The greatest possible area of a triangular region with one vertex at the center of a circle of radius 1 and the other 2 vertices on the circle.

Question

What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius 1 and the other 2 vertices on the circle?

(A)$\frac{1}{2}$

(B)1

(C)$\sqrt2$

(D)$\pi$

(E)$\frac{1+\sqrt2}{4}$

I do not know if it is a right-angled triangle or no? How can I think in this question?

Answer 1

Elaborating more on Mick's comment. Below is a diagram of your problem.

$\hskip{.75in}$

Another formula for the area of a triangle is $$ \frac{1}{2} \cdot a\cdot b \cdot \sin(\theta),$$ where $a$ and $b$ are two adjacent sides and $\theta$ is the angle between $a$ and $b$. We have $a = 1$, $b = 1$ and $\theta = C$, meaning we want to maximize $$ \frac{1}{2} \cdot 1 \cdot 1 \cdot \sin(C) = \frac{1}{2} \cdot \sin(C).$$

Now, the biggest sine can be is 1, and $\sin(C) = 1$ when $C = 90^\circ$. This means you do indeed have a right triangle. Therefore, the maximum area is

$$ \frac{1}{2} \cdot 1 = \frac{1}{2}.$$

Answer 2

Let's think about this for a second. Your total circle has radius $\pi*r^2 = \pi$, so a triangle inside it definitely cannot have the same area $\pi$. Option D is out (always nice to rule out a couple options first, the rest look viable).

now let's think of this as an optimization problem. Assume WLOG that this triangle has one vertex at the point $(1,0)$ (Rotational symmetry on the problem allows us to do this).

Now the area of a triangle is $A = a*b*sin(\Theta)/2$ where $a, b$ are the length of two sides and $\Theta$ is the angle between them. Take the two sides $a, b$ to be the sides of length $1$, so $A = 1*1*sin(\Theta)/2$

Now we know that $A'(\Theta) = 1/2*cos(\Theta)$ and we want to know the maximum possible area, so we look for critical points: $A'(\Theta) = 0 = 1/2*cos(\Theta)$ $cos(\Theta) = 0 \Rightarrow \Theta = \frac{(4i+1)*\pi}{2}$ for $i \in \mathbb{Z}$ (for the sake of simplicity and without losing generality, let's just assume $\Theta = \pi/2$)

Now we know that $A(\Theta)$ has a maximum there (minimums occur when the area is 0 because the two sides go in the same direction, yes those are degenerate triangles).

$$A(\Theta) = 1*1*sin(\pi/2)/2 = 1/2$$

and there you have it, the maximum area occurs in the right triangle with total area 1/2

Answer 3

One doesn't really need calculus for this problem. Without loss of generality, we can take the first two vertices to be $(0,0)$, $(1,0)$ and the third to be some point $(x,y)$ on the unit circle with $y>0$. The resulting triangle has base $1$ and height $y$, so that the area is $y/2$. But the largest possible value of $y$ is $1$, so the max area is $1/2$.