarrassing..................

You are both absolutely correct, and that is the same answer that I have. I don't even know what I mis-read in your initial post SJ, but apologies. It must have been too early this morning and not fully awake !

Well done, and even more so for providing the evidence

No problem, Don. Even though I did arrive at the correct answer, my initial post did contain an error i.e. that the height of the triangle (h) was 100m/Cos30 when, of course, it should have read 100m x Cos30 i.e. 100m x √3/2 or 50m x √3.

sophiebear0_0 posted:

I also get 64.3 metres ?

Hi Peter. Well done. I don't know what I saw this morning that made me think SJ had got it wrong. Embarrasing to say the least !

Cheers, Don