Difference between revisions of "1975 AHSME Problems/Problem 23"
Jiang147369 (talk | contribs) (Created page with "== Problem 23 == In the adjoining figure <math>AB</math> and <math>BC</math> are adjacent sides of square <math>ABCD</math>; <math>M</math> is the midpoint of <math>AB</math>...") |
Jiang147369 (talk | contribs) (→Solution) |
||
Line 23: | Line 23: | ||
== Solution == | == Solution == | ||
− | First, let's draw a few auxiliary lines. Drop altitudes from <math>O</math> to <math>AB</math> and from <math>O</math> to <math>BC</math>. We can label | + | First, let's draw a few auxiliary lines. Drop altitudes from <math>O</math> to <math>AB</math> and from <math>O</math> to <math>BC</math>. We can label the points as <math>J</math> and <math>K</math>, respectively. This forms square <math>OKBJ</math>. Connect <math>OB</math>. |
Without loss of generality, set the side length of the square equal to <math>1</math>. Let <math>NK=x</math>, and since <math>N</math> is the midpoint of <math>CB</math>, <math>KB</math> would be <math>\frac{1}{2} - x</math>. With the same reasoning, <math>MJ=x</math> and <math>JB = \frac{1}{2} - x</math> | Without loss of generality, set the side length of the square equal to <math>1</math>. Let <math>NK=x</math>, and since <math>N</math> is the midpoint of <math>CB</math>, <math>KB</math> would be <math>\frac{1}{2} - x</math>. With the same reasoning, <math>MJ=x</math> and <math>JB = \frac{1}{2} - x</math> | ||
<asy> | <asy> | ||
− | draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)--(2,1)--(2,2)--(1,0) | + | draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)--(2,1)--(2,2)--(1,0)); |
+ | draw((4/3, 0)--(4/3, 2/3)--(2, 2/3)--(2,0)--(4/3, 2/3), dashed); | ||
label("A", (0,0), S); | label("A", (0,0), S); | ||
label("B", (2,0), S); | label("B", (2,0), S); |
Revision as of 17:24, 27 December 2020
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