1972 USAMO Problems/Problem 2

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.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

ACS WASC
ACCREDITED
SCHOOL

Stay Connected

Subscribe to get news and
updates from AoPS, or Contact Us.
© 2015
AoPS Incorporated
Invalid username
Login to AoPS