The video course "Get an A" includes all the topics necessary for the successful passing of the exam in mathematics by 60-65 points. Completely all tasks 1-13 of the Profile USE in mathematics. Also suitable for passing the Basic USE in mathematics. If you want to pass the exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!

Preparation course for the exam for grades 10-11, as well as for teachers. Everything you need to solve part 1 of the exam in mathematics (the first 12 problems) and problem 13 (trigonometry). And this is more than 70 points on the Unified State Examination, and neither a hundred-point student nor a humanist can do without them.

All the necessary theory. Quick solutions, traps and secrets of the exam. All relevant tasks of part 1 from the Bank of FIPI tasks have been analyzed. The course fully complies with the requirements of the USE-2018.

The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.

Hundreds of exam tasks. Text problems and probability theory. Simple and easy to remember problem solving algorithms. Geometry. Theory, reference material, analysis of all types of USE tasks. Stereometry. Cunning tricks for solving, useful cheat sheets, development of spatial imagination. Trigonometry from scratch - to task 13. Understanding instead of cramming. Visual explanation of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. Base for solving complex problems of the 2nd part of the exam.

Proofs of theorems on the properties of a circle circumscribed about a triangle

Midperpendicular to the segment

Definition 1 . Midperpendicular to the segment called, a straight line perpendicular to this segment and passing through its middle (Fig. 1).

Theorem 1. Each point of the perpendicular bisector to the segment is at the same distance from the ends this segment.

Proof . Consider an arbitrary point D lying on the perpendicular bisector to the segment AB (Fig. 2), and prove that triangles ADC and BDC are equal.

Indeed, these triangles are right-angled triangles whose legs AC and BC are equal, while the legs DC are common. From the equality of triangles ADC and BDC, the equality of segments AD and DB follows. Theorem 1 is proved.

Theorem 2 (Reverse to Theorem 1). If a point is at the same distance from the ends of a segment, then it lies on the perpendicular bisector to this segment.

Proof . Let us prove Theorem 2 by the method “by contradiction”. To this end, suppose that some point E is at the same distance from the ends of the segment, but does not lie on the perpendicular bisector to this segment. Let us bring this assumption to a contradiction. Let us first consider the case when points E and A lie on opposite sides of the perpendicular bisector (Fig. 3). In this case, the segment EA intersects the perpendicular bisector at some point, which we will denote by the letter D.

Let us prove that the segment AE is longer than the segment EB . Really,

Thus, in the case when the points E and A lie on opposite sides of the perpendicular bisector, we have obtained a contradiction.

Now consider the case when points E and A lie on the same side of the perpendicular bisector (Fig. 4). Let us prove that the segment EB is longer than the segment AE . Really,

The resulting contradiction completes the proof of Theorem 2

Circle circumscribing a triangle

Definition 2 . A circle circumscribing a triangle, call the circle passing through all three vertices of the triangle (Fig. 5). In this case the triangle is called a triangle inscribed in a circle or inscribed triangle.

Properties of a circle circumscribed about a triangle. Sine theorem

Figure	Picture	Property
Midperpendiculars to the sides of the triangle		intersect at one point .
Center circumscribed about an acute triangle of a circle		Center described about acute-angled inside triangle.
Center circle circumscribed about a right triangle		The center of the described about rectangular midpoint of the hypotenuse .
Center circumscribed about an obtuse triangle of a circle		Center described about obtuse circle triangle lies outside triangle.
		,
Square triangle		S= 2R 2 sin A sin B sin C ,
Radius of the circumscribed circle		For any triangle, the equality is true:

Midperpendiculars to the sides of a triangle

All perpendicular bisectors drawn to the sides of an arbitrary triangle, intersect at one point .

Circle circumscribing a triangle

Any triangle can be circumscribed by a circle. . The center of the circle circumscribed about the triangle is the point at which all the perpendicular bisectors drawn to the sides of the triangle intersect.

Center of a circle circumscribed about an acute triangle

Center described about acute-angled circle triangle lies inside triangle.

Center of a circle circumscribed about a right triangle

The center of the described about rectangular circle triangle is midpoint of the hypotenuse .

Center of a circle circumscribed about an obtuse triangle

Center described about obtuse circle triangle lies outside triangle.

For any triangle, equalities are valid (sine theorem): , where a, b, c are the sides of the triangle, A, B, C are the angles of the triangle, R is the radius of the circumscribed circle.

Area of ​​a triangle

For any triangle, the equality is true: S= 2R 2 sin A sin B sin C , where A, B, C are the angles of the triangle, S is the area of ​​the triangle, R is the radius of the circumscribed circle.

Radius of the circumscribed circle

For any triangle, the equality is true: where a, b, c are the sides of the triangle, S is the area of ​​the triangle, R is the radius of the circumscribed circle.

Proofs of theorems on the properties of a circle circumscribed about a triangle

Theorem 3. All midperpendiculars drawn to the sides of an arbitrary triangle intersect at one point.

Proof . Consider two perpendicular bisectors drawn to the sides AC and AB of the triangle ABC , and denote the point of their intersection with the letter O (Fig. 6).

Since the point O lies on the perpendicular bisector to the segment AC , then, by virtue of Theorem 1, the following equality holds:

Since the point O lies on the perpendicular bisector to the segment AB , then, by virtue of Theorem 1, the following equality holds:

Therefore, the equality is true:

whence, using Theorem 2, we conclude that the point O lies on the perpendicular bisector to the segment BC. Thus, all three perpendicular bisectors pass through the same point, which was to be proved.

Consequence. Any triangle can be circumscribed by a circle. . The center of the circle circumscribed about the triangle is the point at which all the perpendicular bisectors drawn to the sides of the triangle intersect.

Proof . Let's consider the point O, at which all perpendicular bisectors drawn to the sides of the triangle ABC intersect (Fig. 6).

When proving Theorem 3, the following equality was obtained:

from which it follows that the circle centered at the point O and radii OA , OB , OC passes through all three vertices of the triangle ABC , which was to be proved.

For a triangle, both an inscribed circle and a circumscribed circle are always possible.

For a quadrilateral, a circle can only be inscribed if the sums of its opposite sides are the same. Of all the parallelograms, only a rhombus and a square can be inscribed with a circle. Its center lies at the intersection of the diagonals.

A circle can be circumscribed around a quadrilateral only if the sum of the opposite angles is 180°. Of all parallelograms, only about a rectangle and a square can a circle be circumscribed. Its center lies at the intersection of the diagonals.

A circle can be circumscribed around a trapezoid, or a circle can be inscribed in a trapezoid if the trapezoid is isosceles.

Center of the circumscribed circle

Theorem. The center of the circle circumscribed about the triangle is the point of intersection of the perpendicular bisectors to the sides of the triangle.

The center of the circle circumscribed about the polygon is the point of intersection of the midperpendiculars to the sides of this polygon.

Center Inscribed circle

Definition. A circle inscribed in a convex polygon is a circle that touches all sides of this polygon (that is, each of the sides of the polygon is tangent to the circle).

The center of the inscribed circle lies inside the polygon.

A polygon in which a circle is inscribed is called a circumscribed polygon.

A circle can be inscribed in a convex polygon if the bisectors of all its interior angles intersect at one point.

Center of a circle inscribed in a polygon- the point of intersection of its bisectors.

The center of the inscribed circle is equidistant from the sides of the polygon. The distance from the center to any side is equal to the radius of the inscribed circle. By the property of tangents drawn from one point, any vertex of the circumscribed polygon is equidistant from the tangent points lying on the sides emerging from this vertex.

Any triangle can be inscribed in a circle. The center of a circle inscribed in a triangle is called the incenter.

A circle can be inscribed in a convex quadrilateral if and only if the sums of the lengths of its opposite sides are equal. In particular, a circle can be inscribed in a trapezoid if the sum of its bases is equal to the sum of its sides.

A circle can be inscribed in any regular polygon. A circle can also be circumscribed about any regular polygon. The center of the inscribed and circumscribed circles lie at the center of a regular polygon.

For any circumscribed polygon, the radius of the inscribed circle can be found by the formula

Where S is the area of ​​the polygon, p is its semiperimeter.

Regular n-gon - formulas

Formulas for the length of a side of a regular n-gon

1. The formula for the side of a regular n-gon in terms of the radius of the inscribed circle:

2. The formula of the side of a regular n-gon in terms of the radius of the circumscribed circle:

The formula for the radius of the inscribed circle of a regular n-gon

The formula for the radius of the inscribed circle of an n-gon in terms of the side length:

4. The formula for the radius of the circumscribed circle of a regular triangle in terms of the length of the side:

6. The formula for the area of ​​a regular triangle in terms of the radius of the inscribed circle: S = r 2 3√3

7. The formula for the area of ​​a regular triangle in terms of the radius of the circumscribed circle:

4. The formula for the radius of the circumscribed circle of a regular quadrilateral in terms of the side length:

2. The formula of the side of a regular hexagon in terms of the radius of the circumscribed circle: a = R

3. The formula for the radius of the inscribed circle of a regular hexagon in terms of the side length:

6. The formula for the area of ​​a regular hexagon in terms of the radius of the inscribed circle: S = r 2 2√3

7. The formula for the area of ​​a regular hexagon in terms of the radius of the circumscribed circle:

S=	R2 3√3

8. Angle between sides of a regular hexagon: α = 120°

Number value(pronounced "pi") is a mathematical constant equal to the ratio

the circumference of a circle to the length of its diameter, it is expressed as an infinite decimal fraction.

Denoted by the letter of the Greek alphabet "pi". What is pi equal to? In simple cases, it is enough to know the first 3 characters (3.14).

53. Find the length of the arc of a circle of radius R corresponding to the central angle at n°

The central angle based on an arc whose length is equal to the radius of the circle is called a 1 radian angle.

The degree measure of an angle of 1 radian is:

Since the arc is long π R (semicircle), subtends the central angle to 180 ° , then an arc of length R, subtends the angle to π times smaller, i.e.

And vice versa

Because π \u003d 3.14, then 1 rad \u003d 57.3 °

If the angle contains a radian, then its degree measure is

And vice versa

Usually, when denoting the measure of an angle in radians, the name "rad" is omitted.

For example, 360° = 2π rad, write 360° = 2π

The table lists the most common angles in degrees and radians.

CHAPTER VII.

ABOUT THE CIRCLE.

165. Figures inscribed in a circle and described beside it. A polygon whose vertices are on a circle, called. inscribed in a circle; in black The 243rd represents the inscribed triangle, quadrilateral and pentagon.

A polygon whose sides touch a circle is called. described around the circle in black The 244th presents the triangles described. and quadruple.

166. If you want to inscribe some non-correct polygon in a circle, for example. heptagon, then you just need to take 7 arbitrary points A, B, C ... (drawing 245) on the circle and connect them with straight lines.

If you wanted to describe a quadrilateral around a circle, then you should take 4 points on the circle and draw tangents at these points; from the intersection of tangents and is formed described. quadrangular (Ch. 246).

167. Let us now put what needs to be inscribed in the circle of rights. polygonal, eg. pentagon. To do this, you need to divide the circle into 5 equal parts; whole circumference=360°, next in the fifth share it will be 72 °; therefore, we will build at the center of the circle (Ch. 247) along the protractor angle. 72 ° and we will lay the chord AB along the circumference; it fits exactly 5 times, and then a 5-k is formed.

It will be correct, because all its sides are equal to each other; the angles are also equal, since each of them is measured by half of three-fifths of the circle and the next. contains 108°.

If it were necessary to enter the rights. 9-k, then it would be necessary to divide the circle into 9 equal parts, i.e. build at the center corner. at 40°; at 20-ke it is necessary to build a corner. at 18°, etc.

Let's also assume that we need to enter rights. 7-to;
the seventh part of the circle \u003d 360/7 \u003d 51 3/7 \u003d 51 ° 25 "42 6/7". Not only seconds, but also minutes are not marked on the protractor; therefore, such an angle cannot be set aside exactly - we will certainly set aside either more or less than it; hence the last side of the polygon. either less or more of the other sides will come out.

It is much more accurate to inscribe right polygons without the help of a transtorter, but only with a compass and straightedge; but in this way it is possible to inscribe only some polygons, for example. square, 6-room

168. To inscribe a square in a circle, the circle must be divided. into 4 equal parts; and for this it is necessary (Ch. 248) to draw two perpendicular diameters;

if we connect their ends, we get a square, because all its sides are equal to each other, like chords subtending equal arcs; all angles are right, as having a vertex on a circle and resting on the ends of the diameter.

169. To enter the circle of rights. hexagonal., set aside from some point of the circle. (Ch. 249) chord AB = radius; then, having drawn the radii of AO and VO, we obtain an equilateral tr-to AOB; next corner ABO = 60°, and the arc AB will be the sixth of the circle; and therefore the chord AB will be deposited along the circumference. exactly 6 times.

170. Knowing how to enter rights. 6-k, it's easy to enter and right. tr-to. To do this, first divide the circle into 6 equal parts (Ch. 250) at points B, A, C, then connect points A, C and E; we get the correct tr-to ACE, because its sides are equal, since each of them corresponds to an arc that is 1/3 of the circle.

171. The next saosobom can fit every right into the circle with sufficient accuracy. polygonal To enter eg. disposition 9-to, we draw in a circle (black. 251) dia. AB;

we build an equiostor on AB. tr-to ABC; divide AB into 9 equal parts; we connect the vertex C of the tr-ka with the second division point D and continue the straight line CD until it intersects with the circle. in E; chord AE will be deposited around the circumference 9 times.

If it were necessary to enter the rights. 5-k, then it would be necessary to divide the diameter into 5 equal parts (black. 252); for 7 into 7 parts (cher. 253), etc.

172. If anyone is right. polygonal is inscribed in a circle, then the number of sides can be doubled, i.e. enter such rights. many, which would have twice as many sides.

Let eg. ABCDEG (Ch. 254) would be right. 6-to; let's drop perpendiculars from the center O to all sides of the mn-ka; then the arcs AB, BC ... will be divided in half; connecting the division points with the vertices of 6, we get the rights. 12-k. Dropping the perpend. on the sides of this 12, let's write the rights. 24-k, then 48-k, etc.

Thus, using a compass and a ruler, we can enter the correct 6s, 12s, 24s ... as well as squares, 8s, 16s ... into a circle ...

173. With an increase in the number of sides is inscribed. polygonal, the sides themselves will become smaller and smaller, and the perimeter of the mn-ka will more and more approach the circle, so that the circle can be considered as the perimeter of such a right. mn-ka, which has an extremely many sides.

174. If you enter rights in a circle. polygonal, it is easy to describe the rights. polygonal the same number of sides.

Let eg. ABCDE (Ch. 255) will be right. 5-to; we lower from the center to the sides many perpendiculars and draw tangents through the points M, N ..; then it will be right. described 5-k.

It is also possible (Ch. 256) to draw tangents through the vertices of the inscribed polygon.

175. Let's consider, about what figures it is possible to describe a circle. We already know (§ 143) that it is always possible to draw a circle through three points that do not lie on the same straight line; next a circle can be circumscribed around any triangle.

Now let's take a quadrilateral. ABCD (Ch. 257). Let's do the district. through three points A, B, C (we already know how to do this); district this one may also pass through point D, but it may not pass. If it passes through D, then corner. D will contain as many degrees as there are in 1/2 of the arc ABC; and since ug. B is measured 1/2 of the arc ADC, while the arcs ABC and ADC together make up a whole circle, then the angles D and B will total 180 °; but the sum of all angles quadrangular=360°, next and A+C==180°.

So, A circle can only be circumscribed around a quadrilateral in which the sum of the opposite angles is 180°. Thus, it is possible to describe a circle around a rectangle, but not around an oblique parallelogram.

176. About every right. polygonal can describe a circle. Let ABCDEF (Ch. 258) be right. polygon; inside it you can find a point that will be at an equal distance from all its vertices.

To do this, we divide the angles A and B in half by the lines AO and BO; the point of intersection of these lines will be the desired one. We will prove that the lines AO, BO, CO, DO ... are equal to each other.

triangle ABO \u003d OBC, because they have a common side BO, AB \u003d BC, as the sides are right. mn-ka, ug. t = ang. P , as half of angle B; sded. and line AO ​​= CO; but AO \u003d VO, because the tr-to ABO is isosceles, since the corner. t = ang. R , as half equal angles; next, all three lines AO, VO, CO are equal to each other. Comparing the tr-ki BOC and COD, we find that BO = CO = OD ...; next if from O with radius OA, or BO, or OC ... describe a circle, then it will pass through the vertices of all corners of the polygon.

177. If about right. many (black. 259) a circle is described, then the sides AB, BC ... this many will be chords in a circle;

but equal chords are at equal distances from the center; next perpendiculars OM, ON .., lowered from the center O to the sides of the multiple, will be equal to each other, and if we describe a circle from O with a radius of OM or ON .., then it will touch all sides of the multiple at points M, N ... Such a circle is called. inscribed, and its radius is called apothem a lot.

So, in every right. polygonal you can draw a circle.

Thus, the center of the circle described near the mn-ka and inscribed will be the point of intersection of the lines dividing the two corners of the mn-ka in half; described by the radius. the circle will be a line connecting the center with the top of one of the corners of the mn-ka; and inscribed in radius. circle or apothem - a perpendicular, lowered from the center to one of the sides of the mn-ka. The center of the inscribed and circumscribed circles is called. also the center of the correct plural.

178. Questions. 1) What plurals are called. inscribed in a circle? described? 2) Inscribe in a circle some kind of mn-k? describe? 3) How to inscribe in a circle by means of a protractor some rights. mn-to? 4) How to inscribe a square in a circle using a compass and ruler? rights. 6-to? 5) If right. mn-k is inscribed in a circle, then how to enter rights. mn-to, having twice as many sides? 6) If right. mn-k is inscribed in a circle, then how to describe the rights. many of the same number of sides? 7) What is done with the rights perimeter. entered. mn-ka with an increase in the number of sides of it? 8) Is it always possible to describe a circle around the tr-ka? 9) Prove that about everyone is right. can you describe and fit a circle into it? 10) Can a parallelogram be inscribed in a circle? a trapezoid? 11) What is done with the rights perimeter. described. mn-ka with an increase in the number of sides of it?

179. Tasks. 1) Fit in a circle 4-k? 8-to? 10-to? 15th?

2) Describe about a 4-k circle? 7-to? 3-to? 5-to?

3) Inscribe in a circle by transp. rights. 10-in? 15th? 20-k?

4) Inscribe in a circle using a compass and a ruler of rights. 8-to?

5) Describe about the circle by transp. rights. 5-to? 9-to? 10-to?

6) Describe around the circle using a compass and a ruler of rights. 3-to? 6-to? square? 12-to?

7) A circle is described near the tr-ka, and its center is inside the tr-ka; what kind of truck is this? What kind of shopping mall would it be if the center was on the side of the shopping mall? out of tr-ka?

8) Draw such a law with a protractor. 5-k, 8-k, 10-k, so that the radius of the circle described around it = lines t ?

9) Draw such a law with a protractor. 5-k, so that his apothem is equal to this line?

10) On a given straight line, use a protractor to construct the correct 5-k? 8-to? 10-to?

11) A chord is drawn in a circle; perpendiculars are erected from its ends until they meet the circle; meeting points are connected by a straight line; what kind of quadrilateral is it?

12) Glad. circle=3.6 inch; what is the perimeter of the circumscribed square?

13) Prove that the side of a regular track inscribed in a circle is at a distance of half the radius from the center of this circle?

14) A 4-k is inscribed in a circle; its vertices divide the circumference into parts that are in the ratio 4:7:5:11; determine angles 4-ka?

15) Rights are inscribed in the circle. tr-k, and its side is 7 1/2 inches apart. from the center of this circle; determine the radius of the circle?

16) Prove that the interior angle of every right. mn-ka serves as an addition to 180 ° to the angle that will be obtained from the connection of two adjacent vertices of this mn-ka with its center?

17) Prove that if the chord AB (drawing 260) = the radius of the circle O, and AO is the right side. inscribed 10, then by connecting point B with C, we get the right side. inscribed 15th.

On this straight line a build using a compass and ruler: 18) rights. tr-k? 19) square? 20) right. 6-to? 21) right. 8-to? 22) right. 12-to?

By means of a compass and a ruler, construct: 23) a square in rad. r described circle? 24) square by apotheme a ? 25) right. 6-to glad. r description circle? 26) right. 6-to apotheme a ? 27) right. 3-to rad. r description circle? 28) right. 3-to apotheme a ?

29) Inscribe a circle in this rhombus?

30) Describe a circle around a rectangle?

31) A tr-k is inscribed in a circle; one side of it is the diameter, and the other two subtend the arcs, whose ratio is 15:17; determine the angles of the tr-ka?