1982 AMC 12 Problem 9

Attempt Problem 9 of the 1982 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 1982 AMC 12 solutions, or check the answer key.

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

A vertical line divides the triangle with vertices

(0,0),

(1,1)

and

(9,1)

in the

xy

-plane into two regions of equal area. The equation of the line is

x=

2.5

3.0

3.5

4.0

4.5

For

1\le x\le9,

the triangle lies between

y=\frac{x}{9}

and

y=1

Set the accumulated area to one half of the triangle’s total area

The triangle has area

4.

The portion with

0\le x\le1

has area

\int_0^1\left(x-\frac x9\right)\,dx=\frac49.

For a cut

x=t\ge1,

the area to its left is therefore

\frac49+\int_1^t(1-\frac{x}{9})\,dx.

Setting this equal to

2

gives

t-\frac{t^2}{18}=\frac{5}{2},

whose relevant solution is

t=3.

Therefore, the correct answer is B.

Problem 9 in other years