Two squares are inscribed in a semi-circle, as shown below. Find the radius of the semi-circle.

sandwich Aug 18, 2023

#1**+1 **

I had to stare at this one a long time to solve this one! I have created a diagram below with points labeled so that it is easier to follow along.

In the diagram above, notice that OF and OE are both radii of the semicircle. These lengths I have labeled as r. I have labeled L as the length of segment OB. We can now apply Pythagorean's Theorem twice to find the radius of this semicircle.

\(r^2 = {\rm AE}^2 + ({\rm AB} + l)^2 \\ r^2 = 2^2 + (2 + l)^2\) | \(r^2 = {\rm CF}^2 + ({\rm BC}^2 - l)^2 \\ r^2 = 5^2 + (5 - l)^2\) |

Now, we have expressions for both r and L, so we can solve for r, and we are finished. Probably the best way is to solve for L first since that one is easier to solve.

\(2^2 + (2 + l)^2 = 5^2 + (5 - l)^2 \\ 4 + 4 + 4l + l^2 = 25 + 25 - 10l + l^2 \\ 8 + 4l = 50 - 10l \\ 14l = 42 \\ l = 3\)

Now that we have solved for L, we can find the radius.

\(r^2 = 2^2 + (2 + l)^2 \\ r^2 = 2^2 + (2 + 3)^2 \\ r^2 = 4 + 25 \\ r^2 = 29 \\ r = \sqrt{29} \text{ or } r = -\sqrt{29}\)

Since we are dealing with the radius, which is a length, we should reject \(r = -\sqrt{29}\). Therefore, \(r = \sqrt{29} \approx 5.3852\)

The3Mathketeers Aug 19, 2023

#2**+1 **

I tried a similar procedure as you, 3Mathketeers and I got the same answer.....but notice that a square of 5 will not fit inside the semi-circle [a 4 x 5 rectangle will !! ]

In short, I believe that this is an impossible problem as set up by the poster.....

CPhill Aug 19, 2023

#3**+1 **

I apologize for the confusion. As you discovered, the diagram I created is intentionally NOT drawn to scale so that it looks as similar as possible to the original diagram. I figured that as long as points E and F were on the circle, then I could still go ahead and find the radius anyway.

The3Mathketeers
Aug 19, 2023