106.(08四川乐山28题)28．如图(18)，在平面直角坐标系中，的边在轴上，且，以为直径的圆过点．若点的坐标为，，A、B两点的横坐标，是关于的方程的两根．

(1)求、的值；

(2)若平分线所在的直线交轴于点，试求直线对应的一次函数解析式；

(3)过点任作一直线分别交射线、(点除外)于点、．则的是否为定值？若是，求出该定值；若不是，请说明理由．

(08四川乐山28题解析)28．解：(1)以为直径的圆过点，

，而点的坐标为，

由易知，

，····································································································· 1分

即：，解之得：或．

，，

即．···································································································· 2分

由根与系数关系有：

，

解之，．································································································ 4分

(2)如图(3)，过点作，交于点，

易知，且，

在中，易得，··········· 5分

，

，

又，有，

，······································································································· 6分

，

则，即，························································································ 7分

易求得直线对应的一次函数解析式为：．··················································· 8分

解法二：过作于，于，

由，

求得．········································································································ 5分

又，

求得．····························································································· 7分

即，

易求得直线对应的一次函数解析式为：．··················································· 8分

(3)过点作于，于．

为的平分线，．

由，有········································································· 9分

由，有······································································ 10分

，··············································································· 11分

即．···················································································· 12分

105.(08湖南邵阳25题)25．如图(十七)，将含角的直角三角板()绕其直角顶点逆时针旋转解()，得到，与相交于点，过点作交于点，连结．设，的面积为，的面积为．

(1)求证：是直角三角形；

(2)试求用表示的函数关系式，并写出的取值范围；

(3)以点为圆心，为半径作，

①当直线与相切时，试探求与之间的关系；

②当时，试判断直线与的位置关系，并说明理由．

(08湖南邵阳25题解析)25． (1)，

又，································································· 1分

又，···························································· 2分

，

，即是直角三角形；······························································ 3分

(2)在中，，

，

，，

；··············································································· 4分

；··························· 5分

(3)①直线与相切时，则．

，

．

，································································· 6分

又，

是等边三角形，，

，

又；······································································ 7分

②当时，

则有，解之得或；··················································· 8分

(i)当时，，

在中，，，

在中，，········································· 9分

，即，

直线与相离；···························································································· 10分

(ii)当时，

同理可求出：，·············································· 11分

，

直线与相交．···························································································· 12分

104.(08贵州遵义27题)27。(14分)如图(1)所示，一张平行四边形纸片ABCD，AB=10，AD=6，BD=8，沿对角线BD把这张纸片剪成△AB₁D₁和△CB₂D₂两个三角形(如图(2)所示)，将△AB₁D₁沿直线AB₁方向移动(点B₂始终在AB₁上，AB₁与CD₂始终保持平行)，当点A与B₂重合时停止平移，在平移过程中，AD₁与B₂D₂交于点E，B₂C与B₁D₁交于点F，

(1)当△AB₁D₁平移到图(3)的位置时，试判断四边形B₂FD₁E是什么四边形？并证明你的结论；

(2)设平移距离B₂B₁为x，四边形B₂FD₁E的面积为y，求y与x的函数关系式；并求出四边形B₂FD₁E的面积的最大值；

(3)连结B₁C(请在图(3)中画出)。当平移距离B₂B₁的值是多少时，△ B₁B₂F与△ B₁CF相似？

(08贵州遵义27题解析)解：(1) 四边形B₂FD₁E是矩形。

　 因为△AB₁D₁平移到图(3)的，所以四边形B₂FD₁E是一个平行四边形，又因为在平行四边形ABCD中，AB=10，AD=6，BD=8，则有∠ADB是直角。所以四边形B₂FD₁E是矩形。

(2)因为三角形B₁B₂F与三角形AB₁D₁相似，则有B₂F==0.6X,B₁F==0.8x

所以sB₂FD₁E=B₂F×D₁F=0.6X × (8-0.8x)=4.8x-0.48x²

即y=4.8x-0.48x²=12-0.48(x-5)

当x=5时，y=12是最大的值。

(3)要使△ B₁B₂F与△ B₁CF相似，则有　 即

解之得：x=3.6

103.(08云南省卷24题) 24．(本大题满分14分)如图12，已知二次函数图象的顶点坐标为C(1,0)，直线与该二次函数的图象交于A、B两点，其中A点的坐标为(3,4)，B点在轴上.

　 (1)求的值及这个二次函数的关系式；

(2)P为线段AB上的一个动点(点P与A、B不重合)，过P作轴的垂线与这个二次函数的图象交于点E点，设线段PE的长为，点P的横坐标为，求与之间的函数关系式，并写出自变量的取值范围；

(3)D为直线AB与这个二次函数图象对称轴的交点，在线段AB上是否存在一点P，使得四边形DCEP是平行四形？若存在，请求出此时P点的坐标；若不存在，请说明理由.

(08云南省卷24题解析) (1) ∵ 点A(3,4)在直线y=x+m上，

∴ 4=3+m.　　　　　　　　　　　　　　　　 ………………………………(1分)

∴ m=1.　　　　　　　　　　　　　　　　　 ………………………………(2分)

　　　　 设所求二次函数的关系式为y=a(x-1)².　　　 ………………………………(3分)

　　　　 ∵ 点A(3,4)在二次函数y=a(x-1)²的图象上，

　　　　 ∴ 4=a(3-1)²,

　　　　 ∴ a=1.　　　　　　　　　　　　　　　　　 ………………………………(4分)

∴ 所求二次函数的关系式为y=(x-1)².

　 即y=x²-2x+1.　　　　　　　　　　　　　 ………………………………(5分)

(2) 设P、E两点的纵坐标分别为y_P和y_E .

∴ PE=h=y_P-y_E　　　　　　　　　　　　　　 ………………………………(6分)

　　　 =(x+1)-(x²-2x+1)　　　　　　　　　　 ………………………………(7分)

　　　 =-x²+3x.　　　　　　　　　　　　　 　………………………………(8分)

　 即h=-x²+3x (0＜x＜3).　　　　　　　　　 ………………………………(9分)

(3) 存在.　　　　　　　　　　　　　　　　　 ………………………………(10分)

解法1：要使四边形DCEP是平行四边形，必需有PE=DC. …………………(11分)

∵ 点D在直线y=x+1上,

∴ 点D的坐标为(1,2),

∴ -x²+3x=2 .

即x²-3x+2=0 .　　　　　　　　　　　 　　　………………………………(12分)

解之，得　 x₁=2，x₂=1 (不合题意，舍去)　　 ………………………………(13分)

∴ 当P点的坐标为(2,3)时，四边形DCEP是平行四边形.　　 ……………(14分)

解法2：要使四边形DCEP是平行四边形，必需有BP∥CE.　 ………………(11分)

设直线CE的函数关系式为y=x+b.

∵ 直线CE 经过点C(1,0),

∴ 0=1+b,

∴ b=-1 .

∴ 直线CE的函数关系式为y=x-1 .

∴ 　　得x²-3x+2=0.　　　　 ………………………………(12分)

解之，得　 x₁=2，x₂=1 (不合题意，舍去)　 ………………………………(13分)

∴ 当P点的坐标为(2,3)时，四边形DCEP是平行四边形.　 ……………(14分)

102.(08新疆乌鲁木齐23题)23．如图9，在平面直角坐标系中，以点为圆心，2为半径作圆，交轴于两点，开口向下的抛物线经过点，且其顶点在上．

(1)求的大小；

(2)写出两点的坐标；

(3)试确定此抛物线的解析式；

(4)在该抛物线上是否存在一点，使线段与互相平分？若存在，求出点的坐标；若不存在，请说明理由．

(08新疆乌鲁木齐23题解答)23．解：(1)作轴，为垂足，

，半径······················································ 1分

，······································ 3分

(2)，半径

，故，············································ 5分

········································································· 6分

(3)由圆与抛物线的对称性可知抛物线的顶点的坐标为··································· 7分

设抛物线解析式·················································································· 8分

把点代入上式，解得······································································· 9分

····································································································· 10分

(4)假设存在点使线段与互相平分，则四边形是平行四边形········· 11分

且．

轴，点在轴上．·············································································· 12分

又，，即．

又满足，

点在抛物线上······································································································ 13分

所以存在使线段与互相平分．···························································· 14分

101.(08四川南充21题)21．如图，已知平面直角坐标系中，有一矩形纸片，为坐标原点，轴，，现将纸片按如图折叠，为折痕，．折叠后，点落在点，点落在线段上的处，并且与在同一直线上．

(1)求的坐标；

(2)求经过三点的抛物线的解析式；

(3)若的半径为，圆心在(2)的抛物线上运动，

与两坐标轴都相切时，求半径的值．

(08四川南充21题解答)21．解：(1)过作轴于点，如图(第21题图)

在中，，

································· 1分

由对称性可知：

············································································ 2分

点的坐标为····························································································· 3分

(2)设经过的抛物线的解析式为，则

································································································· 4分

解之得

抛物线的解析式为：································································· 5分

(3)与两坐标轴相切

圆心应在第一、三象限或第二、四象限的角平分线上．

即在直线或上·························································································· 6分

若点在直线上，根据题意有

解之得

，

····································································································· 7分

若点在直线上，根据题意有

解之得，

的半径为或．······································································ 8分

96.(08广东佛山25题)25．我们所学的几何知识可以理解为对“构图”的研究：根据给定的(或构造的)几何图形提出相关的概念和问题(或者根据问题构造图形)，并加以研究.

例如：在平面上根据两条直线的各种构图，可以提出“两条直线平行”、“两条直线相交”的概念；若增加第三条直线，则可以提出并研究“两条直线平行的判定和性质”等问题(包括研究的思想和方法).

请你用上面的思想和方法对下面关于圆的问题进行研究：

(1) 如图1，在圆O所在平面上，放置一条直线(和圆O分别交于点A、B)，根据这个图形可以提出的概念或问题有哪些(直接写出两个即可)？

(2) 如图2，在圆O所在平面上，请你放置与圆O都相交且不同时经过圆心的两条直线和(与圆O分别交于点A、B，与圆O分别交于点C、D).

请你根据所构造的图形提出一个结论，并证明之.

(3) 如图3，其中AB是圆O的直径，AC是弦，D是的中点，弦DE⊥AB于点F. 请找出点C和点E重合的条件，并说明理由.

(08广东佛山25题解答)解：(1) 弦(图中线段AB)、弧(图中的ACB弧)、弓形、求弓形的面积(因为是封闭图形)等.　 (写对一个给1分，写对两个给2分)

(2) 情形1　 如图21，AB为弦，CD为垂直于弦AB的直径.　 …………………………3分

结论：(垂径定理的结论之一).　 …………………………………………………………4分

证明：略(对照课本的证明过程给分).　 …………………………………………………7分

情形2　 如图22，AB为弦，CD为弦，且AB与CD在圆内相交于点P.

结论：.

证明：略.

情形3 (图略)AB为弦，CD为弦，且与在圆外相交于点P.

结论：.

证明：略.

情形4　 如图23，AB为弦，CD为弦，且AB∥CD.

结论：　 =　　 .

证明：略.

(上面四种情形中做一个即可，图1分，结论1分，证明3分；

其它正确的情形参照给分；若提出的是错误的结论，则需证明结论是错误的)

(3) 若点C和点E重合，

则由圆的对称性，知点C和点D关于直径AB对称. …………………………………8分

设，则，.………………………………9分

又D是　　 的中点，所以，

即.………………………………………………………10分

解得.……………………………………………………………11分

(若求得或等也可，评分可参照上面的标准；也可以先直觉猜测点B、C是圆的十二等分点，然后说明)

95.(08山东聊城25题)25．(本题满分12分)如图，把一张长10cm，宽8cm的矩形硬纸板的四周各剪去一个同样大小的正方形，再折合成一个无盖的长方体盒子(纸板的厚度忽略不计)．

(1)要使长方体盒子的底面积为48cm²，那么剪去的正方形的边长为多少？

(2)你感到折合而成的长方体盒子的侧面积会不会有更大的情况？如果有，请你求出最大值和此时剪去的正方形的边长；如果没有，请你说明理由；

(3)如果把矩形硬纸板的四周分别剪去2个同样大小的正方形和2个同样形状、同样大小的矩形，然后折合成一个有盖的长方体盒子，是否有侧面积最大的情况；如果有，请你求出最大值和此时剪去的正方形的边长；如果没有，请你说明理由．

(08山东聊城25题解答)(本题满分12分)

解：(1)设正方形的边长为cm，则

．························································································ 1分

即．

解得(不合题意，舍去)，．

剪去的正方形的边长为1cm．·············································································· 3分

(注：通过观察、验证直接写出正确结果给3分)

(2)有侧面积最大的情况．

设正方形的边长为cm，盒子的侧面积为cm²，

则与的函数关系式为：

．

即．······························································································· 5分

改写为．

当时，．

即当剪去的正方形的边长为2.25cm时，长方体盒子的侧面积最大为40.5cm²．········ 7分

(3)有侧面积最大的情况．

设正方形的边长为cm，盒子的侧面积为cm²．

若按图1所示的方法剪折，则与的函数关系式为：

．

即．

当时，．····························· 9分

若按图2所示的方法剪折，则与的函数关系式为：

．

即．

当时，．················································································· 11分

比较以上两种剪折方法可以看出，按图2所示的方法剪折得到的盒子侧面积最大，即当剪去的正方形的边长为cm时，折成的有盖长方体盒子的侧面积最大，最大面积为cm²．

说明：解答题各小题只给了一种解答及评分说明，其他解法只要步骤合理，解答正确，均应给出相应分数．

94.(08广东梅州23题)23．本题满分11分．

如图11所示，在梯形ABCD中，已知AB∥CD， AD⊥DB，AD=DC=CB，AB=4．以AB所在直线为轴，过D且垂直于AB的直线为轴建立平面直角坐标系．

(1)求∠DAB的度数及A、D、C三点的坐标；

(2)求过A、D、C三点的抛物线的解析式及其对称轴L．

(3)若P是抛物线的对称轴L上的点，那么使PDB为等腰三角形的点P有几个?(不必求点P的坐标，只需说明理由)

(08广东梅州23题解答)解： (1) DC∥AB，AD=DC=CB，

　∠CDB=∠CBD=∠DBA，················································································ 0.5分

　　 ∠DAB=∠CBA， ∠DAB=2∠DBA， ············· 1分

∠DAB+∠DBA=90， ∠DAB=60， ··········· 1.5分

　 ∠DBA=30，AB=4， DC=AD=2，　 ·········· 2分

RtAOD，OA=1，OD=，··························· 2.5分

A(-1，0)，D(0， )，C(2， )．　 · 4分

(2)根据抛物线和等腰梯形的对称性知，满足条件的抛物线必过点A(－1，0)，B(3，0)，

故可设所求为　 = (+1)( -3) ······························································ 6分

将点D(0， )的坐标代入上式得， =．

所求抛物线的解析式为　 =　　 ·········································· 7分

其对称轴L为直线=1．······················································································ 8分

(3) PDB为等腰三角形，有以下三种情况：

①因直线L与DB不平行，DB的垂直平分线与L仅有一个交点P₁，P₁D=P₁B，

P₁DB为等腰三角形；　 ·················································································· 9分

②因为以D为圆心，DB为半径的圆与直线L有两个交点P₂、P₃，DB=DP₂，DB=DP₃， P₂DB， P₃DB为等腰三角形；

③与②同理，L上也有两个点P₄、P₅，使得 BD=BP₄，BD=BP₅．　 ······················ 10分

由于以上各点互不重合，所以在直线L上，使PDB为等腰三角形的点P有5个．

93.(08福建南平26题)26．(14分)

(1)如图1，图2，图3，在中，分别以为边，向外作正三角形，正四边形，正五边形，相交于点．

①如图1，求证：；

②探究：如图1，　　　　 ；

如图2，　　　　 ；

如图3，　　　　 ．

(2)如图4，已知：是以为边向外所作正边形的一组邻边；是以为边向外所作正边形的一组邻边．的延长相交于点．

①猜想：如图4，　　　　 (用含的式子表示)；

②根据图4证明你的猜想．

(08福建南平26题解答)(1)①证法一：与均为等边三角形，

，························································································ 2分

且··············································· 3分

，

即························································ 4分

．··················································· 5分

证法二：与均为等边三角形，

，························································································ 2分

且························································································ 3分

可由绕着点按顺时针方向旋转得到··································· 4分

．··························································································· 5分

②，，．········································································ 8分(每空1分)

(2)①········································································································ 10分

②证法一：依题意，知和都是正边形的内角，，，

，即．····························· 11分

．·························································································· 12分

，，······ 13分

，

········································ 14分

证法二：同上可证　 ．··························································· 12分

，如图，延长交于，

，

································ 13分

················· 14分

证法三：同上可证　 ．··························································· 12分

．

，

························································ 13分

即········································································ 14分

证法四：同上可证　 ．··························································· 12分

．如图，连接，

．···································· 13分

即······························· 14分

注意：此题还有其它证法，可相应评分．