29. 解:(1)将图1中的正方形等分成如图的四个小正方形,将这4个转发装置安装在这4个小正方形对角线的交点处,此时,每个小正方形的对角线长为,每个转发装置都能完全覆盖一个小正方形区域,故安装4个这种装置可以达到预设的要求.

····················· (3分)(图案设计不唯一)

(2)将原正方形分割成如图2中的3个矩形,使得.将每个装置安装在这些矩形的对角线交点处,设,则

,得

即如此安装3个这种转发装置,也能达到预设要求.·············································· (6分)

或:将原正方形分割成如图2中的3个矩形,使得的中点,将每个装置安装在这些矩形的对角线交点处,则,即如此安装三个这个转发装置,能达到预设要求.···················································································· (6分)

要用两个圆覆盖一个正方形,则一个圆至少要经过正方形相邻两个顶点.如图3,用一个直径为31的去覆盖边长为30的正方形,设经过交于,连,则,这说明用两个直径都为31的圆不能完全覆盖正方形

所以,至少要安装3个这种转发装置,才能达到预设要求.··································· (8分)

评分说明:示意图(图1、图2、图3)每个图1分.

 

30解:(1)

(2)设存在实数,使抛物线上有一点,满足以为顶点的三角形与等腰直角相似.

为顶点的三角形为等腰直角三角形,且这样的三角形最多只有两类,一类是以为直角边的等腰直角三角形,另一类是以为斜边的等腰直角三角形.

①若为等腰直角三角形的直角边,则

由抛物线得:

的坐标为

代入抛物线解析式,得

抛物线解析式为

②若为等腰直角三角形的斜边,

的坐标为

代入抛物线解析式,得

抛物线解析式为,即

时,在抛物线上存在一点满足条件,如果此抛物线上还有满足条件的点,不妨设为点,那么只有可能是以为斜边的等腰直角三角形,由此得,显然不在抛物线上,因此抛物线上没有符合条件的其他的点.

时,同理可得抛物线上没有符合条件的其他的点.

的坐标为,对应的抛物线解析式为时,

都是等腰直角三角形,

总满足

的坐标为,对应的抛物线解析式为时,

同理可证得:总满足

26. 解:方案一:由题意可得:

到甲村的最短距离为.······································································· (1分)

到乙村的最短距离为

将供水站建在点处时,管道沿铁路建设的长度之和最小.

即最小值为.········································································ (3分)

方案二:如图①,作点关于射线的对称点,则,连接于点,则

.·········································································· (4分)

中,

两点重合.即点.············································· (6分)

在线段上任取一点,连接,则

把供水站建在乙村的点处,管道沿线路铺设的长度之和最小.

即最小值为.··········· (7分)

方案三:作点关于射线的对称点,连接,则

于点,交于点,交于点

为点的最短距离,即

中,

两点重合,即点.

中,.············································· (10分)

在线段上任取一点,过于点,连接

显然

把供水站建在甲村的处,管道沿线路铺设的长度之和最小.

即最小值为.································································ (11分)

综上,供水站建在处,所需铺设的管道长度最短.········ (12分)

25. 解:(1)取中点,联结

的中点,.································· (1分)

.··········································································· (1分)

,得;······································ (2分)(1分)

(2)由已知得.··································································· (1分)

以线段为直径的圆与以线段为直径的圆外切,

,即.·························· (2分)

解得,即线段的长为;······································································· (1分)

(3)由已知,以为顶点的三角形与相似,

又易证得.··············································································· (1分)

由此可知,另一对对应角相等有两种情况:①;②

①当时,

,易得.得;······················································· (2分)

②当时,

.又

,即,得

解得(舍去).即线段的长为2.········································ (2分)

综上所述,所求线段的长为8或2.

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