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Étonné - logic, math, crypto and programming problems

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TPReal:
Ok, I'll help you with the symbols, but still some math (and geometry, as in title) is needed to solve it.
The whole thing looks like: {(x;y): <condition> } and it means: a set of such pairs (x;y) that the <condition> is met. Now the condition: thing a bit like x E <-4;4> means that -4<=x<=4. The next sign (like v upside down) is AND operator, while v (not upside down) is OR operator. And the last part: a thing like a E {b;c} is equivalent to a=b v a=c. And what you have to do is to draw the set in cartesian coordinates, and look at the result.

ivanv:
So the last part would be something like this:

Spoilerx^2+y^2=0.25;x^2+y^2=4 V 25-x^2=(y-3)^2;25-x^2=(y+3)^2

And so if I solve those equations and do a "union" of the pairs I get, I graph those?

I'm just trying to see if I'm following the right path, as I still don't really know what I'm doing :)

Maybe this is as far as I'll get.

TPReal:
Almost right... but look: x E {a;b} is equivalent to x=a v x=b. So the equations you gave are correct, but you should find a union of the results of all four equations! Change both ';' to 'v' in your formula.
In other words - solve each of the four equations you gave separately, and draw all four solutions one on another, thus creating the union of all four of them.

ivanv:
I suck at this... I attached my graph, and it doesn't look like anything...

And here are the equations:

Spoilery = 0.5 * sqrt(1 - 4x^2)
y = -0.5 * sqrt(1 - 4x^2)
y = sqrt(4 - x^2)
y = -sqrt(4 - x^2)
y = 3 + sqrt(25 - x^2)
y = 3 - sqrt(25 - x^2)
y = -3 + sqrt(25 - x^2)
y = -3 - sqrt(25 - x^2)

Am I way off or on to something?

Thanks for such an addicting puzzle!

TPReal:
You attached the image in the wrong way, now everyone can see it :) maybe edit it or attach it in some other way or something. But the image is good (apart from the fact that I can't see there the y = -sqrt(4 - x^2) curve). Now look at the first part of the formula which says that -4<=x<=4 and -2<=y<=2, so take only this rectangular part of the image you have created, and I'm sure you will see clearly what it is!