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National Stage – Oral Phase 20th Philippine Mathematical Olympiad 20 January 2018 JUDGES' COPY EASY 15 seconds, 2 points 1. A line passes through (k; −9) and (7; 3k) and has slope 2k. Find the possible values of k. 9 Answer: or 1 2 3k−(−9) 2 Solution: We have 7−k = 2k, so 3k + 9 = 2k(7 − k) or 2k − 11k + 9 = (2k − 9)(k − 1) = 0. Hence 9 k = 2 or k = 1. 2. Let ∆ABC be a right triangle with legs AB = 6 and BC = 8. Let R and r be the circumradius and the inradius of ∆ABC, respectively. Find the sum of R and r. Answer: 7 Solution: Since ∆ABCp is a right triangle, its circumradius is just half of its hypotenuse. By Pythagorean 2 2 AC theorem, AC = 6 + 8 = 10 and so R = 2 = 10=2 = 5. Now, we can compute the inradius by using the following identity: AB + BC + AC r × = Area (∆ABC) : 2 Since ∆ABC is a right triangle, its area is just (AB × BC)=2 = 48=2 = 24. So that 24 r = = 2: (6 + 8 + 10)=2 Thus, R + r = 5 + 2 = 7. 3. In how many ways can the letters of the word CHIEF be arranged such that I appears at some position after E? Answer: 60 Solution: Start with an empty string that should have 5 letters. We first choose where to put I and E, 5 and there are 2 ways of doing this. Each way has 3! ways to rearrange the remaining letters, so the number of ways is 10 ∗ 6 = 60 ways. Alternatively, note that the letters can be arranged without restriction in 5! ways. Since there are the same number of arrangements where I appears before E as those where E appears before I, the required number is 5!=2 = 60 ways. 4. We say that the constant a is a fixed point of a function f if f(a) = a. Find all values of c such that f(x) = x2 − 2 and g(x) = 2x2 − c share a common fixed point. Answer: 3 and 6 Solution: Note that if f(x) = x, then x2 − x − 2 = 0, or (x − 2)(x + 1) = 0. Thus, the fixed points of f are 2 and −1, and so we wish for either of these to be a fixed point of g. If 2 is a fixed point of g, we must have g(2) = 2 = 8 − c, and so c = 6. If −1 is a fixed point of g, then 2 − c = −1, and c = 3. 5. How many pairs of positive integers (a; b) are there, both not exceeding 10, such that a ≤ gcd(a; b) ≤ b and a ≤ lcm(a; b) ≤ b? Answer: 27 Solution: gcd(a; b) ≥ a =) gcd(a; b) = a. Likewise, lcm(a; b) = b. This can only happen if b is a multiple of a. Hence the answer is 10 + 5 + 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 = 27. Page 1 of 11 (log 3)2 (log 7)2 6. Suppose that loga 125 = log5 3 and logb 16 = log4 7. Find the value of a 5 − b 4 . Answer: −22 Solution: Note that alog5 3 = 125 and blog4 7 = 16. Thus 2 2 a(log5 3) − b(log4 7) = (alog5 3)log5 3 − (blog4 7)log4 7 = 125log5 3 − 16log4 7 = 33 − 72 = −22. 7. Two cards are chosen, without replacement, from a deck of 50 cards numbered 1; 2; 3;:::; 50. What is the probability that the product of the numbers on these cards is divisible by 7? 46 Answer: 175 Solution: 50 Note that there are b 7 c = 7 multiples of 7 from 1 to 50, so there are 43 numbers which are not divisible by 7. The probability that after choosing two cards, the product of these numbers is not divisible by 43 50 43·42 129 129 46 7 is 2 = 2 = 50·49 = 175 . Hence, the probability in question is then 1 − 175 = 175 . 8. A triangle with side lengths 24, 70, 74 is inscribed in a circle. Find the difference between the numerical values of the area and the circumference of the circle in terms of π: Answer: 1295π Solution: Notice that the given triangle is a right triangle with hypotenuse equal to 74 because 122 + 352 = 372 which implies that 242 + 702 = 742: From this, we know that the area and circumference of the circle are 372(π) or 1369π and 74π; respectively. Thus, the difference is equal to 1295π: 9. Find the smallest positive real numbers x and y such that x2 − 3x + 2:5 = sin y − 0:75. 3 π Answer: x = ; y = 2 2 3 2 Solution: The equation above can be rewritten as x − 2 + 1 = sin y: Since the left side is always greater or equal to 1 and the right side is always less than or equal to 1, then both sides must be equal 3 π to 1. This means x = 2 and y = (4k + 1) 2 for all k 2 Z. Choosing k = 0 gives the smallest possible value of y. 10. A standard deck of 52 cards has the usual 4 suits and 13 denominations. What is the probability that two cards selected at random, and without replacement, from this deck will have the same denomination or have the same suit? 5 Answer: 17 Solution: Let A be the event that the 2 chosen cards will have the same denomination; and let B be the event that the 2 chosen cards will have the same suit. Note that A \ B = ;. So that P(A [ B) = P(A) + P(B): Since there are 4 suits to choose from, then there are 4C2 = 6 possible pairs of the same value. There are a total of 13 card values, so that (13)(6) 1 P(A) = = : 52C2 17 Since there are 13 card denominations to choose from, then there are 13C2 = 78 possible pairs of the same suit. There are a total of 4 suits, so that (4)(78) 4 P(B) = = : 52C2 17 1 + 4 5 Thus, (A [ B) = = : P 17 17 Page 2 of 11 a b 1 11. Let a and b be integers for which + = . Find the smallest possible value of jabj. 2 1009 2018 Answer: 504 Solution: Clear denominators to write this as 1009a + 2b = 1. Clearly, a = 1, b = −504 is a solution, and so our solutions are of the form a = 1 + 2k, b = −504 − 1009k. Now, clearly jaj ≥ 1, and jbj ≥ 504, so jabj ≥ 504, and equality is attained when a = 1 and b = −504. p 1 12. Let x be a real number satisfying x2 − 6x + 1 = 0. Find the numerical value of x4 − . x4 p Answer: 8 3 p 1 Solution: Note that x + x = 6. Then 4 1 2 1 2 1 x − = x + x − x4 x2 x2 2 1 1 1 = x + x + x − x2 x x ! 0s 1 1 2 1 1 2 = x + − 2 x + x + − 4 x x @ x A p p p = 4 6 2 = 8 3 13. Factor (a + 1)(a + 2)(a + 3)(a + 4) − 120 completely into factors with integer coefficients. Answer: (a2 + 5a + 16)(a − 1)(a + 6) Solution: We have (a + 1)(a + 2)(a + 3)(a + 4) − 120 = (a + 1)(a + 4)(a + 2)(a + 3) − 120 = (a2 + 5a + 4)(a2 + 5a + 6) − 120 = (a2 + 5a + 5)2 − 1 − 120 = (a2 + 5a + 5)2 − 121 = (a2 + 5a + 5 + 11)(a2 + 5a + 5 − 11) = (a2 + 5a + 16)(a2 + 5a − 6) = (a2 + 5a + 16)(a − 1)(a + 6): 14. Suppose there are 3 distinct green balls, 4 distinct red balls, and 5 distinct blue balls in an urn. The balls are to be grouped into pairs such that the balls in any pair have different colors. How many sets of six pairs can be formed? Answer: 1440 Solution: Since there are 7 green and red balls combined, and 5 blue balls, then there should be one pair with 1 green and 1 red, and all the other pairs must have a blue ball. There are 3 × 4 ways of selecting a green-red pair, and 5! ways of selecting the partners for the blue balls. Thus, there are 3 × 4 × 5! = 1440 possible pairings. 15. This year, our country's team will be participating in the 59th International Mathematical Olympiad, to be held in Cluj-Napoca, Romania. The IMO, which was first held in 1959 also in Romania, has been held annually except in 1980, when it was cancelled due to internal strife in its host country. Which East Asian country was supposed to host the 1980 IMO? Answer: Mongolia Page 3 of 11 AVERAGE 45 seconds, 3 points 1. A regular hexagon is inscribed in another regular hexagon such that each vertex of the inscribed hexagon divides a side of the original hexagon into two parts in the ratio 2 : 1. Find the ratio of the area of the inscribed hexagon to the area of the larger hexagon. 7 Answer: 9 Solution: Without loss of generality, we may assume that the original hexagon has side length 3. Let s then be the side length of the inscribed hexagon. Notice that there are six triangles with sides 1, 2, and s, and the angle between the sides of lengths 1 and 2 is 120◦, as shown below: By the cosine law, we have 1 s2 = 12 + 22 − 2 · 1 · 2 · − = 7 2 p and so s = 7. Since the ratio of the areas of two similar shapes is equal to the square of the ratio p 2 7 7 of their sides, the desired ratio is thus 3 = 9 .
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