[Math] Finding the total area of three squares

areageometryrecreational-mathematics

The picture below is a recent question posted on twitter. Tim Gowers replied that "Got it, but by a fairly brute-force approach."

One way to do it is denoting the sides of the small, medium, large squares as $a$, $a+b$, and $2a+b$, respectively. Then one immediately has
$$
(a+b)+(2a+b)=5\;.\tag{1}
$$
By the inscribed angle theorem, one can then draw a right triangle with hypotenuse of length $5$. By introducing the height with the hypotenuse as the base, one has by similar triangles
$$
\frac{a+b}{3a+b} = \frac{b}{a+b}\;.\tag{2}
$$
It is fairly easy to combine (1) and (2) to get $a=b=1$, which implies that the total area is
$$
1^2+2^2+3^2=14\;.
$$

My solution above assumes that the arc in the picture is a semicircle, which makes the problem relatively easy.

Questions:

  • If one drops the semicircle assumption [added: and assuming only an arc of a circle], can one still get the same answer?

  • If the answer is "no" to the question above, what range of numbers can one have for the area of the three squares?

enter image description here

Best Answer

It is always possible to draw a circular arc through three points, so there is a one-parameter family of solutions if the semicircular requirement is dropped.

  • At one extreme, the middle-sized square has the same size as the largest square, the small square vanishing. Then there are two squares of side length $2.5$, and the area is $5×2.5=12.5$.
  • At the other extreme, the two smaller squares are the same size, their side lengths being half that of the largest square. The two smaller squares are of side $\frac53$ and the larger one of side $\frac{10}3$, for an area of $5×\frac{10}3=\frac{50}3=16.\overline6$.
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