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

(Problem)
(Solution)
Line 2: Line 2:
 
<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 ==
+
<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 11:16, 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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png