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 4 of 5: The right-triangle case

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h\le\frac{AB}{2}\Rightarrow C=90^\circ\text{ is attainable and }\sin C=1

Detailed analysis

If $h\le AB/2$ , a point $C$ at height $h$ can lie on the circle with diameter $AB$ ; equivalently, there is a triangle with $\angle C=90^\circ$ . Since $\sin C\le1$ , this realizes the largest possible sine and hence the maximum altitude product. At equality it is the isosceles right triangle.