1965 IMO Problems/Problem 3

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

1965 IMO (Problems) • Resources
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 4
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1965 IMO Problems/Problem 3

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