# Calculus problem that doesn't seem like calculus

• Jun 7th 2008, 10:05 AM
Bluekitten
Calculus problem that doesn't seem like calculus
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
$\displaystyle (1/2)*sqrt((1/2+x)^2+(1/2-x)^2)*(1/2)$

Quote:

Problem:

Consider a square piece of paper with sides equals to 1 unit. We label the four vertices as ABCD. Now inscribe a quarter circle with radius of 1 unit such that its center is at vertex A. (See figure 1). Next, we fold corner labeled as vertex A to touch the circumference of the quarter circle. We want the fold to create a triangle (the folded part forms a triangle; see figure 2). Note that the end points of the crease have to be on side AB and side AD in order to do this. Our task is to find the exact area of the smallest and the largest of these triangles created in the above process.
http://farm4.static.flickr.com/3081/...ea7d3938_o.jpg

• Jun 7th 2008, 11:47 AM
earboth
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
$\displaystyle (1/2)*sqrt((1/2+x)^2+(1/2-x)^2)*(1/2)$

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.

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

with:
$\displaystyle h = \frac12$

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

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

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

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

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

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

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

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

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

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

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

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