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Problem 1. the Windows on an Old Tram Look Like Shown in the Picture

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Problem 1. the Windows on an Old Tram Look Like Shown in the Picture Problem 1. The windows on an old tram look like shown in the picture. All curves forming the round corners are arcs of a quarter circle with radius 10 cm. A portion of the sliding window is opened 10 cm as you can see in the second picture. The height of the open section is 13 cm. What is the area of the opening in cm2? Result. 130 Solution. It is just the area of the sliding window overlapping the fixed part, which is a rectangle 10 cm × 13 cm. Problem 2. A rectangle is subdivided into nine smaller rectangles as shown in the picture. The number written inside a small rectangle denotes its perimeter. Find the perimeter of the large rectangle. 9 14 10 17 12 Result. 42 Solution. Looking at the picture we notice that the perimeter of the large rectangle equals the sum of the perimeters of the four outer small rectangles with given perimeter minus the perimeter of the middle one. Therefore, the answer is 14 + 9 + 17 + 12 − 10 = 42: Problem 3. Peter deleted one digit from a four-digit prime number and obtained 630. What was the prime number? Result. 6301 Solution. Since the last digit of a (four-digit) prime number cannot be even, the prime number was of the form 630∗. Moreover, the last digit cannot be 5, because the number would be divisible by 5. So the options 1, 3, 7 and 9 are left. But since 630 is divisible by 3, the digits 3 and 9 are impossible. Similarly, 630 is divisible by 7 and so would be 6307. Therefore, the number was 6301. Problem 4. Star architect Pegi wants to build a very modern pentagonal mansion on a rectangular lot of land having side lengths 35 m and 25 m. The floor area of the mansion fits into the lot as can be seen in the picture: (The dots on the boundary mark 5 m distance.) What fraction of the area of the lot does the floor area of the mansion cover? 41 Result. 70 Solution. Since we are only interested in the fraction of areas, we can use 5 m as a unit. By summing up the areas of the three right-angled triangles, we obtain the result 1 5 · 3 6 · 1 4 · 2 41 1 − + + = : 5 · 7 2 2 2 70 1 Problem 5. A square grid of 16 dots, as can be seen in the picture, contains the corners of nine 1 × 1-squares, four 2 × 2-squares, and one 3 × 3-square, for a total of 14 squares whose sides are parallel to the sides of the grid. What is the smallest possible number of dots you can remove so that, after removing those dots, each of the 14 squares is missing at least one corner? Result. 4 Solution. It is necessary to remove four dots, because the four 1 × 1-squares in the corners of the given grid do not share any dots in common. Removing two opposite corners of the grid and two centre dots along the other diagonal provides an example to show that this number is sufficient. Problem 6. Find the digit at the units place of the sum of squares 12 + 22 + 32 + ··· + 20172: Result. 5 Solution. The digits in the unit position of square numbers show up periodically having period 10. We have 12 + 22 + 32 + ··· + 102 = 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 385; so the last digit is 5. Consequently, the last digit of 12 + 22 + ··· + 20102 is 5 due to 201 · 5 = 1005. Furthermore, the last digit of the sum 20112 + 20122 + ··· + 20172 is 0. Altogether, the desired digit is 5. Problem 7. Express the quotient 0:2 0:24 a as a fraction in lowest terms with positive integers a and b. b Note: The overline denotes periodic decimal expansion, for example 0:123 = 0:123123 :::. 11 Result. 12 Solution. The given quotient can be written in the following way: 0:2 0:22 22 · 0:01 11 = = = : 0:24 0:24 24 · 0:01 12 2 Problem 8. Passau has a railway station in the form of a triangle. Anna, Boris, and Cathy observe the railway traffic in Linz, Regensburg, and Waldkirchen, respectively, on the rails coming from Passau. Anna counts 190, Boris 208 and Cathy 72 incoming and outgoing trains in total. How many trains went from Linz to Regensburg or vice versa if no train starts, ends or reverses its direction in Passau? Waldkirchen Cathy Anna Passau Linz Boris Regensburg Result. 163 Solution. We denote the number of trains between Linz and Waldkirchen by r, the one between Linz and Regensburg by w, and the one between Waldkirchen and Regensburg by l. Anna counts all the trains between Linz and the other two cities. Therefore, we obtain the equation r + w = 190. We get the equations l + w = 208 and l + r = 72 similarly. 1 Adding the first two equations and subtracting the third one gives 2w = 190 + 208 − 72 which leads to w = 2 · 326 = 163. Problem 9. Find all positive integers x < 10 000 such that x is a fourth power of some even integer and one can permute the digits of x in order to obtain a fourth power of some odd integer. The result of permuting may not begin with zero. Result. 256 Solution. Assume that x is equal to a4 for some even integer a and that it can be transformed to b4 for some odd integer b. Since 10 000 = 104, both a and b are less than 10. By squaring even one-digit integers and then squaring their last digits, we find out that a4 always ends with 6. Considering the possible values of b, we see that only 54 = 625 and 94 = 6561 contain the digit 6. However, any permutation of digits of 94 would be divisible by 3 (digit sum does not change), so the only possible result would be 64 = 1296, which cannot be permuted to 94. For b4 = 625 we easily find x = a4 = 256 as the only result. 1 Problem 10. In parallelogram ABCD a line through point C meets side AB in point E such that EB = 5 AE. The line segment CE intersects the diagonal BD in point F . Find the ratio BF : BD. D C F A E B Result. 1 : 7 Solution. The triangles EBF and CDF are similar with a scale factor EB : DC = 1 : 6. As a consequence BF : FD = 1 : 6 as well, hence BF : BD = 1 : 7. Problem 11. A big house consists of 100 numbered flats. In every flat, there lives one person or there live two or three persons. The total number of people living in the flats No. 1 to No. 52 is 56 and the total number of people living in the flats No. 51 to No. 100 is 150. How many people live in this house? Result. 200 Solution. Since the maximum number of people in a flat is three, exactly three people live in each flat from No. 51 to No. 100. Therefore, there are 56 − 2 · 3 = 50 people living in the first up to the 50th flat in total. This leads to a total number of 50 + 150 = 200 inhabitants for the entire house. 3 Problem 12. In the first stage, Nicholas wrote the number 3 with a red pencil and 2 with a green pencil on a sheet of paper. In the following stages, he used the red pencil for writing the sum of the two numbers from the previous stage and the green pencil for their (positive) difference. What number did he write in red in the 2017th stage? Result. 3 · 21008 Solution. It is easy to see that in each stage, the red number is greater than the green one. Assuming that in the n-th stage the numbers Rn, Gn are written with red and green, respectively, then in the stage n + 1, we have Rn+1 = Rn + Gn and Gn+1 = Rn − Gn and in the stage n + 2, Rn+2 = Rn+1 + Gn+1 = 2Rn; Gn+2 = Rn+1 − Gn+1 = 2Gn: Therefore, both the numbers are doubled each two stages. Between the 1st and the 2017th stage this happens 1008 times, therefore the result is 3 · 21008. Problem 13. Little Red Riding Hood finds herself at the entrance of the \Rectangular Forest". Starting at point A, she has to reach point B as fast as possible. One possibility is to walk along the edges of the woods which will be 140 m in total. Of course, she knows that according to the triangle inequality a direct path would be shorter. Unfortunately, there is only a path which has the form of a zigzag with two right-angled turns as can be seen in the picture. If she knew that this way was shorter than 140 m, she would dare to take the shortcut. Find the length (in meters) of the zigzag through the forest! B 60 m A 80 m Result. 124 p Solution. By the Pythagorean theorem, the length of the diagonal of the rectangle is 602 + 802 = 100. For example by the cathetus theorem, one can determine the shorter segment cut by the altitude to the diagonal as 602=100 = 36. Now we obtainp 100 − 36 = 64 for the longer segment of the diagonal. Using the altitude theorem, the length of the altitude is 36 · 64 = 48. Altogether, the zigzag path is of length 48 + (64 − 36) + 48 = 124. Problem 14.
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