1965 AMC 12 Problem 12

Question

A rhombus is inscribed in triangle

ABC

in such a way that one of its vertices is

A

and two of its sides lie along

AB

and

AC.

If

AC=6

inches,

AB=12

inches, and

BC=8

inches, the side of the rhombus, in inches, is:

2

3

3\dfrac12

4

5

Let the rhombus side be

s

, with adjacent vertices

\frac{s}{12}

of the way along

AB

and

\frac{s}{6}

along

AC

The opposite rhombus vertex lies on

BC

, so its two affine coefficients sum to

1

Use

A

as the origin and let the vectors to

B,C

be

u,v.

If the rhombus side is

s,

its opposite vertex is

(\frac{s}{12})u+(\frac{s}{6})v.

A point lies on

BC

when these coefficients sum to

1.

Thus

\frac{s}{12}+\frac{s}{6}=1,

giving

s=4.

Therefore, the correct answer is D.

Problem 12 in other years