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, (BB'×CC') × (AC×AB) is a constant; thus for BB'×CC' to be maximum, AB×AC has to me minimum. The problem reduces to finding a location of A for which AB×AC is minimum.
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 AB = AC.
The applet illustrates this point (Hint 2).
Let S be the topmost point of the circumcircle of ΔABC and T the position of A for which TB = TC. S is always above T, or, more accurately, SA' ≥ TA'. The circumcenters of the two triangles BCS and BCT lie on the perpendicular bisector of BC and those of their sides. If O is the circumcenter of ΔABC (and hence of ΔBCS) and Ot that of ΔBCT, then OA' ≥ OtA', implying by the Pythagorean theorem, that, say, BO ≥ BOt, or R ≥ BOt, which tells us that T is the position of A for which the circumradius attains the minimum.
