# Thread: Math game (fun quizzes) ^^

1. ## Re: Math game (fun quizzes) ^^

Ok... I actually plan to create some more original quiz, but for the time being here is this one (not mine) : Prove that the red area is equal to the blue areas Note: the blue areas are the leftovers of the semicircles formed with the corresponding side of the triangle as diameter, while the right-angled triangle itself is inscribed in the larger circle.  Reply With Quote

2. ## Re: Math game (fun quizzes) ^^

Damn it, guys, this is supposed to be a forum with some intellect Anyway, I will give it one more day and then post the answer ^^  Reply With Quote

3. ## Re: Math game (fun quizzes) ^^

Guys, guys, guys... It was actually pretty easy, in that you can treat the critical step as just the analogue of the pythagorean theorem:
-The hypotenuse of the triangle is the diameter of the large circle (that is also obvious since there is a right angle looking at that line=>the arc is a semicircle.
-The other two sides are (noted already) also diameters of their respective, smaller circles. This means they are in analogy to the pyth theorem, since (say) the area of the circles is a constant (pi) multiplied by 1/2side (=r) squared (so the same proportion as a^2+b^2=c^2).
=>The area of the largest circle is equal to the areas of the two smaller circles =>half the area of the large circle is equal to the areas of the two smaller half-circles, and finally:
Half the area of the largest circle is the triangle plus the two empty bits (white) of those two smaller half-circles, but the blue parts of them are also half the area of those circles MINUS the empty bits => the blue areas are the same as the triangle's area.  Reply With Quote

4. ## Re: Math game (fun quizzes) ^^

Ok people, something different (and by all means worse!) this time, since I picked up a book on calculus again. The following problem (which I now can solve) was the reason I first gave up on that book, in the past -roughly a year ago And certainly you can tell it's no fun when this appears in the first damned chapter of a massive book!

Anyway, here you go: (it's part of the problem, but if you can solve this you would be able to solve the rest too) Important note: By "maximum" of two numbers, it is meant that among the two integers one has a higher numerical value. For example, the max of 0,2 is 2, the max of -1,5, is 5 and the max of -5,-2 is -2.  Reply With Quote

5. ## Re: Math game (fun quizzes) ^^

By the way, you don't need to solve it all - it's from a notoriously user-hostile calculus book. Moreover, it is not meant to be solved using calculus! (for it is in that book's first chapter, before the author goes on to non-implicit calculus stuff...).
Any idea on parts would already be good ^^
If not, I will give it, say, 3 days, and then post one solution.  Reply With Quote

6. ## Re: Math game (fun quizzes) ^^

I will give a very nice hint (I know that it is still difficult to solve, but it's a cool part of it, and I personally am enthusiastic about it )
Sometimes, math features an equal amount of treachery that literature will. For example, when you see an absolute value somewhere, you can't help but think it is there so as to ensure the outcome is positive. But it can be a massive diversion, when (due to certain conditions) the outcome would always be positive already, as long as x-y or y-x were chosen...
The absolute value, in that case, only exists to ensure that regardless of whether it is [x-y] or [y-x], the outcome is numerically the same; since [a]=[-a].

Now, the question is, what property are you trying to establish that either x or y have? ^^

As promised, I will provide the full answer in a couple of days.  Reply With Quote

7. ## Re: Math game (fun quizzes) ^^

Mmmm, well, ok ^^
It was problem 13, from Spivak's Calculus.
Basically the fastest way to provide an answer, is to notice that:
a) [a]=[-a]=>[y-x]=[x-y]
b) when you have a max and a min, and only two numbers, one of them is max, the other is min. But regardless of whether x or y is max, [x-y] always give you the same result as max-min. This is because when you subtract the min from the max, the result is always positive, and [x-y] ensures that regardless of [x]>[y] or the other way around, the result is positive => it is max-min.
c) just like [x-y] is written thus as max-min, x+y is written as max+min. So the formula simplifies to (max+min+max-min)/2=max.

I hope you had fun reading this proof I wish more would take part in the thread, but it's ok, I will keep posting stuff.  Reply With Quote

game, math 