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

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== Problem ==
 
== Problem ==
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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:
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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:
  
 
<math>\textbf{(A)}\ 9\qquad
 
<math>\textbf{(A)}\ 9\qquad

Revision as of 01:34, 4 January 2020

Problem

In rectangle $ABCD$ with diagonals $AC$ and $BD$, intersecting at $O$, $BO=4$, $OD = 6$, $AO=8$, $OC=3$, and $AB=6$. The length of $AD$ is:

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

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 (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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