AMC 8 2025 Problem 18

AMC8 2025 年第 18 题

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The circle shown below on the left has a radius of 1 unit. The region between the circle and the inscribed square is shaded. In the circle shown on the right, one quarter of the region between the circle and the inscribed square is shaded. The shaded regions in the two circles have the same area. What is the radius RR , in units, of the circle on the right?
左下图的圆的半径为1个单位,阴影部分为圆和其内接正方形之间的区域。在右下图的圆中,阴影部分为圆和其内接正方形之间区域的四分之一。两个圆中的阴影区域面积相同。请问右下图的圆的半径R是多少个单位?
1
  1. A.2\sqrt2
  2. B.2
  3. C.222\sqrt2
  4. D.4
  5. E.424\sqrt2

正确答案 · Correct Answer

B

解析 · Solution

Solution 1

The area of the shaded region in the circle on the left is the area of the circle minus the area of the square, or (π−2)\big(\pi-2). The shaded area in the circle on the right is 14\dfrac{1}{4} of the area of the circle minus the area of the square, or πR2−2R24\dfrac{\pi R^2-2R^2}{4}, which can be factored as R2(π−2)4\dfrac{R^2(\pi-2)}{4}. Since the shaded areas are equal to each other, we have π−2=R2(π−2)4\pi-2=\dfrac{R^2(\pi-2)}{4}, which simplifies to R2=4R^2=4. Taking the square root, we have R=(B)2R=\text{(B)2} The side length of the smaller square is 2\sqrt{2}, which follows from the Pythagorean Theorem since the square is inscribed in a circle of radius 1, making the diagonal equal to 2. Thus, the area of the square is (2)2=2(\sqrt{2})^2 = 2. The circle in which the square is inscribed has radius 1, so its area is π⋅12=π\pi \cdot 1^2 = \pi. Therefore, the area of the shaded region (the part of the circle not covered by the square) is π−2\pi - 2. Now, consider the larger square, which is inscribed in a circle of radius RR. This square can be divided into four congruent right triangles by drawing lines from the center of the circle to the vertices of the square. Each triangle has legs of length RR and includes a 90∘90^\circ angle at the center. Using the triangle area formula 12absin⁡(C)\displaystyle \frac{1}{2} ab\sin(C) with a = b = R and C=90∘C = 90^\circ, each triangle has area 12R2\displaystyle \frac{1}{2} R^2. Since there are four such triangles, the area of the square is 4⋅12R2=2R2\displaystyle 4 \cdot \frac{1}{2} R^2 = 2R^2. The area of the circle is πR2\pi R^2, so the area of the shaded region (the part of the circle outside the square) is 14(πR2−2R2)=14R2(π−2)\displaystyle \frac{1}{4} (\pi R^2 - 2R^2) = \frac{1}{4} R^2(\pi - 2). We are told that this shaded region also has area equal to π−2\pi - 2, so we set up the equation 14R2(π−2)=π−2\displaystyle \frac{1}{4} R^2(\pi - 2) = \pi - 2. Assuming π−2≠0\pi - 2 \neq 0, we divide both sides by π−2\pi - 2, giving 14R2=1\displaystyle \frac{1}{4} R^2 = 1, which leads to R2=4R^2 = 4, and finally R = 2. Thus, the answer is (B) 2\text{(B) 2}. ~ GREATEST ~

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