Problem:
Squares S1​ and S2​ are inscribed in right triangle ABC, as shown in the figures below. Find AC+CB if area (S1​)=441 and area (S2​)=440.
Solution:
Let a=BC,b=AC. We will first find the hypotenuse c of â–³ABC and the altitude h on c, because these are relatively easy to compute, and because from
c2=a2+b2 and ch=ab
it is easy to find a+b (without first finding a and b, which is harder!).
Consider the ratios of the areas of the smaller triangles surrounding S1​ and S2​, using the fact that all five of these triangles are similar to one another and to △ABC. To simplify the notation, let T denote both △ABC and its area, and similarly, let T1​,T2​,T1​,T2​ and T denote both the triangles indicated in the figures below and their respective areas.
Since T1​T1​​=T2​T2​​=441440​, we find that
Therefore, T=4411​T, and hence the corresponding parts of triang1es T and T are in a linear ratio of 1 to 21.
It follows that c=21440​, since the hypotenuse of T is 440​. Moreover, h=21h~, where h~ denotes the altitude of T on its hypotenuse. Combining the latter equation with the observation h=h~+440​, we find that h=2021​440​,
Notes. More generally, if area (S1​)=p2 and area (S2​)=q2, one can show that
p=ab/(a+b) and q=aba2+b2​/(a2+ab+b2)(1)
Then, if p and q are given, either by solving the equations in (1) simultaneously (in the unknowns a+b and ab) or otherwise, one can show that
a+b=p+(p2/p2−q2​) and ab=p(a+b)(2)
Knowing the values of a+b and ab, one can also determine the values of a and b explicitly with the help of the Quadratic Formula. In the present problem they turn out to be 21(11±311​). Since these two numbers are in the ratio of 10+311​ to 1, the accompanying figures are obviously not drawn to scale.
