Difference between revisions of "1967 AHSME Problems/Problem 32"
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| − | == | + | In quadrilateral <math>ABCD</math> with diagonals <math>AC</math> and <math>BD</math>, intersecting at <math>O</math>, <math>BO=4</math>, <math>OD = 6</math>, <math>AO=8</math>, <math>OC=3</math>, and <math>AB=6</math>. The length of <math>AD</math> is: |
| − | + | <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> | |
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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> | ||
== Solution == | == Solution == | ||
Revision as of 11:16, 13 April 2020
In quadrilateral 
with diagonals 
and 
, intersecting at 
, 
, 
, 
, 
, and 
. The length of 
is:
Solution
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
| 1967 AHSME (Problems • Answer Key • Resources) | ||
| 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. 
