Difference between revisions of "1965 IMO Problems/Problem 3"
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== Problem == | == Problem == | ||
− | Given the tetrahedron <math>ABCD</math> whose edges <math>AB</math> and <math>CD</math> have lengths <math>a</math> and <math>b</math> respectively. The distance between the skew lines <math>AB</math> and <math>CD</math> is <math>d</math>, and the angle between them is <math>\omega </math>. Tetrahedron <math>ABCD</math> is divided into two solids by plane <math>\varepsilon </math>, parallel to lines <math>AB</math> and <math>CD</math>. The ratio of the distances of <math>\varepsilon </math> from <math>AB</math> and <math>CD</math> is equal to <math>k</math>. Compute the ratio of the volumes of the two solids obtained. | + | |
+ | Given the tetrahedron <math>ABCD</math> whose edges <math>AB</math> and <math>CD</math> have lengths <math>a</math> and <math>b</math> respectively. The distance between the skew lines <math>AB</math> and <math>CD</math> is <math>d</math>, and the angle between them is <math>\omega</math>. Tetrahedron <math>ABCD</math> is divided into two solids by plane <math>\varepsilon</math>, parallel to lines <math>AB</math> and <math>CD</math>. The ratio of the distances of <math>\varepsilon</math> from <math>AB</math> and <math>CD</math> is equal to <math>k</math>. Compute the ratio of the volumes of the two solids obtained. | ||
+ | |||
== Solution == | == Solution == | ||
− | {{ | + | |
+ | Let the plane meet <math>AD</math> at <math>X</math>, <math>BD</math> at <math>Y</math>, <math>BC</math> at <math>Z</math> and <math>AC</math> at <math>W</math>. | ||
+ | Take a plane parallel to <math>BCD</math> through <math>WX</math> and let it meet <math>AB</math> in <math>P</math>. | ||
+ | |||
+ | [[File:Prob_1965_3.png|600px]] | ||
+ | |||
+ | 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 <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>AX/AD = BY/BD = BZ/BC.</math> | ||
+ | |||
+ | vol <math>ABWXYZ =</math> vol <math>APWX +</math> vol <math>WXPBYZ.</math> <math>APWX</math> is similar to the | ||
+ | tetrahedron <math>ABCD.</math> The sides are <math>k/(k + 1)</math> times smaller, so | ||
+ | vol <math>APWX = k^3/(k + 1)^3</math> vol <math>ABCD.</math> | ||
+ | |||
+ | The base of the prism <math>WXPBYZ</math> is <math>BYZ</math> which is similar to <math>BCD</math> with | ||
+ | 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 | ||
+ | <math>ABCD,</math> so vol prism <math>= 3 k^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 | ||
+ | hence the ratio is (after a little manipulation) <math>k^2(k + 3)/(3k + 1).</math> | ||
+ | |||
+ | |||
+ | == Remark (added by pf02, November 2024) == | ||
+ | |||
+ | Note that the problem is untypically sloppy or misleading. | ||
+ | It mentions the sizes <math>a, b, d, \omega</math> as if they are needed. | ||
+ | In fact, as the solution above shows, they are not needed either | ||
+ | in expressing the result, or in solving the problem. A thorough | ||
+ | problem solver might worry about not having used all the data in | ||
+ | the problem. | ||
+ | |||
+ | However, one can imagine other solutions, where these quantities | ||
+ | would be used in the process of solving the problem. For example | ||
+ | one could break up the doubly truncated triangular prism <math>ABWXYZ</math> | ||
+ | into pyramids <math>APWX, PXYZW, BPZY</math>. Computing the volume of each | ||
+ | of these pyramids would require all the data in the problem. | ||
+ | The end result should of course be the same, but the thorough | ||
+ | problem solver would not have the uneasy feeling of not having | ||
+ | used all the data in the problem. | ||
+ | |||
+ | |||
+ | == See Also == | ||
+ | {{IMO box|year=1965|num-b=2|num-a=4}} | ||
+ | |||
+ | [[Category:Olympiad Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Latest revision as of 18:09, 10 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)
Remark (added by pf02, November 2024)
Note that the problem is untypically sloppy or misleading.
It mentions the sizes as if they are needed.
In fact, as the solution above shows, they are not needed either
in expressing the result, or in solving the problem. A thorough
problem solver might worry about not having used all the data in
the problem.
However, one can imagine other solutions, where these quantities
would be used in the process of solving the problem. For example
one could break up the doubly truncated triangular prism
into pyramids
. Computing the volume of each
of these pyramids would require all the data in the problem.
The end result should of course be the same, but the thorough
problem solver would not have the uneasy feeling of not having
used all the data in the problem.
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 |