1972 AMC 12 第 26 题

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26.

在图示圆中,MM 是弧 CAB\overset{\frown}{CAB} 的中点,线段 MPMPPP 点垂直于弦 ABAB。若弦 ACAC 的长度为 xx,线段 APAP 的长度为 x+1x+1,则线段 PBPB 的长度等于:

In the circle shown, MM is the midpoint of arc CAB,\overset{\frown}{CAB}, and segment MPMP is perpendicular to chord ABAB at P.P. If the measure of chord ACAC is xx and that of segment APAP is x+1,x+1, then segment PBPB has measure equal to:

3x+23x+2

3x+13x+1

2x+32x+3

2x+22x+2

2x+12x+1

答案:E
知识点:矩形对称性
难度评级:2230
小提示:

MMBB 之间取新点 NN,使弧 MN\overset{\frown}{MN} 等于弧 CA\overset{\frown}{CA}

Copy arc CA\overset{\frown}{CA} to an arc MN\overset{\frown}{MN} ending at a new point NN between MM and BB

大提示:

NQABNQ\perp AB;利用等弧找出一个矩形和相等的水平端线段

Drop NQABNQ\perp AB; use equal arcs to identify a rectangle and equal horizontal end segments

解答:

在弧 MB\overset{\frown}{MB} 上取点 NN,使弧 MN\overset{\frown}{MN}CA\overset{\frown}{CA} 相等,并作 NQABNQ\perp AB。于是 MN=AC=xMN=AC=x。其余两段等弧 AM\overset{\frown}{AM}NB\overset{\frown}{NB} 在弦上的对应投影相等,所以 QB=AP=x+1QB=AP=x+1。又因为 MNPQMNPQ 是矩形,故 PQ=MN=xPQ=MN=x。因此 PB=PQ+QB=x+(x+1)=2x+1 \begin{aligned} PB&=PQ+QB\\ &=x+(x+1)\\ &=2x+1 \end{aligned}\text{。}

因此,正确答案是 E

Choose NN on arc MB\overset{\frown}{MB} so that arcs MN\overset{\frown}{MN} and CA\overset{\frown}{CA} are equal, and drop NQAB.NQ\perp AB. Then MN=AC=x.MN=AC=x. The remaining equal arcs AM\overset{\frown}{AM} and NB\overset{\frown}{NB} give equal corresponding chord projections, so QB=AP=x+1.QB=AP=x+1. Also MNPQMNPQ is a rectangle, hence PQ=MN=x.PQ=MN=x. Therefore PB=PQ+QB=x+(x+1)=2x+1. \begin{aligned} PB&=PQ+QB\\ &=x+(x+1)\\ &=2x+1. \end{aligned}

Therefore, the correct answer is E.

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