Difference between revisions of "1996 AHSME Problems/Problem 30"
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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== | ||
+ | |||
+ | Note that <math>\overarc{AB}</math> is a third of the circumference, therefore, <math>\angle AOB = 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>\theta = \frac{\angle AOB}{2} = 60^{\circ}</math>, therefore, <math>r = \frac72 \cdot \frac{2}{\sqrt{3}} = \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 \frac{\theta}{2} - 4 \cdot (\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]</math> | ||
+ | |||
+ | |||
== See also == | == See also == |
Revision as of 11:14, 30 September 2023
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 is a third of the circumference, therefore,
By the Law of Cosine,
, therefore,
Let be the length of the chord,
By the triple angle formula,
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Last Problem | |
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