Problem 7 of the Belarus Mathematical Olympiad 2004

Let there be given two similar triangles such that the altitudes of the first triangle are equal to the sides of the other. Find the largest possible value of the similarity ratio of the triangles.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Alex Bogomolny, Ph.D.)

Let the first triangle be ABC and the feet from A, B and C to the opposite sides be D, E and F, respectively.

Now let BC = a, AC = b, AB = c and AD = d, BE = e and CF = f.

Assume abc. Because twice the area of triangle ABC = ad = be = cf, our assumption makes fed.

To find the largest possible value of the similarity ratio of the triangles we need to find the largest possible ratio \frac{f}{a}

or the largest possible cos∠FCB.

The similarity of the triangles ADB and CFB gives us \frac{d}{c} = \frac{f}{a} (1)

and the similarity of triangles AEB and AFC, \frac{e}{c} = \frac{f}{b} (2)

And because the altitudes of the first triangle are equal to the sides of the second,

we also have \frac{f}{a} = \frac{e}{b} (3)

From (2), e = \frac{cf}{b} ; substitute it into (3), we then have b = ac

Now the law of cosines gives us b = a + c - 2ac·cos∠ABC

or ac = a + c - 2ac·cos∠ABC

or cos∠ABC = \frac{a + c - ac}{2ac} = \frac{a + c}{2ac} - \frac{1}{2}

But ∠ABC + ∠FCB = 90; therefore, cos∠FCB = sin∠ABC = \sqrt{1 - cos∠ABC}

Hence, cos∠FCB is largest when cos∠ABC is smallest or when \frac{a + c}{2ac} - \frac{1}{2} is smallest, or when \frac{a + c}{2ac} is smallest.

Let x = \frac{a}{c}. Then \frac{a + c}{2ac} = \frac{f(x)}{2}. Function f(x) = x + \frac{1}{x} has no maximum, but has a minimum when f '(x) = 0.

f '(x) = 1 - x^{-2} = 0  gives x^{2} - 1 = 0  so that the only minimum is attained for x = 1, i.e., when a = c. Since, by our assumption, abc, all three sides are equal and both triangles are equilateral. Thus \frac{a}{f} = \frac{2\sqrt{3}}{3}. A pripori, it is not clear which of the two possible ratios is meant to be maximized, the answer is the largest of the two, \frac{a}{f} = \frac{2\sqrt{3}}{3} or \frac{f}{a} = \frac{\sqrt{3}}{2}, which is the former - \frac{2\sqrt{3}}{3} - and is attained for the equilateral triangles.