1079.41 – Word Triangle Arithmetic


In the word triangle below, find four different words of lengths a,b,c,a, b, c, and dd such that a2=bda^2 = bd and ad=b2cad = b^2c.


Solution

One answer is a=SCHOOL,b=OAK,c=OVERCOAT,d=DEMOCRATICa = \mathbf{SCHOOL}, b = \mathbf{OAK}, c = \mathbf{OVERCOAT}, d = \mathbf{DEMOCRATIC}.

There is something that can be done here to cut down on the guessing and checking.

a2b=bd and ad=b2cd=a2b=b2caa3=b3cc=a3b3.\begin{aligned}a^2b =bd \text{ and } ad = b^2c &\leadsto d = \dfrac{a^2}{b} = \dfrac{b^2 c}{a} \\&\leadsto a^3 = b^3 c \\&\leadsto c = \dfrac{a^3}{b^3}.\end{aligned}

Therefore, cc is a perfect cube. So it must be that c=8c = 8 and a=2ba = 2b. Now there are only a few possible values of bb to check, and bb determines everything else.