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Difference between revisions of "1978 USAMO Problems/Problem 4"
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== Solution ==
 
== Solution ==
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(a) Let <math>ABCD</math> be the said tetrahedron, and let the inscribed sphere of <math>ABCD</math> touch the faces at <math>W, X, Y, Z</math>. Then, <math>OW, OX, OY, OZ</math> are normals to the respective faces. We know that the angle between any two normals is equal, so we have <math>|OW|=|OX|=|OY|=|OZ|</math> at equal angles. Now, since <cmath>WX=2\cdot OW\sin\frac{\angle{WOX}}{2},</cmath> and similar for the other sides, we have that <math>WXYZ</math> is a regular tetrahedron. Now, the faces of <math>ABCD</math> are the tangent planes at <math>W</math>, <math>X</math>, <math>Y</math>, and <math>Z</math>. Then, consider a <math>120^{\circ}</math> rotation about <math>OW</math>. The rotation sends <math>X\mapsto Y</math>, <math>Y\mapsto Z</math>, and <math>Z\mapsto X</math>. Thus we have <math>AB=AC</math>, <math>AC=AD</math>, <math>AD=AB</math>, and <math>BC=CD=DB</math>. Performing rotations about the other axes yields that <math>ABCD</math> has equal edges, so it is regular.
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(b) Consider the normals <math>OW, OX, OY, OZ</math>. We can perform a transformation in which we slightly shift <math>X</math>, <math>Y</math>, and <math>Z</math> closer such that <math>\triangle XYZ</math> is still equilateral, and such that all angles between every pair of normals is less that <math>180^{\circ}</math>. Then, shift <math>W</math> such that <math>WX=WY=XY</math>. Then five of the distances are equal but the sixth is not.
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~ tc1729
  
 
== See Also ==
 
== See Also ==

Revision as of 18:23, 14 April 2015

Problem

(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?

Solution

(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.

~ tc1729

See Also

1978 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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