# Thread: Semicircle & Chords (Urgent)

1. ## Semicircle & Chords (Urgent)

Hey guys. I have no idea how to even start this question. Any ideas?? Thanks!

"A semi-circle is dawn outwardly on a chord 'AB' of a circle with center 'O' and units raidus. The point 'C' on this semi-circle which sticks out of the given circle the furthest is on the radius 'ODC' which is perpendcular to 'A'. Of course the Length of 'OC' is dependant on the choice of chord 'AB' in the first place. Determine 'AB' so that 'OC' is maximum"

2. Hello, furnis1!

First, make a sketch . . .
[I'll use Calculus to maximize OC.]

A semi-circle is dawn outwardly on a chord $\displaystyle AB$ of a circle with center $\displaystyle O$ and raidus 1.
The point $\displaystyle C$ on this semi-circle is the furthest from center $\displaystyle O.$
Of course, the length of $\displaystyle OC$ is dependant on the choice of chord $\displaystyle AB.$
Determine $\displaystyle AB$ so that $\displaystyle OC$ is maximum.
Code:
              * * *
*           *  A
*               o
*              / :*   o
1/   :      o
*           / θ  D: *
*       O * - - - + * - - o C
*           \     : *
1\   :      o
*              \ :*  o
*               o
*           *  B
* * *

The circle with center $\displaystyle O$ has radii $\displaystyle OA = OB = 1.$

A semicircle is constructed (outward) with center $\displaystyle D$ and diameter $\displaystyle AB.$
. . It has radii: $\displaystyle DA = DC = DB.$

Let $\displaystyle \angle AOC = \theta,\;\;OC = d$

In right triangle $\displaystyle ADO\!:\;\;OD = \cos\theta\,\text{ and }\,DA = \sin\theta = DC$

Then: .$\displaystyle d \:=\:OD + DC \:=\:\cos\theta + \sin\theta$

To maximize $\displaystyle d$, set $\displaystyle d' = 0$

We have: .$\displaystyle d'\:=\:\sin\theta + \cos\theta \:=\:0\quad\Rightarrow\quad \sin\theta \:=\:\cos\theta \quad\Rightarrow\quad \tan\theta\:=:1$

. . Hence: .$\displaystyle \theta \:=\:45^o$

$\displaystyle OC$ is maximum when $\displaystyle \theta = 45^o$

Then $\displaystyle \angle AOB = 90^o$ and $\displaystyle \Delta AOB$ is an isosceles right triangle.

Since its legs are length 1, its hypotenuse is: .$\displaystyle AB \:=\:\sqrt{2}$

3. great answer prof! thanks for the help - that really cleared it up.