Difference between revisions of "1967 AHSME Problems/Problem 32"

(Problem)
(Solution)
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<math>\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}</math>
 
<math>\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}</math>
  
== Solution ==
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<math>\sqrt{166}</math>
Since the parallel universe seems to have reversed everything, we can see that 2 becomes 3 and 3 becomes 2. This leads us to 100 and 66 because 2*50 and 3*22. Add them together and put one of those check looking hats on the new baby. This gives us our answer.
 
  
 
== See also ==
 
== See also ==

Revision as of 10:16, 13 April 2020

In quadrilateral $ABCD$ with diagonals $AC$ and $BD$, intersecting at $O$, $BO=4$, $OD = 6$, $AO=8$, $OC=3$, and $AB=6$. The length of $AD$ is: $\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$

$\sqrt{166}$

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 31
Followed by
Problem 33
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