1950 AMC 12 Problem 47

Question

A rectangle inscribed in a triangle has its base coinciding with the base

b

of the triangle. If the altitude of the triangle is

h,

and the altitude

x

of the rectangle is half the base of the rectangle, then:

x=\dfrac12h

x=\dfrac{bh}{b+h}

x=\dfrac{bh}{2h+b}

x=\sqrt{\dfrac{hb}{2}}

x=\dfrac12b

The segment across the triangle at height

x

has length

b(1-\frac{x}{h})

The rectangle’s base is

2x

By similarity, the width of the triangle at height

x

above its base is

b(1-\tfrac{x}{h}).

This is the base of the inscribed rectangle. Since the rectangle’s altitude is half its base, its base is

2x.

Hence

2x=b\left(1-\frac{x}{h}\right).

Multiplying by

h

and solving gives

x(2h+b)=bh,

so

x=\dfrac{bh}{2h+b}.

Thus, the correct answer is C.

Problem 47 in other years