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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 that <math>AX/AD = k/(k+1).</math> Similarly
+
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>
Line 21: Line 21:
 
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> Thus
+
<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 $ABCD$ 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 $P$.

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) $k^2(k + 3)/(3k + 1).$


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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