(a) Prove that if the six dihedral (i.e. angles between pairs of faces) of a given tetrahedron are congruent, then the tetrahedron is regular. (b) Is a tetrahedron necessarily regular if five dihedral angles are congruent?
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Tags: AMC, USA(J)MO, USAMO, geometry, 3D geometry, tetrahedron, sphere
tc1729
24.08.2013 20:37
(a) Let $ABCD$ be the said tetrahedron, and let the inscribed sphere of $ABCD$ touch the faces at $W, X, Y, Z$. Then, $OW, OX, OY, OZ$ are normals to the respective faces. We know that the angle between any two normals is equal, so we have $|OW|=|OX|=|OY|=|OZ|$ at equal angles. Now, since \[WX=2\cdot OW\sin\frac{\angle{WOX}}{2},\] and similar for the other sides, we have that $WXYZ$ is a regular tetrahedron. Now, the faces of $ABCD$ are the tangent planes at $W$, $X$, $Y$, and $Z$. Then, consider a $120^{\circ}$ rotation about $OW$. The rotation sends $X\mapsto Y$, $Y\mapsto Z$, and $Z\mapsto X$. Thus we have $AB=AC$, $AC=AD$, $AD=AB$, and $BC=CD=DB$. Performing rotations about the other axes yields that $ABCD$ has equal edges, so it is regular.
(b) Consider the normals $OW, OX, OY, OZ$. We can perform a transformation in which we slightly shift $X$, $Y$, and $Z$ closer such that $\triangle XYZ$ is still equilateral, and such that all angles between every pair of normals is less that $180^{\circ}$. Then, shift $W$ such that $WX=WY=XY$. Then five of the distances are equal but the sixth is not.
Dynamite127
06.02.2014 20:00
tc1729 wrote:
(a) Let $ABCD$ be the said tetrahedron, and let the inscribed sphere of $ABCD$ touch the faces at $W, X, Y, Z$. Then, $OW, OX, OY, OZ$ are normals to the respective faces. We know that the angle between any two normals is equal, so we have $|OW|=|OX|=|OY|=|OZ|$ at equal angles. Now, since \[WX=2\cdot OW\sin\frac{\angle{WOX}}{2},\] and similar for the other sides, we have that $WXYZ$ is a regular tetrahedron. Now, the faces of $ABCD$ are the tangent planes at $W$, $X$, $Y$, and $Z$. Then, consider a $120^{\circ}$ rotation about $OW$. The rotation sends $X\mapsto Y$, $Y\mapsto Z$, and $Z\mapsto X$. Thus we have $AB=AC$, $AC=AD$, $AD=AB$, and $BC=CD=DB$. Performing rotations about the other axes yields that $ABCD$ has equal edges, so it is regular.
(b) Consider the normals $OW, OX, OY, OZ$. We can perform a transformation in which we slightly shift $X$, $Y$, and $Z$ closer such that $\triangle XYZ$ is still equilateral, and such that all angles between every pair of normals is less that $180^{\circ}$. Then, shift $W$ such that $WX=WY=XY$. Then five of the distances are equal but the sixth is not.
This may be well-known but can anyone give a rigorous proof of the fact that every tetrahedron has an insphere?
fmasroor
08.02.2014 00:40
Consider 3 faces that meet at a point. The infinite region bounded by these planes that contains the tetrahedron clearly has a sphere that is tangent to all of them. If it's too big, it will lie completely outside the tetrahedron. Continue making the sphere smaller and smaller; sooner or later it will cross the fourth face, and then later it will cross into the tetrahedron. In the middle somewhere it was tangent the 4th face; this is the insphere. (this also proves that there are 4 exspheres)
huashiliao2020
15.04.2023 07:04
tc1729 wrote:
(a) Let $ABCD$ be the said tetrahedron, and let the inscribed sphere of $ABCD$ touch the faces at $W, X, Y, Z$. Then, $OW, OX, OY, OZ$ are normals to the respective faces. We know that the angle between any two normals is equal, so we have $|OW|=|OX|=|OY|=|OZ|$ at equal angles. Now, since \[WX=2\cdot OW\sin\frac{\angle{WOX}}{2},\]and similar for the other sides, we have that $WXYZ$ is a regular tetrahedron. Now, the faces of $ABCD$ are the tangent planes at $W$, $X$, $Y$, and $Z$. Then, consider a $120^{\circ}$ rotation about $OW$. The rotation sends $X\mapsto Y$, $Y\mapsto Z$, and $Z\mapsto X$. Thus we have $AB=AC$, $AC=AD$, $AD=AB$, and $BC=CD=DB$. Performing rotations about the other axes yields that $ABCD$ has equal edges, so it is regular.
(b) Consider the normals $OW, OX, OY, OZ$. We can perform a transformation in which we slightly shift $X$, $Y$, and $Z$ closer such that $\triangle XYZ$ is still equilateral, and such that all angles between every pair of normals is less that $180^{\circ}$. Then, shift $W$ such that $WX=WY=XY$. Then five of the distances are equal but the sixth is not.
Hmm, I got progress up to your WX=2OWsin(<WOX/2), which one way can follow from LoC: WX=sqrt(2x^2-2xcos<WOX)=2xsqrt((1-cos<WOX)/2)=2xsin<(WOX/2), where the last part follows from sine half angle formula, but I didn't understand how you got WXYZ is a regular tetrahedron. Other than that, nice solution! Could someone clarify? Also, I'm wondering if there are alternate solutions to this.