In a regular triangular pyramid SABC (S is the vertex), E is the midpoint of the slant height of the face SBC. The points F, L and M respectively lie on the edges AB, AC and SC and, AC = 10*AL.
It is known that EFLM is an isosceles trapezoid and the length of its base EF is equal to √7.
Find the volume of the pyramid.
The volume of SABC is e²√2/12 where e is the edge length of SABC.
All we need to do is find the value of e from givens in the problem.
Let
PQ denote the vector from point P to the point Q and
* the vector dot product. Here are the givens:
(1) SABC is a regular triangular pyramid (regular tetrahedron).
(2) E is the midpoint of the slant height of face SBC (if D is the midpoint of BC, then E is the midpoint of SD).
(3) Points F, L, and M respectively lie on edges AB, AC, and SC.
(4) AC = 10*AL (not needed).
(5) EFLM is an isosceles trapezoid (all that is needed is that EF and ML are parallel).
(6) EF = √7.
From (1),
If X,Y are in {A,B,C}; then
SX*SY = e² if X=Y, = e²/2 otherwise.
From (2),
SE = ½
SD = ¼
SB + ¼
SC.
From (3),
SF = x
SA + (1x)
SB, (8)
SL = y
SA + (1y)
SC, and (9)
SM = z
SC (10)
for 0 ≤ x,y,z ≤ 1.
From (7) & (8),
EF =
SF  SE = [x
SA + (1x)
SB]  [¼
SB + ¼
SC]
= x
SA + (¾  x)
SB  ¼
SC. (11)
From (9) & (10),
ML =
SL 
SM = [y
SA + (1y)
SC]  z
SC = y
SA + (1  y  z)
SC. (12)
From (5), (11), and (12),
EF = w
ML for some real w
→
x
SA + (¾  x)
SB  ¼
SC = w[y
SA + (1  y  z)
SC]
→ (x  wy)
SA + (¾  x)
SB  [w(1  y  z)
+ ¼]
SC =
0. (13)
Since the vectors
SA,
SB, and
SC are linearly independent,
all the coefficients in (13) are zero. From (11) we see that
we only need that of
SB (therefore x = ¾). Thus,
EF = ¼(3
SA 
SC).
From (6),
7 = EF² =
EF*EF = (3
SA 
SC)
*(3
SA 
SC)/16
= (9
SA*SA  6
SA*SC +
SC*SC)/16
= e²(9  3 + 1)/16 = 7e²/16
Therefore, e = 4 and
Volume(SABC) = 4³√2/12 = 16√2/3.
QEDEdited on June 15, 2012, 1:36 am
Edited on June 15, 2012, 1:51 am

Posted by Bractals
on 20120615 01:14:35 