Difference between revisions of "1965 IMO Problems/Problem 3"
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Since the distance of <math>AB</math> from <math>WXYZ</math> is <math>k</math> times the distance of <math>CD</math>, | Since the distance of <math>AB</math> from <math>WXYZ</math> is <math>k</math> times the distance of <math>CD</math>, | ||
− | we have that <math>AX = k \cdot XD</math> and hence | + | we have that <math>AX = k \cdot XD</math> and hence <math>AX/AD = k/(k+1).</math> Similarly |
<math>AP/AB = AW/AC = AX/AD.</math> <math>XY</math> is parallel to <math>AB</math>, so also | <math>AP/AB = AW/AC = AX/AD.</math> <math>XY</math> is parallel to <math>AB</math>, so also | ||
<math>AX/AD = BY/BD = BZ/BC.</math> | <math>AX/AD = BY/BD = BZ/BC.</math> | ||
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sides <math>k/(k + 1)</math> times smaller and hence area <math>k^2/(k + 1)^2</math> times | sides <math>k/(k + 1)</math> times smaller and hence area <math>k^2/(k + 1)^2</math> times | ||
smaller. Its height is <math>1/(k + 1)</math> times the height of <math>A</math> above | smaller. Its height is <math>1/(k + 1)</math> times the height of <math>A</math> above | ||
− | <math>ABCD,</math> so vol prism <math>= 3 k^2/(k + 1)^3</math> vol <math>ABCD.</math> | + | <math>ABCD,</math> so vol prism <math>= 3 k^2/(k + 1)^3</math> vol <math>ABCD.</math> |
− | vol <math>ABWXYZ = (k^3 + 3k^2)/(k + 1)^3</math> vol <math>ABCD.</math> | + | |
+ | Thus vol <math>ABWXYZ = (k^3 + 3k^2)/(k + 1)^3</math> vol <math>ABCD.</math> | ||
We get the volume of the other piece as vol <math>ABCD\ -</math> vol <math>ABWXYZ,</math> and | We get the volume of the other piece as vol <math>ABCD\ -</math> vol <math>ABWXYZ,</math> and |
Revision as of 14:15, 7 November 2024
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 at
,
at
,
at
and
at
.
Take a plane parallel to
through
and let it meet
in
.
Since the distance of from
is
times the distance of
,
we have that
and hence
Similarly
is parallel to
, so also
vol vol
vol
is similar to the
tetrahedron
The sides are
times smaller, so
vol
vol
The base of the prism is
which is similar to
with
sides
times smaller and hence area
times
smaller. Its height is
times the height of
above
so vol prism
vol
Thus vol vol
We get the volume of the other piece as vol vol
and
hence the ratio is (after a little manipulation)
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 |