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

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== Solution ==
 
== Solution ==
 
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>.
 
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>.
- [b] lucasxia01[/b]
+
<math>\textbf{-lucasxia01}</math>
  
 
== See also ==
 
== See also ==

Revision as of 14:15, 19 June 2016

Problem

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:

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

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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