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Why is it easy to construct triangles and difficult to construct tetrahedrons?

The sum of interior angles of triangles makes it easy to deal with triangles. What happens if you don't rely on this theorem?

Is there three triangles with angles of 41 °, 76 °and 63 °respectively?

The answer seems simple. We learned in math class that the sum of the interior angles of triangles is 180 °. Because 41 + 76 + 63 = 180, such triangles exist.

But the problem is far more complicated than it seems. The sum theorem of interior angles of triangles tells us that in plane Euclidean geometry, if a triangle is given, the sum of its interior angles is 180 °. But our problem does not give a triangle. On the contrary, our question is whether such triangles exist. The sum theorem of interior angles of triangles does not directly answer this question, but it can help us to construct the triangles we need.

In order to satisfy the sum theorem of interior angles of triangles, each angle of a triangle needs to be less than 180 °. This means that we can always place two corners on the same side of a line segment. For example, we can put an angle of 41 °and an angle of 76 °at both ends of the segment AB.

The two rays from point An and point B must not be parallel. Because Euclidean geometry requires two straight lines that complement each other, that is, 180 °, to be parallel to each other. The angles at point An and point B do not meet this requirement, so the two rays will not be parallel, but will intersect.

We record the intersection of these two rays as point C, and we get another angle at point C. Now we can apply the sum theorem of interior angles of triangles. The third angle must be 180 °- (41 °+ 76 °) = 63 °, so △ ABC is what we expect it to be.

The above argument can be generalized to show that any three angles with a sum of 180 °can form a triangle. Obviously, if we measure it in terms of angles (not radians), we can easily find triangles in which all three angles are rational. First choose two rational numbers x and y that are less than 180, then zonal 180-(xroomy) is also rational. Because of x+y+z=180, these three rational angles can form a triangle.

Although it is so simple to construct planar triangles from rational angles, similar problems in three dimensions are so complex that it took decades for the best mathematicians in the world to solve them. Why do such problems become so difficult when only one dimension is added? To understand this, it is necessary to have a deeper understanding of the interior angles and theorems of triangles.

In three-dimensional space, this problem involves a tetrahedron, which has four triangular sides. You can think of a tetrahedron as a three-dimensional triangle. In two-dimensional space, triangles are the simplest closed graphics with straight boundaries, which can be surrounded by only three line segments. In three dimensions, a tetrahedron is the simplest closed figure surrounded by a straight boundary, which can be constructed from four triangular planes.

The four triangular sides of a tetrahedron are like the three sides of a triangle. But how should the corners correspond? You can imagine that there is a solid angle at each of the four vertices of the tetrahedron, but in this problem we are more concerned about the dihedral angle formed by the intersection of the face and the face.

If you draw two intersecting planes, you will find that there are many angles that can be measured. Which angle should be chosen to represent the angle between the two faces?

The answer is to rotate the two intersecting planes until they look like a two-dimensional corner.

This is what we call dihedral angles.

In a tetrahedron, four faces intersect each other, forming six sides and six dihedral angles. For decades, mathematicians have been trying to figure out what kind of four sides have six reasonable dihedral angles. As mentioned above, if the degree of an angle is rational, then it is a rational angle. This is equivalent to the fact that under the Radian system, the size of the angle is a rational number multiplied by π. (to convert the angle to radians, you need to multiply the angle by π / 180 °, so if an angle is rational under the angle system, it must be a rational number multiplied by π under the Radian system, and vice versa. )

We have seen how easy it is to construct planar triangles from rational angles. But for tetrahedrons, the problem is much more complicated. Consider this simple tetrahedron cut from a corner of the cube.

We can immediately see that the three dihedral angles of the tetrahedron are made up of the faces of the original cube, so they are right angles. It is very convenient to use edges to refer to dihedral angles. In this tetrahedron, the dihedral angles on the edges OA, OB and OC are all right angles.

If the cube is cut at an appropriate angle to make OA=OB=OC, then the dihedral angles with AB, AC and BC as edges should be equal in size. We can cut the cube to OA=OB=OC=1, and then we can calculate the size of the dihedral angle with BC as the edge. The key to measuring the dihedral angle is to make a line segment from the midpoint M of the BC to the O and A points.

If we rotate the tetrahedron and look at the dihedral angle with BC as the edge from the side, this corner will be projected into a ∠ AMO on the plane, and the size of the ∠ AMO is equal to the original dihedral angle. Measuring the size of ∠ AMO requires knowing the length of line segments OA and OM. We already know that OA = 1, and then in order to know the length of OM, we just need to further examine the triangle Δ OCB.

Because ∠ BOC is a right angle, we can use the Pythagorean theorem to get BC = √ 2, and because M is the midpoint of BC, MC = √ 2 prime 2. And Δ OCB is not only a right triangle, but also an isosceles triangle because OB = OC. This means that this is a triangle of 45-45-90 degrees, ∠ OBC and ∠ OCB are both 45 degrees. Δ OCB is an isosceles triangle that ensures that OM is perpendicular to BC, so Δ OMC is also a right triangle. But if ∠ OMC = 90 °and ∠ OCB = 45 °, the sum theorem of triangles tells us that ∠ MOC = 45 °, that is, the small triangle Δ OMC is isosceles, so OM = MC = √ 2max 2.

Now we are finally ready to calculate the size of ∠ AMO

Tan ∠ AMO = 1 / (√ 2 + 2) = 2 / √ 2 = √ 2

In Δ AMO, we know that AO = 1 √ OM = 2 prime 2. In addition, because ∠ AOM is a right angle, we can use trigonometric functions. In a right triangle, the tangent of an angle is the ratio of the length of its opposite side to the length of its adjacent side (right side):

Tan ∠ AMO = 1 / (√ 2 + 2) = 2 / √ 2 = √ 2.

So the size of ∠ AMO is the inverse tangent of √ 2, that is, arctan √ 2, which is an irrational number, so the three dihedral angles of this tetrahedron are irrational, and it is not the rational tetrahedron we are looking for. However, although it is not our goal, this irrational tetrahedron can tell us some important information when looking for a rational tetrahedron.

To understand this, let's approximate the sum of the dihedral angles of the irrational tetrahedron above. Through the calculator or trigonometric function table, we find that ∠ AMO is about 54.74 °.

Now we can sum the six dihedral angles of the tetrahedron OABC: all three right angles are 90 °, and the other three angles are equal to the angles we just calculated, so the sum of the six dihedral angles of the tetrahedron is about 3 × 90 °+ 3 × 54.74 °≈ 434.22 °.

That's the difference. Let's go back to the cube and instead of cutting it in the way OA = OB = OC, we cut a very thin piece from the corner.

The new tetrahedron still has three dihedral angles of 90 °, with OP, OC and OB as edges. But the values of the other three dihedral angles have changed. The corner with BC as the edge looks very small, while the angle with PB and PC as the edge looks not much different from the angle at OB and OC.

In fact, if you keep cutting the tetrahedron thinner and thinner, the point P will be closer to point O, the dihedral angle with BC as the edge will be close to 0 °, and the dihedral angle with PB and PC as the edge will tend to 90 °, so the sum of these angles is approximate:

90 °+ 90 °+ 90 °+ 90 °+ 0 °= 450 °.

As point P approaches point O, the sum of the six dihedral angles of the tetrahedron will approach 450 °. This means that the sum of dihedral angles will change! In the original tetrahedral OABC, the sum of the six dihedral angles was about 434 °, but when we change these angles, their sum also changes. Perhaps on some levels, tetrahedrons can be regarded as three-dimensional versions of planar triangles, but there is one big difference between them: there is no "tetrahedral dihedral angle sum theorem" to guarantee that the sum of these angles is a constant.

This shows that we can only guarantee the dihedral angle of the tetrahedron between 360 °and 540 °. If you are looking for a tetrahedron of rational dihedral angles, this will be a problem. You can't choose five rational corners at random, and then you can be sure that the sixth corner is naturally reasonable. Because unlike triangles, you don't know what the sum of these dihedral angles is.

To make matters worse, you don't even know whether six dihedral angles of any size can form a tetrahedron. Consider five right angles and one acute angle, whose sum is between 450 °and 540 °, which is indeed within the allowable range of the tetrahedron. But there is no tetrahedron made up of such six angles. If five of the six angles are right angles, then one face must have three 90 °dihedral angles. But in this case, these faces cannot be closed to form a tetrahedron: like parallel lines, they never intersect.

A face shared by three right-angled dihedral angles can be part of a prism, but not part of a tetrahedron.

Therefore, the problem of finding all possible rational tetrahedrons is far more complicated than finding five or six rational numbers with a definite sum. In addition, solving this problem requires solving an equation with 105 terms, which comes from a 1976 paper by John Conway and Antonia Jones. Some mathematicians completed this work in 2020, resulting in a complete classification of all rational tetrahedrons.

The interior angles and theorems of triangles are just one of the many reasons to appreciate the elegance and beauty of triangles. For tetrahedrons, the lack of such a theorem precisely shows the beauty and complexity of ascending a dimension.

problem

1. What is the sum of the dihedral angles of a cube?

Answer

A cube has 12 edges, so it has 12 dihedral angles, each of which is 90 degrees, so the sum is 12 × 90 °= 1080 °.

problem

two。 What is the sum of the six dihedral angles of a regular tetrahedron?

Answer

All six dihedral angles are equal, so a suitable right triangle can be made to calculate the size of one of the dihedral angles.

Every face of a regular tetrahedron is an equilateral triangle, so the height of the midline of the side-- the segment from the vertex to the midpoint of the opposite side-- is √ 3 √ 2s, where 1 √ 3 √ 2s is the edge length, which is the oblique edge of the right triangle we need. The center of the base is called the centroid, which is on the midline of the base triangle, 1 inch 3 from the midpoint of the bottom edge of the triangle. The height from the top vertex of the tetrahedron to the center of the base is s, which is a right edge of the right triangle. Therefore, the cosine value of the dihedral angle between the two sides of the regular tetrahedron is (1 √ 3 × 2s) / (√ 3max 2s) = 1 max 3. Because of arccos ≈ 70.53 °, the sum of the six (equivalent) dihedral angles of a regular tetrahedron is about 6 × 70.53 °≈ 423.18 °.

problem

3. Imagine a regular tetrahedron on the desktop. When you press the top vertex down, how does the sum of the six dihedrons change as the tetrahedron is gradually flattened?

Answer

In the process of flattening the tetrahedron, the three dihedral angles connected to the bottom surface gradually become 0 °, and the other three dihedral angles tend to 180 °, so the sum is 3 × 0 + 3 × 180 °= 540 °. This is the upper limit of the sum of dihedral angles of a tetrahedron. In order to reach the minimum value of the sum of dihedral angles, you can push two opposite sides toward each other, the four dihedral angles will become 0 °, and the other two will become 180 °.

problem

4. Can any four angles with a sum of 360 °form a quadrangle?

Answer

Sure. Let these four angles be a, b, c and d, respectively, with a + b + c + d = 360 °. Suppose that both an and b are less than or equal to c and d. C is divided into c "and c", d is divided into d "and d", that is, c = c "+ c", d "+ d", so that a + c "+ d" = 180 °, b + c "+ d" = 180 °(we have enough freedom to do this in many ways). Use these two sets of angles to construct two triangles and resize them so that the opposite sides of an and b are equal in length. Then they are put together, and c is composed of c, b, c and d to form d, so that quadrilaterals with a, b, c, d as angles are obtained.

An interesting question is whether quadrilaterals can always be constructed with a specific set of angles in order.

Original text link:

Https://www.quantamagazine.org/triangles-are-easy-tetrahedra-are-hard-20220131/

The translated content only represents the author's point of view, not the position of the Institute of Physics of the Chinese Academy of Sciences.

This article comes from the official account of Wechat: Institute of Physics, Chinese Academy of Sciences (ID:cas-iop), author: Samuel Velasco, translator: Tibetan idiot, revision: Dannis, Editor: zhenni

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