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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.
  
  
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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>.
 
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>.
 
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>,
 
Since the distance of <math>AB</math> from <math>WXYZ</math> is <math>k</math> times the distance of <math>CD</math>,
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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
 
hence the ratio is (after a little manipulation) <math>k^2(k + 3)/(3k + 1).</math>
 
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.
  
  

Latest revision as of 18:09, 10 November 2024

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 a plane parallel to $BCD$ through $WX$ and let it meet $AB$ in $P$.

Prob 1965 3.png

Since the distance of $AB$ from $WXYZ$ is $k$ times the distance of $CD$, we have that $AX = k \cdot XD$ and hence $AX/AD = k/(k+1).$ 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 $APWX = k^3/(k + 1)^3$ 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 $k^2/(k + 1)^2$ times smaller. Its height is $1/(k + 1)$ times the height of $A$ above $ABCD,$ so vol prism $= 3 k^2/(k + 1)^3$ vol $ABCD.$

Thus vol $ABWXYZ = (k^3 + 3k^2)/(k + 1)^3$ vol $ABCD.$

We get the volume of the other piece as vol $ABCD\ -$ vol $ABWXYZ,$ and hence the ratio is (after a little manipulation) $k^2(k + 3)/(3k + 1).$


Remark (added by pf02, November 2024)

Note that the problem is untypically sloppy or misleading. It mentions the sizes $a, b, d, \omega$ 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 $ABWXYZ$ into pyramids $APWX, PXYZW, BPZY$. 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
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