Difference between revisions of "1996 AHSME Problems/Problem 30"
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==Problem== | ==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 <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | 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 <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | ||
− | < | + | |
− | ==Solution== | + | <math>\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409 </math> |
+ | |||
+ | ==Solution 1== | ||
+ | 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> | ||
+ | import olympiad; import geometry; size(150); defaultpen(linewidth(0.8)); | ||
+ | real angleUnit = 15; | ||
+ | draw(Circle(origin,1)); | ||
+ | pair D = dir(22.5); | ||
+ | pair C = dir(3*angleUnit + degrees(D)); | ||
+ | pair B = dir(3*angleUnit + degrees(C)); | ||
+ | pair A = dir(3*angleUnit + degrees(B)); | ||
+ | pair F = dir(5*angleUnit + degrees(A)); | ||
+ | pair E = dir(5*angleUnit + degrees(F)); | ||
+ | draw(A--B--C--D--E--F--cycle); | ||
+ | dot("$A$",A,A); dot("$B$",B,B); dot("$C$",C,C); dot("$D$",D,D); dot("$E$",E,E); dot("$F$",F,F); | ||
+ | draw(A--D^^B--E^^C--F); | ||
+ | label("$3$",D--C,SW); label("$3$",B--C,S); label("$3$",A--B,SE); label("$5$",A--F,NE); label("$5$",F--E,N); label("$5$",D--E,NW); | ||
+ | </asy> | ||
+ | |||
+ | Furthermore, <math>\angle BAD</math> and <math>\angle BED</math> subtend the same arc, as do <math>\angle ABE</math> and <math>\angle ADE</math>. Hence triangles <math>ABQ</math> and <math>EDQ</math> are similar. Therefore, <cmath>\frac{AQ}{EQ}=\frac{BQ}{DQ}=\frac{AB}{ED}=\frac{3}{5}.</cmath> It follows that <cmath>\frac{\frac{AD-PQ}{2}}{PQ+5} =\frac{3}{5}\quad | ||
+ | \mbox {and}\quad \frac{3-PQ}{\frac{AD+PQ}{2}}=\frac{3}{5}.</cmath> Solving the two equations simultaneously yields <math>AD=360/49,</math> so <math>m+n=\boxed{409}. \blacksquare</math> | ||
+ | |||
+ | ==Solution 2== | ||
All angle measures are in degrees. | All angle measures are in degrees. | ||
Let the first trapezoid be <math>ABCD</math>, where <math>AB=BC=CD=3</math>. Then the second trapezoid is <math>AFED</math>, where <math>AF=FE=ED=5</math>. We look for <math>AD</math>. | Let the first trapezoid be <math>ABCD</math>, where <math>AB=BC=CD=3</math>. Then the second trapezoid is <math>AFED</math>, where <math>AF=FE=ED=5</math>. We look for <math>AD</math>. | ||
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Returning to finding <math>AD</math>, we remember <cmath>\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}\;\text{so}\;AD=\frac{5\sin{3\phi}}{\sin{\theta}}.</cmath> | Returning to finding <math>AD</math>, we remember <cmath>\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}\;\text{so}\;AD=\frac{5\sin{3\phi}}{\sin{\theta}}.</cmath> | ||
Plugging in and solving, we see <math>AD=\frac{360}{49}</math>. Thus, the answer is <math>360 + 49 = 409</math>, which is answer choice <math>\boxed{\textbf{(E)}}</math>. | Plugging in and solving, we see <math>AD=\frac{360}{49}</math>. Thus, the answer is <math>360 + 49 = 409</math>, which is answer choice <math>\boxed{\textbf{(E)}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Let <math>x</math> be the desired length. One can use Parameshvara's circumradius formula, which states that for a cyclic quadrilateral with sides <math>a, b, c, d</math> the circumradius <math>R</math> satisfies <cmath>R^2=\frac{1}{16}\cdot\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)},</cmath> where <math>s</math> is the semiperimeter. Applying this to the trapezoid with sides <math>3, 3, 3, x</math>, we see that many terms cancel and we are left with <cmath>R^2=\frac{27}{9-x}</cmath> Similar canceling occurs for the trapezoid with sides <math>5, 5, 5, x</math>, and since the two quadrilaterals share the same circumradius, we can equate: <cmath>\frac{27}{9-x}=\frac{125}{15-x}</cmath> Solving for <math>x</math> gives <math>x=\frac{360}{49}</math>, so the answer is <math>\fbox{(E) 409}</math>. | ||
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 14:31, 18 June 2017
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 .
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
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