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

The diagram above shows a Rhombus. You need to convince yourself (prove) that it really is a Rhombus and not just any old quadrilateral !!

Alternatively you could prove that the two triangles made bisecting the Rhombus are equilateral triangles as follows :-

Main triangle LMN is equilateral [Symmetry] Hence angle bisectors = side bisectors

The “North East” arm of the centre-line Red Yellow Yellow Red Red circles is inclined 30 deg above the horizontal [Symmetry]

The “South East” arm of the centre-line Red Blue circles is inclined 30 deg below the horizontal [Symmetry]

The angles between the arms of each pair of centre-line RYYRR circles is 120 deg [3 x 120 = 360]

Triangle XYZ, is formed by Quadrilateral WXYZ divided by its short diagonal XZ

Angle X = 60 deg

XY = XZ [Radii (b + c) in each case]

Hence Angle Y = Angle Z = 60 deg

Hence triangle XYZ = Equilateral

Hence 4a + 2b = b + c          [4]          (circle radii are a, b, c, d, e)

.............and just a little bit more algebra to finish off.................

.............and just a little bit more algebra to finish off.................

The Rhombus/Triangle in the last diagram shows:-

4a + 2b = b + c          [4]

I now have four equations, which is enough to solve the puzzle.

[2] and [3] --> b + 2c = 4a + 5b

2a + 2b = c      [5]

Substitute c in [4] --> 4a + 2b = b + 2a + 2b

2a = b          [6]

Substitute in [1] -->  d = 4a + 6a = 10a

Hence Circle D is ten times the diameter of circle A (the smallest)

Eq [6] already shows that Circle B is twice the diameter of Circle A

A couple more substitutions reveals

C is six times the diameter of Circle A and

Circle E is fourteen times the diameter of Circle A

A1; B2; C6; D10; E14

Mrs D has just bought four identical (frictionless) steel balls which she wants me to set up as a centre piece in a restful ornamental sub-garden. They weigh in at 2√6 tones wt each.

She would like them arranged as shown above with three of them resting on a horizontal flat surface and touching each other. The fourth freely sitting in the “hollow” on top of the bottom three. The bottom three will be prevented from rolling apart by spot-welding where they touch.

Allowing for a factor of safety of 3, what tension must each weld be able to withstand ?

_{[now, my maths book gives me the answer, but I haven't tried this one out myself yet !]}

Browsing through my A-Level maths notebook, I noticed that more than 50 years ago I had written this (probably copied off the Blackboard or from a long-forgotten text book……..

"Probability gives us a measure of the likelihood that something will happen. However, it must be appreciated that probability can never predict the number of times that an occurrence actually happens. But being able to quantify the likely occurrence of an event is important because most of the decisions that affect our daily lives are based on likelihoods and not absolute certainties."

So, since it seems very important these days to make good decisions, I thought we might all brush up on “Probabilities”………….

I select four cards at random from a pack of 52 playing cards.

What is the probability that all four cards are Clubs ?

One integer is chosen at random from the set of integers 1,2,3,4,5,6,7,8

What is the probability that it is a prime number?

Two books are taken at random from a shelf containing five paperbacks and four hard backs.

What is the probability that both are paperbacks?

Don Atkinson posted:

Two books are taken at random from a shelf containing five paperbacks and four hard backs.

What is the probability that both are paperbacks?

Wow music not interesting anymore, so that you guys feel obliged to go back to some school problems...

After studying to a masters level of languages and history. Is my livelihood only destined to ridicule Trump and the rest of society as a journalist ?

Bert Schurink posted:

Don Atkinson posted:

Two books are taken at random from a shelf containing five paperbacks and four hard backs.

What is the probability that both are paperbacks?

Wow music not interesting anymore, so that you guys feel obliged to go back to some school problems...

Bert...............this is "THE PADDED CELL"............

.........please allow these two gentlemen in white coats to assist you back towards THE MUSIC ROOM....

Don Atkinson posted:

Two books are taken at random from a shelf containing five paperbacks and four hard backs.

What is the probability that both are paperbacks?

P1 = 5/9

P2 = 4/8

P1 and P2 = 20/72 = 5/18

Don Atkinson posted:

One integer is chosen at random from the set of integers 1,2,3,4,5,6,7,8

What is the probability that it is a prime number?

when I went to school, we were suposed to know that 1 is not a prime number, but 2 is.

I have seen people get this probability value wrong, simply because the underlying facts were confused.

Hopefully that won't be the case on this forum

Spot on winky.

These were the first few probability questions I encountered from years ago. The course work had a steep learning curve after the first few days and I felt confused with Permutations, Combinations and Probability all seeming to roll into one fuzzy moras !

Probability of 4 clubs

(13! x 48!) / (9! x 52!) = 0.002641 = 0.2641%

Well done SJ. Spot-on as usual !

I followed the steps outlined by winky who did well with the books...

P1 = 13/52

P2 = 12/51

P3 = 11/50

P4 = 10/49

Probability of 4 Clubs = P1.P2.P3.P4 = (13.12.11.10)/(52.51.50.49) = 11/(17.5.49)    [by cancelling out]

11/(17.5.49) = 11/4165 = 0.002641

In the (good ?) old days we simply wrote = 0.003 to 3dp

Nowadays, Excel can cope with 48! etc, 52 years ago our school only had four-place Log Tables. 

Don Atkinson posted:

One integer is chosen at random from the set of integers 1,2,3,4,5,6,7,8

What is the probability that it is a prime number?

I recall at school, our maths teacher commenting that deciding how to group data sets was just as important as being able to quantify the probability of an event occurring.

In this example, some of us simply struggled to identify the set of Prime Numbers, with some confusion over numbers 1 and 2.

A bag contains five red and six black balls. The balls are removed one at a time without replacement. If the balls are taken out at random, what is the probability that :-

1. the first two balls removed are red
2. the second ball removed is red.

Don Atkinson posted:

Don Atkinson posted:

One integer is chosen at random from the set of integers 1,2,3,4,5,6,7,8

What is the probability that it is a prime number?

I recall at school, our maths teacher commenting that deciding how to group data sets was just as important as being able to quantify the probability of an event occurring.

In this example, some of us simply struggled to identify the set of Prime Numbers, with some confusion over numbers 1 and 2.

Ok, it was too easy I guess. The Prime numbers are 2,3,5 and 7.

1 is not a prime number.

Don Atkinson posted:

A bag contains five red and six black balls. The balls are removed one at a time without replacement. If the balls are taken out at random, what is the probability that :-

1. the first two balls removed are red
2. the second ball removed is red.

The probability for two reds is 18,18%. For ball number one it is 5/11, for number two 4/10. If you multiply those two, the result is 0,1818.

Don, isn't the answer to the second question dependent on the color of the first? Or are you looking for the 40%?

Hi Mulberry,

Your first answer is spot-on, Brilliant. (2/11).

For the second question, the first ball could be either red or black. But the second ball has to be red. You already know the probability of the first and second ball being red.

I haven't tried it myself, but a "tree" diagram might well help clarify the possible outcomes.

Probability of Red as second ball is 5/11

There are 4 possible outcomes. Red/Red, Red/Black, Black/Black, Black/Red.

Red/Red has a probability of 2/11 whilst the other 3 outcomes each have an equal probability of 3/11. 

Adding probability of Red/Red and Black/Red gives a combined probability of 5/11.

Yes ! Tried the old "tree" diagram and it makes it look............

....................easy/straight-forward/less-difficult/manageable/possible........

Well done sj

drawing out a tree diagram shows the four possible outcomes of picking two balls, and makes the probability of each event easier to generate.

the red-red events generate 2/11 and the others each reduce to 3/11