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334. Point $O$ is located inside triangle $A B C$. Prove that side $A C$ is seen from it at a larger angle than from point $B$. | $\triangle$ Compare triangles $A B C$ and $A O C$. Angles $A$ and $C$ of the second triangle are smaller than the corresponding angles of the first. Since the sum of the angles in both triangles is the same, angle $O$ in triangle $A O C$ will be greater than angle $B$ in triangle $A B C . \triangleleft$
![](https://cdn... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,023 |
343. Prove that symmetry preserves distances: if points $A^{\prime}$ and $B^{\prime}$ are symmetric to points $A$ and $B$, then the distances $A B$ and $A^{\prime} B^{\prime}$ are equal. | $\triangleright$ Visually, this is quite clear: when the sheet is flipped around the axis of symmetry, points $A$ and $B$ coincide with points $A^{\prime}$ and $B^{\prime}$, and the distance between them does not change.
We can prove this by referring to already known geometric facts. Extend $A B$ until it intersects ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,026 |
344. Prove that symmetry preserves not only the lengths of the sides of a triangle but also its angles. | $\Delta$ The third criterion of triangle congruence guarantees that if the lengths of the sides have not changed, then the angles have not changed either. $\triangleleft$
Axial symmetry is often called "mirror" symmetry, because a symmetrical figure looks like a reflection in a mirror surface (in the image, the inscri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,027 |
345. A ray of light emerges from point $A$, reflects off the line $P Q$ at point $L$ according to the law "the angle of incidence equals the angle of reflection," and finally reaches point $B$. Prove that the point $A^{\prime}$, symmetric to point $A$ with respect to $P Q$, lies on the line $B L$ (so that an observer a... | $\triangleright$ Symmetry preserves angles, so the angles formed by the rays $L A$ and $L A^{\prime}$ with the axis of symmetry are equal: $\angle A L P = \angle A^{\prime} L P$. Therefore, the angles $\angle B L Q$ and $\angle A^{\prime} L P$ are also equal, which means that the rays $L A^{\prime}$ and $L B$ continue ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,028 |
348. Points $A$ and $B$ are on the same side of line $l$. Point $A^{\prime}$ is symmetric to point $A$ with respect to line $l$. Segment $A^{\prime} B$ intersects $l$ at point $K$, and point $M$ lies on line $l$ and does not coincide with $K$. Prove that $A K+K B<A M+M B$. | $\Delta$ This is the same problem, but without the "embellishment" about the stream, tourists, etc. Accordingly, we will present the solution. Since $l$ is the perpendicular bisector of $A A^{\prime}$, and points $K$ and $M$ lie on it, we have $A K=A^{\prime} K$ and $A M=A^{\prime} M$. By the triangle inequality (for $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,029 |
349. Lines $l$ and $m$ intersect at point $O$. An arbitrary point $A$ is reflected symmetrically with respect to line $l$, resulting in point $A^{\prime}$. Then point $A^{\prime}$ is reflected symmetrically with respect to line $m$, resulting in point $A^{\prime \prime}$. Prove that the angle $A O A^{\prime \prime}$ do... | $\triangleright$ The angle $A O A^{\prime \prime}$ is divided into four parts, two of which are equal in pairs. Leaving one part from each pair, we get the angle between $l$ and $m . \triangleleft$
In the solution to the previous problem, we used the specific position of point $A$ (if it were at $A^{\prime}$, we would... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,030 |
374. Given a point $A$ and the center of symmetry $O$. How to construct the point $A^{\prime}$, symmetric to $A$ with respect to $O$, using a compass and a straightedge? | $\triangleright$ Draw a straight line through $A$ and $O$ using a ruler, and then use a compass to mark off a segment equal to $AO$ on the other side of $O$. $\triangleleft$ | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,031 |
375. In the figure, points $A$ and $A^{\prime}$ were depicted as centrally symmetric to each other with respect to point $O$. Then point $O$ was erased, and points $A$ and $A^{\prime}$ remained. How can the position of point $O$ be restored? | $\triangleright$ Construct the midpoint of the segment $A A^{\prime} . \triangleleft$
When applying central symmetry to all points of a figure, we obtain a figure centrally symmetric to it. If it coincides with the original, then it is said that the figure has a center of symmetry. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,032 |
377. Prove that central symmetry does not change distances: if points $A^{\prime}$ and $B^{\prime}$ are symmetric to points $A$ and $B$ with respect to point $O$, then the distances $A B$ and $A^{\prime} B^{\prime}$ are equal. | $\triangleright$ Just like with axial symmetry, this is beyond doubt: when the paper is rotated around point $O$, distances do not change, and segment $A B$ transforms into $A^{\prime} B^{\prime}$. However, this can also be proven using the congruence of triangles: triangles $A O B$
 also lies on the boundary of the triangle.
To do this, let's ask ourselves: where can $X^{\prime}$, the point symmetric to $X$ on the boundary of triangle $ABC$, be? In t... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,037 |
389. Prove that if two arcs of a circle are equal in magnitude, then the chords subtending them are equal. | $\triangle$ By connecting the ends of the chords with the center of the circle, we obtain two triangles that are equal by two sides (radii) and the angle between them. $\triangleleft$
(The same can be said about two arcs of the same size in two circles of equal radii.)
 Let $A, B$ and $C$ be points on a circle with center $O$. Then the angle $B A C$ is equal to half the measure of the arc $B C$, that is, half the angle $B O C$.
The an... | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,039 |
392. If $AB$ is the diameter of a circle, and $C$ is any point on the circle, then the angle $ACB$ is a right angle. (a) Derive this statement from the inscribed angle theorem. (b) Prove it without referring to this theorem. | $\Delta$ (a) Angle $ACB$ rests on an arc that constitutes half of the circle. The corresponding central angle will be a straight angle, and its measure is $180^{\circ}$. Therefore, the inscribed angle that rests on this arc is half of $180^{\circ}$, which is $90^{\circ}$.
(b) Connect point $C$ to the center of the cir... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,041 |
393. If $AB$ is the diameter of a circle with center at $O$, and $C$ is any point on the circle, then the angle $COB$ is twice the angle $CAB$. (a) Derive this statement from the inscribed angle theorem. (b) Prove it without referring to this theorem. | $\triangleright$ (a) Indeed, according to this theorem, the angle $\angle C A B$ is equal to half the measure of the arc $B C$, that is, half of the corresponding central angle $C O B$.
(b) The triangle $A O C$ is isosceles, so the angles at its base are equal. Therefore, its exterior angle $C O B$ (which is equal to ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,042 |
395. In quadrilateral $ABCD$, two opposite angles $A$ and $C$ are right angles. Prove that angle $CBD$ is equal to angle $CAD$. | $\triangleright$ The diagonal $B D$ cuts the quadrilateral into two right triangles. Let's construct a circle for which this diagonal will be the diameter. This circle will pass through the vertices of the quadrilateral (recall that in a right triangle, the median to the hypotenuse is equal to half of it). It remains t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,044 |
396. The vertices of an equilateral triangle lie on a circle. Prove that from any point on the circle, one of the sides of the triangle is seen at an angle of $120^{\circ}$, and the other two at an angle of $60^{\circ}$. | $\Delta$ All angles of an equilateral triangle are equal to $60^{\circ}$, so the arcs on which they stand are equal to $120^{\circ}$. Therefore, angles $A M B$ and $B M C$ in the figure are also equal to $60^{\circ}$, and $A M C$ is equal to $120^{\circ} . \triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,045 |
397. If two chords of a circle are parallel, then the arcs enclosed between them are equal in magnitude. | $\triangleright$ Connect the opposite ends of the chords with a segment, we get two equal alternate interior angles. Therefore, the arcs of the circle on which they stand are equal. $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,046 |
398. Prove that if a circle passes through all four vertices of a trapezoid, then the non-parallel sides of the trapezoid are equal (the trapezoid is isosceles). | $\Delta$ According to the previous problem, the arcs enclosed between the bases of the trapezoid are equal. Therefore (problem 389), the chords subtending them, that is, the lateral sides of the trapezoid, are equal. $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,047 |
399. The vertices of the quadrilateral $ABCD$ lie on a circle. Prove that the sum of two opposite angles ($A$ and $C$, as well as $B$ and $D$) is $180^{\circ}$. | $\Delta$ These opposite angles subtend mutually complementary arcs, which together form a complete circle, that is, $360^{\circ}$. Therefore, the sum of the angles is half of $360^{\circ}$, which is $180^{\circ}$. $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,048 |
401. Segments $A B$ and $B C$ are diameters of two circles intersecting at $B$ and one other point. Prove that this second point lies on the line $A C$. | $\triangleright$ Let $D$ be the other point of intersection. The angle $B D A$ is subtended by the diameter of the circle and is therefore a right angle. Similarly, the angle $B D C$ is also a right angle. Thus, the segments $A D$ and $D C$ form a straight angle, meaning they lie on the same straight line, so the point... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,050 |
402. We have seen that from any point on the circumference, its diameter is seen at a right angle. Prove that from points inside the circle it is seen at an obtuse angle, and from points outside the circle - at an acute angle. | $\triangleright$ Let $AB$ be a diameter, and $C$ a point inside the circle. We will prove that angle $ACB$ is obtuse. Extend $AC$ until it intersects the circle at point $D$. Angle $ADB$ will be a right angle, and angle $ACB$ will be an exterior angle of the right triangle, adjacent to its acute angle, and therefore ob... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,051 |
403. Prove that the angle between two intersecting chords $A B$ and $C D$ is equal to half the sum of the arcs $B D$ and $A C$ (enclosed within this angle and the vertical angle to it). | $\triangleright$ Let $O$ be the point of intersection of the chords. Draw the chord $A D$. The angle $B O D$ we are interested in is an exterior angle of triangle $A O D$ and therefore equals the sum of the two non-adjacent angles of this triangle. Both of these angles are inscribed, one is equal to half the arc $A C$,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,052 |
404. The angle $\beta$ between the extensions of two non-intersecting chords $A B$ and $C D$ is equal to half the difference of the measures of arcs $B D$ and $A C$. | $\triangleright$ Let $O$ be the point of intersection of the extensions of the chords. Draw the chord $B C$. In triangle $O B C$, the sum of angle $\beta$ and the inscribed angle $A B C$ is equal to the exterior angle $B C D$, so angle $\beta$ is the difference between these two angles, that is, half the difference of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,053 |
405. A straight line intersects a circle at points $A$ and $B$ and divides the circle into two parts (called segments). Consider one of them, bounded by the chord $A B$ and the arc of the circle. As we know (the inscribed angle theorem), from all points on this arc, the chord $A B$ is seen at the same angle. Prove that... | $\triangleright$ For a right angle, we have already solved this problem and could act in the same way. However, now we can instead refer to two previous problems.
We need to prove that the angle $A O B$ is greater than half the arc $A B$ (the arc lying on the other side of the line is meant). Indeed, according to prob... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,054 |
406. Given two points $A$ and $B$. Draw where the points $C$ are located from which the segment $A B$ is seen at an angle of $30^{\circ}$. | $\checkmark$ Let's construct an equilateral triangle $ABC$ with side $AB$ and a circle centered at point $C$, passing through points $A$ and $B$. The arc $AB$ of this circle is seen from the center at an angle of $60^{\circ}$, and therefore from any point on the circle (lying on the same side of $AB$ as point $C$), the... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,055 |
420. A tangent to a circle was rotated by an angle $\beta$ around the point of tangency, causing it to become a secant. Find the measure of the arc it intercepts. | $\triangleright$ Let $OA$ be the radius drawn to the point of tangency, and $AB$ be the secant. The tangent is perpendicular to the radius, so the angle $OAB$ between the radius and the secant is $90^{\circ}-\beta$. The triangle $OAB$ is isosceles, so the second angle in it is also $90^{\circ}-\beta$, and the third ang... | 2\beta | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,056 |
421. A circle, a triangle with vertices on this circle, and a line tangent to the circle at one of the triangle's vertices are drawn. Identify two pairs of equal angles on this diagram. | $\triangle$ See the figure: both marked angles are equal to half of the marked arc. Similarly, we find another pair of equal angles. $\triangleleft$ | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,057 |
423. Prove that the angle $B M A$ between the tangent $M B$ and the secant $M A C$ is equal to half the difference of the arcs $B C$ and $A B$. | $\triangle$ In triangle $M B A$, angle $M$ is equal to the difference between the external angle $B A C$ (which is half the arc $B C$, since it subtends it) and angle $M B A$ (which is the angle between the tangent and the chord and is equal to half the arc $A B) . \triangleleft$
This problem is a limiting case of pro... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,059 |
425. Two tangents are drawn from point $A$ to a circle. Prove that the segments from point $A$ to the points of tangency are equal. | $\Delta$ Connect the center of the circle with the points of tangency. We get two right triangles, in which the hypotenuse is common, and the legs are equal (as the radii of the circle). Therefore, the other legs, that is, the segments of interest to us, are also equal. $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,060 |
426. Given a circle and a point outside it, draw a tangent to the circle through the point using a compass and a straightedge.
The first reaction of any normal person: why do we need a compass at all?! We can simply place the straightedge so that it passes through the given point and touches the given circle, - and th... | $\Delta$ The first method. Suppose the line is already constructed, and connect the point of tangency $K$ with the center of the circle $O$. This will form a right triangle $A O K$ (angle $K$ is a right angle, as the radius drawn to the point of tangency is perpendicular to the tangent). In this triangle, we know the h... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,061 |
428. Prove that the common external tangents to two circles of equal radius are parallel to each other and to the line of centers (the line passing through the centers of the circles). (A line is called a common external tangent to two circles if it touches them and they lie on the same side of it.) | $\triangleright$ Let's prove that the common tangent is parallel to the line of centers. Connect the centers with the points of tangency. The resulting segments are equal (as radii of equal circles) and parallel (as perpendiculars to the common tangent). Therefore, they are opposite sides of a parallelogram, and the ot... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,063 |
439. Two concentric (with a common center) circles with radii $r$ and $R$ (assume that $r<R$) are given. The smaller circle starts to move, increasing the distance $d$ between the centers of the circles. At what value of $d$ will it touch the larger circle? How many points of intersection will the circles have after th... | $\triangleright$ It is convenient to draw the line of centers - a straight line passing through the centers of the circles, and to consider that the larger circle is stationary while the center of the smaller one moves along this line. It is clear that when the value of $d$ reaches the difference of the radii ($R-r$), ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,065 |
443. A circle intersects a line at two points. Prove that it forms equal angles with the line at these points | $\triangleright$ Following the definition of the angle between a line and a circle, let's draw two tangents at the points of intersection. Together with our line, they form an isosceles triangle (the tangent segments are equal) and therefore the angles at its base are equal. $\triangleleft$
Strictly speaking, we shoul... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,068 |
445. Two circles intersect at two points. Prove that the angles at which they intersect at these points are equal.
 | $\triangleright$ Draw the radii to the points of intersection, and connect the centers of the circles with a segment. Two equal triangles will be formed (by three sides). Therefore, the angles between the radii are equal, and it remains to refer to the previous problem. $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,069 |
446. Two circles are given. We want to draw a common tangent - a line that touches both circles. How many such lines exist? (The answer depends on the ratio of the radii of the circles and the distance between their centers.) | $\triangleright$ Let the radii of the circles be $r$ and $R$ (assuming $rR+r$ there are four tangents - two external and two internal. $\triangleleft$
A common tangent to two circles can be external (both circles lie on the same side of the tangent) or internal (the circles lie on opposite sides).
| 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,070 |
448. Can three non-intersecting (and non-touching) circles and a point outside these circles be drawn so that the circles cover the entire horizon for that point (any ray emanating from the point hits one of the circles)? Can the same be done with two circles? | $\Delta$ Each circle covers the angle of view between two tangents (see figure). This angle is less than $180^{\circ}$, so two circles cannot cover the entire horizon $\left(360^{\circ}\right)$.
Three, however, can. We can divide the entire field of view into three sectors of $120^{\circ}$ each. Each sector can be cov... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,072 |
449. Two points $A$ and $B$ are 8 units apart. How many lines exist that are 5 units away from point $A$ and 3 units away from point $B$? | $\triangleright$ The line $l$ is at a distance $d$ from the point $A$ when it is tangent to the circle of radius $d$ centered at $A$. Therefore, the problem can be restated as follows: given two circles with radii 3 and 5, and the distance between their centers is 8. How many common tangents exist?
We have already sol... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,073 |
450. Construct a common external tangent to two given circles (lying outside each other) using a compass and a straightedge. | $\Delta$ The condition of the problem can be translated into the language of distances: we need to find a line that is at the given distances (radii $r_{1}$ and $r_{2}$, assuming $r_{1}>r_{2}$) from two given points (centers $O_{1}$ and $O_{2}$), and both points must lie on the same side of the line (the external tange... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,074 |
460. Given two points $A$ and $B$. Consider all possible circles passing through both points $A$ and $B$. Where are their centers located? | $\triangle$ If a circle with center $O$ passes through points $A$ and $B$, then the distances $O A$ and $O B$ are equal (as radii of this circle). As we know, in this case, point $O$ lies on the perpendicular bisector of $A B$.
Conversely, any point $O$ on the perpendicular bisector is equidistant from $A$ and $B$, an... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,075 |
461. The teacher gave the following task: there are three wells $A, B, C$; each resident goes to the nearest one. Draw from where people go to each of the wells. One of the students drew the perpendicular bisectors of the sides of triangle $A B C$ (see figure) and wondered where people from the small triangle in the mi... | $\triangle$ His reasoning is correct, but it means that such a triangle does not exist, that is, all three perpendicular bisectors intersect at one point. $\triangleleft$
The same can be said in a slightly different way: | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,076 |
463. Given a triangle. How to construct a circle passing through its vertices? | $\Delta$ The condition of this problem can be reformulated as follows: we need to find a point that is at the same distance from the vertices of the triangle (the center of the desired circle). We have seen that for this, we need to construct the perpendicular bisectors of two sides and find the point of their intersec... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,078 |
465. Given a circle, but its center is not indicated. Restore this center using a compass and a ruler. | $\triangleright$ We already know how to construct a circle passing through three given points - in this construction, the center of the circle is also constructed. So it is enough to take any three points on the circle and perform this construction. $\triangleleft$ | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,080 |
467. The vertices of triangle $ABC$ lie on a circle. The bisector of angle $A$ intersects this circle at point $M$. Prove that point $M$ lies on the perpendicular bisector of side $BC$. | $\triangleright$ The bisector divides the angle into two equal inscribed angles, which, consequently, rest on equal arcs. Therefore, the chords $M B$ and $M C$ are equal. Hence, point $M$ is equidistant from points $B$ and $C$, that is, it lies on the perpendicular bisector of side $B C . \triangleleft$
The statement ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,082 |
468. Prove that the three altitudes of an acute triangle intersect at one point. | $\triangleright$ Draw a line through each vertex of the triangle parallel to the opposite side. Three parallelograms are formed, each of which is divided into two triangles equal to the original one. It is easy to notice that the altitudes of the original triangle are the perpendicular bisectors in the resulting larger... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,083 |
476. In triangle $ABC$, the bisectors of angles $A$ and $B$ intersect at point $M$. Prove that the bisector of angle $C$ also passes through point $M$. | $\Delta$ Points lying on the bisector of an angle are equidistant from its sides. Therefore, point $M$ is equidistant from sides $A B$ and $A C$ (lying on the bisector of angle $A$), and also from sides $A B$ and $B C$ (lying on the bisector of angle $B$). Hence, it is equidistant from all three sides, and thus lies on... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,085 |
477. Given a triangle. Construct a circle lying inside it and touching all three of its sides. | $\triangle$ The center of this circle is equidistant from the sides of the triangle (at a distance equal to the radius of the circle) and therefore lies at the point of intersection of the angle bisectors of the triangle. We know how to construct angle bisectors. $\triangleleft$
The circle lying inside the triangle an... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,086 |
486. Prove that if the vertices of a quadrilateral lie on a circle, then the sums of the opposite angles are $180^{\circ}$. | $\Delta$ This has already been discussed in problem 399: an inscribed angle is equal to half the arc it intercepts; opposite angles of a quadrilateral intercept arcs that together form a full circle $\left(360^{\circ}\right) . \triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,089 |
487. Prove that if a circle touches all four sides of a quadrilateral, then the sums of its opposite sides are equal.
In other words, by drawing four tangents to a circle, we obtain a quadrilateral in which the sums of the opposite sides are equal. | $\triangleright$ Using problem 425 (the equality of tangents to a circle drawn from the same point), we can find four pairs of equal segments in the diagram. Now it is clear that the sum of two opposite sides (any pair) includes all four segments exactly once. $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,090 |
488. Let's draw a circle through the vertices $A, B$ and $C$ of the quadrilateral $ABCD$. Prove that if point $D$ is outside the circle, then the sum of angles $B$ and $D$ is less than $180^{\circ}$, and the sum of angles $A$ and $C$ is greater than $180^{\circ}$. Prove that if point $D$ is inside the circle (as shown ... | $\triangleright$ First, note that the sum of all four angles is $360^{\circ}$. Therefore, if the sum of one pair of angles is greater than $180^{\circ}$, the sum of the second pair is less, and vice versa.
Suppose point $D$ is outside the circle. What will happen to the angles of the quadrilateral if we replace it wit... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,091 |
489. Prove that if the sums of the opposite angles of a quadrilateral are equal to $180^{\circ}$, then a circle can be circumscribed around it. | $\Delta$ This easily follows from the previous problem: if we draw a circle through three of its vertices, the fourth will fall on it. Indeed, it cannot be either inside or outside the circle, since in either case the sum of the opposite angles would be less than or greater than $180^{\circ} . \triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,092 |
491. Prove that if in a convex quadrilateral the sums of the opposite sides are equal, then a circle can be inscribed in it.
(If a circle cannot be inscribed in a hinged quadrilateral, then changing the angles will not help.) | $>$ As a consequence of the previous problem: inscribe a circle so that it touches three sides. Then the fourth side must also touch the circle: otherwise, by the previous problem, the sums of the opposite sides would not be equal. $\triangleleft$
The condition of convexity is essential: one can draw a non-convex quad... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,094 |
492. Prove that a circle can be circumscribed around a convex quadrilateral $ABCD$ if and only if the side $CD$ is seen from points $A$ and $B$ at the same angle. | $\Delta$ If a circle can be circumscribed, then the angles from which side $C D$ is seen from points $A$ and $B$ are inscribed angles subtending the same arc, and therefore are equal. Conversely: let these angles be equal. Draw a circle through points $A, C$, and $D$. Then point $B$ will fall on the same circle. Indeed... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,095 |
500. Find the area of a rectangle with sides 3 and 5. | $\triangleright$ Such a rectangle can be cut into 15 unit squares, so its area is equal to the sum of the areas of these squares, that is, $15 . \triangleleft$ | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,097 |
501. How many square centimeters are in a square meter | $\triangleright$ A square with a side length of 1 meter can be divided into $100 \times 100$ squares with a side length of 1 centimeter, so in one square meter, there are $100 \cdot 100=10000$ square centimeters. $\triangleleft$ | 10000 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,098 |
502. Find the area of a rectangle with sides $1 / 3$ and $1 / 5$.
 | $\triangleright$ If a unit square is cut into 3 parts horizontally and 5 parts vertically, then each part will be a rectangle with sides $1 / 3$ and $1 / 5$. In total, there are 15 parts, they are equal, so the area of each part will be $1 / 15 . \triangleleft$ | \frac{1}{15} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,099 |
503. Find the area of a rectangle with sides $2 / 3$ and $3 / 5$. | $\triangle$ Such a rectangle is composed of $2 \times 3$ rectangles of size $1 / 3 \times 1 / 5$, so its area is $6 / 15 = (2 / 3) \cdot (3 / 5). \triangleleft$
$\|$ The area of a rectangl... | \frac{6}{15}=(\frac{2}{3})\cdot(\frac{3}{5}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,100 |
504. Find the area of a right triangle, both legs of which have a length of 1. | $\triangle$ The diagonal cuts the unit square into two such triangles, so the area of each of them is $1 / 2 . \triangleleft$
 | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,101 |
505. Find the area of a right triangle with legs measuring 3 and 5. | $\triangleright$ Two such triangles can form a $3 \times 5$ rectangle, so the area of each of them is half the area of the rectangle, which is $3 \cdot 5 / 2 . \triangleleft$
The solution... | \frac{15}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,102 |
509. Cut a parallelogram into pieces from which a rectangle can be formed. | $\checkmark$ Place the parallelogram on a horizontal line and drop perpendiculars from the ends of its upper base. It is clear that if we cut off the triangle on the left and attach it to the right, we will get a rectangle. (These two right triangles - the protruding and the missing one - are equal, for example, by hyp... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,106 |
510. The median divides the triangle into two triangles. Prove that these triangles are equal in area by cutting one of them into two parts, from which the second can be assembled. | $\triangle$ Draw two midlines. We will get two triangles, equal to the halves of the parallelogram, and two others that are equal by three sides (each side is half the length of the corresponding side of the larger triangle). $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,107 |
511. Prove that the area of a parallelogram is equal to the product of its side and the height dropped to this side. | $\checkmark$ We can use the solved problem and cut the parallelogram into parts from which we can form a rectangle. But we can do it differently. If we add a triangle $CDE$ ($DE$ is the height dropped from point $D$ to line $AB$, see the figure) to the parallelogram $ABCD$, we get a trapezoid $ABCE$. The same trapezoid... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,108 |
516. On one of two parallel lines, a fixed segment $AB$ is taken, while a point $C$ moves along the other line. Prove that the area of triangle $ABC$ remains constant. | $\triangle$ Indeed, if we drop a perpendicular from point $C$ to the line $A B$, its length does not depend on the position of point $C . \triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,109 |
517. In trapezoid $A B C D$ (bases $A D$ and $B C$), the diagonals intersect at point $O$. Prove that triangles $A O B$ and $C O D$ have equal area. | $\Delta$ This becomes clear if we add triangle $A O D$ to both triangles - we get two triangles $A B D$ and $A C D$, which have the same base $A D$ and equal heights dropped to this base. $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,110 |
518. In triangle $ABC$, a point $M$ is taken on side $AC$, dividing it in the ratio $AM: MC=1: 3$. Find the area of triangle $ABM$, if the area of triangle $ABC$ is 1. | $\triangle$ In triangles $A B C$ and $A B M$, there is a common height dropped from point $B$, and the sides $A C$ and $A M$, onto which it is dropped, differ by a factor of four. Therefore, the area of triangle $A B M$ is $1 / 4 . \triangleleft$
The same reasoning shows that if one side of a triangle is increased or ... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,111 |
520. A circle of radius $r$ is inscribed in a triangle with perimeter $p$ and area $S$. How are these three quantities related? | $\triangleright$ By connecting the center of the circle to the vertices of the triangle, we get three triangles, each of which has a height $r$ dropped to one of the sides of the triangle $(a, b, c)$. The areas of these three parts are respectively $a r / 2, b r / 2$, and $cr / 2$. Adding them up, we get $S=(a+b+c) r /... | \frac{pr}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,112 |
521. By what factor should the side of a square be increased to make its area four times larger? | $\triangleright$ The area of a rectangle with sides $a$ and $b$ is $a b$, so the area of a square with side $a$ is $a \cdot a=a^{2}$. By doubling the side of the square, we increase its area by four times: $(2 a)^{2}=2^{2} a^{2}=4 a^{2} . \triangleleft$ | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,113 |
522. By what factor should the side of a square be increased to double its area? | $\triangle$ When the side of a square is increased by $k$ times, its area increases by $k^{2}$ times: if the side was $a$ and the area $a^{2}$ before, now the side is $k a$ and the area $(k a)^{2}=k^{2} a^{2}$. Therefore, we need to find such a $k$ that $k^{2}$ equals 2. This number (positive) is called the square root... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,114 |
577. Draw an isosceles right triangle on graph paper with the legs equal to the side of a cell, and check the Pythagorean theorem for it: the area of the square built on the hypotenuse is equal to the sum of the areas of the squares built on the legs. | $\triangleright$ The squares built on the legs have a unit area. The square built on the hypotenuse is divided into four triangles, each of which makes up half of a grid square, and thus has an area of 2 - as required. $\triangleleft$ | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,116 |
578. Check the Pythagorean theorem for a right triangle with legs 1 and 2. | $\triangleright$ The squares constructed on the legs have areas of 1 and 4. It remains to find the area of the square constructed on the hypotenuse. This can be done in two ways. First, it can be cut into 4 triangles and a central square. The triangles are equal to the original (legs 1 and 2) and have an area of 1. The... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,117 |
581. On the sides of a right-angled triangle, squares are constructed. The height dropped from the right angle is extended and cuts the square constructed on the hypotenuse into two rectangles. Prove that they are equal in area to the squares constructed on the legs. | $\checkmark$ The solution to this problem is easiest to explain with a sequence of pictures.
We need to prove that each of the shaded rectangles is equal in area to the square constructed on (the corresponding) leg. It is convenient
. Therefore, it divides the equilateral triangle into two right triangles, in which the hypotenuse is $a$, one of the legs is $a / 2$, and the height $h$ is the other leg. Thus, it can be found using the formula:
$$
h=\sqrt{a^{2}-(a / 2)^... | \frac{\sqrt{3}}{2}, | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,121 |
604. Prove the converse statement: if all sides of one triangle are greater than the corresponding sides of another triangle by the same factor, then the corresponding angles of the triangles are equal. | $\triangleright$ Let all sides of triangle $A B C$ be the same multiple of the sides of triangle $A^{\prime} B^{\prime} C^{\prime}$. We will use the following trick: construct triangle $A^{\prime \prime} B^{\prime \prime} C^{\prime \prime}$, where all angles are equal to the corresponding angles of triangle $A B C$, an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,124 |
605. In triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$, angles $A$ and $A^{\prime}$ are equal. Moreover, the ratios of the sides $A^{\prime} B^{\prime}: A B$ and $A^{\prime} C^{\prime}: A C$ are equal. Prove that the triangles are similar. | $\Delta$ Place the smaller triangle (let it be $A^{\prime} B^{\prime} C^{\prime}$) on the larger one so that vertices $A$ and $A^{\prime}$ coincide, $A^{\prime} B^{\prime}$ lies along $A B$, and $A^{\prime} C^{\prime}$ lies along $A C$ (this is possible because angles $A$ and $A^{\prime}$ are equal). If we prove that $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,125 |
606. Which of the following criteria should be applied to prove that any two equilateral triangles are similar? | $\triangleright$ Any of the three will do: all angles are equal to $60^{\circ}$, the ratios of the sides are equal (since in each of the triangles all sides are equal). $\triangleleft$
When solving problems, it is very important to see similar triangles on the diagram. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,126 |
611. From a point $A$ lying outside a circle, a tangent $A X$ is drawn (where $X$ is the point of tangency), as well as a secant intersecting the circle at points $Y$ and $Z$; the points $X, Y$, and $Z$ are connected by segments. Find the similar triangles. | $\triangleright$ Both angles $A X Y$ and $X Z A$ are equal to half the arc $X Y$, and angle $A$ in triangles $A X Y$ and $A X Z$ is common, so they are similar. $\triangleleft$
The last problem can be considered as a limiting case of the penultimate one: when one side of the inscribed quadrilateral tends to zero, the ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,131 |
612. In an acute-angled triangle, heights are dropped from two vertices to the opposite sides. Find two similar triangles. | $\triangleright$ Let in triangle $A B C$ the altitudes $A K$ and $B L$ be drawn. Then, two right triangles $A K C$ and $B L C$ are formed, both containing angle $C$. Therefore, they are similar. $\triangleleft$ | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,132 |
613. In a right triangle, a height is dropped from the vertex of the right angle to the hypotenuse. Find three similar triangles. | $\Delta$ In the figure, it is easy to see three right triangles: the original triangle and the two parts into which the height divides it. Each of these parts shares a common angle with the original triangle and is therefore similar to it. $\triangleleft$ | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,133 |
615. In triangle $ABC$, perpendiculars are dropped from vertices $B$ and $C$ to the angle bisector $AM$ of angle $A$. Find two pairs of similar triangles. | $\triangleright$ Let $B K$ and $C L$ be the altitudes. In the right triangles $A B K$ and $A C L$, the acute angles at vertex $A$ are equal (property of the angle bisector), so they are similar.
On the other hand, the right triangles $B M K$ and $C M L$ are also similar, as they have the same (vertical) angles at vert... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,135 |
616. The bisector $A M$ of triangle $A B C$ is extended to intersect the circumscribed circle at point $K$. Find three triangles in the figure that are all similar to each other. | $\triangle$ We already know that triangles $A B M$ and $M K C$ are similar by three angles - regardless of whether $A M$ is a bisector or not. But if it is, then triangle $A B M$ is also similar to triangle $A K C$ (angles $B$ and $K$ of these triangles subtend the same arc, and the angles at vertex $A$ are equal by th... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,136 |
617. Prove that in a trapezoid, the point of intersection of the diagonals divides both diagonals in the same ratio, equal to the ratio of the bases of the trapezoid. | $\triangle$ Let the diagonals $A C$ and $B D$ of trapezoid $A B C D$ with bases $A D$ and $B C$ intersect at point $O$. Then triangles $A O D$ and $C O B$, as we have seen, are similar. By the property of similar triangles
$$
A O: O C=D O: O B=A D: B C
$$
 B M \\
B M = A B / (k-1... | \begin{aligned}&BM=\frac{AB\cdotBC}{AD-BC},\quadAM=\frac{AB\cdotAD}{AD-BC}\\&CM=\frac{CD\cdotBC}{AD-BC},\quadDM=\frac{CD\cdotAD}{AD-BC}\end{aligned} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,138 |
620. Given a point $A$ lying outside a circle. Through it, a ray is drawn intersecting the circle at points $B$ and $C$. Prove that the product $A B \cdot A C$ does not depend on which ray is drawn. | $\triangleright$ Let another ray be drawn from point $A$, intersecting the circle at points $D$ and $E$. We need to prove that $A B \cdot A C = A D \cdot A E$. By connecting the vertices of the quadrilateral $B C E D$, we see a familiar picture: the opposite sides of the inscribed quadrilateral are extended to intersec... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,140 |
622. Prove that in any triangle, the altitudes are inversely proportional to the sides to which they are dropped (if one side is a certain number of times larger than another, the altitude dropped to it is that many times smaller). | $\triangle$ Let in triangle $A B C$ the altitudes $A K$ and $B L$ be drawn. Right triangles $A K C$ and $B L C$ have a common angle $C$. Therefore, they are similar, and the ratio of the altitudes $A K: B L$ is equal to the ratio of the sides $A C: B C$. In other words, if one side is $k$ times larger than the other, t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,142 |
623. The height of a right triangle, dropped from the vertex of the right angle, divides its hypotenuse into two segments. Prove that
(1) the height is equal to the geometric mean of these segments (the square root of their product);
 lie on one straight line. But the last two points can be found using a ruler, so this line can be constructed. The points where it inte... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,148 |
629. Given two parallel lines and a segment on one of them. How to find its midpoint using only a ruler
保持了原文的换行和格式。 | $\checkmark$ If we take any two points on the other line, we get a trapezoid, and we have reduced the problem to the previous one. $\triangleleft$ | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,149 |
630. The bisector of a triangle's angle divides the opposite side into two segments. Prove that the ratio of these segments is equal to the ratio of the sides adjacent to them. | $\triangleright$ Let $AM$ be the bisector in triangle $ABC$. Drop perpendiculars $BK$ and $CL$ from points $B$ and $C$ to $AM$. Triangles $ABK$ and $ACL$, as we saw in problem 615), are similar, so $BK: CL = AB: AC$. On the other hand, triangles $KBM$ and $LCM$ are also similar, so $BK: CL = BM: MC$. Combining these eq... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,150 |
631. Prove that the square of the length of the bisector $A M$ of triangle $A B C$ is equal to $A B \cdot A C - M B \cdot M C$.
(Using this problem and the previous one, the length of the bisector can be found if the lengths of all sides are known.) | $\triangleright$ We need to prove that $A M^{2}=A B \cdot A C-M B \cdot M C$. Let's make an additional construction from problem 616. Triangles $A B M$ and $A K C$ are similar, so
$$
A B: A M=A K: A C
$$
which means
$$
\begin{aligned}
& A B \cdot A C=A M \cdot A K=A M \cdot(A M+M K)= \\
& \quad=A M^{2}+A M \cdot M K... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,151 |
632. In triangle $ABC$, medians $AK$ and $CM$ are drawn, intersecting at point $O$. Through point $K$, a segment $KN$ is drawn parallel to side $AB$, intersecting median $CM$ at point $N$. Prove that triangle $KON$ is similar to triangle $AOM$, and that triangle $CNK$ is similar to triangle $CMB$. Prove that $AO: OK = ... | $\triangleright$ Since $K N$ is parallel to $A M$, the angles in triangles $K O N$ and $A O M$ are equal, so they are similar by three angles. Similarly for triangles $C N K$ and $C M B$. Since $K$ is the midpoint of $B C$, triangle $C N K$ is half the size of triangle $C M B$, in particular, $K N = \frac{1}{2} M B$. F... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,152 |
662. Point $A$ was moved 13 units to the right and resulted in a point with a coordinate of 8. What was the coordinate of point $A$? | $\triangleright$ Let $A$ have the coordinate $a$. Then $a+13=8$, and $a=8-13=(-5) . \triangleleft$
 | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,153 |
663. The grasshopper jumped from the point with coordinate 8 to the point with coordinate 17.5. Then it made another jump of the same length (in the same direction). At the point with what coordinate did it end up? | $\triangleright$ The length of the jump is $17.5-8=9.5$; another jump will give $17.5+9.5=27 . \triangleleft$ | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,154 |
664. Given two points $A(-3)$ and $B(7)$ (the coordinates are given in parentheses). Find the coordinate of the midpoint of segment $A B$. | $\triangle$ The distance $A B$ is $7-(-3)=10$, half of the distance is 5. Shifting $A$ 5 units to the right (or $B$ 5 units to the left), we get the point with coordinate 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,155 |
666. Which of the points $A(5 / 8)$ and $B(8 / 13)$ is to the left? (The coordinates of the points are given in parentheses; the positive direction is to the right.) | $\triangle$ Coordinates increase from left to right, so we need to compare the fractions $5 / 8$ and $8 / 13$ in terms of size:
$$
\frac{5}{8}=\frac{5 \cdot 13}{8 \cdot 13}=\frac{65}{8 \cdot 13}>\frac{64}{8 \cdot 13}=\frac{8 \cdot 8}{8 \cdot 13}=\frac{8}{13}
$$
This means that the coordinate of point $A$ is greater, ... | B | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,157 |
667. Find the distance between points $A$ (with coordinate $a$) and $B$ (with coordinate $b$). | $\triangle$ To find this distance, one needs to subtract the smaller coordinate from the larger one. Thus, it equals $b-a$ when $b>a$ and $a-b$ when $a>b$. Naturally, the distance is zero when $a=b$. $\triangleleft$
The formula for the distance is conveniently written using the modulus sign:
The absolute value, or mo... | |b-| | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,158 |
668. What is $|a|$, if $a<0$: will it be $a$ or $-a$? | $\triangleright$ It seems tempting to say that the modulus is $a$ (after all, the modulus is a "number without a sign," while $-a$ has a "minus" sign), but this is incorrect. Since $a$ is negative, the modulus must be positive. To find the modulus, one must choose the positive (more precisely, non-negative) number from... | - | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,159 |
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