HoustonHokie

No time to think - need to sleep - but can't avoid the question or the irresitible urge to post first...

picky, picky, picky... Well, I see that others have chimed in while I was typing. I'll submit anyway, but I think it's mostly covered.

Here's my thoughts: What was part b?

I think the one-room house was a spoiler box. The 21 spots were my eyes after opening the spoiler boxes. Kind of like this:

Welcome to the Den jennel. I think you found the right place to put your reply - just remember to use spoilers so that those who haven't yet worked a puzzle can do so without being too tempted to look at your answers! You'll find the Insert: Spoiler box on the left side of the page under Quick Access when you're composing your reply.

Sure - those are fun. There's endless books of those little mysteries - if your fingers don't get worn out from all the typing, I'd say put five or ten up as a single post and let everyone compete like we do with cryptograms or word puzzles, etc. Just make sure they haven't been posted before, or the moderators will get you!

very helpful... now I'll stop looking for 25-letter words to fit! Any more clues forthcoming? Like the presence of vowels in the solution?

[spoiler= ]still can't believe #7 isn't leaning tower of pisa... But maybe its my breakfast when I was in college!

Have to admit that I used a cadd program (although not in 3d like Adam G) to get the area earlier. Let's see - the actual calculation goes something like this: I was assuming the depth of the water was 3D/4 (although given my simple assumption that D=1, (3D)/4 = 3/(4D)). For a depth X greater than D/2, the area is based on the angle theta = pi + 2*arcsin[(X-D/2)/(D/2)], which is the angle of the wetted arc. The area is (theta/2)*(D2/4)+0.5*(X-D/2)*[cos((theta-pi)/2)]. The first term is basically theta/2pi times the area of a circle and the second term accounts for the triangle in the middle that the first term misses. After that little geometry lesson, I think that Adam G got the right answer instead of me because the average end area method assumes a linear reduction in area along the length of the cylinder, which is probably invalid in this case. The area reduces more quickly at the bottom than at the top of the glass. A quick check at the center of the glass would probably confirm that the area is not half that at the bottom of the glass and invalidate the average end area method. Dang! I hate being wrong! Do we need calculus to get the generic solution, or are we missing a trig solution somewhere?

And I'll try to actually use my brain for some of the other 8 digits!

I knew there had to be something simple.

Well, I no longer think that trick works.... I've figured out all the quotes (I think), but don't know how to actually crack the code that made them. Mumbles - if I have the quotes right, the grammar is just fine.