A given tetrahedron $ ABCD$ is isoceles, that is, $ AB=CD$, $ AC=BD$, $ AD=BC$. Show that the faces of the tetrahedron are acute-angled triangles.
Problem
Source: 1972 USAMO Problem 2
Tags: geometry, 3D geometry, tetrahedron, inequalities, parallelogram, triangle inequality
futuremospper
06.03.2010 06:27
It follows immediately that the sum of the face angles at any vertex is $ 180^\circ$. Since the sum of any two angles of a trihedral angle is greater than the third, each angle must be acute. If not, assume that $ \triangle BDC$ is not acute. Denote the midpoint of $ \overline{BC}$ by $ M$. Since $ \triangle ABC \cong \triangle DCB$, $ AM=DM$. Consider the circle with $ M$ as center and $ BC$ as diameter. Since $ \angle BDC$ is nonacute, $ C$ must lie within or on the circle. Thus, $ 2DM \le BC$. By the Triangle Inequality on $ \triangle AMD$, $ AM+MD > AD$. But $ AM=MD$ and $ AD=BC$< so we obtain a contradiction. Hence, the assumption was false and the assertion in the problem is true.
tc1729
20.03.2012 00:18
Suppose $\triangle ABD$ is fixed. By the equality conditions, it follows that the maximal possible value of $BC$ occurs when the four vertices are coplanar, with $C$ on the opposite side of $\overline{AD}$ as $B$. In this case, the tetrahedron is not actually a tetrahedron, so this maximum isn't actually attainable. For the sake of contradiction, suppose $\angle ABD$ is non-acute. Then, $(AD)^2\geq (AB)^2+(BD)^2$. In our optimal case noted above, $ACDB$ is a parallelogram, so \begin{align*} 2(BD)^2 + 2(AB)^2 &= (AD)^2 + (CB)^2 \\ &= 2(AD)^2 \\ &\geq 2(BD)^2+2(AB)^2. \end{align*} However, as stated, equality cannot be attained, so we get our desired contradiction. $\Box$
maXplanK
22.03.2018 16:14
Can anyone show this using vectors ?
cucuoanesh
22.03.2018 16:27
Let ABCD be a tetrahedron .If [ABC]=[ABD]=[ACD]=[BCD] show that ABCD is isosceles .
62861
23.01.2019 21:49
This is also a corollary of Ptolemy's inequality (which is valid in 3D): $AB \cdot CD + AC \cdot BD > AD \cdot BC$ so $AB^2 + AC^2 > BC^2$, et cetera. (Inequality is strict since the points are not coplanar.)