AMC 8 2024 Problem 18
AMC8 2024 年第 18 题
圆★★★☆☆
Three concentric circles centered at O have radii of 1, 2, and 3. Points B and C lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle ∠BOC, as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of ∠BOC in degrees?
以点 O 为圆心的三个同心圆的半径分别为 1、2、3。点 B 和点 C 位于最大圆上。两个较小圆之间的区域被涂上阴影，两个较大圆之间且位于圆心角 ∠BOC 内的区域也被涂上阴影，如图所示。若阴影区域与非阴影区域面积相等，请问 ∠BOC 的度数是多少？
- A.108
- B.120
- C.135
- D.144
- E.150
正确答案 · Correct Answer
A
解析 · Solution
解法1
设 。我们看到阴影区域是内环加上外环的一个 扇形。利用圆的面积公式（ ），我们求得 的面积为 。这简化为 。未阴影部分由最小的圆加上外环的 扇形组成，计算得到 。已知这两部分面积相等。因此， 。解出 得 x=108 。
注意到，对于从外往里数第三个圆环，阴影区域与非阴影区域的面积比等于 与 的比。因此，我们只需要求出阴影区域的面积。最内层圆的半径为 1，第二个圆的半径为 2。所以第一个阴影区域的面积为 。整个圆的面积为 ，因此另一个阴影区域的面积必须是 ，因为非阴影区域和阴影区域的面积相等。所以对于第三个外环，总面积为 ，那么外环的非阴影部分面积为 。如前所述，这两个面积的比就是 与 的比。所以 。设比例为 7x:3x，且 7x+3x = 360，得 x = 36，因此答案为 3x = 108。AMC 8 考试允许使用一把可旋转和拖动的直尺工具。你可以用该工具画一条直线段（我们知道是 180 度），并设所求角为 。我们可以估计 180-x 大约比 小 30 度，所以 x-30=180-x，解得 x=105，因此最接近的答案是 108。
假设所要求的角度是圆周总度数的一个分数 。我们现在列出阴影区域和非阴影区域的面积。半径为1的内圆完全非阴影，因此它对非阴影区域贡献1。（所有面积都是 的倍数，所以我们省略不写。）内环的面积为 ，它对阴影区域贡献。外环的总面积为 ；其中分数 被阴影覆盖，因此外环的阴影部分对阴影区域贡献 ，而其余 1 - x 分数非阴影，所以非阴影部分对非阴影区域贡献 5(1-x)。现在我们建立等式并求解。 1 + 5(1-x) = 3 + 5x 解之得 ，因此度数为 。
设 ， 为内环的面积， 为外环扇形 的面积， 为阴影区域的总面积。那么， 是总面积的一半，所以 。外半径为 内半径为 的环的面积公式为： 因此内环的面积为 外环扇形的面积为
于是我们有：
解出 ：
所以答案是 。
Solution 1
Let . We see that the shaded region is the inner ring plus a sector of the outer ring. Using the formula for the area of a circle ( ), we find that the shaded area is . This simplifies to . The unshaded portion is comprised of the smallest circle plus the sector of the outer ring, which evaluates to . We are told these are equal. Therefore, . Solving for reveals x=108.
Notice that for the 3rd most outer ring of the circle, the ratio of the shaded region to non region is the ratio of to . With that, all we need to do is solve for the shaded region. The inner most circle has radius 1, and the second circle has radius 2. Therefore, the first shaded area has area. The circle has total area , so the other shaded region must have area, as the non and shaded area is equivalent. So for the 3rd outer ring, the total area is , so the non part of the outer ring is . Now as said before, the ratio of these two areas is the ratio of and . So, = . We have 7x:3x where 7x+3x = 360, x = 36, so our answer is 3x = 108, 108. The AMC 8 test allows a ruler tool that you can rotate and drag. You can use the tool to make a straight segment (which we know is 180 degrees), and we let the angle of desire be . We can estimate that 180-x is just about 30 degrees short of itself, so x-30=180-x, solving gives x=105, therefore the closest answer is 108.
Suppose the desired angle is some fraction of the total degree measure of the circle. We now compile a list of the shaded and unshaded areas. The inner circle of radius 1 is completely unshaded, so it contributes 1 to the unshaded area. (Everything will be a multiple of , so we omit it.) The inner annulus has area , which it contributes to the shaded area. The outer annulus has a total area of ; the fraction is shaded, so the shaded portion of the outer annulus contributes to the shaded area, while the other 1 - x fraction is unshaded, so the unshaded portion contributes 5(1-x) to the unshaded area. We now equate and solve. 1 + 5(1-x) = 3 + 5x Upon solving, we find that , so the degree measure is .
Let , be the area of the inner annulus, be the area of the outer annulus sector and the total area of the shaded regions. Then, is half the total area so . The area of an annulus with outer radius and inner radius is given by: So the area of the inner annulus is The area of the outer annulus sector is
Then we have:
Solving for :
So our answer is .
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