It is given a tetrahedron $ABCD$ for which two points of opposite edges are mutually perpendicular. Prove that: (a) the four altitudes of $ABCD$ intersects at a common point $H$; (b) $AH+BH+CH+DH<p+2R$, where $p$ is the sum of the lengths of all edges of $ABCD$ and $R$ is the radii of the sphere circumscribed around $ABCD$. H. Lesov
Problem
Source: Bulgaria 1972 P6
Tags: geometry, 3D geometry, tetrahedron, inequalities, geometrical inequalities
mihaig
21.06.2021 11:21
bump to this splendor
kiyoras_2001
22.06.2024 23:50
First let me fix the wording jasperE3 wrote: It is given a tetrahedron $ABCD$ for which two pairs of opposite edges are mutually perpendicular. Prove that: (a) the four altitudes of $ABCD$ intersect at a common point $H$; (b) $AH+BH+CH+DH<p+2R$, where $p$ is the sum of the lengths of all edges of $ABCD$ and $R$ is the radius of the sphere circumscribed around $ABCD$. H. Lesov (a) is pretty well-known. As for (b), I believe there is a mistake. For tetrahedrons \(ABCD\) with \(H\) inside (or on the border) the given inequality is indeed true:
Suppose that for example \(\angle BDC\ge 90^\circ\). Let \(H_A\) be the orthocenter of \(BCD\). Then \(\angle BAC<\angle BH_AC\le 90^\circ\). Also the projection of \(A\) on \(BC\) coincides with the projection of \(D\) on \(BC\) which lies on the segment \(BC\). Hence the other two angles of \(ABC\) are acute too.
). Let \(H_D\) be the orthocenter of \(ABC\) and let \(H'\) be such that \(\overrightarrow{H_DH'}=-2\overrightarrow{H_DH}\). It is known that \(H'\) lies on the circumsphere of \(ABCD\). Clearly \(H\) is further from \(ABC\) than \(D\). We have \[AH^2 = AH_D^2 + H_DH^2 = AH_D^2 + \frac{H_DH'^2}{4}.\]Consider the tetrahedron \(ABCD\) for which \(ABC\) is equilateral. Let \(D\) vary along the line through \(H_D\) perpendicular to \(ABC\). Then as \(D\) tends to \(H_D\) we have \(AH \sim \frac{H_DH'}{2} \sim R\). Thus \[\sum AH = 4AH\sim 4R\]which means that the inequality \(\sum AH < p + k R\) fails for all \(k<4\). However this inequality is already true for \(k=4\) in general: \[AH^2 = AH_D^2+\frac{H_DH'^2}{4} = AD^2-H_DD^2 + \frac{H_DH'^2}{4} < AD^2 + \frac{H_DH'^2}{4} < \left(AD + \frac{H_DH'}{2}\right)^2<(AD+R)^2\]whence \[\sum AH < p + 4R.\]