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

Latest revision as of 22:42, 27 February 2022

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.

@@ Line 2: / Line 2: @@
 A given tetrahedron <math>ABCD</math> is isosceles, that is, <math>AB=CD, AC=BD, AD=BC</math>. Show that the faces of the tetrahedron are acute-angled triangles.
-==Solution==
+==Solutions==
-Proof by contradiction: Obviously, triangles ACD, DCB, and ABC are all congruent (the three sides are all equal). For the sake of clarity, triangle ABC is the "base", and D is the "pointed" vertex. Assume <DCB is obtuse (this proof will work for any other angle as well). Therefore angles <ABD and <DCA are also obtuse because of the congruence. However, this will mean that AB and AC will be extended outwards, and as these angles approach over 90 degrees, CA and AB will be parallel at 90, and then will extend outwards, making the point A non-existent, which obviously is a contradiction of the fact that ABCD is a tetrahedron.
+===Solution 1===
-==See also==
+Suppose <math>\triangle ABD</math> is fixed.
+By the equality conditions, it follows that the maximal possible value of <math>BC</math> occurs when the four vertices are coplanar, with <math>C</math> on the opposite side of <math>\overline{AD}</math> as <math>B</math>.
+In this case, the tetrahedron is not actually a tetrahedron, so this maximum isn't actually attainable.
+For the sake of contradiction, suppose <math>\angle ABD</math> is non-acute.
+Then, <math>(AD)^2\geq (AB)^2+(BD)^2</math>.
+In our optimal case noted above, <math>ACDB</math> is a parallelogram, so
+<cmath>\begin{align*}
+(BD)^2 + 2(AB)^2 &= (AD)^2 + (CB)^2 \\
+&= 2(AD)^2 \\
+&\geq 2(BD)^2+2(AB)^2.
+\end{align*}</cmath>
+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 <math>AB\leq BC \leq CA</math>. Now assume, for the sake of contradiction, that each face is non-acute; that is, right or obtuse. Consider triangles <math>\triangle ABC</math> and <math>\triangle ABD</math>. They share side <math>AB</math>. Let <math>k</math> and <math>l</math> be the planes passing through <math>A</math> and <math>B</math>, respectively, that are perpendicular to side <math>AB</math>. We have that triangles <math>ABC</math> and <math>ABD</math> are non-acute, so <math>C</math> and <math>D</math> are not strictly between planes <math>k</math> and <math>l</math>. Therefore the length of <math>CD</math> is at least the distance between the planes, which is <math>AB</math>. However, if <math>CD=AB</math>, then the four points <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> are coplanar, and the volume of <math>ABCD</math> would be zero. Therefore <math>CD>AB</math>. However, we were given that <math>CD=AB</math> in the problem, which leads to a contradiction. Therefore the faces of the tetrahedron must all be acute.
+===Solution 3===
+Let <math>\vec{a} = \overrightarrow{DA}</math>, <math>\vec{b} = \overrightarrow{DB}</math>, and <math>\vec{c} = \overrightarrow{DC}</math>. The conditions given translate to
+<cmath>\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*}</cmath>
+We wish to show that <math>\vec{a}\cdot\vec{b}</math>, <math>\vec{b}\cdot\vec{c}</math>, and <math>\vec{c}\cdot\vec{a}</math> are all positive. WLOG, <math>\vec{a}\cdot\vec{a}\geq \vec{b}\cdot\vec{b}, \vec{c}\cdot\vec{c} > 0</math>, so it immediately follows that <math>\vec{a}\cdot\vec{b}</math> and <math>\vec{a}\cdot\vec{c}</math> are positive. Adding all three equations,
+<cmath>\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})</cmath>
+In addition,
+<cmath>\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}) \\
+(\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*}</cmath>
+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.
+==Solution 4==
+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.
+==Solution 5==
+Proof by contradiction: Assume at least one of the tetrahedron's faces are obtuse. WLOG, assume <math>\angle BAC</math> 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 <math>A</math> are congruent to angles in triangle <math>ABC</math>; namely, <math>\angle BAD</math> is congruent to <math>\angle ABC</math>, and <math>\angle CAD</math> is congruent to <math>\angle ACB</math>. Since the internal angles of triangle <math>ABC</math> must add to 180 degrees, so do the angles surrounding point A. Now lay triangle <math>ABC</math> on a flat surface. A diagram would make it clear that from the perspective of an aerial view, the "apparent" measures of <math>\angle BAD</math> and <math>\angle CAD</math> (which are most likely distorted visually, assuming these angles stick up vertically from the flat surface on which triangle <math>ABC</math> lies) can never exceed the true measures of those angles (equality happens when these angles also lie flat on top of triangle <math>ABC</math>). This means that these two angles can never join to form side AD (because <math>\angle BAC</math> is more than the sum of <math>\angle BAD</math> and <math>\angle CAD</math> - a direct consequence of the facts that <math>\angle BAC</math> 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 <math>ABC</math> and the midpoint of <math>BC</math>, which we will call <math>M</math>, we can say that if <math>AM > \frac{BC}{2}</math>, then <math>\angle A < 90</math>.
+Proof: Since <math>M</math> is the midpoint of <math>BC</math>, <math>BM = MC = \frac{BC}{2}</math>. Since it is given that <math>AM > \frac{BC}{2}</math>, we can substitute <math>\frac{BC}{2}</math> to get two inequalities:
+<cmath>AM > CM, \quad AM > BM.</cmath>
+The above inequalities imply that <math>\angle C > \angle CAM</math> and <math>\angle B > \angle BAM</math>. Adding these inequalities and simplifying the RHS, we have that <math>\angle C + \angle B > \angle A</math>. Adding <math>\angle A</math> to both sides, replacing the LHS with <math>180</math> and dividing by <math>2</math> gets us that <math>\angle A < 90</math>. 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 <math>\triangle ABD \cong \triangle BAC</math>. WLOG, assume that <math>\angle ADB</math> is the largest angle in triangle <math>ABD</math>.
+Because <math>\triangle ABD \cong \triangle BAC</math>, the median from <math>D</math> to <math>AB</math> is equal length to the median from <math>C</math> to <math>AB</math>. These points meet at <math>E</math>, the midpoint of <math>AB</math>. By the triangle inequality,
+<cmath>DE + CE > CD.</cmath>By substituting <math>CD</math> with <math>AB</math> (this is a given in the problem) and <math>CE</math> with <math>DE</math>, and then dividing by <math>2</math>, we get that
+<cmath>DE > \frac{AB}{2}.</cmath>By the lemma we showed at the start, this implies that <math>\angle ADB < 90</math>, and since we said that <math>\angle ADB</math> was the largest angle, triangle <math>ADB</math> must be acute. Since all of the faces of this tetrahedron are congruent, then, all of the faces must be acute.
+{{alternate solutions}}
+==See Also==
 {{USAMO box|year=1972|num-b=1|num-a=3}}
+{{MAA Notice}}
 [[Category:Olympiad Geometry Problems]]
+[[Category:3D Geometry Problems]]

1972 USAMO (Problems • Resources)
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