一道超难的数学题俺们全年级的老师都不会,谁会啊?帮个忙

2024-12-27 20:19:26
推荐回答(3个)
回答1:

“等腰梯形”+“AC、BD交于O”+“角ACD=60” => “等边三角形OCD、OAB”
“等边三角形OCD、OAB”+“P、S是OA、OD的中点” => “CD:AB=OS:OP”

S(AOD):S(PQS)=5:4 =>
S(POS):S(PQS)=5:16 =>
[1/2*SIN(120)*OP*OS]:[1/2*SIN(60)*PS^2]=OP*OS/PS^2=5:16 =>
OP*OS/[OP^2+OS^2-2*OP*OS*COS(120)]=5:16 =>
(a^2+b^2+ab)/ab=16/5 (OS=a,OP=b,c=a/b) =>
a^2+b^2-(11/5)*ab=0 =>
c^2-(11/5)c+1=0 =>
c=[11/5 "+/-" sqrt(0.84)]/2=
1.5582575694955840006588047193728 或
0.6417424305044159993411952806272 这两数互为倒数,取小者
即 c=0.6417424305044159993411952806272
(其中△PQS为等边三角形未证明。我很久没接触过这些了,不知道一些使用的公式有没有记错,你自己检查吧。)

回答2:

S(AOD):S(PQS)=5:4 =>
S(POS):S(PQS)=5:16 =>
[1/2*SIN(120)*OP*OS]:[1/2*SIN(60)*PS^2]=OP*OS/PS^2=5:16 =>
OP*OS/[OP^2+OS^2-2*OP*OS*COS(120)]=5:16 =>
(a^2+b^2+ab)/ab=16/5 (OS=a,OP=b,c=a/b) =>
a^2+b^2-(11/5)*ab=0 =>
c^2-(11/5)c+1=0 =>
c=[11/5 "+/-" sqrt(0.84)]/2=
1.5582575694955840006588047193728 或
0.6417424305044159993411952806272 这两数互为倒数,取小者
即 c=0.6417424305044159993411952806272

回答3:

可知三角形AOB,DOC是等边三角形,s是中点,CS垂直于DO,PB垂直与AO,可知角AOB=60°