Problems and solutions

Answer:

$34$

To have one of the numbers as large as possible, the rest has to be as small as possible. Since the numbers are distinct, the least three possible values are $1$, $2$, and $3$. For the average to be equal to $10$, i.e. the sum to be equal to $4\cdot 10=40$, the remaining number has to be $40-(1+2+3)=34$.

Answer:

$505$

Since $4$ is a root, we get $4^2+4m+2020=0$ or $m=-509$, so the equation now reads $x^2-509x+2020=0$ with solutions $4$ and $505$.

Alternatively, if $s$ is the other solution of the given equation, then

and comparing the coefficients of the polynomials yields $4s=2020$ or $s=505$.

Answer:

$7$

As $N$ is larger than $1$ and divides $95-4=91=7\cdot 13$, the smallest possible value is $7$.

Answer:

$54^{\circ }$

Let us label the points as in the picture.

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The size of the interior angle in a regular pentagon is $108^{\circ }$. The triangle $ABC$ is isosceles with $\angle ABC=108^{\circ }$, so

Since $ABH$ is a right-angled triangle with the right angle at vertex $B$,

Answer:

$120$

The sought number of minutes has to be a multiple of all three periods, and since we are looking for the least such number, the answer is the least common multiple of $12$, $10$, and $8$, which is $120$.

Answer:

$8\sqrt{2}$

The fifth blossom is a square with diagonals of length $4$, hence the length of its side is $2\sqrt{2}$ and the perimeter equals $8\sqrt{2}$.

Answer:

$20\%$

Let us denote by $P_1$ and $P_2$ the original heights of the plants. Since they both grew by the same percentage during the week, their new heights are $k{P}_1$ and $k{P}_2$, respectively, for some real number $k>1$ such that $(k-1)\cdot 100\%$ is the sought percentage. Then the measurements imply

Substituting for $P_2$ in the second equation and cancelling $P_1$ in the fraction yields $k^2=1.44$, so $k=1.2$, meaning that the plants grew up by $20\%$.

Answer:

$15$

For each of the three vertices of the large triangle, there are three rhombi “pointing” in direction of the vertex and two $1\times 2$ parallelograms sharing that vertex with the triangle. The picture shows these two types of parallelograms for the top vertex.

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No other types of parallelograms are present in the picture, so in total there are $3\cdot (2+3)=15$ parallelograms.

Answer:

$8$

We are looking for the smallest number $s\ge 505$ of the form $s=27x+36y$, where $x$ and $y$ stand for the numbers of buses of the first and the second type, respectively. Since the greatest common divisor of $27$ and $36$ is $9$, $s$ has to be a multiple of $9$. The smallest multiple of $9$ which is greater or equal to $505$ is $513$ and since $513=27\cdot 3+36\cdot 12$, we conclude that the number of empty seats is $513-505=8$.

Answer:

$10$

In other words, we ask for the number of distinct sequences of the pictures not having the two wolves next to each other. The left-hand wolf can be hung on positions 1, 2, 3, and 4 out of 6, and for each of these positions, the right-hand wolf can be hung to the following positions:

so there are $10$ such sequences in total.

Answer:

$30$

Unfolding the cylinder, we see that during each turn around the cylinder, the string has risen by $6\mathrm{cm}$ while advancing by the circumference $8\mathrm{cm}$ in the horizontal direction. By the Pythagorean theorem, the segment of the string corresponding to one turn is

long. Since there are three turns, we conclude that the total length of the string is $30\mathrm{cm}$.

Answer:

$33$

The last two digits must be zeroes. Furthermore, the first digit of the first factor is 4 and the first digit of the second factor is 7. Since the product is divisible by 100 and consequently by 25, one of the factors has to be divisible by 25 or both factors by 5. There is no two-digit multiple of 25 starting with 4, hence the second factor has to be divisible by 5 and since it cannot end with 0, it equals 75. Moreover, since the product is divisible by 4, the first factor must be divisible by 4 and so it is 48. We have $48\times 75=3600$, where the sum of all digits present is 33.

Answer:

$3566$

Let $x$ and $y$ be the two numbers and $c$ a digit that Ed wrote to (say) the number $x$. Then we have

therefore

It follows that $c$ has to be $5$ since $32099-c$ has to be divisible by $9$. We conclude that $x=3566$ and $y=8779$.

Answer:

$189$

The key observation is that the dots on opposite faces on the large cube always add up to 7, as shown in the left-hand figure below. The large cube has 27 pairs of opposite faces, so no matter how Peter arranges his dice, the total number of dots showing must be $27\cdot 7=189$. One possible arrangement Peter can use is shown in the right-hand figure below.

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Answer:

$5:2$.

The diagonals divide the small $X$-pentomino into four congruent pieces. Furthermore, two such pieces can be glued together to form one square of the larger pentomino, as the picture shows. Thus the large pentomino can be divided into ten such pieces and the sought ratio of areas is $10:4=5:2$.

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Answer:

$5940$

All numbers between $10^3$ and $10^4$ have an even number of digits, hence the sum of digits of every such palindrome is even. Moreover, the palindromes from this range are exactly the numbers of the form $\overline{\mathit{abba}}$, where $a$, $b$ are any digits, $a$ non-zero, which shows that there are $90$ palindromes in that range. In a similar way we infer that there are precisely $900$ palindromes between $10^5$ and $10^6$ and the sum of digits of each of them is even.

The palindromes between $10^4$ and $10^5$ are of the form $\overline{\mathit{abcba}}$ for $a$, $b$, $c$ digits, $a$ non-zero, and the sum of their digits is clearly even if and only if $c$ is even. Hence there are $9$ options for $a$, $10$ for $b$, and $5$ for $c$, which gives $9\cdot 10\cdot 5=450$ sought palindromes in the given range. A similar argument applies to the range from $10^6$ to $10^7$, showing that it contains $4500$ palindromes the sum of whose digits is even.

We conclude that out of all palindromes between $10^3$ and $10^7$, $90+900+450+4500=5940$ have an even sum of digits.

Answer:

$22$

If, for example, all the altos together with three other “skilled” singers fall sick, then there are only nine skilled singer remaining, so there is no way to have ten alto singers. Therefore $S<23$.

On the other hand, observe that if some of the singers fall sick, the situation can only get worse if the sick ones are the “skilled” ones instead of the “ordinary” ones. Therefore, if $22$ singers fall sick, we can assume that $18$ of them are the skilled ones, so only four “ordinary” singers fall sick and it is immediate that at least ten singers remain in each voice. This shows that $S\ge 22$ and consequently, $S=22$.

Answer:

$12$

Using the Pythagorean theorem, we find that the length of a diagonal of the rectangle is

This diagonal is the diameter of the circumscribed circle, the area of which therefore equals $\pi {(5∕2)}^2$. Moreover, the outer half-circles have radii $4∕2$ and $3∕2$, respectively, so the total area of the semicircles is

Consider the whole region formed by the semi-circles and the rectangle; its area is

The shaded region is obtained by removing the circumscribed circle from this all-encompassing region, so its area equals $12$.

Alternatively, we can argue as follows. The Pythagorean theorem gives relation of squares raised above three sides of a right triangle. The same relation holds for half-circles, which we apply to both halves of the given rectangle (divided by a diagonal). Straightforward observation then gives that any two neighbouring grey areas have the sum of half of the rectangle. Therefore the total grey area is equal to that of the rectangle, which is $3\cdot 4=12$.

Answer:

$977$

The sum of the digits of a three-digit number is at most $9+9+9=27$. There are five two-digit primes not greater than $27$, namely $11$, $13$, $17$, $19$, and $23$. The sums of the digits of these primes are $2$, $4$, $8$, $10$, $5$, respectively. Therefore, $p_2=11$ or $p_2=23$ are the only possibilities. The largest three-digit prime number with the digit sum of $23$ is $977$. Since $977>911$, which is the largest three-digit prime having its digit sum equal to $11$, the sought number is $977$.

Answer:

$45^{\circ }$, $67.5^{\circ }$

Let us denote the intersection point by $X$ and the centre of $AC$ by $F$. Observe that the altitude going through $X$ must be the one corresponding to the same side as $X$ belongs to. We now examine all possible locations of $X$.

If $X$ lies on the base $BC$, it must be the intersection of the altitude from $A$ and one of the side axes. From symmetry of triangle $ABC$ we infer that this altitude coincides with the axis of $BC$, so the single-point intersection has to be with one of the remaining axes. The symmetry then implies that it actually intersects the axes of both $AB$ and $AC$ in a single point.

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Since $FX$ is the axis of $AC$, triangle $AXC$ is isosceles. Moreover, we have $\angle AXC=90^{\circ }$ and so $\angle ACB=\angle ACX=45^{\circ }$.

If $X$ lies on $AB$, then it must be the intersection of the altitude from $C$ and the axis of $AC$.

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As in the previous case, triangle $AXC$ is isosceles and right-angled with right angle at $X$, implying that $\angle BAC=\angle XAC=45^{\circ }$. Therefore,

in this case. The case when $X$ lies on $AC$ is completely symmetrical.

Answer:

$3720$

We have

It is easy to see that both $59$ and $61$ are prime numbers, therefore the answer is $1+59+61+3599=3720$.

Answer:

$23$

Molly paid exactly one third of the total costs, that is, all the sausages together costed $28\cdot 3=84$ dollars. Dolly paid for $17$ sausages out of $28$, that is $84\cdot 17∕28=51$ dollars. Since the sisters decided to share the costs equally, she should have paid only $28$ dollars. Thus she should get $51-28=23$ dollars.

Answer:

$\frac{253}{34}$

The gray triangle in the picture is right-angled and shares one angle with the large triangle, hence these two triangles are similar.

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The ratio of lengths of the legs has to be the same for the two triangles, thus we obtain the equation

with the solution $t=253∕34$.

Answer:

$840$

As $11$ and $19$ are primes, one of the three consecutive numbers has to be divisible by $11$ and one, not necessarily a different one, by $19$. Moreover, since the product is positive, all three numbers have to be positive as well. That is, we search for small positive multiples of $11$ and $19$ differing by at most $2$. The smallest are $3\cdot 19=57$ and $5\cdot 11=55$ so we only have to supplement the product by $56$ to get $55\cdot 56\cdot 57=11\cdot 19\cdot 840$. Therefore, $840$ is the sought number.

Answer:

$36^{\circ }$

Let us denote $\mathit{\angle DCP}$ by $x$. Since $D$ and $E$ both lie on the circle with center $C$, $△\mathit{DCE}$ is isosceles with base $\mathit{DE}$. The first given equality $\mathit{\angle PDC}=\mathit{\angle EDC}$ implies that $△\mathit{CDP}$ is congruent to a half of $△\mathit{CDE}$ (in particular to $△\mathit{CDM}$ where $M$ is midpoint of $\mathit{DE}$). Using also the second equality it follows that $\mathit{\angle BCE}=\mathit{\angle ECD}=2\mathit{\angle DCP}(=2x)$ and as these three angles form together a straight angle we have $x+2x+2x=180^{\circ }\Rightarrow x=\mathit{\angle DCP}=36^{\circ }$.

Answer:

$18$

Consider a pair of connected cities. One of them must have one prime more in their prime factorisation. They cannot have the same primes because that would mean the cities would be the same. On the other hand, they cannot have two or more primes difference. To prove that, let $p$, $q$ be not necessarily distinct primes and let cities $a$ and $b=a\cdot p\cdot q$ be connected. Than $a\cdot p$ divides $b$, thus it violates the second condition.

City with number $42$ has prime factorisation $2\cdot 3\cdot 7$. That means that cities with lower index number connected to this city are three, with these prime factorisations: $2\cdot 3$, $2\cdot 7$ and $3\cdot 7$.

The indexes greater than $42$ could be written in the form $42\cdot p$ for some prime $p$. The largest such prime satisfying $42\cdot p<2020$ equals $47$, so $p\in \{2,3,5,7,11,13,17,19,23,29,31,37,41,43,47\}$. This means there are fifteen more cities connected to the city labelled $42$, that is, eighteen in total.

Answer:

$673$

It is clear from the rules that if the blitzer moves from one square to another, it cannot return to the starting square in the following move. However, it can return to that square in two moves, even on a $3\times 3$ chessboard, as the picture shows:

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Therefore if the blitzer gets to one of the squares A1, A3, C2, it can start “cycling” and reach visit each of these squares in its every third move.

If the blitzer starts on B2, then its only possible first move is a rook move to B3 and the subsequent options are A1 or C1 via a knight move. Therefore A1 is reachable from B2 using two moves, but the blitzer cannot directly move to B2 from A1 or C1 to have the cycle of three moves sooner, so we have to use at least two moves to get in such a cycle. That leaves $2018$ moves for cycling the blitzer and since $2018=672\cdot 3+2$, the blitzer can do $672$ cycles. Thus, it visits A1 $673$ times.

Answer:

$2026$

Let us assume that the snail completed exactly $n$ laps during the race. That is, the speed of David was $3$ and that of snail $n$ (in laps per race). Then the rate of change of the margin, i.e. the oriented distance from David to the snail, was $n-3$. Since they met whenever the margin was a natural number, they met $n-3$ times, excluding the beginning. Thus $n=2022$. When they were running in opposite directions, the rate of change of their oriented distance was $n+3=2025$, so they met $2025$ times excluding the start, i.e. $2026$ times in total.

Answer:

$170$

Let us denote $s_i$, $d_i$, $a_i$ number of support, defence and attack items of the level $i$ needed to get one attack item of level 10. We have $s_{10}=d_{10}=0$, $a_{10}=1$ and according to the rules,

for all $i\in \{2,\dots ,10\}$. Using these rules, one can fill in a $10\times 3$ table so that with the exception of the top row $(0,0,1)$, value in every cell is the sum of the two numbers in the row above it, but not in the same column:

Thus the answer is $170$.

An alternative approach is to note that since the roles of support and defence items are interchangeable, actually $s_i=d_i$. It is also easy to prove by induction that $|a_i-s_i|=1$; this holds for $i=10$ and

Finally, we have

so $s_1+d_1+a_1=2^9=512$. These observations together imply that $s_1=d_1=171$ and $a_1=170$.

Answer:

$24\pi$

Let $AC=4$ be the radius of the ball, $BC$ the radius of the cone base and $BD$ the cone height as in the figure below. From the Pythagorean theorem applied to the triangle $ABC$ we get $B{C}^2=A{C}^2-A{B}^2=16-4=12$. From similarity of triangles $ABC$ and $CBD$ we can deduce $\frac{BD}{BC}=\frac{BC}{AB}$. Therefore the volume is

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Answer:

$31$

Let us first recall the basic property of addition: We add the numbers digit-by-digit with the exception that carries must added properly. That is, we add the four pairs of digits and the carries. Since we are adding only two numbers, the carries are at most $1$.

Let us now denote the two numbers by $a$ and $b$ and the sum of digits of any number $n$ by $S(n)$. Then it holds that $S(a+b)=S(a)+S(b)-9\cdot c$, where $c$ is the number of non-zero carries. We now compute $S(a)+S(b)=1+2+3+4+6+7+8+9=40$, so the possible values of $S(a+b)$ are $40$, $31$, $22$, $13$ and $4$.

It is, however, impossible to achieve $40$ since $9$ added to any non-zero digit is greater than or equal to $10$, while $31$ is attainable for instance as follows: $9678+4321=13999;1+3+9+9+9=31$. The result thus reads $31$.

Answer:

$\frac{1}{14}$

Let us consider the half of players containing Bono. Bono will play the final if and only if both of the professional players start in the other half and he beats the other three amateurs in his half.

Let us start with only Bono’s position fixed and assign the two professional players. The probability that both go to the other half is $\frac{4}{7}\cdot \frac{3}{6}$. Using the fact that all the amateurs are evenly matched, the probability that Bono wins two matches against them is $\frac{1}{2}\cdot \frac{1}{2}$. The desired probability is therefore $\frac{4}{7}\cdot \frac{3}{6}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{14}$.

Answer:

$\frac{1}{5248}$

Notice that when $n$ replaces $a$ and $b$, we have

It follows that the sum of the reciprocals of all numbers on the blackboard increases by two in each step. As there were $99$ steps in total, the last number $l$ satisfies the equation

Thus the last number is equal to $\frac{1}{5248}$.

Answer:

$18$

Denote by $H_A$ and $H_E$ the perpendicular projections of $A$ and $E$, respectively, onto line $BC$.

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The right-angled triangles $CA{H}_A$ and $CE{H}_E$ are similar, therefore $CE=3CA$ implies $E{H}_E=3A{H}_A$. From $CD=BD-BC=BC$ we get that the area of triangle $CDE$ is equal to

A similar argument then shows that the areas of triangles $AEF$ and $BFD$ are $2\cdot 4=8$ and $3\cdot 2=6$, respectively, so that the total area is $1+3+8+6=18$.

Answer:

$47$

Let us first denote the numbers of coins from the bags weighted by the king by $a$, $b$, and $c$. First note that the numbers must be pairwise distinct, for if two of them were equal, the two respective bags would be indistinguishable. We shall call such triplets admissible. We search for the smallest possible choice of $a$, $b$, and $c$ such that $a\cdot k+b\cdot l+c\cdot m$ are pairwise distinct numbers for any permutation $(k,l,m)$ of the numbers $10$, $11$ and $12$.

The smallest admissible triplet is $a=0$, $b=1$ and $c=2$. This, however, violates the second condition since $32=2\cdot 10+12=2\cdot 11+10$. The second smallest admissible triplet is $a=0$, $b=1$, and $c=3$, which satisfies the second condition, since

The scale thus has to measure correctly up to 47 grams.

Answer:

$59$

Let us denote the number of Emma’s pineapples on day $i$ at $2$ pm by $s(i)$. Since the first two zeros appear on days $2$ and $5$ and Emma increases every time the number of bought pineapples by one, the distances between two subsequent zeroes in the sequence $s(i)$ form the sequence $3,4,5,\dots$. The zeroes are thus located at positions

Since we want to know where the last zero before $2020$ appears, we solve the quadratic equation $\frac{1}{2}x(x+1)-1=2020$. It has a negative solution (which is irrelevant) and a positive solution $\frac{\sqrt{16169}}{2}-\frac{1}{2}$ lying between $63$ and $64$. Thus the last zero is the 62nd and it is at the position $\frac{63\cdot 64}{2}-1=2015$. The sequence $s(i)$ then continues as follows: $63,62,61,60,59,\dots$, so the sought number $s(2020)$ equals $59$.

Answer:

$8$, $24$

First note that we know the ratio of width of the first and second columns ($30:10$) and of the first and third columns ($18:12$). We also know the ratio of heights of the first, second and fourth rows ($24:18:30$). Moreover, the reduction of the ratios to the lowest terms produces $3:1:2$ and $4:3:5$ for the columns and rows, respectively. Knowing that, we can fill in the table with areas of the corresponding cells as in the figure, where the third row and fourth column are specified only partially with integer parameters $x$ and $y$ to reflect the ratios.

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The ratio of the height of the second and third row could be expressed comparing the areas in the third column as $12:2x$ or from the fourth columns as $3y:12$, see the shaded rectangle above. These two values must, of course, be the same, so we write $12:2x=3y:12$, i.e. $y=24∕x$.

Finally, we compare the total area of the pigsty with the sum of all the boxes to get the equation

which could be rewritten to

The solution is thus $x=4$ or $x=12$, which lead to $y=6$, $?=24$ and $y=2$, $?=8$, respectively.

Answer:

$\frac{5\sqrt{2}}{6}$

Let $A,B,C$ denote the grid points on Philip’s circle. We are given that the radius is less than $\frac{5}{4}$, so the distance between $A$ and $B$ is at most $\frac{5}{2}$, and therefore, up to a rotation, the relative position of $A$ and $B$ is one of the four arrangements shown below:

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In each case, we can find a line of reflectional symmetry for the circle, and the third grid point $C$ must be placed either on this line, or in such a way that its reflection across the line is not a fourth grid point. This makes arrangement 1 impossible.

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Arrangement 2 is possible, as long as we place $C$ on the line of symmetry. Moreover, we must also position $A,C$ and $B,C$ in one of the arrangements 2 through 4, up to rotation. This leads us to Philip’s circle:

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We still need to find the radius, which we can do algebraically. We introduce a coordinate system so that the points $A,B,C$ have coordinates $(1,0)$, $(0,1)$, and $(2,2)$. Substituting these $(x,y)$-values into the general equation of a circle ${(x-h)}^2+{(y-k)}^2=r^2$, we solve for $h$, $k$, and $r$, and arrive at the equation ${(x-\frac{7}{6})}^2+{(y-\frac{7}{6})}^2=\frac{25}{18}$.

Answer:

$720$

Since all colours remain present after the spell, we infer that the spell is just a permutation of the seven original colours. Every such permutation can be decomposed into disjoint oriented cycles of colours such that the permutation simply rotates these cycles. It follows from the third condition in the statement that these cycles cannot be formed by $2$ or $3$ colours, but a $1$-colour cycle is impossible, too. However, it is impossible to distribute the seven colours into more than one cycle so that all the cycles contain at least four colours, so the permutation in question is formed by a single cycle. Finally, there are $6!=720$ such permutations: Fix one of the colours, there are six options to which colour the spell can change it, for this colour there are five options etc., until the last colour gets changed to the first one.

Answer:

$1,6,5,1$

We are looking for an integer $x$ such that $x$ gives remainders $5$ and $1$ when divided by $11$ and $10$, respectively. Since $10$ and $11$ are coprime, by the Chinese remainder theorem, there is precisely one such $x$ among the numbers $0,1,\dots ,109$ (let us call it $x_0$) and all the other solutions are obtained by adding multiples of $10\cdot 11=110$ to $x_0$. A possible easy way to find $x_0$ is to list integers of the form $11k+5$ ($k$ an integer) and pick the one ending with $1$, which turns out to be $71$. This is the number of rotations needed to be done to get the first combination ending with $5,1$. By the above, such a combination will come up again after $110$ rotations (and not earlier), i.e. $110+71=181$ rotations from the beginning. Since $181$ gives remainders $1$ and $6$ when divided by $5$ and $7$, respectively, at that point the lock will show $1,6,5,1$.

Answer:

$144$

Since there are only four mirrors and each ray has to turn at some point, every row and every column has to contain exactly one mirror. This can be achieved in $4!=24$ different ways—we choose the column for the first row in $4$ ways, then the column for the second in $3$ ways and so on. Due to the restrictions on the number of rays going in each direction, we know that there must be two mirrors of each type. Hence having chosen the squares to which the mirrors are to be placed, we have $\left(\genfrac{}{}{0.0pt}{}{4}{2}\right)=6$ ways to choose the orientation. Finally, it is not hard to see that every such configuration has the desired properties, hence there are $24\cdot 6=144$ such configurations.

Answer:

$9$

We see that $x_1=2\cdot 2\cdot 5\cdot 101$ and $x_2=2\cdot 3\cdot 5\cdot 101$. It is easy to see that when a term of the sequence is not divisible by a square of a prime number, all the subsequent terms have this property, too. Therefore, from $x_2$ on we can represent every term $x_n$ as a binary number $b_n$, whose $k$-th digit (from the right) is $1$ if and only if $x_n$ is divisible by the $k$-th prime number. Next, observe that according to the definition of the sequence, $b_{n+1}=b_n+1$ for all $n\ge 2$. We have

and

From the definition of $b_n$, the number of different prime divisors of $x_{2020}$ is just the number of ones in $b_{2020}$, which is $9$.

Answer:

$\frac{13}{36}$

The following figure represents the situation containing all the relevant points: the centroid $S$ of triangle $△ABC$, the midpoints $M_a$, $M_b$, $M_c$ of the sides of $△ABC$, the centroids $D$, $E$, $F$, $G$, $H$, $I$ of the sub-triangles, the midpoints $D^{\prime },E^{\prime },\dots ,I^{\prime }$ of the line segments $M_{b}A,A{M}_c,\dots ,C{M}_b$, respectively.

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Since $A{E}^{\prime }=\frac{1}{4}AB$ and $A{D}^{\prime }=\frac{1}{4}AC$, triangle $△A{E}^{\prime }{D}^{\prime }$ is a homothetic transformation of $△ABC$ with center $A$ and scaling factor $\frac{1}{4}$. Therefore, the area $[A{E}^{\prime }{D}^{\prime }]$ is $\frac{1}{16}[ABC]$. Likewise, we have $[B{G}^{\prime }{F}^{\prime }]=[C{I}^{\prime }{H}^{\prime }]=\frac{1}{16}[ABC]$. Therefore, the area of hexagon $D^{\prime }{E}^{\prime }{F}^{\prime }{G}^{\prime }{H}^{\prime }{I}^{\prime }$ is $\frac{13}{16}[ABC]$. Furthermore, the homothetic transformation with center $S$ and scaling factor $\frac{3}{2}$ maps hexagon $DEFGHI$ to hexagon $D^{\prime }{E}^{\prime }{F}^{\prime }{G}^{\prime }{H}^{\prime }{I}^{\prime }$ and we obtain the area of hexagon $DEFGHI$ as

As a consequence, the area ratio sought is $\frac{13}{36}$.

Answer:

$\frac{1010}{3}$

Adding up the equations for $m=1,\dots ,2019$ we obtain

Using $a_1=a_{2020}$, we can rearrange this as follows:

Thus

Note that such a sequence of real numbers indeed exists: Given $a_1$, the rest of the sequence is determined by the equation in the statement. One can therefore express $a_{2020}$ in terms of $a_1$ only and the condition $a_1=a_{2020}$ then translates to a linear equation in $a_1$; it is not difficult to see that this equation has a solution.

Answer:

$8∕15$

Assume that the strings, which we label $A$, $B$, $C$, $D$, and $E$ are tied on one side of the hand so that $A$ has a free end, $B$ is tied to $C$ and $D$ is tied to $E$. Observe that under such circumstances, we obtain one thread if and only if (on the other side) we tie $A$ to one of $B$, $C$, $D$, $E$ ($4$ options) and subsequently tie the remaining free end of the pair to a string of the other pair ($2$ options)—for example, if $A$ is tied to $B$, then we tie $C$ with $D$ or $E$. Thus there are $4\cdot 2=8$ options.

The total number of ways to tie the knots on the other side is $15$: First we choose the free end ($5$ options) and the rest is determined by pairing one particular end with one of the three remaining ends ($3$ options). We conclude that the probability of forming a single thread is $8∕15$.

Answer:

$619737131179$

Let us consider an oriented graph on 4 vertices labelled $1$, $3$, $7$ and $9$ where two digits are connected with an arrow if and only if the associated two-digit number is a prime. Note that there is a loop at vertex $1$.

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Assume for the moment that there is a walk on the graph moving along the arrows and using every arrow exactly once. Then the largest 2-prime can be obtained by putting $6$ or $8$ in front of the sequence of digits visited by one of such walks. Indeed, it follows from the definition of 2-prime that all its digits except for the first one must be taken from the set $\{1,3,7,9\}$ and no arrow can be used twice. Thus no 2-prime can have more digits than the number of arrows in our graph plus two (one for the first digit and one because we are counting digits), i.e. $12$. Also, the first digit cannot be $9$ or $7$ (it would repeat one of the arrows) and since $61$, $83$, $87$ and $89$ are prime numbers, we do not need smaller digits.

Now we find the walk with the above-mentioned properties which produces the largest possible number. Note that on one hand, there is one more arrow entering vertex $9$ than leaving it while, on the other hand, one more arrow leaving vertex $1$ than entering it. The other two vertices are “balanced” in this sense. It follows that our walk must start in $1$ and end in $9$. From $1$ we move to $9$ as it is the largest possible neighbour, then we continue to $7$ for the same reason, then we cannot return to $9$ (it would terminate the sequence) so we rather move to $3$, then back to $7$ etc. By this sort of greedy algorithm we end up with $19737131179$ and as $81$ is not prime, we conclude that the largest 2-prime is $619737131179$.

Answer:

$\frac{4\sqrt{21}}{7}$

Consider a reflection with respect to the circumcenter $O$ and denote the respective images of points by adding a prime. We observe that points $A^{\prime }$ and $B^{\prime }$ lie on the circumcircle of $△ABC$ and $D^{\prime }=E$ (i.e. $E^{\prime }=D$). The Pythagorean theorem in the right-angled triangle $A{A}^{\prime }{B}^{\prime }$ (note that $A{A}^{\prime }$ is a diameter of the circumcircle) yields

Since $\angle A{B}^{\prime }E=\angle A^{\prime }BD=90^{\circ }$, the Pythagorean theorem in $△A{B}^{\prime }E$ yields

Since $E$ (the image of $D$) lies on $A^{\prime }{B}^{\prime }$ (the image of $AB$) and since the quadrilateral $A{B}^{\prime }C{A}^{\prime }$ is cyclic, it follows that $△A{B}^{\prime }E\sim △A^{\prime }CE$. Hence

and we conclude that

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Alternative solution: The power of point $D$ with respect to the circumcircle of $△ABC$ with radius $r=AO$ equals

(the minus sign is present due to the fact that $D$ lies inside the circle). The Law of Cosines in $△ADO$ yields

Since $cos(\angle AOE)=cos(180^{\circ }-\angle AOD)=-cos(\angle AOD)=-\frac{1}{7}$, the same theorem for $△AOE$ then yields

Similarly as above, from power of point $E$ with respect to the circumcircle of $△ABC$ we obtain

Answer:

$\frac{56}{3}=18\frac{2}{3}$

Let $24$ be the width and $7$ the height of the rectangle. The diagonal going from the lower left to the upper right corner has a slope of $\frac{7}{24}$. When it passes through a square, it cuts off a triangle if and only if it crosses a horizontal side that separates two squares. Since the slope is constant, we can rearrange the line segments of the two triangles that are cut off from the two squares into one big right triangle of width $1$. Then the legs are $1$ and $\frac{7}{24}$ and the hypotenuse is $\sqrt{1+{\left(\frac{7}{24}\right)}^2}=\frac{25}{24}$ long.

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The sum of the perimeters of these two triangles is thus $\frac{56}{24}$. The diagonal crosses a horizontal side exactly six times, plus two big right triangles of the above dimensions are cut off from the first and last squares that the diagonal passes. So the total sum of the perimeters is $8\cdot \frac{56}{24}=\frac{56}{3}$.

Answer:

$6763$

Firstly, in order to be able to move at all from floor $2019$, we must have $2019+n\le 8787\Rightarrow n\le 6768$. Secondly, the condition $d:=\gcd (2020,n)=1$ is necessary, as we can move between floors $a$ and $b$ only if $d\mid a-b$. Bearing in mind that $2020=2^2\cdot 5\cdot 101$, we can eliminate some candidates for the largest $n$: $6768$ divisible by $2$, $6767$ divisible by $101$, $6766$ divisible by $2$, $6765$ divisible by $5$, $6764$ divisible by $2$ and finally $\gcd (6763,2020)=1.$

It remains to prove that $6763$ is elevating. Using the Euclidean algorithm (or Bézout’s identity) we can find integers $x$, $y$ such that $6763x-2020y=1$ and we can assume that $x$, $y$ are even non-negative since adding $2020$ to $x$ and $6763$ to $y$ preserves the equality. Starting in floor $0\le f\le 8786$, we claim that is possible to make a sequence of $x$ moves up and $y$ moves down in such order that we stay in the building and end in floor $f+1$. Indeed, as $2020+6763\le 8787$, we can always move in at least one direction. Furthermore, if we used all of the steps down (resp. up), we must be below (resp. above) floor $f+1$ and making the remaining steps up (resp. down) brings us to floor $f+1$. Similarly it can be shown that we can move one floor down from any floor $1\le f\le 8787$. We conclude that $6763$ is the largest elevating number.

Answer:

$85230$

We number the columns from left to right by $1$ up to $5$ and denote the carries of the respective columns by $c_1,\dots ,c_5$. Observe that $c_1=0$ and $0\le c_i\le 2$ for $i=2,\dots ,5$: Indeed, one cannot get more than $29$ by adding three digits and a carry of size at most $2$ and thus the last inequality follows from an inductive argument. Further note that $c_2\le 1$—as there are two different digits summed up in the second column and $c_3\le 2$—and $c_2\ne 0$ due to the first column, so $c_2=1$ and $G+1=B$. From the fifth column we get $D+E=10$, since $D$ and $E$ cannot be both zeroes, and $c_5=1$. Since $B+R+c_3=c_2\cdot 10+R$, either $B=9$ or $B=8$. Therefore $\mathit{BLUE}$ is the square of an integer $n$ satisfying $90\le n\le 99$ and has four different digits. Eliminating the squares with coincident digits, we are left with $8649$, $9025$, $9216$, $9604$, and $9801$ as possible values for $\mathit{BLUE}$. From $D+E=10$ we can eliminate $9025$, since this would lead to $D=E(=5)$. Furthermore, $9801$ is impossible due to $B=D(=9)$ and $9604$ due to $L=D(=6)$. With the help from the fourth column we can eliminate $9216$, because this would lead to $D=4$ and $E+U+E+1=6+1+6+1=14$ giving $W=4$, which is a contradiction with $D=4$. Therefore the only possible value of $\mathit{BLUE}$ is $8649$.