# Inscribe a semicircle into a square

• Feb 21st 2010, 01:19 PM
Inscribe a semicircle into a square
How do you inscribe a semicircle into a square? And the semicircle's diameter can not be the one of the diagonals of the square. It has to look like the picture below.
• Feb 21st 2010, 03:53 PM
Diagonal
Very messy unless someone else has a solution. I don't even really count this as a solution. However, if you use the following diagram, then from the triangle inside the lower right sector, you can find x in terns of r. Then you have sides of a right triangle that will relate r and d. That is, you can form the ratio of r and d. However, I did a little of the math, and it is VERY messy [I might easily have missed something.] Further, that does not at all imply a drafting solution, but it's the best I can come up with at this time and under my present circumstances.
• Feb 21st 2010, 03:59 PM
Soroban

Quote:

How do you inscribe a semicircle into a square?
And the semicircle's diameter can not be the one of the diagonals of the square.

I believe that the original problem asked for the semicircle of maximum area.
Code:

                C     -  * - - * o * - - - *     :  | *:::::|:::::*  |     : D o:::::::|:::::::* |     :  | *:::::|R:::::::*|     :  |  R*:::|:::::::::|     x  |    * |::::R::::*     :  |      o:-:-:-:-:o B     :  |      O: *:::::::*     :  |      :  *R::::|     :  |      :  45°*::*|     - E o - - - * - - - o *                 F      A

Let $x$ = side of the square.
Let $R$ = radius of the semicircle.
Let $O$ = center of the semicircle.

The semicircle is tangent to the square at $A,B,C,D.$
. . Then: . $OA = OB = OC = OD = R$

In right triangle $OFA\!:\;\angle OAF = 45^o$
. . Hence: . $OF = \frac{R}{\sqrt{2}}$

Since $CO + OF \:=\:x$, we have: . $R + \frac{R}{\sqrt{2}} \:=\:x$

. . Hence: . $R\left(1 + \frac{1}{\sqrt{2}}\right) \:=\:x \quad\Rightarrow\quad R\left(\frac{\sqrt{2}+1}{\sqrt{2}}\right) \:=\:x \quad\Rightarrow\quad R \;=\;\frac{\sqrt{2}}{\sqrt{2}+1}\,x$

Therefore: . $R \;=\;(2-\sqrt{2})x$

• Feb 21st 2010, 06:42 PM
This is the exact problem.
Inscribe a semicircle into the square given below.
And then my teacher put a picture of what it was supposed to look like, which I have attached in my previous post.
I see what you're doing, but how did you get the semicircle's center? Or what would be the first time in constructing this after the square?
O yes, also when you say what R is equal to, how would you go about constructing a segment that multiplies to side x?
• Feb 21st 2010, 08:23 PM
Soroban

Quote:

When you say: . $R \:=\:(2-\sqrt{2})x$
how would you go about constructing a segment that multiplies to side $x$ ?

I've come up with a construction; maybe someone can improve on it.

On a horizontal line, mark off points $A,B,C,D$
. . so that: . $AB \,=\,BC\,=\,CD\,=\,1$
Code:

          1      1      1     - o - - - o - - - o - - - o -       A      B      C      D

At $C$, erect perpendicular $CE$ so that: . $CE = 1$
Code:

                      E                       o    _                       | *  √2                     1|  *                       |    *     - o - - - o - - - o - - - o - -       A      B      C  1  D
Draw $ED\!:\;\;ED = \sqrt{2}$

Using D as center and DE as radius,
draw an arc cutting BC at P.
Code:

                      E                     . o                   .  | *                 .    |  *                 .    |    *     - * - - - * o - - * - - - o -       A      B P    C      D
Hence: . $BP = 2-\sqrt{2}$

Through $A$, draw a line to the upper-right.
On the line, measure off $AQ = x$
Code:

                      *                     *                   *             Q  *               o         x  *  \           *    \         *        \     - o - - - - - o - - o -       A          B    P
Draw $QB.$

Code:

                      *                 R  *                   o             Q  *  \               o    \         x  *  \    \           *    \    \         *        \    \     - o - - - - - o - - o -       A          B    P

Through P, construct $PR \parallel BQ.$

Then: . $QR \:=\:(2-\sqrt{2})x$