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.
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.
Hello, arrowhead566!
I believe that the original problem asked for the semicircle of maximum area.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.Code:C - * - - * o * - - - * : | *:::::|:::::* | : D o:::::::|:::::::* | : | *:::::|R:::::::*| : | R*:::|:::::::::| x | * |::::R::::* : | o:-:-:-:-:o B : | O: *:::::::* : | : *R::::| : | : 45°*::*| - E o - - - * - - - o * F A
Let $\displaystyle x$ = side of the square.
Let $\displaystyle R$ = radius of the semicircle.
Let $\displaystyle O$ = center of the semicircle.
The semicircle is tangent to the square at $\displaystyle A,B,C,D.$
. . Then: .$\displaystyle OA = OB = OC = OD = R$
In right triangle $\displaystyle OFA\!:\;\angle OAF = 45^o$
. . Hence: .$\displaystyle OF = \frac{R}{\sqrt{2}}$
Since $\displaystyle CO + OF \:=\:x$, we have: .$\displaystyle R + \frac{R}{\sqrt{2}} \:=\:x$
. . Hence: .$\displaystyle R\left(1 + \frac{1}{\sqrt{2}}\right) \:=\:x \quad\Rightarrow\quad R\left(\frac{\sqrt{2}+1}{\sqrt{2}}\right) \:=\:x \quad\Rightarrow\quad R \;=\;\frac{\sqrt{2}}{\sqrt{2}+1}\,x$
Therefore: .$\displaystyle R \;=\;(2-\sqrt{2})x$
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?
Hello again, arrowhead566!
I've come up with a construction; maybe someone can improve on it.When you say: .$\displaystyle R \:=\:(2-\sqrt{2})x$
how would you go about constructing a segment that multiplies to side $\displaystyle x$ ?
On a horizontal line, mark off points $\displaystyle A,B,C,D$
. . so that: .$\displaystyle AB \,=\,BC\,=\,CD\,=\,1$Code:1 1 1 - o - - - o - - - o - - - o - A B C D
At $\displaystyle C$, erect perpendicular $\displaystyle CE$ so that: .$\displaystyle CE = 1$Draw $\displaystyle ED\!:\;\;ED = \sqrt{2}$Code:E o _ | * √2 1| * | * - o - - - o - - - o - - - o - - A B C 1 D
Using D as center and DE as radius,
draw an arc cutting BC at P.Hence: .$\displaystyle BP = 2-\sqrt{2}$Code:E . o . | * . | * . | * - * - - - * o - - * - - - o - A B P C D
Through $\displaystyle A$, draw a line to the upper-right.
On the line, measure off $\displaystyle AQ = x$Draw $\displaystyle QB.$Code:* * * Q * o x * \ * \ * \ - o - - - - - o - - o - A B P
Code:* R * o Q * \ o \ x * \ \ * \ \ * \ \ - o - - - - - o - - o - A B P
Through P, construct $\displaystyle PR \parallel BQ.$
Then: .$\displaystyle QR \:=\:(2-\sqrt{2})x$