2010 Euclid Contest. Solutions

Transcription

1 Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 00 Euclid Contest Wednesday, April 7, 00 Solutions 00 Centre for Education in Mathematics and Computing

2 00 Euclid Contest Solutions Page. (a) Solution Since 3 x 7, then 3 x+ 3 x Solution Since 3 x 7 and 7 3 3, then x 3. Therefore, 3 x (b) Since x , then x (c) The lines y x + and y x + both pass through the point B on the y-axis. Since the y-intercept of the line y x + is, then B has coordinates (0, ). Next, we find the x-intercepts of each of the two lines by setting y 0. If y x + and y 0, then x + 0 or x, so A has coordinates (, 0). If y x + and y 0, then 0 x + or x, and so x 4. Thus, C has coordinates (4, 0). Since BO and AC are perpendicular, then we can treat AC as the base of ABC and BO as its height. Note that BO and AC 4 ( ) 6. Therefore, the area of ABC is AC BO (a) Let r, g and b be the masses of the red, green and blue packages, respectively. We are told that r + g + b 60, r + g 5, and g + b 50. Subtracting the second equation from the first, we obtain b Substituting into the third equation, we obtain g 50 b Therefore, the mass of the green package is 5 kg. (b) Suppose that a palindrome p is the sum of the three consecutive integers a, a, a +. In this case, p (a ) + a + (a + ) 3a, so p is a multiple of 3. The largest palindromes less than 00 are 9, 8, 7. Note that 9 and 8 are not divisible by 3, but 7 is divisible by 3. One way to check these without using a calculator is to use the test for divisibility by 3: A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. Therefore, 9 and 8 cannot be the sum of three consecutive integers. The integer 7 can be written as , so 7 is the largest palindrome less than 00 that is the sum of three consecutive integers. (c) Solution Since (x + )(x ) 8, then x 8 or x 9. Thus, (x + x)(x x) x(x + )x(x ) x (x + )(x ) 9(8) 7. Solution Since (x + )(x ) 8, then x 8 or x 9, so x ±3. If x 3, then (x + x)(x x) (3 + 3)(3 3) (9 + 3)(9 3) (6) 7. If x 3, then (x + x)(x x) (( 3) + ( 3))(( 3) ( 3)) (9 3)(9 + 3) 7. In either case, (x + x)(x x) (a) Solution Bea spends 60 minutes flying from H to F, 30 minutes at F, 45 minutes flying from F to G, 60 minutes at G, and then flies from G to H.

3 00 Euclid Contest Solutions Page 3 Thus, her total time is minutes plus the length of time that it takes her to fly from G to H. Since Bea flies at a constant speed, then the ratio of the two distances equals the ratio of the corresponding times. Therefore, HF 60 minutes GF 45 minutes 4 3. Since F GH is right-angled at F, then F GH must be similar to a triangle, and so HG GF 5 3. In particular, this means that the ratio of the times flying H to G and F to G is also 5. 3 Thus, it takes her minutes to fly from G to H. 3 In conclusion, Bea is away from her hive for minutes. Solution Bea spends 60 minutes flying from H to F, 30 minutes at F, 45 minutes flying from F to G, 60 minutes at G, and then flies from G to H. Thus, her total time is minutes plus the length of time that it takes her to fly from G to H. Since Bea flies at a constant speed, then the ratio of the two distances equals the ratio of the corresponding times. Therefore, we can use the Pythagorean Theorem on the times to obtain Time G to H (Time H to F ) + (Time F to G) min since the time is positive. In conclusion, Bea is away from her hive for minutes. (b) Solution Since OP B 90, then OP and P B are perpendicular, so the product of their slopes is. The slope of OP is 4 0 p and the slope of P B is p p 0 4 p 0. Therefore, we need 4 p 4 p 0 p 0p (p )(p 8) 0 6 p(p 0) and so p or p 8. Since each these steps is reversible, then OP B is right-angled precisely when p and p 8. Solution Since OP B is right-angled at P, then OP +P B OB by the Pythagorean Theorem. Note that OB 0 since O has coordinates (0, 0) and B has coordinates (0, 0). Also, OP (p 0) + (4 0) p + 6 and P B (0 p) + (4 0) p 0p + 6. Therefore, (p + 6) + (p 0p + 6) 0 p 0p p 0p + 6 0

4 00 Euclid Contest Solutions Page 4 and so (p )(p 8) 0, or p or p 8. Since each these steps is reversible, then OP B is right-angled precisely when p and p (a) Suppose that Thurka bought x goats and y helicopters. Then 9x + 7y 0. Since x and y are non-negative integers, then 9x 0 so x 0. If x 0, then 7y 0 9x, which does not have an integer solution because is not divisible by 7. If x 9, then 7y 0 9x 30, which does not have an integer solution. If x 8, then 7y 0 9x 49, which does not have an integer solution. If x 7, then 7y 0 9x 68, so y 4. Therefore, 9(7) + 7(4) 0, and so Thurka buys 7 goats and 4 helicopters. (We can check that x 0,,, 3, 4, 5, 6 do not give values of y that work.) (b) Solution Manipulating algebraically, (x + 8) 4 (x + 6) (x + 8) 4 (x + 8) 0 (x + 8) ((x + 8) ) 0 (x + 8) ((x + 8) + )((x + 8) ) 0 Therefore, x 8 or x 0 or x 6. Solution Manipulating algebraically, (x + 8) (x + 0)(x + 6) 0 (x + 8) 4 (x + 6) (x + 8) 4 (x + 8) 0 (x + 8) ((x + 8) ) 0 (x + 8) (x + 6x ) 0 Therefore, x 8 or x 0 or x 6. (x + 8) (x + 6x + 60) 0 (x + 8) (x + 0)(x + 6) 0 Solution 3 Since (x + 8) 4 (x + 6), then (x + 8) x + 6 or (x + 8) (x + 6). From the first equation, x + 6x + 64 x + 6 or x + 4x or (x + 6)(x + 8) 0. From the second equation, x + 6x + 64 x 6 or x + 8x or (x + 0)(x + 8) 0. Therefore, x 8 or x 0 or x 6.

5 00 Euclid Contest Solutions Page 5 5. (a) Solution We use the fact that g(x) g(f(f (x))). Since f(x) x +, then to determine f (x) we solve x y + for y to get y x or y (x ). Thus, f (x) (x ). Since g(f(x)) 4x +, then g(x) g(f(f (x))) g(f( (x ))) 4( (x )) (x ) + (x ) + x x + Solution We use the expressions for f(x) and g(f(x)) to construct g(x). Since f(x) is linear and g(f(x)) is quadratic, then it is likely that g(x) is also quadratic. Since f(x) x +, then (f(x)) 4x + 4x +. Since g(f(x)) has no term involving x, then we subtract f(x) (to remove the 4x term) to get (f(x)) f(x) (4x + 4x + ) (x + ) 4x To get g(f(x)) from this, we add to get 4x +. Therefore, g(f(x)) (f(x)) f(x) +, and so an expression for g(x) is x x +. Solution 3 We use the expressions for f(x) and g(f(x)) to construct g(x). Since f(x) is linear and g(f(x)) is quadratic, then it is likely that g(x) is also quadratic. Suppose that g(x) ax + bx + c for some real numbers a, b, c. Then g(f(x)) g(x + ) a(x + ) + b(x + ) + c a(4x + 4x + ) + b(x + ) + c 4ax + (4a + b)x + (a + b + c) Since we are told that g(f(x)) 4x +, then we can compare coefficients to deduce that 4a 4 and 4a + b 0 and a + b + c. From the first equation, a. From the second equation, b a. From the third equation, c a b. Therefore, an expression for g(x) is x x +. (b) Solution Since the sum of the first two terms is 40 and the sum of the first three terms is 76, then the third term is Since the sum of the first three terms is 76 and the sum of the first four terms is 30, then the fourth term is Since the third term is 36 and the fourth term is 54, then the common ratio in the geometric sequence is Therefore, the fifth term is and the sixth term is

6 00 Euclid Contest Solutions Page 6 Also, the second term is and the first term is Thus, the first six terms of the sequence are 6, 4, 36, 54, 8, 43. Since the first term equals 4 and the common ratio is 3, then the nth term in the sequence ( ) n 3 is 4 3n. n 5 When n 6, this is a fraction whose numerator is odd and whose denominator is even, and so, when n 6, the nth term is not an integer. (An odd integer is never divisible by an even integer.) Therefore, there will be 5 integers in the sequence. Solution Suppose that a is the first term and r is the common ratio between consecutive terms (so that ar is the second term, ar is the third term, and so on). From the given information, a+ar 40 and a+ar+ar 76 and a+ar+ar +ar Subtracting the first equation from the second, we obtain ar 36. Subtracting the second equation from the third, we obtain ar Since ar 3 54 and ar 36, then r ar3 ar Since ar 36 and r 3, then a( 3 ) 36 or 9a 36 or a Since a 6 and r 3 43, then the first six terms of the sequence are 6, 4, 36, 54, 8,. Since the first term equals 4 and the common ratio is 3, then the nth term in the sequence ( ) n 3 is 4 3n. n 5 When n 6, this is a fraction whose numerator is odd and whose denominator is even, and so, when n 6, the nth term is not an integer. (An odd integer is never divisible by an even integer.) Therefore, there will be 5 integers in the sequence. 6. (a) In a triangle, the ratio of the side opposite the 90 to the side opposite the 60 angle is : 3. Note that each of ABC, ACD, ADE, AEF, AF G, and AGH is a triangle. Therefore, AH AG AG AF AF AE AE AD AD AC AC AB. 3 Thus, AH 3 AG ( ) ( ) 3 ( ) 4 ( ) 5 ( ) 6 3 AF 3 AE 3 AD 3 AC 3 AB. (In other words, to get from AB to the length of AH, we multiply by the scaling factor 3 six times.) ( ) 6 Therefore, AH (b) Solution Since AF D is right-angled at F, then by the Pythagorean Theorem, AD AF + F D since AD > 0. Let F AD β. Since ABCD is a rectangle, then BAF 90 β. Since AF D is right-angled at F, then ADF 90 β. Since ABCD is a rectangle, then BDC 90 (90 β) β.

7 00 Euclid Contest Solutions Page 7 A 90 B 90 F D E C Therefore, BF A, AF D, and DF E are all similar as each is right-angled and has either an angle of β or an angle of 90 β (and hence both of these angles). Therefore, AB AF DA 4( and so AB 5) 4 5. DF Also, F E F D F D () and so F E. 4 F A Since ABCD is a rectangle, then BC AD 5, and DC AB 4 5. Finally, the area of quadrilateral BCEF equals the area of DCB minus the area DF E. Thus, the required area is (DC)(CB) (DF )(F E) (4 5)( 5) ()() 0 9 Solution Since AF D is right-angled at F, then by the Pythagorean Theorem, AD AF + F D since AD > 0. Let F AD β. Since ABCD is a rectangle, then BAF 90 β. Since BAF is right-angled at F, then ABF β. Since AF D is right-angled at F, then ADF 90 β. Since ABCD is a rectangle, then BDC 90 (90 β) β. A 90 B 90 F D Looking at AF D, we see that sin β F D AD 5, cos β AF 5 AD 4 5, 5 and tan β F D AF 4. Since AF 4 and ABF β, then AB AF sin β Since F D and F DE β, then F E F D tan β. Since ABCD is a rectangle, then BC AD 5, and DC AB 4 5. Finally, the area of quadrilateral EF BC equals the area of DCB minus the area DF E. Thus, the required area is (DC)(CB) (DF )(F E) (4 5)( 5) ()() 0 9 E C

8 00 Euclid Contest Solutions Page 8 7. (a) Using the facts that 9 3 and 7 3 3, and the laws for manipulating exponents, we have 3 x 9 3 x 7 3 x (3 ) 3 x x 3 3 x x + 3 x 3 3 When two powers of 3 are equal, their exponents must be equal so x + 3 x 3 x 3 x + 3 3x (multiplying by x ) x 3 4x Since x satisfies the equation, then x is a factor of the left side. Using long division or synthetic division, we can factor this out to get (x )(x 3x 3) 0. Using the quadratic formula, the quadratic equation x 3x 3 0 has roots x 3 ± ( 3) 4()( 3) 3 ± Therefore, the solutions to the original equation are x and x 3 ±. (b) To determine the points of intersection, we equate y values of the two curves and obtain log 0 (x 4 ) (log 0 x) 3. Since log 0 (a b ) b log 0 a, the equation becomes 4 log 0 x (log 0 x) 3. We set u log 0 x and so the equation becomes 4u u 3, or u 3 4u 0. We can factor the left side as u 3 4u u(u 4) u(u + )(u ). Therefore, u(u + )(u ) 0, and so u 0 or u or u. Therefore, log 0 x 0 or log 0 x or log 0 x. Therefore, x or x or x Finally, we must calculate the y-coordinates of the points of intersection. Since one of the original curves is y (log 0 x) 3, we can calculate the corresponding values of y by using the fact that y u 3. The corresponding values of y are y and y ( ) 3 8 and y 3 8. Therefore, the points of intersection are (, 0), (, 8) and (00, 8) (a) If Oi-Lam tosses 3 heads, then George has no coins to toss, so cannot toss exactly head. If Oi-Lam tosses, or 0 heads, then George has at least one coin to toss, so can toss exactly head. Therefore, the following possibilities exist: Oi-Lam tosses heads out of 3 coins and George tosses head out of coin Oi-Lam tosses head out of 3 coins and George tosses head out of coins Oi-Lam tosses 0 heads out of 3 coins and George tosses head out of 3 coins We calculate the various probabilities. If 3 coins are tossed, there are 8 equally likely possibilities: HHH, HHT, HTH, THH, TTH, THT, HTT, TTT. Each of these possibilities has probability ( ) 3. Therefore, 8

9 00 Euclid Contest Solutions Page 9 the probability of tossing 0 heads out of 3 coins is 8 the probability of tossing head out of 3 coins is 3 8 the probability of tossing heads out of 3 coins is 3 8 the probability of tossing 3 heads out of 3 coins is 8 If coins are tossed, there are 4 equally likely possibilities: HH, HT, TH, TT. Each of ) 4 these possibilities has probability ( out of coins is. 4 If coin is tossed, the probability of tossing head is. To summarize, the possibilities are. Therefore, the probability of tossing head Oi-Lam tosses heads out of 3 coins (with probability 3 ) and George tosses head 8 out of coin (with probability ) Oi-Lam tosses head out of 3 coins (with probability 3 ) and George tosses head 8 out of coins (with probability ) Oi-Lam tosses 0 heads out of 3 coins (with probability ) and George tosses head 8 out of 3 coins (with probability 3) 8 Therefore, the overall probability is (b) Suppose P AR x and QDR y. R y A x B P Q C D Since P R and P A are radii of the larger circle, then P AR is isosceles. Thus, P RA P AR x. Since QD and QR are radii of the smaller circle, then QRD is isosceles. Thus, QRD QDR y. In ARD, the sum of the angles is 80, so x +(x +40 +y )+y 80 or x+y 40 or x + y 70. Therefore, CP D x y (x + y + 40) (a) (i) Solution LS cot θ cot θ cos θ cos θ sin θ sin θ sin θ cos θ cos θ sin θ sin θ sin θ sin(θ θ) sin θ sin θ sin θ sin θ sin θ sin θ RS as required.

10 00 Euclid Contest Solutions Page 0 Solution LS cot θ cot θ cos θ cos θ sin θ sin θ cos θ sin θ cos θ sin θ cos θ cos θ cos θ sin θ cos θ cos θ ( cos θ ) sin θ RS sin θ as required. (ii) We use (i) to note that on. Thus, sin 8 cot 4 cot 8 and sin 6 cot 8 cot 6 and so S sin 8 + sin 6 + sin sin sin 89 (cot 4 cot 8 ) + (cot 8 cot 6 ) + (cot 6 cot 3 ) + + (cot 048 cot 4096 ) + (cot 4096 cot 89 ) cot 4 cot 89 since the sum telescopes. Since the cotangent function has a period of 80, and 800 is a multiple of 80, then cot 89 cot 9. Therefore, Therefore, α 4. S cot 4 cot 9 cos 4 cos 9 sin 4 sin 9 cos 4 sin sin 4 cos cos 4 sin + sin cos cos cos 4 + sin sin cos ( sin ) + sin sin 4 sin 4

11 00 Euclid Contest Solutions Page (b) Solution We use the notation A BAC, B ABC and C ACB. We need to show that A < (B + C). Since the sum of the angles in ABC is 80, then B + C 80 A, and so this inequality is equivalent to A < (80 A) which is equivalent to 3 A < 90 or A < 60. So we need to show that A < 60. We know that a < (b + c). Thus, a < b + c and so 4a < b + c + bc because all quantities are positive. Using the cosine law in ABC, we obtain a b + c bc cos A. Therefore, 4a < b + c + bc 4(b + c bc cos A) < b + c + bc 4b + 4c 8bc cos A < b + c + bc 4b + 4c 8bc cos A < b + c + bc + 3(b c) (since (b c) 0) 4b + 4c 8bc cos A < b + c + bc + 3b 6bc + 3c 4b + 4c 8bc cos A < 4b + 4c 4bc 8bc cos A < 4bc cos A > (since 8bc > 0) Since a < b + c, then a cannot be the longest side of ABC (that is, we cannot have a b and a c), so A must be an acute angle. Therefore, cos A > implies A < 60, as required. Solution We use the notation A BAC, B ABC and C ACB. We need to show that A < (B + C). Since the sum of the angles in ABC is 80, then B + C 80 A, and so this inequality is equivalent to A < (80 A) which is equivalent to 3 A < 90 or A < 60. So we need to show that A < 60. We know that a < (b + c) which implies a < b + c. a sin A Using the sine law in ABC, we obtain and c a sin C sin A. Therefore, we obtain equivalent inequalities b sin B c a sin B, which gives b sin C sin A a < b + c a < a sin B sin A + a sin C sin A a sin A < a sin B + a sin C (since sin A > 0 for 0 < A < 80 ) sin A < sin B + sin C ( ) ( B + C B C since a > 0. Next, we use the trigonometric formula sin B+sin C sin cos ( ) ( ) B + C B + C Since cos θ for any θ, then sin B + sin C sin sin. ).

12 00 Euclid Contest Solutions Page Therefore, ( ) B + C sin A < sin B + sin C sin ( ) B + C sin A < sin ( ) 80 A sin A < sin 4 sin( A) cos( A) < sin(90 A) sin( A) cos( A) < cos( A) Since 0 < A < 80, then cos( A) > 0, so sin( A) <. Since a < b + c, then a cannot be the longest side of ABC, so A must be an acute angle. Therefore, A < 30 or A < 60, as required. 0. Denote the side lengths of a triangle by a, b and c, with 0 < a b c. In order for these lengths to form a triangle, we need c < a + b and b < a + c and a < b + c. Since 0 < a b c, then b < a + c and a < b + c follow automatically, so only c < a + b ever needs to be checked. Instead of directly considering triangles and sets of triangle, we can consider triples (a, b, c) and sets of triples (a, b, c) with the appropriate conditions. For each positive integer k 3, we use the notation S k to denote the set of triples of positive integers (a, b, c) with 0 < a b c and c < a + b and a + b + c k. In this case, c < a + b and a + b + c k, so c + c < a + b + c k, so c < k or c < k. Also, if 0 < a b c and a + b + c k, then k a + b + c c + c + c, so 3c k or c 3 k. (a) Consider T (0), which is the number of elements in S 0. We want to find all possible triples (a, b, c) of integers with 0 < a b c and c < a + b and a + b + c 0. We need c < 0 5 and c 0. Thus, c 4. 3 Therefore, we need 0 < a b 4 and a + b 6. There are two possibilities: (a, b, c) (, 4, 4) or (a, b, c) (3, 3, 4). Therefore, T (0). Consider T (). We want to find all possible triples (a, b, c) of integers with 0 < a b c and c < a + b and a + b + c. We need c < and c. Thus, c 4 or c 5. 3 If c 4, we need 0 < a b 4 and a + b 7. There is only one possibility: (a, b, c) (3, 4, 4). If c 5, we need 0 < a b 5 and a + b 6. There are three possibilities: (a, b, c) (, 5, 5) or (a, b, c) (, 4, 5) or (a, b, c) (3, 3, 5). Therefore, T () 4. Consider T (). We want to find all possible triples (a, b, c) of integers with 0 < a b c and c < a + b and a + b + c. We need c < and c. Thus, c 4 or c 5. 3 If c 4, we need 0 < a b 4 and a + b 8. There is only one possibility: (a, b, c) (4, 4, 4).

13 00 Euclid Contest Solutions Page 3 If c 5, we need 0 < a b 5 and a + b 7. There are two possibilities: (a, b, c) (, 5, 5) or (a, b, c) (3, 4, 5). Therefore, T () 3. (b) We show that T (m) T (m 3) by creating a one-to-one correspondence between the triples in S m and the triples S m 3. Note that S m is the set of triples (a, b, c) of positive integers with 0 < a b c, with c < a + b, and with a + b + c m. Also, S m 3 is the set of triples (A, B, C) of positive integers with 0 < A B C, with C < A + B, and with A + B + C m 3. Consider a triple (a, b, c) in S m and a corresponding triple (a, b, c ). We show that (a, b, c ) is in S m 3 : Since (a, b, c) is in S m, then c < (m) m. This means that b c m, so a m b c. Therefore, a, b and c are positive integers since a, b and c are positive integers with a b c. Since a b c, then a b c, so 0 < a b c. Since a + b + c m, then c m (a + b) so a + b and c have the same parity. Since c < a + b, then c a + b. (In other words, it cannot be the case that c a + b.) Therefore, c (a ) + (b ) ; that is, c < (a ) + (b ). Since a + b + c m, then (a ) + (b ) + (c ) m 3. Therefore, (a, b, c ) is in S m 3, since it satisfies all of the conditions of S m 3. Note as well that two different triples in S m correspond to two different triples in S m 3. Thus, every triple in S m corresponds to a different triple in S m 3. Thus, T (m) T (m 3). Consider a triple (A, B, C) in S m 3 and a corresponding triple (A +, B +, C + ). We show that (A +, B +, C + ) is in S m : Since (A, B, C) is in S m 3, then A, B and C are positive integers, so A +, B + and C + are positive integers. Since 0 < A B C, then < A+ B + C +, so 0 < A+ B + C +. Since C < A + B, then C + < (A + ) + (B + ) so C + < (A + ) + (B + ). Since A + B + C m 3, then (A + ) + (B + ) + (C + ) m. Therefore, (A +, B +, C + ) is in S m. Note again that two different triples in S m 3 correspond to two different triples in S m. Thus, every triple in S m 3 corresponds to a different triple in S m. Therefore, T (m 3) T (m). Since T (m) T (m 3) and T (m 3) T (m), then T (m) T (m 3). (c) We will use two important facts: (F) T (m) T (m 3) for every positive integer m 3, and (F) T (k) T (k + ) for every positive integer k 3 We proved (F) in (b). Next, we prove (F): Consider a triple (a, b, c) in S k and a corresponding triple (a, b +, c + ). We show that the triple (a, b +, c + ) is in S k+ : Since a, b and c are positive integers, then a, b + and c + are positive integers.

14 00 Euclid Contest Solutions Page 4 Since 0 < a b c, then 0 < a b + c +. Since c < a + b, then c + < a + (b + ). Since a + b + c k, then a + (b + ) + (c + ) k +. Therefore, (a, b +, c + ) is in S k+. Note that, using this correspondence, different triples in S k correspond different triples in S k+. Thus, every triple in S k corresponds to a different triple in S k+. This proves that T (k) T (k + ). Suppose that n N is the smallest positive integer for which T (n) > 00. Then N must be odd: If N was even, then by (F), T (N 3) T (N) > 00 and so n N 3 would be an integer smaller than N with T (n) > 00. This contradicts the fact that n N is the smallest such integer. Therefore, we want to find the smallest odd positive integer N for which T (N) > 00. Next, we note that if we can find an odd positive integer n such that T (n) > 00 T (n ), then we will have found the desired value of n: This is because n and n are both odd, and by property (F), any smaller odd positive integer k will give T (k) T (n ) 00 and any larger odd positive integer m will give T (m) T (n) > 00. We show that N 309 is the desired value of N by showing that T (309) > 00 and T (307) 00. Calculation of T (309) We know that 309 c < 309, so 03 c For each admissible value of c, we need to count the number of pairs of positive integers (a, b) with a b c and a + b 309 c. For example, if c 54, then we need a b 54 and a + b 55. This gives pairs (, 54), (, 53),..., (76, 79), (77, 78), of which there are 77. Also, if c 53, then we need a b 53 and a + b 56. This gives pairs (3, 53),..., (77, 79), (78, 78), of which there are 76. In general, if c is even, then the minimum possible value of a occurs when b is as large as possible that is, when b c, so a 309 c. Also, the largest possible value of a occurs when a and b are as close to equal as possible. Since c is even, then 309 c is odd, so a and b cannot be equal, but they can differ by. In this case, a 54 c and b 55 c. Therefore, if c is even, there are (54 c) (309 c) + 3 c 54 possible pairs (a, b) and so 3 c 54 possible triples. In general, if c is odd, then the minimum possible value of a occurs when b is as large as possible that is, when b c, so a 309 c. Also, the largest possible value of a occurs when a and b are as close to equal as possible. Since c is odd, then 309 c is even, so a and b can be equal. In this case, a (309 c). Therefore, if c is odd, there are (309 c) (309 c) + 3c 307 possible pairs (a, b) and so 3c 307 possible triples. The possible even values of c are 04, 06,..., 5, 54 (there are 6 such values) and the possible odd values of c are 03, 05,..., 5, 53 (there are 6 such values).

15 00 Euclid Contest Solutions Page 5 Therefore, T (309) ( 3 (04) 54) + ( 3 (06) 54) + + ( 3 (54) 54) + ( 3 ) ( 307 (03) + 3 ) ( 307 (05) ) 307 (53) 3( ) ( ) ( )(5) (6)(57) Therefore, T (309) > 00, as required. ( ) Calculation of T (307) We know that 307 c < 307, so 03 c For each admissible value of c, we need to count the number of pairs of positive integers (a, b) with a b c and a + b 307 c. This can be done in a similar way to the calculation of T (309) above. If n is even, there are 3 c 53 possible triples. If n is odd, there are 3c 305 possible triples. The possible even values of c are 04, 06,..., 50, 5 (there are 5 such values) and the possible odd values of c are 03, 05,..., 5, 53 (there are 6 such values). Therefore, T (307) ( 3 (04) 53) + ( 3 (06) 53) + + ( 3 (5) 53) + ( 3 ) ( 305 (03) + 3 ) ( 305 (05) ) 305 (53) 3( ) ( ) ( )(5) (5)(8) Therefore, T (307) < 00, as required. ( ) Therefore, the smallest positive integer n such that T (n) > 00 is n 309. As a final note, we discuss briefly how one could guess that the answer was near N 309. Consider the values of T (n) for small odd positive integers n. In (a), by considering the possible values of c from smallest (roughly n) to largest 3 (roughly n), we saw that T () If we continue to calculate T (n) for a few more small odd values of n we will see that: T (3) T (5) T (7) T (9) T () T (3)

16 00 Euclid Contest Solutions Page 6 The pattern that seems to emerge is that for n odd, T (n) is roughly equal to the sum of the integers from to n, with one out of every three integers removed. 4 Thus, T (n) is roughly equal to of the sum of the integers from to n. 3 4 Therefore, T (n) ( n)( n + ) ( n) 48 n. It makes sense to look for an odd positive integer n with T (n) 00. Thus, we are looking for a value of n that roughly satisfies 48 n 00 or n or n 30. Since n is odd, then it makes sense to consider n 309, as in the solution above.