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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>.~MathJams
 
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
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==1976 AHSME Problems==
 
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{{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}}
 

Revision as of 19:03, 12 July 2020

Problem 3

The sum of the distances from one vertex of a square with sides of length $2$ to the midpoints of each of the sides of the square is

$\textbf{(A) }2\sqrt{5}\qquad \textbf{(B) }2+\sqrt{3}\qquad \textbf{(C) }2+2\sqrt{3}\qquad \textbf{(D) }2+\sqrt{5}\qquad \textbf{(E) }2+2\sqrt{5}$

Solution

The lengths to the side are $1, \sqrt{2^2+1^2}, \sqrt{2^2+1^2}, 1$, respectively. Therefore, the sum is $2+2\sqrt{5}\Rightarrow \textbf{(E)}$.~MathJams

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