A geometry problem by Denton Young

Let the smaller triangle have side lengths a, b, c with a < b < c, and the larger triangle have side lengths b, c, d with b < c < d.

a, b, c, d form a geometric progression, and (a +b) > c. So the ratio is less than 1.618. Since all the sides are integral, a = f^3, b = f^3r (where r is the ratio), c = f^3r^2, and d = f^3r^3.

Since d < 50, d is an integer cube > 2, d must be 27. The ratio r therefore must b 1.5 so that a = 8: b = 12, c = 18.

12 + 18 + 27 = 57

This is a strangely baffling "simple" problem to solve. Since only integers are allowed, small ones less than $50$ , I simply guessed that perhaps all of the sides which are of all different lengths are elements of a geometrical series, and so I tried the ratio $3/2$ since obviously a ratio of $2$ would push it out of bounds and form no real triangles. Voila, the solution pops out $8, 12, 18, 27$ .

Let the triangle side length be a, b, c and another triangle side length be b, c, d.
As these triangles and similar, let the ratio of the length be n, so d = nc = bn^2 = an^3. The first triangle side length is a , an and an^2 while the another is an , an^2 and an^3 .

As any two sides length sum is larger than the third side, so a + an > an^2, an + an^2 > an^3.
a + an > an^2
1 + n > n^2
By trial and error n = 1.5 is a suitable number as all the sides length are integers.

As n = $\frac{3}{2}$ , therefore a = x2^m where m =3.

a = m * 2^3 = 8m.
b = 8m * $\frac{3}{2}$ = 12m
c = 12m * $\frac{3}{2}$ = 18m
d = 18m * $\frac{3}{2}$ = 27m

As d = 27m < 50, therefore m = 1.

So a = 8, b = 12, c = 18, d = 27.
And therefore the perimeter of the larger triangle is 12 + 18 + 27 = 57