1970 IMO Problems/Problem 5

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

Let us show first that angles ADB and ADC are also right. Let H be the intersection of the altitudes of ABC and let CH meet AB at X. Planes CED and ABC are perpendicular and AB is perpendicular to the line of intersection CE. Hence AB is perpendicular to the plane CDE and hence to ED. So BD^2 = DE^2 + BE^2. Also CB^2 = CE^2 + BE^2. Therefore CB^2 - BD^2 = CE^2 - DE^2. But CB^2 - BD^2 = CD^2, so CE^2 = CD^2 + DE^2, so angle CDE = 90°. But angle CDB = 90°, so CD is perpendicular to the plane DAB, and hence angle CDA = 90°. Similarly, angle ADB = 90°. Hence AB^2 + BC^2 + CA^2 = 2(DA^2 + DB^2 + DC^2). But now we are done, because Cauchy's inequality gives (AB + BC + CA)^2 = 3(AB^2 + BC^2 + CA^2). We have equality if and only if we have equality in Cauchy's inequality, which means AB = BC = CA.

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

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