Calculus problem that doesn't seem like calculus

Quote:

Originally Posted by Bluekitten

I can't seem to come up with a formula for the area of this triangle. This should be a calculus problem, but any formula that I can come up with does not require calculus to get the answer. The best formula that I have come up with so far is
$<br /> (1/2)*sqrt((1/2+x)^2+(1/2-x)^2)*(1/2)<br />$

I've attached 2 drawings:

In the left sketch you can see how the triangle changes it's shape when the vertex runs along the quarter circle.

In the right sketch you find all necessary variables
Since the radius of the quarter circle is 1 the length of the arc x is measured in radians.

$A_{triangle} = \frac12 (s+t) \cdot h$

with:
$h = \frac12$

$s = \frac12 \cdot \tan(x)$

$t = \frac12 \cdot \tan\left(\frac{\pi}4 - x\right) = \frac12 \cdot \cot(x)$

Plug in all variables into the equation of $A_{triangle}$ :

$A_{triangle}(x) = \frac12 \left(\frac12 \cdot \tan(x) + \frac12 \cdot \cot(x) \right) \cdot \frac12 = \frac18 \left(\tan(x) + \cot(x) \right)~,~ 0 \leq x \leq \frac{\pi}4$

(Personal remark: The domain of this function is wrong! Probably - but I haven't a proof yet - the domain is $\frac{\pi}6 \leq x \leq \frac{\pi}3$ )

To get the extreme values of A(x) calculate the first derivative of A:

$A'(x)=\frac18 \left(\frac1{(\cos(x))^2} - \frac1{(\sin(x))^2} \right)$

Now solve for x: A'(x) = 0

I've got $x = \frac{\pi}4~\vee~x=\frac{3 \pi}4 \notin \{domain\}$

With $x = \frac{\pi}4$ you get an isosceles right triangle with $A = \frac14$

By trial and error I got $A\left(\frac{\pi}6 \right) = A\left(\frac{\pi}3 \right) \approx 0.288...$

So there you have the smallest triangles at $x = \frac{\pi}4$ and the largest triangles if x approaches the bounds of the domain.