Difference between revisions of "1996 AHSME Problems/Problem 30"
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<math>x = 2 \cdot \frac{7\sqrt{3}}{3} \cdot [3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3] = 2 \cdot \frac{7\sqrt{3}}{3} \cdot (\frac{9\sqrt{3}}{14} - \frac{82\sqrt{3}}{2 \cdot 7^3}) = 9 - \frac{81}{49} = \frac{360}{49}</math> | <math>x = 2 \cdot \frac{7\sqrt{3}}{3} \cdot [3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3] = 2 \cdot \frac{7\sqrt{3}}{3} \cdot (\frac{9\sqrt{3}}{14} - \frac{82\sqrt{3}}{2 \cdot 7^3}) = 9 - \frac{81}{49} = \frac{360}{49}</math> | ||
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+ | Therefore, the answer is <math>\boxed{\textbf{(E) } 409}</math>. | ||
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+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
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+ | ==Solution 5 (Ptolemy's theorem)== | ||
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+ | [[File:1996AHSMEP305.png|500px|center]] | ||
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+ | Note that minor arc <math>\overarc{AB}</math> is a third of the circumference, therefore, <math>\angle AOB = 120^{\circ}</math>. | ||
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+ | By the Law of Cosine, <math>AB = \sqrt{ 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^{\circ}} = \sqrt{ 9 + 25 + 15 } = 7</math> | ||
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+ | By the Ptolemy's theorem of quadrilateral <math>ABDE</math>, <math>AD \cdot BE = AB \cdot DE + BD \cdot AE</math>, <math>AD^2= 3 \cdot 5 + 7^2 = 64</math>, <math>AD = 8</math> | ||
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+ | By the Ptolemy's theorem of quadrilateral <math>ABCD</math>, <math>AC \cdot BD = BC \cdot AD + AB \cdot CD</math>, <math>7AC = 3 \cdot 8 + 3 \cdot 5 = 39</math>, <math>AC = \frac{39}{7}</math> | ||
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+ | By the Ptolemy's theorem of quadrilateral <math>ABCF</math>, <math>AC \cdot BF = AB \cdot CF + BC \cdot AF</math>, <math>AC^2 = 3 \cdot CF + 3 \cdot 3</math>, <math>3CF = (\frac{39}{7})^2 - 9 = \frac{1521-441}{49} = \frac{1080}{49}</math>, <math>CF = \frac{360}{49}</math> | ||
Therefore, the answer is <math>\boxed{\textbf{(E) } 409}</math>. | Therefore, the answer is <math>\boxed{\textbf{(E) } 409}</math>. |
Revision as of 02:33, 1 October 2023
Contents
[hide]Problem
A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to , where and are relatively prime positive integers. Find .
Solution 1
In hexagon , let and let . Since arc is one third of the circumference of the circle, it follows that . Similarly, . Let be the intersection of and , that of and , and that of and . Triangles and are equilateral, and by symmetry, triangle is isosceles and thus also equilateral.
Furthermore, and subtend the same arc, as do and . Hence triangles and are similar. Therefore, It follows that Solving the two equations simultaneously yields so
Solution 2
All angle measures are in degrees. Let the first trapezoid be , where . Then the second trapezoid is , where . We look for .
Since is an isosceles trapezoid, we know that and, since , if we drew , we would see . Anyway, ( means arc AB). Using similar reasoning, .
Let and . Since (add up the angles), and thus . Therefore, . as well.
Now I focus on triangle . By the Law of Cosines, , so . Seeing and , we can now use the Law of Sines to get:
Now I focus on triangle . and , and we are given that , so We know , but we need to find . Using various identities, we see Returning to finding , we remember Plugging in and solving, we see . Thus, the answer is , which is answer choice .
Solution 3
Let be the desired length. One can use Parameshvara's circumradius formula, which states that for a cyclic quadrilateral with sides the circumradius satisfies where is the semiperimeter. Applying this to the trapezoid with sides , we see that many terms cancel and we are left with Similar canceling occurs for the trapezoid with sides , and since the two quadrilaterals share the same circumradius, we can equate: Solving for gives , so the answer is .
Solution 4
Note that minor arc is a third of the circumference, therefore, . Major arc ,
By the Law of Cosine,
, therefore,
Let be the length of the chord,
By the triple angle formula,
Therefore, the answer is .
Solution 5
Note that minor arc is a third of the circumference, therefore, .
,
,
,
Let , ,
, ,
Let be the length of the chord,
By the triple angle formula,
Therefore, the answer is .
Solution 5 (Ptolemy's theorem)
Note that minor arc is a third of the circumference, therefore, .
By the Law of Cosine,
By the Ptolemy's theorem of quadrilateral , , ,
By the Ptolemy's theorem of quadrilateral , , ,
By the Ptolemy's theorem of quadrilateral , , , ,
Therefore, the answer is .
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Last Problem | |
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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