Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.7 AIME, harder Prove it Russia

The base of a pyramid $SABCD$ is a parallelogram $ABCD$ . Prove that for each point inside $SABCD$ , the sum of volumes of tetrahedra $OSAB$ and $OSCD$ is equal to the sum of volumes of tetrahedra $OSBC$ and $OSDA$ .

В основании четырёхугольной пирамиды $SABCD$ лежит параллелограмм $ABCD$ . Докажите, что для любой точки $O$ внутри пирамиды сумма объёмов тетраэдров $OSAB$ и $OSCD$ равна сумме объёмов тетраэдров $OSBC$ и $OSDA$ .

Solution

Let $X$ be the intersection point of the ray $SO$ with the plane $ABCD$ (see Fig. 10). Since the point $O$ lies inside the pyramid, the point $X$ lies inside its base. At the same time, $S_{XAB} + S_{XCD} = S_{XBC} + S_{XDA}$ (one possible proof of this fact can be seen from Fig. 11—each sum equals half the area of the parallelogram $ABCD$ ). Therefore,

$V_{XSAB} + V_{XSCD} = V_{XSBC} + V_{XSDA}, \quad (1)$

since the height of these pyramids, dropped from vertex $S$ , is common.

Similarly,

$V_{XOAB} + V_{XOCD} = V_{XOBC} + V_{XODA}. \quad (2)$

Subtracting equality (2) from equality (1), we obtain the required result.

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