Difference between revisions of "1996 AHSME Problems/Problem 30"
Isabelchen (talk | contribs) (→Solution 4) |
Isabelchen (talk | contribs) (→Solution 5) |
||
Line 75: | Line 75: | ||
==Solution 5== | ==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>. | Note that minor arc <math>\overarc{AB}</math> is a third of the circumference, therefore, <math>\angle AOB = 120^{\circ}</math>. | ||
Line 82: | Line 84: | ||
<math>\sin \frac{120^{\circ}-\alpha}{2} = \frac{\frac52}{r}</math>, <math>\sin (60^{\circ} - \frac{\alpha}{2}) = \frac{5}{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>\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 \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> | |
− | </math>13a = \frac{3\sqrt{3}}{2} \cdot \sqrt{1-a^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 < | + | 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, < | + | 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 < | + | Therefore, the answer is <math>\boxed{\textbf{(E) } 409}</math>. |
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.