Semicircle in square

Geometry Level 3

What is the radius of the largest semicircle that can be enclosed in a square of side length 1?

2 - \sqrt{2}

4 - 2 \sqrt{3}

\frac{1}{2}

\frac{3}{5}

2 solutions

Chew-Seong Cheong
Mar 11, 2015

Due to symmetry, the center of the largest semicircle is on the diagonal of the square with its chord touching two sides of the square as shown in the figure above. If the radius of the semicircle be $r$ , then $r$ increases as the points of contact move left and down, that is when $x$ increases, until the diameter touches the other two sides of the square. Then we note that the diagonal is given by:

$\sqrt{2}r+r =\sqrt{2} \quad \Rightarrow r = \dfrac {\sqrt{2}}{\sqrt{2}+1} = \dfrac {\sqrt{2}(\sqrt{2}-1)}{1} = \boxed{2-\sqrt{2}}$

Of course, the actual work of this question is proving that this is indeed the maximum that can be achieved. There needs to be more done to show that you indeed have the correct scenario.

Calvin Lin Staff - 6 years, 3 months ago

Sir, can you provide us with a detailed solution.

I did the same work and then started thinking about why it should be maximum. The only reason I was able to draw out was based on symmetry. But not very sure if that reasoning works.

Kishlaya Jaiswal - 6 years, 2 months ago

OK, that's right. But can you provide the correct solution here...

Mathews Boban - 6 years, 3 months ago

Matthew, I thought I did. I had edited the solution to include the explanation.

Chew-Seong Cheong - 6 years, 3 months ago

I didn't understand that how have you marked 'r' on the diagonal?

Aditya Sharma - 6 years, 2 months ago

Mariano PerezdelaCruz
Mar 20, 2015

We start considering a circle fully inscribed in square of side L, which is the maximum circle I can inscribe. We draw the two diagonals, taking one of the diagonal, obviously contain the diameter of a semicircle .

Now considering the intersection points of the circle and diagonal we can reduce our primitive square by drawing two parallel to the sides we keep and passing over this intersections then we get a reduced square.

As the angle between diagonals and side is 45 degrees the diameter with little trigonometry effort is 2(2-sqrt(2)) and the radius (2-sqrt(2)).

Semicircle in square

2 solutions

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