problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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321. Spheres and a cube. Once, during transportation, it was required to pack a sphere with a diameter of 30 cm into a cubic box with a side of 32 cm. To prevent the sphere from moving during transportation, 8 identical small spheres had to be placed in the corners of the box. What is the diameter of such a small spher... | 321. Since the center of the smaller sphere of radius $r$ lies on the diagonal and is $16 \sqrt{3}-15-r$ centimeters away from the corresponding corner of the box, to solve the problem we only need to equate this value to $r \sqrt{3}$ and solve the resulting equation for $r$. As a result, we will find the desired diame... | 63-31\sqrt{3}\approx9.308 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,754 |
322. Limit of a Sum. Calculate the value of the expression
$$
\lim _{n \rightarrow \infty} \sum_{k=0}^{n} \frac{k^{2}+3 k+1}{(k+2)!}
$$ | 322.
$$
\begin{gathered}
\sum_{k=0}^{n} \frac{k^{2}+3 k+1}{(k+2)!}=\sum_{k=0}^{n}\left(\frac{1}{k!}-\frac{1}{(k+2)!}\right)= \\
=\left(\frac{1}{0!}+\frac{1}{2!}\right)+\left(\frac{1}{1!}-\frac{1}{3!}\right)+\cdots+\left(\frac{1}{n!}-\frac{1}{(n+2)!}\right)= \\
=\frac{1}{0!}+\frac{1}{1!}-\frac{1}{(n+1)!}-\frac{1}{(n+2... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 41,755 |
323. Interesting numbers. If we sum the reciprocals of all divisors of the number 6, we get $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=2$. Find two other numbers for which the analogous sum is also equal to 2. | 323. Let $1, a, b, \ldots, n$ be the divisors of the number $n$, listed in ascending order; suppose that the equality
$$
\frac{1}{1}+\frac{1}{a}+\frac{1}{b}+\cdots+\frac{1}{n}=2
$$
holds. Multiplying both sides of this equation by $n$, we get
$$
\frac{n}{a}+\frac{n}{b}+\cdots+1=n
$$
where the left side represents t... | 28496 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,756 |
324. An interesting equation. Solve the equation
$$
x^{x+1}+x^{x}=1
$$ | 324. By taking the logarithm of both sides of the given equation, we obtain the equivalent equation
$$
f(x)=x \ln x+\ln (x+1)=0
$$
Since $\ln (1+x)>0$, then for $0<x<\frac{1}{e}$
$$
f^{\prime}(x)=\ln x+1+\frac{1}{1+x}=\ln x e+\frac{1}{1+x}>0
$$
and, therefore, $f(x)$ is monotonically increasing for $x>0$; hence, th... | x\approx0.43605 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,757 |
325. Tetrahedron - from progression. Twelve numbers $a_{i}$ form an arithmetic progression, so that $a_{k}+d=a_{k+1}$. Find the volume of the tetrahedron with vertices at points $\left(a_{1}^{2}, a_{2}^{2}, a_{3}^{2}\right),\left(a_{4}^{2}, a_{5}^{2}, a_{6}^{2}\right)$, $\left(a_{7}^{2}, a_{8}^{2}, a_{9}^{2}\right),\le... | 325. The squares of any three consecutive terms of an arithmetic progression with difference $d$ satisfy the equation
$$
a_{i}^{2}-2\left(a_{i}+d\right)^{2}+\left(a_{i}+2 d\right)^{2}=2 d^{2}
$$
In our case, it takes the form
$$
x-2 y+z=2 d^{2}
$$
Therefore, all four vertices of the tetrahedron are coplanar, and it... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,758 |
327. Diagonals of a cube. It is well known that the diagonals of a square are mutually perpendicular and bisect each other when they intersect. Are similar statements true for an $n$-dimensional cube? | 327. For $n \geqslant 3$, the number of diagonals of an $n$-dimensional cube is $2^{n-1}$, which is greater than $n$. Therefore, all diagonals cannot be mutually perpendicular. The inequality $2^{n-1}>n$ can be easily proven by induction, starting, for example, with $n=3$.
$[\Phi$. Pipp, M. M., 44, 114 (February 1971)... | 2^{n-1}>n | Geometry | proof | Yes | Yes | olympiads | false | 41,760 |
328. Try your hand at combinatorics. Prove that
$$
\sum_{i=0}^{n} \sum_{j=0}^{n}(-1)^{i+j} C_{n}^{i} C_{n}^{j} n^{i-j} j^{i}=n!
$$ | 328. The given double sum can be rewritten as
$$
\sum_{j=0}^{n}(-1)^{j} C_{n}^{j}(n-j)^{n}
$$
But this number represents the $n$-th finite difference, constructed for the function $f(x)=x^{n}$ and the points $x=0,1,2, \ldots, n$, and the value of such a difference, equal to $n!$, is well known.
[R. Gibbs, M. M., 44,... | n! | Combinatorics | proof | Yes | Yes | olympiads | false | 41,761 |
330. Simple pairs. Let $n$ be a natural number greater than 3. Prove that there exist two odd prime numbers $p_{1}$ and $p_{2}$, such that $2 n - p_{1}$ is divisible by $p_{2}$ (the case $2 n - p_{1} < 0$ is not excluded). | 330. I. A somewhat stronger statement can be proven. It turns out that for any natural number $n$, there exist two odd prime numbers $p_{1}$ and $p_{2}$ such that $n-p_{1}$ is divisible by $p_{2}$. Indeed, let us take an arbitrary natural number $n$. Next, choose an odd prime number $p_{1}$ such that $n-p_{1}$ does not... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,763 |
331. Limit Points. Let $A B C$ be an isosceles right triangle with a right angle at vertex $C$. Further, let $P_{0}=A$, $P_{1}=$ midpoint of $B C, P_{2 k}=$ midpoint of $A P_{2 k-1}$ and $P_{2 k+1}=$ midpoint of $B P_{2 k}$ for $k=1,2,3, \ldots$. Show that the sequences $\left\{P_{2 k}\right\}$ and $\left\{P_{2 k+1}\ri... | 331. I. The problem can be generalized to show that in an arbitrary triangle $ABC$, the corresponding limit points divide the side $BC$ into three equal parts. Indeed, the sequences $\left\{P_{2 k}\right\}$ and $\left\{P_{2 k+1}\right\}$ represent infinite subsets of the compact set $ABC$ and, consequently, have limit ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,764 |
332. Property of medians. Let a triangle $A B C$ be given. Further, let $A^{\prime}, B^{\prime}$ and $C^{\prime}$ lie strictly inside the segments $B C, C A$ and $A B$, respectively, and $A A^{\prime}, B B^{\prime}$ and $C C^{\prime}$ intersect at point $G$, with $\frac{A G}{G A^{\prime}}=\frac{B G}{G B^{\prime}}=$ $\f... | 332. We will consider all segments as directed: for example, $A G=-G A$ and so on. Further, we note that points $B, A^{\prime}$ and $C$ are collinear and each is located on one of the sides (or on the extension of a side) of triangle $A G B^{\prime}$. By Menelaus' theorem, we obtain the relation
$$
\frac{A A^{\prime}}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,765 |
333. Without computing the integrals. Show, without computing any of the following integrals, that
$$
2 \int_{-1}^{1} \sqrt{1-x^{2}} d x=\int_{-1}^{1} \frac{1}{\sqrt{1-x^{2}}} d x
$$ | 333. The integral standing on the left side of the given equality represents the area of the region enclosed between the semicircle $y=\sqrt{1-x^{2}}, -1 \leqslant x \leqslant 1$ and the $x$-axis, which is obviously $\pi$. The integral standing on the right side coincides with the length of the same semicircle, that is... | proof | Calculus | proof | Yes | Yes | olympiads | false | 41,766 |
334. Evaluate the error. Can we say that $e^{2737}$ is approximately equal to
$$
\frac{1000!}{70!270!300!220!140!} ?
$$
Euclid Paracelsus Bombast Umbugio wants to use the second number as an approximate value for the first. | 334. $e^{2737} \gg e^{1610} \approx 5^{1000}=(1+1+1+1+1)^{1000}=$
$$
=\sum_{\left(\sum n_{i}=1000\right)} \frac{1000!}{n_{1}!n_{2}!n_{3}!n_{4}!n_{5}!} \gg \frac{1000!}{70!270!300!220!140!}
$$
[Y. Uermer, M. M., 44, 47 (February 1970).] | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,767 |
335. Prime numbers and geometric progression. Can the square roots of three different prime numbers be members of the same geometric progression? | 335. Suppose that the square roots of three different prime numbers $p_{1}, p_{2}$, $p_{3}$ are terms of some geometric progression. Then the equalities $a r^{n_{1}}=\sqrt{p_{1}}$, $a r^{n_{2}}=\sqrt{p_{2}}, a r^{n_{3}}=\sqrt{p_{3}}\left(n_{1}, n_{2}, n_{3}\right.$ - different integers, and we can assume that $n_{1}>n_... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,768 |
336. Prime divisors. Show that the number $(p_{1} \cdot p_{2} \ldots p_{n} + 1)^{2^{k}} - 1$, where $p_{1}, p_{2}, \ldots, p_{n}$ are the first $n$ odd prime numbers, has at least $n+k$ distinct prime divisors. | 336. Let $N=\left(p_{1} p_{2} p_{3} \ldots p_{n}+1\right)^{2^{k}}-1$. Then, since
$$
\begin{aligned}
& a^{2^{k}}-1=(a-1)(a+1)\left(a^{2}+1\right)\left(a^{2^{2}}+1\right) \ldots\left(a^{2^{k-1}}+1\right) \\
& N=\left(p_{1} p_{2} \ldots p_{n}\right)\left[\left(p_{1} p_{2} \ldots p_{n}+1\right)+1\right] \times \\
& \time... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,769 |
337. The number of solutions to the cryptarithm.
| SEND |
| ---: |
| MORE |
| MONEY |
Show that the cryptarithm in a base $b$ number system has exactly $C_{b-8}^{2}$ solutions. | 337. We immediately obtain that $M=1, O=0, S=b-1$, $E+1=N$, and $N+R+1=E+b$ or $N+R=E+b$. Substituting $N$ here, we find that $R=b-2$ or $R=b-1$. But $b-1$ is the value of $S$; therefore, $R=b-2$.
The cryptarithm now takes the form
$$
\begin{array}{ccccc}
& b-1 & E & E+1 & D \\
& 1 & 0 & b-2 & E \\
\hline 1 & 0^{*} ... | C_{b-8}^{2} | Combinatorics | proof | Yes | Yes | olympiads | false | 41,770 |
338. Cubic Equation. Show that all integer solutions of the equation
$$
x^{3}+y^{3}+z^{3}=u^{3}
$$
where $x, y, z, u$ form an arithmetic progression, are multiples of $x=3$, $y=4, z=5, u=6$. | 338. Let $x=a-d, y=a, z=a+d$ and $u=a+2 d$.
Then $(a-d)^{3}+a^{3}+(a+d)^{3}=(a+2 d)^{3}$. Expanding the brackets and combining like terms, we get the equation
$$
2 a^{3}-6 a^{2} d-6 a d^{2}-8 d^{3}=0
$$
or after dividing by 2
$$
a^{3}-3 a^{2} d-3 a d^{2}-4 d^{3}=0
$$
Since $a$ and $d$ are integers, their ratio $\f... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,771 |
339. Position of the Incenter. Let $I$, $O$, and $H$ be the centers of the inscribed, circumscribed circles, and the orthocenter of triangle $A B C$, respectively, where $C > B > A$. Show that the point $I$ must lie inside the triangle $B O H$. | 339. Let $R$ be the radius of the circumscribed circle, and $r$ be the radius of the inscribed circle; $x=\operatorname{tg} \frac{A}{2}, y=\operatorname{tg} \frac{B}{2}, z=\operatorname{tg} \frac{C}{2}; \triangle XYZ$ is the area of triangle $XYZ$.
Regardless of whether angle $C$ is obtuse or acute, angle $OBH = C - A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,772 |
340. Irrational number. Show that the number $\log _{e} 2$ is irrational. | 340. Suppose that $\log _{e} 2$ is rational. Clearly, $\log _{e} 2 \neq$ $\neq 0$; hence, $\log _{e} 2=\frac{p}{q}$, where $p$ and $q$ are integers, $p>0$, and $q \neq 0$. Therefore, $e^{\frac{p}{q}}=2$, or $e^{p}=2^{q}$. This means that $e$ satisfies the equation $x^{p}-2^{q}=0$, which is impossible since the number $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,773 |
341. Greatest Common Divisor. Let $a, m, n$ be positive integers and $n$ be odd. Prove that the greatest common divisor of the numbers $a^{n}-1$ and $a^{m}+1$ does not exceed 2. | 341. Let $d$ be the greatest common divisor of the numbers $a^{n}-1$ and $a^{m}-1$. Then for some integers $k$ and $r$, the equalities $a^{n}=k d+1, a^{m}=r d-1$ hold. Therefore,
$$
a^{m n}=\left(a^{n}\right)^{m}=(k d+1)^{m}=t d+1
$$
for some integer $t$, and
$$
a^{m n}=\left(a^{m}\right)^{n}=(r d-1)^{n}=u d-1
$$
f... | =1or=2 | Number Theory | proof | Yes | Yes | olympiads | false | 41,774 |
342. Simple calculation. Let $\left(x+\frac{1}{x}\right)^{2}=3$; determine what $x^{3}+\frac{1}{x^{3}}$ is equal to. | 342.
$$
\begin{aligned}
x^{3} & +\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)^{3}-3\left(x+\frac{1}{x}\right)= \\
& =\left(x+\frac{1}{x}\right)\left[\left(x+\frac{1}{x}\right)^{2}-3\right]=0
\end{aligned}
$$
[M. Demos, M. M., 45, 102 (February 1972).] | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,775 |
344. May Beetles. In a box sit May beetles. Among them, there are $n$ males and $m$ females. Mr. Entomologist takes beetles out of the box in random order, one by one, without replacement, until he has $k$ males, $1 \leqslant k \leqslant n$. Let $x_{k}$ be the total number of beetles he has taken out of the box. Find t... | 344. If we arrange all $n+m$ beetles in a single row, making sure they don't scatter, the positions for the $n$ males can be chosen in $C_{m+n}^{n}$ ways. Among the first $x-1$ positions, we can place $k-1$ males in $C_{x-1}^{k-1}$ ways, and the remaining $n-k$ males can be placed among the remaining $m+n-x$ positions ... | \frac{k(+n+1)}{n+1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,777 |
345. Gnomonic Magic Squares. A $3 \times 3$ square is called gnomonic-magic if the sums of the numbers forming the $2 \times 2$ squares that remain after removing one of the four "corners" (gnomons) from the original square are equal to each other. Show that in a gnomonic-magic square of the third order, the sums of th... | 345. Let the gnomonic-magic square have the form
$$
\begin{array}{lll}
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{array}
$$
In accordance with the definition,
$$
\begin{gathered}
a_{2}+b_{2}+a_{3}+b_{3}=a_{1}+b_{1}+a_{2}+b_{2}= \\
=b_{1}+c_{1}+b_{2}+c_{2}=b_{2}+c_{2}+b_{3}+c_{3}
\en... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 41,778 |
346. Is the inequality correct? Let $x_{i}$ be the distance from some internal point of the triangle $A_{1} A_{2} A_{3}$ to the side opposite vertex $A_{i}, i=1,2,3$, and let $r$ be the radius of the circle inscribed in this triangle. Prove or disprove the following inequality:
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac... | 346. We will show that the inequality
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}} \geqslant \frac{3}{r}
$$
is generally not true. ${ }^{\star}$ Indeed, let $Q$ be an internal point of a given triangle such that the distances $y_{1}, y_{2}, y_{3}$ from it to the sides of the triangle satisfy the relations
$$
\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 41,779 |
348. Inequality with integrals. Let a continuous function $f(x)$ be defined for $0 \leqslant x \leqslant 1$, such that
$$
0<A \leqslant f(x) \leqslant B
$$
for all $x$ in the given domain. Prove that in this case
$$
A B \int_{0}^{1} \frac{d x}{f(x)} \leqslant A+B-\int_{0}^{1} f(x) d x
$$ | 348. Note that for $0 \leqslant x \leqslant 1$
$$
\frac{(f-A)(f-B)}{f} \leqslant 0
$$
Integrating this inequality from 0 to 1, we obtain the required relation.
[E. Schmeichel, M. M., 45, 229 (March 1972).] | proof | Inequalities | proof | Yes | Yes | olympiads | false | 41,781 |
349. Equation with no solutions. Prove that the Diophantine equation $5^{x}+2=17^{y}$ has no solutions. | 349. From the equality
$$
(3 \cdot 2-1)^{x}+2=(3 \cdot 6-1)^{y}
$$
we find that
$$
(-1)^{x}+2=(-1)^{y}+3 k
$$
from which it is clear that $y$ must be even. On the other hand, from the equality $5^{x}+2=(5 \cdot 3+2)^{y}$ it is seen that $2^{y}-2$ is divisible by 5, which means that $y-1$ is divisible by 4; hence, $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,782 |
351. An "Invisible" Polygon. Find a region bounded by a simple closed polygon such that for any two sides of the given polygon, there exists an interior point from which these sides are visible. However, at the same time, no interior point should allow all sides to be seen simultaneously. | 351. I. Let us divide each side of an isosceles triangle \( T \), whose height is 4, into 4 equal parts. We will cyclically number all vertices and division points from \( A_{1} \) to \( A_{12} \), starting from one of the vertices of triangle \( T \). We will construct three equilateral triangles on the segments \( A_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,784 |
352. Concerning graphics. For what value of $a$ does the graph of the function $a^{x}$ touch the graph of the function $\log _{a} x ?$ | 352. The function $y=\log _{a} x(a>0, a \neq 1, x>0)$ is the inverse of the function $y=a^{x}$. Due to the symmetry of their graphs with respect to the line $y=x$, we conclude that in the case of tangency, both graphs either touch the line $y=x$ or are perpendicular to it. Therefore, at the point of tangency, the follo... | e^{\frac{1}{e}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,785 |
354. Divisibility by 120. Suppose that $a-1$ and $a+1-$ are prime numbers (such a pair is called twin primes), greater than 10. Prove that $a^{3}-4 a$ is divisible by 120. | 354. We will prove a stronger result. For this, note that the number $(a-2)(a-1) a(a+1)(a+2)$ represents the product of five consecutive integers, so one of them is divisible by 3 and one by 5. If $a-1$ and $a+1$ are prime numbers, then $a-2$, $a$, $a+2$ are consecutive even numbers, so at least one of them is divisibl... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,787 |
355. Diagonals of a Polygon. Find the greatest possible number of intersections of the diagonals of a plane convex $n$-gon. | 355. Any four vertices completely determine some quadrilateral. The only point of intersection of the diagonals of such a quadrilateral lies inside the original polygon due to convexity. This point belongs to exactly two diagonals of our $n$-gon. Therefore, the number of diagonal intersections in the original $n$-gon w... | \frac{n(n-1)(n-2)(n-3)}{24} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,788 |
356. A simple cryptarithm. Each letter in the following cryptarithm stands for a definite decimal digit:
$$
3(B I D F O R)=4(F O R B I D)
$$
Restore the original record. | 356. We can write the following relationship:
$$
3[1000(B I D)+(F O R)]=4[1000(F O R)+(B I D)]
$$
Expanding the square brackets, we get
$$
3000(B I D)+3(F O R)=4000(F O R)+4(B I D)
$$
By combining like terms, we arrive at the relationship $2996(B I D)=$ $=3997(F O R)$. Dividing both sides of this equation by 7, we ... | 3(571428)=4(428571) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,789 |
357. A stick inside a hemisphere. A stick of length $d$ is placed in a hemispherical basin with a diameter of $d$. Neglecting the thickness of the stick and assuming that friction is absent, determine the angle that the stick will form with the diameter in the equilibrium position. | 357. I. If we are inventive enough and draw the additional lines shown in Fig. 1, we immediately obtain the following relationships:
$$
\begin{gathered}
y=u \sin \theta \\
v=2 a \sin \theta \\
v^{2}+(u+a)^{2}=4 a^{2}
\end{gathered}
$$
By eliminating \( u \) and \( v \) from these equations, we arrive at the equation
... | \theta\approx3223' | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,790 |
358. Construction with a compass. Let two points $A$ and $B$ be given, which serve as vertices of a square. Find the other two vertices of this square using only one compass. | 358. Let us draw a circle, which we will denote as $A(A B)$, with center at point $A$ and radius $A B$. Starting from point $B$, using the radius $A B$, mark points $C, D, E$ on the circle such that $B C=C D=D E=A B$. Then $B D=\sqrt{3}(A B)$. Draw arcs $B(B D)$ and $E(B D)$, intersecting at point $F$. In this case, $A... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,791 |
359. Arithmetic mean and harmonic mean. In a triangle with sides $a, b, c$, a line connecting the centroid with the incenter is perpendicular to the angle bisector of the angle opposite side $c$. Show that the arithmetic mean of the numbers $a, b, c$ is equal to the harmonic mean of the numbers $a$ and $b$. | 359. Let $G$ be the centroid, $I$ the incenter, $S$ the area, $h_{a}$ and $h_{b}$ the altitudes dropped to sides $a$ and $b$ in triangle $ABC$. Denote by $P$ and $Q$ the points of intersection of line $GI$ with sides $BC$ and $CA$, respectively. Since the sum of the areas of triangles $GPC$ and $GQC$ coincides with the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,792 |
360. Polygon on a chessboard. Let the chessboard consist of squares with a side length of 4. Furthermore, let a regular $4 n$-gon of "radius" $1^{*}$ be thrown onto this board. Determine the probability that this polygon will intersect the side of any square. | 360. We can assume without loss of generality that the center of the polygon falls inside a certain square at a point which is at a distance $l$ from the nearest side of this square. Let us drop a perpendicular from the center of our polygon to this side and consider the angles formed by this perpendicular and the "rad... | \frac{4n}{\pi}\sin\frac{\pi}{4n}-\frac{n}{4\pi}\sin\frac{\pi}{2n}-\frac{1}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,793 |
361. Integers similar to their divisors. For which positive integers of the form $p^{n}-1$ ( $p$ - prime) do all their divisors have the same form? | 361. Note that
1) if $2^{a}$ divides $p^{n}-1$, then $a \leqslant 4$, since $2^{5}+1$ is not a power of a prime number;
2) if $q^{b}$ divides $p^{n}-1$ ($q$ - an odd prime), then $q=2^{k}-1$ and, moreover, $b=1$, since $q^{2}+1$ is even and has at least one non-trivial divisor (and thus cannot be a power of a prime num... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,794 |
362. True or false? Prove or disprove the following statement: a complex number $z$ satisfies the inequality $|z|-\Re z \leqslant \frac{1}{2}$ if and only if $z$ coincides with the product of numbers $a c$, such that $|\bar{c}-a| \leqslant 1$. | 362. The statement is valid and is a consequence of the identity
$$
|a c|-\Re a c=\frac{1}{2}|\bar{c}-a|^{2}-\frac{1}{2}(|c|-|a|)^{2}
$$
The sufficiency of this condition is obvious. To verify its necessity, choose $a$ and $c$ such that $|a|=|c|=|z|^{\frac{1}{2}}$; in this case, the second term on the right-hand side... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 41,795 |
363. Trigonometric identity. Show that $\cos ^{n} \frac{\pi}{n}-\cos ^{n} \frac{2 \pi}{n}+\cdots+(-1)^{n-1} \cos ^{n} \frac{(n-1) \pi}{n}=\frac{n}{2^{n-1}}$.[^20] | 363. Let $\omega=\cos \frac{\pi}{n}+i \sin \frac{\pi}{n}$. By De Moivre's formula, $\omega^{ \pm j}=$ $=\cos \frac{\pi j}{n} \pm i \sin \frac{\pi j}{n}$, and therefore $\cos \frac{\pi j}{n}=\frac{1}{2}\left(\omega^{j}+\omega^{-j}\right)$ and $-1=\omega^{n}$. Consequently,
$$
\begin{gathered}
\sum_{j=0}^{n-1}(-1)^{j} \... | \frac{n}{2^{n-1}} | Algebra | proof | Yes | Yes | olympiads | false | 41,796 |
365. A remarkable property of the triplet. Show that if $m$ and $n$ are positive integers, then the smallest of the quantities $\sqrt[n]{m}$ and $\sqrt[m]{n}$ cannot exceed $\sqrt[3]{3}$. | 365. I. Let $f(x)=x^{\frac{1}{x}}$. It is easy to show that $f(x) \rightarrow 0$ as $x \rightarrow 0, f(x) \rightarrow 1$ as $x \rightarrow \infty, f$ increases when $x$ changes from 0 to $e$, and $f$ decreases when $x$ changes from $e$ to $\infty$.* Therefore, the maximum value, $C$, of the quantities $f(k)$, where $k... | 3^{\frac{1}{3}} | Inequalities | proof | Yes | Yes | olympiads | false | 41,798 |
368. Double Palindrome. Could you find a number that, when written in the decimal system, has the property that if its digits are written in reverse order, the resulting number is twice the given one? | 368. Such a number does not exist. Indeed, let the desired number have the form $\overline{a \ldots b}$; then
$$
\overline{b \ldots a}=2(\overline{a \ldots b})
$$
Further, $a$ can be equal to $1,2,3$ or 4, and the corresponding values of $b$ can be $(2,3),(4,5),(6,7)$ or $(8,9)$. By comparing the last digits of the o... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,801 |
369. Twin Catheti. Prove that the lengths of the two catheti of a right-angled triangle with integer sides cannot be expressed by twin prime numbers. | 369. Suppose that $p$ and $p+2$ are two twin primes such that
$$
p^{2}+(p+2)^{2}=k^{2}
$$
where $k$ is an integer. Then
$$
2 p^{2}+4 p+4=k^{2}
$$
From this, it follows that $k^{2}$, and hence $k$, are even. Let $k=2 n$, then our relation can be rewritten as
$$
2 p^{2}+4 p+4=4 n^{2}
$$
or
$$
p^{2}+2 p+2=2 n^{2}
$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,802 |
370. "Industrial Espionage". One day, the editor of a mathematical journal said to his assistant:
- Here is a cryptarithm where a two-digit number is multiplied by itself. You see, here the given number is written only twice and the final product; the intermediate calculations are omitted.[^21]
The assistant replied:
... | 370. Since the assistant, having learned the parity of the original number, managed to solve the problem, before asking his question, he must have reduced the problem to possible cryptarithms that allow either $n$ odd solutions and one even, or $n$ even and one odd solution, where $n>1$. There are three cryptarithms th... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,803 |
372. Division of a Circle. Four lines intersect at a point $O$ on a plane, and all eight angles are $45^{\circ}$. A circle is placed over this configuration such that point $O$ is inside it. The resulting sectors are shaded alternately. Show that the shaded sectors cover exactly half of the circle. | 372. Let the center of the circle be denoted by $P$, and let $A C$ and $B D$ be two mutually perpendicular chords passing through the point $O$. Suppose further that $O P=a$ and $\theta$ is the angle between $B D$ and $O P$. Then the following relations hold:
\[
\begin{aligned}
\frac{A C}{2} & =\sqrt{r^{2}-a^{2} \cos ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,805 |
373. Cutting a Cube. In a mathematical journal, the following question and answer appeared.
Question: "A carpenter wants to cut a wooden cube with a side of 3 inches into 27 cubes with a side of 1 inch. He can easily do this by making 6 cuts while keeping the pieces together so they don't fall apart. What is the minim... | 373. After the first cut, the cube splits into 2 parts. The larger of these (consisting of 17 one-inch cubes) contains one central cube, for four faces of which another cut is required. After the last of these is made, there will remain at least two one-inch cubes undivided - regardless of any rearrangement of the piec... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,806 |
374. Minimum of a function. Find the minimum value of the function
$$
f(x)=(x+a+b)(x+a-b)(x-a+b)(x-a-b)
$$
without using differential calculus.[^22] | 374. The function $f(x)$ does not change if we replace $a$ with $-a$, $b$ with $-b$, or $x$ with $-x$. Therefore, without loss of generality, we can assume that $x, a$, and $b$ are non-negative and, moreover, that $a \geqslant b$. It is clear that the minimum value of $f(x)$ should be sought where it is negative, that ... | -4a^2b^2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,807 |
375. Polynomial and Integral. Let us define the quantity $T_{k}(x)$ as follows:
$$
T_{k}(x)=\prod_{\substack{i=1 \\ i \neq k}}^{n}(x-i)
$$
Let $P(x)$ be the polynomial of minimal degree that has the property that
$$
P(k)=T_{k}(k), \quad k=1,2, \ldots, n
$$
Show that if we take two integers $s$ and $t$, such that $1... | 375. It is easy to check that
$$
P(x)=\sum_{k=1}^{n} T_{k}(x)=\left[\prod_{k=1}^{n}(x-k)\right]^{\prime}
$$
Therefore,
$$
\int_{s}^{t} P(x) d x=\left.\prod_{i=1}^{n}(x-i)\right|_{s} ^{t}
$$
from which the required equality follows.
[E. Just, M. M., 41, 102 (February 1968).] | proof | Calculus | proof | Yes | Yes | olympiads | false | 41,808 |
376. Amazing Triangle. Find a right triangle whose sides are expressed in integers, and all 9 digits involved in the notation of the sides are different. | 376. If we try to find the triangle we need among those with integer sides that are mutually prime, we will be unsuccessful. However, triangles that satisfy the condition of the problem do exist. For example, we can
take $182(3,4,5)=546,728,910$. Another solution is: $178(3,4,5)=534,712,890$.
[M. Goldberg, M. M., 41, ... | 546,728,910534,712,890 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,809 |
377. Almost a golden rectangle. Consider a rectangle $R$ with sides $x$ and $y, x<y$. Suppose we remove a square with side $x$ from $R$ and obtain a rectangle $R^{\prime}$. It is known that if $R^{\prime}$ is similar to $R$, then $R$ is a "golden rectangle" and $\frac{y}{x}=\frac{1+\sqrt{5}}{2}$. However, suppose that ... | 377. If $x<y-x$ and $R$ is similar to $R^{\prime \prime}$, then
or
$$
\frac{y}{x}=\frac{x}{y-2 x}
$$
$$
\frac{y}{x}=\frac{1+\sqrt{2}}{1}
$$
If $y-x<x$, and $R$ is similar to $R^{\prime \prime}$ and the longer side of $R$ is parallel to the longer side of $R^{\prime \prime}$, then
$$
\frac{y}{x}=\frac{y-x}{2 x-y}
$... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,810 | |
378. The Pancake Problem. A cook is baking pancakes on a round skillet, the diameter of which is 26 units. She places three round pieces of dough of different sizes such that their centers lie on the same straight line and together they cover the entire diameter of the skillet, but - only half of its area. Find the dia... | 378. Since the areas are proportional to the squares of the corresponding diameters, it is sufficient to find 3 different integers, the sum of which is 26 and the sum of whose squares is $676: 2=338$.
Let $x, y, z$ be the diameters of the pancakes. Then
$$
\begin{gathered}
x+y+z=26 \\
x^{2}+y^{2}+z^{2}=338
\end{gathe... | 1,9,16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,811 |
379. Orbit of a spacecraft. Construct the orbit of a spacecraft from any point of which the Earth and the Moon would appear the same in size. | 379. Let the radius of the Earth $E$ be $R$, and the radius of the Moon $M$ be $r$ (see figure). Since from any point on the orbit, the Earth and the Moon
should be visible at the same ang... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,812 |
383. Rational numbers and a polygon. Show that if the number of sides of a convex polygon circumscribed about a certain circle is odd, and the length of each of its sides is expressed by a rational number, then the length of each of the segments into which the sides are divided by the points of tangency is also express... | 383. Let $a_{1}, a_{2}, \ldots, a_{n}$ be the consecutive sides of our polygon, and let $x$ be the segment of $a_{1}$ located between the first vertex and the point of tangency. We express each subsequent segment, located between a vertex and the point of tangency, in terms of $x$. The last segment of side $a_{n}$ is e... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,816 |
384. Integer value. Show that for any positive integer $n$ the number $\frac{(3 n)!}{6^{n} n!}$ is also an integer. | 384. $N=3^{n} n!=3^{n} \cdot 1 \cdot 2 \cdot 3 \cdots n=3 \cdot 6 \cdot 9 \cdots(3 n)$. In the product $(3 n)!=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots$, the factors divisible by three are separated by pairs of adjacent integers, one of which is even. Therefore, the number $(3n)!$, in addition to $N$, contains ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,817 |
386. Which area is larger? For two triangles, the sides are respectively equal to $\sqrt{a^{2}+b^{2}}, \sqrt{b^{2}+c^{2}}, \sqrt{c^{2}+a^{2}}$ and $\sqrt{p^{2}+q^{2}}$, $\sqrt{q^{2}+r^{2}}, \sqrt{r^{2}+p^{2}}$. Which one has a larger area, given that, in addition, $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=p^{2} q^{2}+q^{2} ... | 386. Consider a tetrahedron in which all the dihedral angles at one of the vertices are right angles, and the edges emanating from this vertex are equal to \(a\), \(b\), and \(c\) respectively. The other three edges of the tetrahedron are equal to \(\sqrt{a^{2}+b^{2}}\), \(\sqrt{b^{2}+c^{2}}\), and \(\sqrt{c^{2}+a^{2}}... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,819 |
388. Three Napkins. One lady made 3 round napkins with radii of 2, 3, and 10 inches, respectively. She placed them on a round table so that each napkin touched the other two and the edge of the table. What is the radius of the table top? | 388. If the centers of napkins with radii of 2, 3, and 10 inches are located at points \( C, A \), and \( B \) respectively, then these points form the vertices of a right triangle with sides 5, 12, and 13 inches. Completing the figure \( ABC \) to a rectangle, we denote its fourth vertex as \( O \). From point \( O \)... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,821 |
389. Descartes' Student. A funny story has come down to us about how Descartes proposed the famous Apollonius problem to one of his titled students: to construct a circle tangent to three given circles. In order to express his hidden disdain for the student's pretensions to be capable, he did not warn the poor girl tha... | 389. Suppose the student did not share her teacher's antipathy to negative radii. Then she could reduce the three quadratic equations
$$
\begin{aligned}
& \left(h-h_{1}\right)^{2}+\left(k-k_{1}\right)^{2}=\left(r+r_{1}\right)^{2} \\
& \left(h-h_{2}\right)^{2}+\left(k-k_{2}\right)^{2}=\left(r+r_{2}\right)^{2} \\
& \lef... | notfound | Geometry | proof | Yes | Yes | olympiads | false | 41,822 |
390. Permutation of Digits. Let's take some number, written in an arbitrary numeral system, and rearrange its digits in any order. Prove that the difference between these two numbers is divisible by the number that is one less than the base of the given numeral system. | 390. If the base of the numeral system is $b$, then the given number can be written as
$$
\sum_{i=0}^{n} a_{i} b^{n-i}
$$
Denoting the digit that will stand in the place of $a_{i}$ after the permutation by $a_{i_{p}}$. Then the difference between our two numbers can be written as
$$
\begin{aligned}
& \sum_{i=0}^{n} ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,823 |
391. Fermat's Principle. Apply Fermat's principle to a circular mirror. In other words, if two points $A$ and $B$ are given inside a certain circle, then it is required to find a point $P$ on the circumference such that the value $A P+P B$ takes an extremal value. | 391. This is a well-known problem, the solution of which can be easily obtained by using level lines. For greater generality, we will take an arbitrary closed convex curve, the curvature of which changes continuously.
Consider the family of curves defined by the condition
$$
A P+P B=k \quad \text { (constant) }
$$
T... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,824 |
392. Central Symmetry. Prove that if any line passing through a fixed interior point $O$ of a quadrilateral $A B C D$ divides its perimeter into 2 equal parts, then this quadrilateral is a parallelogram. | 392. I. Draw a line $E F$ through point $O$ that does not pass through any of the vertices of the given quadrilateral. Take two points on the side containing $E$, equidistant from $E$, and draw two more lines through these points and point $O$. These two lines will intersect the side containing point $F$ at two points ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,825 |
Ex. 1. Find the area of a right triangle if the length of the shortest median is 13, and one of the legs is 10. | Ex. 1. 120.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,827 |
Ex. 13. Points $M$ and $N$ are the midpoints of sides $A B$ and $B C$ of square $A B C D$, whose side length is 1. Segments $C M$ and $D N$ intersect at point $P$. Find $A P$. | Ex. 13. 1.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,839 |
Ex. 17. A quadrilateral is inscribed in a circle of radius 25, the diagonals of which are perpendicular and equal to 48 and 40. Find the sides of the quadrilateral. | Ex. 17. $5 \sqrt{10}, 9 \sqrt{10}, 13 \sqrt{10}, 15 \sqrt{10}$. Hint. Find the distances from the center of the circle to the midpoints of the diagonals. | 5\sqrt{10},9\sqrt{10},13\sqrt{10},15\sqrt{10} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,843 |
Ex. 19. In triangle $ABC$, points $M$ and $N$ are chosen on sides $AB$ and $AC$. It is known that $\angle ABC=70^{\circ}, \angle ACB=50^{\circ}, \angle ABN=20^{\circ}$, $\angle ACM=10^{\circ}$. Find $\angle NMC$.
In each of these cases, the generalized sine theorem is used t... | 3,\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,846 |
Ex. 22. (Ptolemy's Theorem.) Prove that the product of the diagonals of a quadrilateral inscribed in a circle is equal to the sum of the products of its opposite sides. | Ex. 22. Let $A B C D$ be a cyclic quadrilateral. Denote the arcs connecting adjacent vertices of the quadrilateral by $2 a, 2 b, 2 c$ and $2 d$. Then the proof of Ptolemy's theorem reduces to proving the identity $\sin (a+b) \sin (b+c)=\sin a \sin c+\sin b \sin d$ under the condition that $a+b+c+d=\pi$. Further, after ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,848 |
Ex. 24. The extensions of the angle bisectors at vertices $P$ and $Q$ of triangle $P Q R$ intersect the circumscribed circle at points $P^{\prime}$ and $Q^{\prime}$, respectively. Find $P^{\prime} Q^{\prime}$, if $P Q=6$, and the radius of the circumscribed circle is 5. | Ex. 24. Answer: $P^{\prime} Q^{\prime}=3 \sqrt{10}$ or $\sqrt{10}$.
Solution. Let the angle $P R Q$ be denoted by $\gamma$. By the Law of Sines, $\sin \gamma=\frac{6}{2 R}=\frac{3}{5}$. Then... | 3\sqrt{10} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,850 |
Ex. 25. The extensions of the altitudes of triangle $ABC$, dropped from vertices $A$ and $B$, intersect the circumcircle of the triangle at points $A'$ and $B'$, respectively. Find $AB$, if $A'B'=12$, and the radius of the circumcircle is 10. | Ex. 25. $AB=6 \sqrt{10}$ or $2 \sqrt{10}$. | 6\sqrt{10} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,851 |
Ex. 27. Triangle $ABC$ is isosceles $(AB = BC)$. On the arc $AC$ of the circumscribed circle, not containing point $B$, an arbitrary point
$M$ is taken. Prove that $MA + MC = 2MB \cos A$. | Ex. 27. Let $\angle A B C=\beta, \angle A B M=\varphi$. Then $\angle C B M=\beta-\varphi$. We use the generalized sine theorem: $M A+M C=2 R(\sin \varphi+\sin (\beta-\varphi))=$ $2 R \cdot 2 \sin \frac{\beta}{2} \cos \frac{2 \varphi-\beta}{2}$. Note that $\angle A=\frac{\pi}{2}-\frac{\beta}{2} \Rightarrow \sin \frac{\b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,853 |
Ex. 36. Prove that in a right triangle, the bisector of the right angle bisects the angle between the median and the altitude drawn from the same vertex.
| Ex. 36. Solution. Let triangle $ABC$ be a right triangle, with angle $C$ being the right angle. Denote $\angle CAB = \alpha, \angle CBA = \beta$. Suppose, for example, $\alpha > \beta$. Draw the median $CM$ and the altitude $CH$. Verify that $\angle MCB = \angle HCA = \beta$. If we subtract equal angles from equal angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,862 |
Ex. 39. Prove that the square of the bisector of a triangle's angle is equal to the product of the adjacent sides minus the product of the segments into which the bisector divides the opposite side.
保留源文本的换行和格式，直接输出翻译结果。
(Note: The last sentence is a note for the translator and should not be included in the translat... | Ex. 39. Proof. Let $A D=\ell-$ be the bisector of angle $A$ of triangle $A B C$ (see the figure on the right). Point $E$ is the intersection point of the extension of the bisector with the circumscribed circle. Note that triangle $A B E$ is similar to triangle $A C D$ by three angles $(\angle C A E=\angle B A E$ and $\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,865 |
Ex. 40. Show that the segments into which the bisector of angle $A$ divides side $BC$ are equal to $\frac{ab}{b+c}$ and $\frac{ac}{b+c}$. Using these formulas and Exercise 39, derive the formula for the length of the bisector in terms of the sides of the triangle: $l_{a}=\frac{2}{b+c} \sqrt{bc p(p-a)}$. | Ex. 40. Let's denote the lengths of the segments into which the bisector $A A^{\prime}$ divides the side $B C$ as $m=B A^{\prime}$ and $n=A^{\prime} C$. Then $m / n=c / b$ and $m+n=a \Rightarrow n=\frac{a b}{b+c}$ and $m=\frac{a c}{b+c}$. Further: $l_{a}^{2}=b c-m n=b c-\frac{a^{2} b c}{(b+c)^{2}}=\frac{4 b c}{(b+c)^{2... | l_{}=\frac{2}{b+}\sqrt{p(p-)} | Geometry | proof | Yes | Yes | olympiads | false | 41,866 |
Ex. 41. In triangle $A B C$, the bisectors $A A^{\prime}$ and $B B^{\prime}$ are drawn. Using the formulas from Exercise 40, show that if $a>b$, then the segments $C A^{\prime}$ and $B A^{\prime}$ are greater than the segments $C B^{\prime}$ and $A B^{\prime}$, respectively. | Ex.41. Using the formulas from the previous exercise, we get that $C B^{\prime}=\frac{a b}{a+c}, A B^{\prime}=\frac{c b}{a+c} \cdot C A^{\prime}>$
$C B^{\prime} \Leftrightarrow \frac{a b}{b+c}>\frac{a b}{a+c} \Leftrightarrow a>b$. Similarly: $B A^{\prime}>A B^{\prime} \Leftrightarrow \frac{a c}{b+c}>\frac{b c}{a+c} \Le... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,867 |
Ex. 42. In triangle $ABC$, the angle bisectors $AA'$ and $BB'$ are drawn. Using exercises 39 and 41, show that if $a > b$, then the angle bisector of angle $A$ is less than the angle bisector of angle $B$. From this, deduce that if two angle bisectors are equal, then the triangle is isosceles. | Ex. 42. In triangle $A B C$, the angle bisectors $A A^{\prime}$ and $B B^{\prime}$ are drawn. According to the formula from Ex. 39, we get $\ell_{a}=b c-C A^{\prime} \cdot B A^{\prime}$, $\ell_{b}=a c-C B^{\prime} \cdot A B^{\prime}$. Since in Ex. 40 the inequalities $C A^{\prime}>C B^{\prime}$ and $B A^{\prime}>A B^{\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,868 |
Ex. 43. Using the formula for the area of a triangle through the product of sides and the sine of the angle between them, prove that $l_{a}=2 b c \cos \frac{A}{2} /(b+c)$. | Ex. 43.
$S_{A B C}=S_{A C A^{\prime}}+S_{A B A^{\prime}} \Rightarrow \frac{b c}{2} \sin A=\frac{b \ell_{a}}{2} \sin \frac{A}{2}+\frac{c \ell_{a}}{2} \sin \frac{A}{2} \Rightarrow l_{a}=\frac{2 b c \cos \frac{A}{2}}{b+c}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,869 |
Ex. 48. Let $ABC$ be an arbitrary triangle, $CP$ be the bisector of angle $C$, and $CQ$ be the bisector of one of the external angles at vertex $C$ (points $P$ and $Q$ lie on line $AB$). Prove that points $P$ and $Q$ harmonically divide segment $AB$ | Ex. 48. Let point $Q$ lie on ray $A B$ beyond point $B$. Draw a line through point $B$ parallel to $A C$, and let $F$ be the point of intersection of this line with line $C Q$. Triangle $C F B$ is isosceles, since $\angle B C F = \angle B F C \Rightarrow B F = B C$. At the same time, $\triangle B F Q \sim \triangle A C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,874 |
Ex. 49. Prove that the square of the bisector of an exterior angle of a triangle is equal to the product of the segments of the opposite side from its ends to the point of intersection with the bisector without the product of the adjacent sides. | Ex. 49. Consider triangle $ABC$. Let $AC=b, BC=a$ and for definiteness $b>a$. Draw $CP$ - the bisector of angle $C$ and $CQ$ - the bisector of the external angle $C$ (points $P$ and $Q$ lie on ray $AB$). We also introduce the following notations: $CQ=\ell^{*}, CP=\ell, AP=m, PB=n, BQ=u, AQ=v$. According to the result o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,875 |
Ex. 50. Prove that the point of intersection of the angle bisectors (which is the center of the inscribed circle) divides each bisector in the ratio equal to the ratio of the sum of the lengths of the adjacent sides to the length of the opposite side, measured from the vertex.
保留源文本的换行和格式，直接输出翻译结果。
Note: The last se... | Ex. 50. Draw a line through vertex $C$ parallel to $A B$, and extend the angle bisectors of angles $A$ and $B$ until they intersect this line. Denote the points of intersection as $E$ and $F$. Note that triangles $A C E$ and $A C F$ are isosceles ($C E=b, C F=a$), therefore, $E F=a+b$.
, $L-$ the point of intersection of the bisector of angle $C$, and $M$ - the point of intersection of the bisector of the external angle at vertex $C$ with the line $A B$. Denote by $\gamma$ half of angle $C$. Triangle $L C M$ is right-angled a... | AC^2+BC^2=4R^2 | Geometry | proof | Yes | Yes | olympiads | false | 41,878 |
Ex. 53. Given a convex quadrilateral $A B M C$, in which $A B=B C, \angle B A M=30^{\circ}$, $\angle A C M=150^{\circ}$. Prove that $A M$ is the bisector of angle $В M C$.
Prove that $A M$ is the bisector of angle $B M C$. | Ex.53. Let's construct point $B^{\prime}$, symmetric to point $B$ with respect to the line $A M$. Triangle $B A B^{\prime}$ is equilateral, since it is isosceles and one of the angles is $60^{\circ}$. Since triangle $A B C$ is isosceles, then $B B^{\prime} C$ is also an isosceles triangle ($B C=B B^{\prime}$). Let the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,879 |
Ex. 57. In triangle $A B C$, the angle bisectors of angles $A$ and $B$ are extended to intersect the circumcircle at points $A^{\prime}$ and $B^{\prime}$ respectively. It is known that $A A^{\prime}=B B^{\prime}$. Show that either triangle $A B C$ is isosceles, or angle $C$ is $60^{\circ}$.
 and $BA'$ (not containing point $C$) are equal to $2\g... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,882 |
Ex. 58. In triangle $ABC$, the sides are known: $AB=5, BC=8, AC=7$. On the bisector of angle $A$ (inside the triangle), a point $O$ is chosen such that the areas of triangles $AOB, AOC, BOC$, taken in the given order, form an arithmetic progression. Find this progression. | Ex.58. The semiperimeter $p$ of triangle $ABC$ is 10. Using Heron's theorem, we find that the area of triangle $ABC$ is $S=10 \sqrt{3}$. From the fact that $S_{AOC}$ is the middle term of the progression, it follows that $S_{AOC}=\frac{10 \sqrt{3}}{3}$. Considering that $S_{AOB}: S_{AOC}=5: 7$, we get that $S_{AOB}=\fr... | \frac{50\sqrt{3}}{21},\frac{70\sqrt{3}}{21},\frac{90\sqrt{3}}{21} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,883 |
Ex. 60. Prove that $m_{a}=\frac{1}{2} \sqrt{2 b^{2}+2 c^{2}-a^{2}}$. | Ex. 60. Let one of the angles that the median $A M=m^{\vec{C}_{1}}$
forms with the line $B C$, for example $\angle A M B$, be denoted by $\phi$. Then, by the cosine rule, we get $c^{2}=m^{2... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,885 |
Ex. 61. Given the sides $b$ and $c$ of a triangle and the median $m_{a}$, determine the other medians. | Ex. 61. $m_{b}=\frac{1}{2} \sqrt{3 b^{2}+6 c^{2}-8 m_{a}^{2}}, m_{c}=\frac{1}{2} \sqrt{6 b^{2}+3 c^{2}-8 m_{a}^{2}}$. | m_{b}=\frac{1}{2}\sqrt{3b^{2}+6^{2}-8m_{}^{2}},m_{}=\frac{1}{2}\sqrt{6b^{2}+3^{2}-8m_{}^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,886 |
Ex. 64. Prove that if two medians are equal, then the triangle is isosceles.
Translate the above text into English, keep the original text's line breaks and format, and output the translation result directly. | Ex. 64. Let in triangle $A B C$ the medians
$A D$ and $B E$ be equal. Then the segments $A M$ and $B M$ are also equal, as they constitute two-thirds of the medians. For the same reason, the segments $D M$ and $M E$ are equal. The angles $A M E$ and $B M D$ between these segments are equal as vertical angles. Therefor... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,889 |
Ex. 65. Prove that $\frac{3 P}{4}<m_{a}+m_{b}+m_{c}<P$, where $P=a+b+c$. | Ex.65. From the triangle inequality, it follows that $A M+M B>A C \Leftrightarrow$ $\frac{2}{3} m_{a}+\frac{2}{3} m_{b}>c$. By changing the letters, we get that $\frac{2}{3} m_{b}+\frac{2}{3} m_{b}>a, \frac{2}{3} m_{a}+\frac{2}{3} m_{c}>b$. Adding the inequalities, we get that $\frac{4}{3}\left(m_{a}+m_{b}+m_{c}\right)... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 41,890 |
Ex. 67. The extensions of the medians $A E$ and $B F$ of triangle $A B C$ intersect the circumcircle of the triangle at points $A_{1}$ and $B_{1}$. It is known that $A A_{1}=B B_{1}$. Prove that either triangle $A B C$ is isosceles, or its sides $a, b$, and $c$ are related by the equation $c^{4}=a^{4}-a^{2} b^{2}+b^{4}... | Ex. 67. Let $A A_{1}=B B_{1}=\ell$. Then $m_{a}\left(\ell-m_{a}\right)=$ $\frac{a^{2}}{4}, m_{b}\left(\ell-m_{b}\right)=\frac{b^{2}}{4}$. Express the medians in terms of the sides of the triangle and, obtaining
in both equations expressions for $\ell$, equate them. | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,892 |
Ex. 70. Prove that the medians, drawn from vertices $A$ and $B$ of triangle $ABC$, are mutually perpendicular if and only if $a^{2}+b^{2}=5 c^{2}$. | Ex. 70. Hint. Use the cosine theorem for triangle $A B M$, where $M$ is the point of intersection of the medians. The sides $A M$ and $B M$ are found using the formulas for medians. | ^{2}+b^{2}=5^{2} | Geometry | proof | Yes | Yes | olympiads | false | 41,895 |
Ex. 71. In triangle $ABC$, the bisector of angle $B$ is equal to the median of side $BC$. Prove that either the triangle is equilateral, or $a^{2}=2(b+c)^{2}$. | Ex. 71. Using the formulas for the length of the median (Ex. $60$) and for the length of the bisector (Ex. 40). $\frac{2\left(b^{2}+c^{2}\right)-a^{2}}{4}=\frac{b c(b+c+a)(b+c-a)}{(b+c)^{2}} \Leftrightarrow$ $2\left(\left(b^{2}+c^{2}\right)-a^{2}\right)(b+c)^{2}=4 b c\left((b+c)^{2}-a^{2}\right)$. After simplification ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,896 |
Ex. 72. The height of a right triangle, dropped to the hypotenuse, is $h$. What is the minimum length that the median, which bisects the larger leg, can have? | Ex. 72. $\frac{3}{2} h$. Solution. Let the segments into which the height $h$ divides the hypotenuse be denoted by $x$ and $y (x \geq y)$. Then $m^{2}=\left(\frac{x}{2}+y\right)^{2}+\left(\frac{h}{2}\right)^{2}=\frac{x^{2}}{4}+x y+y^{2}+\frac{h^{2}}{4}=\frac{5 h^{2}}{4}+\frac{1}{4}\left(x^{2}+\frac{4 h^{4}}{x^{2}}\righ... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,897 |
Ex. 76. It is known that the median $A A_{1}$, the bisector $B B_{1}$, and the altitude $C C_{1}$ intersect at one point. Prove that $\sin C=\cos A \operatorname{tg} B$. | Ex. 76. Let $O$ be the point of intersection of the given lines. Apply Menelaus' theorem to triangle $B C C_{1}$ and the intersecting line $A A_{1}: \frac{A_{1} B}{A_{1} C} \cdot \frac{O C}{O C_{1}} \cdot \frac{A C_{1}}{A B}=1$. By the condition $\frac{A_{1} B}{A_{1} C}=1$. By the property of the angle bisector $\frac{... | \sinC=\cosA\operatorname{tg}B | Geometry | proof | Yes | Yes | olympiads | false | 41,901 |
Ex. 77. The median $A A_{1}$, the bisector $B B_{1}$, and the altitude $C C_{1}$ of triangle $A B C$ intersect at one point, which divides the altitude $C C_{1}$ in the ratio 3 : 1, counting from vertex C. Prove that the median $A A_{1}$ and the bisector $B B_{1}$ of the triangle are perpendicular. | Ex. 77. Let $O-$ be the point of intersection of the given lines. Since $O C=3 O C_{1}$, it follows from the property of the angle bisector that $B C_{1} / B C=1 / 3$, that is, $B C_{1}=\frac{a}{3}$. Denote by $K$ the midpoint of the segment $C C_{1}$. The segment $K A_{1}$ is a midline, its length is half of $B C_{1}$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,902 |
Ex. 82. The orthocenter $H$ of triangle $ABC$ bisects the altitude $AD$. Prove that $\operatorname{tg} B \cdot \operatorname{tg} C=2$. | Ex. 82. From triangle $A D C$, we note that $A D=b \sin C$ and $C D=b \cos C$. The angle $H C D=90^{\circ}-B$, therefore, $A D / 2=b \cos C \operatorname{ctg} B \Rightarrow \frac{b \sin C}{2}=b \cos C \operatorname{ctg} B$. From this, the required equality follows. | \operatorname{tg}B\cdot\operatorname{tg}C=2 | Geometry | proof | Yes | Yes | olympiads | false | 41,907 |
Ex. 83. If $H$ is the orthocenter of triangle $A B C$, then
$$
H A \cdot h_{a}+H B \cdot h_{b}+H C \cdot h_{c}=\frac{a^{2}+b^{2}+c^{2}}{2}
$$ | Ex. 83. Note that a circle can be circumscribed around quadrilateral $A F H E$, and $A H$ is the diameter of this circle. Since triangles $A F E$ and $A B C$ are similar (Ex. 80), the ratio of the diameters of the circumscribed circles is equal to the similarity coefficient, which is $\cos A$ (Ex. 81). Therefore, $A H ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,908 |
Ex. 86. Quadrilateral $A B C D$ is inscribed, $\varphi$ is the angle between the diagonals, $m$ and $n$ are the lengths of the diagonals. Prove that
$$
S=\frac{a d+b c}{2} \sin A=\frac{a b+c d}{2} \sin B=\frac{a c+b d}{2} \sin \varphi
$$ | Ex. 86. Let's divide the quadrilateral $A B C D$ by the diagonal $B D$ into two triangles, the areas of which are equal to $\frac{a d \sin A}{2}$ and $\frac{b c \sin C}{2}=\frac{b c \sin A}{2}$. From this, the first equality follows. Similarly, the second equality can be proven. The third equality follows from Ptolemy'... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,911 |
Ex. 88. Prove that $S=\frac{\sqrt{(a b+c d)(a d+b c)(a c+b d)}}{4 R}$.
Brahmagupta's formula. If a quadrilateral with sides $a, b, c, d$ and semiperimeter $p$ is inscribed in a circle, then
$$
S=\sqrt{(p-a)(p-b)(p-c)(p-d)}
$$
Brahmagupta (598-660) - Indian mathematician and astronomer. | Ex. 88. Using the generalized sine theorem: $\sin A=\frac{m}{2 R}, \sin B=\frac{n}{2 R}$. Substituting into the first and second formulas of Ex. 86, we get that $S^{2}=\frac{(a d+b c)(a b+c d)}{4 R^{2}} m n$. Replacing $m n$ according to Ptolemy's theorem, we obtain the required result. | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,913 |
Ex. 89. A quadrilateral $A B C D$ is inscribed in a circle, the lengths of whose sides are given: $A B=1, B C=2, C D=3, D A=4$. Find its area $and$ the radius of the circumscribed circle. | Ex. 89. $S=2 \sqrt{6}, R=\frac{\sqrt{7} \cdot 55}{4 \sqrt{6}}$. | S=2\sqrt{6},R=\frac{\sqrt{7}\cdot55}{4\sqrt{6}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,914 |
Ex. 90. (Brahmagupta's Problem.) Quadrilateral $A B C D$ is inscribed in a circle. Point $Q$ is the intersection of the mutually perpendicular diagonals $A C$ and $B D$. Point $K$ is the midpoint of side $A B$. Prove that line $K Q$ is perpendicular to $C D$.
![](https://cdn.mathpix.com/cropped/2024_05_21_91ff4e46083e... | Ex. 90. Let's denote the angle $B A C$ as $\alpha$. Then $\angle K Q A = \angle B D C = \alpha$. Since $\angle B Q A = 90^{\circ}$, it follows that $\angle K Q B = \angle D Q P = 90^{\circ} - \alpha$. Therefore, $\angle D P Q = 90^{\circ}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,915 |
Ex. 91. In an isosceles triangle $ABC$, a semicircle is inscribed, the center $O$ of which lies on the base $AB$ of the triangle. An arbitrary tangent $x$ to the semicircle intersects the sides $BC$ and $AC$ at points $M$ and $N$, respectively. Prove that triangles $AON$, $BOM$, and $MON$ are similar. | Ex. 91. Let the angles of triangle $NMC$ be denoted by $\alpha, \beta$ and $\gamma$. Show that the angles of the triangles marked in the problem are equal to $90^{\circ}-\frac{\alpha}{2}, 90^{\circ}-\frac{\beta}{2}, 90^{\circ}-\frac{\gamma}{2}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,916 |
Ex. 99. Through the point of tangency of two circles, two lines are drawn, intersecting one circle at points $A$ and $B$, and the other at points $C$ and $D$. Prove that the lines $A B$ and $C D$ are parallel. | Ex. 99. Let $O$ be the point of tangency of two circles. Choose points $C$ and $D$ such that $O$ lies on segments $A C$ and $B D$. Let $\ell$ be the common tangent drawn at point $O$, and points $M$ and $N$ lie on $\ell$ on opposite sides of $O$, with points $B$ and $M$ lying on opposite sides of line $A C$. Now consid... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,924 |
Ex. 100. Lines $\ell_{1}$ and $\ell_{2}$ are external tangents to two non-intersecting circles, and the distance between the points of tangency on each of these lines is $a$. The internal tangent intersects $\ell_{1}$ and $\ell_{2}$ at points $A$ and $B$. Find the length of segment $A B$. | Ex. 100. a. Note. Use the fact that tangents to a circle drawn from the same point are equal, and divide the segment $A B$ into parts corresponding to the segments of the tangents. | a | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,925 |
Ex. 101. Two circles are constructed on a plane such that each passes through the center of the other. $P$ and $Q$ are the two points of their intersection. A line passing through point $P$ intersects the first circle at point $A$ and the second circle at point $B$ such that $P$ is between $A$ and $B$, and this line fo... | Ex. 101. $S=3+2 \sqrt{3}$. Hint. Show that triangle $A B Q$ is equilateral. | 3+2\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,926 |
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