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 △ABC 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 T1T1=T2T2=441440, we find that
T=T1+T2+440+T=441440(T1+T2+441)+T=441440T+T
Therefore, T=4411T, 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=2021440,
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.
