1976 AMC 12 第 26 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

26.

在所附图中,圆 OO' 的每一点都在圆 OO 的外部。一条内公切线分别与两条外公切线交于 PPQQ。则 PQPQ 的长度

In the adjoining figure, every point of circle OO' is exterior to circle O.O. Let PP and QQ be the points of intersection of an internal common tangent with the two external common tangents. Then the length of PQPQ is

等于内公切线段与外公切线段长度的平均数

the average of the lengths of the internal and external common tangents

当且仅当圆 OOOO' 的半径相等时,才等于一条外公切线段的长度

equal to the length of an external common tangent if and only if circles OO and OO' have equal radii

总是等于一条外公切线段的长度

always equal to the length of an external common tangent

大于一条外公切线段的长度

greater than the length of an external common tangent

等于内公切线段与外公切线段长度的几何平均数

the geometric mean of the lengths of the internal and external common tangents

答案:C
知识点:相切圆切线位似
难度评级:2200
小提示:

在三条切线上分别标出切点

Mark the tangency point on each of the three tangent lines

大提示:

PPQQ 中任一点向同一个圆所作的两条切线段长度相等

From either PP or Q,Q, tangent segments to the same circle have equal lengths

解答:

设内公切线与圆 OOOO' 分别切于 RRSS。设经过 PP 的外公切线与两圆分别切于 XXYY,经过 QQ 的外公切线与两圆分别切于 VVWW。从同一点引出的切线段相等,所以 PR=PX,PS=PY,QR=QV,QS=QW \begin{gathered} PR=PX,\\ PS=PY,\\ QR=QV,\\ QS=QW \end{gathered}\text{。}在内公切线上,PR+QR=PQPR+QR=PQ,且 PS+QS=PQPS+QS=PQ。相加得 PR+QR+PS+QS=2PQ \begin{aligned} PR+QR+PS+QS=2PQ \end{aligned}\text{。}在两条外公切线上,PX+PY=XYPX+PY=XY,且 QV+QW=VWQV+QW=VW。因此 2PQ=XY+VW2PQ=XY+VW。两条外公切线段长度相等,所以 XY=VWXY=VW,从而 PQ=XY=VWPQ=XY=VW

所以正确答案是 C

Let the internal tangent touch OO and OO' at RR and S.S. Let the external tangent through PP touch them at XX and Y,Y, and let the one through QQ touch them at VV and W.W. Equal tangent segments from a point give PR=PX,PS=PY,QR=QV,QS=QW. \begin{gathered} PR=PX,\\ PS=PY,\\ QR=QV,\\ QS=QW. \end{gathered} Along the internal tangent, PR+QR=PQPR+QR=PQ and PS+QS=PQ.PS+QS=PQ. Adding, PR+QR+PS+QS=2PQ. \begin{aligned} PR+QR+PS+QS=2PQ. \end{aligned} Along the external tangents, PX+PY=XYPX+PY=XY and QV+QW=VW.QV+QW=VW. Therefore 2PQ=XY+VW.2PQ=XY+VW. The two external common tangent segments have equal length, so XY=VWXY=VW and hence PQ=XY=VW.PQ=XY=VW.

Therefore, the correct answer is C.

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