Fri Jul 14, 2006 1:58 am
Wow. Thanks for that tip. I'm 99% sure that I've got it. For those still wondering, take the earlier bit about the "170:" and then think back to the Altador Plot for all the rest. 
Mon Jul 17, 2006 6:57 am
I think people are confusing the two Lenny Conundrums. Here is the proof for Conundrums 173 and 174 that the circle gotten MUST be the largest possible circle:
Without loss of generality between the top, left, right, and bottom points, let us place the top of the circle somewhere inside a gray box. From this point, the two arcs of the circle must drop downwards (since it's the top), so they must exit the box through the bottom two corners. The circle is also symmetric through the top point going up-and-down. Therefore, the top of the circle must be on the center of that gray box, to maintain symmetry.
Repeat this reasoning for the right-most or left-most point, and one finds that the circle must be centered both horizontally and vertically.
Without loss of generality, now focus on the top-right quadrant of the circle, and move right from the top. The arc at the top of the circle has a slope of zero. The slope of the circle from here on out must decrease (become more negative).* Thus, we can choose any situation where the slope is < 0 for now, assuming we haven't finished off the top-right quadrant yet (and let's hold off doing that for as long as possible).
Pass the arc through the bottom-right corner of the current box and enter the top-left corner of a new box. Since the circle must somehow exit this box, we now must choose the exit. The top-right corner does not work, as that would require the slope to increase to above zero. The bottom-right corner works. The bottom-left corner forces us to finish the quadrant of the circle first, but we'd rather keep going. The bottom-right corner will be the choice.
However, now we now know something about the arc once we connect the bottom-right and the top-left corners. The straight line connecting these corners has a slope of exactly -1. Since the circle always has a decreasing slope here, the arc bulges "out" towards the upper-right, not "in" - in other words, it has a slope greater (less negative) than -1 in the top-left, then decreases to a slope of less (more negative) than -1 in the bottom-right. Thus, from here on out, we know that the slope must now remain less (more negative) than -1.
In the next box, we again must choose a corner. The top-right still does not work. The bottom-right, now, also no longer works: in order to reach it, the slope must increase back above -1. The bottom-left is now the only choice, so we have to finish the quadrant of the circle in the middle of this box.
Repeat or use symmetry for the other quadrants.
From the center of the circle, draw a line horizontally across and up to one of the corners. Use the Pythagorean Theorem to find the radius.
* To prove that the circle's slope always decreases:
Let's take the equation for the circle, x^2 + y^2 = r^2 (where r is a number), and solve it for y.
y = plus or minus sqrt(r^2 - x^2)
Let's take the top of the circle, which we are interested in.
y = plus sqrt(r^2 - x^2)
dy/dx (the slope) = -x / (sqrt(r^2 - x^2)) = (-x) * (r^2 - x^2)^(-1/2)
d^2 y / (dx)^2 (the rate of change in the slope)
= (-x)[-1/2 * (r^2 - x^2)^(-3/2) * 2x] + (r^2-x^2)^(-1/2)*(-1)
by product and chain rules
= -x^2/[(r^2-x^2)^(3/2)] - 1/sqrt(r^2-x^2)
The first term -x^2/[(r^2-x^2)^(3/2)] must be negative because -x^2 must be negative and the cube of a positive square root must be positive, and negative over positive is negative. The second term -1/sqrt(r^2-x^2) must also be negative, since -1 over a positive square root is negative.
Both terms are negative and added together to create a negative number, so the rate of change in the slope is negative, so the slope always decreases. (This proof works for proving the slope decreases in the top-left quadrant too, where the circle approaches the top.)
[edit] add equation images?
[edit] fix signs for first two eqn images
Without loss of generality between the top, left, right, and bottom points, let us place the top of the circle somewhere inside a gray box. From this point, the two arcs of the circle must drop downwards (since it's the top), so they must exit the box through the bottom two corners. The circle is also symmetric through the top point going up-and-down. Therefore, the top of the circle must be on the center of that gray box, to maintain symmetry.
Repeat this reasoning for the right-most or left-most point, and one finds that the circle must be centered both horizontally and vertically.
Without loss of generality, now focus on the top-right quadrant of the circle, and move right from the top. The arc at the top of the circle has a slope of zero. The slope of the circle from here on out must decrease (become more negative).* Thus, we can choose any situation where the slope is < 0 for now, assuming we haven't finished off the top-right quadrant yet (and let's hold off doing that for as long as possible).
Pass the arc through the bottom-right corner of the current box and enter the top-left corner of a new box. Since the circle must somehow exit this box, we now must choose the exit. The top-right corner does not work, as that would require the slope to increase to above zero. The bottom-right corner works. The bottom-left corner forces us to finish the quadrant of the circle first, but we'd rather keep going. The bottom-right corner will be the choice.
However, now we now know something about the arc once we connect the bottom-right and the top-left corners. The straight line connecting these corners has a slope of exactly -1. Since the circle always has a decreasing slope here, the arc bulges "out" towards the upper-right, not "in" - in other words, it has a slope greater (less negative) than -1 in the top-left, then decreases to a slope of less (more negative) than -1 in the bottom-right. Thus, from here on out, we know that the slope must now remain less (more negative) than -1.
In the next box, we again must choose a corner. The top-right still does not work. The bottom-right, now, also no longer works: in order to reach it, the slope must increase back above -1. The bottom-left is now the only choice, so we have to finish the quadrant of the circle in the middle of this box.
Repeat or use symmetry for the other quadrants.
From the center of the circle, draw a line horizontally across and up to one of the corners. Use the Pythagorean Theorem to find the radius.
* To prove that the circle's slope always decreases:
Let's take the equation for the circle, x^2 + y^2 = r^2 (where r is a number), and solve it for y.
y = plus or minus sqrt(r^2 - x^2)
Let's take the top of the circle, which we are interested in.
y = plus sqrt(r^2 - x^2)
dy/dx (the slope) = -x / (sqrt(r^2 - x^2)) = (-x) * (r^2 - x^2)^(-1/2)
d^2 y / (dx)^2 (the rate of change in the slope)
= (-x)[-1/2 * (r^2 - x^2)^(-3/2) * 2x] + (r^2-x^2)^(-1/2)*(-1)
by product and chain rules
= -x^2/[(r^2-x^2)^(3/2)] - 1/sqrt(r^2-x^2)
The first term -x^2/[(r^2-x^2)^(3/2)] must be negative because -x^2 must be negative and the cube of a positive square root must be positive, and negative over positive is negative. The second term -1/sqrt(r^2-x^2) must also be negative, since -1 over a positive square root is negative.
Both terms are negative and added together to create a negative number, so the rate of change in the slope is negative, so the slope always decreases. (This proof works for proving the slope decreases in the top-left quadrant too, where the circle approaches the top.)
[edit] add equation images?
[edit] fix signs for first two eqn images
Last edited by AySz88 on Wed Jul 19, 2006 5:53 pm, edited 1 time in total.
not a trick
Tue Jul 18, 2006 3:40 am
this problem is acutally a simple math problem. try to figure it out using a formuls...for area...COUGH COUGH 
Re: not a trick
Wed Jul 19, 2006 5:49 pm
ohkanon wrote:this problem is acutally a simple math problem. try to figure it out using a formuls...for area...COUGH COUGH
You're talking about 174, I was talking about 173 (mostly); one wouldn't want to post a solution while a related conundrum was running!
(Actually, now that I think about it, was the answer to 173 available already while 174 was running? That would make 174 ridiculously easy for a LC...)
Thu Jul 20, 2006 12:53 pm
Owww, that math looks hard. you know what else is hard? quantum physics, and stuff with spacetime. i've been looking at the basics of it and it's killing my brain.
so TNT decided to let this week's conundrum go for another week. silly, since it was so easy, and this will just lower the prize to like 3np
so TNT decided to let this week's conundrum go for another week. silly, since it was so easy, and this will just lower the prize to like 3np
Re: not a trick
Thu Jul 20, 2006 1:40 pm
AySz88 wrote:ohkanon wrote:this problem is acutally a simple math problem. try to figure it out using a formuls...for area...COUGH COUGH
You're talking about 174, I was talking about 173 (mostly); one wouldn't want to post a solution while a related conundrum was running!
(Actually, now that I think about it, was the answer to 173 available already while 174 was running? That would make 174 ridiculously easy for a LC...)
Actually, we're on 175 now, so you can talk about either of the two, no problem. Yes, the answer to 173 was available -- but you still had to know what the circle looked like so you'd know the area of the white space to subtract.
As for the current one (175), since the Lenny said he'd be on vacation this week, I was expecting it to run two weeks -- the guy in charge of it is probably really on vacation.