1964 IMO Problems/Problem 6

Problem

In tetrahedron $ABCD$ , vertex $D$ is connected with $D_0$ , the centroid of $\triangle ABC$ . Lines parallel to $DD_0$ are drawn through $A,B$ and $C$ . These lines intersect the planes $BCD, CAD$ and $ABD$ in points $A_1, B_1,$ and $C_1$ , respectively. Prove that the volume of $ABCD$ is one third the volume of $A_1B_1C_1D_0$ . Is the result true if point $D_o$ is selected anywhere within $\triangle ABC$ ?

Solution

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Let $A_{2}$ be the point where line $AD_{0}$ intersects line $BC$

Let $B_{2}$ be the point where line $BD_{0}$ intersects line $AC$

Let $C_{2}$ be the point where line $CD_{0}$ intersects line $AB$

From centroid properties we have:

$|AA_{2}|=3|D_{0}A_{2}|$

$|BB_{2}|=3|D_{0}B_{2}|$

$|CC_{2}|=3|D_{0}C_{2}|$

Therefore,

$\frac{|AA_{2}|}{|D_{0}A_{2}|}=3$

$\frac{|BB_{2}|}{|D_{0}B_{2}|}=3$

$\frac{|CC_{2}|}{|D_{0}C_{2}|}=3$

Since $\Delta D_{0}A_{2}A_{1}\sim \Delta AA_{2}A_{1}$ , then $|AA_{1}|=|DD_{0}| \frac{|AA_{2}|}{|D_{0}A_{2}|}=3|DD_{0}|$

Since $\Delta D_{0}B_{2}B_{1}\sim \Delta BB_{2}B_{1}$ , then $|BB_{1}|=|DD_{0}| \frac{|BB_{2}|}{|D_{0}B_{2}|}=3|DD_{0}|$

Since $\Delta D_{0}C_{2}C_{1}\sim \Delta CC_{2}C_{1}$ , then $|CC_{1}|=|DD_{0}| \frac{|CC_{2}|}{|D_{0}C_{2}|}=3|DD_{0}|$

Since $|AA_{2}|=|BB_{2}|=|CC_{2}|$ and $AA_{1} \parallel BB_{1} \parallel CC_{1} \parallel DD_{0}$ ,

then $\Delta A_{1}B_{1}C_{1}\parallel \Delta ABC$ , and $Area_{\Delta A_{1}B_{1}C_{1}}=Area_{\Delta ABC}$

Let $h_{D}$ be the perpendicular distance from $D$ to $\Delta ABC$

Let $h_{\Delta A_{1}B_{1}C_{1}}$ be the perpendicular distance from $\Delta A_{1}B_{1}C_{1}$ to $\Delta ABC$

$\frac{h_{\Delta A_{1}B_{1}C_{1}}}{h_{D}}=\frac{|AA_{2}|}{|D_{0}A_{2}|}=\frac{|BB_{2}|}{|D_{0}B_{2}|}=\frac{|CC_{2}|}{|D_{0}C_{2}|}=3$

$h_{\Delta A_{1}B_{1}C_{1}}=3h_{D}$

$\frac{1}{3}h_{\Delta A_{1}B_{1}C_{1}}Area_{\Delta A_{1}B_{1}C_{1}}=3\frac{1}{3}h_{D}Area_{\Delta ABC}$

$\frac{h_{\Delta A_{1}B_{1}C_{1}}Area_{\Delta A_{1}B_{1}C_{1}}}{3}=3\frac{h_{D}Area_{\Delta ABC}}{3}$

Since $Volume_{ABCD}=\frac{h_{D}Area_{\Delta ABC}}{3}$ and $Volume_{A_{1}B_{1}C_{1}D_{0}}=\frac{h_{\Delta A_{1}B_{1}C_{1}}Area_{\Delta A_{1}B_{1}C_{1}}}{3}$

then, $Volume_{A_{1}B_{1}C_{1}D_{0}}=3Volume_{ABCD}$

thus, $Volume_{ABCD}=\frac{1}{3}Volume_{A_{1}B_{1}C_{1}D_{0}}$

this proves that the volume of $ABCD$ is one third the volume of $A_1B_1C_1D_0$

The result is NOT true if point $D_o$ is selected anywhere within $\triangle ABC$ as ratios of $\frac{|AA_{2}|}{|D_{0}A_{2}|}$ , $\frac{|BB_{2}|}{|D_{0}B_{2}|}$ , and $\frac{|CC_{2}|}{|D_{0}C_{2}|}$ will have values other than 3 as the point is no longer a centroid. Also, it will make the ratios $\frac{|AA_{2}|}{|D_{0}A_{2}|} \ne \frac{|BB_{2}|}{|D_{0}B_{2}|}\ne \frac{|CC_{2}|}{|D_{0}C_{2}|}$ which means that $\Delta A_{1}B_{1}C_{1} \nparallel \Delta ABC$ and the volume relationship will no longer hold true.

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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1964 IMO Problems/Problem 6

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