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What is the largest area of the semi-circle that can be inscribed in a square of edge length 1 unit?

That is a question posted by harpreet in the topic " maxima & minima" on a Mathematics forum on Orkut.com on 12th July 2009.

If you have insight into this problem, perhaps you would cross-post it to the Orkut forum. If you use Gmail, or Google Mail, then you already have a username for Orkut.com, which is owned by Google.com.

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Replies to This Discussion

We know that the radius is between 1 (side) and (diagonal)

Here is my answer :


How to prove it ?

  1. If we divide the square in two area with the Semicircle radius, we know that the Semicircle must cover partially the greatest area.
  2. radius is between 1 and
  3. The "center" of the SemiCircle must touch at least the 4 sides (or that means that you are not filling the maximum)
  4. Answer should be symmetric since Semicircle is symmetric and square also (weak argument, but usefull as hint)

All together, you know that you have to look for the center on the main diagonal at a minimum distance of 0.5units from two of the 4 sides.

Given that, you can state that the point (4) is not a weak argument :-). If not symmetric one of the two sides is not touched !

It remain the calculus of the area of Semicircle.

I didn't find a smart way to get it area. So, you can calculate it with help of this picture :


Dashed segments are bisector of 45° angles formed by side and diagonal
It might be helpful to look at the problem in a different way. Draw a circle of radius 1 around the orgin (0,0). Identify the semicircle, then draw the four lines that form the square. Find the width of the square, then invert it to find the radius required of the circle. The area of the semi-circle is easily calculated from its radius.
I wonder if this problem would be more or less difficult for a semicircle in other regular polygons.


Provided my algebra is correct, r is (among friends) 0.6 units
Steve, it looks like you have added up four areas that total to the unit area. There is an easier way. Add up two lengths that total to the unit length, or 1. The width of the square, which is 1, has two components. Left of the origin, in the 9 o'clock or negative-x direction, the distance to the edge is r. Right of the origin in the 3 o'clock or x direction the distance is r/sqrt(2), by pythagoras' theorem on the 45 degree triangle. This expression can be factorised as follows to calculate r, the radius of the circle. We both get the same answer for r, in different form.

Now, we just need a volunteer to calculate the area of the semicircle, which was the original question.

The LaTex code for the evaluation is:
r+\frac{r}{\sqrt[]{2}} = 1\\\\
r\:\sqrt[]{2}+r = \sqrt[]{2}\\\\
r(1+\sqrt[]{2}) = \sqrt[]{2}\\\\
r = \frac{\sqrt[]{2}}{1+\sqrt[]{2}}\\\\
r = \frac{(\sqrt[]{2}-1)\:\sqrt[]{2}}{(\sqrt[]{2}-1)(\sqrt[]{2}+1)}\\\\
r = 2 -\sqrt[]{2}\\\\
A_{semicircle}=\frac{\pi r^{2}}{2}\\\\
This code was entered at http://thornahawk.unitedti.org/equationeditor/equationeditor.php to generate the equation.
I have copied our solution over to the original topic "maxima & minima" on a Mathematics community on Orkut.com, acknowledging Mathematics24x7 members as the solvers. Orkut discussions are text only, so I expressed the solution without diagrams or LaTeX markup. The topic on Orkut covers two pages, so click Next to see the second page.
One statement that I made in the Orkut solution was "2-sqrt(2) is bigger than 1/2", but I don't know how to prove that.
Oh yeah...duh...that is even easier!

Colin McAllister said:
Steve, it looks like you have added up four areas that total to the unit area. There is an easier way. Add up two lengths that total to the unit length, or 1. The width of the square, which is 1, has two components. Left of the origin, in the 9 o'clock or negative-x direction, the distance to the edge is r. Right of the origin in the 3 o'clock or x direction the distance is r/sqrt(2), by pythagoras' theorem on the 45 degree triangle. This expression can be factorised as follows to calculate r, the radius of the circle. We both get the same answer for r, in different form.

Now, we just need a volunteer to calculate the area of the semicircle, which was the original question.

The LaTex code for the evaluation is:
r+\frac{r}{\sqrt[]{2}} = 1\\\\
r\:\sqrt[]{2}+r = \sqrt[]{2}\\\\
r(1+\sqrt[]{2}) = \sqrt[]{2}\\\\
r = \frac{\sqrt[]{2}}{1+\sqrt[]{2}}\\\\
r = \frac{(\sqrt[]{2}-1)\:\sqrt[]{2}}{(\sqrt[]{2}-1)(\sqrt[]{2}+1)}\\\\
r = 2 -\sqrt[]{2}\\\\
A_{semicircle}=\frac{\pi r^{2}}{2}\\\\
This code was entered at http://thornahawk.unitedti.org/equationeditor/equationeditor.php to generate the equation.
The problem was to calculate the area of the semicircle. Using the radius (r), just calculated, the area (A) of the semicircle is:
A=pi*r^2/2
A=pi*(2-sqrt(2))*(2-sqrt(2))/2
A=pi*(4-4*sqrt(2)+2)/2
A=pi*(6-4*sqrt(2))/2
A=pi*(3-2*sqrt(2))

It is correct to leave the answer as an expression of irrational numbers, but it can also be approximated to a decimal number:
A=3.141592654*(3-2*Sqrt(2))
A=0.539012085
which is just over one half of the area of the unit square.
This problem is also being discussed from an educational perspective in Straight Edge and Compass Construction For Developmental Math in Dan's Teaching developmental math group.
I describe an alternative approach to the semicircle problem in "Rediscovering Straight-edge and Compass Construction", and recommend it as an experiment in geometry, using just a ruler and a pair of compasses. Drawing the solution is easier than solving it using the algebra of irrational numbers. You could draw it on a PC, by constructive geometry, using the free GeoGebra geometry software.
The linked slideshow demonstrates how I used geometric construction to inscribe a semicircle in a square.
The construction was drawn using Geogebra and the sequence of screen images arranged using OpenOffice Impress.
I have published this as a Powerpoint slideshow on Slideshare.net at:
Inscribe Semicircle In Square by Geometric Construction
This construction is in the public domain.
Linda Fahlberg-Stojanovska has published a paper on compass and straightedge construction, which includes a simpler proof and procedure for inscribing a semicircle in a square. At http://geogebrawiki.pbworks.com/ciit10
From that link, select "Paper in English B/W", and go to section V part B of the paper. The proof, by congruent triangles, is in paragraph 4, and the construction in paragraph 5.
I followed an alternative solution that Linda Fahlberg-Stojanovska described in her recent paper. I used the free Dr Geo geometry software to draw this diagram by straight-edge and compass construction. The sequence of steps are labeled in a small font.

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