Difference between revisions of "1996 AHSME Problems/Problem 30"
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<math>\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409 </math> | <math>\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409 </math> | ||
− | ==Solution 1== | + | ==Solution 1 (Alcumus)== |
In hexagon <math>ABCDEF</math>, let <math>AB=BC=CD=3</math> and let <math>DE=EF=FA=5</math>. Since arc <math>BAF</math> is one third of the circumference of the circle, it follows that <math>\angle BCF = \angle BEF=60^{\circ}</math>. Similarly, <math>\angle CBE =\angle CFE=60^{\circ}</math>. Let <math>P</math> be the intersection of <math>\overline{BE}</math> and <math>\overline{CF}</math>, <math>Q</math> that of <math>\overline{BE}</math> and <math>\overline{AD}</math>, and <math>R</math> that of <math>\overline{CF}</math> and <math>\overline{AD}</math>. Triangles <math>EFP</math> and <math>BCP</math> are equilateral, and by symmetry, triangle <math>PQR</math> is isosceles and thus also equilateral. | In hexagon <math>ABCDEF</math>, let <math>AB=BC=CD=3</math> and let <math>DE=EF=FA=5</math>. Since arc <math>BAF</math> is one third of the circumference of the circle, it follows that <math>\angle BCF = \angle BEF=60^{\circ}</math>. Similarly, <math>\angle CBE =\angle CFE=60^{\circ}</math>. Let <math>P</math> be the intersection of <math>\overline{BE}</math> and <math>\overline{CF}</math>, <math>Q</math> that of <math>\overline{BE}</math> and <math>\overline{AD}</math>, and <math>R</math> that of <math>\overline{CF}</math> and <math>\overline{AD}</math>. Triangles <math>EFP</math> and <math>BCP</math> are equilateral, and by symmetry, triangle <math>PQR</math> is isosceles and thus also equilateral. | ||
<asy> | <asy> | ||
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[[File:1996AHSMEP30.png|400px|center]] | [[File:1996AHSMEP30.png|400px|center]] | ||
− | 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>. Major arc <math>\overarc{AB}</math><math> =240^{\circ}</math>, <math>\angle ACB = 120^{\circ}</math> |
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> | 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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<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> | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(E) } 409}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | [[File:1996AHSMEP302.png|500px|center]] | ||
+ | |||
+ | Note that minor arc <math>\overarc{AB}</math> is a third of the circumference, therefore, <math>\angle AOB = 120^{\circ}</math>. | ||
+ | |||
+ | <math>\sin \frac{\alpha}{2} = \frac{\frac32}{r}</math>, <math>\sin \frac{\alpha}{2} = \frac{3}{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>5 \cdot \sin \frac{\alpha}{2} = 3 \cdot \sin (60^{\circ} - \frac{\alpha}{2})</math> | ||
+ | |||
+ | <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> | ||
+ | |||
+ | 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 <math>x</math> be the length of the chord, <math>\sin \frac{3 \alpha}{2} = \frac{\frac{x}{2}}{r}</math> | ||
+ | |||
+ | By the triple angle formula, <math>\sin \frac{3 \alpha}{2} = 3 \cdot \sin \frac{\alpha}{2} - 4 \cdot \sin(\frac{ \alpha}{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 <math>\boxed{\textbf{(E) } 409}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 6 (Ptolemy's theorem)== | ||
+ | |||
+ | [[File:1996AHSMEP305.png|500px|center]] | ||
+ | |||
+ | Note that major arc <math>\overarc{AE}</math> is two thirds of the circumference, therefore, <math>\angle AFE = 120^{\circ}</math>. | ||
+ | |||
+ | By the Law of Cosine, <math>AE= \sqrt{ 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^{\circ}} = \sqrt{ 9 + 25 + 15 } = 7</math> | ||
+ | |||
+ | By the Ptolemy's theorem of quadrilateral <math>ABDE</math>, <math>AD \cdot BE = AB \cdot DE + BD \cdot AE</math>, <math>AD = BE</math>, <math>AD^2= 3 \cdot 5 + 7^2 = 64</math>, <math>AD = 8</math> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | By the Ptolemy's theorem of quadrilateral <math>ABCF</math>, <math>AC \cdot BF = AB \cdot CF + BC \cdot AF</math>, <math>AC = BF</math>, <math>(\frac{39}{7})^2 = 3 \cdot CF + 3 \cdot 3</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>. |
Latest revision as of 12:09, 20 December 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 (Alcumus)
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 6 (Ptolemy's theorem)
Note that major arc is two thirds 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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