1973 IMO Shortlist Problems/Bulgaria 1


A tetrahedron $\displaystyle ABCD$ is inscribed in the sphere $\displaystyle S$. Find the locus of points $\displaystyle P$, situated in $\displaystyle S$, such that

$\frac{AP}{PA_{1}} + \frac{BP}{PB_{1}} + \frac{CP}{PC_{1}} + \frac{DP}{PD_{1}} = 4,$

where $\displaystyle A_{1}, B_{1}, C_{1}, D_{1}$ are the other intersection points of $\displaystyle AP, BP, CP, DP$ with $\displaystyle S$.


Let $\displaystyle S$ have center $\displaystyle O = (0,0,0)$ and radius $\displaystyle r$. Since the power of $\displaystyle P$ with respect to $\displaystyle S$ is invariant, we may multiply both sides of the condition by that power to obtain

$\displaystyle AP^2 + BP^2 + CP^2 + DP^2 = 4( r^2 - OP^2 )$

We may now use the law of cosines to rewrite the condition thus:

$\begin{matrix} \displaystyle \sum_{\mbox{cyc}}[r^2 + OP^2 - 2 AP \cdot OP \cos (AOP){]} &=& 4 ( r^2 - OP^2 )\\ 4 OP^2 &=& \displaystyle \sum_{\mbox{cyc}}[AP \cdot OP \cos (AOP) {]}\end{matrix}$

If we now let $\displaystyle P = (x,y,z), A = ( a_1 , a_2 , a_3 ), B = ( b_1 , b_2 , b_3 ), C = ( c_1 , c_2 , c_3 ), D = ( d_1 , d_2 , d_3 )$, then we may rewrite the expression (using dot products) thus:

$\displaystyle 4( x^2 + y^2 + z^2 ) = ( a_1 + b_1 + c_1 + d_1 )x + ( a_2 + b_2 + c_2 + d_2 )y + ( a_3 + b_3 + c_3 + d_3 )z$

If we now complete the square for $\displaystyle x, y$, and $\displaystyle z$, it becomes apparent that this is an equation for a sphere centered at the midpoint of the segment with endpoints $\displaystyle O$ and the centroid of $\displaystyle ABCD$ and with radius half the distance from $\displaystyle O$ and the centroid of $\displaystyle ABCD$, which is therefore the desired locus, Q.E.D.

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