Hard Puzzles
I'd appreciate it if after this puzzle is 'done' we continue.
http://img354.imageshack.us/my.php?image=puzzle14qd.gif
Draw a line from A to XY&Z, B to XY&Z and C to XY&Z without any of the lines intersecting.
Which means you can't overlap and say 'it works in 3D'.
Go.
Also, if you claim it's not possible, provide proof :)
do they gotta be striaght?
edit: If not I did it.
Lucky, lucky, lucky, lucky me again! I hardly knew I should use me feet again!
What do you have to say for yourself?
They don't.
And you did it? Pics.
Dammit, now I have to use paint. grrrr
Lucky, lucky, lucky, lucky me again! I hardly knew I should use me feet again!
What do you have to say for yourself?
Informal proof that it's impossible. (This is not rigorous, so I may just be missing a solution, but oh well.) See attached image. It should be clear that you can move the nodes around without affecting anything, since we're not limited in the length or direction of our lines.
Drawing lines from A to X,Y,Z and from B to X,Y,Z will always divide the plane into three areas: two 4-sided polygons which share two sides (blue and green in my attachment), and an area outside those polygons (white).
For proof that it will always form these areas: Take A and B, and any two of X,Y,Z, let's say X and Y, though it doesn't matter. Connecting them as per the original problem will always form a quadrilateral. This should be obvious; if you connect four points with four lines, two lines touching each point, you form a 4-sided polygon AXBY. Now you have one leftover node (Z). You can either place it inside or outside the polygon, nowhere else. If you place it inside, the two lines from A and B will split the polygon into two pieces, which leaves you with two 4-sided polygons sharing two sides. If you place the node outside, the two lines from A and B will form a second polygon which borders the first (sharing two sides). Either way, you end up with two enclosed 4-sided polygons sharing two sides, as per my attachment.
If C is enclosed in AZBY (blue) it can't reach X. If C is enclosed in AYBX (green) it can't reach Z. If C is outside them both (white) it can't reach Y. Forming the two 4-sided polygons from nodes A,B,X,Y,Z or A,C,X,Y,Z or B,C,X,Y,Z all reduce to the above scenario. There are no other possibilities; therefore the original problem is impossible.
EDIT: Clearly if you can show any way that you can connect two of A,B,C with three of X,Y,Z without forming two 4-sided polygons that share two edges, then what I've posted will be disproved.
Last edited by Dr Unne; 05-09-2006 at 05:19 AM.
You forgot one thing.
A new puzzle!
(btw Dr Unne got it)
Aw. Yeah, I linked my math major buddy to it and he told me it was impossible. He signed off before I could ask him for a proof, though. Guess Unne got it first.
Puzzles that ask you to do something that can't be done are annoying, even if the answer is simple. I prefer being challenged to do something that is possible but also difficult. That's a real challenge. None of this "prove that it can't be done" nonsense - I stopped doing that crap when I left High School, thankyaverymuch. No need to spend long times typing out a proof like that above when you could spend your time doing more productive and mind-challenging problems that actually have a solution. Proofs wore the most boring thing about Calculus by a long, long way. I don't care how much intelligence is required, I much prefer a sense of achievement in proving something can be done than proving something can't be done. Give us a real puzzle!
Bow before the mighty Javoo!
That first puzzle very obviously couldn't be done without the Z-axis present. But I agree BoB.I'd find one myself, but I really can't be bothered right now.
Um. I found this on a website. Don't cheat by googling it.Originally Posted by Pureghetto
[qq]After a particularly taxing night of solving logic puzzles and riddles, your brain suddenly shuts down and you find yourself in what you can only hope to be a bizarrely relevant dream. Seated beside a set of pearly gates is the patient-looking figure of St. Peter, white robes and all. "We're glad you could make it, but before you go in, there's just one little question you have to answer," says Peter, and you begin to suspect that this will be a 'thinking' type of question.
"How many angels can dance on the head of a pin?" He raises a hand to point to an inscription on the gates. "The answer lies therein. Each is accounted for, and the further you go, the more you'll get. Take your time; you have plenty of it." After a few cautious steps on the cloud-like floor, you approach the gate and read:
How wondrous too is truth received
For him who'd defy verity
As hope the sick soothes even sates
So benign the uncertain's made[/qq]
I hate it when riddles are mocking me for having English as a second language with words like verity and benign.
Most native English speakers (Americans anyways) likely have no idea what verity means, so don't worry about it.
Is the answer ten? I'm not entirely sure what I'm looking for, but from what I can grasp of the puzzle, that would be a logical solution.
"The most important and recognize player in the history of the country."
Sometimes I wonder what my life would be like if I were as great as Paulo Wanchope.
Dr Unne still hasn't announced whether Cz is right, but I found this:
A monk lives at a temple at the bottom of the hill. Every week, he starts at sunrise and climbs up to the temple at the top of the hill, getting there exactly at sunset. He spends the night at the temple in meditation, then the next morning starts again at sunrise, and walks down the hill to the lower temple, arriving at sunset. He doesn't walk at a steady pace, and often will stop to eat some of the dried fruit he has, or rest, walk slower, etc.
Assuming sunrise and sunset are exactly the same on both days, and that the there is only one path between the temples, will the monk be at the same place at the same time at some point on both days? Prove why he must, or why he won't necessarily, 2 easy ways to explain it.
Looks like someone got frustrated with his limited intelligence.
edit: PG's puzzle:
This puzzle scenario can simply be rephrased as a scenerio where two monks walk from each end to the other, and ask whether or not they cross paths. This is obviously true. Just imagine the red path to be the monk on one day and the blue path to be the monk on the next day. It makes sense.
Last edited by -N-; 05-10-2006 at 02:39 AM.
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