fabpig
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Posts posted by fabpig
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aaagh - can't think
COMMONPLACE
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heart. when it goes down (sinks) you sigh -t. Curiously heavy-hearted, intestingly light-hearted.
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Fully agreed with Bonanova, this problem is a classical paradox...
As suggested, my thoughts on the same....
The situation in this problem is a contradiction.
Logic very well explain us, there can't be a surprise test next week as then it won't be a surprise.
But, in real world, a surprise is a surprise, if logically deducible that won't be surprise any longer.
The way Professor has declared about Surprise quiz was totally contradictory on logical basis, it is like saying ' you will have a surprise quiz, tomorrow (logically not sensible).
Being heard about the such a surprise quiz, the students should have taken it seriously
... should have started preparing their subjects... hehehe...There is a similar paradox. I will like to put here for all you to give a thought...
Once in a town there was a male barber who, every day, shaves every man who doesn't shave himself, and no one else.
Do you think such a Barber could ever exist ?
Again...
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STINKER
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(Assume that to be without the S, Thal)
MUDCAT
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Why thankya!
The good ol' 4th of July. Have a great Independence Day and don't forget.....
WUTIWBYO
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OK, my thinking is this. I'm assuming the cat moves at constant speed and the square is of unit length.
If the path of the cat was a perfect quarter circle, its halfway point would be where the DB diagonal intersected the arc ("X"). For the cat to be looking at the mouse, the tangent at this point must intersect line BC at point Y. YB has to be 0.5 (halfway point for the mouse) for this to be true.
From Pythagorus, we can see XB is ~0.41 [sqrt(2) - radius 1] XY must be the same (tangent at X intersects AB and BC at 45deg). Therefore length YB is sqrt((XY^2) + (XB^2)) =~ 0.58 which is clearly past halfway for the mouse.
So, to extend my thinking, halfway for the cat must be somewhere on the BD diagonal - and he must be looking at the midpoint of BC (new "Y")(halfway mouse). (This bit is more intuitive) - as the cat turns 90deg in total, at halfway he must've turned 45deg. YXB form an isosceles right triangle with YB length 0.5. Again from pythagarus, XB = sqrt((0.5^2)/2) =~ 0.353. so point X is 1.057 from D which is > radius 1.
Or is that a load of old tosh?
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OUTING
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APPLICATIVE -if 1 then E is last because:
APPLICATION = 0 so .........VE = 1
APPROPRIATE = 2, APPROPRIACY = 1, so .........TE = 1
as V/T cannot both be 1, only leaves E.
If 0, that eliminates ..........E, meaning T is 10th.
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W R I N K L Y PRESSED - 1 DETESTS - 0 GROUNDS - 1 BRITTLE - 3 HARPIES - 0 DRAWERS - 1 WRIGGLE - 4 TRICKLE - 4 TRAPPED - 1 PRAYING - 1 BRITISH - 2 WRINKLY - 7 PRESSER - 1 PLAYING - 0 CRINKLE - 5 DRESSED - 1 PARSONS - 0 DRAGGED - 1
Thalia +35
MikeD +5
fabpig -5 from prev game +17. +12 in total
WombatBreath: 992
Thalia: 979
Maquis: 265
TheCube: 261
maurice: 227
'Cat'astrophe: 194
JDave: 179
fabpig: 167
MikeD: 68
sks: 58
curr3nt: 51
Brainiac: 50
Hidden G: 45
Aaryan: 37
Yodell: 25
ShadowAngel7: 15
kristmark1: 10
SMV: 10
benjer3: 5
shakingdavid: 5
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_ R _ _ _ _ _ PRESSED - 1 DETESTS - 0 GROUNDS - 1 BRITTLE - 3 HARPIES - 0 DRAWERS - 1 WRIGGLE - 4 TRICKLE - 4 TRAPPED - 1 PRAYING - 1 BRITISH - 2 PRESSER - 1 PLAYING - 0 CRINKLE - 5 DRESSED - 1 PARSONS - 0 DRAGGED - 1
MikeD +5
fabpig -5 from prev game
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F O R U M CHIPS - 0 STOPS - 0 WORLD - 2 HAIRS - 0 GRASP - 0 NERDY - 1 GNOME - 0 STRAW - 1 ARROW - 1 AHEAD - 0 BRASS - 0 WIELD - 0 PIECE - 0 CARPS - 1 FORUM - 5 Molly Mae +5 MikeD +5 plainglazed +20 fabpig +14
2396 - maurice
1628 - WombatBreath
1497 - Maquis
1363 - plainglazed
1295 - Glycereine
1253 - curr3nt
1134 - fabpig
974 - Quag
648 - Vineetrika
532 - Cherry Lane
521 - araver
521 - Molly Mae
485 - TheCube
386 - Izzy
378 - unreality
305 - dawh
300 - 'cat'astrophe
299 - MikeD
201 - shadow7
199 - Framm
190 - t8t8t8
180 - NickFleming
170 - golfjunkie
170 - benjer3
141 - woon
138 - Thalia
130 - JoeP
121 - JDave
119 - MissKitten
109 - Aaryan
99 - Perry the Platypus
96 - JarZe
93 - Q-cumber
93 - Panther
89 - BlaBlah99
77 - Harvey45
76 - SMV
73 - Brainiac100
66 - MoMoney
64 - abhisk
47 - young scientist
43 - kristmark1
39 - flamebirde
36 - Hirkala
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I agree with Bonanova. I think the apex of the catenary lies at a point on the BD diagonal and the intersection of a line drawn between the midpoints of AB and BC
btw. I don't think negative reputation points are intended for wrong answers: more for rude or insulting statements.
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Remains Of The Day
in New Word Riddles
Posted