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 3 of 4: Sum the three Pythagorean identities

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AB^2+BC^2+CA^2=2(AD^2+BD^2+CD^2)

Detailed analysis

The right angles $ADB$ , $BDC$ , and $CDA$ give $AB^2=AD^2+BD^2$ , $BC^2=BD^2+CD^2$ , and $CA^2=CD^2+AD^2$ . Adding them yields $AB^2+BC^2+CA^2=2(AD^2+BD^2+CD^2)$ .