Difference between revisions of "User talk:1=2"

Revision as of 18:14, 6 July 2011

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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+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.
+**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
+<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>.

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Difference between revisions of "User talk:1=2"

Revision as of 18:14, 6 July 2011