/sci/ - Science & Math

[math]\color{#781b86}{\text{P}}\color{#6c2895}{\text{r}}\color{#6135a3}{\text{e}}\color{#5543b2}{\text{v}}\color{#4a50c0}{\text{i}}\color{#3e5dcf}{\text{o}}\color{#4a6fbe}{\text{u}}\color{#5681ad}{\text{s}}\color{#62939c}{\text{ }}\color{#6ea58b}{\text{T}}\color{#7ab77a}{\text{h}}\color{#85b871}{\text{r}}\color{#90b968}{\text{e}}\color{#9bbb5f}{\text{a}}\color{#a6bc56}{\text{d}}\color{#b1bd4d}{\text{ }}\color{#bab849}{\text{>}}\color{#c3b446}{\text{>}}\color{#ccaf42}{\text{1}}\color{#d5ab3f}{\text{0}}\color{#dea63b}{\text{5}}\color{#dd8b36}{\text{7}}\color{#dc7131}{\text{9}}\color{#dc562b}{\text{8}}\color{#db3c26}{\text{9}}\color{#da2121}{\text{4}}[/math]

bro im begging you can we get the prismrivers

>>10583271
wtf this is very hard to read
go back to the old format pls.

>>10583431
>wtf this is very hard to read
This problem is pathetically easy, so I added the extra visuals to balance things out.

>>10583494
And you are neither funny nor look intelligent. Like all weebs.

>>10583500
>blocks your path

>>10583500
>blocks your path

>>10583271
The square has side length [math] \sqrt5 [/math]
The circle has radius [math] \frac{1}{2} \sqrt{10} [/math]
The rectangle has side lengths [math] 2\sqrt2 [/math] and [math] \sqrt2 [/math]

That's as far as I've gotten...

>>10583271
>find the maximum possible area

do you actually have any free variables to maximize if the second polygon must be a rectangle of area 4 with its corners on the circle? aside from choosing a "long side", i think you're locked in.

>>10583576
the shape of that rectangle is locked in yeah (>>10583559)
but the angle/rotation of the rectangle compared to the square is not

>>10583559
So, if you call the rectangle xy, and you treat the p1->p2 and p2->p3 segments as alternating line segments (since we're given a rectangle, it exhibits parallelogram behavior) called b1 and b2 respectively, we can define the area of the octagon using two formulas.

Note: Brackets are defining the triangle in this case, not a function. That said, the definitions used are based on area.

Using the rectangle it would be defined
4+2[▲b1,b2,x]+2[▲b1,b2,y]

Using the square, it is:
5+4[▲b1,b2,5^(1/2)]

Since they are equivalent values, we get a function reducing down to the following:
[▲b1,b2,x]+[▲b1,b2,y] = 2[▲b1,b2,5^(1/2)]+1/2
But to clean it up while we run it, let's define them as A1, A2, and A3 (respectively).
A1+A2 = 2*A3+1/2
Next stop is to give area functions to these area arguments.

Let's look at A1. Using 1/2*base*height for a triangle's area, we can use x as the base. Same can be said for y or 5^(1/2) in A2 and A3. We can then make h1, h2, and h3 for each of the triangles. Thus,
A1 = 1/2 x h1
A2 = 1/2 y h2
A3 = 1/2 5^(1/2) h3
Now, we want to solve for A3. This will involve some trig on the triangle we defined. Since we have an isosceles triangle made with two radii r,where r=(10^(1/2))/2, and our cord of 5^(1/2), we can make an angle (q) that is arcsin{[(5^(1/2)/2]/[r]}=q. With q, we can use r*cos(q)=h3.
As >>10583576 pointed out, the square's triangles imply a set area. Given that we have a set value of r, we have to have a set value for q, and thus a set value for h3. Since we can go back and take 4*h3+5=Octagon's area, we have a set area for the octagon.

>>10583722
>Since we have an isosceles triangle made with two radii
Shit I done fucked up, that would apply in A1 and A2, not A3. Disregard this, I don't math enough these days.

the corners of the first rectangle are fixed up to rotation around the circle center because it's a square and the vertices must lie on the circle's perimeter.

the corners of the second rectangle are constrained to the segments of the circle's perimeter between odd-numbered vertices. these vertices are the only things you can adjust, and you have to keep the second rectangle area constant at 4.

the maximum area of the octagon is achieved when all vertices are equidistant. so the only way to increase the area of the octagon is moving the even numbered vertices along the perimeter to make the second rectangle more square-like. but you can't do that since the area of that square must be 4. so i think you're just trying to find the area of the octagon, not the "maximum possible" area.

then again, i suck at these geometry problems.

>>10583853
cont.

and i think it just boils down to an easy geometry problem from there. you know the diameter of the circle, and that the octagon is symmetric about your rectangles, or however you say that in math lingo

>>10583853
As I said before, yes the shape of the rectangle is locked in.
It has side lengths 2sqrt(2) and sqrt(2).

But what matters is the rotation of the rectangle compared to the square.

pic related are two examples

>>10583918

ah, so that's your "adjustment". what i said is still true though. you maximize the octagon's area by making the vertices as close to equidistant as you can along the perimeter, so it would have to be closer to your second pic.

>>10583940
yeah it does seem like a good guess that that symmetrical configuration would maximize the octogon's area, but you still have to prove it

>>10583918
I guess next step is do differentiate an area equation based on the variable "a" while using the expanded area equation shown.

>>10583271
This is literally a calculus 1 problem. There is one free variable. The solution is left as an exercise to the reader.

>>10584185
Nvm. It turns out it is even easier than that. From the picture in
>>10584144
It is clear that one has to maximize the areas of the triangles [math]P_2P_3P_4[/math] and [math]P_3P_4P_5[/math] (the other two triangles are copies, you see this by rotating). With the base in the rectangle the height is tallest at the midpoint.
Pretty strange that the exam has such an easy problem.