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

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== 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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If in a parallel universe, apples are orange and oranges are red what is 2+3?
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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>
  
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<math>\textbf{(A)}\ 9\qquad
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<math>\sqrt{166}</math>
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\textbf{(B)}\ 10\qquad
 
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\textbf{(C)}\ 6\sqrt{3}\qquad
 
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\textbf{(D)}\ 8\sqrt{2}\qquad
 
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\textbf{(E)}\ \sqrt{166}</math>
 
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== 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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<math>\textbf{-lucasxia01}</math>
 
  
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After drawing the diagram, we see that we actually have a lot of lengths to work with. Considering triangle ABD, we know values of <math>AB, BD(BD = BO + OD)</math>, but we want to find the value of AD. We can apply stewart's theorem now, letting <math>m = 4, n = 6, AD = X, AB = 6</math>, and we have <math>10 \cdot 6 \cdot 4 + 8 \cdot 8 \cdot 10 = x^2 + 36 \cdot 6</math>, and we see that <math>x = \sqrt{166}</math>, <math>\boxed{E \sqrt{166}}</math>
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1967|num-b=31|num-a=33}}   
 
{{AHSME box|year=1967|num-b=31|num-a=33}}   

Revision as of 18:18, 10 September 2020

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

$\sqrt{166}$

After drawing the diagram, we see that we actually have a lot of lengths to work with. Considering triangle ABD, we know values of $AB, BD(BD = BO + OD)$, but we want to find the value of AD. We can apply stewart's theorem now, letting $m = 4, n = 6, AD = X, AB = 6$, and we have $10 \cdot 6 \cdot 4 + 8 \cdot 8 \cdot 10 = x^2 + 36 \cdot 6$, and we see that $x = \sqrt{166}$, $\boxed{E \sqrt{166}}$

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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