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

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== Problem ==
 
== Problem ==
In rectangle <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:
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If in a parallel universe, apples are orange and oranges are red what is 2+3?
  
 
<math>\textbf{(A)}\ 9\qquad
 
<math>\textbf{(A)}\ 9\qquad

Revision as of 02:38, 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

We note that $BO \cdot DO = AO \cdot CO = 24$. This is the Power of a Point Theorem which only happens to chords in circles. Therefore, we conclude that $ABCD$ is cyclic. We can proceed with similar triangles. Because of inscribed angles, $\triangle ABO \simeq \triangle DCO$ and $\triangle ADO \simeq \triangle BCO$. We find $\frac{CD}{AB} = \frac{3}{4} \implies CD = \frac{9}{2}$ with the first similarity and $\frac{BC}{AD} = \frac{3}{6} \implies BC = \frac{AD}{2}$ with the second similarity. Now, we can apply Ptolemy's theorem which states that in a cyclic quadrilateral, $AB \cdot CD + AD \cdot BC = AC \cdot BD$. We can plug in out values to get $6 \cdot \frac{9}{2} + AD \cdot \frac{AD}{2} = 11 \cdot 10 = 110$. We solve for $AD$ to get $AD = \boxed{\textbf{(E) } \sqrt{166}}$. $\textbf{-lucasxia01}$

See also

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