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:1996AHSMEP305.png|500px|center]] | [[File:1996AHSMEP305.png|500px|center]] | ||
− | Note that major arc <math>\overarc{AE}</math> is two | + | 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 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> |
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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