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suppose that we have Cartesian coordinate system.and suppose that we have three point which depend on parameter $t$,where t belongs to $(0,1)$;points are

$A(cos(3-t),sin(3-t))$

$B(cos(t),sin(t))$

$C(-cos(t),-sin(t))$

goal: find $t$ for which area of triangle $ABC$ is maximum

first of all,i was thinking that we could find length of each side of triangles,for example

$BC=2$

but what about another sides?we can use determinant formula like here

http://people.richland.edu/james/lecture/m116/matrices/applications.html

and goal will be find maximum determinant,but could we it?also i have calculate length of $AB$,which is equal $2*cos(t)*cos(3-t)-2*sin(t)*sin(3-t)$ which is i think

$2*cos(\alpha-\beta)$

or in our case it would be

$2*cos(t-(3-t))=2*cos(2*t-3)$ am on the right way?or could i simplify way of solution? EDITED: so rotation matrix in 2D has form

enter image description here

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The shape and size of the triangle doesn't change when you rotate it. So rotate it so that $B$ comes to lie on $(1,0)$, and $C$ on $(-1,0)$. Where does that place $A$ then? –  Daniel Fischer Jul 15 '13 at 15:55
    
$cos(3),sin(3)$ right? or how rotate? –  dato datuashvili Jul 15 '13 at 15:58
    
@DanielFischer The shape changes. –  Lord Soth Jul 15 '13 at 15:58
    
@LordSoth Huh? If you rotate a figure, you get a congruent figure. –  Daniel Fischer Jul 15 '13 at 16:00
    
@DanielFischer Yea I fell into the same trap. You have $\cos(t)$ and $\cos(3-t)$ one point rotates clockwise, the other rotates counterclockwise, that is why the shape changes. –  Lord Soth Jul 15 '13 at 16:01
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1 Answer

up vote 1 down vote accepted

Hint: All these points lie on the unit circle. In particular, $B$ and $C$ are antipodal, meaning that the line segment $BC$ passes through the origin. Now, forget about point $A$ as in the problem. Where you should put a point $A'$ such that the area $A'BC$ is maximized, where $BC$ is an antipodal line segment? The solution of this problem is to make $A'BC$ a right triangle (You need to prove this). Now, see if your parametric equations form the same right triangle (up to rotations); and by the way, they will.

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but task is about determine $t$ –  dato datuashvili Jul 15 '13 at 16:06
    
I know; I think this is enough hint to keep you going, and you should easily be able to find $t$. Why don't you spend some time on the problem with this hint and let me know if you still have a problem. –  Lord Soth Jul 15 '13 at 16:09
    
i found length of $BC=2$ ,but now i should use Pythagorean theorem to find length of $A'B$ right and point $A'$ itself yes ?,but it will introduce another variables,maybe much more right? –  dato datuashvili Jul 15 '13 at 16:15
    
OK, forget about the question. I give you two antipodal points on the unit circle, say $B$ and $C$. As you said, $|BC| = 2$. Your task is to find a point $A'$, again on the unit circle, such that the area $A'BC$ is maximized. Where you should put the point $A'$ and what is the resulting area? –  Lord Soth Jul 15 '13 at 16:28
    
antipodal means they are on they same line?on diameter? by 180 degree –  dato datuashvili Jul 15 '13 at 16:31
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