Therefore, the maximum area of the triangle is 1/2.

[PsychonautQQ]

What is the greatest possible area of a triangle that has one vertex at the center of a circle with radius 1 and the other 2 vertices on the circumference of the circle? Can anyone give me some insight into this?

I was thinking maybe make the center angel 90 degree's? If so my circle would have an area of 1 I believe. Is this the maximum area?

 

[phinds]

Uh ... you really think a unit circle has an area of 1? You might want to rethink that.

Figure out what the area of a unit circle is. If your central angle is 90 degrees, don't you reckon the area of that pie-segment would be 1/4 of the area of the circle and the area of the triangle would be fairly close to that amount?

How you write an equation for the area of the triangle should be easy but I don't know how you solve it for maximum.

 

[Mark44]

What is the greatest possible area of a triangle that has one vertex at the center of a circle with radius 1 and the other 2 vertices on the circumference of the circle? Can anyone give me some insight into this?

I was thinking maybe make the center angel 90 degree's? If so my circle would have an area of 1 I believe. Is this the maximum area?

No, the circle's area is $\pi$. Your goal is to find the area of an inscribed triangle. If the angle (not "angel" which is something different) at the center of the circle is 90°, then the area of the triangle is 1/2 (isosceles right triangle with legs of length 1). What's to prevent that central angle from being larger than 90°? Can you write an equation that gives the area of a triangle with one vertex at the center of the circle, another at the point (1, 0), and the third at an arbitrary point on the circle?

 

[certainly]

I was thinking maybe make the center angel 90 degree's? If so my circle would have an area of 1 I believe. Is this the maximum area?

Although (as others pointed out), the circle's area is $\pi$, your answer is correct.

Can you write an equation that gives the area of a triangle with one vertex at the center of the circle, another at the point (1, 0), and the third at an arbitrary point on the circle?

Since we already know that the other side also has length 1, I think it would be easier to plug in $a=1,\ b=1$ in Heron's formula.

How you write an equation for the area of the triangle should be easy but I don't know how you solve it for maximum.

Simple, take the derivative of the function describing the area, and let it equal 0 :-)

 

[phinds]

Simple, take the derivative of the function describing the area, and let it equal 0 :-)

DOH ! (Although that could also give you a minimum, depending on what the function looks like ... I don't think it would do that in this particular case).

 

[certainly]

DOH ! (Although that could also give you a minimum, depending on what the function looks like ... I don't think it would do that in this particular case).

It does. In the end you solve for $a^2$ (where $a$ is the third side). Maximum is $+x$, minimum is $-x$.

 

[theodoros.mihos]

For one vertex on the centre of circle, you are right:
Separate the triangle to two orthogonal triangles with vertical lines x and y, so x²+y²=1.
The total triangle area is:
$$ E = xy = x\sqrt{1-x^2} \Rightarrow \frac{dE}{dx} = \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}} = 0 \Rightarrow x=y=\frac{1}{\sqrt{2}} \Rightarrow $$
$$ \tan(\theta/2) = \frac{y}{x} = 1 \Rightarrow \theta=\pi/2 ,\, E_{max}=\frac{1}{2} $$