AMC 8 2026 Problem 6
AMC8 2026 年第 6 题
面积与周长（综合）★★☆☆☆
Peter lives near a rectangular field that is filled with blackberry bushes. The field is 10 meters long and 8 meters wide, and Peter can reach any blackberries that are within 1 meter of an edge of the field. The portion of the field he can reach is shaded in the figure below. What fraction of the area of the field can Peter reach?
Peter 住在一块长方形黑莓地附近。这块地长 10 米，宽 8 米。他能够采摘到距离地块边缘 1 米范围内的所有黑莓。下图中阴影部分表示他能到达的区域。他能到达的区域占整块地面积的几分之几？
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- B.
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- E.
正确答案 · Correct Answer
E
解析 · Solution
解法1
总面积 8×10=80。沿墙 1 英尺宽的范围内不可达，内部不可达矩形为 (10−2)×(8−2)=8×6=48。可达面积 = 80−48=32。占比 。（等价视角：可达比例 = 1 − 。）
重新排列图形：将不可达区域移到角落，可达区域形成 L 形，由两个矩形组成：8×2 和 10×2，重叠 2×2。可达面积 = 16+20−4=32。占比 。
将 1 英尺宽的可达边框切割成四条。上下两条：各 8×1。左右两条（去掉上下行后）：各 (10−2)×1。可达面积 = 8×1×2 + (10−2)×1×2 = 16+16 = 32。
另一种变形：先算四条完整边（8×1×2 + 10×1×2 = 36），再减去四个角被重复计算的部分（1×1×4 = 4）。36−4=32。
Solution 1
Total area = 8 × 10 = 80. The unreachable strip is 1 foot wide along each wall, so the unreachable inner rectangle is (10−2) × (8−2) = 8 × 6 = 48. Reachable area = 80 − 48 = 32. Fraction = . (Equivalently: reachable fraction = 1 − .)
Rearrange the diagram: move the white (unreachable) region to one corner. The reachable (shaded) region forms an L-shape made of two rectangles: 8×2 and 10×2, overlapping in a 2×2 square. Reachable area = 16 + 20 − 4 = 32. Fraction = .
Slice the 1-foot wide reachable border into four strips. Top and bottom: two 8×1 strips. Left and right (after removing top/bottom rows): two (10−2)×1 strips. Total reachable area = 8×1×2 + (10−2)×1×2 = 16 + 16 = 32.
A variation: count all four strips first (8×1×2 + 10×1×2 = 36), then subtract the four corner squares that were double-counted (1×1×4 = 4). Result: 36 − 4 = 32.
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