1989 AMC 12 Problem 6

Attempt Problem 6 of the 1989 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1989 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

a,

b\gt0

and the triangle in the first quadrant bounded by the coordinate axes and the graph of

ax+by=6

has area

6,

then

ab=

3

6

12

108

432

Find the intercepts of the line on the two axes

Use those intercepts as the base and height of the triangle

The intercepts are

\frac{6}{a}

and

\frac{6}{b}.

Therefore the triangle’s area is

\frac12\cdot\frac6a\cdot\frac6b=\frac{18}{ab}=6,

which gives

ab=3.

Thus the correct answer is A.

Problem 6 in other years