Solutions to riddles

Using straightforward calculus, I got that the distance to the goal line should be the geometric mean of the distances from the running line to the two goal posts. That implies that the sum of the angles to the goal posts should be 90 degrees.

"soccer problem"

Lets say the player is at A

Point were his running line intersects the baseline B

The left Goal post C

The right Goal post D

The angle to maximize is then notified as CAD(angle at point A between C and D)

Shalke: Can you desribe what angles you think should be equal using this notation?

Devil: Suppose you mean the angles BCA and BDA you are right :)

Shalke: If you mean angles BAC and BDA you are right :)

I make it 100(1+pi-4*(atan(1/2)) metres, that is about 228.700222 metres. I believe that this is optimal.

To draw this shape, start with a square ABCD, side 100m. Draw arcs, radious 100m, centres A and D. Find E, the midpoint of BC. Draw tangents from E to the arcs. Start at A, swim along the arc until it meets the tangent, then along the tangent, to E, then along the other tangent until it meets the other arc, then along that arc to D.

Nick you are getting very close :-)

Although this can be improved(but not even by a meter)

Marius,

if I did understand the task the distance is 100 m between the two shores.

The swimmer is not on one shore, so he has to swim less than 100 m.

If this is right (not true, because I'm no mathematician), than the smimmer has to swim a smaller distance than 100 m.

So a quarter of a 100 m circle brings him to the shore.

Jupp he doesnt know where he are, he doesnt know the direction to the shore. And after starting swimming he has the worst of luck in choosing his initial direction ;)

Hi Johan

Jupp and Schalke is one person.

I meant neither angle You described. If the forwards aimes at the post, he will not make any goal except by failure.

So he will choose any point between the two posts.

Tasmanian was right with his 45 degree angle. When I read this, I remembered about the 180 degree rule in triangular. (Good that this time in school have passed by. When horror dreamings then they come back in the night.)

Johan,

the case I presumed for the swimmer is the worst case.

Johan,

explaining this. He is only 1 cm away from the shore.

And than You are right. He starts this circle in the dircetion of the shore.

My case is 90 degree better than the worst case. (sigh)

So it's 50 m - or not?

I meant BAC and BAD, but the sum is 90 degrees in both cases (and AB^2 = BC x BD).

50 m radius is meant in the last message.

Jupp, worst case is not starting in the middle of the river. It is when choosing a starting point(as it can be any for the swimmer) he then chose the worst possible direction to start swimming.

My mental problem was putting the center of the circle in the river. So I could not understand all the good hints I got. :( (Beg Your pardon.)

I think that Jupp solution is correct, I figured it myself, but in the "battle" I enjoyed to watch, not to argue. Here's my argumentation: imagine a circle with 100m radius (not 50 cm, nor 100m diameter). If the swimmer goes on the circle only for 1/4 of it (pi*R/2~157m) in the worst case he'll finish on the other shore, otherwise he'll reach even the autoroute :). if you need a starting point to draw the circle, you may find a star to look at it, or if it's cloudy, just fix a goal point and swim in circle....

Draw that 1/4 circle and connect the start and finish point of your drawing. Now imagine that the shore is parallel to this line without intersecting the 1/4 circle. If the shore is close enough to this line, the circle isn't big enough to reach the other side of the river.

5*(5-(1/5))

Replies to Antkam:

POW-riddle: It was stated in the riddle that if anyone said anything except one of the three colors, EVERYONE would be killed. And it is still possible to save everyone except one, as stated earlier! :-)

Scale riddle: This is actually a sub-problem to my second scale riddle, with 4 weightings. With a "known" ball, you start out weighting 5 unknown on the left side and 4 unknown plus the known ball on the right side.

Angels and devils: I really liked your twist - I'll sure use it next time I ask this riddle! :-) My question is as follows: "If I ask the other figure if this door leads to heaven, would his answer be "google"?". If the reply is "yahoo", then it's the door to heaven. If the reply is "google", it's that other, hot place...

re: soccer problem -- i tried not to use calculus, and arrived at an intuitive guess which turns out to be right. (my intuitive formulation was: BAC = CAE where E is the last corner of the rectangle ABDE, i.e., AE is parallel to the goal line.) now my question is: is there a way to prove this is optimal, WITHOUT using calculus? or perhaps using the concepts of calculus (differentials), but without using actual calculus calculations? BTW, Mr Johan Roos, you have the most awesome puzzles! what is your profession? math teacher? :)

re: my own puzzles:

re: my scale problem: marius -- my problem requires you to use only _3_ weighings. i am not sure i understand why you say it is a subcase of your _4_ weighing problem. if you think using 3 weighings is impossible, re-read my exact problem statement again.

re: my google/yahoo angel/devil puzzle, marius is right. i read this variant from a book by raymond smullyan, a logician. i dont have the book with me (someone borrowed it) but IIRC he gave an answer that uses "if and only if"! :)

re: my POW "variation" -- sorry! i didnt read this thread before posting in the original thread! :) :P

Marius's solution to the "angels and devils" riddle is correct. "If and only if" is not required for this one, though, I believe, Smullyan does pose other problems in which you need it.

Work out all of the cases, and you find that Marius's solution works. Regardless of whether it is the angel or the devil who is answering, and regardless of which word means yes and which means no, if you get the opposite word as a reply, it is the door to heaven, but if his reply uses the word you used in your question, it is the door to hell.

I am a big fan of Raymond Smullyan, too, though it's been several years since I read any of his books. Does anyone know, is he still publishing new collections of riddles?

Scale riddle: The reason it's a subproblem of the scale riddle with 4 weightings, is that with 4 weightings, you start out with 39 balls, and put 13 balls on each side. Let's say the scales are equal. You then have 13 balls + lots of balls with "known" weight (although you only need one), and 3 weightings left - which was your puzzle, Antkam. :-)

marius -- my scale puzzle is _14_ balls and _3_ weighings. so it is not a subcase. you keep thinking it is a subcase because you intuitively believe 14 balls and 3 weighings is not possible, so you read my "14" and think "13" instead!! ;) ;) and i dont blame you -- 14 balls and 3 weighings is indeed impossible using your problem statement. however, read my exact problem statement again... my puzzle is a neat little twist that you may not suspect on first glance.

Sorry, Antkam. Yes, I see the difference now - in your puzzle, one does not have to find out if the ball is heavier or lighter than the others. It gets much easier then :-)

Starting out: 5 vs 4 + 1 known. If the scale is NOT equal then, you can continue as in the original problem, and even determine if the ball is heavier or lighter, so let's say these scales are equal.

Of the remaining 5 balls, put 3 on the left side of the scale, and 3 known ones on the right side. Again, if the scales are NOT equal, we're back to the original problem, so let's say they are, again, equal.

Measure one of the remaining 2 balls against a known ball. If they are equal, the last ball is different, but you can't tell if it's heavier or lighter (ofcourse, if they are NOT equal, you can also find out if it's heavier or lighter).

A man at a certain point on earth. He goes 100m to the South, then 100m to the West, and finally 100m to the North, and he arrives where he has started. He is not on the North Pole. Is this possible?

The man is 100m North of the line of latitude that circles that part of the Earth in 100m. That sounds vague. He's in the Southern Hemisphere, close enough to the South pole such that after walking South 100m, his 100m walk west is enough to cicle that section of the Earth.

Yes, if he is whenever on the latitude line which is about 115.9m to the north from the south pole. The latitude line which is about 15.9m to the north from the south pole has length 100m.

Thank you. That was much clearer than how I phrased it.

Re: Coin Problem (two bags with 20 coins each...)

Marius says he can do in 9 weighings and asks if it can

be done in less.

Yes it can.

note 1: If you weigh n coins, then the total weight will

be slightly under n ounces, UNLESS those n coins contain

the heavy coin and do not contain the light coin. In this

case, it will weigh over n ounces.

note 2: The optimal weighings have a very elegant description.

Geography problem: good. Now let's say he's NOT 100m north of the latitude of length 100m. Is that still possible?

It's still possible as long as the man starts 100m North of a latitude line that divides evenly into 100. I have a feeling that this isn't what you mean, but whatever.

100+100/(2*pi*n)meters, n corresponds with the amount of circles he makes going westwards.

100+100/(2*pi*n) meters north of the southpole ofcourse

To mslano: I think this *is* what he means. :) And about my previous message - my intention wasn't to specify your solution (your description was precise enough). You were simple a few seconds faster. :)

mslano: yeah, that was all I meant. Not a very deep puzzle, I admit.

re: glenn's 20-ball-in-each-bag puzzle

i think i can do it in 6 weighings. is that optimal? very nice problem!

re: the soccer problem

i still cannot think of any way to solve it, WITHOUT using at least a little calculus. is there a non-calculus way to solve it?

soccer problem

I dont know of any way to solve this other than calculus.

You can for example express the angle as a function of the players distance to the baseline. Taking the derivative and setting this to zero. After finding this maximum point you can start examining the relationships between the angles at this point.

Re: Answer to bags with 20 coins weighing problem.

Yes, 6 weighings is indeed optimal!

Nice going.

From the first hint, the only way to identify the "bad"

bag is to have a weighing that includes the heavy coin

but does not include the light one. Thus, you need to

find the minimal set of weighings such that at least one

of weighings is guaranteed to contain the heavy coin and

not the light one *regardless* of which of the coins are

heavy and light.

The answer is 6 because 6 choose 3 equals 20. So take the

20 coins from one bag and pair them up with the 20 3-member

subsets of {1,2,3,4,5,6}. The nth weighing will include

exactly those coins whose subset contains n -- set the

scale to 10 ounces since each such weighing includes

exactly 10 coins. Now suppose the selected bag is the

bad one. Since none of the subsets contains another, the subset associated with the heavy coin must contain an element s that is *not* in the subset associated with the light coin. Hence, the s'th weighing will contain the

heavy coin but not the light one.

Note that in *any* set of say w weighings, we can associate

a subset of {1,2,3,...w} with each coin such that the

k'th weighing contains those coins whose subset contains

k. To be a solution to the puzzle, the subsets must

have the property that no subset is contained in another.

Then the optimality of the above solution follows from Sperner's Theorem

Sperner's Theorem: If you have an n-element set, the

maximum number of subsets you can choose with none of the

subsets being included in another is n choose k where

k = n/2 rounded to an integer.

This is "obvious" if the chosen subsets all have to be the

same size; the point is that you can't do any better by

using different sized subsets.

Sperner's theorem implies that 5 weighings are not enough

for 20 coins (5 choose 2 = 10). Since we have a 6 weighing

solution this must be optimal. Note that Sperner's theorem

also implies that 6 weighings are not enough for 21 coins

(since 6 choose 3 = 20 < 21).

I think I found a solution that decreases slightly on Nick Wedd's number of 228.700222 metres. My number is about 227.829 meters.

To draw this shape, start with a recrangle (not a square) ABCD, (vertical) sides AB and CD 100m and (horizontal) sides AD and BC 100+x meters. Draw arcs, radious 100m, from centres A and D. Find E, the midpoint of BC. Draw tangents from E to the arcs (like Nick's solution).

Now draw tangents from A and D to the arcs.

Start at A, swim along the tangent until it meets the circle, then swim along the arc until it meets the other tangent, then along the tangent, to E, and then make the mirror swim from E to D (along two tangents and an arc in ther middle.

Now, if 100+x=100 the solution is exactly Nick'a solution.

If 100+x=200/sqrt(3) then the solution is exactly what my previous post was. (two straight lines that are tangent to circles although id din't have noticed that when I posted it)

And finally if x=4.35901 then the total swimming distance is about 227.829 meters which is optimal (for the way I described). Is it optimal in general?

Nice going yper, your "river man" solution is the optimal one :-)

Thnx Johan.

Two weeks have passed from my riddle #1 and only passenger tried to answer it. So here it is:

yper riddle #1: Why is sex better than a box of chocolates?

yper answer #1: Trick question! Only hex is better than chocolate!

ypers astro generalization

I think flying three edges of a tetrahedron with height 100km would solve it and be similar to the triangle solution for the river problem. Generalizing this tetrahedron to a shape with "yurt curve" edges of ypers river solution maybe gives the best solution.

Solution to volume of remaining part of sphere after

drilling a 6 inch long cylindrical hole straight through

the center.

It helps to draw a picture of the 2-D cross section of

a plane that includes the central axis of the cylinder.

Now draw a right triangle -- one edge goes from the

center of the sphere straight along the axis of the cylinder

to the end of hole, another edge goes from the center of

the sphere to a point where the hole intersects the surface

of the sphere, the third connects the two endpoints and

this edge is a radius of the cylinder. The length of the

hypotenus is R = the radius of the sphere and length of

the other two edges are 3 (half the height of the hole)

and r = the radius of cylinder. So by the pythagorean

theorem, r = square root of (R^2 - 9). The radius of

the planar face of the circular cap is the same as the

radius of the cylinder and the height of each cap is

R-3. We can now substitute into the volume formulas

to get each one in terms of R. Now if we take the

volume of the sphere subtract the volume of the cylinder

and subtract twice the volume of a circular cap (there

is cap on each end of the hole), we find that all the

terms involving R conveniently cancel out and we are

left with 36 Π

Surprisingly, the final answer does not depend on

the size of the original sphere nor on the size of

the cylinder (these two sizes are related).

There is a short "meta-answer" that depends on the fact

that there is indeed a solution.

Suppose there is solution. Since we are not told the

radius of the cylinder, the answer doesn't depend on this

and we can choose any radius we like. Suppose the radius

is 0. Then we don't remove anything and we are left with

a sphere of diameter 6 (radius of 3). Hence, the

remaining volume is 4/3 Π r^3 = 36 Π

re: glenn's sphere with cylindrical hole problem

i think the fact that the volume does not depend on R is very neat. so in some sense, and unfortunately, the neatness is entirely in the puzzle statement itself, not in the solution. :) in other words, this is not so much a neat puzzle, as it is a neat mathematical factoid.

anyway, here is a fun alternative way to prove that the volume does not involve R (or r). imagine the cylindrical hole is lined up vertically, and consider a horizontal plane at height h above the center. the cross section on this plane will be a torus (donut). the outer radius is sqrt(R^2 - h^2) and the inner radius is r. so the torus area is pi*(R^2 - h^2 - r^2). Now, R^2 = 3^2 + r^2, so the torus area is pi*(3^2 - h^2). Integrating over the range h = -3 to +3, we have shown that the volume does not include the term R or r. Then the last part of Glenn's solution (setting r=0) applies.

Geometrically, what happens is that not only is the volume independent of R, but also the area of the cross-sectional torus at height h is same as that of the cross-sectional circle at height h in case of a radius=3 solid sphere (i.e. r=0 case).

This one also needs to get back to the top... :-)

The rope riddle.

Go 18 meter up along one of the ropes and carry the beginning of the rope, you are climbing, with you. Cut the other rope at 18 meter and hold the loose end; so 2 meter is still hanging from the ceiling. Jump over to the short 2 meter rope. Connect de 18 meter rope with the 20 meter rope. Disconnect the 38 meter rope from the ceiling(don't drop it to the floor :-) ) Connect the end of the 2 meter rope to the vacant hook in the ceiling. The 2 meter rope is now on both ends connected to hooks in the ceiling. Dubble the 38 meter rope and hang the middle over the 2 meter rope. Climb down along the dubble rope and pull at one end of the 38 meter rope so it come down. And the way to freedom is there.

correct, except the ropes are not hanging by hooks so you have to tie the short rope into a loop.

Worm Riddle:

Unless I am missing something, the worm uses 304 hours. 300 hours for all the 3x6000 pages times 1 minute pr page, plus 4 hours for the back cover of the first volume, both covers on the second volume and the first cover of the last volume (since it starts at the word A and ends at Zzzax, it doesn't eat through the first and last cover).

Sorry Mark, it takes less than that.

I meant Marius sorry

1 Hint will be posted tomorrow