Difference between revisions of "2012 AMC 12B Problems/Problem 20"
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In the first case, <math>\angle BCD = 90^{\circ}</math>, therefore the area of this trapezoid is <math>\frac{1}{2} (7+11) \cdot 3 = 27</math>. | In the first case, <math>\angle BCD = 90^{\circ}</math>, therefore the area of this trapezoid is <math>\frac{1}{2} (7+11) \cdot 3 = 27</math>. | ||
− | So <math>r_1 + r_2 + r_3 + n_1 + n_2 = 17.5 + 10.666... + 27 + 3 + 5</math>, which is rounded down to 63. | + | So <math>r_1 + r_2 + r_3 + n_1 + n_2 = 17.5 + 10.666... + 27 + 3 + 5</math>, which is rounded down to <math>63... \framebox{D}</math>. |
Revision as of 12:32, 4 December 2012
Problem 20
A trapezoid has side lengths 3, 5, 7, and 11. The sums of all the possible areas of the trapezoid can be written in the form of , where
,
, and
are rational numbers and
and
are positive integers not divisible by the square of any prime. What is the greatest integer less than or equal to
?
Solution
Name the trapezoid , where
is parallel to
,
, and
. Draw a line through
parallel to
, crossing the side
at
. Then
,
. One needs to guarantee that
, so there are only three possible trapezoids:
In the first case, , so
. Therefore the area of this trapezoid is
.
In the first case, , so
. Therefore the area of this trapezoid is
.
In the first case, , therefore the area of this trapezoid is
.
So , which is rounded down to
.