Difference between revisions of "2012 AMC 12B Problems/Problem 20"
Isabelchen (talk | contribs) (→Solution 2) |
Isabelchen (talk | contribs) m (→Solution 2) |
||
(2 intermediate revisions by the same user not shown) | |||
Line 27: | Line 27: | ||
Let the area of the trapezoid be <math>S</math>, the area of the triangle be <math>S_1</math>, the area of the parallelogram be <math>S_2</math>. | Let the area of the trapezoid be <math>S</math>, the area of the triangle be <math>S_1</math>, the area of the parallelogram be <math>S_2</math>. | ||
− | By | + | By [https://en.wikipedia.org/wiki/Heron%27s_formula Heron's Formula] <math>S_1 = \sqrt{\frac{b+c+d-a}{2} \cdot \frac{c+d-a-b}{2} \cdot \frac{a+b+d-c}{2} \cdot \frac{b+c-a-d}{2}}</math> |
− | <math>S_2 = \frac{S_1 \ cdot 2}{c-a} \cdot a = \frac{2aS_1}{c-a}</math> | + | <math>S_2 = \frac{S_1 \cdot 2}{c-a} \cdot a = \frac{2aS_1}{c-a}</math> |
<math>S = S_1 + S_2 = S_1(1+\frac{2a}{c-a}) = S_1 \cdot \frac{c+a}{c-a} = \frac14 \cdot \frac{c+a}{c-a} \cdot \sqrt{(b+c+d-a)(c+d-a-b)(a+b+d-c)(b+c-a-d)}</math> | <math>S = S_1 + S_2 = S_1(1+\frac{2a}{c-a}) = S_1 \cdot \frac{c+a}{c-a} = \frac14 \cdot \frac{c+a}{c-a} \cdot \sqrt{(b+c+d-a)(c+d-a-b)(a+b+d-c)(b+c-a-d)}</math> | ||
Line 62: | Line 62: | ||
<math>S = \frac14 \cdot \frac{18}{4} \cdot \sqrt{(3+11+5-7)(11+5-7-)(7+5+5-11)(3+11-7-5)} = 27</math> | <math>S = \frac14 \cdot \frac{18}{4} \cdot \sqrt{(3+11+5-7)(11+5-7-)(7+5+5-11)(3+11-7-5)} = 27</math> | ||
− | <math>\frac{35}{2} + \frac{32}{3} + 27 + | + | Thus the answer is <math>\frac{35}{2} + \frac{32}{3} + 27 + 3 + 5</math>, which rounds down to <math>\boxed{\textbf{(D) } 63}</math> |
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
Latest revision as of 07:14, 27 December 2022
Problem 20
A trapezoid has side lengths 3, 5, 7, and 11. The sum 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 1
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, by Law of Cosines, , so
. Therefore the area of this trapezoid is
.
In the second case, , so
. Therefore the area of this trapezoid is
.
In the third case, , therefore the area of this trapezoid is
.
So , which rounds down to
.
Solution 2
Let the area of the trapezoid be , the area of the triangle be
, the area of the parallelogram be
.
If ,
,
,
If ,
,
,
, which is impossible as
If ,
,
,
, which is impossible as
If ,
,
,
, which is impossible as
If ,
,
,
If ,
,
,
Thus the answer is , which rounds down to
Video Solution
https://youtu.be/8w1vrsD2urs ~Math Problem Solving Skills
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.