Difference between revisions of "1972 USAMO Problems/Problem 2"
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===Solution 2=== | ===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 | + | 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=== | ===Solution 3=== | ||
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\end{align*}</cmath> | \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. | 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}} | {{alternate solutions}} | ||
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Latest revision as of 23:42, 27 February 2022
Contents
Problem
A given tetrahedron is isosceles, that is,
. Show that the faces of the tetrahedron are acute-angled triangles.
Solutions
Solution 1
Suppose is fixed.
By the equality conditions, it follows that the maximal possible value of
occurs when the four vertices are coplanar, with
on the opposite side of
as
.
In this case, the tetrahedron is not actually a tetrahedron, so this maximum isn't actually attainable.
For the sake of contradiction, suppose is non-acute.
Then,
.
In our optimal case noted above,
is a parallelogram, so
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 . Now assume, for the sake of contradiction, that each face is non-acute; that is, right or obtuse. Consider triangles
and
. They share side
. Let
and
be the planes passing through
and
, respectively, that are perpendicular to side
. We have that triangles
and
are non-acute, so
and
are not strictly between planes
and
. Therefore the length of
is at least the distance between the planes, which is
. However, if
, then the four points
,
,
, and
are coplanar, and the volume of
would be zero. Therefore
. However, we were given that
in the problem, which leads to a contradiction. Therefore the faces of the tetrahedron must all be acute.
Solution 3
Let ,
, and
. The conditions given translate to
We wish to show that
,
, and
are all positive. WLOG,
, so it immediately follows that
and
are positive. Adding all three equations,
In addition,
Equality could only occur if
, which requires the vectors to be coplanar and the original tetrahedron to be degenerate.
Solution 4
Suppose for the sake of contradiction that is not acute. Since all three sides of triangles
and
are congruent, those two triangles are congruent, meaning
. Construct a sphere with diameter
. Since angles
and
are both not acute,
and
both lie on or inside the sphere. We seek to make
to satisfy the conditions of the problem. This can only occur when
is a diameter of the sphere, since both points lie on or inside the sphere. However, for
to be a diameter, all four points must be coplanar, as all diameters intersect at the center of the sphere. This would make tetrahedron
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 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
are congruent to angles in triangle
; namely,
is congruent to
, and
is congruent to
. Since the internal angles of triangle
must add to 180 degrees, so do the angles surrounding point A. Now lay triangle
on a flat surface. A diagram would make it clear that from the perspective of an aerial view, the "apparent" measures of
and
(which are most likely distorted visually, assuming these angles stick up vertically from the flat surface on which triangle
lies) can never exceed the true measures of those angles (equality happens when these angles also lie flat on top of triangle
). This means that these two angles can never join to form side AD (because
is more than the sum of
and
- a direct consequence of the facts that
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 and the midpoint of
, which we will call
, we can say that if
, then
.
Proof: Since
is the midpoint of
,
. Since it is given that
, we can substitute
to get two inequalities:
The above inequalities imply that
and
. Adding these inequalities and simplifying the RHS, we have that
. Adding
to both sides, replacing the LHS with
and dividing by
gets us that
. 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
. WLOG, assume that
is the largest angle in triangle
.
Because
, the median from
to
is equal length to the median from
to
. These points meet at
, the midpoint of
. By the triangle inequality,
By substituting
with
(this is a given in the problem) and
with
, and then dividing by
, we get that
By the lemma we showed at the start, this implies that
, and since we said that
was the largest angle, triangle
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.
See Also
1972 USAMO (Problems • Resources) | ||
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.