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Are you guys smart enough to solve this simple problem?
>>
>>9863641
we are smart enough to not do your homework for you.
>>
no frick you do it yourself nigg
>>
>>9863641
27.
>>
>>9863641
28
>>
>>9863641
this is a joke, right?
>>
>>9863748
This. It's simply unsolvable.

Not because it's somehow difficult, but because of lack of conditions to get the answer.

OP need to kys asap
>>
>>9863641
Let's get started. If I read the picture correctly then:
O = (0,0)
A = (1,0)
C1 = {(x,y) | x^2 + y^2 = 1}
P = (cos(theta), sin(theta))
Q = (-cos(theta), -sin(theta))
R = (cos(theta), -sin(theta))
So far so good but how are C2, M and N defined?
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>>9863641
Theta is a scalar so the cross product is 0.

Then the limit evaluates to 0/0 or indeterminate.

So then you have to apply a different limit test. Eyeballing it, I'd say that T(theta) is non linear so the limit is 0.
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>>9863748
>>9863758
>>9863772
this, we don't know anything about the smaller circle
>>
>>9863641
OP here.

Let P be (cos theta, sin theta)

then R is (cos theta, -sin theta) and Q (-cos theta, - sin theta) and PQ is the diameter of C2.
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>>9863819
PQ is already the diameter of C1, and C2 looks smaller to me...
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>>9863819
>>9863849
>PQ
*QR.

Also M, N are bisectors.
>>
Just put in on geogebra and see how it behaves. It's probably going to be zero though since the triangle deform itself once close to 0.
>>
>>9863856
>bisectors
Intersection point, right? They don't look bisectors at all. Just KYS and never come back.

Also do your homework by your own.
>>
The answer is 2.
>>
4. Simple.
>>
>>9863641
>simple problem
kys
>>
I'll fuck with it, let t=theta for typing sake, intersecting the line QP and C_2 and solve the quadratic you get M as cos(2t)e^(it)
>>
what kind of stupid question is this? the answer is 0 for theta going to 0
>>
Alright, N is (cos^2(t),sin(t)(cos(t)-1)) by the same method, all points now have locations as function of t
>>
Lengths of small triangle sides are

PR=2sin(t)
RN=cos(t)sqrt(2-2cos(t))
NP=sqrt(2(cos(t)-1)(cos^2(t)-2))
>>
Big triangle side lengths are

MN=RN
NC=cos(t)sqrt(2+2cos(t))
CM=2cos^2(t)
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>>9865053
>>9865094
You need the coordinates of the corners rather than the side-lengths if you want to calculate the area.
https://en.wikipedia.org/wiki/Shoelace_formula
>>
Was thinking I'd use herons to get area but it's way faster to use 1/2bh, length from M to QN is cos^2(t)sqrt(2-2cos(t))

T(t)=cos(t)sin(t)(1-cos(t))
S(t)=cos^3(t)sin(t)
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>>9865121
Yeah I always forget that's an option

Anyway apply the hospital twice to the limit and get 2

>>9864210
Can't believe any of the throwaway answers were right
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>>9865205
Wrong.

Cross product of theta^2 and T(theta) is 0.

Impossible to get 2 out of it.
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>>9863641
i cant find the 3. point to specify the smaller circle, i require more information
>>
>>9863641
chinese SAT
do your homework alone, chigga
>>
>>9863641
1.
>>
>>9863875
>by your own
*on your own
lmao
>>
P, Q are on s^2 : x^2+y^2+z^2=4 and P1, P2, Q1, Q2 are foot of perpendicular of P,Q. Find the maximum value of the function f(x) = 2|PQ|^2-|P1Q1|^2-|P2Q2|^2

Try this one.
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File: fasfsa.png (38 KB, 463x330)
38 KB
38 KB PNG
>>9869379
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>>9869381
>>9869379
kys




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