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

(The original problem was not hard enough.)
(Better solution.)
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== Solution ==
 
== Solution ==
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We note that <math>BO \cdot DO = AO \cdot CO = 24</math>. This is the Power of a Point Theorem which only happens to chords in circles. Therefore, we conclude that <math>ABCD</math> is cyclic. We can proceed with similar triangles. Because of inscribed angles, <math>\triangle ABO \simeq \triangle DCO</math> and <math>\triangle ADO \simeq \triangle BCO</math>. We find <math>\frac{CD}{AB} = \frac{3}{4} \implies CD = \frac{9}{2}</math> with the first similarity and <math>\frac{BC}{AD} = \frac{3}{6} \implies BC = \frac{AD}{2}</math> with the second similarity. Now, we can apply Ptolemy's theorem which states that in a cyclic quadrilateral, <math>AB \cdot CD + AD \cdot BC = AC \cdot BD</math>. We can plug in out values to get <math>6 \cdot \frac{9}{2} + AD \cdot \frac{AD}{2} = 11 \cdot 10 = 110</math>. We solve for <math>AD</math> to get <math>AD = \boxed{\textbf{(E) } \sqrt{166}}</math>.
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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 99 because 2*50 and 3*33. Add them together and put one of those check looking hats on the new baby. This gives us our answer.
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<math>\textbf{-lucasxia01}</math>
 
  
 
== See also ==
 
== See also ==

Revision as of 01:49, 4 January 2020

Problem

If in a parallel universe, apples are orange and oranges are red what is 2+3?

$\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$

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 99 because 2*50 and 3*33. 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

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