1975 AHSME Problems/Problem 23
Problem 23
In the adjoining figure and
are adjacent sides of square
;
is the midpoint of
;
is the midpoint of
; and
and
intersect at
. The ratio of the area of
to the area of
is
Solution
First, let's draw a few auxiliary lines. Drop altitudes from to
and from
to
. We can label the points as
and
, respectively. This forms square
. Connect
.
Without loss of generality, set the side length of the square equal to
. Let
, and since
is the midpoint of
,
would be
. With the same reasoning,
and
We can also see that is similar to
. That means
.
Plugging in the values, we get: . Solving, we find that
. Then,
. The area of
and
together would be
. Subtract this area from the total area of
to get the area of
.
So, . The question asks for the ratio of the area of
to the area of
, which is
.
The answer is . ~jiang147369
See Also
1975 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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