IMC_2012_web_solutions

We note that b a 1 also each of the first 94 numbers

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We note that b a 1 . Also, each of the first 94 numbers in the list is between 1 and a , and each of the last 94 numbers is at least b. So if we replace the first 94 numbers by 1, and the last 94 numbers by b , we obtain the sequence of numbers     95 94 ..... , , , , , , 1 ..., 1, 1, 1, b b b b a (1) whose sum does not exceed 477, the sum of the original sequence. Therefore 477 95 94 b a (2) As a 1 , it follows that 477 95 94 95 95 b a b , hence 382 95 477 95 b and therefore P Q R S P Q S R T
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12 95 382 b . Therefore, since b is an integer, 4 b . When 4 b , it follows from (2) that 477 380 94 a , giving 3 a . This shows that the maximum possible values for a and b are 3 and 4, respectively. We can see that these values are possible, as, if we substitute these values in (1), we obtain a sequence of numbers with sum 477 4 95 3 1 94 . So 3.5 is the maximum possible value of the median. Extension Problem 22.1 What is the maximum possible value of the median number of cups of coffee bought per customer on a day when the Sundollars Coffee Shop sells 201 cups of coffee to 100 customers, and every customer buys at least one cup of coffee? 23. In the triangle PQR , 2 PS ; 1 SR ; 0 45 PRQ ; T is the foot of the perpendicular from P to QS and 0 60 PST . What is the size of QPR ? A 45 0 B 60 0 C 75 0 D 90 0 E 105 0 Solution: C In the triangle PST , 0 90 PTS and 0 60 PST . Therefore 0 30 TPS and the triangle PST is half of an equilateral triangle. It follows that 1 2 1 PS ST . Therefore triangle RST is isosceles, and hence SRT STR . By the Exterior Angle Theorem, PST SRT STR . Therefore 0 30 SRT STR . Hence 0 0 0 15 30 45 SRT PRQ QRT . Using the Exterior Angle Theorem again, it follows that QRT TQR STR , and hence 0 0 30 45 QRT STR TQR 0 15 . Therefore the base angles of triangle TQR are equal. Hence TQR is an isosceles triangle, and so RT QT . We also have that the base angles in triangle TPR are both equal to 0 30 , and so RT PT . Therefore PT RT QT . So PTQ is an isosceles right-angled triangle. Therefore 0 45 QPT . Finally, we deduce that 0 0 0 75 30 45 TPS QPT QPR . P Q R T S 60 0 45 0 2 1 P S Q R T 60 0 15 0 15 30 0 30 0 30 0 2 1 45 0 45 0
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13 24. All the positive integers are written in the cells of a square grid. Starting from 1, the numbers spiral anticlockwise. The first part of the spiral is shown in the diagram. Which number is immediately below 2012? A 1837 B 2011 C 2013 D 2195 E 2210 Solution: D The key to the solution is to note that the squares of the odd numbers occur on the diagonal leading downwards and to the right from the cell which contains the number 1, and the squares of the even numbers occur on the diagonal which leads upwards and to the left of the cell which contains the number 4.
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