Difference between revisions of "1965 IMO Problems/Problem 3"
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== Solution == | == Solution == | ||
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| + | Let the plane meet AD at X, BD at Y, BC at Z and AC at W. Take plane parallel to BCD through WX and let it meet AB in P. | ||
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| + | Since the distance of AB from WXYZ is k times the distance of CD, we have that AX = k·XD and hence that AX/AD = k/(k+1). Similarly AP/AB = AW/AC = AX/AD. XY is parallel to AB, so also AX/AD = BY/BD = BZ/BC. | ||
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| + | vol ABWXYZ = vol APWX + vol WXPBYZ. APWX is similar to the tetrahedron ABCD. The sides are k/(k+1) times smaller, so vol APWX = k3(k+1)3 vol ABCD. The base of the prism WXPBYZ is BYZ which is similar to BCD with sides k/(k+1) times smaller and hence area k2(k+1)2 times smaller. Its height is 1/(k+1) times the height of A above ABCD, so vol prism = 3 k2(k+1)3 vol ABCD. Thus vol ABWXYZ = (k3 + 3k2)/(k+1)3 vol ABCD. We get the vol of the other piece as vol ABCD - vol ABWXYZ and hence the ratio is (after a little manipulation) k2(k+3)/(3k+1). | ||
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| + | == See Also == | ||
| + | {{IMO box|year=1965|num-b=2|num-a=4}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
[[Category:3D Geometry Problems]] | [[Category:3D Geometry Problems]] | ||
Latest revision as of 10:52, 30 September 2022
Problem
Given the tetrahedron 
whose edges 
and 
have lengths 
and 
respectively. The distance between the skew lines and is 
, and the angle between them is 
. Tetrahedron is divided into two solids by plane 
, parallel to lines and . The ratio of the distances of from and is equal to 
. Compute the ratio of the volumes of the two solids obtained.
Solution
Let the plane meet AD at X, BD at Y, BC at Z and AC at W. Take plane parallel to BCD through WX and let it meet AB in P.
Since the distance of AB from WXYZ is k times the distance of CD, we have that AX = k·XD and hence that AX/AD = k/(k+1). Similarly AP/AB = AW/AC = AX/AD. XY is parallel to AB, so also AX/AD = BY/BD = BZ/BC.
vol ABWXYZ = vol APWX + vol WXPBYZ. APWX is similar to the tetrahedron ABCD. The sides are k/(k+1) times smaller, so vol APWX = k3(k+1)3 vol ABCD. The base of the prism WXPBYZ is BYZ which is similar to BCD with sides k/(k+1) times smaller and hence area k2(k+1)2 times smaller. Its height is 1/(k+1) times the height of A above ABCD, so vol prism = 3 k2(k+1)3 vol ABCD. Thus vol ABWXYZ = (k3 + 3k2)/(k+1)3 vol ABCD. We get the vol of the other piece as vol ABCD - vol ABWXYZ and hence the ratio is (after a little manipulation) k2(k+3)/(3k+1).
See Also
| 1965 IMO (Problems) • Resources | ||
| Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
| All IMO Problems and Solutions | ||
