Difference between revisions of "1996 AHSME Problems/Problem 30"
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==Solution 5== | ==Solution 5== | ||
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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>. | 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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<math>\sin \frac{120^{\circ}-\alpha}{2} = \frac{\frac52}{r}</math>, <math>\sin (60^{\circ} - \frac{\alpha}{2}) = \frac{5}{2r}</math> | <math>\sin \frac{120^{\circ}-\alpha}{2} = \frac{\frac52}{r}</math>, <math>\sin (60^{\circ} - \frac{\alpha}{2}) = \frac{5}{2r}</math> | ||
− | <math>\frac{\sin \frac{\alpha}{2}}{\sin (60^{\circ} - \frac{\alpha}{2}) } = \frac{\frac{3}{2r}}{\frac{5}{2r}} = \frac35</math> | + | <math>\frac{\sin \frac{\alpha}{2}}{\sin (60^{\circ} - \frac{\alpha}{2}) } = \frac{\frac{3}{2r}}{\frac{5}{2r}} = \frac35</math>, <math>5 \cdot \sin \frac{\alpha}{2} = 3 \cdot \sin (60^{\circ} - \frac{\alpha}{2})</math> |
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− | <math>5 \cdot \sin \frac{\alpha}{2} = 3 \cdot \sin (60^{\circ} - \frac{\alpha}{2})</math> | ||
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− | < | + | <math>5 \cdot \sin \frac{\alpha}{2} = 3 ( \sin 60^{\circ} \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2} \cos 60^{\circ} = 3 ( \frac{\sqrt{3}}{2} \cdot \cos \frac{\alpha}{2} - \frac12 \cdot \sin \frac{\alpha}{2})</math> |
− | + | <math>13 \cdot \sin \frac{\alpha}{2} = \frac{3\sqrt{3}}{2} \cdot \cos \frac{\alpha}{2}</math> | |
− | </math>13a = \frac{3\sqrt{3}}{2} \cdot \sqrt{1-a^2}<math> | + | Let <math>\sin \frac{\alpha}{2} = a</math>, <math>\cos \frac{\alpha}{2} = \sqrt{1-a^2}</math>, <math>13a = \frac{3\sqrt{3}}{2} \cdot \sqrt{1-a^2}</math> |
− | < | + | <math>169a^2 = 27-27a^2</math>, <math>196a^2=27</math>, <math>\sin \frac{\alpha}{2} = a = \sqrt{\frac{27}{196}} = \frac{3 \sqrt{3}}{14}</math> |
− | Let < | + | Let <math>x</math> be the length of the chord, <math>\sin \frac{3 \theta}{2} = \frac{\frac{x}{2}}{r}</math> |
− | By the triple angle formula, < | + | By the triple angle formula, <math>\sin \frac{3 \theta}{2} = 3 \cdot \sin \frac{\theta}{2} - 4 \cdot \sin(\frac{ \theta}{2})^3 = 3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3</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> |
− | Therefore, the answer is < | + | Therefore, the answer is <math>\boxed{\textbf{(E) } 409}</math>. |
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
Revision as of 11:50, 30 September 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 .
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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