# Pythagoras/trigonometry question

Does anyone know how to answer this question?

A right-angled triangle is to be constructed with hypotenuse (the longest side) of length one metre in such a way as to maximize the perimeter of the triangle (the sum of the lengths of its sides). Determine the lengths of the other two sides.

• So, you should show your thoughts on it: Is it a calculus problem? Write formulas. What should be maximized? What methods does the text describe for maximizing? – GEdgar Apr 23 '13 at 14:40

Hint: you should have found out by now that in such a right triangle, the legs are $\sin(\theta)$ and $\cos(\theta)$ where $\theta$ is a fixed non-right angle of the triangle.

The perimeter is then $P(\theta)=\sin(\theta)+\cos(\theta)+1$. Work to maximize this on the interval $[0,\pi/2)$.

(While looking for critical points of $P(\theta)$, you should probably not even need a calculator to determine $\theta$. Strictly speaking, you should also take some steps to confirm that it's a maximum and not a minimum!)

Let's call the lengths of two legs $x$ and $y$. Since the triangle is right and the hypotenuse has length $1$, $x$ and $y$ are related by the equation $x^2 + y^2 = 1$. We need both $x,y\geq 0$.

We would like to maximize the perimeter, $P = 1 + x + y$, subject to this constraint. We will be able to write $y$ as a function of $x$ using the constraint, so in fact we have $$P(x) = 1 + x + y(x).$$

Now this is a straightforward calculus exercise in optimization. Can you take it from here?

$AB=1$, $\angle BCA=90^{\circ}$
Find the maximum of the perimeter of $\triangle ABC$, $(AB+BC+CA)$

Sol)
Let $E$ be the midpoint of the arc $AB$ and think of the circle $\Omega (E,\overline{EA})$
$D$ is the intersection of $AC$ and $\Omega$
$2\angle ADB=\angle AEB=\angle ACB=\angle ADB+\angle DBC$
$\therefore$ $\angle CDB=\angle DBC$, where $BC=CD$

$AB+BC+CA=AB+CD+CA=AB+AD\leq AB+AF$
The equality hold when $C=E$
Thus, $BC=CA=\dfrac{1}{\sqrt{2}}$