1976 AMC 12 第 26 题

先试着解答 1976 AMC 12 第 26 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1976 AMC 12 解答,或核对答案。

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 O′O' 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 PP 和 QQ。则 PQPQ 的长度
In the adjoining figure, every point of circle O′O' 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 OO 和 O′O' 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 O′O' 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
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:
从 PP 或 QQ 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 OO 和 O′O' 分别切于 RR 和 SS。设经过 PP 的外公切线与两圆分别切于 XX 和 YY,经过 QQ 的外公切线与两圆分别切于 VV 和 WW。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 O′O' 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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