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That is amazingly useful.
∫3x² dx=x³+c
Sorry, felt like trying it out...
4.3 posts per day compared to your 10!
And I see you're almost up to 1000 posts!
We need another party!
Make that past 1000 posts.
And I missed it!
The case where the rectangle has one side on the hypotenuse may be handled similarly. Is the maximum area in that case <. =, or > ab/4? I won't spoil the fun by giving that away!
Nor will I. It's annoying how after you do lots of complicated work to find the answer, you can tell just by looking though!
Here :
^
/ \
/ /\
/ / \
/ -----X \
/ \ \
/ \ \
/ \ \
/ \ \
-----------------------------
Hmm... not the best picture ever, but does it help?
Also, what do you NOW understand about NIH's avatar?
I'm intrigued, how did Zach edit my post without the tell-tale "Last edited by Zach" appearing?
He even fooled MathsIsFun!
If agoulding is still here, I've got an idea.
Me and loads of friends had a free period (thankyou A-levels!) and so we went to the nearby park.
It was very hot and so we decided to all climb a tree to get the combination of fun and shade.
Anyway, we managed to get thirteen people and a football up there at once, so you could say that the capacity of the tree was thirteen people.
A possible demonstration would be to see how many children could get up trees of different sizes. Too bad about health and safety... How about a car or a small room in the school? A phone box?
I heard the race version except the english cat was called four three two.
So... four three two won!
Substitute in s = (a+b+c)/2:
A = sqrt( ((a+b+c)/2)((a+b+c)/2-a)((a+b+c)/2-b)((a+b+c)/2-c) )
Yuck.
Simplifying: A= sqrt( (a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16)
Take out the denominator: A=sqrt ((a+b+c)(-a+b+c)(a-b+c)(a+b-c)/4
I think I've taken the wrong route, but anyway...
Multiply out the brackets: A= sqrt((-a^4+2a^2*b^2+2a^2*c^2+2b^2*c^2-b^4-c^4))/4
Quite a few things cancelled out there, or it would be even more horrible!
To prove that A isn't a whole number we need to prove that the horrible thing inside the brackets isn't 4 times a square number.
Anyone?
Similar to NIH's thing, pick a point on the equilateral triangle and draw a line parallel to the side adjacent clockwise to the side that you are drawing the line to. (Complicated description, hope you can decode that!) Do the same thing for the other two, and the sum of the lengths of the lines is equal to the length of one of the sides. You can draw lines parallel to the anticlockwise side instead and it'll still work.
The numeral converter gives IVX as 14, but it's already been proven wrong once...
You can laugh in dreams. Unfortunately, you usually also laugh in real life, which wakes you up. Stupid funny dreams!
Nope, someone just put that in the way because I said that I thought the murderer was [ Ma loo ga wooga ]
If I was Zach, I'd complain about the lack of an é, but instead I'll just say that the official answer is 200, meaning that everyone was right. Yay!
1) Lovely psychadelic evil eye thingy. (May help in deducing the murderer?)
2) When do people start dying?
[ Ma loo ga wooga ]
Sorry, have I made it too obviouS?
By the way, I tried MIM on the numeral converter thingy, and it gave me 2001.
"Math Error"
But, "imaginary numbers" are defined as 'J'
or even 'i'
What if the villagers don't know what he took, but try to deduce what he took...
Feel free to moderate this if I've made it too obvious and spoiled everyone's fun.
Isn't the rule that x^0=1 when x =/= 0? (Sorry for the rubbish non-equal sign)
Don't worry, binomials are even easier than quadratics!
You had: -7x^2+27x=0
Factorise the x: x(-7x+27)=0
Solve: x=0 or -27/7.
You already used stuff that is much harder than this to get where you got to, so this shouldn't be any problem.
If you're still confused (which you shouldn't be) then you can still use the quadratic equation, taking c=0.
1) Sorry for thinking that you were trying to cheat.
2) Of course advanced calculus isn't boring! One of the reasons why I used it!
3) Um, thats it, really.
The moral? Zach shouldn't take exams.
For the extension, I used advanced calculus that I won't bore you with now (unless someone requests it), but basically, if the triangle has perpendicular sides of lengths a and b, then the biggest rectangle will always have sides of length a/2 and b/2, subject to the proof being proved that I mentioned previously. This obviously means that the rectangle will have an area of ab/4, which means that its area will also be half of that of the triangle.
First I was wary about helping, because as you used phrases such as 'as an extension' and 'generalise', it looked suspiciously like something that you had to do for school. However, due to a combination of the facts that it has now been quite a long time, so if it was work you would most probably have handed it in by now, and you gave that cool trigonometry help thingy on the 'This is Cool' board, and it was such an interesting puzzle, I decided to help.
Thinking about the corner opposite the corner of the square that is also in the corner of the triangle, we know that it must have the same x and y co-ordinates (because it's a square) and that the point must lie somewhere on the diagonal line of the triangle (because it's a big square). Using MathsIsFun's graph idea, the equation of the diagonal line in the triangle would be y=-4/3x+4. We also know that the square's opposite corner must lie somewhere on y=x, so we have 2 simultaneous equations.
Substitute in y=x and you get x=-4/3x+4.
Add 4/3x: 7/3x=4
Divide by 7/3: x=4/ (7/3)=12/7.
This means that the largest square has a length of 12/7 units and an area of 144/49, or 2 44/49, square units.
Unfortunately, this post is in the realm of the chocolate teapot until someone proves that the biggest square must have the same corner as the triangle and I'm terrible at proofs. The pigeon-hole principle might do it, but I'm not exactly sure how to use it on this.
Thanks for the offer, but it's unfair for the murderer if there's an extra person guessing that everyone else knows not to guess. I'll be in the next one, if there is one. Don't kill me!
The problem with this is that some people will want to play and say "Can I join in?" in a whingy voice even though you have clearly said that you only allowed people to join up until yesterday. Allow me to demonstrate...
Can I join in?