# Geometric Optimization

from the Asian Pacific Mathematical Olympiad

The applet below illustrates a Problem 3 from the 2009 Asian Pacific Mathematical Olympiad:

Consider all the triangles ABC which have a fixed base BC and whose altitude from A is a constant h. For which of these triangles is the product of its altitudes a maximum?

What if applet does not run? |

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Copyright © 1996-2018 Alexander Bogomolny### Solution by

Vo Duc Dien

Let A', B' C' be the feet of the altitudes from A, B, C, respectively. The problem asks for the product AA'×BB'×CC' to be maximum. But AA' is constant, so BB'×CC' must be maximum.

The area of the triangle ABC is also constant since base BC is fixed.

Twice the area of triangle ABC = AA'×BC = BB'×AC = CC'×AB. From there, BB'×AC×CC'×AB, which is the square of twice the area of triangle ABC, is also constant. Regrouping,

Let R be the radius of the circumcircle of ΔABC, and a, b and c as the lengths of its sides, as usual. There is a formula

4R Area( ΔABC) = abc.

But, as we know, Area( ΔABC) is fixed, so the product abc of the three sides and the circumradius R attain minimum simultaneously. The minimum is achieved for an isosceles triangle when

The applet illustrates this point (Hint 2).

What if applet does not run? |

Let S be the topmost point of the circumcircle of ΔABC and T the position of A for which _{t} that of ΔBCT, then OA' ≥ O_{t}A', implying by the Pythagorean theorem, that, say, _{t},_{t},

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