A Fistful of Brain Teasers

Don Atkinson posted:

notnaim man posted:

Total frustration, I have a pile of envelopes with numbers scribbled on. I got 347, then 375. Then, I have been told there is a similar puzzle about getting to see a mistress without being seen en route every day of the year, so without counting again, 365?

Tantalising, Frustrating, annoying............That's what some of these Brain Teasers are all about Also a bit of fun andgreat satisfaction when you get there !

Take a careful look at the "13" Node. ie there are 13 different routes to that specific Node.

They are quite easy to follow. and en-route you will have passed through Nodes that could have been reached by 5, 3 and another 5 routes respectively. (You might have noticed, there is only 1 route to the NW corner and there is only 1 route to the SE corner).

There is a "pattern" developing, with a Pub at the end.

(but sorry, the number of routes isn't 345, 347, 365 or 375)

More Frustration ! or Great satisfaction ?

Decided to have one more go: 321

So, for the person with a lover, they’d have been seen more than once 44 or 45 times.

Glad you didnt use a grid of 6x6, as the answer would have been a rather large 1683

Got there in the end (Pun intended !)......well done

And yes, 1683 for the next row/column, (but that bit was easy )

Innocent Bystander posted:

I’m struggling as usual to upload a sketch, necessary for the explanation below to make sense. This is my last attempt - I’m providing the link as well in case it still doesn’t show.

 

https://imgur.com/a/WyfFD

Angles A,A are identical because sides X,X identical
Similarly B,B and C,C
Perpendicular from centre to any side bisects it for same reason.

Cosine rule:
Cos (A+C)   =  (15^2 + 13^2 - 14^2 ) /  2*15*13     = 0.50769
(A+C) = 59.49°

Total internal angles of triangle = 180° = 2A+2B+2C
So A+B+C=90°
So the other angle in one of the rt angle triangles with angle B and side length 7 = A+C (the angle found above)

Sin 59.49   =  7/x
∴ X =  7/sin 59.49      = 8.097 miles

  

Don, do you leave your bike at the station and return the same way, or, more sensibly, take it on the train with you to allow you to use it at your destination, with the flexibility when returning to choose the route home according to wind, either for maximum exercise or minimum effort depending how the day went

IB,

Your approach to this particular puzzle has inspired a new puzzle concerning circles and polygons (eg a triangle).

I'll post it later today - the training airfield is closed due to snow !

3,4,5 Triangles et al

Now I guess we have all heard of the “3,4,5” triangle ?

It’s a right angled triangle. The sides are all integers.

3^2 = 9;    4^2 = 16;  9+16 = 25, the root of which is 5

All nice and tidy.

The next such triangle has sides 5,12 and 13. But I guess many of us knew that as well ?

So, the question is…..

What’s the next one ?

……….And the next ?

…………… And the next ?

………………..In fact, how many such triangles are there ?

PS Note: I skipped the 6,8,10 triangle and others of a similar nature, because that’s just the same as the 3,4,5 triangle using a different scale !

PPS: this not the "inspirational" puzzle I referred to above ! But I hope it's not too dull and boring !!

There’s an infinite number, though as the third side diminishes in proportion to the other two, and they only ever differ by 1 (in the lowest common denominator), the triangles will tend towards being a straight line (and in any practical application, the accuracy of measuring the two long sides is likely to diminish as the lengths become extreme, and so would tend to become varable in relation to one another, with the result that you wouldn’t be able to rely on which is the longer side or which is the right angle, both then approximating 90°)

3,4,5
5,12,13
7,24,25
9,40,41
11,60,61
13,84,85
15,112,113
17,144,145
19,180,181
21,220,221
23,264,265
25,312,313
27,364,365
29,420,421
31,480,481
33,544,545
35,612,613

etc. (lowest common denominators only given)

This is a printout from Excel.

I didn't type each number into the spreadsheet as a copy from some book.

I created a simple set of formulae to generate  the length of each of the three sides.

Anybody able to provide the formulae ?

Well, My formula has just revealed to me that there is another series with the longer sides differing by 2 and the short side incrementing by 4 instead of 2. Does your formula find them?

8,15,17
12,35,37
16,63,37
20,99,101
24,143,145
28,195,197
32,255,257
36,323,325

etc

Does your formula find them, Don?

This is my formula:

Sides X,Y,Z      for X any length >1

Y=( X^2  -1 )/2
Z=( X^2  +1 )/2                   [ or more simply, Z=Y+1 ]

If side X is integer, all the odd lengths of side X give integers for Y & Z. Even integral lengths of side X result in semi integrals and the three side lengths need doubling to make them integers (e,g for X=4, Y=7.5 & Z=8.5, doubling gives 8,15,17)

Innocent Bystander posted:

Well, My formula has just revealed to me that there is another series with the longer sides differing by 2 and the short side incrementing by 4 instead of 2. Does your formula find them?

8,15,17
12,35,37
16,63,37
20,99,101
24,143,145 formulae
28,195,197
32,255,257
36,323,325

etc

Does your formula find them, Don?

Generally, yes !

The formulae are based on inputting two numbers, normally integers. Call them A and B with B > A

The three sides of the triangle are X, Y and Z

X = B² - A²

Y = 2(A + B)

Z = A² + B²

If the inputs (A, B) are 1, 2..............you get the 3,4,5 triangle

If the inputs are 2,3........................you get the 5,12,13 triangle etc

Inputting (say) 1½, 2½ etc, you get triangles with non-integer sides, but these can be easily turned into integers, usually by doubling.

Inputting (say) 1,3, you get triangles with a bigger difference in the two long sides than unity.

The diagram shows three regular polygons, an equilateral triangle, a square and a regular pentagon, each inscribed within a circle radius R.

What's the area of each of these polygons ?

I make it 1.299R^2 circle, 2R^2 square, 2.377R^2 pentagon.

haven't double-checked, but relatively seems right so will do!

So far, so good - subject to that double-check............

.........using Excel..........

Not double checking. 

My working in case of interest ( though as usual may not show, so link included, first the triangle and pentagon:

https://imgur.com/a/SUIfK

And the square: 

Square diagonal = 2R
Side = sqrt ( ( (2R)^2 ) /2)
Area = side^2
= ( (2R)^2 )/2
= 2R^2

Innocent Bystander posted:

Not double checking. 

My working in case of interest ( though as usual may not show, so link included, first the triangle and pentagon:

https://imgur.com/a/SUIfK

 

And the square: 

Square diagonal = 2R
Side = sqrt ( ( (2R)^2 ) /2)
Area = side^2
= ( (2R)^2 )/2
= 2R^2

 

ah! when you said:-

"....so will do" in your previous post. I thought you meant "so, I will double-check". I didn't want to interupt that !

I now presume that you meant, "It looks kinda right, so it will do, and I won't bother with a double-check"

Anyway, your figures were spot-on ! - Calculator, Excel or Trig tables doesn't matter !

Ambiguity (though not intended on this occasion) is great for keeping options open!

The next stage is..........

Can you generate a formula to calculate the area of ANY "n" sided regular polygon (all sides equal length) , inscribed within a circle radius R ?

And thus show that as "n" gets very large, the area approaches that of a circle = πR²  ?

PS. You might have already done this on your way to the solution of the first three polygons !!

Again, an Excel spreadsheet generated using my particular formula. (hidden, obviously !)

Other formulae would be more than acceptable !!

Well, the way I did the pentagon gives the formula - and I had nearly done the triangle that way anyway. So:

Area =n * R Sin (360/2n) * R Cos (360/2n ) 
          = n * Sin(180/n) * Cos(180/n) * R^2

Or in radians = n * Sin(pi/n) * Cos(pi/n) * R^2
As n gets large Sin n tends to 0, Cos n tends to 1,when simplifies to: pi * R^2

Innocent Bystander posted:

Well, the way I did the pentagon gives the formula - and I had nearly done the triangle that way anyway. So:

Area =n * R Sin (360/2n) * R Cos (360/2n ) 
          = n * Sin(180/n) * Cos(180/n) * R^2

Or in radians = n * Sin(pi/n) * Cos(pi/n) * R^2
As n gets large Sin n tends to 0, Cos n tends to 1,when simplifies to: pi * R^2

Hi IB,

This bit leads to a difficulty !! (but otherwise, you are so close )

First, your formula (in radians or in degrees - it doesn't matter) is fundamentally the same as mine. (However, I didn't use the formula in the format you have presented. I made a substitution, which somewhat simplified the formula, but doesn't change the output.)

Let me outline the "difficulty" you have introduced.

Your formula "could" be written:-

Area = a * b * c * d  where

a = n   (ie the number of sides)

b = Sin(π/n)

c = Cos(π/n)

d = R²

As you rightly say as n gets large, "b" tends towards "Zero" (but this is where the difficulty arises). In any equation involving the product of (say) four numbers, if any ONE term (in this case "b") tends towards Zero, then the product (in this case the Area) also tends towards Zero. This is the aspect that needs to be reconsidered.

For sure, element "c" tends toward a value of 1 as "n" gets very large, and in doing so, eliminates the embeded need for "n" in that element.

Element "a" also tends towards infinity, but rather than disappear, begins to cause problems. However, if element "b" is sorted out, so will this element !

Of course, I could just explain how I managed to sort out that pesky little "Sin" element, but I know it's much more satisfying for people to "discover" these things for themselves, even if a few "prompts" are required along the way !

In terms of  maximising accuracy by avoiding finding sine of numbers very close to zero with  polygons having very large numbers of sides, what is called for is pythagorus combined with trig using cosine to find the triangle height.

Area = n.R.Cos(180/n).√( R² -  (R.Cos(180/n) )²)

or in radians for Excel

          = n.R.Cos(π/n).√( R² - (R.Cos(π/n) )² )

N.B. with my original formula Excel maintains accuracy to 5 figures with up to at least 10^6 sides, which is more than I can visualise!

Don Atkinson posted:

The next stage is..........

Can you generate a formula to calculate the area of ANY "n" sided regular polygon (all sides equal length) , inscribed within a circle radius R ?

And thus show that as "n" gets very large, the area approaches that of a circle = πR²  ?

PS. You might have already done this on your way to the solution of the first three polygons !!

Hi IB,

You had already achieved the first part of the task with your formula:-

"....... in radians (Area) = n * Sin(pi/n) * Cos(pi/n) * R^2........"      (I have the same formula)

However.......

I indicated thatthe second part of the task wasn't actually resolved with your statement about that formula :-

"..............As n gets large, Sin n tends to 0, Cos n tends to 1,when simplifies to: pi * R^2..........."

If n --> ∞(infinity), and Sin(pi/n) --> 0, and Cos(pi/n) --> 1

the formula reduces to :-

Area = "infinity" times "zero" times "one" times "R²"....................and "infinity" x "zero" is meaningless.

But there is a neat "approximation" that can be substituted for Sin(pi/n) for large "n"..........

Moving away from the simplicity and certainty of our O-Level Geometry................

...............to the more unpredictable world of English and its use as a language of confusion and obfuscation...

Replace each word below with a four-letter word (no ! not THAT four-letter word ). Each new word works as part of a chain, with the last two letters of each word forming the first two letters of the next.

Develop

Was in debt

Modify

Irritate

Cut

Frank

Concludes

Clearly, I do have a list of such words, but there could be countless other lists. Let's see.

Don Atkinson posted:

Moving away from the simplicity and certainty of our O-Level Geometry................

...............to the more unpredictable world of English and its use as a language of confusion and obfuscation...

Replace each word below with a four-letter word (no ! not THAT four-letter word ). Each new word works as part of a chain, with the last two letters of each word forming the first two letters of the next.

 

Develop Grow

Was in debt Owed

Modify Edit

Irritate Itch

Cut Chop

Frank Open

Concludes Ends

Clearly, I do have a list of such words, but there could be countless other lists. Let's see.

Crikey Eoink, that was fast ! the ink didn't have time to dry .....well done !