1970 IMO Problems/Problem 5

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Problem

In the tetrahedron $ABCD$, angle $BDC$ is a right angle. Suppose that the foot $H$ of the perpendicular from $D$ to the plane $ABC$ in the tetrahedron is the intersection of the altitudes of $\triangle ABC$. Prove that

$( AB+BC+CA )^2 \leq 6( AD^2 + BD^2 + CD^2 )$.

For what tetrahedra does equality hold?

Solution

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1970 IMO (Problems) • Resources
Preceded by
Problem 4
1 2 3 4 5 6 Followed by
Problem 6
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