2014中考数学压轴题精编----安徽篇1.(安徽省)如图,已知△ABC ∽△A 1B 1C 1,相似比为k (k >1),且△ABC 的三边长分别为a 、b 、c (a >b >c ),△A 1B 1C 1的三边长分别为a 1、b 1、c 1. (1)若c =a 1,求证:a =kc ;(2)若c =a 1,试给出符合条件的一对△ABC 和△A 1B 1C 1,使得a 、b 、c 和a 1、b 1、c 1都是正整数,并加以说明;(3)若b =a 1,c =b 1,是否存在△ABC 和△A 1B 1C 1,使得k =2?请说明理由.1.解(1)证:∵△ABC ∽△A 1B 1C 1,且相似比为k (k >1),∴1a a=k ,∴a =ka 1 又∵c =a 1,∴a =kc ·················································································· 3分 (2)解:取a =8,b =6,c =4,同时取a 1=4,b 1=3,c 1=2 ······························ 8分此时1a a =1b b =1c c=2,∴△ABC ∽△A 1B 1C 1且c =a 1 ····································· 10分注:本题也是开放型的,只要给出的△ABC 和△A 1B 1C 1符合要求就相应赋分. (3)解:不存在这样的△ABC 和△A 1B 1C 1.理由如下: 若k =2,则a =2a 1,b =2b 1,c =2c 1 又∵b =a 1,c =b 1,∴a =2a 1=2b =4b 1=4c∴b =2c ································································································· 12分 ∴b +c =2c +c =3c <4c =a ,而b +c >a故不存在这样的△ABC 和△A 1B 1C 1,使得k =2. ··········································· 14分注:本题不要求学生严格按反证法的证明格式推理,只要能说明在题设要求下k =2的情况不可能即可.2.(安徽省B 卷)如图,Rt △ABC 内接于⊙O ,AC =BC ,∠BAC 的平分线AD 与⊙O 交于点D ,与BC 交于点E ,延长BD ,与AC 的延长线交于点F ,连结CD ,G 是CD 的中点,连结OG . (1)判断OG 与CD 的位置关系,写出你的结论并证明; (2)求证:AE =BF ; (3)若OG ·DE =3(2-2),求⊙O 的面积.B C AA 1 a b cB 1C 1 a 1 b 1c 1 A C B F D EO G2.解:(1)猜想:OG ⊥CD .证明:如图,连结OC 、OD ,则OC =OD .∵G 是CD 的中点∴由等腰三角形的性质,有OG ⊥CD . ················· 2分(2)证明:∵AB 是⊙O 的直径,∴∠ACB =90°.而∠CAE =∠CBF (同弧所对的圆周角相等). 在Rt △ACE 和Rt △BCF 中∵∠ACE =∠BCF =90°,AC =BC ,∠CAE =∠CBF ∴Rt △ACE ≌Rt △BCF .(ASA )∴AE =BF . ············································································ 6分(3)解:如图,过点O 作BD 的垂线,垂足为H ,则H 为BD 的中点.∴OH =21AD ,即AD =2OH . 又∠CAD =∠BAD ,∴CD =∠BD ,∴OH =OG . 在Rt △BDE 和Rt △ADB 中∵∠DBE =∠DAC =∠BAD ,∴Rt △BDE ∽Rt △ADB . ∴AD BD =DBDE,即BD 2=AD ·DE . ∴BD 2=AD ·DE =2OG ·DE =6(2-2). ······································ 8分 又BD =FD ,∴BF =2BD .∴BF 2=4BD 2=24(2-2).……………………………………①. ······· 9分 设AC =x ,则BC =x ,AB =2x .∵AD 是∠BAC 的平分线,∴∠F AD =∠BAD . 在Rt △ABD 和Rt △AFD 中∵∠ADB =∠ADF =90°,AD =AD ,∠F AD =∠BAD ∴Rt △ABD ≌Rt △AFD .(ASA ) ∴AF =AB =2x ,BD =FD .∴CF =AF -AC =2x -x =(2-1)x . 在Rt △BCF 中,由勾股定理,得BF 2=BC 2+CF 2=x2+[(2-1)x ]2=2(2-2)x2.…………②. ······ 10分由①、②,得2(2-2)x2=24(2-2).∴x2=12,∴x =32或32-(舍去).∴AB =2x =2·32=62.∴⊙O 的半径长为6. ····························································· 11分 ∴S ⊙O =π·(6)2=6π. ·························································· 12分3.(安徽省B 卷)已知:抛物线y =ax2+bx +c (a ≠0)的对称轴为x =-1,与x 轴交于A 、B 两点,与y轴交于点C ,其中A (-3,0)、C (0,-2).ACBF D E HO G(1)求这条抛物线的函数表达式.(2)已知在对称轴上存在一点P ,使得△PBC 的周长最小.请求出点P 的坐标.(3)若点D 是线段OC 上的一个动点(不与点O 、点C 重合).过点D 作DE ∥PC 交x 轴于点E ,连接PD 、PE .设CD 的长为m ,△PDE 的面积为S .求S 与m 之间的函数关系式.试说明S 是否存在最大值,若存在,请求出最大值;若不存在,请说明理由.3.解:(1)由题意得⎪⎪⎩⎪⎪⎨⎧2 0 391 2----+===c c b a a b································································· 2分解得a =32,b =34,c =-2.∴这条抛物线的函数表达式为y =32x2+34x -2 ·································· 4分(2)如图,连结AC 、B C .由于BC 的长度一定,要使△PBC 的周长最小,必须使PB +PC 最小. 点B 关于对称轴的对称点是点A ,AC 与对称轴x =-1的交点即为所求的点P . 设直线AC 的表达式为y =kx +b ,则⎩⎨⎧203--+ ==b b k ················································· 6分 解得k =-32,b =-2. ∴直线AC 的表达式为y =-32x -2 ······························把x =-1代入上式,得y =-32×(-1)-2=-34. ∴点P 的坐标为(-1,-34) ························································· 8分 (3)S 存在最大值,理由如下:∵DE ∥PC ,即DE ∥AC ,∴△OED ∽△OAC .∴OD OE =OC OA ,即m OE -2=23,∴OE =3-23m ,∴AE =23m .方法一: 连结OPS =S △POE +S △POD -S △OED=21×(3-23m )×34+21×(2-m )×1-21×(3-23m )×(2-m ) =-43m2+23m ········································································· 10分 ∵-43<0,∴S 存在最大值. ······················································· 11分 ∵S =-43m2+23m =-43(m -1)2+43 ∴当m =1时,S 最大=43. ··························································· 12分 方法二:S =S △OAC-S △OED-S △P AE-S △PCD=21×3×2-21×(3-23m )×(2-m )-21×23m ×34-21×m ×1 =-43m2+23m ········································································· 10分 以下同方法一.4.(安徽省芜湖市)(本小题满分12分)如图,BD 是⊙O 的直径,OA ⊥OB ,M 是劣弧上一点,过M 点作⊙O 的切线MP 交OA 的延长线于P 点,MD 与OA 交于N 点. (1)求证:PM =PN ; (2)若BD =4,P A =23AO ,过B 点作BC ∥MP 交⊙O 于C4.解:(1)证明:连接OM ··································· 1分∵MP 是⊙O 的切线,∴OM ⊥MP ∴∠OMD +∠DMP =90°∵OA ⊥OB ,∴∠OND +∠ODM =90° 又∵∠MNP =∠OND ,∠ODM =∠OMD ∴∠DMP =∠MNP ,∴PM =PN ··············· 4分 (2)解:设BC 交OM 于点E ,∵BD =4,∴OA =OB =21BD =2 ∴P A =23AO =3,∴PO =5 ································································· 5分 ∵BC ∥MP ,OM ⊥MP ,∴OM ⊥BC ,∴BE =21BC ··································· 7分 ∵∠BOM +∠MOP =90°,在Rt △OMP 中,∠MPO +∠MOP =90° ∴∠BOM =∠MPO ,又∵∠BEO =∠OMP =90° ∴△OMP ∽△BEO ,∴OP OM =BOBE······················································· 10分得:52=2BE ,∴BE =54,∴BC =58··················································· 12分 5.(安徽省芜湖市)如图,在平面直角坐标系中放置一矩形ABCO ,其顶点为A (0,1)、B (-33,1)、C (-33,0)、O (0,0).将此矩形沿着过E (-3,1)、F (-334,0)的直线EF 向右下方翻折,B 、C 的对应点分别为B ′、C ′.(1)求折痕所在直线EF 的解析式;(2)一抛物线经过B 、E 、B ′三点,求此二次函数解析式;(3)能否在直线EF 上求一点P ,使得△PBC 周长最小?如能,求出点P 的坐标;若不能,说明理由.5.解:(1)由于折痕所在直线EF 过E (-3,1)、F (-334,0) ∴tan ∠EFO =3,直线EF 的倾斜角为60° ∴直线EF 的解析式为:y -=tan60°[x -(-3)]化简得:y =3x +4. ············································································ 3分 (2)设矩形沿直线EF 向右下方翻折后,B 、C 的对应点为B ′(x 1,y 1),C ′(x 2,y 2) 过B ′作B ′A ′⊥AE 交AE 所在直线于A ′点∵B ′E =BE =32,∠B ′EF =∠BEF =60° ∴∠B ′EA ′=60°,∴A ′E =3,B ′A ′=3∴A 与A ′重合,B ′在y 轴上,∴x 1=0,y 1=-2,即B ′(0,-2)【此时需说明B ′(x 1,y 1)在y 轴上】 ························································ 6分 设二次函数的解析式为:y =ax2+bx +c∵抛物线经过B (-33,1)、E (-3,1)、B ′(0,-2)∴⎩⎨⎧27a -33b +c =13a -3b +c =1c =-2解得⎪⎪⎪⎩⎪⎪⎪⎨⎧233431--- ===c b a∴该二次函数解析式为:y =-31x2-334x -2 ·············································· 9分 (3)能,可以在直线EF 上找到P 点,连接B ′C 交EF 于P 点,再连接BP 由于B ′P =BP ,此时点P 与C 、B ′在一条直线上,故BP +PC =B ′P +PC 的和最小由于为BC 定长所以满足△PBC 周长最小. ················································ 10分 设直线B ′C 的解析式为:y =kx +b 则⎪⎩⎪⎨⎧b k b +==--3302 解得⎪⎩⎪⎨⎧2932--==b k ∴直线B ′C 的解析式为:y =-932x -2 ········································· 12分又∵点P 为直线B ′C 与直线EF 的交点∴⎪⎩⎪⎨⎧43 2932+==--x x y y 解得⎪⎪⎩⎪⎪⎨⎧1110 31118--==y x ∴点P 的坐标为(-31118,-1110) ························································· 14分 6.(安徽省合肥一中自主招生)已知:甲、乙两车分别从相距300(km )的M 、N 两地同时出发相向而行,其中甲到达N 地后立即返回,图1、图2分别是它们离各自出发地的距离y (km )与行驶时间x (h )之间的函数图象.(1)试求线段AB 所对应的函数关系式,并写出自变量的取值范围;(2)当它们行驶到与各自出发地的距离相等时,用了29h ,求乙车的速度; (3)在(2)的条件下,求它们在行驶的过程中相遇的时间.6.解:(1)设线段AB 所对应的函数关系式为y =kx +b把(3,300),(427,0)代入得⎩⎪⎨⎪⎧300=3k +b 0=427k +b 解得⎩⎪⎨⎪⎧k =-80b =540yh 图1y h图2∴线段AB 所对应的函数关系式为y 甲=-80x +540 ···························· 5分 自变量x 的取值范围是3<x ≤427(或3≤x ≤427,下同) ··················· 7分 (2)∵x =29在3<x ≤427中,∴把x =29代入y 甲=-80x +540中得y 甲=180 ∴乙车的速度为29180=40(km/h ) ·················································· 12分 (3)由题意知有两次相遇方法一:①当0≤x ≤3时,100x +40x =300,解得:x =715 ····························· 16分 ②当3<x ≤427时,(540-80x )+40x =300,解得:x =6 ···················· 20分 综上所述,当它们行驶了715小时或6小时时,两车相遇 方法二:设经过x 1小时两车首次相遇 则40x 1+100x 1=300,解得:x 1=715··············································· 16分 设经过x 2小时两车第二次相遇则80(x 2-3)=40x 2,解得:x 2=6 ·················································· 20分7.(安徽省合肥一中自主招生)如图1,在△ABC 中,AB =BC ,且BC ≠AC ,在△ABC 上画一条直线,若这条直线..既平分△ABC 的面积,又平分△ABC 的周长,我们称这条线为△ABC 的“等分积周线”. (1)请你在图1中用尺规作图作出一条△ABC 的“等分积周线”;(2)在图1中过点C 能否画出一条“等分积周线”?若能,说出确定的方法;若不能,请说明理由; (3)如图2,若AB =BC =5cm ,AC =6cm ,请你找出△ABC 的所有“等分积周线”,并简要说明确定的方法.7.解:(1)图略,作线段AC 的中垂线BD 即可 ················································· 2分 (2)不能如图1,若直线CD 平分△ABC 的面积 那么S △ADC=S △DBC∴21AD ·CE =21BD ·CE ∴AD =BD ····································· 5分A BC图1DEA B C 图2 A B C 图1∵AC ≠BC ,∴AD +AC ≠BD +BC∴过点C 不能画出一条“等分积周线” ············································ 7分 (3)①若直线经过顶点,则AC 边上的中垂线即为所求线段 ························ 8分②若直线不过顶点,可分以下三种情况: (a )直线与BC 、AC 分别交于E 、F ,如图2所示过点E 作EH ⊥AC 于点H ,过点B 作BG ⊥AC 于点G 易求得BG =4,AG =CG =3 设CF =x ,则CE =8-x 由△CEH ∽△CBG ,可得EH =54(8-x ) 根据面积相等,可得21·x ·54(8-x )=6 ················ 10分 ∴x =3(舍去,即为①)或x =5∴CF =5,CE =3,直线EF 即为所求直线 ······································ 12分 (b )直线与AB 、AC 分别交于M 、N ,如图3所示由(a )可得AM =3,AN =5,直线MN 即为所求直线 (仿照上面给分) ·············································· 15分 (c )直线与AB 、BC 分别交于P 、Q ,如图4所示过点A 作AY ⊥BC 于点Y ,过点P 作PX ⊥BC 于点X由面积法可得AY =524设BP =x ,则BQ =8-x 由相似,可得PX =2524x据面积相等,可得21·2524x ·(8-x )=6 ·················· 17分 ∴x =2148+>5(舍去)或x =2148- 而当BP =2148-时,BQ =2148+>5,舍去∴此种情况不存在 ····································································· 19分 综上所述,符合条件的直线共有三条 ············································· 20分 (注:若直接按与两边相交的情况分类,也相应给分)8.(安徽省合肥一中自主招生)如图,在Rt △ABC 中,∠C =90°,AC =3cm ,BC =4cm ,点P 以一定的速度沿AC 边由A 向C 运动,点Q 以1cm/s 的速度沿CB 边由C 向B 运动,设P 、Q 同时运动,且当一点运动到终点时,另一点也随之停止运动,设运动时间为t (s ). (1)若点P 以43cm/s 的速度运动 ①当PQ ∥AB 时,求t 的值;②在①的条件下,试判断以PQ 为直径的圆与直线AB 的位置关系,并说明理由.A BC图2E FG HABC图3 MNAB C图4PQ X Y(2)若点P 以1cm/s 的速度运动,在整个运动过程中,以PQ 为直径的圆能否与直线AB 相切?若能,请求出运动时间t ;若不能,请说明理由.8.解:(1)①如图1,当PQ ∥AB 时,有AC PC=CBCQ ·············· 2分 即3433t=4t ,解得:t =2 ∴当t =2秒时,PQ ∥AB ································ 5分 ②解法1:如图2,当t =2秒时,PQ ∥AB ,此时PQ 为△ACB 的中位线,PQ =25······························· 6分取PQ 的中点M ,则以PQ 为直径的圆的圆心为M , 半径为21PQ ················································ 8分 过点M 、C 向AB 作垂线,垂足分别为N 、H 则CH =512,MN =21CH =56·························· 10分 ∵MN <21PQ ,∴直线AB 与以PQ 为直径的圆相交 ··············································· 12分解法2:如图3,当t =2秒时,PQ ∥AB ,此时PQ 为 △ACB 的中位线,取PQ 的中点M ,分别过点M 、C 向 AB 作垂线,垂足分别为N 、H ,CH 交PQ 于点G ,连接CM∵MN =21CH ,即MN =GH =CG在Rt △CGM 中,GC <MC ,∴MN <MC∴直线AB 与以PQ 为直径的圆相交 ················ 12分 解法3:如图4,当t =2秒时,PQ ∥AB ,此时PQ 为△ACB 的中位线,过点Q 向AB 作垂线,垂足为N , 则Rt △BNQ ∽Rt △BCA ,∴AB BQ =AC NQ ,即52=3NQ , ∴NQ =56AC BPQ 图1AC B PQ 图3M HNACBPQ 图4MNAC BPQ 图2 M HNA CB PAC B 备用图由平行线间的距离处处相等可知,点M 到AB 的距离为56,小于21PQ ∴直线AB 与以PQ 为直径的圆相交 ·············································· 12分 (2)解法1:如图5,取PQ 的中点M ,作MN ⊥AB 、PG ⊥AB 、QH ⊥AB ,垂足分别为N 、G 、H则由Rt △APG ∽Rt △ABC ,得PG =54t ············ 14分 由Rt △BHQ ∽Rt △BCA ,得HQ =53(4-t) ········· 16分 此时MN 是梯形PGHQ 的中位线,∴MN =56+10t ··············································· 20分当PQ 2=4MN 2时,以PQ 为直径的圆与直线AB 相切即(3-t)2+t2=4(56+10t)2···························· 26分解得:t 1=3,t 2=4927 ··································································· 30分 解法2:如图6,取PQ 的中点M ,作MH ⊥AB 、MG ⊥AC 、MN ⊥BC ,垂足分别为H 、G 、N 连接AM 、BM 、CM由S △ABC=S △ACM+S △BCM +S △ABM 可得:21×3×2t +21×4×21(3-t)+21×5×MH =21×3×4 解得:MH =56+10t当PQ 2=4MN 2时,以PQ 为直径的圆与直线AB 相切即(3-t)2+t2=4(56+10t)2···························· 26分解得:t 1=3,t 2=4927 ····································· 30分 解法3:如图7,取PQ 的中点M ,作MH ⊥AB 、MN ⊥BC ,垂足分别为H 、N ,延长NM 交AB 于点G ,则MN =21PC =21(3-t),NQ =21CQ =2t ,∴NB =4-2t 由Rt △BGN ∽Rt △BAC ,得GN =3-83t ,∴GM =3-83t -21(3-t)=23+81t又∵Rt △GMH ∽Rt △ABC ,∴BC MH =AB GM ,即4MH =58123t解得:MH =56+10t当PQ 2=4MN 2时,以PQ 为直径的圆与直线AB 相切即(3-t)2+t2=4(56+10t)2···························· 26分解得:t 1=3,t 2=4927 ····································· 30分B图7图5图69.(安徽省蚌埠二中自主招生)青海玉树发生7.1级强震后,为使人民的生命财产损失降到最低,部队官兵发扬了连续作战的作风。