# An Assortment Of Puzzles

phgao
Posers and Puzzles 27 Apr '05 06:48
1. 27 Apr '05 06:48
1. If a + (1/(b+(1/c))) = 37/16

Then what does a + b + c = ?

2. What is the largest possible size of an angle of a triangle formed by joing the midpoints of three edges of a cube?

3. There are 10 couples at a party. Each pair of people shake hands, except that no woman shakes hands with either her husband or another woman. How many handshakes are there?
2. 27 Apr '05 11:34
Originally posted by phgao
1. If a + (1/(b+(1/c))) = 37/16

Then what does a + b + c = ?

2. What is the largest possible size of an angle of a triangle formed by joing the midpoints of three edges of a cube?

3. There are 10 couples at a party. Each pair of people shake hands, except that no woman shakes hands with either her husband or another woman. How many handshakes are there?
1. 10 (2+3+5)
3. 27 Apr '05 11:40
Originally posted by kody magic
1. 10 (2+3+5)
Looks good to me!
4. 27 Apr '05 11:42
The answer is corect but would you mind posting how it was achieved?
5. 27 Apr '05 12:39

Kody has sent me this:
Not seen this before and solved through a combination of workings and assumptions.

Workings would be:

37/16=2.3125
So, a+(1/(b+(1/c)))=2.3125
1/(b+(1/c))=2.3125-a

Now as the left is obviously going to be a fraction (as it is 1/something) I assumed a=2, so:

1/(b+(1/c))=0.3125

Again, 1/c is going to be a fraction, so 1/b must be b=3, as 0.3125 is just under 0.3333 (or 1/3), so:

1/(3+(1/c))=0.3125
1=0.3125(3+(1/c))
1=0.9375+0.3125(1/c)
0.0625=0.3125(1/c)
0.2=1/c
1=0.2c=5

Not very scientific, but it got there!

But the thing that I dont quite comprehend is how b=3 was got. You can get a=2 as 1/(b+(1/c)) is a fraction but how can you just say that b=3, i'm stuck to find the reasoning there as 5/16 is already a fraction, you cant just change it into 0.3125 and say if b=3 it is only slightly bigger than .3125.

If we did it that way, then when we get 1/(b+(1/c) = 5/16 we can just flip it to get: b+1/c = 16/5
b+1/c = 3.2
And you can just say that b=3 and c=5, and i'm sure that isnt the way to get the values. Can anyone else find a way that doesnt use guesses even though they are fairly obvious.
6. 27 Apr '05 14:353 edits
Originally posted by phgao
1. If a + (1/(b+(1/c))) = 37/16

Then what does a + b + c = ?
1)
You should really define what type of numbers a,b,c are.
Are they reals, (positive) fractions, (positive) integers, etc?

2) I can't see one greater than 90 degrees.

3)
There are (20 x 19)/2 = 190 ways for everybody to shake hands with each other.
There are (10 x 9)/2 = 45 ways for the 10 women (not) to shake hands with each other.
There are 10 ways for the women (not) to shake hands with their respective husbands.

Hence required number of ways = 190 - 45 - 10 = 135

In general, with 2n couples number of ways = 3n(n-1)/2

7. PBE6
Bananarama
27 Apr '05 15:271 edit
Originally posted by phgao
1. If a + (1/(b+(1/c))) = 37/16

Then what does a + b + c = ?

2. What is the largest possible size of an angle of a triangle formed by joing the midpoints of three edges of a cube?

3. There are 10 couples at a party. Each pair of peo ...[text shortened]... ither her husband or another woman. How many handshakes are there?
1. Yeah, THUDandBLUNDER is right about defining what kind of answer you're looking for. I assumed you wanted positive integers, because they're nice and round, and the stipulation usually gives you the info you need to answer the question. Here's how I did it:

a + 1/(b + 1/c)) = (abc + a + c)/(bc + 1) = 37/16

I assumed that the numerators and denominators were the same, and equated them.

bc + 1 = 16
bc = 15

abc + a + c = 37
a(15) + a + c = 37
16a + c = 37

Then I tried values for "a". Letting a = 1 gave c = 21, which seemed odd for this type of question. However, letting a = 2 gave c = 5, and subsequently b = 3. Nice round numbers, so I was pretty sure I had your solution in mind.

a + b + c = 2 + 3 + 5 = 10.

2. I'm not sure, but I think the maximum angle is 90 degrees. You can create a right angle triangle by starting at any mid-point and moving to the mid-point opposite the first on the same face, and then connecting point 1 to point 3. Picking a different point for point 2 results in a longer triangle with the same base. If you imagine a right angle triangle laying on it's hypotenuse like so --> /\, stretching the triangle upwards will shrink the angle at the top (the 90 degree angle). So any elongation will result in a smaller angle. That's why I think the largest possible angle is 90 degrees.

3. All 10 men will shake hands with each other, which results in (10 choose 2) handshakes, and each woman will shake hands with 9 men (not her husband). So the total is:

(10 choose 2) + 10*9 = 45 + 90 = 135 handshakes.
8. 27 Apr '05 17:48
Originally posted by THUDandBLUNDER

In general, with 2n couples number of ways = 3n(n-1)/2

That is, with n couples (not 2n).
9. 27 Apr '05 19:19
Originally posted by phgao
3. There are 10 couples at a party. Each pair of people shake hands, except that no woman shakes hands with either her husband or another woman. How many handshakes are there?
It's impossible to tell, because there's no information on whether all couples consist of one man and one woman.
10. PBE6
Bananarama
27 Apr '05 19:49
Originally posted by Nordlys
It's impossible to tell, because there's no information on whether all couples consist of one man and one woman.
That's true! I wonder how we missed that? Oh, I know - we're all dead below the waist on this forum.
11. 27 Apr '05 20:032 edits
Originally posted by PBE6
That's true! I wonder how we missed that?
We missed it because the obvious implication is that we are speaking of husbands and wifes only.
Or should we show our puzzles to a bunch of hair-splitting, nit-picking forensic lawyers before posting them?
12. PBE6
Bananarama
27 Apr '05 20:121 edit
Originally posted by THUDandBLUNDER
We missed it because the obvious implication is that we are speaking of husbands and wifes only.
Or should we show our puzzles to a bunch of hair-splitting, nit-picking forensic lawyers before posting them?
Technically, the terms "husband" and "wife" apply to married homosexuals, too. And I've watched enough "L Word" and "Queer As Folk" that I should have snarked out Nordlys' post myself.

As far as consulting with RHP legal staff before posting, I prefer boxers to briefs, legal or otherwise. Besides, I think the hair-splitting, nit-picking forensic lawyers only tell you if you have crabs or not.

13. 27 Apr '05 20:572 edits
Originally posted by PBE6
Technically, the terms "husband" and "wife" apply to married homosexuals, too. And I've watched enough "L Word" and "Queer As Folk" that I should have snarked out Nordlys' post myself.

I prefer to apply the principle of Occam's razor, especially if the alternatives leads to no answer at all.

I didn't particularly mean RHP lawyers.
Any lawyer would be happy to sniff out double meanings etc.
At the right price, of course.

14. 27 Apr '05 21:22
Originally posted by THUDandBLUNDER
We missed it because the obvious implication is that we are speaking of husbands and wifes only.
Or should we show our puzzles to a bunch of hair-splitting, nit-picking forensic lawyers before posting them?
Well, if you can do the hair-splitting and nit-picking yourself, that would be fine, too. ðŸ˜‰ Seriously, you have to be prepared for trick questions in a forum like this, so if I encounter a puzzle which seems trivial, I'll start looking for less obvious answers. So in this context, I think nit-picking is indeed a good idea.
15. 28 Apr '05 08:09
One more thing b4 I go,

The answer to Q2, is 120 degrees not 90 degrees, note: the largest possible size of an angle of a triangle in a Square is 90 degrees!

Also thanks to PEB6 for explaining the question.

The answer to C is correct.

Plus the 'couple' thing is explained in the Q- ...no woman shakes hands with either her husband...!! So why the confusion?

Here's another

There are n prisoners in a row of locked cells. With the return of the King from the Crusades, a partial amnesty is declared and it works like this. When the prisoners are still asleap, the jailer walks past the cells n times, each time he turns the lock in every cell (so that every cell is now open). On the second pass he turns the lock on every secod cell (meaning that these cells are now locked again). On the third pass, he turns the lock on every third cell, and so on. In general, on the kth pass, he turns the lock on every kth cell. The question is, which cells are unlocked at the ned of the process so that the prisoner is free to go?