Increasing the length of an edge by is the same as multiplying it by . The volume of the cube is now , after multiplication it will be . There is an increase by , which is .

9

We want all intersections of the lines to be distinct, lie in the ring and every line to be a secant of the inner circle.

Since is divisible by , it follows . Since it is divisible by , implying that .

25

The number of students is divisible by and is larger than (Pigeonhole principle). It is easy to verify that is sufficient.

Consider what is left after eating all the pieces with chocolate layer. It will be a rectangular prism out of all pieces without chocolate. There must be then pieces of cake without chocolate on it. From the total amount of pieces with chocolate we easily count the desired percentage: .

We aim to find . Let be the foot of the perpendicular dropped from onto . As is right and isosceles we have and the rest follows from the Pythagorean theorem.

2012

It suffices to go through ways to vary the left-hand side and see if the right-hand side can be changed accordingly. The process can be speeded up by various observations (for instance the result has to start with 2, hence the first number is 4 and the second one starts with 5).

300

Without loss of generality let . Then . This value is achieved for .

5

Adding the given equations we get

45

Let us order the trees by their distance to the gardener and check each one for visibility. If a tree is not crossed out, we mark it as visible and cross out all the trees aligned with it.

We shall distinguish the cases according to the number of black faces involved in the coloring. For zero black faces we have one option as well as for one black face. For two black faces there are two ways—-either the faces are neighbouring or they are not. For three black faces there are again two colorings (either these three faces share a common vertex or they do not). For four, five, and six the situation is analogous to the one with two, one, and zero black faces, respectively (we can place the white faces instead of the black ones). Altogether, we have ways to color the cube.

72

Focus on trapezoid and observe that is the midpoint of its side . Hence if we denote the foot of perpendicular dropped from onto by then is the midline of the trapezoid and its area is thus .

1028

If is even then is always a square. If is odd then is a square if and only if itself is a square. As the answer is .

Observe that the red vertex traces circular arcs. Twice it traces of a circumference of a circle with radius 1 and once it remains fixed. Hence the total length of its trajectory equals .

The number has to be divisible by so it ends with two zeroes. Moreover, it has to be divisible by 9 (), so the number of 1's is a multiple of nine. The smallest positive integer with these properties is .

28

Bill's age is the arithmetic mean of the positive integer and its square . Therefore all possible values of are , , , , , and . The only square among them is , hence .

or in reverse order.

For the two corners to coincide we have to fold about the line passing through the center of the rectangle perpendicular to one of its diagonals.

Focus on the resulting pentagon consisting of two congruent right triangles and an isosceles triangle in the middle. The diagonal splits the isosceles triangle into two congruent right triangles. Using the ratio of the areas we learn that in fact all four right triangles forming the pentagon are congruent. Hence the angle by their common vertex is . The ratio of the side lengths of the rectangle is readily obtained invoking the symmetry about the fold line.

16

The number has to be divisible by , thus it must end with or . Also, it has to be divisible by , so its digit sum has to be divisible by , which implies that after dividing by the sum of the first two digits gives remainder (if the last digit is ) or (if the last digit is ). The remainder can be achieved as a sum of numbers with remainders , (the order of the digits matters, so we have combinations). Similarly the remainder can be achieved only as , (again combinations).

Denote by , , the centers of the smaller circles and by the center of . Observe that , , are vertices of an equilateral triangle with side length . Its altitudes are of length , and they intersect at . However, in equilateral triangle the altitudes are at the same time the medians. Hence . The radius of is then the sum of and the radius of the smaller circle.

Let for an integer and a digit . Then . The digit in the tens place is odd if and only if the digit in the tens place of the number is odd. Therefore is or . The last digit in both cases is .

19

The following is a palindromic -chain:

.

-chain

Adding the equations yields . Since we are looking for , , positive, after dividing by and taking the square root, we have . By substituting back to the original equations we get , , , from which we easily obtain the solution .

0

Draw the chords symmetrical to the two we already have about the center of the circle and observe that if the two areas are to be the same then one of the chords has to be a diameter of the circle.

The officer has to walk around the first building either in the very beginning or in the very end of his route (in any of the two possible directions) which gives us 4 options. For any of the remaining two buildings he may independently decide if he walks around them clockwise or counterclockwise. These decisions determine the route uniquely, hence the answer is .

A line parallel to the shorter leg passing through another vertex of the trapezoid splits it into a parallelogram and a triangle with side lengths 3, , and , which is therefore isosceles and right. The area is now simply .

Let us assign to each ordering a value that denotes the number of (not necessary neighbouring) pairs such that the taller person stands behind the shorter one. Every step decreases by at most one and if then there exists a step that decreases it. In the beginning, is at most (if every pair is switched, i.e. if the people are in the opposite order). Thus, the ”worst” ordering requires steps.

59

If he can pay some price, he can also pay this price plus a multiple of . Thus, it is enough to observe the multiples of and their remainders modulo . , , , , , , . From on, he can pay all the amounts, as the remainders repeat themselves. Thus, the last price he cannot pay is . (Should he be able to pay this amount, it would hold . The left-hand side modulo is , while the right-hand side is not as the first multiple of () with remainder is as shown above.)

A shape with a given area has the smallest perimeter if it is a circle. So all that needs to be done is to cut out the circle with area consisting of one fourth of the big circle and dividing the rest into three parts with equal area.

Let be a common solution of both equations. By subtracting the equations we get . So either or . If , (substitute back). In the latter case the equations do not have any real roots.

Let be a number that we get after crossing out the first digit and be the crossed out digit. The original number is then . It should hold that , which can be simplified to . However, the left-hand side is nonzero and divisible by , while the right-hand side is not for .

In order to maximize the area, the side of length is surely the shorter leg. From the Pythagorean theorem we get . We want to maximize . That is the same as to minimize the difference between the lengths of the hypotenuse and the longer leg. If , the parity does not match. If , we get a simple equation for .

Consider the possible radii of the excircle. For , , and we see that the orthocentre coincides with one of the vertices. For and we can construct the desired triangles.

14

Use the formula several times to get . But for the , all the other factors in the product are divisible by exactly once, since for every positive integer , gives a remainder when divided by .

Let us observe the highest number that satisfies the condition of different nonzero digits which is , but it has persistence only because there is a five and an even number, so that in the second step we obtain . To obtain the result, it suffices to get rid of , and switch the last two digits to satisfy the evenness.

Let us consider a quadrilateral in which points and merge into one point. Then is a midline in and the ratio is obviously .

270

Let us imagine that we are looking at the cube from above and count the area of the surfaces oriented upwards. In the depth of cm (the side of the cube) there are cm, in the depth of both and cm there are cm. The situation is the same from all directions.

For each trajectory there exist a trajectory of the same length that first moves only right and then only upwards, so it suffices to consider such trajectories only. The longest trajectory must clearly pass through the centre. Let us denote the length of the trajectory travelled to the right from the centre of the circle and the length of the trajectory that we will travel from the centre upwards. As the longest route will end at the circumference of the circle, it holds that . We want to maximize , which is equivalent to maximizing . From the trivial inequality we have . The equality occurs when we have equality for , so . Adding the distance to the centre of the circle, we get the result .

The remainder of the product is the same as the product of the remainders of the factors, for example . In order to make the final remainder equal to , the number can be divisible by neither nor , so we have to exclude all the two's and three's from . After that we are left with the number , which gives the remainder . Each factor gives a remainder either or , so we just have to exclude one number with remainder . The smallest such number is .

49

We divide the lines into three groups: those that pass through the center of the cube (), those that pass through the center of the side but not the center of the cube ( for each side), and those lines on which the edges lie (). We conclude that there are lines altogether.

It is clear that . Altogether, there were points allocated each day, that gives a total sum of points. However, that is also equal to . Comparing these two expressions, we get . Trying out several possibilities, we find out that for we can find a way in which the contestants could receive points during the competition and for we cannot find such a way. The sum of all is then .

For the sake of simplicity, consider an infinite -dimensional space that we cut by planes. When the space is cut in some way, the next plane adds as many new parts as it had cut through. Let us take the lines that are intersections of the older planes with the new plane. The number of the new pieces is the same as the number of the parts that the lines are dividing the new plane into. Thus, , where is the result for intersections and is the number of new parts, i.e. the number of parts into which we can divide a plane by lines. By drawing on paper we can find the maximum values for , , and , which are , , a . For we get the result .

For curious souls, this is a generalization: .

We have two types of solids: A regular tetrahedron with side length and a tetrahedral pyramid with all sides of length . The base area of the tetrahedral pyramid is and the height can be obtained from the Pythagorean theorem as , so its volume is . Similarly, the base area of the tetrahedron (equilateral triangle) is and the height , so its volume is . When we sum it up, we get .

\textit*{Other solution:} Watch what happens if you inscribe the star into a cube.

The probability that the hitchhiker is not picked during the first minutes is . If the probability of him not being picked during minutes is , then for the minutes it is . Hence and the probability of catching a car is .

For each mistake we find the probability that it won't stay till the end. The probability that the mistake will stay days is , and the probability that some mistake will not stay days is . The events of not spotting a mistake are independent, so we can multiply: .

Let denote the number of red cards, the number of blue cards and the number of yellow cards. Each red card adds points to the overall score ( for itself and for each blue card) and each yellow card adds points. Therefore is the overall score. For , the maximum score is . For , the score is always . For , it is worth to swap all red cards for yellow cards. Therefore we choose for and get the overall score in this case. Respecting the condition , we get the maximum at and , which yields the score .

First observe that both rolls have a symmetric distribution, i.e., the probability that we get on a -sided die is the same as the probability that we get (from to ). Similarly, the probability that we get on three -sided dice is the same as the probability that we get (from to ). By a similar line of thought we may conclude that the probability of Matthew winning is the same as the probability of CD winning. Thus, the answer is , where is the probability of a draw. And that is , because whatever CD rolls, Matthew always has a chance that he rolls the same.

During each route, the traveler gathers each of the numbers , , , at least twice (one time for the column and one time for the row). Furthermore, for each except for and , there exists a route that crosses the -th row exactly once and -th column exactly once. Thus, the will occur in the result exactly once. As for the , each traveler's number contains at least three 's. (The traveler must step on the bottom right corner and on the previous where there is , too.) It is easy to verify that there is a route that does not contain more than three 's.

From the fact that the area of the triangle is half of the product of its height and length of its side we get that the triangle will be similar to the triangle with sides , , . We can express one of the altitudes using its sides. For example, we can use the Heron's formula for sides of length , a and then compare this with the area . Thus, the , from which it follows that and perimeter is then .

Denote , then . By solving the quadratic equation for we get , and because is rational, also for some rational. After simple manipulation we have . In order for to be an integer, has to be an integer as well: If it were with , coprime, then and would divide , which is a contradiction. Since , must be odd. Moreover, from we have . Thus, is the solution.

First we choose the tile in the middle to be (, and , ). Denote by , the numbers written in the little square at the coordinates and , respectively ( being the upper left corner square). Observe that and . If the couple equals one of , , , there is only one way to fill the rest of the maxi-square. In the case , there are four ways. As there are possible middle tiles, the result is .

1010309

Let us focus on the condition that the number increased by 1 is divisible by . From this it immediately follows that the last digit has to be . Because of the divisibility by three, the digit sum of the balloon is of the form and also it should be twice the number of digits from the statement. So the digit sum is even, of the form and greater than , thus it can only be or , to which correspond numbers of digits and , respectively. Since odd and even digits alternate, the smallest possible solutions relaxing the condition on divisibility by are and with digit sums and . The digit sum in the second case is odd, so there is no way of achieving the digit sum of . In the first case we have to increase the digit sum by , which means that we have to increase one digit by . We can easily check that the only solution (taking divisibility by back into consideration) is the number .

9376

We seek such that for some integer . As divides the right-hand side, it must divide the left-hand side; thus divides either or , since they are coprime. Similarly divides either or . The former divisibility implies that is of the form , where and . The remainder of when divided by is , therefore the remainder of when divided by is . Hence from the latter divisibility. It is then readily seen that has digits if and only if , , ; we can easily verify that this satisfies all the conditions.

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We establish a one-to-one correspondence between -digit palindromes and -digit numbers—-the number corresponds to the palindrome for a non-zero digit and arbitrary digits and . The total sum of the palindromes can be then computed digit-wise; each digit adds , each digit adds and each adds to the total sum. Each possible value of appears in the total sum times and each possible value of or appears times. The total sum is thus .

First we rewrite the problem to get rid of the condition of , , , being odd. Let , and likewise for , , . Then there is a one-to-one correspondence between quadruples of odd positive integers satisfying and quadruples of nonnegative integers satisfying . The solution is now a standard exercise on combinations with repetitions allowed; the answer is (each quadruple corresponds to a sequence of 47 ”balls” and 3 ”separators”).

Let be the number satisfying the conditions above. The number created by subsequently writing two times is equal to for a suitable integer . Therefore, if we let , we obtain , which is a perfect square. It remains to choose in such a way that has eleven digits and its decimal representation starts with . The only possible choice is .

132

Let be one of the triangles (, ). Let be the incenter of and its circumcenter. Since , the angle is right and the radii and of the inscribed and the circumscribed circle of , respectively, are found as and . There is only one diameter of the circle different from that is tangent to the incircle of ; it is the one symmetric to with respect to the line . Note that lies on the perpendicular bisector of . Indeed, this bisector passes through , and since its distance from is , its distance from is . The other triangle is then symmetrical to w.r.t. and and . Finally, the intersection of our two triangles can be computed by taking the area of and subtracting the areas of three small triangles similar to ; their side lengths are 9, 12, 15 (the triangle at ), 6, 8, 10 (at ), and 3, 4, 5 (at ), respectively. Thus the area of the intersection is

2040

Let us paint the ten cells one after another. Denote the cells (columns by letters). Without loss of generality, the cells , are painted (from symmetry, it is sufficient to multiply the answer by ) and also the cell (multiply by four). If the cell is painted, it remains to determine the number of complying paintings of grid, which is . If a different cell in the second row is painted, w.l.o.g. let it be (multiply by 3), and in the third column (again multiply by 3). If is now painted, there is only one possibility how to finish (, , , are painted). Otherwise there are four possibilities (without loss of generality in the third row is painted and in the fourth column). To sum up, the overall number of colorings is

Let be the value we are minimizing. If and then and using triangle inequalities it is easy to show that if is closer to exactly , or vertices than , then . Next, let us assume that the remoteness of is measured to and the remotenesses of , , are measured to , i.e., let us minimize the value . The minimum is attained when and is the Fermat point of the triangle , that is if . (The required property of the Fermat point can be verified by rotating the triangle around by into a triangle and comparing the length of the broken line to the length of the segment ). Finally, we compute using the Pythagorean theorem. It turns out to be less than two so it is the answer.

The place No. 1 can be taken as the last one only if place No. 2 is taken as the first one (with the probability of ) and place No.  right after it. Then some places from to will be filled until gaps of size and remain. In that moment the probability of place No. 1 to be taken as the last one is equal to the inverse of number of the places, since every place has equal probability of being taken as the last one. Note that by looking only at sizes of the gaps and placing cars into the widest ones, we always end up with the same number of gaps of sizes and . Thus we can uniquely determine their count.

Let be the number of places which remain free, if the cars park on places in a row with first and last places already taken. The first car will park in the middle, which splits the task into the cases and . Thus we have a recurrence relation with initial conditions and . After a bit of work and computing values for small we can derive closed form:

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Since

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