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10.194. Point $C_{1}$ is the midpoint of side $A B$ of triangle $A B C$; angle $\operatorname{COC}_{1}$, where $O$ is the center of the circle circumscribed around the triangle, is a right angle. Prove that $|\angle B-\angle A|=90^{\circ}$. | ## Solution.
Let $\angle A=\alpha, \angle B=\beta, \angle C=\gamma$. For definiteness, let $\beta>\alpha$ (Fig. 10.4). $\angle CAB$ is inscribed in a circle with center at point $O$ and subtends the same arc $BC$ as the central angle $\angle COB$, therefore
, $M$ and $K$ being the points of intersection of the circle with the specified sides of the square, and $P$ and $E$ being the points of intersection with the sides $BC$ and $CD$. By the pr... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,805 |
10.196. Given a triangle $A B C$, in which $2 h_{c}=A B$ and $\angle A=75^{\circ}$. Find the measure of angle $C$. | ## Solution.
Let the height $CD$ be denoted by $h$, and the segment $AD$ by $x$ (Fig. 10.6). We have $\angle ACD = 90^{\circ} - 75^{\circ} = 15^{\circ}$. Draw $AF$ such that $\angle CAF = \angle ACD = 15^{\circ}$. Then $\angle AFD = 30^{\circ}$, and from $\triangle ADF$ we get $AF = FC = 2x$, $DF = x \sqrt{3}$. But $D... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,806 |
10.197. A circle is inscribed in a right-angled triangle with sides 6, 8, and $10 \mathrm{~cm}$. Lines are drawn through the center of the circle parallel to the sides of the triangle. Calculate the lengths of the segments of the sides of the triangle cut off by the constructed lines. | Solution.
Fig. 10.7
Let's find the radius of the inscribed circle. We have $r=\frac{S}{p}=\frac{0.5 \cdot 6 \cdot 8}{0.5(6+8+10)}=\frac{24}{12}=2$ cm. Next, since $\triangle A B C \sim \tr... | \frac{3}{2},\frac{8}{3},\frac{25}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,807 |
10.198. The bisectors of the obtuse angles at the base of the trapezoid intersect on the other base of the trapezoid. Find all sides of the trapezoid if its height is 12 cm, and the lengths of the bisectors are 15 and 13 cm. | ## Solution.
Let in trapezoid $A B C K$ (Fig. 10.8) $B C \| A K, B D$ and $C D$ be the bisectors of $\angle A B C$ and $\angle B C K, D E$ - the height, $D E=12$ cm, $B D=15$ cm, $C D=13$ cm. From $\triangle B E D \quad\left(\angle B E D=90^{\circ}\right): B E=\sqrt{B D^{2}-D E^{2}}=9$ cm. From the right
 $BC \| AD, BC=4$ cm, $AD=16$ cm. Since a circle is circumscribed around the given trapezoid, then $AB=CD$. Since a circle can be inscribed in the given trapezoid, then $AD+BC=AB+CD=2AB; AB=\frac{AD+BC}{2}=10$ cm. $BK$ is the height of the trapezoid. Then $AK=\frac{AD-BC}{2}=6$... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,809 |
10.200. A rhombus with side $m$ is inscribed in a triangle so that one angle is common to both, and the opposite vertex of the rhombus lies on the side of the triangle, dividing this side into segments of length $p$ and $q$. Find the sides of the triangle. | Solution.
In $\triangle A B C$ (Fig. 10.10)
Fig. 10.10, an inscribed rhombus $A D E F$ is given, where $B E=p, E C=q, D E=F E=m$, then $B C=p+q . \triangle D B E \sim \triangle A B C$ (by... | p+\frac{(p+q)}{p};\frac{(p+q)}{q} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,810 |
10.201. Given a triangle $A B C$ such that $A B=15 \text{~cm}, B C=12$ cm and $A C=18 \text{~cm}$. Calculate in what ratio the incenter of the triangle divides the angle bisector of angle $C$. | ## Solution.
Let $C K$ be the bisector of $\angle C, O$ be the center of the circle inscribed in $\triangle A B C$, the intersection point of its bisectors (Fig. 10.11). Let $B K=x$ cm, $x>0$. Then $A K=(15-x)$ cm. By the property of the bisector $\frac{B K}{B C}=\frac{A K}{A C}$; $\frac{x}{12}=\frac{15-x}{18} ; x=6$.... | 2:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,811 |
10.202. Given an isosceles triangle with the base equal to \(a\) and the lateral side equal to \(b\). Prove that the center of the inscribed circle divides the angle bisector at the base in the ratio \((a+b): b\), measured from the vertex of the angle. | Solution.
In $\triangle ABC$ (Fig. 10.12) $AB = BC = b, AC = a, O$ is the center of the inscribed circle, $AM$ is the bisector of $\angle BAC$, and $BO$ is the bisector of $\angle ABM$. Then, $\frac{AO}{OM} = \frac{AB}{BM} = \frac{BC}{BM} = \frac{BM + MC}{BM} = 1 + \frac{MC}{BM} = 1 + \frac{AC}{AB} = 1 + \frac{a}{b} =... | \frac{b}{b} | Geometry | proof | Yes | Yes | olympiads | false | 48,812 |
10.203. From a point on the circumference, two chords of lengths 9 and $17 \mathrm{~cm}$ are drawn. Find the radius of the circle if the distance between the midpoints of these chords is 5 cm. | Solution.
From point $B$ of the circle (Fig. 10.13), chords $A B=9$ cm and $B K=17$ cm, $M N=5$ cm are drawn. Then $M N$ is the midline of $\triangle A B K$, $A K=2 M N=10$ cm. The desired radius $R$ will be found as the radius of the circumcircle of $\triangle A B K: R=\frac{A B \cdot A K \cdot B K}{4 S_{\triangle A ... | 10\frac{5}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,813 |
10.204. From a point on the circumference, two chords of lengths 10 and 12 cm are drawn. Find the radius of the circle if the distance from the midpoint of the shorter chord to the longer chord is 4 cm. | Solution.
From point $B$ of the circle (Fig. 10.14), chords $AB=10$ cm and $BC=12$ cm are drawn, $M$ is the midpoint of $AB$, $MN \perp BC$, $MN=4$ cm. Drop a perpendicular $AK$ from point $A$ to segment $BC$. In $\triangle ABK - MN$ is the midline. Then $AK=2MN=8$ cm. From $\triangle MNB \quad (\angle MNB=90^{\circ})... | 6.25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,814 |
10.205. A circle with a radius of 5 cm is inscribed in some angle. The length of the chord connecting the points of tangency is 8 cm. Two tangents, parallel to the chord, are drawn to the circle. Find the sides of the resulting trapezoid. | Solution.
The circle with center $O$ (Fig. 10.15) touches the sides of angle $E$ at points $K$ and $L, K L=8 \text{ cm}, O K=5 \text{ cm}, A B\|C D\| K L$. Draw the diameter of the circle $M N \| K L, P$ is the intersection point of $E O$ and $K L, \angle K P O=90^{\circ}, \triangle K P O \sim$ $\sim \triangle O K M-$... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,815 |
10.206. What integers represent the sides of an isosceles triangle if the radius of the inscribed circle is $3 / 2 \mathrm{~cm}$, and the radius of the circumscribed circle is $25 / 8$ cm | Solution.
In $\triangle ABC$, $AB = BC$, $BD$ is the altitude, $O_1$ is the center of the inscribed circle, and $O_2$ is the center of the circumscribed circle (see Fig. 10.16). The radii of these circles are $r = \frac{3}{2}$ cm and $R = \frac{25}{8}$ cm, respectively. $E$ is the point of intersection of the ray $BD$... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,816 |
10.207. A rectangle with a perimeter of 24 cm is inscribed in a triangle with sides 10, 17, and 21 cm, such that one of its sides lies on the largest side of the triangle. Find the sides of the rectangle. | Solution.
In $\triangle ABC$, we have $AC=21 \text{ cm}, AB=10 \text{ cm}, BC=17$ cm, vertices $L$ and $Q$ of rectangle $MNQL$ lie on $AC$, $M$ on $AB$, and $N$ on $BC$.
Fig. 10.17
Fig. 1... | 5\frac{7}{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,817 |
10.208. Perpendiculars are drawn from the vertex of the acute angle of a rhombus to the lines containing the sides of the rhombus to which this vertex does not belong. The length of each perpendicular is 3 cm, and the distance between their bases is $3 \sqrt{3}$ cm. Calculate the lengths of the diagonals of the rhombus... | Solution.
Since $\triangle A E F$ is isosceles (Fig. 10.18), the bisector $A M$ is perpendicular to $E F$ and lies on the diagonal of the rhombus. We find $A M^{2}=A F^{2}-M F^{2}=9-\frac{27}{4}=\frac{9}{4}$, i.e., $A M=\frac{3}{2}$ (cm). In $\triangle A C F$, we have $\angle F=90^{\circ}$ and $F M \perp A C \Rightarr... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,818 |
10.210. Find the radius of the circle circumscribed around an isosceles trapezoid with bases 2 and 14 and a lateral side of $10 \mathrm{~cm}$. | Solution.
The circle circumscribed around trapezoid $ABCD$ is also circumscribed around $\triangle ACD$ (Fig. 10.20), and such a circle is unique. Its radius is found using the formula $R=\frac{abc}{4S}$. We have $S=\frac{1}{2} AD \cdot BE$, where $BE=\sqrt{AB^2 - AE^2} = \sqrt{10^2 - 6^2} = 8$; hence, $S_{\triangle A... | 5\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,819 |
10.211. On the larger leg of a right triangle, a circle is constructed with this leg as its diameter. Determine the radius of this circle if the smaller leg of the triangle is 7.5 cm, and the length of the chord connecting the vertex of the right angle with the point of intersection of the hypotenuse and the circle is ... | Solution.
In $\triangle ABC$ (Fig. 10.21) $\angle ACB=90^{\circ}, BC>AC, AC=7.5$ cm, $N$ is the intersection point of the circle mentioned in the condition and the hypotenuse $AB, CN=6$ cm. $\angle CNB$ is an inscribed angle and subtends the diameter. Therefore, $\angle CNB=90^{\circ}$. From $\triangle ANC \quad (\ang... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,820 |
10.212. The vertices of a rectangle inscribed in a circle divide it into four arcs. Find the distances from the midpoint of one of the larger arcs to the vertices of the rectangle, if its sides are 24 and 7 cm. | ## Solution.
Since $AC$ is the diameter of the circle (Fig. 10.22), then $R=0.5 \sqrt{24^{2}+7^{2}}=12.5$ cm. In $\triangle B O F$, we have $O F=\sqrt{O B^{2}-B F^{2}}=\sqrt{12.5^{2}-12^{2}}=3.5$ cm; hence, $M F=12.5-3.5=9$ cm, $M K=12.5+3.5=16$ cm. From $\triangle M B F$ and $\triangle M A K$, we find the required di... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,821 |
10.213. The center of a semicircle inscribed in a right triangle, such that its diameter lies on the hypotenuse, divides the hypotenuse into segments of 30 and 40 cm. Find the length of the arc of the semicircle, enclosed between the points of tangency with the legs. | Solution.
Draw radii $O D$ and $O E$ to the points of tangency (Fig. 10.23). We have $O D=O E=C E=C D$, i.e., $E C D O$ is a square. Let $R$ be the radius of the circle; then the length of the arc $E D$ is $\frac{\pi R}{2}$. Since $\triangle A E O \sim \triangle O D B$,
we have $\frac{A E}{O D}=\frac{A O}{O B}=\frac{3... | 12\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,822 |
10.214. An isosceles triangle with an acute angle of $30^{\circ}$ at the base is circumscribed around a circle of radius 3. Determine the sides of the triangle. | Solution.
Draw the radius $O E \perp B C$ (Fig. 10.24). Since $\angle O B E=\frac{1}{2} \angle A B C=60^{\circ}$, then $\angle B O E=30^{\circ}$ and $B E=\frac{1}{2} B O$. From $\triangle B E O$ we find $B O^{2}=\frac{1}{4} B O^{2}+9$, from which $B O=2 \sqrt{3}$. In $\triangle A D B$ we have $A B=2 B D$. But $B D=B O... | 4\sqrt{3}+6,6\sqrt{3}+12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,823 |
10.215. In a right-angled triangle, the medians of the legs are $\sqrt{52}$ and $\sqrt{73}$. Find the hypotenuse of the triangle. | ## Solution.
In $\triangle A B C$ (Fig. 10.25) $\angle A C B=90^{\circ}, B P$ and $A E$ are medians, $B P=\sqrt{52}$, $A E=\sqrt{73}$. Let $B C=x, A C=y$. Then $A B=\sqrt{x^{2}+y^{2}}$.
From $\triangle A C E\left(\angle A C E=90^{\circ}\right):$
$$
A C^{2}+C E^{2}=A E^{2} ; y^{2}+\frac{x^{2}}{4}=73
$$
From $\triang... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,824 |
10.216. Two circles, with radii of 4 and 8, intersect at a right angle. Determine the length of their common tangent. | Solution.
Let $A$ be the point of intersection of the circles with centers $O$ and $O_{1}$ (Fig. 10.26), $O A=8 \text{ cm}, O_{1} A=4 \text{ cm}$, $M N$ - their common tangent. According to the problem, the circles intersect at a right angle, so the tangents to them at point $A$ are perpendicular to each other. Theref... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,825 |
10.217. What necessary and sufficient conditions must a trapezoid satisfy to have both an inscribed and a circumscribed circle? | Solution.
For a circle to be inscribed in a trapezoid and for a circle to be circumscribed around it, it is necessary and sufficient that the trapezoid is isosceles and the lateral side equals the half-sum of the bases.
Necessity.
Let \(ABCD\) be a trapezoid, around which a circle is circumscribed with center at poi... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,826 |
10.218. A line parallel to the bases of a trapezoid passes through the point of intersection of its diagonals. Find the length of the segment of this line, enclosed between the lateral sides of the trapezoid, if the bases of the trapezoid are 4 and 12 cm. | ## Solution.
In trapezoid $ABCD$ (Fig. 10.28) $BC \| AD, BC=4$ cm, $AD=12$ cm, $O$ is
Fig. 10.28
. We will prove that $A F = B D$ as chords subtending equal arcs. Indeed, $\cup A C + \cup B D = 180^{\circ}$ (since $A B \perp C D$), and $\cup A C + \cup A F = 180^{\circ}$ (since $C F$ is a diameter). Therefore, $\cup A F = \cup B D$ and $A F = B D$. In the right tri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 48,828 |
10.220. Show that the sum of the distances from any point taken on the side of an equilateral triangle to its two other sides is a constant value. | ## Solution.
Let $D$ be an arbitrary point on side $AB$ of the equilateral $\triangle ABC$ (Fig. 10.30). Drop perpendiculars $DM$ and $DN$ from point $D$ to sides $AC$ and $BC$. Let $DM = h_1$, $DN = h_2$, and $AC = a$. Then $S_{\triangle ADC} = \frac{1}{2} a h_1$, $S_{\triangle BDC} = \frac{1}{2} a h_2$, $S_{\triangl... | h_1+h_2=\frac{\sqrt{3}}{2} | Geometry | proof | Yes | Yes | olympiads | false | 48,829 |
10.221. Two sides of a triangle are 6 and $8 \mathrm{~cm}$. The medians drawn to these sides are perpendicular to each other. Find the third side of the triangle. | ## Solution.
In $\triangle A B C$ (Fig. 10.31) $A C=6$ cm, $B C=8$ cm, $B K$ and $A N$ are medians. $O$ is the point of their intersection, $A N \perp B K$. Let $O N=x$ cm, $O K=y$ cm. Then $A O=2 x$ cm, $B O=2 y$ cm. From $\triangle B O N\left(\angle B O N=90^{\circ}\right): B O^{2}+O N^{2}=B N^{2} ;$ $4 y^{2}+x^{2}=... | 2\sqrt{5}\mathrm{~} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,830 |
10.222. Circles of radii $R$ and $r$ touch each other externally. The lateral sides of an isosceles triangle are their common tangents, and the base touches the larger of the circles. Find the base of the triangle. | Solution.
$O$ and $O_{1}$ are the centers of the given circles, $\triangle A B C$ is the triangle mentioned in the problem, $A C=B C$ (Fig. 10.32). $C K$ is the altitude of $\triangle A B C, O$ and $O_{1}$ lie on $C K . F$ and $E$ are the points of tangency of the circles with centers $O_{1}$ and
^{2}=(60-c)^{2}$ or $a^{2}+2 a b+b^{2}=3600-120 c+c^{2} \quad\left(^{*}\right.$ ). But $a^{2}+b^{2}=c^{2}$, and $6 c=0.5 a b$ (the area of the triangle). Substituting these expressio... | 15,20,25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,832 |
10.224. Given an isosceles triangle with a base of 12 cm and a lateral side of $18 \mathrm{~cm}$. What segments need to be laid off from the vertex of the triangle on its lateral sides, so that by connecting their ends, a trapezoid with a perimeter of 40 cm is obtained? | ## Solution.
By the condition in $\triangle A B C$ (Fig. 10.33) we have: $A B=B C=18, A C=12$. Let $B D=B E=x$. Then $A D=E C=18-x$. The perimeter of trapezoid $A D E C$ is: $P=A C+D E+2 A D=40$, from which $D E=40-(2 A D+A C)=2 x-8$. $\triangle A B C \sim$ $\sim \triangle D B E$, therefore, $\frac{B D}{D E}=\frac{A B... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,833 |
10.225. Two circles of different radii touch each other externally. Find the angle determined by the chords connecting the point of contact of the circles with the points of contact of their common external tangent. | Solution.
Let $AB$ be the common external tangent of the given circles, $C$ be the point of tangency, and $D$ be the intersection point of $AB$ and the common tangent of the circles at point $C$ (Fig. 10.34). Then $DB = DC$, $DA = DC$, hence $\angle DCB = \angle DBC$, $\angle DAC = \angle DCA$. From $\triangle ACB: \a... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,834 |
10.226. A circle is inscribed in a right-angled triangle. The point of tangency divides the hypotenuse in the ratio 2:3. Find the sides of the triangle if the center of the inscribed circle is $\sqrt{8}$ cm away from the vertex of the right angle. | Solution.
Let the radius of the circle be $r$, and the length of the hypotenuse be $5x$. Then (Fig. 10.35) $BC = 3x + r, AC = 2x + r$, since $BK = BL, AK = AM$. Since $\angle OCM = 45^{\circ}$, we have $r = OC / \sqrt{2} = \sqrt{8} / \sqrt{2} = 2 \text{ cm}$. For the area of triangle $ABC$, we have the expression $S =... | 6,8,10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,835 |
10.227. Inside an equilateral triangle, a point $M$ is taken, which is at distances $b, c, d$ from its sides. Find the height of the triangle. | Solution.
Let point $M$ be at a distance $b$ from side $AB$ of the equilateral $\triangle ABC$, at a distance $c$ from $BC$, and at a distance $d$ from $AC$ (Fig. 10.36). Let $AB = a$, and $h$ be the height of $\triangle ABC$.
 $B K^{2}=A B \cdot D B=9 \cdot 4=36$, from which $B K=6 \text{ cm}, K C=3 \text{ cm}$. We draw the radius $O K$ to the point of tangency and $O N \perp A B$. Then $A N=N D=\frac{5}{2} \text{ cm}$. Since $\triangle A N O \sim \triangle O K C$ (right trian... | \frac{15}{\sqrt{11}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,837 |
10.229. A parallelogram with sides of 3 cm and 5 cm and a diagonal of 6 cm is inscribed in a triangle. Find the sides of the triangle, given that the diagonals of the parallelogram are parallel to the lateral sides of the triangle, and the smaller of its sides lies on the base of the triangle. | Solution.
Let in $\triangle A B C D K=E F=3 \text{cm}, D E=5 \text{cm}, D F=6$ cm, $D K \| E F$, $O$ - the point of intersection of the diagonals $D F$ and $E K$ of the parallelogram $D E F K$ (Fig. 10.38). Let $E O=x$ cm. Then $D F^{2}+E K^{2}=2\left(D E^{2}+E F^{2}\right)$; $36+4 x^{2}=2(25+9) ; x=2 \sqrt{2}$. OEBF ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,838 |
10.230. The height, base, and sum of the lateral sides of the triangle are 24, 28, and 56 cm, respectively. Find the lateral sides. | Solution.
The area of the triangle $S=\frac{24 \cdot 28}{2}=336 \mathrm{~cm}^{2}$. Let one of the lateral sides be $x$ cm, then the second one is $(56-x)$ cm. The semi-perimeter of the triangle
. Let $E D=x$. Then, using the equality $A B+C D=B C+A D$, we get $2 r+C D=\frac{4 r}{3}+\frac{4 r}{3}+x$, from which $C D=\frac{2 r}{3}+x$. In $\triangle C E D$ we have $C D^{2}=C E^{2}+E D^{2} \Leftrightarrow\left(\frac{2 r}{3}+x\right)^{2}=4 r^{2}+x^{2}$, from ... | 4r,\frac{10r}{3},2r | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,840 |
10.233. Find the midline of an isosceles trapezoid with height $h$, if the lateral side is seen from the center of the circumscribed circle at an angle of $120^{\circ}$. | ## Solution.
In trapezoid $ABCD$ (Fig. 10.41) $BC \| AD, AB=CD, O$ is the center of the circumscribed circle, $\angle AOB=120^{\circ}$. Then the inscribed $\angle ADB=\frac{1}{2} \angle AOB=60^{\circ}$. $BE$ is the height of the trapezoid, $BE=h$. Since the trapezoid is isosceles, the length
; then $K L=\sqrt{K O^{2}-L O^{2}}$. $L O=B E=18-13=5(\mathrm{~cm}) \Rightarrow K L=\sqrt{13^{2}-5^{2}}=12$ (cm), $B K=13-12=1$ (cm). We have found that the side of the square is divided into segments of 1 and 17 cm.
Answer: 1 and 17 cm. | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,843 |
10.235. In an isosceles triangle, the angle at the base is $72^{\circ}$, and the bisector of this angle has a length of $m$. Find the lengths of the sides of the triangle. | Solution.
Let in $\triangle ABC$, $AB=BC$, $\angle BAC=\angle BCA=72^{\circ}$, $AD$ is the bisector of $\angle BAC$, $AD=m$ (Fig. 10.43). $\angle BAD=\angle DAC=36^{\circ}$, $\angle ABC=180^{\circ}-2\angle BAC=36^{\circ}$, $\angle ADC=180^{\circ}-(\angle DAC+\angle BCA)=72^{\circ}$. Therefore, $\triangle ADB$ and $\tr... | ;\frac{(1+\sqrt{5})}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,844 |
10.236. In an isosceles triangle, the angle at the vertex is $36^{\circ}$, and the bisector of the angle at the base is $\sqrt{20}$. Find the lengths of the sides of the triangle. | Solution.
Refer to the diagram from the previous problem. In $\triangle ABC$, $AB = BC$, $\angle ABC = 36^\circ$, $AD$ is the angle bisector of $\angle BAC$, $AD = \sqrt{20}$. $\angle BAC = \angle BCA = 72^\circ$, $\angle BAD = \angle DAC = 36^\circ$, $\angle ADC = \angle BAD + \angle ABC = 72^\circ$. Therefore, $\tri... | 2\sqrt{5};5+\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,845 |
10.237. The diagonals of a quadrilateral are equal, and the lengths of its midlines are $p$ and $q$. Find the area of the quadrilateral. | Solution.
Since $K L, L M, N M, K N$ are the midlines of the corresponding triangles (Fig. 10.44), then $K L=N M=\frac{1}{2} A C, K N=L M=\frac{1}{2} B D$. By the condition $A C=B D \Rightarrow K L M N-$ is a rhombus, the area of which is equal to $\frac{1}{2} p q$. Let the required area be $S$. Then $S=S_{\triangle A... | pq | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,846 |
10.238. The larger base of the trapezoid is twice as large as its smaller base. A line parallel to the bases is drawn through the point of intersection of the diagonals. Find the ratio of the height of each of the two resulting trapezoids to the height of the given trapezoid. | Solution.
Let in trapezoid $ABCD$ (Fig. 10.45) $AD \| BC, AD=2BC, O$ be the intersection point of $AC$ and $BD$, and $MN$ be the line mentioned in the problem. Draw through point $O$ the height $KL$ of trapezoid $ABCD$. Then $KO$ is the height of trapezoid $MBCN$ and $\triangle BOC$, and $OL$ is the height of trapezoi... | \frac{1}{3};\frac{2}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,847 |
10.239. Find the radius of the circle, in the segment of which, corresponding to a chord of length $6 \mathrm{~cm}$, a square with a side of $2 \mathrm{~cm}$ is inscribed. | Solution.
Let $r$ be the desired radius. In $\triangle A O K$ (Fig. 10.46), we have $O K=\sqrt{r^{2}-9}$, and in $\triangle O B N$, we have $O N^{2}+B N^{2}=O B^{2}$ or $(O K+2)^{2}+1=r^{2}$. Therefore, $r^{2}-9+4 \sqrt{r^{2}-9}+4+1=r^{2}$, from which $r^{2}-9=1$, i.e., $r=\sqrt{10}$ cm.
Answer: $\sqrt{10}$ cm. | \sqrt{10} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,848 |
10.240. The length of the base of an isosceles triangle is $12 \mathrm{~cm}$, and the side lengths - $18 \mathrm{cm}$. Heights are drawn to the lateral sides of the triangle. Calculate the length of the segment whose ends coincide with the bases of the heights. | ## Solution.
Let in $\triangle A B C$ (Fig. 10.47) $A B=B C=18 \mathrm{~cm}, A C=12 \mathrm{~cm}, A N, C M$, $B D$ - altitudes. $A D=\frac{1}{2} A C=6$ cm. $M N \| A C$ ( $\triangle A B C$ - isosceles), then $\triangle M B N \sim \triangle A B C \cdot \triangle B C D \sim \triangle A N C$ (both are right triangles, $\... | \frac{28}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,849 |
10.241. In an isosceles triangle with a lateral side equal to $b$, the angle bisectors of the angles at the base are drawn. The segment of the line between the points of intersection of the bisectors with the lateral sides is equal to $m$. Determine the base of the triangle. | Solution.
Since $\triangle A B C$ is isosceles, then $C E=A D$ and $D E \| A C$ (Fig. 10.48).
From the similarity of triangles $A B C$ and $D B E$, it follows that $\frac{m}{A C}=\frac{B D}{b}$. Considering that $C D$ is the bisector, we have $\frac{B D}{D A}=\frac{B C}{A C}$. Let $A C=x, B D=y$. Then we have $\left\... | \frac{}{b-} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,850 |
10.242. The base of an isosceles triangle is 8 cm, and the lateral side is 12 cm. Find the length of the segment connecting the points of intersection of the angle bisectors at the base with the lateral sides of the triangle. | Solution.
In $\triangle ABC$ (Fig. 10.49): $AB=BC=12, AC=8, AN$ and $CM$ are angle bisectors. Let $MN=x$. As in problem 10.240, $MN \| AC$. Since
Fig. 10.50
$\angle ACM = \angle NCM$ and $... | 4.8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,851 |
10.243. Inside an angle of $60^{\circ}$, there is a point located at distances of $\sqrt{7}$ and $2 \sqrt{7}$ cm from the sides of the angle. Find the distance from this point to the vertex of the angle. | Solution.
Let point $O$ be the vertex of the angle ($\angle O=60^{\circ}$), $AB$ and $AC$ be the distances from point $A$ to the sides, $AB=\sqrt{7}$ cm, $AC=2\sqrt{7}$ cm (Fig. 10.50). $D$ is the intersection point of line $AB$ and ray $OC$, $\angle D=30^{\circ}$. From $\angle ACD$ ($\angle ACD=90^{\circ}$): $AD=2AC=... | \frac{14\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,852 |
10.244. A circle with a radius of 3 cm is inscribed in a triangle. Calculate the lengths of the sides of the triangle if one of them is divided by the point of tangency into segments of 4 and $3 \mathrm{~cm}$. | Solution.
Let $O$ be the center of the circle with radius 3 inscribed in $\triangle ABC$ (Fig. 10.51). $M, N, K$ are the points of tangency of this circle with the sides $AB, BC, AC$ respectively. $BN=3 \text{ cm}, NC=4 \text{ cm}$. Then $ON \perp BC, OM \perp AB$, $ON=OM=3 \text{ cm}$. $BM=BN=3 \text{ cm}$. Therefore... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,853 |
10.245. Three circles are inscribed in a corner - a small, a medium, and a large one. The large circle passes through the center of the medium one, and the medium one through the center of the small one. Determine the radii of the medium and large circles if the radius of the smaller one is $r$ and the distance from it... | Solution.
Let $O_{1}, O_{2}, O_{3}$ be the centers of the circles mentioned in the problem, inscribed in the angle $B A C$ (Fig. 10.52), $O_{1} A=a . E, F, K$ - the points of tangency of the small, medium, and large circles with the side $A C$ of the angle, respectively. Then $O_{1} E=r, O_{1} E \perp A C, O_{2} F \pe... | \frac{}{-r};\frac{^{2}r}{(-r)^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,854 |
10.246. The center of the circle inscribed in a right trapezoid is at a distance of 8 and $4 \mathrm{cm}$ from the ends of the lateral side. Find the midline of the trapezoid. | ## Solution.
In trapezoid $ABCD$ (Fig. 10.53) $BC \| AD, AB \perp AD, O$ is the center of the inscribed circle, $OC=4 \text{ cm}, OD=8 \text{ cm}$. Since by the condition the circle with center $O$ touches the sides $\angle BCD$ and $\angle ADC$, then $CO$ and $OD$ are the bisectors of these angles. But $\angle BCD + ... | \frac{18\sqrt{5}}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,855 |
10.247. The bases of two equilateral triangles with sides $a$ and $3a$ lie on the same straight line. The triangles are located on opposite sides of the line and do not have any common points, and the distance between the nearest ends of their bases is $2a$. Find the distance between the vertices of the triangles that ... | Solution.
Let in the equilateral triangles $A B C$ and $D F E$ (Fig.10.54) $A C=a$, $D E=3 a, C D=2 a$. Since $\angle B C D=\angle F D C=120^{\circ}$, then $B C \| D F, K$ is the intersection point of the line $D F$ and the line passing through point $B$ parallel to $A C$. Therefore, $B K D C$ is a parallelogram, $B K... | 2\sqrt{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,856 |
10.248. To two externally tangent circles with radii $R$ and $r$, a secant is constructed such that the circles intercept three equal segments on it. Find the lengths of these segments. | Solution.
Let the desired length be $2x$. Then $AB = 4x$, $AO_1 = \sqrt{R^2 - x^2}$, $BO_2 = \sqrt{r^2 - x^2}$ (Fig. 10.55). Draw $O_2C \parallel AB$. In $\triangle O_1O_2C$, we have $O_1C = \sqrt{O_1O_2^2 - O_2C^2} \Rightarrow \sqrt{R^2 - x^2} - \sqrt{r^2 - x^2} = \sqrt{(R + r)^2 - 16x^2}^{*}$.
Multiplying both side... | \frac{1}{2}\sqrt{\frac{14Rr-R^2-r^2}{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,857 |
10.249. Prove that the distance from the orthocenter (the point of intersection of the altitudes) to the vertex of the triangle is twice the distance from the center of the circumscribed circle to the side opposite this vertex. | Solution.
Let $O$ be the center of the circumcircle of $\triangle ABC$, and $O_{1}$ be its orthocenter (Fig. 10.56). Construct $\triangle LMN$, whose sides are the midlines of the given triangle. The altitudes of $\triangle LMN$ intersect at point $O$, since these altitudes are perpendicular to the sides of $\triangle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 48,858 |
10.250. On the segment $A B$, a point $M$ is taken, and on the segments $A M$ and $M B$, squares are constructed on one side of the line $A B$, the circumcircles of which intersect at point $N$. Prove that the line $A N$ passes through the vertex of the second square and that the triangle $A N B$ is a right triangle. | ## Solution.
Connect the vertices of the squares $A$ and $C, B$ and $D$ (Fig. 10.57). Extend $BD$ until it intersects with $AC$. Denote the point of intersection as $N$ and show that it coincides with the point of intersection of the circumcircles of the squares. Indeed, since $\triangle A C M = \triangle B D M$, then... | proof | Geometry | proof | Yes | Yes | olympiads | false | 48,859 |
10.251. Five circles are inscribed in an angle of $60^{\circ}$ such that each subsequent circle (starting from the second) touches the previous one. By what factor is the sum of the areas of all five corresponding circles greater than the area of the smallest circle? | ## Solution.
Let $A$ be the vertex of the angle. $O_{i}$ and $r_{i}$ be the center and radius of the $i$-th circle $(i=1, \ldots, 5)$. Since $\frac{1}{2} \angle A=30^{\circ}$, then $A O_{i}=2 r_{i}, A O_{i-1}=2 r_{i-1}$, from which $A O_{i}=A O_{i-1}+r_{i-1}+r_{i}$ or $2 r_{i}=2 r_{i-1}+r_{i-1}+r_{i}$, i.e., $r_{i}=3 ... | 7381 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,860 |
10.252. The sides of a triangle are in the ratio $5: 4: 3$. Find the ratio of the segments of the sides into which they are divided by the point of tangency of the inscribed circle. | ## Solution.
Since in $\triangle A B C$ (Fig. 10.58) $A B: A C: B C=5: 4: 3$, then $\angle C=90^{\circ}$. Let $B C=3 x$. Then $A C=4 x, A B=5 x$. The radius of the inscribed circle in a right triangle $r=\frac{B C+A C-A B}{2}=x$. $O$ is the center of the inscribed circle, $M, N, K$ are the points of tangency with the ... | 1:3;1:2;2:3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,861 |
10.253. For a triangle with sides 26, 28, and 30 cm, find the product of the radii of the circumscribed and inscribed circles. | ## Solution.
Let the sides of the triangle be: $a=26 \text{ cm}, b=28$ cm, $c=30$ cm. The semi-perimeter of the triangle $p=\frac{26+28+30}{2}=42$ cm. Let $S$ be the area of the triangle. The radius of the circumscribed circle $R=\frac{a b c}{4 S}$. The radius of the inscribed circle $r=\frac{S}{p}$. Then $R \cdot r=\... | 130^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,862 |
10.254. In triangle $ABC$, medians $AL$ and $BM$ are drawn, intersecting at point $K$. Vertex $C$ lies on the circle passing through points $K, L, M$. The length of side $AB$ is $a$. Find the length of median $CN$. | Solution.
Since $CN$ is the median, then $CK=\frac{2}{3} CN$ (Fig. 10.59). By connecting points $L$ and $M$, we obtain the midline $LM$; therefore, $\frac{LM}{AB}=\frac{CF}{CN}=\frac{1}{2}$, i.e., $CF=\frac{1}{2} CN, LM=\frac{a}{2}, LF=FM=\frac{1}{2} AN=\frac{a}{4}, FK=\frac{1}{6} CN$. We have
, $O_{1}$ is the center of a circle that touches the chords $A B$ and $A C$ and the given circle at points $M, D$, and $K$ respectively. Then $O_{1} D \perp A C, O_{1} M \perp A B$. $\angle B A C$ is an inscribed and right angle. Therefore, $B ... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,864 |
10.256. The lengths of two sides of an acute triangle are $\sqrt{13}$ and $\sqrt{10} \mathrm{~cm}$. Find the length of the third side, knowing that this side is equal to the height drawn to it. | Solution.
Let in $\triangle A B C$ (Fig. 10.61) $A B=\sqrt{10}$ cm, $B C=\sqrt{13}$ cm, height $B D=A C$. Since $\triangle A B C$ is an acute-angled triangle, point $D$ lies on the segment $A C$. Let $B D=A C=x$ cm. From $\triangle A D B\left(\angle A D B=90^{\circ}\right)$: $A D=\sqrt{A B^{2}-B D^{2}}=\sqrt{10-x^{2}}... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,865 |
10.257. Through point $P$ of the diameter of a given circle, a chord $A B$ is drawn, forming an angle of $60^{\circ}$ with the diameter. Calculate the radius of the circle if $A P=a$ and $B P=b$. | Solution.
Let $MN$ be the diameter of the circle with center $O$, passing through point $P$ (Fig. 10.62), $\angle APO=60^{\circ}$. Drop the perpendicular $OK$ to the chord $AB$. Then $BK=\frac{1}{2}AB=\frac{a+b}{2}; PK=BK-BP=\frac{a+b}{2}-b=\frac{a-b}{2}$. In $\triangle OKP$, $\angle OKP=90^{\circ}, \angle KOP=30^{\ci... | \sqrt{^2-+b^2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,866 |
10.258. The distances from a point $M$, lying inside triangle $A B C$, to its sides $A C$ and $B C$ are 2 cm and 4 cm, respectively. Calculate the distance from point $M$ to line $A B$, if $A B=10 \mathrm{~cm}, B C=17 \mathrm{~cm}$, $A C=21 \mathrm{~cm}$. | ## Solution.
Let the required distance be $x$ (Fig. 10.63). Then $S_{\triangle A M B}=0.5 \cdot 10 x=5 x$. $S_{\triangle A M B}=S_{\triangle A B C}-S_{\triangle A M C}-S_{\triangle B M C}$. Let's find these areas: $S_{\triangle A B C}=\sqrt{24 \cdot 14 \cdot 3 \cdot 7}=$ $=84 \text{ cm}^{2}, S_{\triangle A M C}=0.5 \c... | 5.8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,867 |
10.259. On the segment $A C$ of length 12 cm, a point $B$ is constructed such that $A B=4 \text{ cm}$. On the segments $A B$ and $A C$ as diameters, semicircles are constructed in the same half-plane with boundary $A C$. Calculate the radius of the circle that is tangent to the constructed semicircles and $A C$. | ## Solution.
Points $O_{1}$ and $O_{2}$ are the centers of semicircles with diameters $A B$ and $A C$ and radii $R_{1}=2$ cm and $R_{2}=6$ cm, respectively (Fig. 10.64). $O_{3}$ is the center of the circle with the unknown radius $x$ cm, $x>0$. Then $O_{1} O_{3}=(x+2)$ cm,
 $AC=48$ cm, $BE$ is the altitude, $BE=8.5$ cm, $O$ is the center of the inscribed circle, $M, N, K$ are the points of tangency of this circle with the sides $AB, BC, AC$ of the triangle, respectively. Then $ON \perp BC$, $ON=4$ cm, $AM=AK$, $CK=CN$, $BM=BN$. The area of triang... | 5\, | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,869 |
10.261. In an isosceles triangle \(ABC (AB = BC)\), a point \(D\) is taken on side \(BC\) such that \(BD: DC = 1: 4\). In what ratio does the line \(AD\) divide the height \(BE\) of triangle \(ABC\), counting from vertex \(B\)? | Solution.
Let $A D$ intersect $B E$ at point $F$ (Fig. 10.66). Draw $D K \| A C$. Since $\triangle B C E \sim \triangle B D K$, we have $\frac{B K}{B E}=\frac{K D}{E C}=\frac{B D}{B C}=\frac{1}{5}$, from which $B K=\frac{1}{5} B E, K D=\frac{1}{5} E C$. Since $\triangle K F D \sim \triangle A F E$, then $\frac{K D}{A ... | 1:2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,870 |
10.262. In a right-angled triangle, the height drawn to the hypotenuse is $h$; the radius of the inscribed circle is $r$. Find the hypotenuse. | Solution.
Let $a$ and $b$ be the legs, $c$ be the hypotenuse, and $p$ be the semiperimeter of the given triangle. $r=\frac{a+b-c}{2}=\frac{a+b+c}{2}-c=p-c ; p=r+c$. The area of the triangle $S=\frac{1}{2} c h=p r$. Then $\frac{1}{2} c h=(r+c) r ; c(h-2 r)=2 r^{2} ; c=\frac{2 r^{2}}{h-2 r}$. Answer: $\frac{2 r^{2}}{h-2... | \frac{2r^{2}}{-2r} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,871 |
10.263. The medians of a triangle are equal to $5, \sqrt{52}$, and $\sqrt{73}$ cm. Prove that the triangle is a right triangle. | Solution.
Let the sides of the triangle be denoted as $a, b, c$, and the medians to these sides as $m_{a}, m_{b}, m_{c}\left(m_{a}=5, m_{b}=\sqrt{52}, m_{c}=\sqrt{73}\right)$. According to the formula expressing a side of the triangle in terms of its medians, we have:
$$
a=\frac{2}{3} \sqrt{2\left(m_{b}^{2}+m_{c}^{2}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 48,872 |
10.264. Show that in any right-angled triangle, the sum of the semiperimeter and the radius of the inscribed circle is equal to the sum of the legs. | ## Solution.
Let $a, b, c, p$ be the legs, hypotenuse, and semi-perimeter of the triangle, respectively, and $r$ be the radius of the inscribed circle. Then $p+r=$ $=\frac{a+b+c}{2}+r=\frac{a+b}{2}+\frac{c}{2}+r$. Consider the sum $\frac{c}{2}+r$. Since the area of the triangle $S=\frac{a b}{2}=p r$, we have $r=\frac{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 48,873 |
10.265. Show that in any right-angled triangle, the sum of the diameters of the circumscribed and inscribed circles is equal to the sum of its legs. | ## Solution.
Let $a, b$ be the legs, $c$ be the hypotenuse, $R$ be the radius of the circumscribed circle, and $r$ be the radius of the inscribed circle. As is known, $R=\frac{c}{2}$, $r=\frac{a+b-c}{2}$, then $2 R+2 r=c+2 \frac{a+b-c}{2}=c+a+b-c=a+b$. This is what we needed to prove.
. Let $A E=m_{a}, B F=m_{b}$. Then in $\triangle B M E$ and $\triangle A M F$ respectively we have $\frac{m_{a}^{2}}{9}+\frac{4 m_{b}^{2}}{9}=\frac{a^{2}}{4}, \frac{4 m_{a}^{2}}{9}+\frac{m_{b}^{2}}{9}=\frac{b^{2}}{4}$. Add... | \sqrt{\frac{^{2}+b^{2}}{5}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,875 |
10.267. On the segment $A B$, a point $C$ is taken, and on the parts $A C$ and $C B$ of the segment $A B$ as diameters, semicircles are constructed. Prove that the sum of the lengths of these semicircles does not depend on the position of point $C$ on the segment $A B$. | ## Solution.
Let the length of segment $AB$ be $2l$. Denote the radius of one of the circles as $x$; then the radius of the second circle is $l-x$. The sum of the lengths of the semicircles is $L=\pi x + \pi(l-x) = \pi l$, i.e., it does not depend on $x$. This is what we needed to prove.
. Draw $D L \| A B$ and drop perpendiculars from points $D$ and $E$ to $A B$. Since triangles $A D C$ and $C E B$ are equilateral, $D L=F M=F C+C M=\frac{x}{2}+\frac{l-x}{2}=\frac{l}{2}$, $E L=\frac{(l-x) \sqrt{3}}{2}-\frac{x \sqrt{3}}{2}=\frac{(l-2 x) \sqrt{3}}{2}$. T... | at\the\midpoint\of\segment\A\B | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,877 |
10.269. The heights of a triangle are 12, 15, and 20 cm. Prove that the triangle is a right triangle. | ## Solution.
Let the area of the triangle be $S$. Then its sides are: $a=\frac{2 S}{15}$, $b=\frac{2 S}{20}$, $c=\frac{2 S}{12}$. We have $a^{2}+b^{2}=4 S^{2}\left(\frac{1}{15^{2}}+\frac{1}{20^{2}}\right)=4 S^{2}\left(\frac{1}{225}+\frac{1}{400}\right)=$ $=\frac{4 S^{2}}{25}\left(\frac{1}{9}+\frac{1}{16}\right)=\frac{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 48,878 |
10.270. Find the ratio of the sum of the squares of all medians of a triangle to the sum of the squares of all its sides. | ## Solution.
Let $a, b, c$ be the sides of a triangle, and $m_{a}, m_{b}, m_{c}$ be the medians drawn to these sides. We use the formulas:
$$
m_{a}=\frac{1}{2} \sqrt{2\left(b^{2}+c^{2}\right)-a^{2}}, m_{b}=\frac{1}{2} \sqrt{2\left(a^{2}+c^{2}\right)-b^{2}}, m_{c}=\frac{1}{2} \sqrt{2\left(a^{2}+b^{2}\right)-c^{2}}
$$
... | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,879 |
10.272. The numbers $m_{1}, m_{2}$, and $m_{3}$ represent the lengths of the medians of a certain triangle. Show that if the equality $m_{1}^{2}+m_{2}^{2}=5 m_{3}^{2}$ holds, then the triangle is a right triangle. | ## Solution.
Let the sides of the triangle be denoted by $a, b, c$. Then $4 m_{1}^{2} = 2(b^{2} + c^{2}) - a^{2}, 4 m_{2}^{2} = 2(a^{2} + c^{2}) - b^{2}$ (see additional relations $2^{\circ}$. Adding these equations, we get $4(m_{1}^{2} + m_{2}^{2}) = 4c^{2} + a^{2} + b^{2}$. By the problem's condition, $m_{1}^{2} + m... | ^{2}=^{2}+b^{2} | Geometry | proof | Yes | Yes | olympiads | false | 48,880 |
10.275. Through the intersection point of the diagonals of a trapezoid, a line parallel to the bases is drawn, intersecting the lateral sides at points $M$ and $N$. Prove that $M N=2 a b /(a+b)$, where $a$ and $b$ are the lengths of the bases. | ## Solution.
Let $A D=a, B C=b$ (Fig. 10.70). Since $\triangle M B O \sim \triangle A B D$, we have $\frac{M O}{a}=\frac{B O}{B D}$, thus $M O=a \cdot \frac{B O}{B D}$. Similarly, $\triangle O N D \sim \triangle B C D$, so $\frac{O N}{b}=\frac{O D}{B D}$, $O N=b \cdot \frac{O D}{B D}$. Therefore, $M N=M O+O N=\frac{a ... | \frac{2}{+b} | Geometry | proof | Yes | Yes | olympiads | false | 48,883 |
10.276. A right triangle $ABC$ is divided by the altitude $CD$, drawn to the hypotenuse, into two triangles $BCD$ and $ACD$. The radii of the circles inscribed in triangles $BCD$ and $ACD$ are 4 and $3 \text{ cm}$, respectively. Find the distance between their centers. | ## Solution.
Let $O_{2}$ be the center of the circle inscribed in $\triangle B C D$, and $O_{1}$ be the center of the circle inscribed in $\triangle A C D$ (Fig. 10.71), $N$ and $M$ be the points of tangency of these circles with the line $A B$. Then $O_{2} N \perp A B, O_{1} M \perp A B, O_{2} N=r_{2}=4$ cm, $O_{1} N... | 5\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,884 |
10.277. Find the bisector of the right angle of a triangle with legs equal to $a$ and $b$. | ## Solution.
In $\triangle A B C$ (Fig. 10.72) $\angle C=90^{\circ}, C B=a, C A=b, C K-$ is the bisector of $\angle B C A$. Let $C K=x . S_{\triangle A B C}=S_{\triangle C B K}+S_{\triangle A C K} ; \frac{1}{2} A C \cdot C B=$ $=\frac{1}{2} A C \cdot C K \sin \angle A C K+\frac{1}{2} B C \cdot C K \sin \angle B C K$. ... | \frac{\sqrt{2}}{+b} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,885 |
10.278. Given a square with side length $a$. Determine the sides of an equal-area isosceles triangle, where the sum of the lengths of the base and the height dropped to it is equal to the sum of the lengths of the two legs. | Solution.
Let $x$ be the base, $z$ be the lateral side, and $y$ be the height of the sought triangle drawn to the base. According to the problem, we have $\left\{\begin{array}{l}\frac{1}{2} x y=a^{2}, \\ x+y=2 z,\end{array}\right.$ By the Pythagorean theorem (Fig. 10.73) $y^{2}+\frac{x^{2}}{4}=z^{2}$. Thus, we have th... | \sqrt{3},\frac{5\sqrt{3}}{6},\frac{5\sqrt{3}}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,886 |
10.279. Points $M, N, P, Q$ are the midpoints of sides $A B, B C, C D$ and $D A$ of rhombus $A B C D$. Calculate the area of the figure that is the intersection of quadrilaterals $A B C D, A N C Q$ and $B P D M$, if the area of the rhombus is $100 \mathrm{~cm}^{2}$. | ## Solution.
Let $E$ and $F$ be the points of intersection of segment $AN$ with $DM$ and $BP$, respectively, and $K$ and $L$ be the points of intersection of segment $CQ$ with $BP$ and $DM$, respectively (Fig. 10.74). $\triangle AQC = \triangle ANC$ (by two sides and the included angle), therefore, $\angle QCA = \angl... | 20^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,887 |
10.280. Determine the angles of an isosceles triangle if its area is related to the area of a square constructed on the base as $\sqrt{3}: 12$. | ## Solution.
Let $A C D K$ be the given square (Fig. 10.75), $A B C$ be the given triangle, $A B = B C$, and $B P$ be the height of $\triangle A B C$. Denote $A C = a$, $B P = h$, $S_{\triangle A B C} = \frac{1}{2} a h$, and $S_{A C D K} = a^{2}$. According to the problem, $\frac{\frac{a h}{2}}{a^{2}} = \frac{\sqrt{3}... | 30,30,120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,888 |
10.281. A quadrilateral with angles of $120, 90, 60$, and $90^{\circ}$ is inscribed in a circle. The area of the quadrilateral is $9 \sqrt{3}$ cm $^{2}$. Find the radius of the circle, given that the diagonals of the quadrilateral are perpendicular to each other. | Solution.
Let $ABCD$ be the quadrilateral mentioned in the problem (Fig. 10.76), $\angle ABC = \angle ADC = 90^\circ$, $\angle BAD = 120^\circ$, $\angle BCD = 60^\circ$. $P$ is the intersection point of $AC$ and $BD$, and $AC \perp BD$. Since $\angle ABC = 90^\circ$, $AC$ is the diameter of the given circle. $BP$ is t... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,889 |
10.282. In triangle $A B C$, a line $D E$ is drawn parallel to the base $A C$. The area of triangle $A B C$ is 8 sq. units, and the area of triangle $D E C$ is 2 sq. units. Find the ratio of the length of segment $D E$ to the length of the base of triangle $A B C$. | Solution.
We have $S_{\triangle D E C}=\frac{1}{2} D E \cdot H K=2$ (Fig. 10.77) and $H K=\frac{4}{D E} ; S_{\triangle A B C}=$ $=\frac{1}{2} A C \cdot B H=8$, from which $B H=\frac{16}{A C}$. Since $\triangle A B C \sim \triangle D B E$, then $\frac{D E}{A C}=\frac{B K}{B H}$.
Further, $B K=B H-H K$ and, therefore,
... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,890 |
10.283. The area of a right-angled triangle is $24 \mathrm{~cm}^{2}$, and the hypotenuse is 10 cm. Find the radius of the inscribed circle. | Solution.
Let $a$ and $b$ be the legs of the triangle. We have the system $\left\{\begin{array}{l}a b=48, \\ a^{2}+b^{2}=100,\end{array} \Rightarrow\right.$ $\Rightarrow a^{2}+2 a b+b^{2}=196, a^{2}-2 a b+b^{2}=4, \Rightarrow a+b=14, a-b=2$ and, consequently, $a=8$ (cm), $b=6$ (cm). Since $S=p r$, then $24=12 r, r=2$ ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,891 |
10.284. A square and an equilateral triangle are described around a circle of radius $R$, with one side of the square lying on a side of the triangle. Calculate the area of the common part of the triangle and the square. | Solution.
$\triangle A B C$ and square $E D D_{1} E_{1}$ (Fig. 10.78) are the ones mentioned in the problem, $B N$ is the altitude of $\triangle A B C$, $O$ is the center of the given circle, $M$ and $F$ are the points of intersection of $A B, M_{1}$ and $F_{1}$ are the points of intersection of $B C$ with the sides o... | \frac{R^{2}\sqrt{3}(6\sqrt{3}-4)}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,892 |
10.285. In a circle of radius $R$, two parallel chords are drawn on opposite sides of the center, one of which subtends an arc of $60^{\circ}$, the other $120^{\circ}$. Find the area of the part of the circle enclosed between the chords. | ## Solution.
The area of the segment with an arc of $60^{\circ}$ is $S_{1}=\frac{\pi R^{2}}{6}-\frac{R^{2} \sqrt{3}}{4}$, and the area of the segment with an arc of $120^{\circ}$ is $S_{2}=\frac{\pi R^{2}}{3}-\frac{R^{2} \sqrt{3}}{4}$. The desired area is
$$
\begin{gathered}
S=\pi R^{2}-S_{1}-S_{2}=\frac{R^{2}(\pi+\s... | \frac{R^{2}(\pi+\sqrt{3})}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,893 |
10.286. Two circles of radii $r$ and $3r$ touch each other externally. Find the area of the figure enclosed between the circles and their common external tangent. | ## Solution.
Let $O_{1}$ and $O_{2}$ be the centers of circles with radii $r$ and $3r$ respectively (Fig. 10.79), $AB$ be their common external tangent, $A$ and $B$ be the points of tangency. Then $O_{1}A \perp AB, O_{2}B \perp AB, O_{1}A = r, O_{2}B = 3r, O_{1}O_{2} = 4r$. $P$ is the point of tangency of the given ci... | \frac{r^{2}(24\sqrt{3}-11\pi)}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,894 |
10.287. Find the area of a triangle inscribed in a circle of radius 2 cm, if two angles of the triangle are $\pi / 3$ and $\pi / 4$. | ## Solution.
Since $\angle A C B=\frac{\pi}{4}$ (Fig. 10.80), then $A B=2 \sqrt{2}$ cm (side of the inscribed square). Draw $B D \perp A C$. Considering that $\angle A B D=\frac{\pi}{6}$, we have $A D=\frac{1}{2} A B=\sqrt{2}$ (cm); further, since $\triangle B D C$ is isosceles, then
$D C=B D=\frac{\sqrt{3}}{2} A B=\f... | \sqrt{3}+3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,895 |
10.288. Find the area of a trapezoid whose diagonals are 7 and 8 cm, and the bases are 3 and 6 cm. | Solution.
Let in trapezoid $A B C D$ (Fig. 10.81) $B C \| A D, B C=3 \text{ cm}, A D=6$ cm, $A C=7 \text{ cm}, B D=8$ cm. Draw $C E \| B D$. Then $A E=A D+D E=$ $=A D+B C=9$ cm, $C E=B D$. Therefore, trapezoid $A B C D$ and triangle $A C E$ have equal areas (they have equal heights and the base of the triangle is equa... | 12\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,896 |
10.289. A circle is inscribed in a rhombus with side $a$ and an acute angle of $60^{\circ}$. Determine the area of the quadrilateral whose vertices are the points of tangency of the circle with the sides of the rhombus. | Solution.
The radius of the circle inscribed in the rhombus $ABCD$ (Fig. 10.82) is $R=\frac{a \sqrt{3}}{4}$, since $\angle A=60^{\circ}$. The quadrilateral $KLMN$ is a rectangle, as its angles subtend the diameter of the circle.
. Then $O_{1} O_{2} = R + r$, $O_{1} A \perp A B$, $O_{1} A = R$, $O_{2} B = r$, $O_{1} A \parallel O_{2} B... | \frac{\piR^{2}r^{2}}{(\sqrt{R}+\sqrt{r})^{4}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,898 |
10.291. The center of the circle inscribed in a right trapezoid is 1 cm and 2 cm away from the ends of the lateral side. Find the area of the trapezoid. | Solution.
Let in trapezoid $ABCD$ (Fig. 10.84) $BC \| AD, AB \perp AD, O$ - the center
of the angle bisectors of $\angle BCD$ and $\angle ADC$. Since $\angle BCD + \angle ADC = 180^\circ$,... | 3.6^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,899 |
10.293. A circle of radius $R$ is inscribed in some angle, and the length of the chord connecting the points of tangency is $a$. Two tangents parallel to this chord are drawn, as a result of which a trapezoid is formed. Find the area of this trapezoid. | ## Solution.
Let $L$ and $M$ be the points of tangency (Fig. 10.86); then $S L = S M$, from which $A B = C D$, since $A D \parallel B C \parallel L M$. Draw $O K \perp L M$ and $B H \perp A D$. The required area $S = \frac{1}{2}(A D + B C) B H$. For the circumscribed trapezoid, we have $A D + B C = A B + C D = 2 A B$;... | \frac{8R^{3}}{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,901 |
10.295. The legs of a right triangle are 6 and 8 cm. A circle is drawn through the midpoint of the smaller leg and the midpoint of the hypotenuse, touching the hypotenuse. Find the area of the circle bounded by this circle. | ## Solution.
Given in $\triangle ABC$ (Fig. 10.88) $\angle ABC=90^{\circ}, AB=6$ cm, $BC=8$ cm, points $E$ and $D$ are the midpoints of $AB$ and $AC$, respectively, and $O$ is the center of the circle mentioned in the problem. $AC=\sqrt{AB^{2}+BC^{2}}=10$ cm. Then $ED$ is the midline of $\triangle ABC$, $ED=\frac{1}{2... | \frac{100\pi}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,902 |
10.296. Find the area of a right-angled triangle if the radii $R$ and $r$ of the circumscribed and inscribed circles are given. | Solution.
Let $a, b, c$ be the legs and hypotenuse of the given triangle, respectively. $c=2R$, $r=\frac{a+b-c}{2}=\frac{a+b-2R}{2}$, $a+b=2(R+r)$. But $a^{2}+b^{2}=c^{2}$, then $(a+b)^{2}-2ab=4R^{2}$, $2ab=(a+b)^{2}-4R^{2}=4(R+r)^{2}-4R^{2}=8Rr+4r^{2}$, $ab=2(2Rr+r^{2})=2r(2R+r)$. The area of the triangle $S=\frac{1}... | r(2R+r) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,903 |
10.298. Determine the area of the segment if its perimeter is $p$, and the arc contains $120^{\circ}$. | ## Solution.
Let the radius of the circle be $R$. Since the arc of the segment contains $120^{\circ}$, its length is $\frac{2 \pi R}{3}$, and the chord spanning this arc (the side of the inscribed regular triangle) is $R \sqrt{3}$. Therefore, the perimeter of the segment is $p=\frac{2 \pi R}{3}+R \sqrt{3}$, from which... | \frac{3p^{2}(4\pi-3\sqrt{3})}{4(2\pi+3\sqrt{3})^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,905 |
10.300. The side of a regular triangle is equal to $a$. Determine the area of the part of the triangle that lies outside a circle of radius $a / 3$, the center of which coincides with the center of the triangle. | ## Solution.
The desired area $S=S_{1}-S_{2}+3 S_{3}$, where $S_{1}=\frac{a^{2} \sqrt{3}}{4}$ is the area of the triangle, $S_{2}=\frac{\pi a^{2}}{9}$ is the area of the circle, and $S_{3}$ is the area of the segment cut off by the triangle from the circle. The chord of this segment is $\frac{a}{3}$; therefore, $S_{3}... | \frac{^{2}(3\sqrt{3}-\pi)}{18} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,907 |
10.301. Find the ratio of the area of a square inscribed in a segment with an arc of $180^{\circ}$ to the area of a square inscribed in a segment of the same circle with an arc of $90^{\circ}$. | ## Solution.
Let $A B C D$ be a square inscribed in a segment with an arc of $180^{\circ}$ (Fig. 10.91.1), and $A_{1} B_{1} C_{1} D_{1}$ be a square inscribed in a segment of the same circle with an arc of $90^{\circ}$ (Fig. 10.91.2). Let $R$ be the radius of the circle, $A B=x, A_{1} B_{1}=y$. Consider the square $A ... | 10:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,908 |
10.303. In triangle $ABC$, medians $BD$ and $CE$ are drawn; $M$ is their point of intersection. Prove that triangle $BCM$ is equal in area to quadrilateral $ADME$. | ## Solution.
Draw the third median $A K$ of triangle $A B C$ (Fig. 10.93), and drop perpendiculars $B F$ and $M N$ to $A C$. Since $M D = B D / 3$, then $M N = B F / 3$. $S_{\triangle A M C} = \frac{1}{2} A C \cdot M N = \frac{1}{2} A C \cdot \frac{1}{3} B F = \frac{1}{3} S_{\triangle A B C}$. $M D$ is the median of $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 48,910 |
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