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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==Solution 2== | ==Solution 2== | ||
+ | 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>. | ||
+ | |||
+ | Since <math>ABCD</math> is an isosceles trapezoid, we know that <math>\angle BAD=\angle CDA</math> and, since <math>AB=BC</math>, if we drew <math>AC</math>, we would see <math>\angle BCA=\angle BAC</math>. Anyway, <math>\widehat{AB}=\widehat{BC}=\widehat{CD}</math> (<math>\widehat{AB}</math> means arc AB). Using similar reasoning, <math>\widehat{AF}=\widehat{FE}=\widehat{ED}</math>. | ||
Let <math>\widehat{AB}=2\phi</math> and <math>\widehat{AF}=2\theta</math>. Since <math>6\theta+6\phi=360</math> (add up the angles), <math>2\theta+2\phi=120</math> and thus <math>\widehat{AB}+\widehat{AF}=\widehat{BF}=120</math>. Therefore, <math>\angle FAB=\frac{1}{2}\widehat{BDF}=\frac{1}{2}(240)=120</math>. <math>\angle CDE=120</math> as well. | Let <math>\widehat{AB}=2\phi</math> and <math>\widehat{AF}=2\theta</math>. Since <math>6\theta+6\phi=360</math> (add up the angles), <math>2\theta+2\phi=120</math> and thus <math>\widehat{AB}+\widehat{AF}=\widehat{BF}=120</math>. Therefore, <math>\angle FAB=\frac{1}{2}\widehat{BDF}=\frac{1}{2}(240)=120</math>. <math>\angle CDE=120</math> as well. | ||
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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 | + | ==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>. | 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>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | [[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>. 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> | ||
+ | |||
+ | <math>\frac{\angle AOB}{2} = 60^{\circ}</math>, therefore, <math>r = \frac{\frac{AB}{2}}{\sin 60^{\circ}} = \frac{\frac{7}{2}}{ \frac{\sqrt{3}}{2} } = \frac{7\sqrt{3}}{3}</math> | ||
+ | |||
+ | <math>\sin \frac{\theta}{2} = \frac{\frac32}{r} = \frac{3}{2r} = \frac{3}{2 \cdot \frac{7\sqrt{3}}{3}} = \frac{3 \sqrt{3}}{14}</math> | ||
+ | |||
+ | 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, <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 <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>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
== See also == | == See also == |
Latest revision as of 13:09, 20 December 2023
Contents
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 |
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