Problems and solutions

Answer:

$4022$

As the sum is odd, one of the primes must be even and therefore equal to 2. Thus $2013=2+2011$ is the only candidate which happens to work as 2011 is indeed a prime. The result then is $2\cdot 2011=4022$.

Answer:

$\frac{2}{3}\pi$

Denote the areas of the three parts as in the figure.

Combining the obvious $A=C$ (symmetry) with the given $B=A+C$ yields $B=2A$. Thus the area of the intersection is equal to two thirds of the area of the left circle, that is $\frac{2}{3}\pi$.

Answer:

1

First we place the yellow pins and our only option is to place them on the hypotenuse of the triangle. By the same argument we now have an only option to place the four red pins (again on the hypotenuse of the triangle formed by empty spots). Continuing with this argument we see that there is a unique arrangement which satisfies the given conditions.

Answer:

$3558$

Since $600=2^3\cdot 3\cdot 5^2$, we may only use the digits $1$, $2$, $3$, $4$, $5$, $6$, $8$. Using the digit $1$ only increases the number, so we will do better without it. Also, we must use two fives as it is the only way to get $5^2$. The product of the remaining digits then must be $24$, thus one digit is not enough and from the pairs $3\cdot 8$ and $4\cdot 6$, the former contains the smallest digit and leads to the solution, which is $3558$.

Answer:

$33$

We multiply the two equations and obtain

Answer:

$513315$

We start by analyzing the last condition. The only four-digit power of $11$ is $1331$, therefore the sought number is of the form $\overline{a1331a}$ for some digit $a$. Since it is divisible by $9$, so must be also its sum of digits $2a+8$. Since $2a+8$ is even and less than or equal to $26$, the only choice is $2a+8=18$, i.e. $a=5$ and the answer is $513315$.

Answer:

$1:3$

Since $C$ trisects the arc $\mathit{AB}$, we can draw point $E\in k$ such that $A$, $C$, $E$, and $B$ (in this order) are consecutive vertices of a regular hexagon.

Now if $S$ is the center of $k$, triangle $\mathit{ASC}$ is equilateral with $D$ the midpoint of $\mathit{SA}$ (altitudes coincide with medians in equilateral triangles). Then we easily obtain $\mathit{AD}:\mathit{DB}=1:3$.

Answer:

$84$

Note that Jim ate at most $3\cdot 14=42$ chips while Tim ate at least $2\cdot 21=42$. Since they both started with the same amount of chocolate chips, it must have been $42$, and the answer is $42+42=84$.

Answer:

$16$

From each of the letters N, á, b, o we can continue reading in exactly two ways. Therefore, we can read the word Náboj in $2^4=16$ ways.

Answer:

$6$

If two truth-tellers sat next to one another, the one on the left would not call his right neighbour a liar. For the same reason two liars cannot sit side by side. It remains to check that an arrangement in which truth-tellers and liars alternate satisfies the condition, making the number $6$ our answer.

Answer:

$0$

For a coloring to be invariant under the $90^{\circ }$ rotation, its four corner tiles must have the same color and the same holds for the remaining four non-central tiles. Thus we either need eight tiles of one color or two quadruplets of tiles with the same color. Since we have neither of that available, the answer is 0.

Answer:

$48\%=\frac{12}{25}$

Denote by $D$ the total number of married couples on the island. Then the total numbers of males and females are $\frac{5}{2}D$, $\frac{5}{3}D$, respectively. Thus, we have $\frac{5}{2}D+\frac{5}{3}D=\frac{25}{6}D$ inhabitants from which $2D$ are married. The fraction of married islanders is then

Answer:

$14\sqrt{3}=\frac{42}{\sqrt{3}}\text{cm}$

Any equilateral triangle lying between a pair of parallel lines can be suitably enlarged so that two of its vertices lie on opposite borderlines. From all such equilateral triangles the longest side length is obtained from one with the whole edge on one of the borderlines.

Then if the width of the strip is $21\text{cm}$, the largest fitting equilateral triangle has altitude $21\text{cm}$. The side-length is then $\frac{21\text{cm}}{\sin 60^{\circ }}=14\sqrt{3}\text{cm}$. Since $14\sqrt{3}<14\cdot 2<29.7$, this triangle fits inside the rectangle.

Answer:

$10$

The fold-lines are captured in the following figure.

Altogether, we see 10 squares: the whole paper, the square connecting the midpoints of the edges and in each of these two, we see four smaller squares.

Answer:

$3^5=243$

Observe that each pentagon must have the center of the figure in its interior. We have three choices for each of the five sides (outer, middle or inner line), thus there are $3^5=243$ pentagons.

Answer:

$1808$

Denote the numbers by $a$, $b$. Since adding a zero digit is the same as multiplying by ten, we only have to solve the following system of equations:

Subtracting them we learn $a=205$. Plugging it in the first one we get $b=1808$, which is the answer.

Answer:

$3.5$

Since $3^2+5^2<7^2$, the angle opposite the $7$-side is obtuse. Thus the triangle can be covered with a circle of radius $3.5$. On the other hand, no smaller circle can cover the longest side, so $3.5$ is our final answer.

Answer:

$12$

If we denote the numbers of each girl’s lollipops by $L$, $M$, $N$, and $S$, respectively, we know that $L+M\ge 41$, $L+N\ge 41$, and $L+S\ge 41$. Summing these inequalities we get $2L+(L+M+N+S)\ge 123$. Since $L+M+N+S=100$, we learn $2L\ge 23$, or $L\ge 12$ ($L$ is a positive integer).

On the other hand, the distribution $L=12$, $M=N=29$, $S=30$ satisfies the conditions of the problem.

Answer:

$\frac{5}{8}=0.625$

Denote by $S$ the intersection of the diagonals and by $h$ the length of the $B$-altitude in $△\mathit{ABC}$. The area of $\mathit{ABCD}$ can be found as $\mathit{AC}\cdot h$ which gives $h=80∕16=5$. The sought-after value is then $\sin \mathit{\angle ASB}=\sin \mathit{\angle CSB}=\frac{h}{\mathit{BS}}=\frac{5}{8}$.

Answer:

${(-1)}^{-4}+{(-3)}^{-2}=\frac{10}{9}$

In order to obtain a positive number, we need to take even exponents. Also, since ${(-2)}^{-4}={(-4)}^{-2}<{(-1)}^{-2}$, we will take the exponents $-2$ and $-4$. Further, as a negative power is decreasing, we will take $-1$ and $-3$ as bases. Out of the remaining two options, ${(-1)}^{-4}+{(-3)}^{-2}=\frac{10}{9}$ has greater value.

Answer:

$\frac{11}{18}$

For an angle to form a graph of a function, it must not contain two points with the same $x$-coordinate. We may assume that the vertex of the angle is at point $(0,0)$. We will rotate the angle by one full circle (clockwise) and determine when it is a graph of a function. We choose to start in a position in which one leg coincides with the positive ray of the $y$ axis and the other lies to the right. Observing the breaking point after $70^{\circ }$ rotation, we see that the answer is $\frac{110}{180}=\frac{11}{18}$.

Answer:

$0.04=\frac{1}{25}$

We are in fact asking about numbers from $1$ to $100$ which, when repeatedly multiplied by $2$, become a multiple of $100$. Such number must be a multiple of 25 and all four of these ($25$, $50$, $75$, $100$) work. The answer is $\frac{4}{100}=0.04=\frac{1}{25}$.

Answer:

$1510$

Odd numbers can be written as $2k+1={(k+1)}^2-k^2$ and multiples of four as $4k={(k+1)}^2-{(k-1)}^2$. Numbers of the form $4k+2$ are not differential since $a^2-b^2=(a+b)(a-b)$ is a product of two positive integers of the same parity, therefore either odd or multiple of $4$. There are $503$ numbers of the form $4k+2$ from $1$ to $2013$, hence the answer is $1510$.

Answer:

$16$

We look at the internal angles of the polygons by vertex $A$. One of them is $90^{\circ }$, another is $120^{\circ }$, and since the three angles add up to $360^{\circ }$ ($A$ is in the interior), the remaining angle is $150^{\circ }$, which corresponds to a regular dodecagon (12 vertices). Since each pair of polygons shares a side, the perimeter of $M$ is $4+6+12-(3\cdot 2)=16$.

Answer:

$2^{-50}$

Imagine Mike writes the numbers in pairs: First he randomly chooses two available numbers and then their order. The smaller of them going first thus has probability $\frac{1}{2}$ (regardless of the chosen numbers) and thus for 50 pairs the probability is ${\left(\frac{1}{2}\right)}^{50}=2^{-50}$.

Answer:

$23$

Since we can add only red or white paint, the resulting paint has to contain exactly three tins of blue. Blue is not contained in pink paint, thus the ratio $1:2$ in cyan implies that there need to be exactly $9$ tins of cyan paint ($3$ blue and $6$ white). Finally, from the ratio $2:1$ in the desired paint we infer that it has to be mixed from $27$ tins ($9$ of which are mixed to produce cyan and $18$ to pink). As we have already mixed four tins, we have to add $27-4=23$ tins to complete the paint.

Answer:

$\frac{9}{10}$

The magician takes a white rabbit before a gray one with the probability of $\frac{3}{4}$, hence the hat contains three times as many white rabbits as the gray ones. In like manner, there are three times as many gray rabbits as the black ones. Therefore the hat is inhabited by nine times as many white rabbits as the black ones and the sought-after probability is $\frac{9}{10}$.

Answer:

$37$

We add $a+b$ to both sides of the equation and get $50(a+2b)=2013+(a+b)$. The left-hand side of the equation is divisible by $50$, and so must be the right-hand side. It follows that $a+b=50k-13$ for some $k\in \mathbb{N}$.

If $a+b\ge 87$, then $49a+99b>49a+49b\ge 49\cdot 87>2013$, which is not possible. Thus the only possibility is $a+b=37$.

Answer:

$6$

Point $D$ belongs to the circle with diameter $\mathit{ZY}$, the center of which we denote by $O$. This circle has radius $1$ and by Pythagorean Theorem we have $\mathit{PO}=\sqrt{3^2+4^2}=5$. Hence from triangle inequality we obtain $\mathit{PD}\le \mathit{PO}+\mathit{OD}=6$. The equality is attained if $P$, $O$, $D$ lie on one line, so the maximum distance is $6$.

Answer:

$11$, $53$

Denote by $K$ the number of boxes that contain $x$ apples ($K\le 20$). In the remaining $20-K$ boxes we have four apples and hence

We have $K\le 20$, hence the only possible choices are $K=1$ and $K=7$. From $K=1$ we easily obtain $x=53$, and $K=7$ corresponds to $x=11$.

Answer:

$\sqrt{\frac{2015}{2011}}$

From the given equality we obtain $a^2+b^2=2013\mathit{ab}$. Using this identity we have

From $a>b>0$ we know that the expression $\frac{a+b}{a-b}$ is positive and hence using the two relations above yields

Answer:

$144=3\cdot 2\cdot 4!$

Denote values in the individual squares as in the picture.

It follows from the problem statement that $A+B+C+D+E+F+G=1+2+3+4+5+6+7=28$ and $A+B=D+E+F+G=\frac{1}{2}(28-C)$. Therefore $C$ must be even.

If $C=2$, then $A+B=13=6+7$, which gives two alternatives how to fill in the squares in the left column. For each of them we can reorder the numbers in the bottom row in $4!$ ways. If $C=4$, then $A+B=12=5+7$ ($6+6$ is not legitimate since identical numbers are disallowed), and if $C=6$, then $A+B=11=4+7$ ($5+6$ is also illegitimate since $C=6$ already), which gives in both cases $2\cdot 4!$ possibilities of finishing the table. Hence we obtain altogether $3\cdot 2\cdot 4!=144$ possibilities.

Answer:

$12$

Pythagorean Theorem for right triangles $\mathit{ADC}$ and $\mathit{ADB}$ gives $C{D}^2=A{C}^2-A{D}^2$ and $B{D}^2=A{B}^2-A{D}^2$. After subtracting, we obtain

Since $D$ lies on the segment $\mathit{BC}$, we also have

hence $\mathit{CD}-\mathit{BD}=12$.

Answer:

$6$ minutes

We denote the speed of the tram by $t$, the speed of the pedestrian by $c$ and the distance between two trams by $d$. From the given information we deduce that

By summing up the equations and dividing by $2$ we obtain

It follows that the tram travels the distance $d$ in 6 minutes, which implies that the interval between two trams is 6 minutes.

Answer:

$148=\left(\genfrac{}{}{0.0pt}{}{11}{3}\right)-17$

We can pick a point triplet out of the given eleven ones in

ways. It remains to compute the number of point triplets lying on a line and to subtract the resulting number from the one above. There are $4+1+4=9$ horizontal triplets, $3+3+1+1=8$ transversal triplets, hence the number of triangles is $165-17=148$.

Answer:

$6^{10}\cdot {(10!)}^3$

The way of eating the sweets can be uniquely determined in the following way: For each row we choose the order of its sweets in which Thomas will eat them and also choose the order of the rows in which he will switch between them.

For each row, we have $10!$ possibilities to choose the order of sweets, which gives us ${(10!)}^3$ possibilities for three rows.

Now we need to count the number of the order of rows. We know that if there is the same number of sweets in all rows, then Thomas can choose any of them, then he needs to choose one of the remaining two (choosing the same row would result in having two sweets less in this row) and then he has to choose the remaining one. After three steps he has again the same number of sweets in all three rows. It follows that it is enough to choose ten times the order of the three rows, which can be done in ${(3!)}^{10}=6^{10}$ different ways.

Thomas can eat the box of chocolates in $6^{10}\cdot {(10!)}^3$ different ways.

Answer:

$5$

Any doubling number with six digits can be written as $1001\cdot k$ for some positive integer $k$. Conversely, for any three digit positive integer $k$ the formula $1001\cdot k$ gives us a six-digit doubling number.

As $2013=3\cdot 11\cdot 61$ and $1001$ is divisible by $11$, but not by $3$ or $61$, it follows that the six digit integer is doubling if and only if the corresponding three digit $k$ in $1001\cdot k$ is a multiple of $3\cdot 61=183$. We have exactly five three digit $k$ divisible by $183$ and hence we have five six-digit doubling numbers that are divisible by $2013$.

Answer:

$\frac{5}{16}=\frac{\left(\genfrac{}{}{0.0pt}{}{6}{3}\right)}{2^6}$

Let us count the number of paths from the upper left to the lower right corner and the probability that the robot will go along one such path. Each path consists of three steps down and three steps right which gives us $\left(\genfrac{}{}{0.0pt}{}{6}{3}\right)$ possible paths. The probability that the robot will follow this path is always $2^{-6}$ as in each of the steps he must choose the correct direction. The overall probability is thus $2^{-6}\cdot \left(\genfrac{}{}{0.0pt}{}{6}{3}\right)$.

Answer:

$\frac{70}{37}$

It is easy to see (as the sum of the digits is divisible by three) that both numerator and denominator are divisible by three. By division we get $212121210=3\cdot 70707070$ and $112121211=3\cdot 37373737$. Now observe that $70707070=70\cdot 1010101$ and $37373737=37\cdot 1010101$, which implies that the basic form of our fraction is $\frac{70}{37}$.

Answer:

$13$

It is sufficient to find when is $\mathit{DX}+\mathit{XC}$ an integer. Let us consider $C^{\prime }$, $D^{\prime }$ such that $A$, $B$ are midpoints of $D{D}^{\prime }$, $C{C}^{\prime }$ respectively. Then $\mathit{DX}+\mathit{XC}=\mathit{DX}+X{C}^{\prime }$. The right-hand side attains its minimum when $X$ is the midpoint of $\mathit{AB}$ and maximum when $X$ is equal to $A$ or $B$.

Using Pythagorean Theorem we compute $D{C}^{\prime }=\sqrt{30^2+40^2}=50$ and $A{C}^{\prime }=\sqrt{30^2+20^2}=\sqrt{1300}$, from which we have $36<A{C}^{\prime }<37$. So, if $X$ moves from $A$ to $B$, then $\mathit{DX}+\mathit{XC}$ first decreases from a number a little larger than $20+36=56$ ($X=A$) to $50$ ($X$ is the midpoint of $\mathit{AB}$) and then increases to a number a little larger than $56$ ($X=B$). Hence, it is integer for $6+1+6=13$ positions of the point $X$.

Answer:

in the reverse order (eventually complemented by a circular shift), i.e. $r_{11},r_{10},r_9,\dots ,r_1$ or $r_{10},r_9,\dots ,r_1,r_{11}$, …, $r_1,r_{11},r_{10},\dots ,r_2$

Observe that the table is symmetric with respect to the other diagonal. To make it symmetric with respect to the marked diagonal it is sufficient to reflect it over the horizontal axis.

Remark: For this table there are no solutions except the eleven mentioned above.

Answer:

$254$

Denoting $A_n=a_n^3$ our problem is equivalent to finding the smallest integer $k\ge 2$ such that $A_2\cdot A_3\cdots A_k>4^3=64$. We have

and hence

It remains to solve the inequality

for integer $k$. Substracting $2k$ from both sides and multiplying them by $1∕k$ yields equivalent inequality $k+\frac{1}{k}>254$, the smallest integer solution of which is $k=254$.

Answer:

$20$

Suppose that between the slices with consecutive numbers we have exactly $k-1$ other slices, that is by moving by $k$ slices we get from slice with number $1$ to slice with number $2$, from slice $2$ to slice $3$ and so on. All these movements have to be in the same direction, otherwise we would get from slice $i$ to the previous slice $i-1$ and not to $i+1$. From the slice $n$ we get in this way necessarily to slice number $1$, because any other $i$ has in distance $k$ slices with numbers $i+1$ and $i-1$. By moving with the step $k$ slices we will finally pass through the whole pizza. Hence there is $s$ such that moving $s$ times by $k$ slices we will end exactly at the neighboring slice. Thus we get

which implies that $7-(-13)=20$ is divisible by $n$. There is a slice with number $17$ and hence $n\ge 17$ which implies that the only solution is $n=20$.

Answer:

$144$

We denote by $r$ and $s$ the number of rows and columns in the lecture hall. From the instructions we have $\mathit{rs}=11r+3s+2$, which is equivalent to

The numbers in the brackets are either $5$ and $7$, or $1$ and $35$ in some order. By writing down all four possibilities we can find that the smallest number $\mathit{rs}$ corresponds to the case $r-3=5$, $s-11=7$, when $\mathit{rs}=8\cdot 18=144$. It remains to show that we can really put the students in the lecture hall as requested, which can be seen in the picture.

Answer:

$9\sqrt{3}=\frac{27}{\sqrt{3}}$

Denote by $[\mathit{XY}Z]$ the area of triangle $\mathit{XY}Z$.

Denote by $M$ the intersection of $\mathit{TL}$ with $k$ ($M\ne T$). Since $T$ is the center of homothety with factor $\frac{4}{3}$ sending $k$ to $l$, points $L$ and $M$ correspond and thus $\mathit{TL}=\frac{4}{3}\mathit{TM}$ and $[\mathit{TKL}]=\frac{4}{3}[\mathit{TKM}]$ (the triangles share an altitude from $K$). Hence it suffices to maximize the area of $\mathit{TKM}$ inscribed in circle $k$ with radius $3$. From all such triangles the equilateral one has the greatest area, namely

Hence the area of the corresponding triangle $\mathit{TKL}$ is

Answer:

$32$

Bob can double the smallest distance between any two numbers in each step by removing every other element. In this way he wins at least $2^5=32$ crowns. Alice can halve the difference of the largest and the smallest number in each step by removing upper half (or lower half) of all remaining numbers. This ensures that she loses at most $1024∕2^5=32$ crowns. Therefore if they both play in the best way, Bob will get $32$ crowns.

Answer:

$\frac{6}{5}=1.2$-times

We denote by $100m$ the original average IQ of students at Matfyz and $100v$ the original average IQ of students at SOU. Further, we denote by $p$ the average IQ of students who changed from Matfyz to SOU. The average IQ at Matfyz increased by $2\%$ and thus the average IQ of remaining students at Matfyz is $102m$. From the ratio $5:1$ of remaining students and from the original average $100m$ we obtain the equation $100m=\frac{5}{6}\cdot 102m+\frac{1}{6}p$, which is the same as $p=90m$. Analogously the average IQ at SOU was originally $100v$ and with the new students it has risen to $102v$. From the ratio $3:1$ we have the equation $102v=\frac{3}{4}\cdot 100v+\frac{1}{4}p$, which implies $p=108v$. Altogether this gives us

Answer:

$\frac{\pi }{14}$

Let us focus on points $A_1$, $A_4$, $A_{14}$ and $A_{11}$.

Since $11=4+7$, the line segment $A_4{A}_{11}$ is a diameter of the circle $k$. On the other hand we have $4-1=3=14-11$, so $A_1{A}_4{A}_{11}{A}_{14}$ is an isosceles trapezoid and its bases $A_4{A}_{11}$ and $A_1{A}_{14}$ are parallel. The area of the triangle $A_1{A}_4{A}_{14}$ is therefore the same as the area of the triangle $A_{1}O{A}_{14}$, where $O$ is the center of $k$. The area to compute is then equal to the area of the sector $A_{1}O{A}_{14}$, i.e. one fourteenth of the area of the disc.

Answer:

Bill, $\left(\genfrac{}{}{0.0pt}{}{12}{6}\right)=924$ subsets more

Consider an arbitrary 12-element subset $B$ and assume there exists an integer $i$ such that $B$ contains exactly one of the numbers $2i-1$, $2i$. Let us take the smallest such $i$ and construct a 12-element set $f(B)$, which contains the same elements as $B$ with the only exception: It contains the other element from the couple $2i-1$, $2i$.

It is easy to see that $f(f(B))=B$ and that by applying $f$ on a Bill’s subset we obtain Carol’s subset and vice versa. So, $f$ is a bijection between Bill’s and Carol’s subsets, if we consider just those subsets for which there exists an $i$ from the previous paragraph. It remains to count the remaining subsets, for which such $i$ does not exist.

In each of the remaining subsets, there must be exactly six odd numbers and six even numbers which are successors of the odd ones. Sum of the elements of such subsets is always even (hence they belong to Bill’s list) and the number of such subsets is $\left(\genfrac{}{}{0.0pt}{}{12}{6}\right)$.

Answer:

$1594$

Let us assume without loss of generality that $b+c=800$. At least one of the numbers $a$, $b+c-a=800-a$, $a+b+c=800+a$ is divisible by three, so it can be prime number only if it is equal to three. Since $800+a>800$, there are only two possibilities, $a=3$ or $800-a=3$.

If $a=3$, then we have $3+(b-c)\ge 2$ and simultaneously $3-(b-c)\ge 2$, so $|b-c|\le 1$. This is impossible, since $b+c=800$.

So, we know that $800-a=3$, i.e. $a=797$. The largest among Jack’s prime numbers is $a+b+c=797+800=1597$. Since $b+c=800$, none of the Jack’s primes is even. Therefore, the smallest number is $800-a=3$. The difference is then $1597-3=1594$.

Let us remark that such numbers exist, we can take $a=797$, $b=223$, $c=577$.

Answer:

$\frac{1}{1007}$

Imagine the bar is in one piece. Helen marked it randomly with two stains, whereas Alex marked it randomly with $2012$ breaks. The bar thus contains altogether $2014$ marks, two of which are stains. The total number of possibilities of which marks can be stains is $\left(\genfrac{}{}{0.0pt}{}{2014}{2}\right)=1007\cdot 2013$. The stains are on the same piece (or shard) if and only if they are neighbouring marks, which occurs in $2013$ cases. The resultant probability is

Answer:

$3456=2^5\cdot 5!-2^4\cdot 4!$

Since $0+1+\cdots +9=9\cdot 5$, the numbers under investigation must be divisible by nine, hence even by $99999$. Let us denote $A$ and $B$ the numbers consisting of the former and the latter five-tuple of the investigated number respectively. We have

Given that $A$, $B$ are five-digit positive integers less than $99999$, we observe

Therefrom we obtain the necessary and sufficient condition on $A$, $B$ for divisibility of the investigated ten-digit number by $99999$: For $i=1,\dots ,5$, the $i$-th digit of $B$ is a complement of $i$-th digit of $A$ into nine. We couple the available digits into five pairs

We know that these pairs must be used in certain order ($5!$ options) and at the same time, in each pair we may choose, which digit will be put in $A$ and which in $B$ ($2^5$ options). However, we have to remove the choices including a zero as the first digit in $A$ – that is $4!$ options of redistributing the remaining pairs and $2^4$ options of redistributing their digits between $A$ and $B$. The amount of desired numbers is thus $5!\cdot 2^5-4!\cdot 2^4.$

Answer:

$3^{2014}-2^{2014}$

Define the polynomial $Q(x)={\sum }_{k=0}^{2013}\left(\genfrac{}{}{0.0pt}{}{x}{k}\right)2^k$. Its degree is 2013 and moreover, for any $x\in \{0,\dots ,2013\}$ the binomial theorem yields

The polynomial $P(x)-Q(x)$ is of degree 2013 and it possesses 2014 roots, whereby it must be zero. Hence $P(x)=Q(x)$ and it only remains to compute

Answer:

$149.1^{\circ }$

Denote by $E$ the image of $B$ in reflection over $\mathit{AD}$.

Then $\mathit{AE}=\mathit{AB}=\mathit{AC}$ and $\mathit{\angle EAC}=\mathit{\angle BAC}-2\cdot \mathit{\angle BAD}=60^{\circ }$, entailing that the triangle $\mathit{AEC}$ is equilateral and $\mathit{CE}=\mathit{CA}$. In addition, $\mathit{DE}=\mathit{DB}=\mathit{DA}$ due to symmetry, and therefore $\mathit{CD}$ is the perpendicular bisector of $\mathit{AE}$ and $\mathit{\angle ACD}=\frac{1}{2}\mathit{\angle ACE}=30^{\circ }$. Now it is simple to use the nonconvex quadrilateral $\mathit{ABDC}$ for computing

Answer:

$180625$

Let $X$ be the sought-after number. First we deduce the digit to be removed must be the second one.

Proceeding by contradiction, assume the two initial digits have remained. Out of an $n$-digit number $X$, the removal produces an $(n-1)$-digit number $X^{\prime }$. The number $10\cdot X^{\prime }$ has then $n$ digits again, the two initial ones being the same as in $X$, yet $X\ne X^{\prime }$ because the former does not end with a zero. This is a contradiction – subtracting two multiples of an $(n-1)$-digit number cannot produce an $(n-2)$-digit number.

Now we write $X=a\cdot 10^{n+1}+b\cdot 10^n+c$, where $a$ and $b$ are digits ($a\ne 0$) and $c<10^n$ is a number with a nonzero terminal digit. Removal of the second digit produces $X^{\prime }=a\cdot 10^n+c$. Hence, for a suitable $k\in \mathbb{N}$, it follows that

At this point we observe $k<20$. Indeed, provided $k\ge 20$, the number $X$ would have a greater initial digit than $X^{\prime }$, which is impossible. Let us modify the equality into

Given that the left-hand side is divisible both by $2^n$ and $5^n$, the right-hand side must share the same property. The number $c$ does not end with a zero, hence $k-1$ has to be divisible by at least one of the prime numbers $2$, $5$ in their full power. Seeing that $k<20$, we conclude $n\le 4$ (by reason of $2^5>20$ and even $5^2>20$), resulting in $X$ having at most $6$ digits. On the other hand, for $n=4$ it must be $k-1=16$, yielding

In order for the right-hand side to be nonnegative, the only option is $a=1$ ($a$ and $b$ are digits). As for $b$, we may choose $b=8$, $b=9$, the latter of which we disregard for $c$ would end with a zero. With $b=8$, we calculate $c=5^4=625$ and finally verify that $X=1\cdot 10^5+8\cdot 10^4+625=180625$ really is a solution to the problem.

Answer:

$3$

If any of $a$, $b$, $c$ is zero, then all the numbers are zero, which contradicts their mutual distinctness. Similarly, the numbers $a$, $b$, $c$ cannot be three.

We use the third equality to express $c$ in the first and the second equation and then we modify the second expression $b=(\mathit{ab}-2b-2)a$ into $b(a^2-2a-1)=2a$. Since the right-hand side is nonzero, the same must hold also on the left; it is therefore legitimate to divide by $a^2-2a-1$ and hence to express $b$ by means of $a$. We insert the result into the first expression. After some modifications we obtain

hence $a$ is a root of the polynomial $P(x)=x^3+3x^2-9x-3$. By analogy, we derive that $b$ and $c$ are roots of the same polynomial as well. Since $a$, $b$, $c$ are mutually distinct, we infer $P(x)=(x-a)(x-b)(x-c)$. Comparing coefficients of the absolute term, we get $\mathit{abc}=3$.

Answer:

$\frac{2}{7}$

Without loss of generality assume $a>b>c$. The corresponding altitudes and medians then satisfy $v_a<v_b<v_c$ and $t_a<t_b<t_c$ respectively. At the same time, $v_a<t_a$, $v_b<t_b$ and $v_c<t_c$, which implies $v_b=t_a$ and $v_c=t_b$.

Let us denote by $M$ the midpoint of the side $\mathit{BC}$ and by $M_0$ the projection of $M$ to the side $\mathit{AC}$. In the right triangle $\mathit{AM}{M}_0$ we observe $M{M}_0=\frac{1}{2}{v}_b=\frac{1}{2}{t}_a=\frac{1}{2}\mathit{AM}$, hence $\mathit{\angle MAC}=30^{\circ }$. Denoting by $N$ the midpoint of the side $\mathit{AC}$, we obtain similarly $\mathit{\angle NBA}=30^{\circ }$.

Now let $G$ be the centroid of triangle $\mathit{ABC}$. Consider the equilateral triangle $A_{1}B{C}_1$ with $\mathit{BN}$ being one of its medians. The point $A_1$ fulfills $\mathit{\angle NB}{A}_1=30^{\circ }$ as well as $\mathit{\angle G}{A}_{1}N=30^{\circ }$ and yet it is distinct from the point $A$ (the triangle $\mathit{ABC}$ is scalene). “The real” point $A$ must be therefore the other intersection of the ray $B{A}_1$ with the arc $G{A}_{1}N$, that is, the midpoint of the line segment $B{A}_1$. Consequently $\mathit{\angle BAC}=120^{\circ }$ and $\mathit{AC}:\mathit{AB}=2$.

In a triangle with an angle $\alpha =120^{\circ }$ and side lengths $\mathit{AB}=1$, $\mathit{AC}=2$ we employ the Law of Cosines to calculate $a=\sqrt{1^2+1\cdot 2+2^2}=\sqrt{7}$ and $t_c=\sqrt{1∕4+1+4}=\frac{1}{2}\sqrt{21}$, then the area $S=\frac{1}{2}\cdot 1\cdot 2\cdot \sin 120^{\circ }=\frac{1}{2}\sqrt{3}$ and finally $v_a=2S∕a=3∕\sqrt{21}$. Putting everything together we thus obtain