Solving C4 problems from the word mathematics (start). IV.
Time to the exam is less and less trial EGE It is more and more often, the nerves of schoolchildren and their teachers are tensioning all stronger. On the eve of the opening of the "intensive preparation" season for graduation and entrance exams, I suggest you to practice in solving C4 problems from the benefit developed by Mio to prepare schoolchildren to the exam in mathematics. The tasks are given with solutions, however, it would be useful to solve them first independently.
Option 3. Triangle ABC inscribed in the circle of radius 12. It is known that AB \u003d 6 I. BC. \u003d 4. Find AC.
Decision:
From the sinus theorem for a triangle ABC We have:
From the main trigonometric identity Find that:
Then on the cosine theorem for the triangle ABC We have for both cases:
Answer: √35 ± √15.
Option 5. In a triangle ABCheld heights BM.and CN., O.- Center inscribed circle. It is known that BC \u003d.24 , Mn \u003d.12. Locate the circle radius described near the triangle BOC..
Decision:
Two possible cases: ∠a - sharp and ∠a - stupid
Two cases are possible:
1) Let ∠ A. - acute (left drawing). We prove that triangles Amn. and ABC Like. Indeed, points B., N., M. and C. Lying on a circle with a diameter BC., therefore, ∠ NMB. = ∠NCB.from rectangular triangles BAM and BNC.:
∠Amn. = 90 0 — ∠NMB,∠B \u003d.90 0 —
∠NCB., of which, obviously, it follows that ∠ Amn.=
∠B.besides ∠ A.- Common for both triangles, therefore, they are similar to two corners.
Of rectangular triangle Amb.: cos∠. A. = AM./AB ANC.: cos∠. A. = AN./AC.The same relationships are obviously the ratios of the parties in such triangles Amn. and ABCwhat follows that COS∠ A \u003d nm./BC \u003d.1/2, which means ∠ A \u003d.60 0, since the sum of the corners in the triangle is 180 0, ∠ B +.∠C \u003d.
120 0. The center inscribed in the triangle of the circle lies, as is known, at the point of intersection of its bisector. From this we conclude that:
∠OBC +.∠O. CB \u003d.
1/2 · (∠ B +.∠C) \u003d.
60 0, which means ∠ Boc \u003d.120 0. By the sinus theorem for a triangle BOC. We have: BC./ Sin∠. BOC. =
2R.where R. R. = 8√3.
2) Let now ∠ A. - stupid (right drawing). From a rectangular triangle ABM. Find that Cos∠ BAM = AM./AB, from a rectangular triangle CAN Find that Cos∠ CAN \u003d AN./AC. ∠BAM \u003d.∠C. AN. since they are vertical means AM./AB = AN./AC \u003d COS∠. BAM \u003d COS∠. BAs Since the last front corners are adjacent. So triangles ABC and ANM. Like the corner and two proportional parties. The similarity ratio is COS∠ BAc \u003d Mn. /BC \u003d. -1/2, and the corner itself ∠ BAc \u003d 120 0 .
Further reasoning are similar. Since the sum of the corners in the triangle is 180 0, ∠ B +.∠C \u003d.
60 0. The center inscribed in the triangle of the circle lies at the point of intersection of its bisector, so:
∠OBC +.∠O. CB \u003d.
1/2 · (∠ B +.∠C) \u003d.
30 0, which means ∠ Boc \u003d.150 0. By the sinus theorem for a triangle BOC. We have: BC./ Sin∠. BOC. =
2R.where R.- The desired radius described near the triangle of the circle. From here: R. = 24.
Answer: 8√3 or 24.
Option 8. The perimeter of an in-beamed trapezium is 52. It is known that in this trapezion you can enter the circle, and the side is divided by a touch point in relation to 4: 9. Direct, passing through the center of the circle and the vertex of the trapezium, cuts off from the trapezion of the triangle. Find the attitude of this triangle area to the trapezoid area.
Decision:
Figure to solve the C4 task with a trapezium
By the theorem on the segments of tangents KB. = BP. = PC. = CQ. = 4x., QD. = DL = LA = AK = 9x., then the perimeter of the trapezium is 4 · (9 x. + 4x.) \u003d 52, from where x. \u003d 1. From here we calculate the sides AB = CD \u003d 13 and base BC. = 8, AD \u003d 18. then AH. = (AD — BC.) / 2 \u003d 5. From a rectangular triangle BHA. According to Pythagora theorem we find the height of the trapez Bh. \u003d 12, sin∠ A. \u003d sin∠. D. \u003d 12/13. The area of \u200b\u200bthe trapezium is then equal S. = (BC. + AD) · Bh./2 = 156.
Depending on what direct is referred to in terms of the problem, two cases are possible:
1) Let this direct passes through the vertex containing a smaller base of the trapezoid (in the drawing is straight BM.). The center inscribed in the corner of the circle lies on its bisector, that is, ∠ ABM. = ∠MBC., ∠MBC. = ∠Amb. (as the liar with parallel straight lines BC., AD And Sale BM.) means ∠ ABM. = ∠Amb. and triangle ABM. - Isol, AM. = AB \u003d 13. Then the area of \u200b\u200bthe triangle ABM. \u003d 0.5 · AB · AM. · Sin∠. A. \u003d 0.5 · 13 · 13 · 12/13 \u003d 78, and the desired ratio is 78/156 \u003d 1/2.
2) Now let the direct referring to the condition, passes through a vertex containing a smaller base of the trapezium (in the drawing is straight AN.). Perform additional construction: I will prolong the base BC. And straight AN. before intersection at point Y.. Similarly, we prove that the triangle ABY. - Isol, AB = By = 13, CY. = By — BC. \u003d 5. Triangles CNY and And. Like two corners (∠ And. = ∠CNY like vertical, ∠ Cya. = ∠Yad. how will the lies underly with parallel straight lines BC., AD And Sale AY.) So DN. : NC. = AD : CY. \u003d 18: 5, then DN. = 18/23 CD = 18/23 AB \u003d 234/23. Then the area of \u200b\u200bthe triangle ADN. \u003d 0.5 · AD · DN. · Sin∠. D. \u003d 0.5 · 18 · 234/23 · 12/13 \u003d 1944/23, and the desired relation is 162/299.
Answer: 1/2 or 162/299.
Sergey Valerievich
Sections: Mathematics
At the final lessons on the geometry of the time to break the tasks around the course as a whole, practically does not remain. A B. Kim Eger Traditionally, tasks are included, the solution of which requires knowledge of planimeuria on the topic "inscribed and described circles." Therefore, the proposed material will help not only to recall this topic, but also to systematize the previously obtained knowledge to solve the planimetric tasks to the inscribed and described circles, as well as to prepare for solving such tasks in the USE. It is assumed that the student at least at the minimum level owns the whole course of school geometry (planimetry).
The first and most important stage of the decision of the geometric problem is to build a drawing. It is impossible to learn to solve sufficiently meaningful tasks, without working out strong skills for the manufacture of "good" drawings, without working out habits (even reflex) - not to start solving the task until "big and beautiful" drawing is made. As the main method of solving geometrical problems, an algebraic method is put forward with the compilation of the subsequent algorithm. Putting the algebraic method into the chapter of the corner, it is necessary to warn against excessive passion of algebra and a score, not to forget that we are talking Yet about geometric tasks, and therefore, working on the task, you should look for geometric features, learn to watch and see geometry. Highlighting two terms defining the ability to solve geometrical tasks- drawing plus method, add the third - possession of certain theorems and reference tasks known to geometric facts.
I. The necessary theorems and reference tasks for the circle inscribed in the triangle and the quadrangle, and the circle described near the triangle and the quadrangle. ( Attachment 1 )
II. Solving tasks on ready-made drawings (conveniently use the codecope).
In this case, the students verbally explain the course of solving problems, formulate theorems and reference tasks used in solving tasks on ready-made drawings.
Ready drawing |
Dano |
Decision |
| AB \u003d BC. | Tanner segments are: Bm \u003d bk \u003d 5 AB \u003d BC \u003d 12 Mc \u003d Cn \u003d 7, AC \u003d 14, AK \u003d AN \u003d 7, Pabc \u003d 12 + 12 + 14 \u003d 38 Answer: P ABC \u003d 38 |
|
| AB \u003d 6, |
Tanner sections are equal: av \u003d sun 1)  , 2) AB \u003d Sun, because In - Bissektris 3) ABC - equilateral, pabc \u003d 6 3 \u003d 18 Answer: P ABC \u003d 18 |
|
|
AD - Diameter of the circle, AB \u003d 3, VD \u003d 4. 1. Prove: NM AD 2. R \u003d? |
1. Because AD - diameter, then DB An and AC DN, i.e. AC and DB - altitude AND, then NK - height, because They intersect at one point. So NM AD. 2. ad \u003d \u003d 5, R \u003d Answer: R \u003d 2.5 |
| R \u003d? | AC - Diameter of the circle and hypotenuse of rectangular ABC, R \u003d \u003d 1.5 Answer: R \u003d 1.5 |
|
| AB \u003d 24, OK \u003d 5. |
O is the intersection point of middle perpendiculars to the parties. BKO - rectangular, VK \u003d AK \u003d 12, Ko \u003d 5, at \u003d \u003d 13 \u003d r Answer: R \u003d 13 |
III. Solving tasks.
1. Find the perimeter of the rectangular triangle, if the radius of the inscribed circle is 2 cm, and the hypotenuse is 13 cm.
 |
Let am \u003d an \u003d x, then AC \u003d X + 2, CB \u003d 2 + 13 - X \u003d 15 - X (x + 2) 2 + (15 - x) 2 \u003d 169 x 2 - 13x + 30 \u003d 0 x 1 \u003d 10, x 2 \u003d 3; AC \u003d 6, CB \u003d 12; P \u003d 30 cm Answer: P \u003d 30 cm. |
2. The radius inscribed in the rectangular triangle of the circle is 3 cm, o - center inscribed circle ,,. Find a triangle area.
 |
JSC - Bissektris, AKO - rectangular, sin \u003d sin 30 o \u003d  , AO \u003d 6, AN \u003d AK \u003d \u003d 3, AC \u003d 3 + 3, TG 60 O \u003d, CB \u003d  S ABC \u003d.  = Answer: S \u003d cm2. |
3. Triangle perimeter 84. The touch point of the inscribed circle divides one of the sides to the segments 12 and 14. Find the radius of the inscribed circle and the ABC area, if the ov \u003d 18, o is the center of the inscribed circle.
4. In an equally chained triangle, the distance from the center of the inscribed circle to the vertex of a no equal angle is 5 cm. Bore is 10 cm. Find the radius of the inscribed circle.
 |
OB \u003d 5,  ,OM \u003d OB. . =  , Bh \u003d 5 + r, AH \u003d 2R, AHB - rectangular,  4R 2 \u003d 100 - (5 + R) 2, R 2 + 2R - 15 \u003d 0, R 1 \u003d - 5, R 2 \u003d 3 Answer: R \u003d 3 cm. |
5. The basis of an equally sized triangle, inscribed in a radius circle of 5 cm, is 6 cm. Find the perimeter of the triangle.
| AHO - rectangular: oh \u003d 4, bh \u003d 4 + 5 \u003d 9, AB \u003d BC \u003d \u003d P \u003d. Answer: P \u003d cm. |
6. The perimeter of the ABC triangle is 72 cm. AB \u003d BC, AB: AC \u003d 13:10. Find the radius described near the triangle of the circle.
AB + BC + AC \u003d 72,  ,  AC \u003d 20, AB \u003d BC \u003d \u003d 26, BH \u003d 24 Bn \u003d na \u003d 13,  , R \u003d.  Answer: R \u003d cm. |
7. The basis of a stupid iscessed triangle is equal to 24 cm, and the radius of the described circle is 13 cm. Find the side side of the triangle.
8. The circle, the diameter of which serves the ABS triangle, passes through the intersection point of the median of this triangle. Find the ratio of the length of the side of the AC to the length of the median spent on it.
 |
AO \u003d OC \u003d R \u003d OM, BM \u003d 2R, Bo \u003d 3r, Answer:. |
9. Locate the equal trapezium area described near the circle with a radius 4, if it is known that the lateral side of the trapezium is equal to 10.
 |
S ABCD \u003d. Because Circle inscribed, then AB + CD \u003d AD + BC \u003d 20 H \u003d 2R \u003d 8,  , S abcd \u003d 10 8 \u003d 80 Answer: 80. |
10. Dan Rhombd ABCD. The circle described near the ABD triangle crosses the large diagonal of the AC rhombus at the point E. Find CE if AB \u003d, BD \u003d 16.
IV. Tasks for self-decide.
1. The radius of the circle, inscribed in the rectangular triangle, is 2 cm, and the radius of the circle described is 5 cm. Find a larger triangle catache.
Answer: (6; 8).
2. Near an equifiable triangle with the base of the AC and an angle at the base of the 75th describes a circle with the center of O. Find its radius if the area of \u200b\u200bthe triangle is equal to 16.
Answer: (8).
3. Find the roundabout radius included in the ABC's acute triangle if the height BH is 12 and it is known that.
Answer: (4).
4. One of the cathets of the rectangular triangle is 15, and the projection of the second category on the hypotenuse is 16. Locate the circle diameter described near this triangle.
Answer: (25).
5. A circumference is inscribed in an equally chaired triangle. In parallel, its base of the AU was carried out tangent to the circle, crossing the sides at points D and E. Find the circle radius if DE \u003d 8, AC \u003d 18.
Answer: (6).
6. Near the ABC triangle is described. The median of the AM triangle is extended to the intersection with a circle at point K. Find the AC side if AM \u003d 18, MK \u003d 8, BK \u003d 10.
Answer: (15).
7. The circle inscribed in an equilibrium triangle concerns its side sides at points K and A. Point K divides the side of this triangle on the segments 15 and 10, counting from the base. Find the length of the CA length.
Answer: (12).
8. The angle in the ABS triangle is 60 o, the radius of the circle described about ABC is 2. to find the radius of the circle passing through points A and C and the center of the circle inscribed in the ABC.
Answer: (2).
9. The sides of the triangle are equal to 5, 6 and 7. Find the ratio of segments to which the bisector of the larger angle of this triangle is divided by the center of the circle inscribed in the triangle.
Answer: (11: 7).
10. The radius of the circle, inscribed in the rectangular triangle, is equal to the durability of its cathets. Find the ratio of a larger category to a smaller.
. Find the hypotenuse and the radius of the circle described near the triangle.If all sides of the polygon touch the circle, the circumference is called inscribed in a polygon, and polygon - described Near this circle. In Figure 231, the EFMN quadrilator is described near the circle with the center O, and the DKMN quadroller is not described near this circumference, as the DK side does not apply to the circle.
Fig. 231.
In Figure 232, the ABC triangle is described near the circle with the center of O.
Fig. 232.
We prove the theorem about the circle inscribed in the triangle.
Theorem
Evidence
Consider an arbitrary triangle ABC and denote the letter on the intersection point of its bisector. Cut out from the point of perpendicular OK, OL and Oh oh, respectively, to the parties of AV, Sun and Ca (see Fig. 232). Since the point is equidistant from the side of the ABC triangle, then Ok \u003d OL \u003d Ohm. Therefore, the circle with the center of radius ok passes through points K, L and M. The sides of the ABC triangle touches this circle at points to, L, M, since they are perpendicular to the radii OK, OL and OM. So, a circle with the center of radius ok is inscribed in the ABC triangle. Theorem is proved.
Note 1.
Note that only one circle can enter into a triangle.
In fact, let's say that in a triangle you can enter two circles. Then the center of each circle is equidistant of the sides of the triangle and, it means that the point of crossing the bisector of the triangle is coincided, and the radius is equal to the distance from the point of the side of the triangle. Consequently, these circles coincide.
Note 2.
Let us turn to Figure 232. We see that the ABC triangle is made up of three triangles: ABO, and SAO. If in each of these triangles to take for the base of the side of the ABC triangle, then the radius of the circle inscribed in the ABC triangle will be height. Therefore, the area s triangle ABC is expressed by the formula
In this way,
Note 3.
In contrast to the triangle not in every quadril can enter the circle.
Consider, for example, a rectangle, in which adjacent sides are not equal, that is, a rectangle that is not square. It is clear that in such a rectangle you can "place" a circle relating to three of its parties (Fig. 233, a), but it is impossible to "put" a circle so that it concerns all four of its parties, that is, you can not enter the circle. If you can enter a circle into a quadricle, then its parties have the following wonderful property:
Fig. 233.
This property is easy to install, using Figure 233, b, on which the same letters are marked with equal segments of tangents. In fact, AV + CD \u003d a + b + c + d, aircraft + ad-a + b + c + d, therefore av + cd \u003d aircraft + ad. It turns out that the opposite statement is also true.
Described circle
If all the tops of the polygon lie on the circle, the circumference is called described Near the polygon, and a polygon - inscribed In this circle. In Figure 234, the ABCD quadril is entered into a circle with the center of Oh, and the AECD quadrilator is not inscribed in this circle, since the vertex e does not lie on the circle.
Fig. 234.
The ABC triangle in Figure 235 is inscribed in a circle with the center of O.
Fig. 235.
We prove the theorem about the circle described near the triangle.
Theorem
Evidence
Consider an arbitrary triangle ABC. Denote by the letter on the intersection point of the middle perpendicular to its parties and carry out the segments of OA, OB and OS (Fig. 235). Since the point is equidistant from the vertices of the ABC triangle, then about A \u003d OS \u003d OS. Therefore, the circle with the center of the OA radius passes through all three vertices of the triangle and, it means described near the ABC triangle. Theorem is proved.
Note 1.
Note that near the triangle can only be described by one circle..
In fact, we assume that near the triangle you can describe two circles. Then the center of each of them is equal to its vertices and therefore coincides with the point of intersection of the middle perpendiculars to the sides of the triangle, and the radius is equal to the distance from the point of the triangle vertices. Consequently, these circles coincide.
Note 2.
In contrast to the triangle about the quadril can not always describe the circle.
For example, it is impossible to describe a circle near a rhombus that is not a square (explain why). If you can describe a circle about a quadril, then its corners have the following wonderful property:
This property is easy to install if refer to Figure 236 and use the inserted corner theorem. Indeed,
where follows
Fig. 236.
It turns out true and the opposite:
Tasks
689. In an equally chained triangle, the base is 10 cm, and the side side is 13 cm. Find the radius of the circle inscribed in this triangle.
690. Find the base of an anose-free triangle if the center inscribed in it divides the height conducted to the base in relation to 12: 5, counting from the vertex, and the side side is 60 cm.
691. The point of touching the circle inscribed in an equilibrium triangle, divides one of the lateral sides to segments equal to 3 cm and 4 cm, counting from the base. Find the perimeter of the triangle.
692. A circle is inscribed in the ABC triangle, which concerns the parties of the AV, Sun and Ca at points P, Q and R. Find the AR, RV, BQ, QC, SV, RA, if Av \u003d 10 cm, Sun \u003d 12 cm, sa \u003d 5 cm.
693. In a rectangular triangle, the circle of radius is inscribed in the perimeter of the triangle, if: a) hypotenuse is 26 cm, R \u003d 4cm; b) the touch point divides the hypotenuse on segments equal to 5 cm and 12 cm.
694. Locate the circle diameter, inscribed in the rectangular triangle, if the triangle hypotenosis is equal to C, and the amount of cathets is equal to m.
695. The sum of the two opposite sides of the quadrilateral described is 15 cm. Find the perimeter of this quadril.
696. Prove that if you can enter a circle in the parallelograms, this parallelogram is rhombus.
697. Prove that the area of \u200b\u200bthe described polygon is equal to half the work of its perimeter on the radius of the inscribed circle.
698. The sum of the two opposite sides of the quadrilateral described is 12 cm, and the radius inscribed in it is 5 cm. Find the area of \u200b\u200bthe quadricle.
699. The sum of the two opposite sides of the described quadrilateer is 10 cm, and its area is 12 cm 2. Find the roundabout radius, inscribed in this quadril.
700. Prove that in any rhombus you can enter a circle.
701. Instruct three triangles: acute, rectangular and stupid. In each of them, enter the circle.
702. The triangle of the ABC is inscribed into the circle so that AV is the circle diameter. Find the corners of the triangle, if: a) bc \u003d 134 °; b) ac \u003d 70 °.
703. Invoices of ABC is chained triangle with the base of the aircraft. Find the corners of the triangle if the sun \u003d 102 °.
704. The circle with the center O is described near the rectangular triangle. a) Prove that the point is the middle of the hypotenuse. b) find the sides of the triangle if the circle diameter is equal to D, and one of acute corners The triangle is equal to α.
705. Near the rectangular triangle ABC with a direct angle with a circle described. Find the radius of this circle if: a) ac \u003d 8 cm, sun \u003d 6 cm; b) ac \u003d 18 cm, ∠b \u003d 30 °.
706. Find the side of the equilateral triangle, if the radius of the circumference described near it is 10 cm.
707. The angle, an opposite base of an in-beamed triangle is 120 °, the side side of the triangle is 8 cm. Find the circle diameter described near this triangle.
708. Prove that you can describe the circle: a) near any rectangle; b) near any anaulic trapezium.
709. Prove that if about the parallelogram can describe the circle, then this parallelogram is a rectangle.
710. Prove that if the circle can be described near the trapezion, then this trapezium is free.
711. Inscribe three triangles: stupid, rectangular and equilateral. For each of them, build the circle described.
