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2016年中考数学压轴题精编—甘肃篇2016年中考数学压轴题精编—甘肃篇12016年中考数学压轴题精编—甘肃篇1．（甘肃省兰州市）如图1，已知矩形ABCD的顶点A与点O重合，AD、AB分别在x轴、y轴上，且AD＝2，AB＝3；抛物线y＝－x2＋bx＋c经过坐标原点O和x轴上另一点E（4，0）．（1）当x取何值时，该抛物线有最大值，最大值是多少？（2）将矩形ABCD以每秒1个单位长度的速度从图1所示的位置沿x轴的正方向匀速平行移动，同时一动点P也以相同的速度．．．．．从点A出发向B匀速移动．设它们运动的时间为t秒（0≤t≤3），直线AB与该抛物线的交点为N（如图2所示）．①当t＝411时，判断点P是否在直线ME上，并说明理由；②以P、N、C、D为顶点的多边形面积是否可能为5，若有可能，求出此时N点的坐标；若无可能，请说明理由．1．解：（1）∵抛物线y＝－x2＋bx＋c经过坐标原点O（0，0）和点E（4，0）∴c＝0，b＝4∴抛物线的解析式为y＝－x2＋4x·············································1分∵y＝－x2＋4x＝－(x－2)2＋4∴当x＝2时，该抛物线有最大值，最大值是4·····························2分（2）①点P不在直线ME上∵M（2，4），E（4，0），设直线ME的关系式为y＝kx＋b则4k＋b＝02k＋b＝4解得k＝－2b＝8所以直线ME的关系式为y＝－2x＋8·········································3分由已知条件易得，当t＝411时，OA＝AP＝411∴P（411，411）····································································4分∴当x＝411时，y＝－2×411＋8＝25≠411∴当x＝411时，点P不在直线ME上·········································5分②以P、N、C、D为顶点的多边形面积可能为5∵点A在x轴的非负半轴上，且N在抛物线上，∴OA＝AP＝t∴点P，N的坐标分别为（t，t）、（t，－t2＋4t）···························6分MBCDExOy(A)图1MExOy图2NBCDAP2016年中考数学压轴题精编—甘肃篇2016年中考数学压轴题精编—甘肃篇2∴AN＝－t2＋4t（0≤t≤3）∴AN－AP＝(－t2＋4t)－t＝－t2＋3t＝t(3－t)≥0∴PN＝－t2＋3t····································································7分（ⅰ）当PN＝0，即t＝0或t＝3时，以点P，N，C，D为顶点的多边形是三角形，此三角形的高为AD∴S＝21DC·AD＝21×3×2＝3（ⅱ）当PN≠0时，以点P，N，C，D为顶点的多边形是梯形∵PN∥CD，AD⊥CD∴S＝21(CD＋PN)·AD＝21(3－t2＋3t)×2＝－t2＋3t＋3·················8分当－t2＋3t＋3＝5时，解得t＝1或t＝2·······································9分因为1、2都在0≤t≤3范围内，故以P、N、C、D为顶点的多边形面积为5综上所述，当t＝1或t＝2时，以点P，N，C，D为顶点的多边形面积为5当t＝1时，N点的坐标为（1，3）···········································10分当t＝2时，N点的坐标为（2，4）···········································11分说明：（ⅱ）中的关系式，当t＝0和t＝3时也适合（故在阅卷时没有（ⅰ），只有（ⅱ）也可以，不扣分）2．（甘肃省天水市、庆阳市、定西市、白银市、嘉峪关市等九市联考）如图，抛物线与x轴交于A（－1，0）、B（3，0）两点，与y轴交于点C（0，－3），设抛物线的顶点为D．（1）求该抛物线的解析式与顶点D的坐标；（2）以B、C、D为顶点的三角形是直角三角形吗？为什么？（3）探究坐标轴上是否存在点P，使得以P、A、C为顶点的三角形与△BCD相似？若存在，请指出符合条件的点P的位置，并直接写出点P的坐标；若不存在，请说明理由．2．解：（1）设该抛物线的解析式为y＝ax2＋bx＋c由抛物线与y轴交于点C（0，－3），可知c＝－3即抛物线的解析式为y＝ax2＋bx－3····························································1分OABxyCD2016年中考数学压轴题精编—甘肃篇2016年中考数学压轴题精编—甘肃篇3把A（－1，0）、B（3，0）代入，得a－b－3＝09a＋3b－3＝0解得a＝1，b＝－2∴抛物线的解析式为y＝x2－2x－3······························································3分∵y＝x2－2x－3＝(x－1)2－4∴顶点D的坐标为（1，－4）·····································································4分（2）以B、C、D为顶点的三角形是直角三角形·································5分理由如下：如图1，过点D分别作x轴、y轴的垂线，垂足分别为E、F在Rt△BOC中，OB＝OC＝3，∴BC2＝18····················································6分在Rt△CDF中，DF＝1，CF＝OF－OC＝4－3＝1，∴CD2＝2···························7分在Rt△BDE中，DE＝4，BE＝OB－OE＝3－1＝2，∴BD2＝20··························8分∴BC2＋CD2＝CE2，∴△BCD为直角三角形···············································9分（3）如图2，连接AC，可知Rt△COA∽Rt△BCD，得符合条件的点为O（0，0）·········································································10分过A作AP1⊥AC交y轴正半轴于P1，可知Rt△CAP1∽Rt△COA∽Rt△BCD求得符合条件的点为P1（0，21）·······························································11分过C作CP2⊥AC交x轴正半轴于P2，可知Rt△P2CA∽Rt△COA∽Rt△BCD求得符合条件的点为P2（9，0）································································12分∴符合条件的点有三个：O（0，0），P1（0，21），P2（9，0）3．（甘肃省陇南市）如图，△ABC中，∠ACB＝90°，∠A＝30°，AB＝4，将一个30°角的顶点P放在AB边上滑动，保持30°角的一边平行于BC，且交边AC于点E，30°角的另一边交射线BC于点D，连结ED．（1）当四边形PEDC为平行四边形时，求AP的长；OABxyCD图1EFOABxyCD图2P2P12016年中考数学压轴题精编—甘肃篇2016年中考数学压轴题精编—甘肃篇4（2）设AP＝x，动角和△ABC重叠部分的图形的面积为y，求y与x之间的函数关系式，并写出自变量x的取值范围；（3）△PED能否成为等腰三角形？若能，求出AP的长；若不能，请说明理由．3．解：（1）设AP＝x，则BP＝4－x∵∠ACB＝90°，∠A＝30°，AB＝4，∴BC＝2，∠B＝60°∵PE∥BC，∴∠APE＝∠B＝60°，∠AEP＝∠C＝90°∴PE＝21x···································································1分∴∠BPD＝90°，∠BDP＝30°，∴BD＝2BP＝8－2x∴CD＝BD－BC＝8－2x－2＝6－2x·····································2分∵PE∥CD，∴当PE＝CD时，四边形PEDC为平行四边形∴21x＝6－2x，解得x＝512故此时AP的长为512·······················································4分（2）当PD过直角顶点C时，EC＝PE·tan30°＝21x·33＝63x又EC＝AC－AE＝23(4－x)，∴63x＝23(4－x)，解得x＝3·································································5分ⅰ）如图1，当0＜x≤3时，设PD与AC的交点为O，则动角和△ABC重叠部分的图形为Rt△POE∴y＝21PE·OE＝21·21x·21x·33＝243x2即y＝243x2（0＜x≤3）··················································6分ⅱ）如图2，当3＜x＜4时，动角和△ABC重叠部分的图形为直角梯形PDCE∵DC＝BC－BD＝2－(8－2x)＝2x－6∴y＝21(DC＋PE)·EC＝21(2x－6＋21x)·23(4－x)＝－835x2＋34x－36BADCPE30°BAC备用图BADCPE30°图1OBADCPE30°图2BADCPE30°图3APE30°2016年中考数学压轴题精编—甘肃篇2016年中考数学压轴题精编—甘肃篇5即y＝－835x2＋34x－36（3＜x＜4）···························8分（3）能．设AP＝xⅰ）当点D在BC上时若PD＝ED，如图3，则∠PED＝∠DPE＝30°，∴∠EDC＝30°∴DC＝23ED，∴2x－6＝23·3(4－x)解得x＝724···································································9分若PD＝PE，如图4，则3(4－x)＝21x解得x＝113848··························································10分若PE＝DE，则PE2＝DE2＝DC2＋EC2∴(21x)2＝(2x－6)2＋[23(4－x)]2整理得：3x2－20x＋32＝0，解得x1＝4（舍去），x2＝38（舍去）································································11分ⅱ）当点D在BC延长线上时，如图5∵∠PED＞90°，∴若△PED为等腰三角形，只能PE＝DE∴∠PDE＝∠DPE＝30°，∴∠PED＝120°，∴∠CED＝30°∴DE＝2CD，∴21x＝2[2((4－x)－2]解得x＝38·····························································································12分综上所述，当AP＝724或113848或38时，△PED能成为等腰三角形4．（甘肃省张掖市）如图，抛物线y＝ax2－5ax＋4经过△ABC的三个顶点，BC∥x轴，点A在x轴上，点C在y轴上，AB平分∠CAO，P是抛物线对称轴上的动点．（1）求该抛物线的解析式；（2）若P在x轴下方，且△PAB是直角三