A Fistful of Brain Teasers

Clive B posted:

30sq.cm

Well done Clive !

The area of the yellow square is 25m²

What is the area of the visible parts of the combined blue circle and red square ie those parts coloured red and blue ?

Of course, I would like to see the proof of your answer.

My drawing might not be deadly accurate. The blue circle is tangential to the red square and the yellow square just touches the blue circle at each corner.

Don Atkinson posted:

The circle has centre O, and TX is a tangent touching the circle at X

The area of the triangle OXT is …… ?

....... yes, I know, it's dead easy !

Proof

xt  = sqrt(13**2 - 5**2)  by Pythagoras  = 12

area = 0.5 * base * height = 0.5 * ox * xt = 30

Don Atkinson posted:

The area of the yellow square is 25m²

What is the area of the visible parts of the combined blue circle and red square ie those parts coloured red and blue ?

Of course, I would like to see the proof of your answer.

My drawing might not be deadly accurate. The blue circle is tangential to the red square and the yellow square just touches the blue circle at each corner.

Proof

 circle radius = 2.5 sqrt (2) 

red square side = circle diameter = 5 * sqrt(2)

area between red and yellows squares = 50 - 25 = 25

Don Atkinson posted:

If the double slashes mean equal length then 58

Phil

Filipe posted:

Don Atkinson posted:

The circle has centre O, and TX is a tangent touching the circle at X

The area of the triangle OXT is …… ?

....... yes, I know, it's dead easy !

Proof

xt  = sqrt(13**2 - 5**2)  by Pythagoras  = 12

area = 0.5 * base * height = 0.5 * ox * xt = 30

Nice summary, and it agrees with Clive’s response. Well done !

Don Atkinson posted:

The area of the yellow square is 25m²

What is the area of the visible parts of the combined blue circle and red square ie those parts coloured red and blue ?

Of course, I would like to see the proof of your answer.

My drawing might not be deadly accurate. The blue circle is tangential to the red square and the yellow square just touches the blue circle at each corner.

Red= 50   (Side = diam of circle = diagonal of yellow (Pythag))

Blue = π*((√50)/2)²  = 39.27

Visible blue = 39.27-25 = 14.27

Visible yellow = 50-39.27 = 10.73

Filipe posted:

Don Atkinson posted:

The area of the yellow square is 25m²

What is the area of the visible parts of the combined blue circle and red square ie those parts coloured red and blue ?

Of course, I would like to see the proof of your answer.

My drawing might not be deadly accurate. The blue circle is tangential to the red square and the yellow square just touches the blue circle at each corner.

Proof

 circle radius = 2.5 sqrt (2) 

red square side = circle diameter = 5 * sqrt(2)

area between red and yellows squares = 50 - 25 = 25

The answer is correct - well done.

There is a neat way to illustrate this.

Innocent Bystander posted:

Don Atkinson posted:

The area of the yellow square is 25m²

What is the area of the visible parts of the combined blue circle and red square ie those parts coloured red and blue ?

Of course, I would like to see the proof of your answer.

My drawing might not be deadly accurate. The blue circle is tangential to the red square and the yellow square just touches the blue circle at each corner.

Red= 50   (Side = diam of circle = diagonal of yellow (Pythag))

Blue = π*((√50)/2)²  = 39.27

Visible blue = 39.27-25 = 14.27

Visible yellow = 50-39.27 = 10.73

I think you mean “visible red” rather than “visible yellow” in the last line. But the answer is correct.

Don Atkinson posted:

Filipe posted:

Don Atkinson posted:

The area of the yellow square is 25m²

What is the area of the visible parts of the combined blue circle and red square ie those parts coloured red and blue ?

Of course, I would like to see the proof of your answer.

My drawing might not be deadly accurate. The blue circle is tangential to the red square and the yellow square just touches the blue circle at each corner.

Proof

 circle radius = 2.5 sqrt (2) 

red square side = circle diameter = 5 * sqrt(2)

area between red and yellows squares = 50 - 25 = 25

The answer is correct - well done.

There is a neat way to illustrate this.

Rotate yellow through 45 and it’s corners bisect the sides of the red square. The circle then becomes irrelevant because the rotation is an equivalence transformation. The area is four of the triangles while the total area is 8 triangles. 

Yep ! Nicely spotted.

If you rotate the circle and yellow square by 45 deg, the complicated arithmetic vanishes and we can see that the red square is twice the area of the yellow square and hence the visible blue and red area is 25 sqm

Don Atkinson posted:

Another simple one, otherwise there is a risk of only IanG and Eoink taking part !

A firm sells stationery in packages as follows :-

A notebook and two biros cost 48p; a notebook and 4 pencils cost 58p, a notebook, a pencil and 3 felt-tipped pens cost 64p.

How much should you expect to pay for a package containing a notebook, a pencil, a biro and a felt-tipped pen ?

Looks like this one was overlooked or........

.....it looked too easy/difficult  

Don Atkinson posted:

Don Atkinson posted:

Another simple one, otherwise there is a risk of only IanG and Eoink taking part !

A firm sells stationery in packages as follows :-

A notebook and two biros cost 48p; a notebook and 4 pencils cost 58p, a notebook, a pencil and 3 felt-tipped pens cost 64p.

How much should you expect to pay for a package containing a notebook, a pencil, a biro and a felt-tipped pen ?

Looks like this one was overlooked or........

.....it looked too easy/difficult  

3 equations and 4 unknowns is impossible to solve

Filipe posted:

Don Atkinson posted:

Don Atkinson posted:

Another simple one, otherwise there is a risk of only IanG and Eoink taking part !

A firm sells stationery in packages as follows :-

A notebook and two biros cost 48p; a notebook and 4 pencils cost 58p, a notebook, a pencil and 3 felt-tipped pens cost 64p.

How much should you expect to pay for a package containing a notebook, a pencil, a biro and a felt-tipped pen ?

Looks like this one was overlooked or........

.....it looked too easy/difficult  

3 equations and 4 unknowns is impossible to solve

Not necessarily, when there is a constraint that the terms and solutions must be integers. Start by thinking that the notebook price is in the set 46, 44, 42.... and also in the set 54, 50, 46, 42.... and also in the set 61, 58, 55, 52..... There may be only one notebook price that is in every set. There could be more than one solution, however.

If non-integer prices are allowed, it is a bit harder, if not impossible.

Filipe, winky is providing a very clear insight to the problem. Given the pricing structure is in English pence, there are no half pence these days, although there were when I first encountered this puzzle and perhaps you will find a solution more easily if you accept half pence.

The two wheels on each axle of a railway wagon are rigidly connected together.

When an engine negotiates a bend one would expect one of the wheels to skid (because one wheel has to go
further than the other).

Neither wheel skids, however, even if the rails aren't banked.

Why?

PS I don't think that the solution usually quoted completely eliminates wheel skid at all speeds, but it certainly plays a significant part.

55p for the stationary package I think. 

winkyincanada posted:

Filipe posted:

Don Atkinson posted:

Don Atkinson posted:

Another simple one, otherwise there is a risk of only IanG and Eoink taking part !

A firm sells stationery in packages as follows :-

A notebook and two biros cost 48p; a notebook and 4 pencils cost 58p, a notebook, a pencil and 3 felt-tipped pens cost 64p.

How much should you expect to pay for a package containing a notebook, a pencil, a biro and a felt-tipped pen ?

Looks like this one was overlooked or........

.....it looked too easy/difficult  

3 equations and 4 unknowns is impossible to solve

Not necessarily, when there is a constraint that the terms and solutions must be integers. Start by thinking that the notebook price is in the set 46, 44, 42.... and also in the set 54, 50, 46, 42.... and also in the set 61, 58, 55, 52..... There may be only one notebook price that is in every set. There could be more than one solution, however.

If non-integer prices are allowed, it is a bit harder, if not impossible.

Good thinking to restrict to positive intergers. This establishes fibre tip > 6, and  = 7 yields biro = 5, pencil = 5, pad = 38 and f = 8 yields 7, 6, 34 both of which yield 55! So I would guess there an infinite number of solution if some items are sold at a loss which still yield the price of one of each 55.

Trick question! Essentially the solutions form a space (5D?) with a set of constraints given by the defining relationships. What’s the name describing it [@mention:1566878603968902]?

Phil

Ian G. posted:

55p for the stationary package I think. 

Spot-on with the thinking !

Filipe posted:

winkyincanada posted:

Filipe posted:

Don Atkinson posted:

Don Atkinson posted:

Another simple one, otherwise there is a risk of only IanG and Eoink taking part !

A firm sells stationery in packages as follows :-

A notebook and two biros cost 48p; a notebook and 4 pencils cost 58p, a notebook, a pencil and 3 felt-tipped pens cost 64p.

How much should you expect to pay for a package containing a notebook, a pencil, a biro and a felt-tipped pen ?

Looks like this one was overlooked or........

.....it looked too easy/difficult  

3 equations and 4 unknowns is impossible to solve

Not necessarily, when there is a constraint that the terms and solutions must be integers. Start by thinking that the notebook price is in the set 46, 44, 42.... and also in the set 54, 50, 46, 42.... and also in the set 61, 58, 55, 52..... There may be only one notebook price that is in every set. There could be more than one solution, however.

If non-integer prices are allowed, it is a bit harder, if not impossible.

Good thinking to restrict to positive intergers. This establishes fibre tip > 6, and  = 7 yields biro = 5, pencil = 5, pad = 38 and f = 8 yields 7, 6, 34 both of which yield 55! So I would guess there an infinite number of solution if some items are sold at a loss which still yield the price of one of each 55.

Trick question! Essentially the solutions form a space (5D?) with a set of constraints given by the defining relationships. What’s the name describing it [@mention:1566878603968902]?

Phil

Well Phil, looks like you got there in the end. Well done.

I'll leave winky (or others) to respond to your name request ! (I don't know the answer)

Don Atkinson posted:

The two wheels on each axle of a railway wagon are rigidly connected together.

When an engine negotiates a bend one would expect one of the wheels to skid (because one wheel has to go
further than the other).

Neither wheel skids, however, even if the rails aren't banked.

Why?

PS I don't think that the solution usually quoted completely eliminates wheel skid at all speeds, but it certainly plays a significant part.

The wheels are tapered like shallow truncated cones. They move relative to the track so the outside wheel has an effectively greater diameter than the inside wheel during a turn. Super elevation of the track (banking) assists a bit, but the taper is the main thing. The taper also helps stop the wheel flanges from hitting/rubbing the rails when the train is going straight.

13325 units for A to F

fatcat posted:

20

It’s more than 20.

Ian G. posted:

13325 units for A to F

It’s less than 13325 miles