1987 AHSME Problems/Problem 30
Problem
In the figure, has
and
. A line
, with
on
and
, divides
into two pieces of equal area.
(Note: the figure may not be accurate; perhaps
is on
instead of
The ratio
is
Solution
First we show that is on
, as in the given figure, by demonstrating that if
, then
has more than half the area, so
is too far to the right. Specifically, assume
and drop an altitude from
to
that meets
at
. Without loss of generality, assume that
, so that
(as they have the same height, and area is
times the product of base and height), which is
, as required. Thus we must move
to the left, scaling
by a factor of
such that
. Thus
, which is answer
.
See also
1987 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Last Question | |
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