The position of a point in space can be specified by two of its orthogonal projections, for example, horizontal and frontal, frontal and profile. The combination of any two orthogonal projections allows you to find out the value of all coordinates of a point, build a third projection, and determine the octant in which it is located. Let's consider several typical problems from the descriptive geometry course.
According to a given complex drawing of points A and B, it is necessary:
Let us first determine the coordinates of point A, which can be written in the form A (x, y, z). Horizontal projection of point A - point A ", having coordinates x, y. Draw from point A" perpendiculars to axes x, y and find A х, A у, respectively. The x coordinate for m. A is equal to the length of the segment A x O with a plus sign, since A x lies in the region positive values the x-axis. Taking into account the scale of the drawing, we find x = 10. The coordinate y is equal to the length of the segment A y O with a minus sign, since m. A y lies in the region negative values y-axis. Taking into account the scale of the drawing y = –30. Frontal projection of point A - point A "" has coordinates x and z. Let us drop the perpendicular from A "" to the z-axis and find A z. The z-coordinate of point A is equal to the length of the segment A z O with a minus sign, since A z lies in the region of negative values of the z-axis. Taking into account the drawing scale z = –10. Thus, the coordinates of point A are (10, –30, –10).
The coordinates of point B can be written as B (x, y, z). Consider the horizontal projection of point B - m. B ". Since it lies on the x-axis, then B x = B" and the coordinate B y = 0. The abscissa x of point B is equal to the length of the segment B x O with a plus sign. Taking into account the scale of the drawing x = 30. Frontal projection of point B - point B˝ has coordinates x, z. Let's draw a perpendicular from B "" to the z-axis, so we find B z. The applicate z of point B is equal to the length of the segment B z O with a minus sign, since B z lies in the region of negative values of the z-axis. Taking into account the scale of the drawing, we determine the value z = –20. So the B coordinates are (30, 0, -20). All the necessary constructions are shown in the figure below.
Building projections of points
Points A and B in the plane П 3 have the following coordinates: A "" "(y, z); B" "" (y, z). In this case, A "" and A "" "lie in the same perpendicular to the z-axis, since they have a common z-coordinate. Similarly, B" "and B" "" lie on the common perpendicular to the z-axis. To find the profile projection of point A, we put the value of the corresponding coordinate found earlier along the y-axis. In the figure, this is done using an arc of a circle of radius A y O. After that, draw a perpendicular from A y until it intersects with the perpendicular restored from point A "" to the z-axis. The intersection point of these two perpendiculars defines the position of A "" ".
Point B "" "lies on the z-axis, since the y-ordinate of this point is zero. To find the profile projection of point B in this problem, you just need to draw a perpendicular from B" "to the z-axis. The intersection point of this perpendicular with the z-axis is B "" ".
Determining the position of points in space
Visualizing a spatial layout made up of projection planes P 1, P 2 and P 3, the arrangement of octants, as well as the order of transformation of the layout into diagrams, one can directly determine that point A is located in the third octant, and point B lies in the plane P 2.
Another option for solving this problem is the method of exclusions. For example, the coordinates of point A are (10, -30, -10). The positive abscissa x allows us to judge that the point is located in the first four octants. A negative y-ordinate indicates that the point is in the second or third octants. Finally, a negative applicate z indicates that m. A is located in the third octant. The above reasoning is clearly illustrated by the following table.
| Octants | Coordinate signs | ||
| x | y | z | |
| 1 | + | + | + |
| 2 | + | – | + |
| 3 | + | – | – |
| 4 | + | + | – |
| 5 | – | + | + |
| 6 | – | – | + |
| 7 | – | – | – |
| 8 | – | + | – |
Point B coordinates (30, 0, -20). Since the ordinate of m. B is equal to zero, this point is located in the plane of projections P 2. A positive abscissa and a negative applicate point B indicate that it is located on the border of the third and fourth octants.
Construction of a visual image of points in the system of planes P 1, P 2, P 3
Using a frontal isometric projection, we have built a spatial layout of the III octant. It is a rectangular trihedron, whose faces are the planes P 1, P 2, P 3, and the angle (-y0x) is 45 º. In this system, the segments along the x, y, z axes will be plotted in full size without distortion.
We will start constructing a visual image of point A (10, -30, -10) with its horizontal projection A ". Putting the corresponding coordinates along the abscissa and ordinate axes, we find the points A x and A y. Intersection of perpendiculars reconstructed from A x and A y respectively to the axes x and y determines the position of point A ". Setting aside from A "segment AA" parallel to the z-axis towards its negative values, the length of which is 10, we find the position of point A.
A visual image of point B (30, 0, -20) is constructed in a similar way - in the plane P2 along the x and z axes, you need to postpone the corresponding coordinates. The intersection of the perpendiculars reconstructed from B x and B z will determine the position of point B.
Duration: 1 lesson (45 minutes).
Class: 6th grade
Technologies:
- Microsoft multimedia presentation Office PowerPoint, Notebook;
- the use of an interactive whiteboard;
- student handouts created with Microsoft Office Word and Microsoft Office Excel.
annotation:
On the topic "Coordinates" in thematic planning 6 hours are allotted. This is the fourth lesson on Coordinates. At the time of the lesson, students have already become familiar with the concept of "coordinate plane" and the rules for constructing a point. Knowledge is updated in the form frontal survey... In revision lessons, all students are included in different kinds activities. In this case, all channels of perception and reproduction of the material are used.
The assimilation of the theory is also tested in the course of oral work (the task is to solve the crossword puzzle, in which quarter is the point). Additional assignments are provided for strong students.
The lesson uses multimedia equipment and an interactive whiteboard to demonstrate presentations and assignments in Microsoft Office PowerPoint and Notebook. For creating test items and handout material were used: Microsoft Office Excel, Microsoft Office Word.
Using your interactive whiteboard expands your presentation options. In Notebook software, students can independently move objects to the desired location. V Microsoft program Office PowerPoint has the ability to set the movement of objects, so there is a physical minute for the eyes.
The lesson uses:
- examination homework;
- frontal work;
- individual work of students;
- presentation of the student's report;
- performing oral and written exercises;
- work of students with an interactive whiteboard;
- independent work.
Lesson summary.
Target: to consolidate the skills of finding the coordinates of the marked points and to build points according to the given coordinates.
Lesson Objectives:
educational:
- generalization of knowledge and skills of students on the topic "Coordinate plane";
- intermediate control of students' knowledge and skills;
developing:
- development of the communicative competence of students;
- development of students' computational skills;
- development logical thinking;
- developing students' interest in the subject through unconventional form conducting a lesson;
- development mathematically competent speech, students' outlook;
- skill development independent work with a tutorial and additional literature;
- development of aesthetic feelings of students;
educational:
- fostering discipline in organizing work in the classroom;
- education of cognitive activity, a sense of responsibility, a culture of communication;
- education of accuracy when performing constructions.
During the classes.
- Organizing time.
Greet students by communicating the topic and purpose of the lesson. Checking the readiness of the class for the lesson. The task is set: to repeat, generalize, systematize knowledge on the announced topic.
2. Updating knowledge.
Verbal counting.
1) Individual work: several people do the work on the cards.
2) Working with the class: calculate examples and form a word. The table is on the interactive whiteboard's screen, and the letters are written into the table using a marker from the interactive whiteboard.
Pupils take turns going to the blackboard and writing down the letters. It turns out the word "Prometheus". One of the students, who prepared a report in advance, explains what this word means. (The ancient Greek astronomer Claudius Ptolemy, who used latitude and longitude as coordinates already in the 2nd century.)
Frontal work.
The task "Solve the crossword puzzle" will help you remember the basic concepts on the topic "Coordinate plane".
The teacher shows a crossword puzzle on the interactive whiteboard screen and asks the students to solve it. Students use electronic markers to write down words in a crossword puzzle.
1. Two coordinate lines form a coordinate….
2. Coordinate straight lines are coordinate….
3. What is the angle formed at the intersection of the coordinate lines?
4. What is the name of a pair of numbers that determine the position of a point on the plane?
5. What is the name of the first number?
6. What is the name of the second number?
7. What is the name of the segment from 0 to 1?
8. How many parts is the coordinate plane divided by coordinate lines?
3. Consolidation of skills and abilities to build a geometric figure according to the specified coordinates of its vertices.
Construction of geometric shapes. Working with a textbook in notebooks.
- №1054а “Construct a triangle if the coordinates of its vertices are known: A (0; -3), B (6: 2), C (5: 2). Specify the coordinates of the points at which the sides of the triangle intersect the x-axis. "
- Construct a quadrilateral ABCD if A (-3; 1), B (1; 1), C (1; -2), D (-3; -2). Determine the type of quadrilateral. Find the coordinates of the intersection of the diagonals.
4. Physiotherapy for the eyes.
On the slide, students should follow the movement of the object with their eyes. At the end of the physical minute, a question is asked about geometric shapes resulting from eye movement.
5. Control over the skills to build points on the coordinate plane according to the specified coordinates.
Independent work. Competition of artists.
The coordinates of the points are recorded on the slide. Also cards are printed for each student. If you correctly mark the points on the coordinate plane and connect them sequentially, you get a picture. Each student completes the task independently. After completing the work, it opens correct drawing on the screen. Each student receives an assessment for independent work.
6. Homework.
- No. 1054b, No. 1057a.
- Creative task: Draw a point-by-point pattern on the coordinate plane and record the coordinates of these points.
7. Summing up the lesson.
Questions to students:
- What is a coordinate plane?
- What are the coordinate axes OX and OY called?
- What is the angle formed when the coordinate lines intersect?
- What is the name of a pair of numbers that determine the position of a point on a plane?
- What is the first number called?
- What is the second number called?
Literature and Resources:
- G.V. Dorofeev, SB Suvorova, IF Sharygin “Mathematics. 6kl "
- Maths. Grade 6: Lesson plans (according to the textbook by G.V. Dorofeev, etc.)
- http://www.pereplet.ru/nauka/almagest/alm-cat/Ptolemy.htm
When constructing a point along the specified coordinates, it must be remembered that, in accordance with the drawing rules, the scale along the axis Oh decreases in 2 times in comparison with the scale along the axes OU and Oz.
1. Build a point: A (2; 1; 3) x A = 2; for A = 1; z A = 3
a) usually, first of all, they build a projection of a point onto a plane Ooh. Mark points x A = 2 and for A = 1 and draw through them straight lines parallel to the axes Oh and OU. The point of their intersection has coordinates (2;1; 0) Point built A 1 (2; 1; 0.)
A (2; 1; 3)
0
for A = 1
x A = 2 at
A 1 (2; 1; 0) 0 for A = 1at
NS x A = 2 A 1 (2; 1; 0)
NS
b) further from point A 1 (2; 1; 0) restore the perpendicular to the plane Ooh (draw a straight line parallel to the axis Оz ) and lay on it a segment equal to three: z A = 3.
2. Build a point: B (3; - 2; 1) x B = 3; for B = -2; Z B = 1
z
for B = - 2
B (3; -2; 1) O at
B 1 (3; -2) x B = 3
NS
3. Construct a point C (-2; 1; 3 ) z C (-2; 1; 3)
X A = -2; Y A = 1; Z A = 3
x C = - 2 C 1 (-2; 1; 0)
y A = 1 y
4. Given a cube. A ... D 1, whose edge is equal to 1 ... The origin is the same as the point V, ribs VA, VS and BB 1 coincide with the positive rays of the coordinate axes. Name the coordinates of all other vertices of the cube. Calculate the diagonal of the cube.
z
AB = BC = BB 1 BD 1 = =
В 1 (0; 0; 1) С 1 (0; 1; 1) = =
A 1 (1; 0; 1) D 1 (1; 1; 1)
B (0; 0; 0) C (0; 1; 0) y
A (1; 0; 0) D (1; 1; 0)
5. Build dots A (1; 1; -1) and B (1; -1; 1). Does the line segment intersect the coordinate axis? coordinate plane? Does the line segment pass through the origin? Find the coordinates of the intersection points, if any. z The points lie in a plane perpendicular to the axis Oh.
Line intersects axis Oh
and plane hoy
at the point
B (1; -1; 1) 
0(0;0;0)
C (1; 0; 0)
A (1; 1; -1)
6.Find the distance between two points: A (1; 2; 3) and B (-1; 1; 1).
a)AB = = = = 3
b)C (3; 4; 0) and D (3; -1; 2).
СD = = =
In space, to determine the coordinates of the midpoint of the segment, the third coordinate is entered.
B (x B; y B; z B)
WITH( ; ; )
A (x A; y A; z A)
7.Find coordinates WITH midpoints of segments: a)AB, if A (3; - 2; - 7), B (11; - 8; 5),
x M = = 7; for M = = - 5; z M = = - 1; C (7; - 5; - 1)
8. Point coordinates A (x; y; z). Write the coordinates of the points symmetrical to the given one with respect to:
b) coordinate lines
v) origin
a) If point A 1
symmetric to the given relative to the coordinate plane hoy,
then the difference in
point coordinates will only be in the coordinate sign z: A 1 (x; y; -z).
point A 2 Ohz, then A 2 (x; -y; z).
point A 3 symmetric to the given relative to the plane Oyz, then A 2 (-x; y; z).
b) If point A 4
symmetric with respect to the coordinate line Oh,
then the difference in
point coordinates will only be in coordinate signs at
and z: A 4 (x; -y; -z).
point A 5 OU, then A 5 (-x; y; -z).
point A 6 symmetric to a given relative to a straight line Оz, then A 6 (-x; -y; z).
v) If point A 7 is symmetric to the given one with respect to the origin, then A 6 (-x; -y; -z).
CONVERSION OF COORDINATES
The transition from one coordinate system to another is called transformation of the coordinate system.
We will consider two conversion cases coordinate system, and derive formulas for the relationship between the coordinates of an arbitrary point of the plane in different systems coordinates. (The method of transforming the coordinate system is similar to transforming graphs).
1.Parallel transfer... In this case, the position of the origin of coordinates changes, but the direction of the axes and the scale remain unchanged.
If the origin goes to a point 0 1 with coordinates 0 1 (x 0; y 0), then for point M (x; y) relationship between system coordinates x0y and x 0 0y 0 expressed by the formulas:
x = x 0 + x "
y = y 0 + y "
The formulas obtained make it possible to find the old coordinates by the known new NS" and at " and vice versa.
y M (x; y) M (x "; y")
0 1 (x 0; y 0), x "
x 0 x "
2.Rotating coordinate axes... In this case, both axes are rotated by the same angle, and the origin and scale remain unchanged.
M (x; y)
y 1 x 1
Point coordinates M in the old system M (x; y) and M (x "; y") - in the new one. Then the polar radius in both systems is the same, and the polar angles are respectively equal + and , where - polar angle at new system coordinates.
According to the formulas for the transition from polar to rectangular coordinates, we have:
x = rcos ( + ) x = rcos Cos - rsin Sin
y = rsin ( + ) y = rcos Sin + rsin Cos
But rcos = x " and rsin = y ", therefore
x = x "· cos - y "· sin
y = x "· sin + y "· cos
Answer the questions in writing:
- What is called a rectangular coordinate system on a plane? in space?
- Which axis is called the applicate axis? Ordinate? Abscissa?
- What is the notation for unit vectors on coordinate axes?
- What is called an ortom?
- How is the length of the segment specified by the coordinates of its ends calculated in a rectangular coordinate system?
- How are the coordinates of the midpoint of the segment specified by the coordinates of its ends calculated?
- What is called a polar coordinate system?
- What is the relationship between the coordinates of a point in rectangular and polar coordinate systems?
Complete tasks:
1. At what distance from the coordinate planes is the point A (1; -2; 3)
2. At what distance is the point A (1; -2; 3) from coordinate lines a)OU; b) OU; v)Oz;
3. What condition are satisfied by the coordinates of points in space, equally distant:
a) from two coordinate planes Ooh and Oyz; AB
b) from all three coordinate planes
4. Find the coordinates of the point M midpoint AB, A (-2; -4; 1); B (0; -1; 2) and name the point symmetrical to the point M, relatively a) axes Oh
b) axes OU
v) axes Oz.
5. Given a point B (4; - 3; - 4). Find the coordinates of the bases of the perpendiculars dropped from a point on the coordinate axes and coordinate planes.
6.On axis OU find a point equidistant from two points A (1; 2; - 1) and B (-2; 3; 1).
7. In plane Ohz find a point equidistant from three points A (2; 1; 0); B (-1; 2; 3) and C (0; 3; 1).
8. Find the lengths of the sides of the triangle ABC and its area , if the coordinates of the vertices : A (-2; 0; 1), B (8; - 4; 9), C (-1; 2; 3).
9. Find the coordinates of the projected points A (2; -3; 5); B (3; -5;); WITH(- ; - ; - ).
10. Given points A (1; -1; 0) and B (-3; - 1; 2). Calculate the distance from the origin to the given points.
VECTORS IN SPACE. BASIC CONCEPTS
All quantities that are dealt with in physics, technology, everyday life are divided into two groups. The first are fully characterized by their numerical value: temperature, length, mass, area, work. Such quantities are called scalar.
Other quantities such as force, speed, displacement, acceleration, etc. determined not only by their numerical value, but also by their direction. Such quantities are called vector, or vectors. A vector quantity is geometrically depicted as a vector.
Vector-it is a directed straight line segment, i.e. segment having
a certain length and direction.
Construct complex drawings of points: A(15,30,0), V(30,25,15), WITH(30,10,15), D(15,30,20)
We will divide the solution of the problem into four stages.
1. A(15,30,0); x A= 15 mm ; y A= 30mm ; zA= 0.
What do you think, if the point A coordinate z A= 0, then what position does it occupy in space?
It looks like a complex drawing of a point A plotted at given coordinates
If a point has one coordinate equal to zero, then the point belongs to one of the projection planes. V in this case the point has no height: z= 0, hence the point A lies in the plane N 1.
In a complex drawing, the original (i.e. the point itself A) is not depicted, there are only its projections.
2. V(30,25,15) and WITH(30,10,15).
At the second stage, let's combine the construction of two points.
x B= 30mm; x C= 30mm
y B= 35mm; y C= 10mm
z B= 15mm; z C= 15mm
At points V and WITH: x B = x C= 30mm, z B = z C= 15mm
a) Coordinates NS points are the same, therefore, in the P 1 - P 2 system, the projections of the points lie on the same communication line (Fig. 1.2),
b) Coordinates z points coincide, (both points are equally distant from N 1 by 15mm,) i.e. they are located at the same height, therefore at P 2 the projections of the points coincide: IN 2=(C 2).
v) To determine visibility relative to P 2 look at fig. 1.3. The observer sees the point V which covers the point WITH, i.e. point V located closer to the observer, therefore on P 2 it is visible. (See M1 - 13 & 16).
In system P 2 P 3 the projections of the points also lie on the same communication line and the visibility is determined by the arrow (Fig. 1.2).
Points V and WITH- are called frontally competing.
3. D(15,30,20); x D= 15mm; y D= 30mm; z D= 20mm.
a) In this complex drawing (Fig. 1.4), three projections of the point are built D (D 1,D 2,D 3).
All three coordinates have numerical values, nonzero, so the point does not belong to any projection plane.
b) Compatible spatial image A and D(fig. 1.5). In system P 1 -P 2 point projections A and D lie on the same line of communication, only a point D above the point A, hence D- visible, and A- invisible (visible on N 1 the point that is located above)
At the fourth, final stage, we connect all three fragments of complex drawings of points A, B, C,D in one general.
Points A and D- called horizontally competing.
Construct traces of the plane given by ∆BCD and determine the distance from point A to given plane the right triangle method(coordinates of points A, B, C and D see Table 1 of the Tasks section);
1.2. An example of the task number 1
The first task presents a set of tasks on topics:
1. Orthographic projection, Monge plot, point, line, plane: by known coordinates of three points B, C, D build horizontal and frontal projections of the plane given by ∆ BCD;
2. Traces of a straight line, traces of a plane, properties of belonging to a straight plane: construct traces of the plane given by ∆ BCD;
3. General and specific planes, intersection of a straight line and a plane, perpendicularity of a straight line and a plane, intersection of planes, right triangle method: determine the distance from the point A to the plane ∆ BCD.
1.2.1. By the known coordinates of three points B, C, D we construct the horizontal and frontal projections of the plane given by ∆ BCD(Figure 1.1), for which it is necessary to construct horizontal and frontal projections of the vertices ∆ BCD, and then connect the projections of the vertices of the same name.
It is known that trail of the plane is called the straight line obtained as a result of the intersection of a given plane with the projection plane .
Near the plane general position 3 tracks: horizontal, frontal and profile.
In order to build traces of a plane, it is enough to construct traces (horizontal and frontal) of any two straight lines lying in this plane and connect them to each other. Thus, the trace of the plane (horizontal or frontal) will be uniquely determined, since through two points on the plane (in this case, these points will be traces of straight lines), you can draw a straight line, and moreover, only one.
The basis for this construction is property of belonging to a straight plane: if a straight line belongs to a given plane, then its traces lie on the traces of the same name on this plane .
The trace of a straight line is the point of intersection of this straight line with the plane of projections .
The horizontal trace of a straight line lies in the horizontal plane of the projections, the frontal one - in the frontal plane of the projections.
Consider building horizontal track straight DB, for which it is necessary:
1. Continue frontal projection straight line DB before crossing the axis X, intersection point M 2 is a frontal projection of a horizontal track;
2. From point M 2 restore the perpendicular (the line of the projection connection) before its intersection with the horizontal projection of the straight line DB M 1 and will be the horizontal projection of the horizontal trace (Figure 1.1), which coincides with the trace itself M.
Similarly, the construction of the horizontal trace of the segment is carried out. SV straight: point M '.
To build frontal footprint segment CB direct, it is necessary:
1. Continue horizontal projection straight line CB before crossing the axis X, intersection point N 1 is a horizontal projection of the frontal track;
2. From point N 1 restore the perpendicular (line of projection connection) before its intersection with the frontal projection of the straight line CB or its continuation. Intersection point N 2 and will be the frontal projection of the frontal trace, which coincides with the trace itself N.
By connecting the dots M ′ 1 and M 1 by a straight line segment, we obtain the horizontal trace of the plane απ 1. Point α x of intersection of απ 1 with the axis X called vanishing point ... To construct the frontal trace of the plane απ 2, it is necessary to connect the frontal trace N 2 with the vanishing point of the tracks α x
Figure 1.1 - Construction of plane traces
The algorithm for solving this problem can be presented as follows:
- (D 2 B 2 ∩ OX) = M 2 ;
- (MM 1 ∩ D 1 B 1) = M 1 = M;
- (C 2 B 2 ∩ OX) = M ′ 2 ;
- (M ′ 2 M ′ 1 ∩ C 1 B 1) = M ′ 1 = M ′;
- (CB∩ π 2) = N 2 = N;
- (MM ′) ≡ απ 1;
- (α x N) ≡ απ 2.
1.2.2. To solve the second part of the first task, you need to know that:
- distance from point A to the plane ∆ BCD is determined by the length of the perpendicular restored from this point to the plane;
- any straight line is perpendicular to a plane if it is perpendicular to two intersecting straight lines lying in this plane;
- on the diagram of the projection of a straight line, perpendicular plane, are perpendicular to the oblique projections of the horizontal and front of this plane or to the traces of the same name of the plane (Fig. 1.2) (see in the lectures the Theorem on the perpendicular to the plane).
To find the base of the perpendicular, it is necessary to solve the problem of intersection of a straight line (in this problem, such a straight line is the perpendicular to the plane) with the plane:
1. To enclose the perpendicular in the auxiliary plane, which should be taken as the plane of a particular position (horizontally projecting or frontally projecting, in the example, horizontally projecting γ is taken as an auxiliary plane, that is, perpendicular to π 1, its horizontal trace γ 1 coincides with a horizontal projection of the perpendicular);
2. Find the line of intersection of a given plane ∆ BCD with auxiliary γ ( MN in fig. 1.2);
3. Find the point of intersection of the line of intersection of the planes MN perpendicular (point TO in fig. 1.2).
4. To determine the true value of the distance from the point A to a given plane ∆ BCD should use the right triangle method: the true value of the segment is the hypotenuse of a right-angled triangle, one leg of which is one of the projections of the segment, and the other is the difference in distances from its ends to the plane of projections in which the construction is being carried out.
5. Determine the visibility of the perpendicular sections by the method of competing points. For example - points N and 3 to determine visibility at π 1, point 4 , 5 - to determine the visibility at π 2.
Figure 1.2 - Building a perpendicular to the plane
Figure 1.3 - An example of design control task №1
Video example of task # 1
1.3. Task options 1
| Option | Coordinates (x, y, z) of points | |||
|---|---|---|---|---|
| A | V | WITH | D | |
| 1 | 15; 55; 50 | 10; 35; 5 | 20; 10; 30 | 70; 50; 40 |
| 2 | 80; 65; 50 | 50; 10; 55 | 10; 50; 25 | 75; 25; 0 |
| 3 | 95; 45; 60 | 130; 40; 50 | 40; 5; 25 | 80; 30; 5 |
| 4 | 115; 10; 0 | 130; 40; 40 | 40; 5; 25 | 80; 30; 5 |
| 5 | 55; 5; 60 | 85; 45; 60 | 100; 5; 30 | 50; 25; 10 |
| 6 | 55; 5; 60 | 70; 40; 20 | 30; 30; 35 | 30; 10; 10 |
| 7 | 60; 10; 45 | 80; 45; 5 | 35; 0; 15 | 10; 0; 45 |
| 8 | 5; 0; 0 | 35; 0; 25 | 20; 0; 55 | 40; 40; 0 |
| 9 | 50; 5; 45 | 65; 30; 10 | 30; 25; 55 | 20; 0; 20 |
| 10 | 60; 50; 35 | 40; 30; 0 | 30; 15; 30 | 80; 5; 20 |
| 11 | 65; 35; 15 | 50; 0; 30 | 20; 25; 25 | 5; 0; 10 |
| 12 | 75; 65; 50 | 45; 10; 35 | 60; 20; 10 | 10; 65; 0 |
| 13 | 95; 0; 15 | 85; 50; 10 | 10; 10; 10 | 55; 10; 45 |
| 14 | 45; 40; 40 | 80; 50; 10 | 10; 10; 10 | 55; 10; 45 |
| 15 | 80; 20; 30 | 55; 30; 60 | 15; 10; 20 | 70; 65; 30 |
| 16 | 75; 35; 35 | 55; 30; 60 | 25; 10; 20 | 70; 65; 30 |
| 17 | 75; 65; 50 | 45; 5; 55 | 5; 45; 10 | 70; 20; 0 |
| 18 | 65; 15; 20 | 40; 5; 60 | 0; 5; 25 | 60; 60; 20 |
| 19 | 70; 20; 10 | 45; 15; 60 | 5; 10; 20 | 60; 65; 10 |
| 20 | 20; 50; 45 | 10; 20; 10 | 55; 50; 10 | 80; 0; 60 |
| 21 | 0; 5; 50 | 50; 50; 40 | 5; 55; 10 | 45; 5; 0 |
| 22 | 55; 50; 65 | 45; 55; 5 | 0; 10; 45 | 70; 0; 40 |
| 23 | 65; 5; 15 | 40; 60; 10 | 0; 20; 5 | 60; 20; 60 |
| 24 | 50; 20; 45 | 45; 60; 30 | 5; 20; 10 | 60; 30; 5 |
| 25 | 55; 15; 40 | 40; 50; 25 | 5; 15; 10 | 50; 40; 10 |
| 26 | 15; 45; 40 | 10; 25; 5 | 20; 10; 30 | 65; 40; 35 |
| 27 | 70; 30; 30 | 55; 30; 60 | 20; 5; 15 | 65; 60; 25 |
| 28 | 90; 0; 15 | 80; 45; 10 | 10; 10; 10 | 50; 10; 45 |
| 29 | 110; 10; 0 | 120; 35; 30 | 35; 5; 20 | 70; 20; 5 |
| 30 | 45; 40; 40 | 80; 45; 10 | 10; 10; 10 | 55; 10; 40 |
