problem stringlengths 32 5.4k | answer stringlengths 1 319 | solution stringlengths 3 12.9k | difficulty float64 1 10 |
|---|---|---|---|
Given a quadrilateral $A B C D$ with an area of 1. From its internal point O, perpendiculars OK, OL, OM, and ON are dropped to the sides $A B, B C, C D$, and $D A$ respectively. It is known that $A K \geq K B, B L \geq L C, C M \geq M D$, and $D N \geq N A$. Find the area of the quadrilateral $K L M N$. | \frac{1}{2} | From two inclined lines drawn from one point, the one with the larger projection is larger. Therefore, from the inequalities given in the problem, it follows that $O A \geq O B \geq O C \geq O D \geq O A$. Hence, $O A=O B=O C=O D$, which means $O$ is the center of the circle circumscribed around quadrilateral $A B C D$... | 2 |
10. Let positive real numbers $a, b, c$ satisfy $a^{2}+4 b^{2}+9 c^{2}=4 b+12 c-2$, find the minimum value of $\frac{1}{a}+\frac{2}{b}+\frac{3}{c}$. | 6 | $$
a^{2}+4 b^{2}+9 c^{2}=4 b+12 c-2 \Rightarrow a^{2}+(2 b-1)^{2}+(3 c-2)^{2}=3,
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(1+1+1)\left[a^{2}+(2 b-1)^{2}+(3 c-2)^{2}\right] \geqslant(a+2 b+3 c-3)^{2} \\
\Rightarrow(a+2 b+3 c-3)^{2} \leqslant 9 \Rightarrow 0<a+2 b+3 c \leqslant 6 .
\end{array}
$$... | 6.5 |
Given the nine-sided regular polygon $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8 A_9$, how many distinct equilateral triangles in the plane of the polygon have at least two vertices in the set $\{A_1,A_2,\dots,A_9\}$?
$\text{(A) }30 \qquad \text{(B) }36 \qquad \text{(C) }63 \qquad \text{(D) }66 \qquad \text{(E) }72$ | 66 | Each of the $\binom{9}{2} = 36$ pairs of vertices determines two equilateral triangles, for a total of 72 triangles. However, the three triangles $A_1A_4A_7$, $A_2A_5A_8$, and $A_3A_6A_9$ are each counted 3 times, resulting in an overcount of 6. Thus, there are $\boxed{66}$ distinct equilateral triangles. | 5.25 |
Konyagin S.V.
Given a square with side length 1. From it, four corners are cut off - four triangles, each of which has two sides along the sides of the square and these sides are $1 / 3$ of their lengths. With the resulting 8-sided polygon, the same operation is performed: from each vertex, a triangle is cut off, two ... | \frac{5}{7} | Let $M_{0}$ be the original square, and $M_{1}, M_{2}, M_{3}, \ldots$ be the polygons obtained from $M_{0}$ by sequentially cutting off corners. It is convenient to also consider the polygon $N_{k}$, whose vertices are the midpoints of the sides of $M_{k} (k=0,1,2, \ldots)$.
Let $A$ be an arbitrary vertex of the polyg... | 5 |
6. A uniformly growing uniform grassland is divided into two unequal parts, $\mathrm{A}$ and $\mathrm{B}$. A herd of cows first eats the grass on plot A, finishing it in exactly 7 days. Then the herd continues to eat the grass on plot B at the same speed, finishing it in 4 days, at which point plot A has just recovered... | 105:44 | 【Answer】105:44
【Analysis】A The grassland from being eaten clean to growing back to the original amount of grass takes 4 days. Assuming A grassland grows 1 unit of grass per day, then in 4 days it grows 4 units. The original amount of grass in A grassland is 4 units. The speed at which the cow eats grass is $(4+7) \div ... | 7.6 |
3.11 Apples of the first grade for a total of 228 rubles and apples of the second grade for a total of 180 rubles were delivered to the store. During unloading, the delivered apples were accidentally mixed. It was found that if all the apples are now sold at one price - 90 kopecks lower than the price per kilogram of f... | \text{85 kg} | 3.11 Let's construct the following table:
| Apple Variety | Cost, rub | Quantity, tons | Price, rub/kg |
| :---: | :---: | :---: | :---: |
| First | 228 | $x$ | $\frac{228}{x}$ |
| Second | 180 | $x+5$ | $\frac{180}{x+5}$ |
According to the problem, we have the equation
$\left(\frac{228}{x}-0.9\right)(2 x+5)=408$,
... | 3 |
308. The scale division value of the amperemeter is 0.1 A. The readings of the amperemeter are rounded to the nearest whole division. Find the probability that an error exceeding $0.02 \mathrm{~A}$ will be made during the reading. | 0.6 | Solution. The rounding error of the measurement can be considered as a random variable $X$, which is uniformly distributed in the interval between two adjacent integer divisions. The density of the uniform distribution $f(x)=1 /(b-a)$, where $(b-a)$ is the length of the interval in which the possible values of $X$ are ... | 2 |
According to the Gregorian calendar, 97 leap days must be inserted over 400 consecutive years. After how many years will there be a 1-day difference between the Gregorian calendar and the "precise" calendar, if the length of a year is 365 days 5 hours 48 minutes 46 seconds? | 3323 | Regardless of the number of years, the Gregorian calendar offers 97 days every 400 years, which is $97 \cdot 24=2328$ hours, compared to 5 hours, 48 minutes, and 46 seconds 400 times over. It is reasonable to count the minutes as $\frac{4}{5}$ of an hour, thus in 400 years, $400 \cdot\left(5+\frac{4}{5}\right)=$ 2320 h... | 5.25 |
20. [10] There exist several solutions to the equation
$$
1+\frac{\sin x}{\sin 4 x}=\frac{\sin 3 x}{\sin 2 x},
$$
where $x$ is expressed in degrees and $0^{\circ}<x<180^{\circ}$. Find the sum of all such solutions. | 320^{\circ} | Answer: $320^{\circ}$
Solution: We first apply sum-to-product and product-to-sum:
$$
\begin{array}{c}
\frac{\sin 4 x+\sin x}{\sin 4 x}=\frac{\sin 3 x}{\sin 2 x} \\
2 \sin (2.5 x) \cos (1.5 x) \sin (2 x)=\sin (4 x) \sin (3 x)
\end{array}
$$
Factoring out $\sin (2 x)=0$,
$$
\sin (2.5 x) \cos (1.5 x)=\cos (2 x) \sin (3 x... | 10 |
4. Given a regular quadrilateral pyramid $P-A B C D$ with all edges of equal length. Taking $A B C D$ as one face, construct a cube $A B C D-E F G H$ on the other side of the pyramid. Then, the cosine value of the angle formed by the skew lines $P A$ and $C F$ is | \frac{2+\sqrt{2}}{4} | 4. $\frac{2+\sqrt{2}}{4}$.
Assume the edge lengths of the quadrilateral pyramid $P-ABCD$ are all 2. Taking the center $O$ of the square $ABCD$ as the origin, and the directions of $\overrightarrow{DA}$, $\overrightarrow{DC}$, and $\overrightarrow{OP}$ as the positive directions of the $x$-axis, $y$-axis, and $z$-axis,... | 4 |
(1) The range of the function $f(x)=\sqrt{\frac{x+1}{x^{2}+4 x+7}}$ is $\qquad$ | \left[0, \frac{\sqrt{6}}{6}\right] | (1) $\left[0, \frac{\sqrt{6}}{6}\right]$ | 3 |
$650$ students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?
$\mathrm{(A)} \frac{2}... | \frac{5}{2} | The answer is $\dfrac{\text{number of students who preferred spaghetti}}{\text{number of students who preferred manicotti}}$
So,
$\frac{250}{100}$
Simplify,
$\frac{5}{2}$
The answer is $\boxed{\textbf{(E)}\ \dfrac{5}{2}}$ | 3.2 |
7. Let $[a]$ denote the greatest integer not exceeding $a$, for example: $[8]=8, [3.6]=3$.
Some natural numbers can be expressed in the form $[x]+[2 x]+[3 x]$, such as 6 and 3:
$$
\begin{array}{c}
6=\left[\frac{5}{4}\right]+\left[2 \times \frac{5}{4}\right]+\left[3 \times \frac{5}{4}\right], \\
3=[0.8]+[2 \times 0.8]+[... | 1347 | $1347$ | 6 |
Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$. Find the degree measure of $\angle BAG$. | 58 | Let $O$ be the center of the circle with $ABCDE$ on it.
Let $x$ be the degree measurement of $\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}$ in circle $O$
and $y$ be the degree measurement of $\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}$ in circle $C$.
$\angle ECA$ is, therefore, $5y$ by way... | 4.2 |
I2.1 If the $50^{\text {th }}$ power of a two-digit number $P$ is a 69-digit number, find the value of $P$.
(Given that $\log 2=0.3010, \log 3=0.4771, \log 11=1.0414$.)
I2.2 The roots of the equation $x^{2}+a x-P+7=0$ are $\alpha$ and $\beta$, whereas the roots of the equation $x^{2}+b x-r=0$ are $-\alpha$ and $-\beta... | 4 | Reference: 1995 HG5 . . $37^{100} \ldots$ 157-digit number, $37^{15} \ldots n$-digit ....
$$
\begin{array}{l}
P^{50}=y, 10<P \leq 99,10^{68} \leq y<10^{69} \\
P=y^{\frac{1}{50}} ; 10^{68 \div 50}<P<10^{69 \div 50} \\
1.34<\log P<1.38 \\
\log 22=\log 2+\log 11=1.3424 ; \log 24=3 \log 2+\log 3=1.3801 \\
\log 22<\log P<\l... | 3 |
$A$, $B$, $C$ are three piles of rocks. The mean weight of the rocks in $A$ is $40$ pounds, the mean weight of the rocks in $B$ is $50$ pounds, the mean weight of the rocks in the combined piles $A$ and $B$ is $43$ pounds, and the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds. What is the gr... | 59 | Let pile $A$ have $A$ rocks, and so on.
The total weight of $A$ and $C$ can be expressed as $44(A + C)$.
To get the total weight of $B$ and $C$, we add the weight of $B$ and subtract the weight of $A$: $44(A + C) + 50B - 40A = 4A + 44C + 50B$
Therefore, the mean of $B$ and $C$ is $\frac{4A + 44C + 50B}{B + C}$, which i... | 3.4 |
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying:
$$
f(x(1+y))=f(x)(1+f(y))
$$ | $f(x) = 0 \text{ or } f(x) = x$ | The zero constant function is a solution, but we exclude this case. By setting $x=0$, we get $f(0)=0$, and by setting $y=-1$, we get $f(-1)=-1$. By setting $x=-1$ and $y=-\frac{1}{2}$, we get $f\left(-\frac{1}{2}\right)=-\frac{1}{2}$. Finally, by setting $x=-\frac{1}{2}$ and $y=1$, we get $f(1)=1$. Then, by setting $x=... | 5.75 |
In the diagram, $A B C F$ is a rectangle with $A B=30 \mathrm{~cm}$ and $A F=14 \mathrm{~cm}$. Points $E$ and $D$ are on $F C$ so that $F E=5 \mathrm{~cm}$ and the area of quadrilateral $A B D E$ is $266 \mathrm{~cm}^{2}$. The length of $D C$ is
(A) $17 \mathrm{~cm}$
(B) $16 \mathrm{~cm}$
(C) $19 \mathrm{~cm}$
(D) $20 ... | 17 \text{ cm} | Solution 1
Since $A B$ and $E D$ are parallel, quadrilateral $A B D E$ is a trapezoid.
We know that $A B=30 \mathrm{~cm}$.
Since $A B C F$ is a rectangle, then $F C=A B=30 \mathrm{~cm}$.
Suppose that $D C=x \mathrm{~cm}$.
Then $E D=F C-F E-D C=(30 \mathrm{~cm})-(5 \mathrm{~cm})-(x \mathrm{~cm})=(25-x) \mathrm{cm}$... | 2.4 |
Ben participates in a prize draw. He receives one prize that is equally likely to be worth $\$ 5, \$ 10$ or $\$ 20$. Jamie participates in a different prize draw. She receives one prize that is equally likely to be worth $\$ 30$ or $\$ 40$. What is the probability that the total value of their prizes is exactly $\$ 50$... | \frac{1}{3} | Since there are two possible prizes that Jamie can win and each is equally likely, then the probability that Jamie wins $\$ 30$ is $\frac{1}{2}$ and the probability that Jamie wins $\$ 40$ is $\frac{1}{2}$. If Jamie wins $\$ 30$, then for the total value of the prizes to $\$ 50$, Ben must win $\$ 20$. The probability t... | 1 |
## Task 4
Pair the following numbers so that they always add up to the same sum.
$\begin{array}{llllllll}43 & 202 & 100 & 175 & 98 & 257 & 125 & 200\end{array}$ | (43, 257), (100, 200), (98, 202), (125, 175) | The sum is always 300. Pairs are (43, 257), (100, 200), (98, 202), (125, 175) | 1.4 |
In 1972, filling the tank of a small car cost $R \$ 29.90$, and in 1992, it cost $\$ 149.70$ to fill the same tank. Which of the following values best approximates the percentage increase in the price of gasoline over this 20-year period?
(a) $20 \%$
(b) $125 \%$
(d) $300 \%$
(d) $400 \%$
(e) $500 \%$ | 400 \% | The increase in value was
$$
149.70-29.90=119.80 \text { reais }
$$
which corresponds to:
$$
\frac{119.80}{29.90} \times 100 \% = 400.66 \%
$$
The correct option is (d). | 2.4 |
4. The polynomial $p(x)$ has the following property: for every triplet of integers $a, b, c$ such that $a+b+c=2022$ we have that $p(a)+p(b)+p(c)=p(674)$. It is also known that $p(0)=-2696$. What is the value of $p(2022)$?
(A) -2696
(B) 674
(C) 5392
(D) 8088
(E) It is not possible to determine it with the given data. | 5392 | 4. The answer is $\mathbf{( C )}$. Substituting $a=b=c=674$ (integers that indeed satisfy $a+b+c=$ 2022) we get $3 p(674)=p(674)$, which means $p(674)=0$. Substituting then $a=b=0$ and $c=2022$ we obtain
$$
2 p(0)+p(2022)=p(674)=0 \Rightarrow p(2022)=-2 p(0)=5392 \text {. }
$$ | 4 |
20. Let $a$ and $b$ be real numbers such that $17\left(a^{2}+b^{2}\right)-30 a b-16=0$. Find the maximum value of $\sqrt{16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9}$.
(2 marks)
設 $a$ 、 $b$ 為實數 , 使得 $17\left(a^{2}+b^{2}\right)-30 a b-16=0$ 。 求 $\sqrt{16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9}$ 的最大值。 | 7 | 20. 7
20. Rewrite the given equation as $(a+b)^{2}+16(a-b)^{2}=16$. Hence we may let $a+b=4 \cos x$ and $a-b=\sin x$. Note that
$$
\sqrt{16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9}=\sqrt{(4 a-2 b)^{2}-3(4 a-2 b)+9}=\sqrt{\left(4 a-2 b-\frac{3}{2}\right)^{2}+\frac{27}{4}} .
$$
Since $4 a-2 b=(a+b)+3(a-b)=4 \cos x+3 \sin x$, wh... | 7 |
14. A ball was added to an urn containing one white ball - either white or black (with equal probabilities of selection). After this, one ball was randomly drawn from the urn. It turned out to be white. What is the conditional probability that the remaining ball in the urn is also white? | \frac{2}{3} | Solution. Let
$$
\begin{aligned}
A & =\{\text { a white ball was added to the urn }\}, \\
C & =\{\text { a white ball was drawn }\}, \\
D & =\{\text { a white ball remains }\} .
\end{aligned}
$$
Note that $A=C D$. Then
$$
\begin{aligned}
\mathrm{P}(D \mid C)=\frac{\mathrm{P}(D C)}{\mathrm{P}(C)}=\frac{\mathrm{P}(A)}... | 2 |
LVIII OM - I - Problem 2
Determine all pairs of positive integers $ k $, $ m $, for which each of the numbers $ {k^2+4m} $, $ {m^2+5k} $ is a perfect square. | \{(1, 2), (9, 22), (8, 9)\} | Suppose the pair $ (k,m) $ satisfies the conditions of the problem.
If the inequality $ m\geq k $ holds, then
$$(m+3)^2=m^2+6m+9>m^2+5m\geq m^2+5k>m^2,$$
and since $ m^2+5k $ is a square of an integer, it follows that one of the equalities $ m^2+5k=(m+1)^2 $ or $ m^2+5k=(m+2)^2 $ must hold.
If $ m^2+5k=(m+1)^2=m^2+2... | 5 |
$$
\begin{array}{l}
\text { 7. If } f(x)=x^{6}-2 \sqrt{2006} x^{5}-x^{4}+x^{3}-2 \sqrt{2007} x^{2}+2 x-\sqrt{2006} \text {, then } f(\sqrt{2006}+ \\
\sqrt{2007})= \\
\end{array}
$$ | \sqrt{2007} | 7. $\sqrt{2007}$ Hint: $f(x)=x^{4}\left(x^{2}-2 \sqrt{2006} x-1\right)+x\left(x^{2}-2 \sqrt{2007} x+1\right)+x-\sqrt{2006}-\sqrt{2007}$ $+\sqrt{2007}, f(\sqrt{2006}+\sqrt{2007})=\sqrt{2007}$. | 5.4 |
16. On the sides of the equilateral triangle $A B C$, which have length 1, three squares are constructed, as shown in the figure. What is the perimeter of the triangle $A^{\prime} B^{\prime} C^{\prime}$?
(A) $3 \sqrt{4+\sqrt{3}}$
(D) $\frac{3}{2} \sqrt{3}$
(E) $\frac{3}{2}+\frac{3}{4} \sqrt{3}$
(C) $3 \sqrt{1+2 \sqrt{3... | 3 \sqrt{4+\sqrt{3}} | (16) The correct answer is $(A)$.
Triangle $A^{\prime} B^{\prime} C^{\prime}$ is equilateral for evident reasons of symmetry (or, if you prefer, by observing the congruence of triangles $A^{\prime} B^{\prime} A, B^{\prime} C^{\prime} B, C^{\prime} A^{\prime} C$).
Let $D$ be the orthogonal projection of $A^{\prime}$ o... | 3.4 |
5. On the number line, points with integer coordinates are painted red and blue according to the following rules: a) points whose coordinate difference is 7 must be painted the same color; b) points with coordinates 20 and 14 should be painted red, and points with coordinates 71 and 143 - blue. In how many ways can all... | 8 | Answer. In eight ways.
Solution. From part a), it follows that the coloring of all points with integer coordinates is uniquely determined by the coloring of the points corresponding to the numbers $0,1,2,3,4,5$, and 6. The point $0=14-2 \cdot 7$ must be colored the same as 14, i.e., red. Similarly, the point $1=71-10 ... | 6.67 |
2. Usually, Dima leaves home at $8:10$ AM, gets into Uncle Vanya's car, who delivers him to school by a certain time. But on Thursday, Dima left home at 7:20 and ran in the opposite direction. Uncle Vanya waited for him and at $8:20$ drove after him, caught up with Dima, turned around, and delivered him to school 26 mi... | 8.5 | # Answer: 8.5 times.
## Solution:
The car was on the road for 16 minutes longer than usual, due to spending 8 minutes catching up to Dima and 8 minutes returning home. The car caught up with Dima at 8:28, and during the 68 minutes from 7:20 to 8:28, Dima ran the same distance that the car traveled in 8 minutes, i.e.,... | 8.5 |
29. [18] Compute the remainder when
$$
\sum_{k=1}^{30303} k^{k}
$$
is divided by 101 . | 29 | Answer: 29 The main idea is the following lemma:
Lemma. For any non-negative integer $n$ and prime $p, \sum_{k=n+1}^{n+p^{2}-p} k^{k} \equiv 1(\bmod p)$.
Proof. Note that $a^{b}$ depends only on the value of $a(\bmod p)$ and the value of $b(\bmod p-1)$. Since $p$ and $p-1$ are relatively prime, the Chinese Remainder Th... | 6 |
A four-meter-long rod has a $20 \mathrm{~kg}$ weight attached to one end, and $30,40,50 \mathrm{~kg}$ weights are attached at one, two, and three meters from this end, respectively. At the other end of the rod, a $60 \mathrm{~kg}$ weight is attached. Where should the rod be supported so that it remains in balance? The ... | 2 \frac{1}{2} \text{ m} | In equilibrium, the algebraic sum of static moments is $0\left(\sum P p=0\right)$. Let $x$ be the distance from the support point to the end of the rod where the $20 \mathrm{~kg}$ weight hangs, then
$$
\begin{gathered}
20 x+(x-1) 30+(x-2) 40+(x-3) 50+(x-4) 60=0 \\
20 x+30 x-30+40 x-80+50 x-150+60 x-240=0 \\
200 x=500 ... | 1.8 |
884. Find the length of the cardioid $x=2 a \cos t-a \cos 2 t, y=$ $-2 a \sin t-a \sin 2 t$. | 16a | Solution. We apply formula (1); based on the given parametric equations of the cardioid and the formula for the differential of the arc of a plane curve (Chapter 5, §), we transform the line integral in formula (1) into an ordinary integral with the variable $t$.
$$
\begin{gathered}
\dot{x}=-2 a \sin t+2 a \sin 2 t, \... | 3.75 |
How many numbers less than 100 are divisible by 2, 3, or 5? | 74 | Here there are many possibilities.
- $A_{2}=\left\lfloor\frac{100}{2}\right\rfloor=50$ numbers less than 100 divisible by 2.
- $A_{3}=\left\lfloor\frac{100}{3}\right\rfloor=33$ numbers less than 100 divisible by 3
- $A_{5}=\left\lfloor\frac{100}{5}\right\rfloor=20$ numbers less than 100 divisible by 5
- $A_{2,3}=\... | 1.33 |
8. Among the 9 natural numbers $1,2,3, \cdots, 9$, if 3 numbers are randomly selected, let $x$ be the number of pairs of adjacent numbers among these 3 numbers (for example: if the 3 numbers selected are $1,2,3$, then there are two pairs of adjacent numbers, 1,2 and 2,3, at this time the value of $x$ is 2), then the ma... | \frac{2}{3} | Solve for $x=0,1,2$, and $P(x=0)=\frac{C_{7}^{3}}{C_{9}^{3}}=\frac{5}{12}, P(x=1)=\frac{6 \times 2+5 \times 6}{C_{9}^{3}}=\frac{1}{2}, P(x=2)=\frac{7}{C_{9}^{3}}=\frac{1}{12}$. Therefore, $E(x)=\frac{1}{2}+\frac{1}{6}=\frac{2}{3}$. | 2 |
49. Given the polynomials $A=10 x^{2}-6 x y+3 y^{2}, B=a x^{2}+b x y+c y^{2}, C=15 y^{4}-36 x y^{3}+62 x^{2} y^{2}-$ $20 x^{3} y$. If these three polynomials satisfy $A \times B=C$, then $a=$ $\qquad$ ,$b=$ $\qquad$ , $c=$ . $\qquad$ | 0, -2, 5 | Reference answer: $0,-2,5$ | 3.25 |
10. [8] The polynomial $f(x)=x^{2007}+17 x^{2006}+1$ has distinct zeroes $r_{1}, \ldots, r_{2007}$. A polynomial $P$ of degree 2007 has the property that $P\left(r_{j}+\frac{1}{r_{j}}\right)=0$ for $j=1, \ldots, 2007$. Determine the value of $P(1) / P(-1)$. | \frac{289}{259} | Answer: $\frac{289}{259}$. For some constant $k$, we have
$$
P(z)=k \prod_{j=1}^{2007}\left(z-\left(r_{j}+\frac{1}{r_{j}}\right)\right) .
$$
Now writing $\omega^{3}=1$ with $\omega \neq 1$, we have $\omega^{2}+\omega=-1$. Then
$$
\begin{array}{c}
P(1) / P(-1)=\frac{k \prod_{j=1}^{2007}\left(1-\left(r_{j}+\frac{1}{r_{j... | 8.8 |
Suppose that $P(z), Q(z)$, and $R(z)$ are polynomials with real coefficients, having degrees $2$, $3$, and $6$, respectively, and constant terms $1$, $2$, and $3$, respectively. Let $N$ be the number of distinct complex numbers $z$ that satisfy the equation $P(z) \cdot Q(z)=R(z)$. What is the minimum possible value of ... | 1 | The answer cannot be $0,$ as every nonconstant polynomial has at least $1$ distinct complex root (Fundamental Theorem of Algebra). Since $P(z) \cdot Q(z)$ has degree $2 + 3 = 5,$ we conclude that $R(z) - P(z)\cdot Q(z)$ has degree $6$ and is thus nonconstant.
It now suffices to illustrate an example for which $N = 1$: ... | 1.67 |
8. [6] $A B C D$ is a convex quadrilateral such that $A B<\underline{A D}$. The diagonal $\overline{A C}$ bisects $\angle B A D$, and $m \angle A B D=130^{\circ}$. Let $E$ be a point on the interior of $\overline{A D}$, and $m \angle B A D=40^{\circ}$. Given that $B C=$ $C D=D E$, determine $m \angle A C E$ in degrees. | 55 | Answer: $\mathbf{5 5}$. First, we check that $A B C D$ is cyclic. Reflect $B$ over $\overline{A C}$ to $B^{\prime}$ on $\overline{A D}$, and note that $B^{\prime} C=C D$. Therefore, $m \angle A D C=m \angle B^{\prime} D C=m \angle C B^{\prime} D=180^{\circ}-m \angle A B^{\prime} C=180^{\circ}-m \angle C B A$. Now $m \a... | 5.8 |
9. (16 points) Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=a, a_{2}=b, 2 a_{n+2}=a_{n+1}+a_{n}$. If $\lim _{n \rightarrow \infty}\left(a_{1}+a_{2}+\right.$ $\left.\cdots+a_{n}\right)=4$, find the values of $a, b$.
保留源文本的换行和格式，直接输出翻译结果。 | $a=6, b=-3$ | 9. From $2 a_{n+2}=a_{n+1}+a_{n}$, we can get
$$
a_{n+2}-a_{n+1}=-\frac{1}{2}\left(a_{n+1}-a_{n}\right).
$$
Thus, $\left\{a_{n+1}-a_{n}\right\}$ is a geometric sequence with the first term $a_{2}-a_{1}=b-a$ and the common ratio $-\frac{1}{2}$, so $a_{n+1}-a_{n}=\left(-\frac{1}{2}\right)^{n-1}(b-a)$.
Taking $n$ as $1,2... | 5 |
$2 \cdot 109$ If $x^{4}+a x^{2}+b x+c=0$ has roots $1, 2, 3$, and another one, then $a+c$ equals
(A) 35.
(B) 24.
(C) -12.
(D) -61.
(E) -63.
(17th American High School Mathematics Examination, 1966) | -61 | [Solution] . Since the sum of the four roots of the equation is 0 (the opposite number of the coefficient of $x^{3}$), the fourth root of the equation is -6.
From $(x-1)(x-2)(x-3)(x+6)=x^{4}-25 x^{2}+60 x-36=0$, we know
$$
a+c=-25-36=-61 \text {. }
$$
Therefore, the answer is $(D)$. | 4 |
11.5. In a zoo, there live chameleons of three colors: $x$ chameleons of yellow color, 2022 chameleons of white color, and 100 chameleons of pink color. Chameleons can meet each other, but only two at a time. If two chameleons of the same color meet, they do not change their color. If two chameleons of different colors... | x \equiv 0 \pmod{3}, x \ge 0 | ## Solution.
Assume that at a certain moment in the zoo, there are: $a$ brown chameleons, $b$ white chameleons, and $c$ pink chameleons, i.e., we have the triplet $(a, b, c)$. After the meeting of any two chameleons, this triplet changes to one of the following three:
a) $(a-1, b-1, c+2)$, or
b) $(a-1, b+2, c-1)$, o... | 5 |
20. Let $a_{1}, a_{2}, \ldots$ be a sequence satisfying the condition that $a_{1}=1$ and $a_{n}=10 a_{n-1}-1$ for all $n \geq 2$. Find the minimum $n$ such that $a_{n}>10^{100}$. | 102 | 20. Ans: 102
Note that from $a_{n}=10 a_{n-1}-1$, we have
$$
a_{n}-\frac{1}{9}=10\left(a_{n-1}-\frac{1}{9}\right)
$$
for all $n \geq 2$. Thus,
$$
a_{n}-\frac{1}{9}=10^{n-1}\left(a_{1}-\frac{1}{9}\right)=10^{n-1} \frac{8}{9}
$$
for all $n \geq 1$. Therefore
$$
a_{n}=\frac{\left(1+8 \times 10^{n-1}\right)}{9} .
$$
Obse... | 8.75 |
7.A, B, C represent digits from 1 to 9, $\overline{A B C}+\overline{B C A}+\overline{C A B}+\overline{A B}+\overline{B C}+\overline{C A}+A+B+C=861$. What is the sum of all three-digit numbers $\overline{A B C}$ that satisfy this condition? $\qquad$ | 3885 | 【Answer】 3885 | 5.67 |
13.438 By mixing $2 \mathrm{~cm}^{3}$ of three substances, 16 g of the mixture was obtained. It is known that $4 \mathrm{r}$ of the second substance occupies a volume that is $0.5 \mathrm{~cm}^{3}$ larger than $4 \mathrm{r}$ of the third substance. Find the density of the third substance, given that the mass of the sec... | 4 \text{ g/cm}^3 | Solution. Let $x, y, z$ be the densities of three substances. Then, according to the condition $\left\{\begin{array}{l}2 x+2 y+2 z=16, \\ \frac{4}{y}-\frac{4}{z}=0.5, \quad \Rightarrow z=4 . \\ y=2 x\end{array}\right.$
Answer: 4 g $/ \mathrm{cm}^{3}$. | 3.67 |
Bakayev E.v.
Inside an isosceles triangle $\$ \mathrm{ABC} \$$, a point $\$ K \$$ is marked such that $\$ C K=\mathrm{AB}=\mathrm{BC} \$$ and $\angle K A C=30^{\circ}$. Find the angle $\$ A K B \$$.
# | 150^{\circ} | 
line $BC$
$ be the sum of the elements in $A$. There are exactly 613 -element subsets $A$ of $\{1, \ldots, 23\}$ with $s(A)=36$. Find the number of 3 -element subsets of $\{1, \ldots, 23\}$ with $s(A)<36$. | 855 | 8. Ans: 855
The map $\{a, b, c\} \rightarrow\{24-a, 24-b, 24-c\}$ is a bijection from the set of 3-element subsets of $\{1, \ldots, 23\}$ with $s(A)36$. The number of 3 -element subsets of $\{1, \ldots, 23\}$ is $\binom{23}{3}=1771$. Therefore, the number of 3-element subsets of $\{1, \ldots, 23\}$ with $s(A)<36$ is
$... | 8 |
19. Design a packaging box with a square base to transport four different sizes of chess sets, where the base of each chess box is also square, with side lengths of 21 cm, 12 cm, 14 cm, and 10.5 cm, respectively. To ensure that the packaging box can completely cover the base regardless of which size of chess set it is ... | 84 | 19. Solution: To ensure that each type of chess box can completely cover the bottom of the packaging box, the side length of the bottom of the packaging box should be a common multiple of the side lengths of the bottom of each chess box. Therefore, the smallest side length of the box bottom is the least common multiple... | 1.5 |
10. (12 points) There are now two animals, mice and rabbits, which grow in the following ways: every month, the number of mice doubles from the previous month, and the number of rabbits becomes the sum of the numbers from the previous two months (the second month and the first month have the same number). For example: ... | 5 | 【Answer】Solution: According to the problem:
To minimize the number of rabbits, the number of mice is at least 1;
The number of mice doubles in the second month, quadruples in the third month, and so on, until it becomes 64 times in the seventh month, which is 64.
The number of rabbits is 1 in the first month, 1 in the ... | 3.4 |
Problem 3. Determine the functions that satisfy the conditions
$$
x y \lg (x y) \leq y f(x)+x f(y) \leq f(x y) \text{, for all } x, y>0 \text{. }
$$ | f(x) = x \lg x | Solution. For $x=y=1$, from the statement we get $0 \leq 2 f(1) \leq f(1)$, hence $f(1)=0$.
For $y=1$, the relation from the statement leads to $x \lg x \leq f(x)$, for any $x>0$.
For $y=\frac{1}{x}$, from the statement we obtain $0 \leq \frac{f(x)}{x}+x f\left(\frac{1}{x}\right) \leq 0$, so $f\left(\frac{1}{x}\right... | 4 |
Task 2 - 260712 In the material issuance of a company, the keys of twelve padlocks have been mixed up due to a mishap.
Since only one of the twelve keys fits each padlock and only one padlock fits each key, which cannot be distinguished from each other externally, it must be determined which key belongs to which padlo... | 66 | From the total of twelve padlocks, exactly one fits each key. If we take one of the twelve keys and try to find out which of the twelve padlocks it fits, in the worst case, it could happen that after eleven attempts, the lock that the key fits has not yet been tried.
Since the key does belong to one of the locks, it m... | 3.67 |
1. Find the bases $z$ of all number systems in which the four-digit number $(1001)_{z}$ is divisible by the two-digit number $(41)_{z}$. | 5 | 1. Since the digit 4 appears in the notation of a two-digit number, it necessarily follows that $z \geq 5$. From the expanded notations $(1001)_z = z^3 + 1$ and $(41)_z = 4z + 1$, we are looking for exactly those natural numbers $z \geq 5$ for which the number $z^3 + 1$ is a multiple of the number $4z + 1$. Using the E... | 5 |
123. Rationalize the denominator of the fraction:
$$
\frac{1}{\sqrt[5]{16}+\sqrt[5]{8}+\sqrt[5]{4}+\sqrt[5]{2}+1}
$$ | \sqrt[5]{2}-1 | $\triangle$ Let's use the identity
$$
a^{5}-b^{5}=(a-b)\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right)
$$
(see § 1, item 1.1). Let $a=\sqrt[5]{2}, b=1$; then $a^{5}-b^{5}=2-1=1$. Therefore, the conjugate factor for the denominator of the given fraction is $\sqrt[5]{2}-1$.
Answer: $\sqrt[5]{2}-1$. | 3 |
4. At a table, there are four people: Luca, Maria, Nicola, and Paola. Each of the four always lies or never lies. Moreover, they do not like to talk about themselves, but rather about their friends; so when asked who among them always lies, their answers are:
Luca: "every girl is always honest"
Maria: "every boy is a... | 2 | 4. The answer is $\mathbf{( C )}$. Luca certainly cannot be telling the truth, since the two girls contradict each other, so they cannot both be telling the truth. If Maria were telling the truth, Nicola would be lying; therefore, it is not true that there is a truthful girl, and in particular, Maria is not, which is a... | 3 |
10. (10 points) A rectangular chessboard composed of unit squares of size $m \times n$ (where $m, n$ are positive integers not exceeding 10), has a chess piece placed in the unit square at the bottom-left corner. Two players, A and B, take turns moving the piece. The rules are: move up any number of squares, or move ri... | 90 | 【Analysis】When $M=N$, the first player A must lose; when $M$ is not equal to $N$, A can use a unique strategy to return to the starting point of the board, making B go first, and B will lose. There are 10 cases where A will lose, and $100-10=90$ cases where A will win.
【Solution】Assume A goes first, and B goes second,... | 2.2 |
Spencer was walking home from school when he realized he had forgotten his homework. He walked back to the school, picked up his homework and then walked home. The graph shows his distance from home at different times. In total, how far did he walk?
(A) $2800 \mathrm{~m}$
(B) $1000 \mathrm{~m}$
(C) $800 \mathrm{~m}$
(D... | 1400 \mathrm{~m} | Spencer starts $1000 \mathrm{~m}$ from home and walks to a point $800 \mathrm{~m}$ from home, a distance of $200 \mathrm{~m}$. He then walks to a point $1000 \mathrm{~m}$ from home, for a distance of another $200 \mathrm{~m}$. Finally, he walks home, a distance of $1000 \mathrm{~m}$. So Spencer has walked a total of $2... | 3.4 |
Excellent student Polycarp and poor student Kolya were forming the smallest 5-digit number consisting of different even digits. Polycarp formed his number correctly, while Kolya made a mistake. However, it turned out that the difference between Kolya's number and the correct answer was less than 100. What numbers did P... | 20486 | Think about what the first two digits of Polycarp's number and the last two digits of Kolya's number should be.
## Solution
If we proceed as in the previous problem, Polycarp should have formed the number 02468, but the first digit cannot be zero, so Polycarp formed the number 20468. Let's try to find Kolya's number.... | 1.5 |
5. A company's working hours are from 8:30 AM to 5:30 PM. During this period, the hour and minute hands of the clock overlap times. | 9 | 【Analysis】Method one: At 8:30, the hour hand is between 8 and 9, and the minute hand points to 6. At 9:00, the hour hand points to 9, and the minute hand points to 12, the minute hand has surpassed the hour hand. Therefore, between 8:30 and 9:00, the minute hand and the hour hand overlap once.
Similarly, between 9:00 a... | 4 |
10. Let $A$, $B$, and $C$ be three distinct points on $\odot O$, and $\angle A O B=120^{\circ}$, point $C$ lies on the minor arc $\overparen{A B}$ (point $C$ does not coincide with $A$ or $B$). If $\overrightarrow{O C}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}(\lambda, \mu \in \mathbf{R})$, then the range o... | (1,2] | 10. $(1,2]$.
Connect $O C$, intersecting $A B$ at point $D$. Let $\overrightarrow{O D}=m \overrightarrow{O C}$. Then, by the problem, $\overrightarrow{O D}=m \lambda \overrightarrow{O A}+m \mu \overrightarrow{O B}$, and $m \lambda+m \mu=1 \Rightarrow \lambda+\mu=\frac{1}{m}$.
Assume the radius of the circle is 1, and... | 9.2 |
253. The digits of a three-digit number form a geometric progression with different terms. If this number is decreased by 200, the result is a three-digit number whose digits form an arithmetic progression. Find the original three-digit number. | 842 | $\triangle$ The number of three-digit numbers with distinct digits forming a geometric progression is small, and all of them can be easily found by enumeration:
$124,421,139,931,248,842,469,964$.
We will discard the numbers 124 and 139, as they are less than 200. Subtract 200 from the remaining numbers and find the n... | 3.6 |
4.2. 12 * Given that $x, y, z$ are positive numbers, and satisfy $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(x+z)$.
保留源文本的换行和格式，翻译结果如下：
4.2. 12 * Given that $x, y, z$ are positive numbers, and satisfy $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(x+z)$. | 2 | From the arithmetic mean being greater than or equal to the geometric mean, we get
$$
(x+y)(x+z)=y z+x(x+y+z) \geqslant 2 \sqrt{y z \cdot x(x+y+z)}-2,
$$
and when $x=\sqrt{2}-1, y=z=1$, the above inequality holds with equality. Therefore, the minimum value of the expression $(x+y)(x+z)$ is 2. | 4 |
10. As shown in the figure, in isosceles $\triangle ABC$, $AB=AC$, point $P$ is on the altitude $AD$ of $BC$, and $\frac{AP}{PD}=\frac{1}{3}$. The extension of $BP$ intersects $AC$ at $E$. If $S_{\triangle ABC}=21$, then $S_{\triangle DEC}=$ $\qquad$ . | 9 | $9$ | 6 |
A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 47 | Since we are asked to find $\tan \theta$, we can find $\sin \theta$ and $\cos \theta$ separately and use their values to get $\tan \theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$... | 4.8 |
Martin wrote the pair of integers $(1011,1012)$ on the board. Then, every minute, if the pair $(a, b)$ is written on the board, he erases it and replaces it, according to his choice, with one of the pairs $(b, a),(b+1, a-1)$ or $(b-2, a+2)$, with the constraint of only writing pairs where both numbers are non-negative.... | \{(a, 2013-a) \mid a \in \mathbb{Z}, 0 \le a \le 2013\} | No operation changes the sum of the integers written on the board, so Martin can only write pairs of the form ( $a, 2013-a$ ), with $0 \leqslant a \leqslant 2013$.
Furthermore, if Martin starts from a pair $(a, b)$ for which $a \geqslant 1$, he can successively write the pairs $(b+1, a-1)$ and $(a-1, b+1)$. Thus, by in... | 7.6 |
\section*{Problem 19 - V01219}
Draw the ellipse
\[
9 x^{2}+25 y^{2}=225
\]
and determine graphically and analytically the points where the focal rays are perpendicular to each other. | \left( \pm \frac{5 \sqrt{7}}{4}, \pm \frac{9}{4} \right) | }
Rearranging the equation of the ellipse yields
\[
\frac{x^{2}}{25}+\frac{y^{2}}{9}=1
\]
i.e., the semi-axes \(a=5\) and \(b=3\). The linear eccentricity is thus \(e=\sqrt{a^{2}-b^{2}}=4\). The foci have the coordinates \(F_{1}(-4,0)\) and \(F_{2}(4,0)\).
The construction is carried out with compass and straighted... | 3.67 |
A sequence consists of the digits $122333444455555 \ldots$ such that each positive integer $n$ is repeated $n$ times, in increasing order. Find the sum of the 4501st and 4052nd digits of this sequence. | 13 | Note that $n$ contributes $n \cdot d(n)$ digits, where $d(n)$ is the number of digits of $n$. Then because $1+\cdots+99=4950$, we know that the digits of interest appear amongst copies of two digit numbers. Now for $10 \leq n \leq 99$, the number of digits in the subsequence up to the last copy of $n$ is $$1+2+3+\cdots... | 6 |
13.192. There are three vessels containing unequal amounts of liquid. To equalize these amounts, three pourings were made. First, $1 / 3$ of the liquid was poured from the first vessel into the second, then $1 / 4$ of the liquid that ended up in the second vessel was poured into the third, and finally, $1 / 10$ of the ... | 12 \text{ l}, 8 \text{ l}, \text{ and } 7 \text{ l} | Solution.
Let's form a table:
| Cо- container | Initial amount of liquid | After the first transfer | After the second transfer | After the third transfer |
| :---: | :---: | :---: | :---: | :---: |
| 1 | $x$ l | $\frac{2}{3} x$ l | $\frac{2}{3} x \pi$ | $\frac{2}{3} x+\frac{1}{10}\left(z+\frac{1}{4}\left(y+... | 3 |
8. A sequence of numbers starts from 0 and follows the rule below, continuing up to 2020. This sequence contains $\qquad$ numbers.
$$
0 \rightarrow 1 \rightarrow 3 \rightarrow 6 \rightarrow 10 \rightarrow 11 \rightarrow 13 \rightarrow 16 \rightarrow 20 \rightarrow 21 \rightarrow 23 \rightarrow 26 \text {..... }
$$ | 809 | $809$ | 6.67 |
8. The tiny but precious Dodecahedral Diamond is located 2 meters from the south wall and 3 meters from the west wall of a rectangular room whose north and south walls are 4 meters long and the east and west walls are 3 meters long. A thief drops from the ceiling inside the room and touches the floor one meter from the... | \sqrt{13} | 8. The answer is (D). Let $L$ be the point on the floor where the thief "lands", $D$ the point where the diamond is located; let $D_{E}, D_{W}, D_{N}, D_{S}$ be the symmetrical points of $D$ with respect to the east, west, north, and south walls $(E, W, N, S$ in the figure). Imagine that the thief touches the wire (to ... | 6 |
A positive integer $N$ is a palindrome if the integer obtained by reversing the sequence of digits of $N$ is equal to $N$. The year 1991 is the only year in the current century with the following 2 properties:
(a) It is a palindrome
(b) It factors as a product of a 2-digit prime palindrome and a 3-digit prime palindrom... | 4 | Solution by e_power_pi_times_i
Notice that all four-digit palindromes are divisible by $11$, so that is our two-digit prime. Because the other factor is a three-digit number, we are looking at palindromes between $1100$ and $2000$, which also means that the last digit of the three-digit number is $1$. Checking through ... | 3.75 |
Now, 4 different colors of flowers are to be planted. Each section will plant one color of flower, and adjacent sections cannot plant the same color of flower. The number of different planting methods is $\qquad$ (answer with a number).
Translate the above text into English, please keep the original text's line breaks... | 120 | 9. 120 Solution 1 As shown in the figure, number the 6 parts of the flower bed.
According to the different planting schemes for zones $1, 2, 3, 4$, consider two cases.
(1) If zones $1, 2, 3, 4$ are planted with 4 different colors of flowers, there are $A_{4}^{4}$ methods. Then arrange zone 5: if zone 5 is the same col... | 10 |
7. From a long piece of wire, isosceles triangles are formed in the following way:
A segment of length $10 \mathrm{~cm}$ and two segments of length $8 \mathrm{~cm}$ are measured and used to form an isosceles triangle, without cutting the wire. Then, the formation of isosceles triangles from the remaining wire continue... | $189 \text{ cm}^2$ | 7. a) All triangles that fit the described configuration are isosceles triangles where the leg is 2 cm shorter than the base.
If we denote the length of the base by $a$, and the length of the legs by $b$, the triangle inequality must hold: $2b > a$.
1 POINT
Thus, we have the following triangles:
| $a$ | $b$ | $2b >... | 5 |
9. (2004 Slovenia Mathematical Olympiad) Find all integer solutions to the equation $a^{b}=a b+2$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | (2,3), (1,-1), (-1,3) | 9. Solution: If $b1$, the number $a^{b}$ is not an integer.
For $a=1$, we get $1=b+2, b=-1$.
For $a=-1$, we get $(-1)^{b}=-b+2$, this equation has no solution. In fact, when $b>0$, the number $a^{b}$ is divisible by $a$, so $a$ divides 2, because $a^{b}-a b=2$, therefore, $a=-2$, $-1,1,2$.
If $a=-2$, we have $(-2)^{b}... | 9 |
6. What is the greatest area that a rectangle can have, the coordinates of whose vertices satisfy the equation $|y-x|=(y+x+1)(5-x-y)$, and whose sides are parallel to the lines $y=x$ and $y=-x$? Write the square of the found area in your answer. $\quad(12$ points $)$
# | 432 | # Solution.
Substitution: $x_{1}=x+y, y_{1}=y-x$. This substitution increases all dimensions by a factor of $\sqrt{2}$. We have $\left|y_{1}\right|=\left(x_{1}+1\right)\left(5-x_{1}\right), \quad S\left(x_{1}\right)=4\left(x_{1}-2\right)\left(x_{1}+1\right)\left(5-x_{1}\right), x_{1} \in(2 ; 5)$. $S^{\prime}\left(x_{1... | 7 |
8. (10 points) Cars A and B start from locations $A$ and $B$ respectively at the same time, heading towards each other. They meet after 3 hours. Car A then turns around and heads back to $A$, while Car B continues on. After Car A reaches $A$ and turns around to head towards $B$, it meets Car B again after half an hour.... | 7.2 | 【Analysis】After meeting, A still needs 3 hours to return to A's place. At the second meeting, the distance A is from the meeting point is equal to the distance A can travel in 2.5 hours. B takes 3.5 hours to travel this distance, so the speed ratio of A to B is 7:5. A and B meet after 3 hours, so B alone would need $3 ... | 4 |
# Problem 8. (5 points)
In triangle $ABC$, $AB=13$ and $BC=15$. A point $D$ is taken on side $AC$ such that $AD=5$ and $CD=9$. The bisector of the angle adjacent to angle $A$ intersects line $BD$ at point $E$. Find $DE$. | \frac{15}{2} | Answer: $7.5 \| 15 / 2$
## Solution:
Let's calculate the area of triangle $ABC$ using Heron's formula: $S=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{21 \cdot 6 \cdot 7 \cdot 8}=84$ (where $p$ is the semi-perimeter of the triangle, and $a, b, c$ are its sides). Then the length of the height $BH$, dropped to side $AC$, is $\frac{2S... | 6.33 |
9. Eleonora plays with a die and a (stopped) clock that initially shows 12. She rolls the die 2008 times and moves the hour hand forward by as many hours as the result. What is the probability that the hour hand will be horizontal at the end?
(A) 0
(B) $\frac{1}{2008}$
(C) $\frac{1}{1004}$
(D) $\frac{1}{12}$
(E) $\frac... | \frac{1}{6} | 9. The answer is $\mathbf{( E )}$. The hour hand is horizontal if the clock shows 3 or 9. Let $K$ be the hour indicated by the clock just before the last die roll. If $K$ is one of the 6 numbers between 3 and 8, there is exactly 1 outcome of the die out of 6 that would allow reaching 9, and there is none that would all... | 9.25 |
What is the smallest root of the equation $\frac{|x-1|}{x^{2}}=6$?
(a) $-\frac{1}{3}$
(b) $-\frac{1}{2}$
(c) $\frac{1}{3}$
(d) $\frac{1}{4}$
(e) $\frac{3}{2}$ | -\frac{1}{2} | The correct option is (b).
- Case 1: $x \geq 1$. In this case, $x-1 \geq 0$ and, therefore, $|x-1|=x-1$. The given equation takes the form $(x-1) / x^{2}=6$, or $6 x^{2}-x+1=0$. This equation has no real roots because $\Delta=(-1)^{2}-4 \times 6 \times 1=1-24$ is negative. Therefore, we have no solutions $x$ greater t... | 2 |
25. A drinks carton is formed by arranging four congruent triangles as shown. $Q P=R S=4 \mathrm{~cm}$ and $P R=$ $P S=Q R=Q S=10 \mathrm{~cm}$.
What is the volume, in $\mathrm{cm}^{3}$, of the carton?
A $\frac{16}{3} \sqrt{23}$
B $\frac{4}{3} \sqrt{2}$
C $\frac{128}{25} \sqrt{6}$
D $\frac{13}{2} \sqrt{23}$
E $\frac{8}... | \frac{16}{3} \sqrt{23} | Solution
A
The lengths in this question are given in terms of centimetres. However, for convenience, we will ignore these units in the calculations until we reach the final answer.
The drinks carton is in the shape of a pyramid. We therefore use the fact that the volume of a pyramid is given by the formula
$$
\text { ... | 2.8 |
31. Let $x, y, z$ be real numbers such that
$$
x+y+z=1 \quad \text { and } \quad x^{2}+y^{2}+z^{2}=1
$$
Let $m$ denote the minimum value of $x^{3}+y^{3}+z^{3}$. Find $9 m$. | 5 | 31. Answer: 5 .
It is clear that $|x|,|y|,|z| \leq 1$. Note that
$$
0=x+y+z-x^{2}-y^{2}-z^{2}=x(1-x)+y(1-y)+z(1-z) .
$$
Without loss of generality, assume that $z \leq 0$. Then $x=(1-y)+(-z) \geq 0$ and similarly $y \geq 0$. Since
$$
1-z^{2}=x^{2}+y^{2} \geq \frac{(x+y)^{2}}{2}=\frac{(1-z)^{2}}{2},
$$
we have $-1 / 3 ... | 5 |
A4. The value of the expression $\frac{\left(1-\sin 30^{\circ}\right)\left(1+\cos 60^{\circ}\right)}{\left(1+\cos 30^{\circ}\right)\left(1-\sin 60^{\circ}\right)}:\left(2+\tan 45^{\circ}\right)^{2}$ is equal to:
(A) $\frac{1}{9}$
(B) $\frac{1}{4}$
(C) $\frac{1}{3}$
(D) $\frac{1}{2}$
(E) 1 | \frac{1}{3} | A4. We calculate: $\frac{\left(1-\sin 30^{\circ}\right)\left(1+\cos 60^{\circ}\right)}{\left(1+\cos 30^{\circ}\right)\left(1-\sin 60^{\circ}\right)}:\left(2+\tan 45^{\circ}\right)^{2}=\frac{\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)}{\left(1+\frac{\sqrt{3}}{2}\right)\left(1-\frac{\sqrt{3}}{2}\right)}:(2+1)^{2... | 1.8 |
4. (10 points) Arrange $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}$ and the average of these 6 fractions in ascending order, then this average value is in the $\qquad$th position. | 5 | 4. (10 points) Arrange $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}$ and the average of these 6 fractions in ascending order, then this average value is in the $\qquad$th position.
【Analysis】First, find the average of these 6 fractions, then arrange them to get the result.
【Solution】So... | 3.8 |
A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?
$\textbf{(A)}\frac{1}{8} \qquad \textbf{(B)}\frac{1}{4} \qquad \textbf{(C)}\frac{3}{8} \qquad \textbf{(D)}\frac{1}{2} \qquad \textbf{(E)}\frac{3}{4}$ | \frac{3}{8} | There are $2^3 = 8$ ways to flip the coins, in order.
There are two ways to get exactly two consecutive heads: HHT and THH.
There is only one way to get three consecutive heads: HHH.
Therefore, the probability of flipping at least two consecutive heads is $\boxed{\textbf{(C)}\frac{3}{8}}$. | 1.25 |
Let $A_{1}, A_{2}, \ldots, A_{m}$ be finite sets of size 2012 and let $B_{1}, B_{2}, \ldots, B_{m}$ be finite sets of size 2013 such that $A_{i} \cap B_{j}=\emptyset$ if and only if $i=j$. Find the maximum value of $m$. | \binom{4025}{2012} | In general, we will show that if each of the sets $A_{i}$ contain $a$ elements and if each of the sets $B_{j}$ contain $b$ elements, then the maximum value for $m$ is $\binom{a+b}{a}$. Let $U$ denote the union of all the sets $A_{i}$ and $B_{j}$ and let $|U|=n$. Consider the $n$ ! orderings of the elements of $U$. Note... | 8 |
1. What is the value of $\frac{2020}{20 \times 20}$ ?
A 10.1
B 5.5
C 5.1
D 5.05
E 0.55 | 5.05 | Solution
D
Note: In the absence of a calculator, the best way to tackle this question is to first do some cancelling and then the division. There is more than one way to do this. In our method we twice divide the numerator and denominator by 10 .
$$
\frac{2020}{20 \times 20}=\frac{202}{2 \times 20}=\frac{20.2}{2 \times... | 2 |
$3 \cdot 2$ If $n$ is a real number, then the system of equations
$$
\left\{\begin{array}{l}
n x+y=1, \\
n y+z=1, \\
x+n z=1 .
\end{array}\right.
$$
has no solution if and only if the value of $n$ is
(A) -1 .
(B) 0 .
(C) 1 .
(D) 0 or 1 .
(E) $\frac{1}{2}$.
(24th American High School Mathematics Examination, 1973) | -1 | [Solution] If $n \neq -1$, using the method of addition, subtraction, or substitution, we can solve to get
$$
(x, y, z)=\left(\frac{1}{n+1}, \frac{1}{n+1}, \frac{1}{n+1}\right) \text {, }
$$
This is the solution to the known system of equations.
If $n=-1$, adding the two sides of the three equations yields $0=3$, a co... | 3 |
The numbers $1, 2, 3, 4, 5, 6, 7,$ and $8$ are randomly written on the faces of a regular [octahedron](https://artofproblemsolving.com/wiki/index.php/Octahedron) so that each face contains a different number. The [probability](https://artofproblemsolving.com/wiki/index.php/Probability) that no two consecutive numbers,... | 085 | Choose one face of the octahedron randomly and label it with $1$. There are three faces adjacent to this one, which we will call A-faces. There are three faces adjacent to two of the A-faces, which we will call B-faces, and one face adjacent to the three B-faces, which we will call the C-face.
Clearly, the labels for... | 5.25 |
Example 2 Let $A, B, C$ be the three interior angles of a triangle, find the maximum value of $\sin A+\sin B+\sin C$.
untranslated text remains the same as requested. However, if you need the entire text to be translated, please let me know! | \frac{3 \sqrt{3}}{2} | Assuming $A \leqslant B \leqslant C$, then $A \leqslant 60^{\circ} \leqslant C$, so by the lemma, we have
$$\sin A+\sin B+\sin C \leqslant \sin 60^{\circ}+\sin \left(A+C-60^{\circ}\right)+\sin B$$
Assuming again that $A+C-60^{\circ} \leqslant B$, then because $\left(A+C-60^{\circ}\right)+B=A+B+C-$ $60^{\circ}=120^{\ci... | 2.75 |
Example 4 (14th Irish Mathematical Olympiad, 2001) Find the smallest positive integer $a$ such that there exists a positive odd integer $n$ satisfying $2001 \mid\left(55^{n}+a \cdot 32^{n}\right)$. | 436 | Since $2001=87 \times 23$, it follows from the problem statement that there exists a positive odd number $n$ such that $87 \mid\left(55^{n}+a \cdot 32^{n}\right)$ and $23 \mid\left(55^{n}+a \cdot 32^{n}\right)$. Therefore,
$$
0 \equiv 55^{n}+a \cdot 32^{n} \equiv(-32)^{n}+a \cdot 32^{n}=32^{n}(a-1)(\bmod 87) \text {, h... | 6 |
142. The quality control department checks 900 parts for standardization. The probability that a part is standard is 0.9. Find, with a probability of 0.95, the boundaries within which the number $m$ of standard parts among those checked will be contained. | $792 \leqslant m \leqslant 828$ | The solution is as follows. Given $n=900, p=0.9, q=0.1$. Therefore,
$$
2 \Phi(\varepsilon \sqrt{900 /(0.9 \cdot 0.1)})=0.95, \text { or } \Phi(100 \varepsilon)=0.475
$$
From the table in Appendix 2, we find $\Phi(1.96)=0.475$, so $100 \varepsilon=1.96$. Hence, $\varepsilon \approx 0.02$.
Thus, with a probability of ... | 1.75 |
5. The real numbers $x$ and $y$ satisfy the equation $(6 x-5 y)^{4}+(4 y-3)^{6}=0$. What is the value of $x+y$?
(A) 1
(B) $10 / 7$
(C) $4 / 3$
(D) $13 / 9$
(E) $11 / 8$ | \frac{11}{8} | (5) The correct answer is $(E)$.
Since the expression is the sum of two squares, it vanishes if and only if both quantities are zero. It follows that $y=\frac{3}{4}$ and $6 x-\frac{15}{4}=0$, i.e., $x=\frac{5}{8}$. Therefore, $x+y=\frac{3}{4}+\frac{5}{8}=\frac{11}{8}$. Question proposed by Paolo Francini. | 3.4 |
1. Real numbers $a, b, c$ are such that $a+1 / b=9, b+1 / c=10$, $c+1 / a=11$. Find the value of the expression $a b c+1 /(a b c)$. | 960 | Answer: 960.
Sketch of the solution. By multiplying the equations, expanding the brackets, and grouping, we get: $a b c + 1/(a b c) + a + 1/b + b + 1/c + c + 1/a = 990$. From this, $a b c + 1/(a b c) = 990 - 9 - 10 - 11 = 960$. | 7 |
Penta chooses 5 of the vertices of a unit cube. What is the maximum possible volume of the figure whose vertices are the 5 chosen points? | \frac{1}{2} | Label the vertices of the cube $A, B, C, D, E, F, G, H$, such that $A B C D$ is the top face of the cube, $E$ is directly below $A, F$ is directly below $B, G$ is directly below $C$, and $H$ is directly below $D$. We can obtain a volume of $\frac{1}{2}$ by taking the vertices $A, B, C, F$, and $H$. To compute the volum... | 3 |
3. Let the complex numbers be
$$
z_{1}=\sin \alpha+2 \mathrm{i}, z_{2}=1+\cos \alpha \cdot \mathrm{i}(\alpha \in \mathbf{R}) \text {. }
$$
Then the minimum value of $f=\frac{13-\left|z_{1}+\mathrm{i} z_{2}\right|^{2}}{\left|z_{1}-\mathrm{i} z_{2}\right|}$ is . $\qquad$ | 2 | 3. 2 .
$$
\begin{array}{l}
\text { Let }\left|z_{1}-\mathrm{i} z_{2}\right|=\sqrt{(\sin \alpha+\cos \alpha)^{2}+1^{2}} \\
=\sqrt{2+\sin 2 \alpha}=t .
\end{array}
$$
Then $t \in[1, \sqrt{3}]$, and at this time we have
$$
\begin{array}{l}
\left|z_{1}+\mathrm{i} z_{2}\right|^{2}=(\sin \alpha-\cos \alpha)^{2}+3^{2} \\
=10... | 3 |
10. [8] A real number $x$ is chosen uniformly at random from the interval $[0,1000]$. Find the probability that
$$
\left\lfloor\frac{\left\lfloor\frac{x}{2.5}\right\rfloor}{2.5}\right\rfloor=\left\lfloor\frac{x}{6.25}\right\rfloor .
$$ | \frac{9}{10} | Answer:
$\frac{9}{10}$
Solution: Let $y=\frac{x}{2.5}$, so $y$ is chosen uniformly at random from [0,400]. Then we need
$$
\left\lfloor\frac{\lfloor y\rfloor}{2.5}\right\rfloor=\left\lfloor\frac{y}{2.5}\right\rfloor .
$$
Let $y=5 a+b$, where $0 \leq b<5$ and $a$ is an integer. Then
$$
\left\lfloor\frac{\lfloor y\rfloo... | 8 |
6. The sum of three fractions is $\frac{83}{72}$, where their numerators are in the ratio $5: 7: 1$. The denominator of the third fraction is in the ratio $1: 4$ to the denominator of the first fraction, and the denominator of the second fraction to the denominator of the third fraction is in the ratio $3: 2$. Determin... | \frac{5}{24}, \frac{7}{9}, \frac{1}{6} | 6. Let $\frac{x}{a}, \frac{y}{b}$, and $\frac{z}{c}$ be the sought fractions.
Then $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=\frac{83}{72}$ and $x: y: z=5: 7: 1$,
so it follows that $x=5k, y=7k, z=k$.
1 POINT
Furthermore, from $c: a=1: 4$ it follows that $a=4c$.
1 POINT
From $b: c=3: 2$ it follows that $b=\frac{3}{2}c... | 4 |
In the quadrilateral $A B C D$, the points $K$ and $L$ are the closer to $A$ trisection points of sides $A B$ and $A D$, respectively, and the points $N$ and $M$ are the closer to $C$ trisection points of sides $C B$ and $C D$, respectively. Determine the ratio of the areas of quadrilaterals $A B C D$ and $K L M N$. | \frac{9}{4} | We need to distinguish two cases based on whether the quadrilateral $A B C D$ is convex or concave.
1988-11-377-1.eps
1. Figure
If the quadrilateral $A B C D$ is convex, then the area of the quadrilateral $K L M N$ can be calculated by subtracting the areas of the triangles $A L K, D M L, C N M$, and $B K N$ from th... | 4 |
If $y=a+\frac{b}{x}$, where $a$ and $b$ are constants, and if $y=1$ when $x=-1$, and $y=5$ when $x=-5$, then $a+b$ equals:
$\textbf{(A)}\ -1 \qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | 11 | Plugging in the x and y values, we obtain the following system of equations:
\[\begin {cases} 1 = a - b \\ 5 = a - \frac{b}{5} \end {cases}\]
We can then subtract the equations to obtain the equation $4 = 0.8b$, which works out to $b = 5$.
Plugging that in to the original system of equations:
\[\begin {cases} 1 = a - 5... | 1 |
(2) Indicate the relationship between the graph of the function $y=|x|$ and the graph of the function $y=\sqrt{x^{2}-2|x|+1}$. | $y=||x|-1|$ | (2) The function $y=\sqrt{x^{2}-2|x|+1}$ is equivalent to $y=|| x|-1|$. Its graph can be obtained by first shifting the graph of the function $y=|x|$ down by 1 unit, and then reflecting the part of the resulting graph that is below the $x$-axis about the $x$-axis. | 2 |
5. According to national regulations, only adults who are 18 years old or older but not over 70 years old are eligible to apply for a motor vehicle driver's license. Li Ming, a sixth-grade student, said: "My dad has a driver's license, and the product of his age and his birth month and day is 2975." How old is Li Ming'... | 35 | 5. According to national regulations, only adults who are 18 years old or older and no more than 70 years old are eligible to apply for a motor vehicle driver's license. Li Ming, a sixth-grade elementary school student, said: "My dad has a driver's license, and the product of his age and his birth month and day is 2975... | 5 |
11 Limit $\lim _{n \rightarrow \infty}\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right) \cdots\left(1-\frac{1}{n^{2}}\right)=$ $\qquad$ | \frac{1}{2} | $11 \frac{1}{2}$ Hint: Since
$$
\begin{aligned}
\prod_{k=2}^{n}\left(1-\frac{1}{k^{2}}\right) & =\prod_{k=2}^{n}\left(1-\frac{1}{k}\right) \prod_{k=2}^{n}\left(1+\frac{1}{k}\right) \\
& =\prod_{k=2}^{n} \frac{k-1}{k} \prod_{k=2}^{n} \frac{k+1}{k}=\frac{1}{n} \cdot \frac{n+1}{2}
\end{aligned}
$$
Therefore, the required... | 3 |
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