Difference between revisions of "1975 AHSME Problems/Problem 23"
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The answer is <math>\boxed{\textbf{(C)}\ \frac{2}{3}}</math>. ~[https://artofproblemsolving.com/wiki/index.php/User:Jiang147369 jiang147369] | The answer is <math>\boxed{\textbf{(C)}\ \frac{2}{3}}</math>. ~[https://artofproblemsolving.com/wiki/index.php/User:Jiang147369 jiang147369] | ||
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+ | ==See Also== | ||
+ | {{AHSME box|year=1975|num-b=22|num-a=24}} | ||
+ | {{MAA Notice}} |
Latest revision as of 16:48, 19 January 2021
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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