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Ex. 102. A circle touches the legs of a right triangle and the circle circumscribed around it. Prove that its radius is twice the radius of the circle inscribed in the triangle. | Ex. 102. Let's apply the coordinate method. Place the origin at vertex $C$, the x-axis along ray $C A$, and the y-axis along ray $C B$. Denote the radius of the inscribed circle by $r$, the circumradius by $R$ $(R=c / 2)$, and the radius of the third circle, which touches the coordinate axes and the circumcircle, by $\... | \rho= | Geometry | proof | Yes | Yes | olympiads | false | 41,927 |
Ex. 103. A diameter divides a circle into two parts, one of which contains a smaller inscribed circle touching the larger circle at point $M$, and the diameter at point $K$. The ray $MK$ intersects the larger circle a second time at point
. Moreover, $\angle D A B=180^{\circ}-\angle C B D$. At the same time, $\angle C B D=\angle B D E$. From this, the statement to be proved follows. | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,931 |
Ex. 107. Two circles intersect at points $Q_{1}$ and $Q_{2}$. Points $A$ and $B$ lie on the first circle, and rays $A Q_{2}$ and $B Q_{1}$ are drawn through them, with $A Q_{2}$ intersecting the second circle again at point $C$. On the arc $Q_{1} Q_{2}$ of the first circle, lying inside the second circle, a point $F$ i... | Ex. 107. Hint. Show that $\angle Q N P = \angle Q F P$, that is, the segment $Q P$ is seen from points $Q$ and $F$ at the same angle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,932 |
Ex. 108. A circle with center at point $O$ is inscribed in triangle $A B C$. A second circle with center at point $P$ touches rays $A B$ and $A C$ and side $B C$ externally. It is known that angle $B A C$ is equal to $\alpha$, and the sum of the distances from point $O$ to vertices $B$ and $C$ is $t$. Find the sum of t... | Ex. 108. Answer: $m \operatorname{ctg} \frac{\pi-\alpha}{4}$. Solution. Let the distance $O P$ be denoted by $d$. Angles $O C P$ and $O B P$ are right angles (see the figure), hence the quadrilateral $O C P B$ can be inscribed in a circle, in which $O P=d$ is the diameter.
$. The length of $A B$ is $n$. The segment $M N$ with endpoints on the sides of the angle is tangent to both circles. Find the difference in the areas of triangles $B M N... | Ex. 109. $\frac{n^{2} \sin \alpha}{4}$. | \frac{n^{2}\sin\alpha}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,934 |
Ex. 110. On a plane, a rhombus $A B C D$ and two circles circumscribed around triangles $B C D$ and $A B D$ are constructed. Ray $B A$ intersects the first circle at point $P$ (different from $B$), and ray $P D$ intersects the second circle at point $Q$. It is known that $P D=1$ and $D Q=2+\sqrt{3}$. Find the area of t... | Ex. 110. 1/2.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,935 | |
Ex. 111. Prove that:
a) $\operatorname{tg} \frac{A}{2} \leq \frac{a}{2 h_{a}}$
b) $a^{2}=b^{2}+c^{2}-4 S \operatorname{ctg} A$
c) $\sin \frac{A}{2} \leq \frac{a}{2 \sqrt{b c}}$
d) $\operatorname{ctg} A+\operatorname{ctg} B+\operatorname{ctg} C=\frac{a^{2}+b^{2}+c^{2}}{4 S}$. | Ex. 111. a) Using the formulas $\operatorname{tg} \frac{A}{2}=\frac{1-\cos A}{\sin A}$ and $h_{a}=\frac{2 S}{a}=\frac{b c \sin A}{a}$, we obtain the equivalent inequality $(1-\cos A) b c \leq a^{2}$. Applying the cosine theorem to this inequality, we get $(b-c)^{2} \geq 0$. b) After applying the formula $S=\frac{1}{2} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,936 |
Ex. 112. Prove that
a) $a^{2}+b^{2}=2 c^{2} \Rightarrow \quad \cos C \geq \frac{1}{2}, \quad \operatorname{ctg} A+\operatorname{ctg} B=2 \operatorname{ctg} C$;
b) $\operatorname{tg} \frac{A}{2}+\operatorname{tg} \frac{B}{2}+\operatorname{tg} \frac{C}{2} \leq \operatorname{ctg} A+\operatorname{ctg} B+\operatorname{ctg} ... | Ex. 112. a) Applying the cosine theorem, we get that $\cos C=\frac{c^{2}}{2 a b}$. Since $2 a b \leq$ $a^{2}+b^{2}=2 c^{2}$, it follows that the inequality $\cos C \geq \frac{1}{2}$ holds. Further, applying the formulas from Ex. $111 b$), we get that $\operatorname{ctg} C=\frac{c^{2}}{4 S}$ and $\operatorname{ctg} A+\o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,937 |
Ex. 113. Prove the inequalities and verify that the estimates are accurate:
a) $0<\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \leq \frac{1}{8}$
b) $-1<\cos A \cos B \cos C \leq \frac{1}{8}$;
b) $1<\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2} \leq \frac{3}{2}$
c) $2<\cos \frac{A}{2}+\cos \frac{B}{2}+\cos \f... | Ex. 113. Hint. Use the auxiliary statement. Lemma. If $\varphi$ is a fixed angle, $0<\varphi<180^{\circ}$ and $\alpha+\beta=\varphi$, then the expressions $\sin \alpha+\sin \beta, \sin \alpha \sin \beta, \cos \alpha+\cos \beta$ and $\cos \alpha \cos \beta$ reach their maximum value when $\alpha=\beta$. Using the lemma,... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 41,938 |
$$
\begin{aligned}
S & =r R(\sin A+\sin B+\sin C)=4 r R \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}= \\
& =\frac{R^{2}}{2}(\sin 2 A+\sin 2 B+\sin 2 C)=2 R^{2} \sin A \sin B \sin C
\end{aligned}
$$
Exercise 114. Prove that
$$
\begin{aligned}
S & =r R(\sin A+\sin B+\sin C)=4 r R \cos \frac{A}{2} \cos \frac{B}{2}... | Ex. 114. The first equality follows from the formula $S=p r$ and the Law of Sines, the second is proved as a conditional identity by direct transformation, the third follows from the relation $S_{A B C}=S_{A O B}+S_{A O C}+S_{B O C}$, where $O$ is the center of the circumscribed circle, with the sign of one of the term... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,939 |
Ex. 118. A circle with center on side $AB$ of triangle $ABC$ touches sides $AC$ and $BC$. Find the radius of the circle, given that it is expressed as an integer, and sides $AC$ and $BC$ are equal to 5 and 3. | Ex. 118. Answer: $r=1$. Solution. By equating the expressions for the area of triangle $ABC$, we get that $\frac{a+b}{2} r=\frac{a b}{2} \sin C$. From this, it follows that $r=\frac{a b}{a+b} \sin C$. In our case, $r=\frac{15}{8} \sin C$. If $r \geq 2$, then $\sin C>1$. This solves the problem. | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,943 |
Ex. 119. The lengths of the sides of a triangle are integers. It is known that the height, drawn to one of the sides, divides it into integer segments, the difference of which is 7. For what smallest value of the length of this side is the height, drawn to it, also an integer? | Ex. 119. 25. Solution. Let $A B C$ be the considered triangle, $B D$ be the height. Denote $A D=x, D C=y, A D=h$. Let $y>x \Rightarrow y-x=7 \Rightarrow x=\frac{b-7}{2}$. Expressing the height in two ways, we get the equality $a^{2}-y^{2}=c^{2}-x^{2} \Rightarrow$ $(y-x)(y+x)=(a-c)(a+c) \Rightarrow 7 b=(a-c)(a+c)$. Sinc... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,944 |
Ex. 121. The area of an integer triangle is also expressed as an integer equal to the semiperimeter. Show that this is possible in the only case when the sides of the triangle are $3,4,5$. | Ex. 121. Using Heron's formula and the condition of the problem, we write that $4(a+b+c)=(a+b-c)(a+c-b)(b+c-a)$. Suppose now that two sides of the triangle are equal, for example, $a=b$. Then $4(2a+c)=(2a-c)c^2$. If $c$ is odd, the left side is even, and the right side is odd, hence there are no solutions. If $c=2k$, t... | 3,4,5 | Geometry | proof | Yes | Yes | olympiads | false | 41,946 |
Ex. 124. Prove that if in an integer triangle the sine of one of the angles is a rational number, and is represented as an irreducible fraction, then the cosine of the same angle is also represented as an irreducible fraction with the same denominator, and the denominator of both fractions is an odd number, while the n... | Ex. 124. Let $\sin A=\frac{m}{n}$, where $m$ and $n$ are coprime. The cosine of the same angle is always a rational fraction. Let $\cos A=\frac{k}{l}$, where $k$ and $l$ are coprime. Then $m^{2} l^{2}+n^{2} k^{2}=n^{2} l^{2} \Rightarrow m^{2} l^{2}=n^{2}\left(l^{2}-k^{2}\right)$. Since $m$ and $n$ have no common factor... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,949 |
Ex. 126. Prove that the sides of any rectangular integer triangle are proportional multiples of the sides of a primitive triangle with legs \(a\) and \(b\) and hypotenuse \(c\), which are expressed by the formulas (up to a permutation of the legs)
\[
a=2 m n, \quad b=m^{2}-n^{2}, \quad c=m^{2}+n^{2}
\]
where \(m\) an... | Ex. 126. First, note that if the sides (numbers expressing the lengths of the sides) are mutually prime, then they are also pairwise mutually prime. In particular, this means that only one of the sides is even. It is not difficult to verify that this can only be the leg. Let this be the leg $a$. Write the Pythagorean t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,951 |
Ex. 127. Show that for Pythagorean numbers the following statements are true:
a) the numbers in a Pythagorean triple are pairwise coprime;
b) one of the legs is an even number, and it is divisible by 4;
c) one of the legs is divisible by 3;
d) one of the sides is divisible by 5;
e) the hypotenuse is an odd number ... | Ex. 127. Statements $a$) and b) immediately follow from the assumptions of mutual simplicity. Let's prove statement $c$). If $m$ or $n$ is divisible by 3, then $a^{\prime}$ is divisible by 3. If neither of these numbers is divisible by 3, then their squares, when divided by 3, have a remainder of 1, and therefore, $b^{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,952 |
Ex. 128. Show that if the sine and cosine of some angle are rational numbers, represented as irreducible fractions, then these fractions have the same denominator, which is an odd number of the form $4k+1$. The numerators are coprime numbers of different parity, and the even numerator is divisible by 4. | Ex. 128. Let's show that $\sin \alpha$ and $\cos \alpha$ have a common denominator. Let $\sin \alpha=\frac{a}{c}, \cos \alpha=\frac{b}{d}, \operatorname{LCM}(c, d)=n, n=c k_{1}=d k_{2}$, where $k_{1}$ and $k_{2}$ are coprime natural numbers. Then $a^{2} k_{1}^{2}+b^{2} k_{2}^{2}=n^{2} \Rightarrow b$ is divisible by $k_... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,953 |
Ex. 129. In an integer-sided triangle, two sides are equal to 10. Find the third side, given that the radius of the inscribed circle is an integer. | Ex. 129. Answer: 12. Solution. Let the third side be denoted by $a$, and the angle subtending it by $\alpha$. Then $\sin \alpha=\frac{10+10+a}{10 \cdot 10} \cdot r$. At the same time, $1 \leq a \leq 19 \Rightarrow r \leq 4$. If $\sin \alpha=1$, then $x=5, r=4$, but a triangle with such data does not exist. According to... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,954 |
Ex. 130. Prove that the radius of the circle circumscribed around a simple triangle cannot be an integer.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Ex. 130. Let $a$ be the odd side. By the Law of Sines, $\sin A=\frac{a}{2 R}-$ a fraction, where after simplification the denominator remains an even number, which contradicts the statement of the previous exercise. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,955 |
Ex. 131. Prove that if the diameter of the circle circumscribed around a simple triangle is an integer, then
a) it is odd, and has the form $2 R=4 n+1$;
b) the triangle has two odd sides, and one even side, which is divisible by $4$;
c) the semiperimeter is also an integer;
d) the area is an even integer;
e) the s... | Ex. 131. Using, as in exercise 130, the sine theorem, we get that $2 R$ is an odd number of the required form. By the cosine theorem, $\cos A=\frac{(b+c-a)(b+c+a)}{2 b c}$. If all three sides are odd or only one is odd, then the numerator of the fraction is odd and, consequently, with any reduction, the denominator wil... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,956 |
Ex. 140. A circle is inscribed in a trapezoid with integer sides. The midline of the trapezoid divides it into two parts, the areas of which are 15 and 30. Find the radius of the inscribed circle.
## Several problems | Ex. 140. 5/2. Solution. Let $A D=a$ be the larger base, and $B C=b$ be the smaller base. By calculating the ratio of the areas of the upper and lower trapezoids, we find that $a=5 b$ and $b h=15$, where $h$ is the height of the trapezoid. It follows that $h$ is a rational number.
Assume for definiteness that $A B \leq... | \frac{5}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,958 |
Ex. 142. Angle $A$ in triangle $A B C$ is equal to $\alpha$. A circle passing through $A$ and $B$ and tangent to $B C$ intersects the median to side $B C$ (or its extension) at point $M$, different from $A$. Find $\angle B M C$. | Ex. 142. Answer: $180^{\circ}-\alpha$. Hint. Note that $\angle C B M=\angle B A D$, and $\angle B C M=\angle C A D$ (due to the similarity of triangles $C M D$ and $A D C$, where $D$ is the midpoint of $B C$). | 180-\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,959 |
Ex. 147. In a convex quadrilateral, the point of intersection of the diagonals divides each of them in the ratio $1: 2$. The quadrilateral formed by sequentially connecting the midpoints of the sides is a square with a side of 3. Find the perimeter of the original quadrilateral. | Ex. 147. $4 \sqrt{5}+6 \sqrt{2}$. | 4\sqrt{5}+6\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,963 |
Ex. 158. Show that in an acute-angled triangle $S=p^{*} R$, where $p^{*}$ is the semiperimeter of the triangle formed by the bases of the altitudes (the billiard triangle). | Ex. 158. It follows from the formulas $S=\frac{R^{2}}{2}(\sin 2 A+\sin 2 B+\sin 2 C)$ (Ex. 114) and $p^{*}=\frac{R}{2}(\sin 2 A+\sin 2 B+\sin 2 C)$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,971 |
Ex. 160. Consider an arbitrary triangle $ABC$. Find a point on the plane for which the sum of the squares of the distances to the vertices of the triangle is the smallest. | Ex. 160. The required point is the point of intersection of the medians. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,973 |
Ex. 165. Inside an angle of $90^{\circ}$, a point is given, located at distances of 8 and 1 from the sides of the angle. What is the minimum length of the segment passing through this point, with its ends lying on the sides of the angle? | Ex. 165. $5 \sqrt{5}$. | 5\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,976 |
Ex. 170. Find the smallest radius of a semicircle in which a triangle with the given sides can be inscribed such that two vertices of the triangle lie on the semicircle, and the third vertex lies on the diameter of the semicircle.
a) $7,8,9$
b) $9,15,16$ | Ex. 170. a) $R=6$; b) $R=9$. | )6;b)9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,978 |
## 1. On a Hunt
Three hunters shoot at a rabbit simultaneously. The chances of the first hunter succeeding are rated as 3 out of 5, the second hunter's chances are 3 out of 10, and finally, the third hunter's chances are only 1 out of 10. What is the probability that the rabbit will be killed? | 1. If the hare escapes from the hunters, it means that the first hunter missed (2 chances out of 5), as did the second (7 out of 10), and the third (9 out of 10). Therefore, the probability of a miss is $2 / 5 \cdot 7 / 10 \cdot 9 / 10=126 / 500$. Consequently, the hare will be shot with a probability of $1-126 / 500=0... | 0.748 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,979 |
## 2. As many boys as girls
Consider all possible families with two children. Half of such families are "lucky," i.e., the number of boys in the family matches the number of girls. Is the same true for families with four children? (Assume that the birth of a boy and a girl is equally likely.) | 2. We will call the "distribution" of children in a family a sequence of four children characterized only by their gender and listed in order of seniority. For example: "boy, girl, boy, boy," which we will briefly write as "BGBB".
Families with four children correspond to 16 types of equally probable distributions:
B... | \frac{3}{8} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,980 |
## 4. On a Garden Bench
There were five of them: three boys and two girls. It was spring. They had nothing to do. They sat down next to each other on a garden bench to warm up in the sun, and they arranged themselves purely by chance. Is it more likely that the two girls will be separated from each other or that they ... | 4. The number of ways to choose two seats out of five on a bench for two girls is
$$
C_{5}^{2}=\frac{5!}{3!2!}=10
$$
It is obvious (why?), that four of these ways allow the girls to sit next to each other.
Thus, we have 4 chances out of 10 that the girls will be next to each other, and 6 chances out of 10 that they ... | \frac{4}{10} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,982 |
## 5. Black and White Balls
An urn contains 5 white and 5 black balls. Then, 3 balls are randomly drawn one by one. This is done in two ways: 1 - the drawn ball is returned to the urn, 2 - the balls are not returned. In which of these two cases is there a higher probability of drawing one white and two black balls? | 5. In the first case (drawing a ball with replacement), we have:
$P_{\mathbf{i}}=$ (number of variants where one of the three balls is white) $\times 1 / 2^{3}=3 / 8=9 / 24$.
In the second case (drawing without replacement), we get:
$P_{2}=$ [number of ways to choose one (white) ball from five $\times$ number of way... | \frac{10}{24} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,983 |
## 7. At the Dentist's Office
In the waiting room, two women and ten men are waiting for their turn. At their disposal are eight copies of the latest magazine and four copies of the morning newspaper. In how many ways can they distribute the newspapers and magazines among themselves, if both women insist on reading th... | 7. If both women will be reading a newspaper, then the corresponding number of distribution options for newspapers and magazines will be equal to the number of ways in which the remaining two copies of the newspaper can be distributed among ten men, i.e.
$$
C_{10}^{2}=\frac{10!}{2!8!}=45
$$
If the two women will be r... | 255 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,985 |
## 8. On the Café Terrace
One beautiful spring evening, Dupont and Durand were playing dice on the café terrace. They took turns throwing two dice. If the sum was 7, Durand won a point, and if the sum was 8, Dupont won. On whom would you bet if you had to place a wager? | 8. No matter what number comes up on the first die, Durand, when throwing the second, knows that he has one chance in six of winning (if a 1 comes up after a 6, a 2 after a 5, etc.), whereas if a 1 comes up on the first die, Dupont has no chance of winning. Therefore, when making a bet, it's better to bet on Durand. | Dur | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,986 |
## 9. A Group of Young People
The group consists of three boys and three girls. Each of the boys is loved by one of the three girls, and each girl is loved by one of the boys; but one of the girls sadly noted that in their group, no one is loved by the person they love themselves. Is this sad circumstance really so un... | 9. Let's denote the girls by the letters $F_{1}, F_{2}, F_{3}$, and the three guys by $-G_{1}, G_{2}, G_{3}$. We will calculate the desired probability using the scheme («tree») illustrated in the figure.
The total number of outcomes corresponding to the "sad variant" of unrequited love turns out to be $3 \cdot 52=156... | 0.214 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,987 |
## 10. Auto Rally
This was an extremely dangerous auto rally. It started with a small and very narrow bridge, where one out of five cars fell into the water. Then came a terrible sharp turn, where three out of ten cars ended up in the ditch. Further along the way, there was such a dark and winding tunnel that one out ... | 10. For a car to successfully complete an auto rally, it must not get into an accident on the bridge (which happens in 4 out of 5 cases), nor on a curve (7 out of 10), nor in a tunnel (9 out of 10), nor, finally, on a sandy road (3 out of 5). Thus, we get
$$
\frac{4 \cdot 7 \cdot 9 \cdot 3}{5 \cdot 10 \cdot 10 \cdot 5... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,988 |
## 11. How Many Are We?
To help you answer this question, let's say that the probability that at least two of us have the same birthday is less than $1 / 2$, but this would be false if one more person joined us. | 11. The probability that two people do not share the same birthday is $364 / 365$. For three people, the probability that no two of them share the same birthday is (364/365)$\cdot$(363/365), and so on. In the case of $n$ people, the same probability is
$$
(364 / 365) \cdot(363 / 365) \cdot \ldots \cdot(365-n+1) / 365
... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,989 |
## 12. Convoy of Ships
In 1943, in one American port, there were four escort ships, seven cargo ships, and three aircraft carriers. From these ships, a transport convoy was formed to deliver food and ammunition to Europe. At the head was an escort ship, followed by three cargo ships, then an aircraft carrier, and last... | 12. The number of ways to choose the first escort ship is 4. The selection of three cargo ships can be carried out
$$
C_{3}^{3}=\frac{7!}{3!4!}=35
$$
ways.
The number of ways to choose an aircraft carrier is 3. The last escort ship can be chosen in 3 ways (since only three unused ships remain).
Thus, there are $4 \... | 1260 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,990 |
## 14. Lacewing
One fine summer day, François was looking for Béatrice in Cabourg. Where could she be? Perhaps on the beach (one chance in two) or on the tennis court (one chance in four), or maybe in a café (also one chance in four). If Béatrice is on the beach, which is large and crowded, François has one chance in ... | 14. The probability of meeting is:
on the beach $(1 / 2) \cdot(1 / 2)=1 / 4$,
on the court $(1 / 4) \cdot(2 / 3)=1 / 6$,
in the cafe $(1 / 4) \cdot(1)=1 / 4$.
The total probability of meeting is $4 / 6=2 / 3$.
Therefore, the probability that François will not find Béatrice is $1-2 / 3=1 / 3$.
However, the probabi... | \frac{3}{4} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,992 |
## 15. The Emir and His Oil
My emirate encompasses a vast desert, in the middle of which I have built my palace. It also includes territorial waters along the coastline, which I value greatly. The territorial waters are quite extensive (covering an area that reaches a third of the desert's area), and part of my oil fi... | 15. Let \( S \) be the total area of the emirate. The area of its marine part is \( S / 4 \).
Let \( x \) be the fraction of the total area that
106
is occupied by the sea without oil-bea... | \frac{2}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,993 |
## 16. Playing Cards
Take any card at random from my deck, and then put it back. Do this three times. You have 19 chances out of 27 to draw at least one face card (king, queen, or jack) this way, because the proportion of face cards in the deck is...
Please complete the last sentence. | 16. Let $p$ be the unknown proportion of cards with figures in the deck.
Then the probability of not drawing a card with a figure in one draw from the deck is $1-p$. The probability of not drawing a card with a figure in three draws is $(1-p)^{3}$. In this case, the probability of drawing at least one card with a figu... | \frac{1}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,994 |
## 17. Sociological Survey
It is known that in a certain group of people, $42\%$ have never skied, $58\%$ have never flown on an airplane, and $29\%$ have both skied and flown on an airplane. In which case do we have a higher chance: meeting someone who has skied among those who have never flown on an airplane, or mee... | 17. If $42 \%$ of the people in the considered group have never skied, then $58 \%$ have skied. Since $29 \%$ of the people have skied and flown on an airplane, it follows that $29 \%$ have skied but have not flown on an airplane.
Therefore, the first probability we are interested in, which is the ratio of the number ... | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,995 |
## 18. Expected Lifespan
An ethnographer found that in a certain primitive tribe he studied, the distribution of the lifespan of the tribe members could be described as follows: $25 \%$ of them only lived up to 40 years, $50 \%$ died at 50 years, and $25 \%$ - at 60 years. He then randomly selected two individuals to ... | 18. If the one of the two members of the tribe who lives longer reaches 40 years, then obviously both "subjects" will live only to 40 years; the corresponding probability is
$$
25 \cdot 25 \% = 1 / 4 \cdot 1 / 4 = 1 / 16
$$
If the "long-liver" reaches 50 years, this means that either the first of the two will live on... | 53 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,996 |
## 19. Red and Green
On a large one-way street, two traffic lights are located one after another; each traffic light is designed so that the period when the green light is on constitutes two-thirds of the total operating time of the traffic light. A motorist noticed that when he, moving at a normal speed, passes the f... | 19. Let's consider what a sequence of traffic light signals might be. There are four possibilities: "both green"; "the first is green, the second is red"; "the first is red, the second is green"; "both are red". Let $p_{1}, p_{2}, p_{3}$, and $p_{4}$ be the probabilities of the respective four events. Since each traffi... | \frac{1}{2} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,997 |
## 20. Stewardesses
To serve a transatlantic flight of an airliner, three stewardesses are required, who are chosen by lot from 20 girls competing for these positions. Seven of them are blondes, the rest are brunettes. What is the probability that among the three selected stewardesses there will be at least one blonde... | 20. If among the three chosen girls there is at least one blonde and one brunette, this means that the cases where all stewardesses turned out to be brunettes
$$
\left(\text { probability of this: } \frac{13 \cdot 12 \cdot 11}{20 \cdot 19 \cdot 18}\right)
$$
or all - blondes
$$
\left(\text { probability of this: } \... | 0.718 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,998 |
## 21. Sunday Pie
On Sunday, as usual, Monsieur Dupont went for a walk. Meanwhile, his wife was trying in vain to decide whether he would remember to buy a pie and whether she would have to go to the bakery herself. In the end, she asked her son: “Do you know if your father intended to buy a pie?”
However, the child,... | 21. Let the disappointment associated with buying two pies instead of one be evaluated by Madame Dupont as having a value of $\alpha$. Then, putting herself in her son's position, she should evaluate the disappointment caused by not buying any pies at all as having a value of $2 \alpha$. If she goes to the pastry shop ... | \alpha/2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,999 |
## 22. We Need Boys
The ruler of a certain country, for purely military reasons, wanted there to be more boys than girls among his subjects. Therefore, he decreed that there should not be more than one girl in any family. As a result, in this country, among the children of each woman, the last and only the last was a ... | 22. Let $N$ be the number of mothers who no longer have children. How many of them have girls? One each, i.e., $N$. And how many have boys? Half of them have no boys at all, since the probability that the first child born is a girl is $1 / 2$. A quarter of the mothers have one boy, because $1 / 4 = 1 / 2 \cdot 1 / 2$ i... | \frac{1}{2} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,000 |
## 28. Passenger Ship
During a long voyage of one passenger ship, it was noticed that at each dock a quarter of the passenger composition is renewed, that among the passengers leaving the ship, only one in ten boarded at the previous dock, and finally, that the ship is always fully loaded.
Determine what fraction of ... | 28. Let the ship travel from the $n$-th pier to the ( $n+1$ )-th pier. It is known that:
1) a quarter of the total number of passengers appeared on the ship at the $n$-th pier;
2) among the passengers who left the ship at the $n$-th pier, every tenth boarded at the ( $n-1$ )-th pier;
3) a quarter of the total number of... | \frac{21}{40} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,006 |
## 29. Umbrella
One Saturday evening, two brothers, Jean and Pierre, exchanged their observations about the behavior of their neighbor, Madame Martin. They agreed that every Sunday she goes out of the house once (and only once), that in two out of three cases she takes an umbrella with her, and therefore even on good ... | 29. Since Madame Martin takes an umbrella with her one time out of two when the weather is good, and good weather occurs once every two days, the event "she takes an umbrella with her when the weather is good" happens once every four days $(1 / 2 \cdot 1 / 2=1 / 4)$. However, it is known that Madame Martin generally ta... | Pierre | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,007 |
## 31. Paul and Carolina
Paul and Carolina have arranged a date under the Triumphal Arch between 11 o'clock and noon. Each of them will arrive at some (random) moment during this time interval; according to their agreement, if the other does not appear within a quarter of an hour, the one who arrived first will not wa... | 31. Consider a Cartesian coordinate system on a plane. We will mark Paul's arrival time on the x-axis and Carolina's arrival time on the y-axis. Then each point on the plane corresponds to a specific arrival time of one and the other at the agreed place. Let the origin of coordinates on each axis correspond to 11 o'clo... | \frac{7}{16} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,009 |
## 32. Rain and Good Weather
In a certain region, rainy days account for a quarter of all days. In addition, it has been observed that if it rains on a certain day, it will rain the next day in two out of three cases. What is the probability that the weather will be good on a given day if there was no rain the day bef... | 32. Let's consider the (n-1)th and nth days. There are four possible scenarios, the probabilities of which we will calculate.
## 116
The probability that it rains on both of these two days: \(1 / 4 \cdot 2 / 3 = 1 / 6\).
The probability that it rains on the nth day but not on the (n-1)th day: \(1 / 4 - 1 / 6 = 1 / 1... | \frac{8}{9} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,010 |
## 33. In Search of a Job
To find a job after demobilization, soldier Maurice began sending letters to various companies where people of his specialty can be employed. He believes that each of his applications has a one in five chance of being accepted, and he stops sending letters as soon as he finds that he has at l... | 33. The probability that Maurice will remain unemployed after sending $n$ letters is $(1-1 / 5)^{n}=(4 / 5)^{n}$. Therefore, the probability of finding a job is $1-(4 / 5)^{n}$.
Maurice will stop writing when $n$ becomes such that this probability is not less than $3 / 4$, i.e., when the inequality
$$
(4 / 5)^{n} \le... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,011 |
## 34. Family Breakfast
Every Sunday, a married couple has breakfast with their mothers. Unfortunately, each spouse's relationship with their mother-in-law is quite strained: both know that there are two out of three chances of getting into an argument with their mother-in-law upon meeting. In the event of a conflict,... | 34. Two out of three times, the husband argues with his mother-in-law, and one out of three times (since $1 / 2 \cdot 2 / 3=1 / 3$) this leads to a family quarrel.
Two out of three times, the wife argues with her mother-in-law, and one out of three times this leads to a family quarrel.
The proportion of Sundays when ... | \frac{4}{9} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,012 |
## 35. Being Late for Work
Maurice goes to work either by his own car (and then due to traffic jams, he is late half of the time), or by subway (and then he is late only one time out of four). If on any given day Maurice arrives at work on time, he always uses the same mode of transport the next day as he did the prev... | 35. Let $p_{n}$ be the probability that Maurice goes to work by his car on the $n$-th day. Then the probability that he will take the metro on the $n$-th day is $1-p_{n}$.
Maurice will go by car on the $n$-th day if he went by car on the $(n-1)$-th day and did not get late (which happens once in two times) or if he we... | \frac{1}{3} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,013 |
## 36. Daily Newspapers
In a small town where many vacationers spend their holidays, $28 \%$ of the adult vacationers read "Mond", $25 \%$ read "Figaro", and $20 \%$ read "Orore". Additionally, $11 \%$ of the vacationers read both "Mond" and "Figaro", $3 \%$ read both "Mond" and "Orore", and $2 \%$ read both "Figaro" ... | 36. Let $x$ be the desired percentage. Denote by $V$ the set of all vacationers, by $M$ the set of "Mond" readers, by $F$ the set of "Figaro" readers, and finally by $A$ the set of "Aurore" readers. To determine the desired percentage, we will use the Venn diagram provided below.
![](https://cdn.mathpix.com/cropped/20... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,014 |
## 37. International Conference
Four French, two representatives from the Republic of Ivory Coast, three English and four Swedes gathered at a conference to discuss forestry issues. The conference participants from the same country initially sat next to each other on a bench with 13 seats during the opening ceremony, ... | 37. The number of ways in which 4 French people can sit next to each other on a bench is 4!; representatives from Ivory Coast can be arranged in 2! ways; the English - in 3! ways, and the Swedes - in 4! ways.
From this, we get that the number of ways in which all 13 people can sit on a bench, given that the order of t... | 165888 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,015 |
## 38. "Sensible" or "Sensitive"
On an English test, I had to translate a text into English. When translating one phrase, I wondered which word would be better to use: sensible (reasonable) or sensitive (perceptive). If I choose one of these words at random, I have a one in two chance of making a mistake. I could look... | 38. Let $\alpha$ denote the measure of inconvenience associated with an incorrect translation. Without cheating, I risk making a mistake one out of two times (the measure of inconvenience being $\alpha$), which results in an average inconvenience of $\alpha / 2$. By cheating, I risk getting caught (the measure of incon... | 24/50\cdot\alpha | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,016 |
## 39. Memories of the Trip
During the trip, I saw Romanesque cathedrals, triumphal arches, waterfalls, and medieval castles. I photographed half of these tourist attractions. I saw three times as many cathedrals as triumphal arches, and as many medieval castles as waterfalls. One quarter of the photographs I took dep... | 39. Let $x$ be the number of photos of castles, $y$ be the number of photos of triumphal arches, $z$ be the number of photos of waterfalls, and $t$ be the number of photos of cathedrals. Out of all the sights seen, the following were not photographed: 0 triumphal arches, $(3 y - t)$ cathedrals, $x$ castles, and $(2 x -... | \frac{1}{4} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,017 |
## 40. At Lunch
Ten married couples are having lunch together. Five people are chosen by lot to set the table. What is the probability that among these five people there will not be a single married couple? | 40. The number of ways to choose 5 people out of 20 is
$$
C_{20}^{5}=\frac{20!}{5!15!}
$$
The number of ways to choose such that no pair is represented in its entirety is
$$
C_{10}^{5} \cdot 2^{5}=25 \cdot \frac{10!}{5!5!}
$$
where $C_{10}^{5}$ is the number of ways to choose 5 pairs out of 10, and $2^{5}$ is the n... | \frac{168}{323} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,018 |
## 41. English Dictionary
In the class, there are 30 students. On each lesson, Miss Blackwick, the English teacher, assigns them to memorize a page of an English-French dictionary for the next class. At each lesson, she quizzes one student out of three, and among those questioned, there is always one out of ten who wa... | 41. Since one student out of three is called each time and among the called students, one out of ten was asked in the previous lesson, one student out of 30 is called for two consecutive days. Therefore, the risk of being punished by Miss Blackwick is one chance in 30. This means that a negligent student will be asked ... | \frac{11}{30} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,019 |
## 42. Twins
Anna and Brigitte, two little girls, are twins; they look like each other as much as two drops of water. At school, they sit at the same desk; one of them always sits on the left, and the other on the right. Each of them claims that she is Brigitte. The teacher knows from experience that, generally speaki... | 42. If Anna is sitting on the left and both girls call themselves Brigitte, this means that the girl sitting on the left[^24] is lying, while the girl on the right is telling the truth; the probability of this is
$$
1 / 5 \cdot 3 / 4=3 / 20
$$
But since both of them call themselves Brigitte, one of them is definitely... | \frac{3}{7} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,020 |
## 43. The Big Race
In the big car race of the year, only cars of three brands participated: "Jaguar", "Ford", and "Maserati." If the odds against Ford losing are 2 to 2, and the odds against Jaguar losing are 5 to 1, then what should be the odds against Maserati losing?
(Note: "losing" here means "not winning," i.e.... | 43. The probability of "Ford" losing is $2 /(2+2)=$ $=1 / 2$. Therefore, the probability of "Ford" winning is also $1 / 2$. The probability of "Jaguar" losing is estimated at $5 /(5+1)=5 / 6$. Therefore, the probability of "Jaguar" winning is $1-5 / 6=1 / 6$.
![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad... | 2to1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,021 |
## 1. Different Voting Methods
In one electoral district, 100,000 voters are casting their votes. Three candidates are running: Montoran, Ajuda-Pinto, and the Usselian vidam ${ }^{1}$.
The residents of the district are divided into four groups, each of which evaluates the proposed candidates differently.
The first g... | 1. If the election is held in one round, Montoran will receive 33,000 votes, Ajuda-Pinto - 30,000, and the Ussel warden - 37,000 votes. Therefore, the Ussel warden will be elected.
In a two-round election, the two candidates who receive the most votes in the first round will compete, i.e., the Ussel warden and Montora... | Ajuda-Pinto | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,022 |
## 3. At the Club
Five vacationers met in a club one day and started talking about where they live.
Amelie. I live in Acapulco, as does Benoit, and Pierre lives in Paris.
Benoit. I live in Brest, as does Charles. Pierre lives in Paris.
Pierre. I, like Amelie, do not live in France; Melanie lives in Madrid.
Melanie... | 3. Suppose Pierre does not live in Paris. Then the two other statements by Amélie and Benoît are true. This means that Benoît lives in Acapulco and Brest simultaneously, which is impossible. Therefore, our initial assumption is incorrect, i.e., Pierre lives in Paris.
So Pierre is lying when he says he does not live in... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,025 |
## 6. Black Cap, White Cap
"Come here," says Mrs. Pickwick to her three pupils. "Look, I have five caps: three white and two black. Now, while you stand with your eyes closed, I will put one cap on each of you. When you open your eyes again, each of you will be able to see the caps of the others, but of course, not yo... | 6. Let $E_{1}, E_{2}$, and $E_{3}$ denote the three students of Mrs. Pickwick.
The first one, $E_{1}$, reasoned as follows: My two companions have white hats. Mine is either white or black. If it is black, then $E_{2}$ should reason as follows: “$E_{1}$ has a black hat, and $E_{3}$ has a white one. Therefore, mine mus... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,029 |
## 9. Shooting Competition
Three friends competed in shooting at a target.
- "I bet," said the first before the competition began, "that at least one of you two will miss the target on the first try."
- "I bet," replied the second, "that if you manage to hit the target on your first shot, you will win your bet."
- "A... | 9. There are three different assumptions.
$C_{1}$: The first shooter hit the target with the first shot;
$C_{2}$: The second shooter hit the target with the first shot;
$C_{3}$: The third shooter hit the target with the first shot.
The second shooter said that if $C_{1}$ is true, then $C_{2}$ and $C_{3}$ cannot bot... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,032 |
## 10. Women's Superstitions
Every mother of seven children who knows English wears a bun, even if she wears glasses. Every woman who wears glasses has seven children or speaks English. No woman who does not have seven children wears glasses unless she has a bun. Every woman who has seven children and wears glasses kn... | 10. From the fifth sentence, we learn that if a woman is a mother of seven children, then she does not wear a wig. The first sentence states that every mother of seven children who knows English wears a wig: therefore, no mother of seven children speaks English. However, the fourth sentence asserts that every mother of... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,033 |
## 11. Medical Congress
At the international dermatology congress, English, German, and French doctors gathered. The number of French doctors was twice that of the Germans, who, in turn, were twice as many as the English. Two completely different methods for treating pityriasis Gilbert were proposed at the congress, a... | 11. Let $x$ be the number of English delegates, then the number of German delegates was $2 x$, and the number of French delegates was $4 x$. Let $y$ be the number of Germans who approved Simon's method.
; thus, this woman is Madame Duchemin. Therefore, the therapist's wife could only be Ma... | Dr.Deschamps | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,035 |
## 13. Motorcycle Racing
In one of the races, an equal number of Italian and Swedish riders participated. The former were mainly short brunettes, the latter - tall blondes. However, there were exceptions. For example, one fifth of the Swedish riders were short brunettes. On the other hand, it is known that in the tota... | 13. Since one fifth of the Swedish drivers were short brunettes, four fifths of them were tall blondes. Therefore, two out of every five drivers were simultaneously tall, blonde, and Swedish.
At the same time, it is known that among all the drivers, for every two tall blondes, there were three short brunettes. This me... | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,036 |
## 14. Cousins
Three brothers, Pierre, Paul, and Jacques, gathered all their children in a country house during their vacation. Here is what each of the children said:
I s a b e l l e. I am 3 years older than Jean. T h e r e s a. My father is named Jacques. I v. I am 2 years older than Isabelle.
M a r y. I prefer to... | 14. In total, there are eight children: five girls and three boys. Due to the last two statements, two brothers each have three children, and the third (Father François) has two. According to the second and sixth statements, Jacques has a daughter named Thérèse and two sons. Therefore, François is not Jacques' son. Thu... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,037 |
## 15. Should the Baby Be Fed?
One Saturday evening, a young married couple went to the movies, leaving their two children, aged three and six months, in the care of their grandmother. The grandmother, who took her duties very seriously, pondered whether the baby had been fed for the last time that day. She had no pri... | 15. Let $\alpha$ denote the "harm measure" for an infant receiving two portions of food instead of one; then the harm of not being fed at all is estimated to be $2 \alpha$.
In the second strategy, the grandmother has one chance in four of making a mistake, whether she feeds the infant or not. Therefore, she risks caus... | strategyispreferable | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,038 |
## 16. Pre-Election Discussion
Before the election to the legislative assembly, three brothers had a long discussion, which they concluded with the following rather complicated statements.
P i e r r e. If Jean votes for Dubois, I will vote for Dupont. But if he votes for Durand, I will vote for Dubois. On the other h... | 16. Since all three brothers voted for different candidates, there are six possibilities:
1) Pierre - for Dupont; Jacques - for Dubois, Jean - for Durand
2) Pierre - for Dupont, Jacques - for Durand, Jean - for Dubois
3) Pierre - for Durand, Jacques - for Dupont, Jean - for Dubois;
4) Pierre - for Durand, Jacques - for... | PierrevotedforDubois,Jacques-forDur,Jean-forDupont | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,039 |
## 17. Martians
Very complex observations have shown that the planet Mars is a desert, except for two large cities: Marsopolis (whose inhabitants never lie) and Mars-City (whose inhabitants never tell the truth). Martians move freely between the two cities,
 lies systematically, another (the accomplice) sometimes lies and sometimes tells the truth, and the last one (the innocent suspect) never lies. The interrogation began with questions about the profession of each o... | 18. Alfred said: "As for Bertrand, if you ask him, he will tell you that he is a painter." This is true. Therefore, Alfred is not a thief. If he is an accomplice, then the statements of Bertrand and Charles are completely opposite. But this is not the case, since both of them claim that Alfred is a piano tuner. Therefo... | Theaccompliceworksaninsuranceagent | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,041 |
## 19. Robbery
To escape after committing a robbery, the criminal must cross three rivers in succession. After each bridge, there was a fork in the road, and one could go right, straight, or left. The police managed to catch an accomplice and interrogated him to find out where the criminal had fled. Here is the answer... | 19. Since the criminal used each of the three directions only once, there are six different escape routes:
1) right, left, straight;
2) right, straight, left;
3) left, right, straight;
4) left, straight, right;
5) straight, right, left;
6) straight, left, right.
But if the first possibility is accepted, all three mes... | left,right,straight | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,042 |
## 20. Lying Journalists
Three journalists observed a person during breakfast and made the following notes. Ju l y. First, he drank whiskey, then he ate duck with oranges and dessert. Finally, he drank coffee.
Ja c k. He did not drink an aperitif. He ate a pie and a pear "Helen of Troy."
D j i m. First, he drank whi... | 20. Jules and Jim each made four statements. Jacques made three statements. There is no overlap in the information provided by Jules and Jacques. Jules and Jim have three statements in common, while Jacques and Jim have one in common. It is clear from this that Jacques is the one who constantly lies, Jules speaks only ... | strawberry\sherbet | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,043 |
## 22. Liars
Jean and Pierre sometimes lie. Jean says to Pierre: "When I don't lie, you don't lie either." Pierre replies: "And when I lie, you lie."
Is it possible that during this conversation, one of them was lying and the other was not? | 22. Let $J$ denote the statement "Jean is lying," and $P$ the statement "Pierre is lying." Jean, thus, told Pierre: "If $J$ is not true, then $P$ is not true." Pierre replied: "If $P$ is true, then $J$ is true." Each of these two statements means only that it is impossible for $J$ to be false and $P$ to be true at the ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,045 |
## 23. Paper Fish
Three teachers were chatting peacefully, sitting on a bench during the break. They were so engrossed in their conversation that they didn't notice the mischievous children attaching paper fish to their backs. When they stood up from the bench, all three began to laugh. Each of them happily thought th... | 23. Let $A, B$ and $C$ be three teachers. $A$ said to herself: «$B$ sees that $C$ is laughing. But $B$ does not know that she has a fish on her back. Therefore, if I did not have a fish on my back, $B$ should be surprised why $C$ is laughing and deduce from this that she also has a fish on her back. But this is not hap... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,046 |
## 28. High Jump
Three girls were practicing high jumping before a physical education test. The bar was set at a height of $1,20 \mathrm{M}$.
"I bet," says the first girl to the second, "that my jump will be successful if and only if yours is unsuccessful."
Suppose the second girl said the same to the third, and the... | 28. Suppose that none of the three girls lost the bet. What does this mean? If the first girl successfully jumped the high jump, then the second one failed (the first bet); it follows (the second bet) that the third girl also jumped successfully; but then (the third bet) the first one failed, which is impossible. There... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,051 |
## 29. Eiffel Tower and Flower Hat
If every American woman from Minnesota wears a flower hat while visiting the Eiffel Tower, and if every visitor to the Eiffel Tower who wears a flower hat is an American woman from Minnesota, can we conclude that all American women from Minnesota who wear a flower hat visit the Eiffe... | 29. Of course not: it is quite possible that a certain American woman from Minnesota wears a hat with flowers but does not visit the Eiffel Tower. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,052 |
## 30. New City
The New City is built in the form of 12 rectangular blocks of houses, separated by streets; there is a mailbox at each corner of a block and at each street intersection. The blocks touch each other with entire sides or have only one common corner.
We can observe that for each of these configurations, the number of mailboxes is 26. In fact, this number doe... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,053 |
## 31. Visit to Paris
Aunt Amelia visited Paris as a guest of her three little nephews and walked a lot with them around the city. After this, each of them told the following.
First nephew. We went up the Eiffel Tower, but we did not visit Montparnasse, however, we visited the Arc de Triomphe.
Second nephew. We went... | 31. Let $E$ denote the statement: they visited the Eiffel Tower; $A$: they visited the Arc de Triomphe; $M$: they visited Montparnasse, and finally, $P$: they visited the Ballroom. The nephews of Aunt Amelie claimed that:
the first: $E$ and $A$, but not $M$;
the second: $E$ and not $A, M$ and not $P$;
the third: not... | AuntAmelietookhernephewstotheEiffelTower,Montparnasse,theArcdeTriomphe,butdidnottakethemtotheBallroom | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,054 |
## 32. Xenon, Zephyr, and Enofa
Two friends at the races were speculating about the outcome of the next race. One of them bet on Xenon and Enofa, while the other bet on Zephyr.
- I bet,- said the second,- that if my horse finishes in the top three, then Xenon will also be in the top three.
- And I,- replied the first... | 32. Let $X, Y$ and $Z$ denote the following three statements:
$X$ - Xenon is in the top three;
$Y$ - Enofa is in the top three;
$Z$ - Zephyr is in the top three.
Since the second bet was not lost, if one of the statements, $X$ or $Y$, turned out to be true, then the first bet was lost. But if the first bet was lost... | Zephyr | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,055 |
## 33. Five Prisoners' Wives
Above the valley looms a formidable castle. Along one of its walls, towering like a high cliff, are five towers, each containing a prisoner. Two of them are inveterate liars, two others never lie, and the last one can either lie or tell the truth.
One fine day, as the guard was distributi... | 33. Let's consider several possible options sequentially.
The translation is provided as requested, maintaining the original text's line breaks and format. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,056 |
## 1. Highway in the Basque Country
The Biarritz - Bilbao highway is approximately 150 km long. In my "Mercedes," I drive it 25 minutes faster than my wife in her car. One day, we set off simultaneously: she from Bilbao, and I from Biarritz. When we met, I noticed that the difference in the distances we had left to tr... | 1. Let $x$ be the distance from the meeting point to Bilbao. Then the distance to Biarritz is $150-x$. The difference between these distances, $150-2x$, is 25, because according to the problem, the difference in the time intervals required for each of us to reach the final destination is exactly 25 minutes. But if
$$
... | 62.5, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,057 |
## 3. Before the Competition
A group of cyclists decided to train on a small road before the competition. All of them are riding at a speed of 35 km/h. Then one of the cyclists suddenly breaks away from the group and, moving at a speed of 45 km/h, covers 10 km, after which he turns back and, without slowing down, rejo... | 3. Let $x$ be the time that has passed since the cyclist broke away from the group until his return. Clearly, this cyclist has traveled a distance of $45 x$ km in this time, while the rest of the group has only traveled $35 x$ km. However, the sum of these two distances consists of the distance the cyclist traveled alo... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,059 |
## 4. From Bordeaux to Saint-Jean-de-Luz
Two cyclists set out simultaneously from Bordeaux to Saint-Jean-de-Luz (the distance between these cities is approximately 195 km). One of the cyclists, whose average speed is 4 km/h faster than the second cyclist, arrives at the destination 1 hour earlier. What is the speed of... | 4. Let $v$ (km/h) be the desired speed, and $t$ (h) be the time it takes for the cyclist moving at the higher speed $v$ to cover the entire distance.
The speed of the second cyclist, who is moving slower, is $v-4$. The time spent by him on the entire journey is $t+1$. Thus, we have
$$
195=v t=(v-4)(t+1),
$$
from whi... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,060 |
## 5. The Story of the Vagabonds
At the very moment when Pierrot left the "Commercial" bar, heading for the "Theatrical" bar, Jeanno was leaving the "Theatrical" bar, making his way to the "Commercial" bar. They were walking at a constant (but different) speed. When the vagabonds met, Pierrot proudly noted that he had... | 5. Let $x$ (m) - be the distance between the bars we are looking for; $d$ - the distance traveled by Pierrot by the time of the meeting; $V$ (m/s) - Pierrot's speed, and $v$ - Jeanno's speed before the fight.
The sum of the distances traveled by the vagrants is
$$
d+(d-200)=x, \quad \text { from which } \quad x=2 d-2... | 1000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,061 |
## 6. Invited Evening
To celebrate her husband being awarded the Legion of Honor, his wife, a well-known lady from the Paris region, decided to order cookies from the best pastry shop in Paris.
- Our evening, - she clarifies on the phone, - will start exactly at 6 PM. I want the fresh cookies to be delivered precisel... | 6. Let $v$ be the speed of interest (but currently unknown), $t$ the corresponding delivery time, and $d$ (km) the distance the car must travel. We obtain the following system of equations:
$$
\text { 1) } t=d / v, \text { 2) } t+1 / 4=d / 20, \text { 3) } t-1 / 4=d / 60 \text {. }
$$
From equations (2) and (3), it f... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,062 |
## 7. The Bicycle Rider in Love
A young cyclist set off to his fiancée, intending to give her a bouquet of flowers. Then he returned back. Since on the way to his fiancée he was holding the bouquet in one hand, his average speed was only 17 km/h, but on the return trip it increased to 23 km/h.
 d \approx 0.102 d$.
Therefore, the required average speed is approximat... | 19.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,063 |
## 8. The Emir and His Chauffeur
Emir Ben Sidi Mohammed always drives from his palace to the airport at the same speed along a magnificent highway that runs through the desert. If his chauffeur increased the average speed of the car by $20 \mathrm{km} /$ h, the Emir would save 2 minutes, and if the chauffeur drove at ... | 8. Let $I$ (km) be the unknown distance; $v$ (km/h) the speed of the emir's car, and finally, $t$ (h) the time it takes for the emir to travel from the palace to the airport. We obtain the following system of equations:
$$
v=I / t ; \quad v+20=I /(t-2 / 60) ; \quad v-20=I /(t+3 / 60) .
$$
Excluding the first equation... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,064 |
## 9. By car and by bicycle
Jean and Jules set off simultaneously from city $A$ to city $B$, one by car and the other by bicycle.
After some time, it turned out that if Jean had traveled three times more, he would have had to travel twice less than he currently has left, and that if Jules had traveled half as much, h... | 9. Let Jean have traveled a distance $x$, and let $x^{\prime}$ be the distance he has left to travel; similarly, let Jules have traveled a distance $y$, and let $y^{\prime}$ be the distance he has left to travel. In this case,
$$
x+x^{\prime}=3 x+x^{\prime} / 2, \quad y+y^{\prime}=y / 2+3 y^{\prime}
$$
Therefore,
$$... | Jean | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,065 |
## 10. Escalator in the Metro
Usually, I go up the escalator in the metro. I calculated that when I walk up the stairs of the moving upward escalator, I climb 20 steps myself, and the entire ascent takes me exactly $60 \mathrm{c}$. My wife walks up the stairs more slowly and only climbs 16 steps; therefore, the total ... | 10. Let $x$ be the unknown number of steps on the stationary escalator. When the escalator is working, it rises at a speed of $x-20$ steps in 60 seconds (this follows from calculations related to my ascent) or at a speed of $x-16$ steps in 72 seconds (which follows from how my wife ascends). Thus, the speed of the esca... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,066 |
## 11. Polar Expedition
A polar explorer set out from one point to another on a sled pulled by a team of five dogs. But after 24 hours, two of the dogs died, causing the explorer's speed to decrease to $3 / 5$ of the original speed, and he was delayed by two days.
- Oh! - exclaimed the polar explorer. - If the two do... | 11. If two dogs had run 120 km more, the polar explorer's delay would have been reduced by 24 hours (48-24). Therefore, if these dogs had run 240 km further, the polar explorer's delay would have been 48-24-24=0 hours, i.e., he would have arrived on time; hence, the distance from the place where the dogs fell to the de... | 320 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,067 |
## 12. Punctual Wife
Every Saturday between 15 and 16, I play tennis with my friend Philippe. My wife arrives to pick me up in the car precisely at 16:10. One day, Philippe fell ill. Unaware of this, I set out for tennis as usual. However, at 15:05, realizing that Philippe wasn't coming, I gathered my things and start... | 12. On the day when Philip fell ill, my wife drove the car for 10 minutes less than usual; that means she drove 5 minutes less in each direction. Therefore,
I met her at 4:05 PM, instead of 4:10 PM as usual. But then I walked for 60 minutes to cover the distance my wife drives in 5 minutes. Thus, I walk 12 times slowe... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,068 |
## 13. The Scout and the Drummer
On the occasion of the village festival, a procession was organized that stretched for 250 m; the scouts led the procession, and the musicians brought up the rear. Soon after the march began, the youngest scout remembered that he had not tied his neckerchief, which was left with his fr... | 13. Let $x$ be the unknown speed of the procession. When the scout ran to the end of the procession, he moved relative to the procession (considered stationary!) at a speed of $10+x$; when he returned, he moved relative to the procession at a speed of $10-x$. Since the length of the procession is 250 m, the total time ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,069 |
## 14. Janine and Monica
One day, Janine and Monica went swimming in a small river, starting from the same place, but Janine swam against the current, while Monica swam with the current. It turned out that Monica had forgotten to take off her large wooden beads, and they immediately slipped off her neck and floated do... | 14. The speeds of the girls relative to (stationary!) water are the same. But relative to the water in the river, the beads remain stationary - they move with the same speed as the water (the speed of the current). The swimmers cover the same distances relative to the water and after half an hour meet at the point wher... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,070 |
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