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Difference between revisions of "1972 USAMO Problems/Problem 2"

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Equality could only occur if <math>\vec{a} = \vec{b} + \vec{c}</math>, which requires the vectors to be coplanar and the original tetrahedron to be degenerate.
 
Equality could only occur if <math>\vec{a} = \vec{b} + \vec{c}</math>, which requires the vectors to be coplanar and the original tetrahedron to be degenerate.
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==Soluiton 4==
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Suppose for the sake of contradiction that <math>\angle BAC</math> is not acute. Since all three sides of triangles <math>BAC</math> and <math>CDB</math> are congruent, those two triangles are congruent, meaning <math>\angle BDC=\angle BAC>90^{\circ}</math>. Construct a sphere with diameter <math>BC</math>. Since angles <math>BAC</math> and <math>BDC</math> are both not acute, <math>A</math> and <math>D</math> both lie on or inside the sphere. We seek to make <math>AD=BC</math> to satisfy the conditions of the problem. This can only occur when <math>AD</math> is a diameter of the sphere, since both points lie on or inside the sphere. However, for <math>AD</math> to be a diameter, all four points must be coplanar, as all diameters intersect at the center of the sphere. This would make tetrahedron <math>ABCD</math> degenerate, creating a contradiction. Thus, all angles on a face of an isosceles tetrahedron are acute.
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 21:01, 25 January 2016

Problem

A given tetrahedron $ABCD$ is isosceles, that is, $AB=CD, AC=BD, AD=BC$. Show that the faces of the tetrahedron are acute-angled triangles.

Solutions

Solution 1

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.

Solution 2

It's not hard to see that the four faces are congruent from SSS Congruence. Without loss of generality, assume that $AB\leq BC \leq CA$. Now assume, for the sake of contradiction, that each face is non-acute; that is, right or isosceles. Consider triangles $\triangle ABC$ and $\triangle ABD$. They share side $AB$. Let $k$ and $l$ be the planes passing through $A$ and $B$, respectively, that are perpendicular to side $AB$. We have that triangles $ABC$ and $ABD$ are non-acute, so $C$ and $D$ are not strictly between planes $k$ and $l$. Therefore the length of $CD$ is at least the distance between the planes, which is $AB$. However, if $CD=AB$, then the four points $A$, $B$, $C$, and $D$ are coplanar, and the volume of $ABCD$ would be zero. Therefore $CD>AB$. However, we were given that $CD=AB$ in the problem, which leads to a contradiction. Therefore the faces of the tetrahedron must all be acute.

Solution 3

Let $\vec{a} = \overrightarrow{DA}$, $\vec{b} = \overrightarrow{DB}$, and $\vec{c} = \overrightarrow{DC}$. The conditions given translate to \begin{align*} \vec{a}\cdot\vec{a} &= \vec{b}\cdot\vec{b} + \vec{c}\cdot\vec{c} - 2(\vec{b}\cdot\vec{c}) \\ \vec{b}\cdot\vec{b} &= \vec{c}\cdot\vec{c} + \vec{a}\cdot\vec{a} - 2(\vec{c}\cdot\vec{a}) \\ \vec{c}\cdot\vec{c} &= \vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} - 2(\vec{a}\cdot\vec{b}) \end{align*} We wish to show that $\vec{a}\cdot\vec{b}$, $\vec{b}\cdot\vec{c}$, and $\vec{c}\cdot\vec{a}$ are all positive. WLOG, $\vec{a}\cdot\vec{a}\geq \vec{b}\cdot\vec{b}, \vec{c}\cdot\vec{c} > 0$, so it immediately follows that $\vec{a}\cdot\vec{b}$ and $\vec{a}\cdot\vec{c}$ are positive. Adding all three equations, \[\vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} + \vec{c}\cdot\vec{c} = 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} + \vec{b}\cdot\vec{c})\] In addition, \begin{align*} (\vec{a} - \vec{b} - \vec{c})\cdot(\vec{a} - \vec{b} - \vec{c})&\geq 0 \\ \vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} + \vec{c}\cdot\vec{c}&\geq 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} - \vec{b}\cdot\vec{c}) \\ 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} + \vec{b}\cdot\vec{c})&\geq 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} - \vec{b}\cdot\vec{c}) \\ \vec{b}\cdot\vec{c}&\geq 0 \end{align*} Equality could only occur if $\vec{a} = \vec{b} + \vec{c}$, which requires the vectors to be coplanar and the original tetrahedron to be degenerate.

Soluiton 4

Suppose for the sake of contradiction that $\angle BAC$ is not acute. Since all three sides of triangles $BAC$ and $CDB$ are congruent, those two triangles are congruent, meaning $\angle BDC=\angle BAC>90^{\circ}$. Construct a sphere with diameter $BC$. Since angles $BAC$ and $BDC$ are both not acute, $A$ and $D$ both lie on or inside the sphere. We seek to make $AD=BC$ to satisfy the conditions of the problem. This can only occur when $AD$ is a diameter of the sphere, since both points lie on or inside the sphere. However, for $AD$ to be a diameter, all four points must be coplanar, as all diameters intersect at the center of the sphere. This would make tetrahedron $ABCD$ degenerate, creating a contradiction. Thus, all angles on a face of an isosceles tetrahedron are acute.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1972 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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