IMO 1970 · Problem 5

In the tetrahedron \(ABCD\), angle \(BDC\) is a right angle. Suppose that the foot \(H\) of the perpendicular from \(D\) to the plane \(ABC\) 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?

Official text of the problem. Official problem paper (PDF)

Answer

Equality holds precisely when the three edges \(DA,DB,DC\) are mutually perpendicular and have equal lengths.

Solution

Take \(D\) as the origin and write \[ \mathbf a=\overrightarrow{DA},\qquad \mathbf b=\overrightarrow{DB},\qquad \mathbf c=\overrightarrow{DC},\qquad \mathbf h=\overrightarrow{DH}. \] We first prove that \(\mathbf a,\mathbf b,\mathbf c\) are pairwise perpendicular.

Since \(DH\) is perpendicular to the plane \(ABC\), \[ \mathbf h\cdot(\mathbf b-\mathbf c)=0. \] Also, because \(H\) lies on the altitude from \(A\), \[ (\mathbf a-\mathbf h)\cdot(\mathbf b-\mathbf c)=0. \] Adding gives \[ \mathbf a\cdot\mathbf b=\mathbf a\cdot\mathbf c. \] Applying the same argument to the altitude from \(B\) yields \[ \mathbf b\cdot\mathbf a=\mathbf b\cdot\mathbf c. \] Thus all three pairwise scalar products are equal. The hypothesis \(\angle BDC=90^\circ\) gives \(\mathbf b\cdot\mathbf c=0\), so \[ \mathbf a\cdot\mathbf b =\mathbf a\cdot\mathbf c =\mathbf b\cdot\mathbf c=0. \] These equations remain valid regardless of the location of \(H\).

Consequently, by the Pythagorean theorem, \[ AB^2=AD^2+BD^2,\qquad BC^2=BD^2+CD^2,\qquad CA^2=CD^2+AD^2. \] The Cauchy–Schwarz inequality now gives \[ \begin{aligned} (AB+BC+CA)^2 &\leq 3(AB^2+BC^2+CA^2)\\ &=6(AD^2+BD^2+CD^2), \end{aligned} \] as required.

Equality holds if and only if \(AB=BC=CA\). By the three Pythagorean identities, this is equivalent to \(AD=BD=CD\). Together with the perpendicularity already proved, this gives the stated characterization.

To verify that these equality configurations satisfy all the hypotheses, take, for any \(t>0\), \[ D=(0,0,0),\quad A=(t,0,0),\quad B=(0,t,0),\quad C=(0,0,t). \] The plane \(ABC\) is \(x+y+z=t\), so the perpendicular foot from \(D\) is \[ H=(t/3,t/3,t/3). \] Here \[ \overrightarrow{AH}\cdot\overrightarrow{BC} =(-2t/3,t/3,t/3)\cdot(0,-t,t)=0, \] and the corresponding identities hold cyclically. Hence \(H\) is the orthocentre of \(ABC\); also \(\angle BDC=90^\circ\). All three sides of \(ABC\) equal \(t\sqrt2\), so equality indeed holds.

The solution was checked against John Scholes' IMO solution archive (Kalva).