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BCHL85
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#1
Report Thread starter 15 years ago
#1
I found that there are not many problems about that here. So let's start with some ...

1)A.BCD is the tetrahedron. M, N is the mid points of BC and AD.
AB = CD = 2a. Angle between AC and CD is 60 degree.
(P) is the plane passes through MN and parallel to AB.
Find the area that (P) makes with the tetrahedron.
(it's not so hard question)

2)S.ABC is tetrahedron with ABC is the right isosceles triangle at B, BC = BA = a, SA is perpendicular to plane (ABC) and SA = rt(2).
(P) is the plane that passes through the mid-point of SB and perpendicular to SB.
O is the mid point of BC.
d is the line passes through O and perpendicular to the plane (ABC).
Show how to find the intersect point K of d and plane (P).
Find size of OK.

I'll post more ... if someone is interested in
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misshn
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#2
Report 15 years ago
#2
(Original post by BCHL85)
I found that there are not many problems about that here. So let's start with some ...

1)A.BCD is the tetrahedron. M, N is the mid points of BC and AD.
AB = CD = 2a. Angle between AC and CD is 60 degree.
(P) is the plane passes through MN and parallel to AB.
Find the area that (P) makes with the tetrahedron.
(it's not so hard question)
I tried and just wondered if there would be any mistake in this question, especially the line angle between Ac and CD is 60 degree. Could it be AB and CD? If so then the answer is rt(3).(a^2)/2.
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misshn
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#3
Report 15 years ago
#3
(Original post by BCHL85)
2)S.ABC is tetrahedron with ABC is the right isosceles triangle at B, BC = BA = a, SA is perpendicular to plane (ABC) and SA = rt(2).
(P) is the plane that passes through the mid-point of SB and perpendicular to SB.
O is the mid point of BC.
d is the line passes through O and perpendicular to the plane (ABC).
Show how to find the intersect point K of d and plane (P).
Find size of OK.
Let M,N,P be the mid points of SB, SC and AC respectively.
BC is perpendicular(perp.) to AB; BC is perp. to SA=>BC is perp. to SB
MN//BC =>MN is perp. to SB => (P) passes through M,N.
ON//SB=>ON is perp. to BC; OK is perp. to BC =>NK is perp. to BC and MN
=> K is on the line through N in (P) which is perp. to MN
Besides, OK is perp. to (ABC) so OK//SA//NP =>O,K,P,N are coplanar.
Consider the trapezium OKPN(OK//NP and perp. to OP)
NK is perp. to SB=>NK is perp. to ON => Angle ONK is right angle.
So K is the intersection of the line through N, perp. to ON and the line through O, parallel to ON in (ONP).

To find the size of OK: As PN//OK, angle PNO = angle NOK =>Triangle PNO is similar to triangle NOK
=>ON/PN=OK/ON=>OK=(ON^2)/PN
PN=SA/2=1/rt(2) and ON=rt(PN^2+OP^2)=rt(1/2+a^2/4)
=>OK=(a^2+2)/4rt(2)

Is it right?
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