User talk:1=2

1966 IMO Problem 3 Progress

Let the tetrahedron be $ABCD$ , and let the circumcenter be $O$ . Therefore $OA=OB=OC=OD$ . It's not hard to see that $O$ is the intersection of the planes that perpendicularly bisect the six sides of the tetrahedron. Label the plane that bisects $AB$ as $p_{AB}$ , and define $p_{AC}$ , $p_{BC}$ , $p_{AD}$ , $p_{BD}$ , and $p_{CD}$ similarly. Now consider $p_{AB}$ , $p_{BC}$ , and $p_{CA}$ . They perpendicularly bisect three coplanar segments, so their intersection is a line perpendicular to the plane containing $ABC$ . Note that the circumcenter of $ABC$ is on this line.

- solving easier version**

I shall prove that given triangle $ABC$ , the minimum value of $PA+PB+PC$ is given when $P$ is the circumcenter of $ABC$ . It's easy to see that, from the Pythagorean Theorem, that if $P$ is not on the plane containing $ABC$ , then the projection $P'$ of $P$ onto said plane satisfies

$P'A+P'B+P'C<PA+PB+PC$

so it suffices to consider all points $P$ that are coplanar with $ABC$ .

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