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>. | ||
| − | - | + | <math>\textbf{-lucasxia01}</math> |
== See also == | == See also == | ||
Revision as of 13:15, 19 June 2016
Problem
In quadrilateral 
with diagonals 
and 
, intersecting at 
, 
, 
, 
, 
, and 
. The length of 
is:
Solution
We note that 
. This is the Power of a Point Theorem which only happens to chords in circles. Therefore, we conclude that is cyclic. We can proceed with similar triangles. Because of inscribed angles, 
and 
. We find 
with the first similarity and 
with the second similarity. Now, we can apply Ptolemy's theorem which states that in a cyclic quadrilateral, 
. We can plug in out values to get 
. We solve for to get 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
