Problem 5

In tetrahedron

ABCD

\angle BDC=90^\circ

. The foot

H

of the perpendicular from

D

to plane

ABC

is the intersection of the altitudes of

\triangle ABC

. Prove that

(AB+BC+CA)^2\le6(AD^2+BD^2+CD^2)

. For what tetrahedra does equality hold?

Step 1 of 4: Use the orthocenter to introduce E

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E=CH\cap AB,\quad AB\perp ED,\quad BC^2-BD^2=CE^2-DE^2

A tetrahedron for the orthocenter construction

Detailed analysis

Let $E=CH\cap AB$ . The planes CDE and ABC are perpendicular, and $AB$ is perpendicular to their intersection $CE$ ; hence $AB\perp$ plane $CDE$ and $AB\perp ED$ . Therefore $BD^2=DE^2+BE^2$ and $BC^2=CE^2+BE^2$ , so $BC^2-BD^2=CE^2-DE^2$ .