View source for 1973 IMO Shortlist Problems/Bulgaria 1

== Problem ==

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

<center>
<math> \frac{AP}{PA_{1}} + \frac{BP}{PB_{1}} + \frac{CP}{PC_{1}} + \frac{DP}{PD_{1}} = 4, </math>
</center>

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

== Solution ==

Let <math>\displaystyle S </math> have center <math>\displaystyle O = (0,0,0)</math> and radius <math>\displaystyle r </math>.  Since the [[power of a point | power]] of <math>\displaystyle P </math> with respect to <math>\displaystyle S </math> is invariant, we may multiply both sides of the condition by that power to obtain

<center>
<math>\displaystyle AP^2 + BP^2 + CP^2 + DP^2 = 4( r^2 - OP^2 ) </math>
</center>

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

<center>
<math> \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} </math>
</center>

If we now let <math>\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 ) </math>, then we may rewrite the expression (using [[scalar product | dot products]]) thus:

<center>
<math> \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 </math>
</center>

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


{{alternate solutions}}

== Resources ==

* [[1973 IMO Shortlist Problems]]
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=586975#586975 Discussion on AoPS/MathLinks]


[[Category:Olympiad Geometry Problems]]