High School Mathematics Solid Geometry Plus Combinatorial Problems

There are c(12,3) ways to select three points from 12 points.

There are c(4,3) ways to select three points from four points.

There are c(3,3) ways to select three points from three points.

So to know how many eligible three points are coplanar, as long as the selection method to subtract the selection method from the four points is the selection, and all the selections are c(12,3).

There are three possibilities for four-point coplanarity.

Four of the six faces are six of the six faces.

There are three types of red sections shown.

Twelve types of blue sections are illustrated.

A total of 3+6+12=21 species.

That is, there are 21 x c(4,3) ways to select three points from the four points of coplanarity, and the difference between it and c(12,3) is the result.

i.e. c(12,3)-21 x c(4,3)=220-84=136

There are 12 points in total, either 3 or 4 points are coplanar. I think of 3 points like this, c(12,1)xc(4,1)xc(4,1)+c(12,1)*c(2,1)*(c(4,1)+c(2,1)))- classify and discuss each point, analyze the conditions that make up the surface (a line segment and a point), take two points on one surface as a line segment, and then select another feasible point as the condition for forming the surface. 4 points can subtract 3 points from all faces.

We've also dealt with it several times, I think your requirements are higher, I guess you're the same as me, I want to ask by the way, you must not be a high school student, your question is very difficult to answer, today I made a ** to give you, do not do well please advise, I don't expect you to be satisfied, make a friend.

This ball should be stuck in the corner.

Let the radius be r, and the distance from the center of the ball to the angle is the root number 3

There is a small ball in the middle, and if you draw a diagram, you can see that from the tangent point to the center of the small ball is r, and from the center of the small ball to the angle, it is the root number 3r

So the equation is 1 + (root number 3 + 1) r = root number 3

The relationship between a sphere with a radius of 1 and a cube with a radius of 2 is 2 and the cube is shown below

As you can see from the diagram above, the space left for the ball is only the 8 corners of the cube, and they are all the same size. If you want the ball to be inside the cube and be the largest, you can only tangent to the three sides of the cube and the big ball at the same time. In this case, the center of the ball should be on a long diagonal with a length of 2 and a number 3.

On the diagonal, the length of 1 + 3 has been used up by the big ball, so only 2 3-(1 + 3) = 3-1 is left.

The length of the diagonal occupied by the small ball is similar to that of the large ball, which is r+r root number 3 = root number 3-1

Since the formula you gave r (1 + root number 3) = root number 3-1 should be proposed by the friend above, the graph without removing the auxiliary line is as follows:

A partial enlargement of the figure after adding a small ball is as follows:

The first question was done upstairs, and I did the second question first.

The main idea of the proof is to use the line plane to push the line plane to the line plane parallel, and then push the line plane parallel to the plane and then push the line plane to be parallel.

Specific proof: do DE focus H, connect HF, HG

From H, G is the midpoint to obtain EG parallel AE, and there is AE BC (not explained) EG surface BCD (skipping a step, the same below).

The same goes for HF surface BCD

fh ge=g, fh and ge face fgh

Face FGH Face BCD

and fg face fgh

FG Parallel Plane BCD

can provide you with an idea. I don't use standard mathematical language to write it, because it might be clearer.

1.Draw a square on paper.

Obviously you need (45 * 30 * 18) small cuboids to put this cube together. This makes your cube look as if it has been crossed a lot.

2.A coordinate system is established with one of the vertices of the cube as the origin.

Now, the "lattice" of these cuboids, that is, the interface of the cuboids, has an exact coordinate number. These coordinates are multiples of 2, multiples of 3, and multiples of 5. It's easy to see that all of these junctions are integers.

3.Now draw a diagonal line of the cube from the origin.

It extends from (0, 0, 0) to (90, 90, 90), or we call it (a, a, a). Now ask: when might it pass through the box?

Naturally, it is possible when a is an integer. So let's count in 3 directions:

From the direction of the edge length of 2, it passes through 45 faces. (90 / 2)

From the direction of the edge length of 3, it passes through 30 faces. (90 / 3)

From the direction where the edge length is 5, it passes through 18 faces. (90 / 5)

Our diagonal line will then pass through the faces of 45 + 30 + 18 = 93 small cuboids.

4.Of course, this is not the answer, because it is possible for a diagonal to cross 2 faces at the same time at a certain point (a, a, a). For example, (6, 6, 6), (10, 10, 10).

In fact, here the diagonal line passes through one of the edges of some box. We're going to subtract these extra faces.

2 * 3 = 6, 90 6 = 15: The diagonal passes through 15 of these edges.

2 * 5 = 10, 90 10 = 9: The diagonal line passes through 9 of these edges.

3 * 5 = 15, 90 15 = 6: The diagonal line passes through 6 of these edges.

So the number becomes 93 - 15 - 9 - 6 = 63.

5.But that's still not the answer. Obviously, we also need to calculate the case of crossing 3 faces at the same time, such as (30, 30, 30). In fact, the diagonal line passes through one of the vertices of a certain box here. We're going to add that number back.

2 * 3 * 5 = 30, 90 30 = 3: The diagonal passes through 3 of these vertices.

So the number ends up being 63 + 3 = 66

6.Obviously, you can't find a case where a diagonal line crosses more than 4 faces at the same time, because we are talking about a box with up to three faces sharing a vertex. So this is the end of our calculations.

So the answer is 66, choose option B. The expression for the whole is :

Solution: Draw a diagram of the square ABCD-A1B1C1D1 to (1) connect MN and A1C1, take the midpoint G of B1C1, and connect MG so MG is the median line of A1C1.

Then mg= 2 2a

Connecting GN gives A1C1 = 2A, GN = B1B = A

Because the plane b1c1cb plane abcd, gn bc, mgn is a right triangle.

Therefore, the tangent of the angle between Mn and A1C1 is gn mg=A2 2A= 2A(2) to connect DB and D1B1

Take the midpoints e and o of d1d and d1b1

If eo is connected, then a1o=1 2a1c1=2:2aa

eo=1/2db1=√3/2a

a1e = a1d1 +d1e = a +1 2a = 5 2a because a1o +eo = a1e

So a1oe is a right triangle.

So the angle between db1 and a1c1 is 90°

The tangent of the angle between Mn and A1C1 = root number 2.

The angle between db1 and a1c1 = 90°

Vertical PA, BC Vertical Plane, BC Vertical AD .

2.Triangular pyramid volume = SADC * BC3 = 4 * 4 3 =16 3

The 3 angular bisector line intersects AB and point M, connects DM, in the plane PCM, passes P as a DM parallel line, and the point of the intersection CM extension line is the Q,

Split the pyramid.

S = area of pyramid b-efa1 + area of pyramid d1-efa1.

Easy to find s=a3 18

Can you explain the topic completely and correctly?