Direct ( MN), which has only one with its circle common point (A), called tangent to the circle.
The common point is called in this case point of contact.
Possibility of existence tangent, and, moreover, drawn through any point circle, as a point of tangency, is proven as follows theorem.
Let it be required to carry out circle with center O tangent through the point A. To do this from the point A, as from the center, we describe arc radius A.O., and from the point O, as the center, we intersect this arc at the points B And WITH a compass solution equal to the diameter of the given circle.
After spending then chords O.B. And OS, connect the dot A with dots D And E, at which these chords intersect with a given circle. Direct AD And A.E. - tangents to a circle O. Indeed, from the construction it is clear that triangles AOB And AOC isosceles(AO = AB = AC) with bases O.B. And OS, equal to the diameter circle O.
Because O.D. And O.E.- radii, then D - middle O.B., A E- middle OS, Means AD And A.E. - medians, drawn to the bases of isosceles triangles, and therefore perpendicular to these bases. If straight D.A. And E.A. perpendicular to the radii O.D. And O.E., then they - tangents.
Consequence.
Two tangents drawn from one point to a circle are equal and form equal angles with the straight line connecting this point to the center.
So AD=AE and ∠ OAD = ∠OAE because right triangles AOD And AOE, having a common hypotenuse A.O. and equal legs O.D. And O.E.(as radii), are equal. Note that here the word “tangent” actually means “ tangent segment” from a given point to the point of contact.
1. Two tangents from one point.
Let two tangents $$AM$$ and $$AN$$ be drawn to a circle with center at point $$O$$, points $$M$$ and $$N$$ lie on the circle (Fig. 1).
By definition of tangent $$OM \perp AM$$ and $$ON \perp AN$$. In right triangles $$AOM$$ and $$AON$$, the hypotenuse $$AO$$ is common, the legs $$OM$$ and $$ON$$ are equal, which means $$\Delta AOM = \Delta AON$$. From the equality of these triangles it follows that $$AM=AN$$ and $$\angle MAO = \angle NAO$$. Thus, if two tangents are drawn from a point to a circle, then:
1.1$$(\^{\circ}$$. !} the tangent segments from this point to the tangent points are equal;
1.2$$(\^{\circ}$$. !} a straight line passing through the center of the circle and given point, bisects the angle between the tangents.
Using property 1.1$$(\^{\circ}$$, легко решим следующие две задачи. (В решении используется тот факт, что в каждый треугольник можно вписать окружность).!}
Based on $$AC$$ isosceles triangle$$ABC$$ is located at point $$D$$, with $$DA = a$$, $$DC = b$$ (Fig. 2). Circles inscribed in triangles $$ABD$$ and $$DBC$$ are tangent to line $$BD$$ at points $$M$$ and $$N$$, respectively. Find the segment $$MN$$.
.
$$\triangle$$ Let $$a > b $$. Let us denote $$x = MN$$, $$y = ND$$, $$z = BM$$.
By the property of tangents $$DE = y$$, $$KD = x + y $$, $$AK = AP = a - (x + y)$$, $$CE = CF = b - y$$, $ $BP = z$$, and $$BF = z + x$$. Let's express sides(Fig. 2a): $$AB = z+a-x-y$$, $$BC=z+x-b-y$$. By condition $$AB=BC$$, so $$z+a-x -y = z+x+b-y$$. From here we find $$x=\frac((a-b))(2)$$, i.e. $$MN=\frac((a-b))(2)$$. If $$a \lt b$$, then $$MN=\frac((b-a))(2)$$. So $$MN=\frac(1)(2)|a-b|$$. $$\blacktriangle$$
ANSWER
$$\frac(|a-b|) (2)$$
Prove that in a right triangle the sum of the legs is equal to twice the sum of the radii of the inscribed and circumscribed circles, i.e. $$a+b=2R+2r$$.
$$\triangle$$ Let $$M$$, $$N$$ and $$K$$ be the points of tangency between the sides of the right triangle $$ABC$$ (Fig. 3), $$AC=b$$, $$BC=a$$, $$r$$ - radius of the inscribed circle, $$R$$ - radius of the circumscribed circle. Recall that the hypotenuse is the diameter of the circumscribed circle: $$AB=2R$$. Further, $$OM \perp AC$$, $$BC \perp AC$$, therefore, $$OM \parallel BC$$, similarly to $$ON \perp BC$$, $$AC \perp BC$$, means $$ON \parallel AC$$. A quadrilateral $$MONC$$ is by definition a square, all its sides are equal to $$r$$, so $$AM = b - r$$ and $$BN = a - r$$.
By the property of tangents $$AK=AM$$ and $$BK=BN$$, therefore $$AB = AK + KB = a+b-2r$$, and since $$AB=2R$$ , then we get $$a+b=2R+2r$$. $$\blacktriangle$$
Property 1.2$$(\^{\circ}$$ сформулируем по другому: !} The center of a circle inscribed in an angle lies on the bisector of that angle.
A trapezoid $$ABCD$$ with bases $$AD$$ and $$BC$$ is described around a circle with center at point $$O$$ (Fig. 4a).
a) Prove that $$\angle AOB = \angle COD = $$90$$(\^{\circ}$$ .!}
b) Find the radius of the circle if $$BO = \sqrt(5)$$ and $$AO = 2 \sqrt(5)$$. (Fig. 4b)
$$\triangle$$ a) The circle is inscribed in the angle $$BAD$$, by property 1.2$$(\^{\circ}$$ $$AO$$ - биссектриса угла $$A$$, $$\angle 1 = \angle 2 = \frac{1}{2} \angle A$$; $$BO$$ - биссектриса угла $$B$$, $$\angle 3 = \angle 4 = \frac{1}{2} \angle B$$. Из параллельности прямых $$AD$$ и $$BC$$ следует, что $$\angle A + \angle B = 180^{\circ}$$,поэтому в треугольнике $$AOB$$ из $$\angle 1 + \angle 3 = \frac{1}{2} (\angle A + \angle B) = 90^{\circ}$$ следует $$\angle AOB = 90^{\circ}$$.!}
Similar to $$CO$$ and $$DO$$ bisectors of angles $$C$$ and $$D$$ of a trapezoid, $$\angle COD = 180^(\circ) - \frac(1)(2)(\ angle C + \angle D) = 90^(\circ)$$.
b) The triangle $$AOB$$ is right-angled with legs $$AO = 2 \sqrt(5)$$ and $$BO = \sqrt(5)$$. Find the hypotenuse $$AB=\sqrt(20+5) = 5$$. If a circle touches side $$AB$$ at point $$K$$, then $$OK \perp AB$$ and $$OK$$ is the radius of the circle. By the property of a right triangle, $$AB \cdot OK = AO \cdot BO$$, whence $$OK = \frac(2\sqrt(5)\cdot \sqrt(5))(5) = 2$$. $$\blacktriangle$$
ANSWER
2. The angle between a tangent and a chord with a common point on the circle.
Recall that the degree measure of an inscribed angle is equal to half degree measure the arc on which it rests.
Theorem 1. The measure of the angle between a tangent and a chord having a common point on a circle is equal to half the degree measure of the arc enclosed between its sides.
$$\square$$ Let $$O$$ be the center of the circle, $$AN$$ be the tangent (Fig. 5). Let us denote the angle between the tangent $$AN$$ and the chord $$AB$$ as $$\alpha$$. Let's connect the points $$A$$ and $$B$$ to the center of the circle.
Thus, the degree measure of the angle between the tangent and the chord is equal to half the degree measure of the arc $$AnB$$, which is enclosed between its sides, and, therefore, the angle $$BAN$$ is equal to any inscribed angle subtended by the arc $$AnB$$ . (Similar arguments can be made for the angle $$MAB$$). $$\blacksquare$$
Point $$C$$ lies on the circle and is separated from the tangents drawn from point $$M$$ to the circle at distances $$CS = a$$ and $$CP = b$$ (Fig. 6). Prove that $$CK = \sqrt(ab)$$.
$$\triangle$$ Let's draw the chords $$CA$$ and $$CB$$. The angle $$SAC$$ between the tangent $$SA$$ and the chord $$AC$$ is equal to the inscribed angle $$ABC$$. And the angle $$PBC$$ between the tangent $$PB$$ and the chord $$BC$$ is equal to the inscribed angle $$BAC$$. We obtained two pairs of similar right triangles $$\Delta ASC \sim\Delta BKC$$ and $$\Delta BPC \sim \Delta AKC$$. From the similarity we have $$\dfrac(a)(AC)=\dfrac(x)(BC)$$ and $$\dfrac(b)(BC)=\dfrac(x)(AC)$$, which implies $ $ab=x^2$$, $$x=\sqrt(ab)$$. (If the projection of the point $$C$$ onto the line $$AB$$ lies outside the segment $$AB$$, the proof does not change much). (Ch. etc.) $$\blacktriangle$$
Reception applied in the solution - drawing the “missing” chords - often helps in problems and theorems with a circle and a tangent, such as, for example, in the proof of the following theorem "about tangent and secant".
Theorem 2. If from one point $$M$$ a tangent $$MA$$ and a secant $$MB$$ are drawn to a circle, intersecting the circle at point $$C$$ (Fig. 7), then the equality $$MA is valid ^2 = MB \cdot MC$$, i.e. if a tangent and a secant are drawn from a point $$M$$ to a circle, then the square of the tangent segment from the point $$M$$ to the point of tangency equal to the product the lengths of the secant segments from the point $$M$$ to the points of its intersection with the circle.
$$\square$$ Let's draw the chords $$AC$$ and $$AB$$. The angle $$MAC$$ between the tangent and the chord is equal to the inscribed angle $$ABC$$, both measured by half the degree measure of the arc $$AnC$$. In triangles $$MAC$$ and $$MBA$$, the angles $$MAC$$ and $$MBA$$ are equal, and the vertex angle $$M$$ is common. These triangles are
are similar, from the similarity we have $$MA/MB = MC/MA$$, which implies $$MA^2 = MB \cdot MC$$. $$\blacksquare$$
The radius of the circle is $$R$$. From the point $$M$$ a tangent $$MA$$ and a secant $$MB$$ are drawn, passing through the center $$O$$ of the circle (Fig. 8). Find the distance between point $$M$$ and the center of the circle if $$MB = 2MA$$.
$$\triangle$$ Let us denote the required distance $$x: \: x=MO$$, then $$MB = x+R$$, $$MC=x-R$$ and by condition $$MA=MB/2= (x+R)/2$$. By the tangent and secant theorem, $$(x+R)^2/4=(x+R)(x-R)$$, from which, reducing by $$(x+R)$$, we get $$(x+R )/4=x-R$$. We easily find $$x = \dfrac(5)(3)R$$. $$\blacktriangle$$
ANSWER
$$\dfrac(5)(3)R$$
3. Property of circle chords.
It is useful to prove these properties yourself (it is better reinforced), you can analyze the proofs from the textbook.
1.3$$(\^{\circ}$$. Диаметр, перпендикулярный хорде, делит её пополам. Обратно: диаметр, проходящей через середину хорды (не являющуюся диаметром) перпендикулярен ей. !}
1.4$$(\^{\circ}$$. Равные хорды окружности находятся на !} equal distance from the center of the circle. Conversely: equal chords are located at equal distances from the center of the circle.
1.5$$(\^{\circ}$$. !} Arcs of a circle enclosed between parallel chords, are equal (Fig. 9 will suggest the path of proof).
1.6$$(\^{\circ}$$. Если две хорды $$AB$$ и $$CD$$ пересекаются в точке $$M$$, то $$AM \cdot MB = CM \cdot MD$$, т. е. произведение длин отрезков одной хорды равно произведению длин отрезков другой хорды (на рис. 10 $$\Delta AMC \sim \Delta DMB$$). !}
Let us prove the following statement.
1.7$$(\^{\circ}$$. !} If in a circle of radius $$R$$ the inscribed angle subtended by a chord of length $$a$$ is equal to $$\alpha$$, then $$a = 2R\textrm(sin)\alpha$$.
$$\blacksquare$$ Let in a circle of radius $$R$$ the chord $$BC = a$$, the inscribed angle $$BAC$$ subtend the chord $$a$$, $$\angle BAC = \alpha$$ (Fig. 11 a,b).
Let's draw the diameter $$BA^(")$$ and consider right triangle$$BA^(")C$$ ($$\angle BCA^(")= 90^(\circ)$$, based on the diameter).
If the angle $$A$$ is acute (Fig. 11a), then the center $$O$$ and the vertex $$A$$ lie on the same side of the straight line $$BC$$, $$\angle A^(") = \angle A$$ and $$BC = BA^(") \cdot \textrm(sin)A^(")$$, i.e. $$a=2R\textrm(sin)A^(")$ $.
If the angle $$A$$ is obtuse, the center $$O$$ and the vertex $$A$$ lie along different sides from the straight line $$BC$$ (Fig. 11b), then $$\angle A^(") = 180^(\circ) - \angle A$$ and $$BC = BA^(") \cdot \textrm (sin)A^(")$$, i.e. $$a=2R\textrm(sin)(180-A^("))=2R\textrm(sin)A^(")$$.
If $$\alpha = 90^(\circ)$$, then $$BC$$ is the diameter, $$BC = 2R = 2R\textrm(sin)90^(\circ)$$.
In all cases the equality $$a=2R\textrm(sin)A^(")$$ is true. $$\blacktriangle$$
So, $$\boxed(a = 2R\textrm(sin)\alpha)$$ or $$\boxed(R = \dfrac(a)(2\textrm(sin)\alpha))$$. (*)
Find the radius of the circle circumscribed about the triangle $$ABC$$, in which $$AB = 3\sqrt(3)$$, $$BC = 2$$ and angle $$ABC = 150^(\circ)$$.
$$\triangle$$ In the circle circumscribed about the triangle $$ABC$$, the angle $$B$$ subtended by the chord $$AC$$ is known. From the proven formula it follows $$R = \dfrac(AC)(2\textrm(sin)B)$$.
Let us apply the cosine theorem to the triangle $$ABC$$ (Fig. 12) and take into account that
$$\textrm(cos)150^(\circ) = \textrm(cos)(180^(\circ)-30^(\circ)) = -\textrm(cos)30^(\circ) = -\ dfrac(\sqrt(3))(2)$$, we get
$$AC^2 = 27+4+2\cdot 3\sqrt(3) \cdot 2 \cdot \dfrac(\sqrt(3))(2) = 49,\: AC=7$$.
We find $$R = \dfrac(AC)(2\textrm(sin)150^(\circ)) = \dfrac(7)(2\textrm(sin)30^(\circ)) = 7$$. $$\blacktriangle$$
ANSWER
We use the property of intersecting chords to prove the following theorem.
Theorem 3. Let $$AD$$ be the bisector of triangle $$ABC$$, then
$$AD^2 = AB\cdot AC - BD\cdot CD$$ , i.e. If$$AB=c,\: AC=b,\: BD=x,\:DC=y$$ , That$$AD^2 = bc-xy$$ (Fig. 13a).
$$\square$$ Let us describe a circle around the triangle $$ABC$$ (Fig. 13b) and denote the point of intersection of the continuation of the bisector $$AD$$ with the circle as $$B_1$$. Let us denote $$AD = l $$ and $$DB_1 = z $$. The inscribed angles $$ABC$$ and $$AB_1C$$ are equal, $$AD$$ is the bisector of angle $$A$$, so $$\Delta ABD \sim \Delta AB_1C$$ (at two angles). From the similarity we have $$\dfrac(AD)(AC) = \dfrac(AB)(AB_1)$$, i.e. $$\dfrac(l)(b) = \dfrac(c)(l+z) $$, whence $$l^2=bc-lz$$. By the property of intersecting chords, $$BD\cdot DC = AD \cdot DB_1$$, i.e. $$xy=lz$$, so we get $$l^2=bc-xy$$ . $$\blacksquare$$
4. Two tangent circles
To conclude this section, we will consider problems with two tangent circles. Two circles that have a common point and a common tangent at that point are called tangent. If the circles are located on the same side of a common tangent, they are called relating internally(Fig. 14a), and if located on opposite sides of the tangent, then they are called relating externally(Fig. 14b).
If $$O_1$$ and $$O_2$$ are the centers of circles, then by definition of tangent $$AO_1 \perp l$$, $$AO_2 \perp l$$, therefore, in both cases common pointtouch lies on the line of centers.
Two circles of radii $$R_1$$ and $$R_2$$ ($$R_1 > R_2$$) are internally tangent at point $$A$$. Through point $$B$$ lying on larger circumference, a straight line is drawn tangent to the smaller circle at point $$C$$ (Fig. 15). Find $$AB$$ if $$BC = a$$.
$$\triangle$$ Let $$O_1$$ and $$O_2$$ be the centers of the larger and smaller circles, $$D$$ be the intersection point of the chord $$AB$$ with the smaller circle. If $$O_1N \perp AB$$ and $$O_2M \perp AB$$, then $$AN=AB/2$$ and $$AM=AD/2$$ (since the radius perpendicular to the chord divides it in half). From the similarity of triangles $$AO_2M$$ and $$AO_1N$$ it follows that $$AN:AM = AO_1:AO_2$$ and, therefore, $$AB:AD = R_1:R_2$$.
By the tangent and secant theorem we have:
$$BC^2 = AB\cdot BD = AB (AB-AD) = AB^2(1 - \dfrac(AD)(AB))$$,
i.e. $$a^2 = AB^2(1-\dfrac(R_2)(R_1))$$.
So $$AB = a \sqrt(\dfrac(R_1)(R_1-R_2))$$. $$\blacktriangle$$
Two circles of radii $$R_1$$ and $$R_2$$ are externally tangent at point $$A$$ (Fig. 16). Their common external tangent touches the larger circle at point $$B$$ and the smaller circle at point $$C$$. Find the radius of the circle circumscribed by triangle $$ABC$$.
$$\triangle$$ Let's connect the centers $$O_1$$ and $$O_2$$ with the points $$B$$ and $$C$$. By definition of a tangent, $$O_1B \perp BC$$ and $$O_2C \perp BC$$. Therefore, $$O_1B \parallel O_2C$$ and $$\angle BO_1O_2 + \angle CO_2O_1 = 180^(\circ)$$. Since $$\angle ABC = \dfrac(1)(2) \angle BO_1A$$ and $$\angle ACB = \dfrac(1)(2) \angle CO_2A$$, then $$\angle ABC + \ angle ACB = 90^(\circ)$$. It follows that $$\angle BAC = 90^(\circ)$$ , and therefore the radius of the circle circumscribed about a right triangle is $$ABC$$ , equal to half hypotenuse $$BC$$.
Let's find $$BC$$. Let $$O_2K \perp O_1B$$, then $$KO_2 = BC,\: O_1K = R_1-R_2,\: O_1O_2 = R_1+R_2$$. Using the Pythagorean theorem we find:
$$KO_2 = \sqrt(O_1O_2^2 - O_1K^2)= 2\sqrt(R_1R_2), \: \underline(BC = 2\sqrt(R_1R_2) )$$.
So, the radius of the circle circumscribed about the triangle $$ABC$$ is equal to $$\sqrt(R_1R_2)$$. In the solution $$R_1 > R_2$$, for $$R_1
ANSWER
$$\sqrt(R_1R_2)$$
Tangent segments to a circle drawn from one point are equal and equal equal angles with a straight line passing through this point and the center of the circle. PROOF. A. 3. B. 4. 1. 2. S. O. By the theorem about the tangent property, angles 1 and 2 are right angles, therefore triangles ABO and ACO are right-angled. They are equal, because have a common hypotenuse OA and equal legs OV and OS. Therefore, AB = AC and angle 3 = angle 4, which is what needed to be proven.
Slide 4 from the presentation "Circle" geometry. The size of the archive with the presentation is 316 KB.Geometry 8th grade
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Most often, it is geometric problems that cause difficulties for applicants, graduates, and participants mathematical olympiads. If you look at the 2010 Unified State Exam statistics, you can see that geometric problem About 12% of participants started C4, but only 0.2% of participants received full marks, and overall the task turned out to be the most difficult of all those proposed.
Obviously, the sooner we offer schoolchildren beautiful or unexpected solutions to problems, the more likely to interest and captivate seriously and for a long time. But how difficult it is to find interesting and complex tasks at the 7th grade level, when the systematic study of geometry is just beginning. What can be offered to a student interested in mathematics who knows only the signs of equality of triangles, the properties of adjacent and vertical angles? However, one can introduce the concept of a tangent to a circle, as a straight line that has one common point with the circle; assume that the radius drawn to the point of contact is perpendicular to the tangent. Of course, it is worth considering all possible cases of arrangement of two circles and common tangents to them, which can be drawn from zero to four. By proving the theorems proposed below, you can significantly expand the set of problems for seventh graders. At the same time, simultaneously prove important or simply interesting and fun facts. Moreover, since many statements are not included in the school textbook, they can be discussed in circle classes and with graduates when repeating planimetry. These facts turned out to be relevant last academic year. Since many diagnostic work itself Unified State Exam work contained a problem for the solution of which it was necessary to use the property of the tangent segment proved below.
T 1
Segments of tangents to a circle drawn from
equal to one point (Fig. 1)
This is the theorem that you can first introduce to seventh graders.
In the process of proof, we used the sign of equality of right triangles and concluded that the center of the circle lies on the bisector of the angle BSA.
Along the way, we remembered that the bisector of an angle is locus points of the interior region of an angle equidistant from its sides. The solution to a far from trivial problem is based on these facts, accessible even to those just beginning to study geometry.
1. Angle bisectors A, IN And WITH convex quadrilateral ABCD intersect at one point. Rays AB And DC intersect at a point E, and the rays
Sun And AD at the point F. Prove that a non-convex quadrilateral AECF the sums of the lengths of opposite sides are equal.
Solution (Fig. 2). Let ABOUT– point of intersection of these bisectors. Then ABOUT equidistant from all sides of the quadrilateral ABCD, that is
is the center of a circle inscribed in a quadrilateral. By theorem 1
the following equalities are true: AR = A.K.,
ER = E.P., F.T. = FK. Let's add the left and right sides term by term and get the correct equality:
(AR + ER) + F.T. = (A.K. +FK) + E.P.; A.E. + (F.C. + C.T.) = A.F. + (EU + PC). Because ST = RS, That AE + F.C. = A.F. + EU, which was what needed to be proven.
Let us consider a problem with an unusual formulation, for the solution of which it is sufficient to know the theorem 1 .
2. Is there n-a triangle whose sides are sequentially 1, 2, 3, ..., n, into which a circle can be inscribed?
Solution. Let's say this n-gon exists. A 1 A 2 =1, …, A n-1 A n= n– 1,A n A 1 = n. B 1 , …, B n – corresponding points touch. Then by Theorem 1 A 1 B 1 = A 1 B n< 1, n – 1 < A n B n< n. By the property of tangent segments A n B n= A n B n-1 . But, A n B n-1< A n-1 A n= n – 1. Contradiction. Therefore no n-gon satisfying the conditions of the problem.
T 2 Amounts opposing sides quadrilateral circumscribed about
circles are equal (Fig. 3)
Schoolchildren, as a rule, easily prove this property of the described quadrilateral. After proving the theorem 1 , it is a training exercise. We can generalize this fact - the sums of the sides of a circumscribed even triangle, taken through one side, are equal. For example, for a hexagon ABCDEF right: AB + CD + EF = BC + DE + FA.
Solution (Fig. 1). Since the quadrilaterals ABEF and ECDF are cyclic, then by Theorem 2 P ABEF = 2(AB + EF) and P ECDF = 2(CD + EF), by condition
P ABEF – P ECDF = 2(AB + EF) – 2(CD + EF) = 2p. AB – CD = p. AB = a + p.
a tangent to the circle is drawn intersecting the segments AB And AC at points M And R respectively. Prove that the perimeter of a triangle AMR and the magnitude of the angle MPA do not depend on the choice of point X.
Solution (Fig. 5). By Theorem 1 MV = MX and RS = RH. Therefore, the perimeter of the triangle AMR equal to the sum of the segments AB And AC. Or double tangent drawn to the excircle for a triangle AMR . The value of the MOP angle is measured by half the angle VOS, which does not depend on the choice of point X.
Solution (Fig. 6). Method one (algebraic). Let AK = AN = x, Then BK = BM = c – x, CM = CN = a – c + x. AC = AN + NC, then we can create an equation for x: b = x + (a – c + x). Where
.
Method two (geometric). Let's look at the diagram. Segments of equal tangents, taken one at a time, add up to the semi-perimeter
triangle. Red and green make up a side A. Then the segment we are interested in x = p – a. Of course, the results obtained coincide.
. Z Note that from the reference problem 1
follows that CM = p Δ ABC. b + x = p; x = p – b. The resulting formulas have application in the following problems. 4. Find the radius of a circle inscribed in a right triangle with legs a, b and hypotenuse With. Solution (Fig. 8). T ok how OMCN - square, then the radius of the inscribed circle is equal to the tangent segment CN.
.
5. Prove that the points of tangency of the inscribed and excircle with the side of the triangle are symmetrical about the middle of this side.
Solution (Fig. 9). Note that AK is a tangent segment of the excircle for a triangle ABC. According to formula (2)
. VM- line segment
tangent to the incircle for a triangle ABC. According to formula (1)
. AK = VM, and this means that the points K and M equidistant from the middle of the side AB, Q.E.D.
6. Two common external tangents and one internal tangent are drawn to two circles. The internal tangent intersects the external tangents at points A, B and touches the circles at points A 1 And IN 1 . Prove that AA 1 = BB 1.
Solution (Fig. 10). Stop... What is there to decide? This is just a different formulation of the previous problem. Obviously, one of the circles is inscribed and the other is excircle for a certain triangle ABC. And the segments AA 1 and BB 1 correspond to segments AK And VM tasks 5. It is noteworthy that the task proposed at All-Russian Olympiad schoolchildren in mathematics, is solved in such an obvious way.
7. The sides of the pentagon in the order of traversal are 5, 6, 10, 7, 8. Prove that a circle cannot be inscribed in this pentagon.
Solution (Fig. 11). Suppose that in a pentagon ABCDE you can inscribe a circle. Moreover, the parties AB, B.C., CD, DE And EA are equal to 5, 6, 10, 7 and 8, respectively. Let us mark the tangent points in sequence – F, G, H, M And N. Let the length of the segment A.F. equal to X.
Then B.F. = FD – A.F. = 5 – x = B.G.. G.C. = B.C. – B.G. = = 6 – (5 – x) = 1 + x = CH. And so on: HD = DM = 9 – x; M.E. = EN = x – 2, AN = 10 – X.
But, A.F. = AN. That is 10 - X = X; X= 5. However, the tangent segment A.F. cannot equal side AB. The resulting contradiction proves that a circle cannot be inscribed in a given pentagon.
8. A circle is inscribed in a hexagon; its sides in the order of circumambulation are 1, 2, 3, 4, 5. Find the length of the sixth side.
Solution. Of course, we can designate a tangent segment as X, As in previous task, write an equation and get the answer. But, it is much more efficient and effective to use a note to the theorem 2 : the sums of the sides of a circumscribed hexagon, taken through one another, are equal.
Then 1 + 3 + 5 = 2 + 4 + X, Where X– unknown sixth side, X = 3.
Solution (Fig. 12). Since the lengths of all sides are integers, the fractional parts of the lengths of the segments are equal BT, B.P., DM, DN, A.K. And AT. We have AT + TV= 1, and fractional parts of segment lengths AT And TB are equal. This is only possible when AT + TV= 0.5. By theorem 1
VT + VR.
Means, VR= 0.5. Note that the condition CD= 3 turned out to be unclaimed. Obviously, the authors of the problem assumed some other solution. Answer: 0.5.
10. In a quadrangle ABCD AD = DC, AB = 3, BC = 5. Circles inscribed in triangles ABD And CBD touch a segment BD at points M And N respectively. Find the length of the segment MN.
Solution (Fig. 13). MN = DN – DM. According to formula (1) for triangles DBA And DBC accordingly, we have:
11. Into a quadrangle ABCD you can inscribe a circle. Circles inscribed in triangles ABD And CBD have radii R And r respectively. Find the distance between the centers of these circles.
Solution (Fig. 13). Since by condition the quadrilateral ABCD inscribed, by theorem 2 we have: AB + DC = AD + BC. Let's use the idea of solving the previous problem. . This means that the points of contact of the circles with the segment DM match up. The distance between the centers of the circles is equal to the sum of the radii. Answer: R+r.
In fact, it has been proven that the condition is in a quadrilateral ABCD you can inscribe a circle, equivalent to the condition - in convex quadrilateral ABCD circles inscribed in triangles ABC And ADC touch each other. The opposite is true.
It is proposed to prove these two mutually inverse statements in the following problem, which can be considered a generalization of this one.
12. In a convex quadrilateral ABCD (rice. 14) circles inscribed in triangles ABC And ADC touch each other. Prove that circles inscribed in triangles ABD And BDC also touch each other.
13. In a triangle ABC with the parties a, b And c on the side Sun point marked D so that circles inscribed in triangles ABD And ACD touch a segment AD at one point. Find the length of the segment BD.
Solution (Fig. 15). Let's apply formula (1) for triangles ADC And A.D.B., calculating DM two 
Turns out, D– point of contact with the side Sun circle inscribed in a triangle ABC. The opposite is true: if the vertex of a triangle is connected to the tangency point of the inscribed circle on opposite side, then the circles inscribed in the resulting triangles touch each other.
14. Centers ABOUT 1 , ABOUT 2 and ABOUT 3 three non-intersecting circles of the same radius are located at the vertices of a triangle. From points ABOUT 1 , ABOUT 2 , ABOUT 3, tangents to these circles are drawn as shown in the figure.
It is known that these tangents, intersecting, formed a convex hexagon, the sides of which are painted red and blue. Prove that the sum of the lengths of the red segments is equal to the sum of the lengths of the blue ones.
Solution (Fig. 16). It is important to understand how to use the fact that given circles have equal radii. Note that the segments BR And DM are equal, which follows from the equality of right triangles ABOUT 1 BR And O 2 B.M.. Likewise D.L. = D.P., FN = FK. We add the equalities term by term, then subtract from the resulting sums identical segments of tangents drawn from the vertices A, WITH, And E hexagon ABCDEF: AR And A.K., C.L. And C.M., EN And E.P.. We get what we need.
Here is an example of a problem in stereometry, proposed at the XII International Mathematical Tournament for High School Students “Cup in Memory of A. N. Kolmogorov”.
16. Given a pentagonal pyramid SA 1 A 2 A 3 A 4 A 5 . There is a sphere w, which touches all edges of the pyramid and another sphere w 1, which touches all sides of the base A 1 A 2 A 3 A 4 A 5 and continuations of the lateral ribs SA 1, SA 2, SA 3, SA 4, SA 5 beyond the tops of the base. Prove that the top of the pyramid is equidistant from the vertices of the base. (Berlov S. L., Karpov D. V.)
since the tangent segments are equal. Let C i A i = a i. Then p SAiAi +1 = s+a i +a i+1, and from the equality of the perimeters it follows that a 1 = a 3 = a 5 = a 2 = a 4, from where S.A. 1 = S.A. 2 = S.A. 3 = S.A. 4 = S.A. 5 .
17. Unified State Exam. Diagnostic work 8.12.2009, S–4. Given a trapezoid ABCD, the foundations of which BC = 44,AD = 100, AB = CD= 35. Circle tangent to lines AD And A.C., touches the side CD at the point K. Find the length of the segment CK.BDC and BDA, touch the sides ВD at points E And F. Find the length of the segment E.F..
Solution. Two cases are possible (Fig. 20 and Fig. 21). Using formula (1) we find the lengths of the segments DE And DF.
In the first case AD = 0,1AC, CD = 0,9A.C.. In the second - AD = 0,125AC, CD = 1,125A.C.. We substitute the data and get the answer: 4.6 or 5.5.
Problems for independent solution/
1. Perimeter isosceles trapezoid, circumscribed about the circle is equal to 2 rub. Find the projection of the trapezoid's diagonal onto the larger base. (1/2r)
2. Open bank Unified State Exam problems mathematics. AT 4. To a circle inscribed in a triangle ABC (Fig. 22), three tangents are drawn. The perimeters of the cut triangles are 6, 8, 10. Find the perimeter given triangle. (24)
3. Into a triangle ABC circle is inscribed. MN – tangent to the circle, MÎ AC, NÎ BC, BC = 13, AC = 14, AB = 15. Find the perimeter of the triangle MNC. (12)
4. To a circle inscribed in a square with side a, a tangent is drawn intersecting its two sides. Find the perimeter of the cut triangle. (A)
5. A circle is inscribed in a pentagon with sides A, d, c, d And e. Find the segments into which the point of tangency divides the side equal to A.
6. A circle is inscribed in a triangle with sides 6, 10 and 12. A tangent is drawn to the circle so that it intersects two long sides. Find the perimeter of the cut triangle. (16)
7. CD– median of the triangle ABC. Circles inscribed in triangles ACD And BCD, touch the segment CD at points M And N. Find MN, If AC – Sun = 2. (1)
8. In a triangle ABC with the parties a, b And c on the side Sun point marked D. To circles inscribed in triangles ABD And ACD, a common tangent is drawn intersecting AD at the point M. Find the length of the segment AM. (Length AM does not depend on the position of the point D And
equal to ½ ( c + b – a))
9. A circle of radius is inscribed in a right triangle A. The radius of the circle tangent to the hypotenuse and extensions of the legs is equal to R. Find the length of the hypotenuse. ( R–a)
10. In a triangle ABC the lengths of the sides are known: AB = With, AC = b, Sun = A. A circle inscribed in a triangle touches a side AB at the point C 1. The excircle touches the extension of the side AB per point A at the point C 2. Determine the length of the segment C 1 C 2. (b)
11. Find the lengths of the sides of the triangle divided by the point of tangency of the inscribed circle of radius 3 cm into segments of 4 cm and 3 cm (7, 24 and 25 cm in a right triangle)
12. Soros Olympiad 1996, 2nd round, 11th grade. Given a triangle ABC, on the sides of which points are marked A 1, B 1, C 1. Radii of circles inscribed in triangles AC 1 B 1, BC 1 A 1, SA 1 B 1 equal in r. Radius of a circle inscribed in a triangle A 1 B 1 C 1 equals R. Find the radius of a circle inscribed in a triangle ABC. (R +r).
Problems 4–8 are taken from the problem book by Gordin R.K. “Geometry. Planimetry." Moscow. Publishing house MCNMO. 2004.
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