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Difference between revisions of "1965 IMO Problems/Problem 3"

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== Problem ==
 
== Problem ==
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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 ==
  
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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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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 plane parallel to <math>BCD</math> through <math>WX</math> and let it meet <math>AB</math> in <math>P</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 that <math>AX/AD = k/(k+1).</math> Similarly
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<math>AP/AB = AW/AC = AX/AD.</math> <math>XY</math> is parallel to <math>AB</math>, so also
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<math>AX/AD = BY/BD = BZ/BC.</math>
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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
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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
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<math>ABCD,</math> so vol prism <math>= 3 k^2/(k + 1)^3</math> vol <math>ABCD.</math>  Thus
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vol <math>ABWXYZ = (k^3 + 3k^2)/(k + 1)^3</math> vol <math>ABCD.</math>
  
−
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.
+
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>
  
−
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 14:10, 7 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 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 $AX = k \cdot XD$ 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 $APWX = k^3/(k + 1)^3$ 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) $k^2(k + 3)/(3k + 1).$


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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