A Fistful of Brain Teasers

Filipe posted:

Hope half miles are not allowed on the first leg!

Correct ! Half miles are NOT allowed !

Filipe posted:

Filipe posted:

The first four unique triples are (3,4,5), (5,12,13), (8,15,17), (7,24,25) so the product of their hypotenuses is 27,625.

It’s 5,525 dividing by 5

Ian G offered that solution last night (see at the top of this page. Sorry !

I think I already indicated that the answer is more that 50 but less than 1000.

I have never been able to find a simple "formula" for this problem. But once you have found the first two, the other two seem to follow fairly easily.

Sadly, none of your four triangles feature in my list

32,126,130 (1,63,65)*2

50,120,130 (5,12,13)*10

66,112,130 (33,56,65)*2

78,104,130 (3,4,5)*26

That's really well done Phil. ! i'm impressed !

I'm going to give you 8/10 because you are so close to getting the final answer.

Remember, AF has to be the smallest integer value that works, with all the other dimensions also integers.

So, the answer lies between 50 and 130.

I want to go back to the stationary store questions briefly since I have found a way that avoids all the trial and error searching. 

Take N,B,P,FTP as the price of notebook, biro, pencil and felt tipped pen. The information given provides the equations

 N+2B = 48

N+4P = 58

N+P+3FTP =64

3 equations 4 unknowns so no simple general solution.

We are asked only to find the cost for N+B+P+FTP, so let’s try and see how we go.

If we add all three equations we get

3N + 2B + 5P + 3FTP = 170

 which  for reasons that will become clear we can write as  

 3N + 2B + 2P + 3P + 3FTP = 170

 Subtracting the 1^stfrom the second original equations gives 4P-2B=10 or 2P=B+5

 Now using this to substitute for the 2P we get

 3N + 3B + 3P + 3FTP = 165

or

 N +  B  +P + FTP = 55 as required.

So there are a multitude of solutions with prices for each object between 1p and 52p. 

Ian G. posted:

I want to go back to the stationary store questions briefly since I have found a way that avoids all the trial and error searching. 

Take N,B,P,FTP as the price of notebook, biro, pencil and felt tipped pen. The information given provides the equations

 N+2B = 48

N+4P = 58

N+P+3FTP =64

3 equations 4 unknowns so no simple general solution.

We are asked only to find the cost for N+B+P+FTP, so let’s try and see how we go.

If we add all three equations we get

3N + 2B + 5P + 3FTP = 170

 which  for reasons that will become clear we can write as  

 3N + 2B + 2P + 3P + 3FTP = 170

 Subtracting the 1^stfrom the second original equations gives 4P-2B=10 or 2P=B+5

 Now using this to substitute for the 2P we get

 3N + 3B + 3P + 3FTP = 165

or

 N +  B  +P + FTP = 55 as required.

So there are a multitude of solutions with prices for each object between 1p and 52p. 

Brilliant !

13,84,85

36,77,85

40,75,85 (8,15,17)*5

51,68,85 (3,4,5)*17

I needed to calculate more base triples and low and behold!

85 is still too high !

So now, we are between 50 and 85 !

Getting there !

Have another think about your previous solution at 130, but make sure the arithmetic is correct.

39,52,65

25,60,65

16,63,65

33,56,65

Looks like you cracked it Phil. Well done !

Don Atkinson posted:

Looks like you cracked it Phil. Well done !

I needed the primitive Pythagorean triples (a,b,c are coprime - only common divisor is 1) and then find the lowest linear combination of four of them.

Filipe posted:

Don Atkinson posted:

Looks like you cracked it Phil. Well done !

I needed the primitive Pythagorean triples (a,b,c are coprime - only common divisor is 1) and then find the lowest linear combination of four of them.

Yep ! Not an easy one. But you and Ian had a good crack at it. Well done.

Find the marked angles a, b, and c. O is the centre of the circle.

This and the next few are designed to be very simple (after the trauma of the Road AF and the Notebook drama!).

Basically, he who can find his O-Level maths book the quickest, will be the winner !

A potato merchant has to be able weigh quantities from one pound to one hundred and twenty-one pounds, to the nearest pound.

To do this, what is the minimum number of weights he requires and how heavy should each weight be?

Don Atkinson posted:

Find the marked angles a, b, and c. O is the centre of the circle.

This and the next few are designed to be very simple (after the trauma of the Road AF and the Notebook drama!).

Basically, he who can find his O-Level maths book the quickest, will be the winner !

Is AB = BC in this case?

ursus262 posted:

Don Atkinson posted:

Find the marked angles a, b, and c. O is the centre of the circle.

This and the next few are designed to be very simple (after the trauma of the Road AF and the Notebook drama!).

Basically, he who can find his O-Level maths book the quickest, will be the winner !

Is AB = BC in this case?

Hi ursus262,

You should be able to answer that question when you have solved the angles a, b, c.

(The drawing isn't entirely accurate ! but the quoted angles are the correct angles.)

Assuming BOD is a straight line, a=55 b = 125 and c=65 methinks

Ian G. posted:

Assuming BOD is a straight line, a=55 b = 125 and c=65 methinks

55 (half the centre angle), c = 60 (similar triangles) , 130 (360 degrees - ...)

Filipe, my logic is:

OB and OA are the same length and the angle OAB = 60 so OAB must be an equilateral triangle with three angles = 60. 

Thus the angle BOC has to be 110-60 = 50. ( and so the figure is not symmetric about BOD)

The triangle BOC is isoscoles with apex angle 50 so the  two base angles are 65 each.

So c=65 and b= 60 +65 = 125.

a is 55 by either adding up other triangles now that you know b and c or half the central angle.

a is 50 degrees, b is 130, and c is 60.

Hi Ian, BA is not known so OAB is isosceles rather than equilateral.

I think the potato one might be 7 - 1, 2, 4, 8, 16, 32, 64.

Hi all,

Just to re-inforce what I said earlier......... the drawing isn't entirely accurate, ie it might LOOK symetrical etc, but it isn't. I haven't marked any "equal" lines.

A, B, C and D are all on the perimeter of the circle. O is the centre of the circle.

And as I implied earlier, this one came straight out of my O-Level geometry book

There is a trap because there are inconsistencies. I think Don has incorrectly set OAB to 60 when it should be 62.5. Hence c 62.5 also. b = 125.

Naughty teacher!

Phil

Filipe posted:

Hi Ian, BA is not known so OAB is isosceles rather than equilateral.

You don't need to know BA:

OB and OA are the same and the base angle OAB is 60 degrees so angle ABO is also 60 so it must be equilateral.