Trickly interview question

Q.
A square is inscribed in a circle. A smaller square is drawn; it shares a side with the large square, and its other two corners touch the circle.
 What is the ratio of the large square’s area to the small square’s area?

 

I found this problem online, and the person who posted it said it was asked during a job interview. I can admit I could not solve it, but it is interesting and I wanted to share it. Can you figure it out?


Answer:

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The key is seeing a specific right triangle that relates the sides of the squares to each other and the radius of the circle.
Let r be the radius of the circle. Draw a diameter of the circle to two corners of the large square. The resulting triangle is an isosceles right triangle with a hypotenuse 2r.



Now we have a right triangle with half the small square’s side (s/2), the small square’s side plus half the large square’s side (s + r/√2), and a hypotenuse equal to the radius (r).
Using the Pythagorean theorem, we have:
(s/2)² + (s + r/√2)² = r²
The radius of the large circle is an unknown constant. We can treat the above equation as a quadratic equation in s. So let us expand the above and solve for s.
s²/4 + s² + rs√2 + r²/2 = r²
(5/4)s² + rs√2 – r²/2 = 0
5s² + rs(4√2) – 2r² = 0
(5s – r√2)(s + r√2) = 0
We have two possible solutions:
s = r√2/5
or
s = –r√2
Since the small square’s side length is positive, we can reject the second solution. This means we have:
s = r√2/5
The large square’s side was equal to r√2, so this means the large square’s side is 5 times as large. The ratio of the areas is the square of the ratio of the sides. This means the large square’s area is 25 times as large as the small square’s.

2nd method:
 use the intersecting chord theorem! When two chords intersect, the product of the divided lengths for one chord equals the product for the other chord.

Extend the right side of the small square vertically to the circle. Label the side of the large square as chord AB and label the vertical chord as CD, and they intersect at point E, as in the following diagram.



If the small square has side s, and x is the length of EB, then we have:
AE(EB) = CE(ED)
(x + s)x = s(2x + 2s)
x² + xs = 2xs + 2s²
x² – xs – 2s² = 0
(x – 2s)(x + s) = 0
For x to be a positive value, we have the only solution x = 2s.
The large square has a side 2x + s = 5s. This is 5 times larger than the small square, so its area will be 25 times as large.

3rd method: 

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