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黄金分割专项练习30题

黄金分割专项练习30题
黄金分割专项练习30题

黄金分割专项练习

1.定义:如图1,点C在线段AB上,若满足AC2=BC?AB,则称点C为线段AB的黄金分割点.如图2,△ABC中,AB=AC=1,∠A=36°,BD平分∠ABC交AC于点D.

(1)求证:点D是线段AC的黄金分割点;

(2)求出线段AD的长.

2.如图,用长为40cm的细铁丝围成一个矩形ABCD(AB>AD).

(1)若这个矩形的面积等于99cm2,求AB的长度;

(2)这个矩形的面积可能等于101cm2吗若能,求出AB的长度,若不能,说明理由;

(3)若这个矩形为黄金矩形(AD与AB之比等于黄金比),求该矩形的面积.(结果保留根号)

9.在数学上称长与宽之比为黄金分割比的矩形为黄金矩形,如在矩形ABCD中,当时,称矩形ABCD

为黄金矩形ABCD.请你证明黄金矩形是由一个正方形和一个更小的黄金矩形构成.

10.如图,设AB是已知线段,在AB上作正方形ABCD;取AD的中点E,连接EB;延长DA至F,使EF=EB;以线段AF为边作正方形AFGH.则点H是AB的黄金分割点.

为什么说上述的方法作出的点H是这条线段的黄金分割点,你能说出其中的道理吗请试一试,说一说.

12.已知AB=2,点C是AB的黄金分割线,点D在AB上,且AD2=BD?AB,求的值.

14.五角星是我们常见的图形,如图所示,其中,点C,D分别是线段AB的黄金分割点,AB=20cm,求EC+CD的长.

15.人的肚脐是人的身高的黄金分割点,一般来讲,当肚脐到脚底的长度与身高的比为时,是比较好看的黄金身段.一个身高的人,他的肚脐到脚底的长度为多少时才是黄金身段(保留两位小数)

17.如图,点P是线段AB的黄金分割点,且AP>BP,设以AP为边长的正方形面积为S1,以PB为宽和以AB为长的矩形面积为S2,试比较S1与S2的大小.

18.如图,在平行四边形ABCD中,E为边AD延长线上的一点,且D为AE的黄金分割点,即,BE交DC于点F,已知,求CF的长.

20.(如图1),点P将线段AB分成一条较小线段AP和一条较大线段BP,如果,那么称点P为线段AB的黄金分割点,设=k,则k就是黄金比,并且k≈.

(1)以图1中的AP为底,BP为腰得到等腰△APB(如图2),等腰△APB即为黄金三角形,黄金三角形的定义为:满足≈的等腰三角形是黄金三角形;类似地,请你给出黄金矩形的定义:;

(2)如图1,设AB=1,请你说明为什么k约为;

(3)由线段的黄金分割点联想到图形的“黄金分割线”,类似地给出“黄金分割线”的定义:直线l将一个面积为S的图形分成面积为S1和面积为S2的两部分(设S1<S2),如果,那么称直线l为该图形的黄金分割线.(如

21.在人体躯干(脚底到肚脐的长度)与身高的比例上,肚脐是理想的黄金分割点,即比例越接近,越给人以美感.张女士原来脚底到肚脐的长度与身高的比为,她的身高为,她应该选择多高的高跟鞋穿上看起来更美(精确到十分位)

23.如图,用纸折出黄金分割点:裁一张正方的纸片ABCD,先折出BC的中点E,再折出线段AE,然后通过折叠使EB落到线段EA上,折出点B的新位置B′,因而EB′=EB.类似地,在AB上折出点B″使AB″=AB′.这时B″就是AB的黄金分割点.请你证明这个结论.

25.如图,在△ABC中,点D在边AB上,且DB=DC=AC,已知∠ACE=108°,BC=2.

(1)求∠B的度数;

(2)我们把有一个内角等于36°的等腰三角形称为黄金三角形.它的腰长与底边长的比(或者底边长与腰长的比)等于黄金比.

①写出图中所有的黄金三角形,选一个说明理由;

②求AD的长;

③在直线AB或BC上是否存在点P(点A、B除外),使△PDC是黄金三角形若存在,在备用图中画出点P,简要说明画出点P的方法(不要求证明);若不存在,说明理由.

28.折纸与证明﹣﹣﹣用纸折出黄金分割点:

第一步:如图(1),先将一张正方形纸片ABCD对折,得到折痕EF;再折出矩形BCFE的对角线BF.

第二步:如图(2),将AB边折到BF上,得到折痕BG,试说明点G为线段AD的黄金分割点(AG>GD)

29.三角形中,顶角等于36°的等腰三角形称为黄金三角形,如图1,在△ABC中,已知:AB=AC,且∠A=36°.(1)在图1中,用尺规作AB的垂直平分线交AC于D,并连接BD(保留作图痕迹,不写作法);

(2)△BCD是不是黄金三角形如果是,请给出证明;如果不是,请说明理由;

(3)设,试求k的值;

(4)如图2,在△A1B1C1中,已知A1B1=A1C1,∠A1=108°,且A1B1=AB,请直接写出的值.

30.如图1,点C将线段AB分成两部分,如果,那么称点C为线段AB的黄金分割点.某研究小组在进行课题学习时,由黄金分割点联想到“黄金分割线”,类似地给出“黄金分割线”的定义:直线l将一个面积为S的图形分成两部分,这两部分的面积分别为S1,S2,如果,那么称直线l为该图形的黄金分割线.

(1)研究小组猜想:在△ABC中,若点D为AB边上的黄金分割点(如图2),则直线CD是△ABC的黄金分割线.你认为对吗为什么

(2)请你说明:三角形的中线是否也是该三角形的黄金分割线

(3)研究小组在进一步探究中发现:过点C任作一条直线交AB于点E,再过点D作直线DF∥CE,交AC于点F,连接EF(如图3),则直线EF也是△ABC的黄金分割线.请你说明理由.

(4)如图4,点E是平行四边形ABCD的边AB的黄金分割点,过点E作EF∥AD,交DC于点F,显然直线EF是平行四边形ABCD的黄金分割线.请你画一条平行四边形ABCD的黄金分割线,使它不经过平行四边形ABCD各边黄金分割点.

黄金分割专项练习30题参考答案:

1.(1)证明:∵AB=AC=1,

∴∠ABC=∠C=(180°﹣∠A)=(180°﹣36°)=72°,∵BD平分∠ABC交AC于点D,

∴∠ABD=∠CBD=∠ABC=36°,

∴∠BDC=180°﹣36°﹣72°=72°,

∴DA=DB,BD=BC,

∴AD=BD=BC,

易得△BDC∽△ABC,

∴BC:AC=CD:BC,即BC2=CD?AC,

∴AD2=CD?AC,

∴点D是线段AC的黄金分割点;

(2)设AD=x,则CD=AC﹣AD=1﹣x,

∵AD2=CD?AC,

∴x2=1﹣x,解得x1=,x2=,

即AD的长为

2.解:(1)设AB=xcm,则AD=(20﹣x)cm,

根据题意得x(20﹣x)=99,

整理得x2﹣20x+99=0,解得x1=9,x2=11,

当x=9时,20﹣x=11;当x=11时,20﹣11=9,

而AB>AD,

所以x=11,即AB的长为11cm;

(2)不能.理由如下:

设AB=xcm,则AD=(20﹣x)cm,

根据题意得x(20﹣x)=101,

整理得x2﹣20x+101=0,

因为△=202﹣4×101=﹣4<0,

所以这个矩形的面积可能等于101cm2;

(3)设AB=xcm,则AD=(20﹣x)cm,

根据题意得20﹣x=x,

解得x=10(﹣1),

则20﹣x=10(3﹣),

所以矩形的面积=10(﹣1)?10(3﹣)=(400﹣800)cm2.3.解:(1)∵∠A=36°,AB=AC,

∴∠ABC=∠ACB=72°,

∵BD平分∠ABC,

∴∠CBD=∠ABD=36°,∠BDC=72°,

∴AD=BD,BC=BD,

∴△ABC∽△BDC,

∴=,即=,

∴AD2=AC?CD.

∴点D是线段AC的黄金分割点.

(2)∵点D是线段AC的黄金分割点,

∴AD=AC,

∵AC=2,

∴AD=﹣1

4.解:(1)腰与底之比为黄金比为黄金比如图,

(2)作法:①画线段AB作为三角形底边;

②取AB的一半作AB的垂线AC,连接BC,在BC上取CD=CA.

③分别以A点和B点为圆心、以BD为半径划弧,交点为E;

④分别连接EA、EB,则△ABE即是所求的三角形.

(3)证明:设AB=2,则AC=1,BC=,AE=BE=BD=BC﹣CD=﹣1,=.

5.解:(1)由于P为线段AB=2的黄金分割点,

则AP=2×=﹣1,

或AP=2﹣(﹣1)=3﹣;

(2)如图,点P是线段AB的一个黄金分割点.

6.解:(1)设AC=x,则BC=AB﹣AC=1﹣x,

∵AC2=BC?AB,

∴x2=1×(1﹣x),

整理得x2+x﹣1=0,

解得x1=,x2=(舍去),

所以线段AC的长度为;

(2)设线段AD的长度为x,AC=l,

∵AD2=CD?AC,

∴x2=l×(l﹣x),

∴x1=,x2=(舍去),

∴线段AD的长度AC;

(3)同理得到线段AE的长度AD;

上面各题的结果反映:若线段AB分成两条线段AC和BC(AC>BC),且使AC是AB和BC的比例中项(即AB:AC=AC:BC),则C点为AB的黄金分割点

∵在△ABC中,AB=AC,∠A=36°,

∴∠ABC=∠ACB==72°.

∵∠1=∠2,

∴∠1=∠2=∠ABC=36°.

∴在△BDC中,∠BDC=180°﹣∠2﹣∠C=72°,∴∠C=∠BDC,

∴BC=BD.

∵∠A=∠1,

∴AD=BC.

∵△ABC和△BDC中,∠2=∠A,∠C=∠C,

∴△ABC∽△BDC,

∴,

又∵AB=AC,AD=BC=BD,

∴,

∴AD2=AC?CD,即D是AC的黄金分割点

8.证明:∵AB=AC,∠A=36°,

∴∠ABC=(180°﹣36°)=72°,

∵BD平分∠ABC,交于AC于D,

∴∠DBC=×∠ABC=×72°=36°,

∴∠A=∠DBC,

又∵∠C=∠C,

∴△BCD∽△ABC,

∵AB=AC,

∴=,

∵AB=AC=2,BC=﹣1,

解得AD=,

AD:AC=():2.

∴点D是线段AC的黄金分割点.

9.证明:在AB上截取AE=BC,DF=BC,连接EF.

∵AE=BC,DF=BC,

∴AE=DF=BC=AD,

又∵∠ADF=90°,

∴四边形AEFD是正方形.

BE=,

∴,

∴矩形BCFE的宽与长的比是黄金分割比,矩形BCFE是黄金矩形.∴黄金矩形是由一个正方形和一个更小的黄金矩形构成.

10.解:设正方形ABCD的边长为2,

在Rt△AEB中,依题意,得AE=1,AB=2,

由勾股定理知EB===,

∴AH=AF=EF﹣AE=EB﹣AE=﹣1,

HB=AB﹣AH=3﹣;

∴AH2=()2=6﹣2,

AB?HB=2×(3﹣)=6﹣2,

2

11.证明:(1)∵∠A=36°,∠C=72°,

∴∠ABC=180°﹣36°﹣72°=72°,

∵∠ADB=108°,

∴∠ABD=180°﹣36°﹣108°=36°,

∴△ADB是等腰三角形,

∵∠BDC=180°﹣∠ADC=180°﹣108°=72°,

∴△BDC是等腰三角形,

∴AD=B D=BC.

(2)∵∠DBC=∠A=36°,∠C=∠C,

∴△ABC∽△BDC,

∴BC:AC=CD:BC,

∴BC2=AC?DC,

∵BC=AD,

∴AD2=AC?DC,

∴点D是线段AC的黄金分割点.

12.解:∵D在AB上,且AD2=BD?AB,

∴点D是AB的黄金分割点

而点C是AB的黄金分割点,

∴AC=AB=﹣1,AD=AB﹣AB=AB=3﹣或AD=﹣1,AC=3﹣,∴CD=﹣1﹣(3﹣)=2﹣4,

∴==或==.

13.解:矩形ABFE是黄金矩形.

∵A D=BC,DE=AB,

∴==﹣1==.

∴矩形ABFE是黄金矩形.

∴AD=AB=10﹣10,

∵EC+CD=AC+CD=AD,

∴EC+CD=(10﹣10)cm.

15.解:设他的肚脐到脚底的长度为xm时才是黄金身段,

根据题意得x:=,

即x=×≈(m).

答:他的肚脐到脚底的长度为时才是黄金身段.

16.解:(1)在Rt△APD中,AP=1,AD=2,由勾股定理知PD===,∴AM=AF=PF﹣AP=PD﹣AP=﹣1,

DM=AD﹣AM=3﹣.

故AM的长为﹣1,DM的长为3﹣;

(2)点M是AD的黄金分割点.

由于=,

∴点M是AD的黄金分割点.

17.解:∵点P是线段AB的黄金分割点,且AP>BP,

∴AP2=BP×AB,

又∵S1=AP2,S2=PB×AB,

∴S1=S2.

18.解:∵四边形ABCD为平行四边形,

∴∠CBF=∠AEB,∠BCF=∠BAE,

∴△BCF∽△EAB,

∴,即,

把AD=,AB=+1代入得,=,

解得:CF=2.

故答案为:2.

19.解:矩形EFDC是黄金矩形,

证明:∵四边形ABEF是正方形,

又∵,

∴,

即点F是线段AD的黄金分割点.

∴,

∴,

∴矩形CDFE是黄金矩形.

20.解:

(1)满足≈的矩形是黄金矩形;

(2)由=k得,BP=1×k=k,从而AP=1﹣k,

由得,BP2=AP×AB,

即k2=(1﹣k)×1,

解得k=,

∵k>0,

∴k=≈;

(3)因为点P是线段AB的黄金分割点,所以,

设△ABC的AB上的高为h,则

∴直线CP是△ABC的黄金分割线.

(4)由(2)知,在BC边上也存在这样的黄金分割点Q,则AQ也是黄金分割线,设AQ与CP交于点W,则过点W 的直线均是△ABC的黄金分割线,故黄金分割线有无数条.

设选择的高跟鞋的高度是xcm,则根据黄金分割的定义得:=,

解得:x≈.

故她应该选择左右的高跟鞋穿上看起来更美.

22.解:设正方形ABCD的边长为2a,

在Rt△AEB中,依题意,得AE=a,AB=2a,

由勾股定理知EB==a,

∴AH=AF=EF﹣AE=EB﹣AE=(﹣1)a,

HB=AB﹣AH=(3﹣)a;

∴AH2=(6﹣2)a2,

AB?HB=2a×(3﹣)a=(6﹣2)a2,

∴AH2=AB?HB,

所以点H是线段AB的黄金分割点.

23.证明:设正方形ABCD的边长为2,

E为BC的中点,

∴BE=1

∴AE==,

又∵B′E=BE=1,

∴AB′=AE﹣B′E=﹣1,

∴AB″

∴点B″是线段AB的黄金分割点.

24.证明:∵正方形ABCD的边长为2,E为BC的中点,

∴BE=1

∵EF=BE=1,

∴AF=AE﹣EF=﹣1,

∴A M=AF=﹣1,

∴AM:AB=(﹣1):2,

∴点M是线段AB的黄金分割点.

25.解:(1)∵BD=DC=AC.

则∠B=∠DCB,∠CDA=∠A.

设∠B=x,则∠DCB=x,∠CDA=∠A=2x.

又∠BOC=108°,

∴∠B+∠A=108°.

∴x+2x=108,x=36°.

∴∠B=36°;

(2)①有三个:△BDC,△ADC,△BAC.

∵DB=DC,∠B=36°,

∴△DBC是黄金三角形,

(或∵CD=CA,∠ACD=180°﹣∠CDA﹣∠A=36°.∴△CDA是黄金三角形.

或∵∠ACE=108°,

∴∠ACB=72°.又∠A=2x=72°,

∴∠A=∠ACB.

∴BA=BC.

∴△BAC是黄金三角形.

②△BAC是黄金三角形,

∴,

∵BC=2,∴AC=﹣1.

∵BA=BC=2,BD=AC=﹣1,

∴AD=BA﹣BD=2﹣(﹣1)=3﹣,

ⅰ)以CD为底边的黄金三角形:作CD的垂直平分线分别交直线AB、BC得到点P1、P2.ⅱ)以CD为腰的黄金三角形:以点C为圆心,CD为半径作弧与BC的交点为点 P3.

26.证明:在正方形ABCD中,取AB=2a,

∵N为BC的中点,

∴NC=BC=a.

在Rt△DNC中,.

又∵NE=ND,

∴CE=NE﹣NC=(﹣1)a.

∴.

故矩形DCEF为黄金矩形.

27.解:(1)

(2)CM=AB(4分)

28.证明:如图,连接GF,设正方形ABCD的边长为1,则DF=.

在Rt△BCF中,BF==,

则A′F=BF﹣BA′=﹣1.

设AG=A′G=x,则GD=1﹣x,

在Rt△A′GF和Rt△DGF中,有A'F2+A'G2=DF2+DG2,

解得x=,

即点G是AD的黄金分割点(AG>GD).

29.解:(1)如图所示;

(2)△BCD是黄金三角形.

证明如下:∵点D在AB的垂直平分线上,∴AD=BD,

∴∠ABD=∠A.

∵∠A=36°,AB=AC,

∴∠ABC=∠C=72°,

∴∠ABD=∠DBC=36°.

又∵∠BDC=∠A+∠ABD=72°,

∴∠BDC=∠C,

∴BD=BC,

∴△BCD是黄金三角形.

(3)设BC=x,AC=y,

由(2)知,AD=BD=BC=x.

∵∠DBC=∠A,∠C=∠C,

∴△BDC∽△ABC,

整理,得x2+xy﹣y2=0,

解得.

因为x、y均为正数,所以.

(4).

理由:延长BC到E,使CE=AC,连接AE.

∵∠A=36°,AB=AC,

∴∠ACB=∠B=72°,

∴∠ACE=180°﹣72°=108°,

∴∠ACE=∠B1A1C1.

∵A1B1=AB,

∴AC=CE=A1B1=A1C1,

∴△ACE≌△B1A1C1,

∴AE=B1C1.

由(3)知,

∴,,

∴.30.解:(1)直线CD是△ABC的黄金分割线.理由如下:设△ABC的边AB上的高为h.

则,,,∴,.

又∵点D为边AB的黄金分割点,

∴,

∴.

故直线CD是△ABC的黄金分割线.

∴,即,

故三角形的中线不可能是该三角形的黄金分割线.

(3)∵DF∥CE,

∴△DFC和△DFE的公共边DF上的高也相等,

∴S△DFC=S△DFE,

∴S△ADC=S△ADF+S△DFC=S△ADF+S△DFE=S△AEF,S△BDC=S四边形BEFC.

又∵,

∴.

因此,直线EF也是△ABC的黄金分割线.(7分)

(4)画法不惟一,现提供两种画法;

画法一:如答图1,取EF的中点G,再过点G作一条直线分别交AB,DC于M,N点,则直线MN就是平行四边形ABCD 的黄金分割线.

画法二:如答图2,在DF上取一点N,连接EN,再过点F作FM∥NE交AB于点M,连接MN,则直线MN就是平行四边形ABCD的黄金分割线.

(9分

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