Difference between revisions of "User talk:1=2"
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+ | 1966 IMO Problem 3 Progress | ||
+ | Let the tetrahedron be <math>ABCD</math>, and let the circumcenter be <math>O</math>. Therefore <math>OA=OB=OC=OD</math>. It's not hard to see that <math>O</math> is the intersection of the planes that perpendicularly bisect the six sides of the tetrahedron. Label the plane that bisects <math>AB</math> as <math>p_{AB}</math>, and define <math>p_{AC}</math>, <math>p_{BC}</math>, <math>p_{AD}</math>, <math>p_{BD}</math>, and <math>p_{CD}</math> similarly. Now consider <math>p_{AB}</math>, <math>p_{BC}</math>, and <math>p_{CA}</math>. They perpendicularly bisect three coplanar segments, so their intersection is a line perpendicular to the plane containing <math>ABC</math>. Note that the circumcenter of <math>ABC</math> is on this line. | ||
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+ | **solving easier version** | ||
+ | |||
+ | I shall prove that given triangle <math>ABC</math>, the minimum value of <math>PA+PB+PC</math> is given when <math>P</math> is the circumcenter of <math>ABC</math>. It's easy to see that, from the Pythagorean Theorem, that if <math>P</math> is not on the plane containing <math>ABC</math>, then the projection <math>P'</math> of <math>P</math> onto said plane satisfies | ||
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+ | <cmath>P'A+P'B+P'C<PA+PB+PC</cmath> | ||
+ | |||
+ | so it suffices to consider all points <math>P</math> that are coplanar with <math>ABC</math>. |
Revision as of 18:14, 6 July 2011
1966 IMO Problem 3 Progress
Let the tetrahedron be , and let the circumcenter be
. Therefore
. It's not hard to see that
is the intersection of the planes that perpendicularly bisect the six sides of the tetrahedron. Label the plane that bisects
as
, and define
,
,
,
, and
similarly. Now consider
,
, and
. They perpendicularly bisect three coplanar segments, so their intersection is a line perpendicular to the plane containing
. Note that the circumcenter of
is on this line.
- solving easier version**
I shall prove that given triangle , the minimum value of
is given when
is the circumcenter of
. It's easy to see that, from the Pythagorean Theorem, that if
is not on the plane containing
, then the projection
of
onto said plane satisfies
so it suffices to consider all points that are coplanar with
.