T = V - we'll let other people get the rest. I'm glad the Translator is back

[Honestants and swindlecants XI]

When the Randomcant considers the answers that the Honestant and Swindlecant would give, how exactly does he evaluate those answers? I see two possibilities:

1. The Randomcant compares the answers he would give if he were an Honestant vs. if he were a Swindlecant.

2. The Randomcant compares the answers the other two gods would give if you turned and asked them the question instead of him.

The reason I ask (and I may be barking up the wrong tree here) is that I'm not sure how the Randomcant would answer the question "Are you sitting in the middle?" or any other question that depends on the relative order of the three gods. Assuming the Randomcant was on one end with the Swindlecant in the middle and the Honestant on the other end, his answer to "Are you sitting in the middle?" with the first method would be yes (H = no, S = yes => R = yes), but his answer with the second method would be no (H = no, S = no => R = no)

The OP states that every step was either 3 feet or 2 feet, depending on the person, and that the distance was just 200 feet (not 204). The race could have been 200 feet in just one direction and the outcome would have been the same. Trap Sprung. The OP doesn't say how it was done, just that it was.

Exactly. Having to turn around makes the difference between 67 or 68 total steps for the firefighter, but the nurse still wins either way. Now, if they had traveled 204 feet (102 each way), that would be different - they would tie.

The best egg for him to choose is the next to last egg (the one that is 990 feet away), so he doesn't have to go as far to get his 49th egg. I'm not sure who wins yet - that will take a few more calcs than I have time for right now.

For each leg of the race, the firefighter has to take 100/3 = 33.33 => 34 steps, or 68 steps total.

For each leg of the race, the nurse has to take 100/2 = 50 steps, or 100 steps total.

She takes 3 steps for every 2 of his, so, because 100/68 < 3/2, the nurse wins.

1. double "t" in sentence.

2. misteaks instead of mistakes

3. there are two, not three mistakes in the sentence

A tidy answer? Maybe...

OK, based on some info from previous posts and some math, we can say (given a triangle with unit length sides):

The height of the triangle is √3/2.

The distance along the perpendicular bisector from one edge of the triangle to the center of the opposite internal circle is 1/2.

The distance from one vertex of the triangle to the center of either opposite internal circle is √2/2 (the hypotenuse of a triangle whose legs are each 1/2).

The radius of the internal circles is half the distance from the center of the circle to the nearest vertex of the triangle = (√3-1)/4 or (√3/2 - 1/2)/2.

OK now to the construction:

Call the vertices of the triangle A, B, and C.

1. Create a circle about C radius CB. Create a circle about B with same radius. Mark the point of intersection as D.

2. Create a circle about D radius DC. Mark the intersection with circle B as point E.

3. Create a circle about E radius EC (this radius is double the height of the triangle = √3). Create a circle about A radius AD (also length √3). Mark the intersection as point F.

4. Notice that the length of BF is √2 (one leg of a right triangle with hypotenuse = √3 and other leg = 1). Construct the midpoint (as above) from B to F, and mark it as G.

5. Now, length of BG is √2/2. Create a circle about B radius BG. This circle will pass through the center of both internal circles opposite from B.

6. Create a circle about A and C also radius BG. This marks the centers of all three internal circles.

7. Construct the midpoint from any interior circle's center to the nearest vertex. This midpoint is on the selected interior circle. You can use this radius to create all three circles.

All in all, I don't think I'd call it easy peasy...

Nice... Finally!

I pulled out the old engineering compass and tried to make it work. The proof is in the doing, right? And the OP says you have a compass (on steel, as modified, to prevent folding), not a computerized drafting tool that will only draw circles.

After 3 attempts, I gave up. When you combine the 2 methods and get the full procedure, it takes a dozen circle constructions to simply identify the centers of the small circles, and that's only because several of the points required to find point G as the midpoint of BF have already been defined and you can skip several steps in the first bisection procedure. By that point, I was far enough off that the centers were in the wrong place and I didn't even try the second bisection procedure (requiring 7 more constructions) to get the radius of the small circles. And I started with the largest triangle that I could (2" sides) and still make all the points that I required with a 6" compass (largest theoretical radius is 5.66").

So it seems we may have 3 equally valid procedures, given the difficulty of using a real-life compass to construct these tangent circles:

1. the procedure I suggested in post 19

2. the procedure I suggested in post 4, after using sihyunie's bisection procedure (instead of folding) to find the midpoint of one side of the triangle

3. jb_riddler's procedure in post 34

If I had the worthless computerized drafting tool which could only draw circles, I think all three are valid, because the tool would pick out intersections or tangents with exact precision. With a compass, I doubt either of the three will actually get you the three small circles, unless you maybe have a very large bar compass and the means to keep the point and the pencil perpendicular to the plane of the triangle.

It would be interesting to introduce an assumed error into the puzzle and figure out which is the best method to use in real life. For example, assume:

1. the pencil thickness and compass point are both about 1/50 of an inch thick (based on 0.5 mm pencil lead, meaning the distance plotted could be 1/25 of an inch away from true distance)

2. the location of a marked intersection is also about 1/50 of an inch from it's true location (true center of shape defined by two intersecting 1/50 inch-thick arcs)

3. the location of a marked tangent is about 1/20 of an inch from it's true location (true point of tangency, if one exists, or point of minimum distance between non-intersecting circles, or maximum distance between intersecting circles)

For each method, figure out how close the circle centers are likely to be to their theoretical locations, how close the circle radius is likely to be to the theoretical value, and how close the circles are likely to being tangent with each other and with the sides of the triangle.

It should be possible to see how error could propogate through each step of the construction and somehow weigh the relative benefits of each construction method. But that would not be an easy peasy problem, either.

All in all, a very good puzzle... but not at all easy peasy!

Well HH, do you want the honor of putting a tidy answer in there and call it a wrap? - I thought one of two young ladies may have put this t'bed before you came back to it.

I thought they might, too - but I guess they've left this one alone. Actually, I've still got a little snag, or I would finish it up.

Obviously, one of the missing pieces was a method to bisect one side of the triangle, which we've got a procedure for now.

But I still need a means for cleanly getting the center of the circles. After bisecting a triangle edge, I have a method for putting one arc through the center of the circles, but it would be really good to have another arc to go through them so I can define the centers based on that intersection and not have to rely on any eyeballed tangents. And to be really precise, I'd like to have a means for getting the exact radius of the small circles by intersections without trying to be tangent to the triangle edge.

I know a radius of S1 * √2 or S1 * (√3 - 1) / 2 will do to find the center of the circles if I try to use one of the triangle corners as the center of my construction arc. But I don't have one of those distances yet (nor can I find a power of 2 multiple of them, which I could eventually subdivide). I'd prefer to use the S1 * √2 distance, because then I'd have a real intersection and not the common point of two tangent arcs. There may be other distances from other points of intersection from the circles I've constructed so far, but none that are working for me yet.

The radius of the small circles is S1 * (√3 - 1) / 4, which is half of S1 * (√3 - 1) / 2 noted above, so I could get that, assuming I could find the center of the circles (with precision). So, I guess it's not quite done yet - at least, not as done as I'd like...

y'alls is a possessive pronoun

all y'alls is plural possessive

doubt that's what you were going for, though

besides, it had an apostrophe before going possessive

in general, I think most indefinite pronouns will work. Adding to the list from other replies:

one's

another's

something's

somebody's

anything's

anybody's

Thanks for the sketch - and the drawing program tip.

And especially thanks for the final pieces to a puzzle that's bugged me for four months now. Way to go

An interesting approach - normally the bisection is done by drawing arcs either side using both end points of the line and connecting the line - normally requiring a straight edge - you would be winging it with a best guess and free hand use of a pencil on paper or scribe on steel....

How to get that pinpoint accuracy?

I think we can finally get there... It takes a while, but we can get there. Check out the answer to this topic: Going around in circles, but getting where you want to go. The elusive bisection of a line using only a compass is solved, and I think that's all we were missing here

yeah - there's a ton

a Java applet I found produced 225 words. We could be at this for a while... Of course, having cheated, I won't post any, but will await everyone else finding the other 215 words

Either 3:00 or 9:00.

At no other times are the hands at right angles and aligned with minute marks.

So 3:31 is definitely an answer. In 4 minutes, it will be 3:35, and it will be (35-30)*60 = 300 seconds after the half hour (3:30), which is 10 times the 30 minutes it was before 3:30 at the last time the hour and minute hands were 90 degrees apart (3:00). But couldn't 9:31 also be an answer? Or does the phrase "this afternoon" disqualify so late a time?

When a ten year old boy Tim wanted to know the current time, his uncle Jonathan looked at his 12 hour analog wristwatch, and replied, "At the next four minutes, it will be exactly ten times as many seconds after the half -hour as it was minutes before the same half- hour last time the two hands were in right angles this afternoon."

Determine the current time, assuming that the above conversation took place after 12:00 p.m. but before 12:00 a.m.

Three questions:

1. Does the phrase "at the next four minutes" indicate that the uncle is referring to the next even increment of 4 minutes (i.e., 12:04, 12:08, 12:12, etc.) or does it mean four minutes from the current time?

2. Do you mean whole minutes/seconds, or do you consider the decimal portion of those numbers as well?

3. Are the two hands in referred to the hour and minute, or the minute and second?

Prove that the distance from where Allen and Bernard meet to where they cross the stream is the same as the distance from Allen's house to where they cross the stream.

I get that the distance from where Allen and Bernard meet to where they cross the stream is the same as the distance from Allen's house to where they meet, both distances being 0.26 miles. And here's a diagram with my solution:

When Bernard meets Allen he has covered half of distance he will travel to Chuck's house, and he has gone equal the distance from his house to Edward's.

2.88784 miles. ABC is an equilateral right triangle, which is bisected by E. BE=CE=1, and BC=2. Chuck's house is sqrt(2) due North of Allen. Allen travels sqrt(2)-1 miles east before meeting Bernard. Then they travel sqrt((sqrt(2)-1)^2+sqrt(2)^2) NorthEast to Chucks, and then Allen returns home.

I think you have the distance between Bernard's house and the meeting point as 1 mile, but the distance between that meeting point and Chuck's house is larger than 1 mile, which doesn't agree with the above point from the OP.

Bernard should travel 2 miles to Chuck's, and Allen, because he leaves his house after Bernard leaves his, should travel more than 1 mile but less than 2 to get to Chuck's. I don't read that CE=1 (necessarily) from the OP. If you place point D as the meeting point for Allen and Bernard, I read that BD = CD = BE = 1.

After cranking the numbers some more, I now think that Allen walks 0.260 + 1.000 + 0.965 = 2.225 miles.

Allen lives 0.965 miles from Chuck; Bernard lives 1.587 miles from Chuck; Allen & Bernard live 1.260 miles from each other. Edward lives 1 mile from Bernard (given), 0.766 miles from Allen, and 0.587 miles from Chuck. The point where Allen & Bernard meet is 0.260 miles from Allen's house and 1 mile from Bernard's & Chuck's houses.

But we know that Bernard has travelled only 1 mile (equal the distance from his house to Edward's) when he meets Allen.

I had 1 mile from Allen to Edward, not Bernard to Edward

So that means the distances from Bernard to Edward, Bernard to the point where he meets Allen, and from that meeting point to Chuck are all 1 mile. You'd think that would make the equations easier, but it didn't. Anywhooo... now I get:

Allen walks 0.579 + 1 + 0.815 = 2.394 miles.

Allen lives 0.815 miles from Chuck; Bernard lives 1.414 miles from Chuck; Allen & Bernard live 1.579 miles from each other. Edward lives 1 mile from Bernard (given), 1.222 miles from Allen, and 0.414 miles from Chuck. The point where Allen & Bernard meet is 0.579 miles from Allen's house and 1 mile from Bernard's & Chuck's houses.

Quick clarification request.

When they meet, do you mean Allen has covered half the distance he will travel or Bernard has covered half the distance? And from "his house to Edwards," does this refer to Allen or Bernard?

My assumption was that both referred to Bernard, as it appears Bernard is the subject of the initial clause and Allen is the object. I think good grammar would keep Bernard as the subject throughout. Although I hate relying on grammar to figure out a geometry problem...

3.017 miles = 0.327 + 1.365 + 1.325.

Allen lives 1.325 miles from Chuck; Bernard lives 2.150 miles from Chuck; Allen & Bernard live 1.692 miles from each other. Edward lives 1 mile from Allen (given), 1.365 miles from Bernard, and 0.784 miles from Chuck. The point where Allen & Bernard meet is 0.327 miles from Allen's house and 1.365 miles from Bernard's & Chuck's houses. I'm sure there's other pertinent information, but that's enough for me.

[Geometric "proof"]

Your mistake is in assuming that A1 > A1a. If A1 = A1a, then A1b = 0, and that changes the look of the diagram (imagine keeping only the part of the diagram above the dotted line). S1 would not extend below the dotted line and the cyan line could "swivel" about the intesection of the two red S2 lines, creating a whole range of triangles where the given conditions are met, but are not congruent. They would have an angle A1, sides S1 & S2, but A2 on one triangle would equal A3 on the other, and vice versa. Most importantly, S3 would not be equal for the two triangles thus created.

a interacts with 2 Ps because...well let's see if this explains it...

Spoiler for pic:

it interacts, but it doesn't have to be next to each other to interact - just that there's more than one p...

So are you saying that the case of being a double letter is special? If the word were 'apropos', would the 'a' interact with 1 or 2 'p's (it's hard to find a word with an 'a' and two 'p's that are not double letters

)? I guess what I'm looking for is: what is an 'interaction'? Is it that the 'a' sound is next to the 'p' sound and you count how many 'p's make that 'p' sound adjacent to the 'a'? Or does it just mean that there are two 'p's in a word and at least one of them is next to an 'a'? Or is it simply that there are two 'p's in a word that also has an 'a', regardless of where they are situated relative to each other? (I'm so confused )

That being said, I think that IDoNotExist has got the only answer possible: parade

There are three different ways the workspaces can be assigned: 1, 2, or 3 men in offices. Because we're not talking about specific individuals, I think that each would be equally likely, and the case we're dealing with would have a probability of 1/3. But, I'm probably wrong (at least 50% chance of that

).

And, thinking about that assertion for a moment before I hit the Add Reply button, I think I probably am wrong. If the men are M1, M2, and M3 and the women are W1 and W2, then I think you can come up with 10 different ways to arrange them in workspaces:

Office: M1 M2 M3 Cube: W1 W2

Office: M1 M2 W1 Cube: M3 W2

Office: M1 M2 W2 Cube: M3 W1

Office: M1 M3 W1 Cube: M2 W2

Office: M1 M3 W2 Cube: M2 W1

Office: M2 M3 W1 Cube: M1 W2

Office: M2 M3 W2 Cube: M1 W1

Office: M1 W1 W2 Cube: M2 M3

Office: M2 W1 W2 Cube: M1 M3

Office: M3 W1 W2 Cube: M1 M2

6 of these arrangements have 2 men and 1 woman in offices, so the probability is 60%. There, I think that's more right

the letters are how many times that letter interacts with another. It's kind of confusing.

Basicly...

say the word was apple. under 'a' and 'a', there would be nothing, because a is the only a there, it doesn't interact with any other 'a's. But there would be a '2' under 'a' and 'p' because a interacts with two p's - the first and the second. under 'p' and 'p', there would also be a '2' because the first 'p' interacts with the 2nd, and the 2nd 'p' interacts with the first.

Easier??

I'm not sure I see how the 'a' in 'apple' interacts with 2 'p's. I understand how it could interact with 1 'p' - the one it's immediately adjacent to. But I don't get how it interacts with the second one... For the word 'apple', I would think that 'a' interacts 1 time with 'p'; 'p' interacts 1 time with 'a', 2 times with 'p', and 1 time with 'l'; 'l' interacts 1 time with 'p' and 1 time with 'e', and 'e' interacts 1 time with 'l'. Is that a good application of the rules?

Okay - starting with the basic information given:

A ball 40 foot in diameter is 40/(2*3.14159) = 6.37' in radius. The volume of a 6.37' radius sphere is (4/3)*(3.14159)*(6.37³) = 1080.76 cf. Now, take the volume and divide by the length of twine and you get 1080.76 / 7827737 * 144 = 0.0199 inch² for the area occupied by a cross-section of twine.

Here's where it gets tricky. The twine is presumably circular, so I'd be tempted to divide by pi/4 and take the square root to get the diameter of a circle with area = 0.0199 inch². But when laid side by side and packed in layers, I think each piece of twine will occupy a square area, rather than circular, so there will be tiny gaps between each consecutive strand of twine. There's probably some other optimal (hexagonal?) packing to be considered here, as well as what happens when you add a layer of twine, but I don't want to mess with it right now.

Assuming then that each strand of twine occupies a square cross-section, the diameter of the twine is sqrt(0.0199) = 0.141 inch, or about 9/64 inch.

Consider the excellent diagram that Furikan made:

http://brainden.com/forum/uploads/monthly_03_2009/post-14039-1236807711.gif

You are saying that (with rounding)

a = 5000M

s= 5002M

h = 60M

No way.

It really is. The radius of the arc segment is really large - I get about 50.8 km. If you cut a 5 km chord across a 50 km circle, the distance between the two at the midpoint is 0.06 km, or 60 m. Keep in mind the bulge height is only about 1% of the pipe length.

Of course, in real life, the pipe would have squirmed around like a snake before it finally split at the joints - oil everywhere.