AMC12 2002 B

AMC12 2002 B · Q5

AMC12 2002 B · Q5. It mainly tests Arithmetic sequences basics, Polygons.

Let $v, w, x, y,$ and $z$ be the degree measures of the five angles of a pentagon. Suppose that $v < w < x < y < z$ and $v, w, x, y,$ and $z$ form an arithmetic sequence. Find the value of $x$.

设 $v, w, x, y,$ 和 $z$ 是一个五边形的五个角的度数。假设 $v < w < x < y < z$ 且 $v, w, x, y,$ 和 $z$ 构成一个等差数列。求 $x$ 的值。

(A) 72 72

(B) 84 84

(D) 108 108

(E) 120 120

Answer

Correct choice: (D)

正确答案：（D）

Solution

The sum of the degree measures of the angles of a pentagon (as a pentagon can be split into $5- 2 = 3$ triangles) is $3 \cdot 180 = 540^{\circ}$. If we let $v = x - 2d, w = x - d, y = x + d, z = x+2d$, it follows that \[(x-2d)+(x-d)+x+(x+d)+(x+2d) = 5x = 540 \Longrightarrow x = 108 \ \mathrm{(D)}\] Note that since $x$ is the middle term of an arithmetic sequence with an odd number of terms, it is simply the average of the sequence. You can always assume the values are the same so $\frac{540}{5}=108$ Let $v$, $w$, $x$, $y$, $z$ be $v$, $v + d$, $v+2d$, $v+3d$, $v+4d$, respectively. Then we have \[v + w + x + y + z = 5v + 10d = 180^{\circ} (5 - 2) = 540^{\circ}\] Dividing the equation by $5$, we have \[v + 2d = x = 108^{\circ} \mathrm {(D)}\]

五边形的内角和（因为五边形可以分成 $5- 2 = 3$ 个三角形）为 $3 \cdot 180 = 540^{\circ}$。令 $v = x - 2d, w = x - d, y = x + d, z = x+2d$，则 \[(x-2d)+(x-d)+x+(x+d)+(x+2d) = 5x = 540 \Longrightarrow x = 108 \ \mathrm{(D)}\] 注意，由于 $x$ 是项数为奇数的等差数列的中项，它就是该数列的平均数。你也可以直接认为这些值的平均数相同，所以 $\frac{540}{5}=108$ 令 $v$, $w$, $x$, $y$, $z$ 分别为 $v$, $v + d$, $v+2d$, $v+3d$, $v+4d$。则有 \[v + w + x + y + z = 5v + 10d = 180^{\circ} (5 - 2) = 540^{\circ}\] 两边同除以 $5$，得 \[v + 2d = x = 108^{\circ} \mathrm {(D)}\]

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