Difference between revisions of "1976 AHSME Problems/Problem 3"
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| − | The lengths to the side are <math>1, \sqrt{2^2+1^2}, \sqrt{2^2+1^2}, 1</math>, respectively. Therefore, the sum is <math>2+2\sqrt{5}\Rightarrow \textbf{(E)}</math>. | + | The lengths to the side are <math>1, \sqrt{2^2+1^2}, \sqrt{2^2+1^2}, 1</math>, respectively. Therefore, the sum is <math>2+2\sqrt{5}\Rightarrow \textbf{(E)}</math>.~MathJams |
==1976 AHSME Problems== | ==1976 AHSME Problems== | ||
{{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}} | {{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}} | ||
Revision as of 20:03, 12 July 2020
Problem 3
The sum of the distances from one vertex of a square with sides of length 
to the midpoints of each of the sides of the square is
Solution
The lengths to the side are 
, respectively. Therefore, the sum is 
.~MathJams
1976 AHSME Problems
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