Problem 3

Consider all triangles

ABC

with a fixed base

AB

and whose altitude from

C

is a constant

h

. For which of these triangles is the product of its three altitudes a maximum?

Step 5 of 5: The high-altitude case is isosceles

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h>\frac{AB}{2}\Rightarrow C<90^\circ,\quad\sin C\text{ is maximal}\Longleftrightarrow AC=BC

Detailed analysis

If $h>AB/2$ , the angle at $C$ is necessarily acute. With $AB$ and the altitude fixed, reflecting or varying the horizontal position of $C$ shows that $\angle C$ is largest when $C$ is above the midpoint of $AB$ , i.e. when $AC=BC$ . For acute angles this maximizes $\sin C$ , so the isosceles triangle is the required one.