Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1965 IMO Problems/Problem 3"
(→‎Solution)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
−
{{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.
 
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.
  
Line 9: Line 9:
  
 
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).
 
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 ==  
 
== See Also ==  

Revision as of 09:52, 30 September 2022

Problem

Given the tetrahedron $ABCD$ whose edges $AB$ and $CD$ have lengths $a$ and $b$ respectively. The distance between the skew lines $AB$ and $CD$ is $d$, and the angle between them is $\omega$. Tetrahedron $ABCD$ is divided into two solids by plane $\varepsilon$, parallel to lines $AB$ and $CD$. The ratio of the distances of $\varepsilon$ from $AB$ and $CD$ is equal to $k$. 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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1965_IMO_Problems/Problem_3&oldid=178563"