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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[[File:1996AHSMEP305.png|500px|center]] | [[File:1996AHSMEP305.png|500px|center]] | ||
− | Note that | + | 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> | + | 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^2= 3 \cdot 5 + 7^2 = 64</math>, <math>AD = 8</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>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> | + | 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>. | Therefore, the answer is <math>\boxed{\textbf{(E) } 409}</math>. |
Latest revision as of 12:09, 20 December 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 (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 |
Followed by Last Problem | |
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