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 obtuse. 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.

Solution 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.

Solution 5

Proof by contradiction: Assume at least one of the tetrahedron's faces are obtuse. WLOG, assume $\angle BAC$ is an obtuse angle. Using SSS congruence to prove that all four faces of the tetrahedron are congruent also shows that the angles surrounding point $A$ are congruent to angles in triangle $ABC$ ; namely, $\angle BAD$ is congruent to $\angle ABC$ , and $\angle CAD$ is congruent to $\angle ACB$ . Since the internal angles of triangle $ABC$ must add to 180 degrees, so do the angles surrounding point A. Now lay triangle $ABC$ on a flat surface. A diagram would make it clear that from the perspective of an aerial view, the "apparent" measures of $\angle BAD$ and $\angle CAD$ (which are most likely distorted visually, assuming these angles stick up vertically from the flat surface on which triangle $ABC$ lies) can never exceed the true measures of those angles (equality happens when these angles also lie flat on top of triangle $ABC$ ). This means that these two angles can never join to form side AD (because $\angle BAC$ is more than the sum of $\angle BAD$ and $\angle CAD$ - a direct consequence of the facts that $\angle BAC$ is obtuse and all three angles add up to 180 degrees), so the tetrahedron with obtuse triangle faces is impossible.

Solution 6

Lemma: given triangle $ABC$ and the midpoint of $BC$ , which we will call $M$ , we can say that if $AM > \frac{BC}{2}$ , then $\angle A < 90$ . Proof: Since $M$ is the midpoint of $BC$ , $BM = MC = \frac{BC}{2}$ . Since it is given that $AM > \frac{BC}{2}$ , we can substitute $\frac{BC}{2}$ to get two inequalities: $AM > CM, \quad AM > BM.$ The above inequalities imply that $\angle C > \angle CAM$ and $\angle B > \angle BAM$ . Adding these inequalities and simplifying the RHS, we have that $\angle C + \angle B > \angle A$ . Adding $\angle A$ to both sides, replacing the LHS with $180$ and dividing by $2$ gets us that $\angle A < 90$ . This is our desired inequality, so we are done. Note that all faces of this tetrahedron are congruent, by SSS. In particular, we will use that $\triangle ABD \cong \triangle BAC$ . WLOG, assume that $\angle ADB$ is the largest angle in triangle $ABD$ . Because $\triangle ABD \cong \triangle BAC$ , the median from $D$ to $AB$ is equal length to the median from $C$ to $AB$ . These points meet at $E$ , the midpoint of $AB$ . By the triangle inequality, $DE + CE > CD.$ By substituting $CD$ with $AB$ (this is a given in the problem) and $CE$ with $DE$ , and then dividing by $2$ , we get that $DE > \frac{AB}{2}.$ By the lemma we showed at the start, this implies that $\angle ADB < 90$ , and since we said that $\angle ADB$ was the largest angle, triangle $ADB$ must be acute. Since all of the faces of this tetrahedron are congruent, then, all of the faces must be acute.

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

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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