Difference between revisions of "1996 AHSME Problems/Problem 30"
Isabelchen (talk | contribs) (→Solution 5) |
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<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> | <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 \ | + | 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 \ | + | 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> | <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> |
Revision as of 11:50, 30 September 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
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 .
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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