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2009年全国各地中考数学压轴题专辑参考答案及评分标准（一）1．解：（1）①直线FG1与直线CD的位置关系为互相垂直．··································1分证明：如图1，设直线FG1与直线CD的交点为H．∵线段EC、EP1分别绕点E逆时针旋转90°依次得到线段EF、EG1．∴∠P1EG1＝∠CEF＝90°，EG1＝EP1，EF＝EC．∵∠G1EF＝90°－∠P1EF，∠P1EC＝90°－∠P1EF．∴∠G1EF＝∠P1EC，∴△G1EF≌△P1EC．∴∠G1FE＝∠P1CE＝90°，∴∠EFH＝90°，∴∠FHC＝90°．∴FG1⊥CD．···············································································2分②按题目要求所画图形见图1，直线G1G2与直线CD的位置关系为互相垂直．······················································4分（2）∵四边形ABCD是平行四边形，∴∠B＝∠ADC．∵AD＝6，AE＝1，tanB＝34．∴DE＝5，tan∠EDC＝tanB＝34．可得CE＝4由（1）可得四边形FECH为正方形∴CH＝CE＝4①如图2，当P1点在线段CH的延长线上时∵FG1＝CP1＝x，P1H＝x－4．∴S△P1FG1＝21FG1·P1H＝21x(x－4)＝21x2－2x即y＝21x2－2x（x＞4）．·······························5分②如图3，当P1点在线段CH上（不与C、H两点重合）时∵FG1＝CP1＝x，P1H＝4－x．∴S△P1FG1＝12FG1·P1H＝21x(4－x)＝－21x2＋2x即y＝－21x2＋2x（0＜x＜4）．···································6分③当P1点与H点重合时，即x＝4时，△P1FG1不存在．·······················7分综上所述，y与x之间的函数关系式及自变量x的取值范围是y＝21x2－2x（x＞4）或y＝－21x2＋2x（0＜x＜4）．·····································8分ADBCEF图1HP1G2G1P2ADBCEF图2HP1G1ADBCEF图3HP1G12．解：（1）∵A（－6，0），C（0，34），∴OA＝6，OC＝34．设DE与y轴交于点M．∵DE∥AB，∴△DMC∽△AOC．∴OAMD＝COCM＝CACD＝21，∴CM＝32，MD＝3．同理可得EM＝3，∴OM＝63．∴D点的坐标为（3，36）．·······················································2分（2）由（1）可得点M的坐标为（0，36）．∵DE∥AB，EM＝MD．∴y轴所在直线是线段ED的垂直平分线．∴点C关于直线DE的对称点F在y轴上．∴ED与CF互相垂直平分，∴CD＝DF＝FE＝EC．∴四边形CDFE为菱形，且点M为其对称中心．作直线BM，设BM与CD、EF分别交于点P、点Q．易知△FQM≌△CPM，∴FQ＝CP．∵FE＝CD，∴QE＝PD．∵EC＝DF∴QE＋EC＋CP＋PQ＝PD＋DF＋FQ＋QP．∴直线BM将四边形CDFE分成周长相等的两个四边形．∵点B（6，0）和点M（0，36）在直线y＝kx＋b上∴3606＝＝＋bbk解得363＝＝－bk∴直线BM的解析式为y＝－3x＋36．······································4分（3）确定G点位置的方法：过A点作AH⊥BM于点H，则AH与y轴的交点即为所求的G点．∵OB＝6，OM＝36，∴∠OBM＝60°，∴∠BAH＝30°．在Rt△OAG中，OG＝AO·tan∠BAH＝32．∴G点的坐标为（0，32）．（或G点的位置为线段OC的中点）········7分3．解：（Ⅰ）如图1，折叠后点B与点A重合，则△ACD≌△BCD．设点C的坐标为（0，m）（m＞0），则BC＝OB－OC＝4－m．于是AC＝BC＝4－m．在Rt△AOC中，由勾股定理，得AC2＝OC2＋OA2．即(4－m)2＝m2＋22，解得m＝23．∴点C的坐标为（0，23）．··········································4分（Ⅱ）如图2，折叠后点B落在边OA上的点为B′，则△B′CD≌△BCD．11AByxOCEDPQMFGHABOyx图1CD由题设OB′＝x，OC＝y，则B′C＝BC＝OB－OC＝4－y．在Rt△B′OC中，由勾股定理，得B′C2＝OC2＋OB2．∴(4－y)2＝y2＋x2，即y＝－81x2＋2．····························6分∵点B′在边OA上，∴0≤x≤2．∴解析式y＝－81x2＋2（0≤x≤2）为所求．∵当≤x≤2时，y随x的增大而减小．∴y的取值范围为23≤y≤2．··························································7分（Ⅲ）如图3，折叠后点B落在边OA上的点为B′′，且B′′D∥OB，则∠OCB′′＝∠CB′′D．又∵∠CBD＝∠CB′′D，∴∠OCB′′＝∠CBD．∴CB′′∥BA，∴Rt△COB′′∽Rt△BOA．∴OAOB＝OBOC，∴OC＝2OB′′．·····································9分在Rt△B′′OC中，设OB′′＝x0（x0＞0），则OC＝2x0．由（Ⅱ）知，2x0＝－81x02＋2，解得x0＝－8±54．∵x0＞0，∴x0＝－8＋54＝54－8．∴点C的坐标为（0，58－16）．·················································10分4．解：（Ⅰ）∵y1＝x，y2＝x2＋bx＋c，y1－y2＝0．∴x2＋(b－1)x＋c＝0．·································································1分将α＝31，β＝21分别代入x2＋(b－1)x＋c＝0，得02112103113122＝＋)－(＋)(＝＋)－(＋)(cbcb解得6165＝＝－cb∴函数y2的解析式为y2＝x2－65x＋61．···········································3分（Ⅱ）由已知，得AB＝62，设△ABM的高为h．则S△ABM＝21AB·h＝122h＝3121，即2h＝1441．根据题意，|t－T|＝2h．由T＝t2＋61t＋61，得|－t2＋65t－61|＝1441．当－t2＋65t－61＝1144时，解得t1＝t2＝125；当t2－65t＋61＝1144时，解得t3＝1225－，t4＝1225＋．ABOyx图2CDB′ABOyx图3CDB″∴t的值为125，1225－，1225＋．··················································6分（Ⅲ）由已知，得α＝α2＋bα＋c，β＝β2＋bβ＋c，T＝t2＋bt＋c．∴T－α＝(t－α)(t＋α＋b)，T－β＝(t－β)(t＋β＋b)．α－β＝(α2＋bα＋c)－(β2＋bβ＋c)，整理得(α－β)(α＋β＋b－1)＝0．∵0＜α＜β＜1，∴α－β≠0，∴α＋β＋b－1＝0．∴α＋b＝1－β＞0，β＋b＝1－α＞0．又0＜t＜1，∴t＋α＋b＞0，t＋β＋b＞0．∴当0＜t≤α时，T≤α＜β；当α＜t≤β时，α＜T≤β；当β＜t＜1时，α＜β＜T．··························································10分5．解：（1）∵点B与点A（1，0）关于原点对称∴点B的坐标为（－1，0）·····························································1分∵直线y＝x＋b经过点B，∴－1＋b＝0．∴b＝1·························································································2分∴直线BD的方程为y＝x＋1…………①直线CM的方程为y＝4…………②联立①②解得x＝3，y＝4．∴点D的坐标为（3，4）．······························································4分（2）点O与点D之间的距离为|OD|＝2243＝5····································5分因点P在x轴的正半轴上，故可设其坐标为（x0，0）．①当△POD是以OD为底的等腰三角形时，则OP2＝DP2．即x02＝(x0－3)2＋42，解得x0＝625．此时点P的坐标为（625，0）··························································6分②当△POD是以OP为底的等腰三角形时|OD|＝|PD|＝5，|OP|＝6此时点P的坐标为（6，0）·····························································7分③当△POD是以DP为底的等腰三角形时|OP|＝|OD|＝5此时点P的坐标为（5，0）·····························································8分综上分析，当△POD是等腰三角形时，满足条件的点P有3个，其坐标分别为（625，0）、（6，0）、（5，0）．（3）当以PD为半径的圆P与圆O外切时．若点P的坐标为（6，0），则圆P的半径PD＝5，圆心距PO＝6．此时圆O的半径r＝1·····································································9分若点P的坐标为（5，0），则圆P的半径PD＝22435＋)－(＝52，圆心距PO＝5．此时圆O的半径r＝5－52···························································10分若点P的坐标为（625，0），则圆P的半径PD＝625，圆心距PO＝625．此时圆O不存在···········································································11分综上所述，所求圆O的半径等于1或5－52．·································12分6．解：（1）∵∠ABC＝90°，AD∥BC，∴∠A＝90°．∵AD∥BC，∴∠ADB＝∠DBC．∵AB＝AD＝2，∴∠ADB＝∠ABD＝45°．∴∠ABD＝∠DBC＝45°，∴∠PBC＝45°．·········································1分∵PCPQ＝ABAD，AB＝AD，点Q与点B重合，∴PB＝PQ＝PC．∴∠PBC＝∠PCB＝45°···································································2分∴∠BPC＝90°···············································································3分∴△PBC是等腰直角三角形∴PC＝22BC＝223······················