AMC12 2012 B

AMC12 2012 B · Q20

AMC12 2012 B · Q20. It mainly tests Area & perimeter, Geometry misc.

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 $r_1\sqrt{n_1}+r_2\sqrt{n_2}+r_3$, where $r_1$, $r_2$, and $r_3$ are rational numbers and $n_1$ and $n_2$ are positive integers not divisible by the square of any prime. What is the greatest integer less than or equal to $r_1+r_2+r_3+n_1+n_2$?

一个梯形边长为 3、5、7 和 11。所有可能梯形面积之和可写成 $r_1\sqrt{n_1}+r_2\sqrt{n_2}+r_3$ 的形式，其中 $r_1$, $r_2,$ 和 $r_3$ 是有理数，$n_1$ 和 $n_2$ 是不能被质数的平方整除的正整数。$r_1+r_2+r_3+n_1+n_2$ 的最大整数小于或等于该值的是多少？

(A) 57 57

(B) 59 59

(D) 63 63

(E) 65 65

Answer

Correct choice: (D)

正确答案：（D）

Solution

Name the trapezoid $ABCD$, where $AB$ is parallel to $CD$, $AB<CD$, and $AD<BC$. Draw a line through $B$ parallel to $AD$, crossing the side $CD$ at $E$. Then $BE=AD$, $EC=DC-DE=DC-AB$. One needs to guarantee that $BE+EC>BC$, so there are only three possible trapezoids: \[AB=3, BC=7, CD=11, DA=5, CE=8\] \[AB=5, BC=7, CD=11, DA=3, CE=6\] \[AB=7, BC=5, CD=11, DA=3, CE=4\] In the first case, by Law of Cosines, $\cos(\angle BCD) = (8^2+7^2-5^2)/(2\cdot 7\cdot 8) = 11/14$, so $\sin (\angle BCD) = \sqrt{1-121/196} = 5\sqrt{3}/14$. Therefore the area of this trapezoid is $\frac{1}{2} (3+11) \cdot 7 \cdot 5\sqrt{3}/14 = \frac{35}{2}\sqrt{3}$. In the second case, $\cos(\angle BCD) = (6^2+7^2-3^2)/(2\cdot 6\cdot 7) = 19/21$, so $\sin (\angle BCD) = \sqrt{1-361/441} = 4\sqrt{5}/21$. Therefore the area of this trapezoid is $\frac{1}{2} (5+11) \cdot 7 \cdot 4\sqrt{5}/21 =\frac{32}{3}\sqrt{5}$. In the third case, $\angle BCD = 90^{\circ}$, therefore the area of this trapezoid is $\frac{1}{2} (7+11) \cdot 3 = 27$. So $r_1 + r_2 + r_3 + n_1 + n_2 = 17.5 + 10.666... + 27 + 3 + 5$, which rounds down to $\boxed{\textbf{(D)}\ 63}$.

命名梯形为 $ABCD$，其中 $AB$ 平行于 $CD$，$AB<CD$，且 $AD<BC$。过 $B$ 作一条与 $AD$ 平行的直线，与边 $CD$ 交于 $E$。则 $BE=AD$，$EC=DC-DE=DC-AB$。需要保证 $BE+EC>BC$，因此只有三种可能的梯形： \[AB=3, BC=7, CD=11, DA=5, CE=8\] \[AB=5, BC=7, CD=11, DA=3, CE=6\] \[AB=7, BC=5, CD=11, DA=3, CE=4\] 第一种情况，由余弦定理，$\cos(\angle BCD) = (8^2+7^2-5^2)/(2\cdot 7\cdot 8) = 11/14$，所以 $\sin (\angle BCD) = \sqrt{1-121/196} = 5\sqrt{3}/14$。因此该梯形面积为 $\frac{1}{2} (3+11) \cdot 7 \cdot 5\sqrt{3}/14 = \frac{35}{2}\sqrt{3}$。第二种情况，$\cos(\angle BCD) = (6^2+7^2-3^2)/(2\cdot 6\cdot 7) = 19/21$，所以 $\sin (\angle BCD) = \sqrt{1-361/441} = 4\sqrt{5}/21$。因此该梯形面积为 $\frac{1}{2} (5+11) \cdot 7 \cdot 4\sqrt{5}/21 =\frac{32}{3}\sqrt{5}$。第三种情况，$\angle BCD = 90^{\circ}$，因此该梯形面积为 $\frac{1}{2} (7+11) \cdot 3 = 27$。所以 $r_1 + r_2 + r_3 + n_1 + n_2 = 17.5 + 10.666... + 27 + 3 + 5$，向下取整为 $\boxed{\textbf{(D)}\ 63}$。

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