2011 Fermat Contest (Grade 11)

Transcription

1 Te CENTRE for EDUCATION in MATHEMATICS and COMPUTING 011 Fermat Contest (Grade 11) Tursday, February 4, 011 Solutions 010 Centre for Education in Matematics and Computing

2 011 Fermat Contest Solutions Page 1. Evaluating, = = 0 9. Answer: (D). If y = 77, ten 7y = 7y = 7(77) = = 8. Answer: (A) 3. Since te area of te rectangle is 19 and its lengt is 4, ten its widt is 19 4 = 8. Terefore, its perimeter is = 64. Answer: (A) 4. Since n + 9 = 5, ten n + 9 = 5 = 65. Tus, n = 65 9 = Since P RS is equilateral, ten all tree of its angles equal 60. In particular, RSP = 60. Since QS = QT, ten QST is isosceles and so T SQ = ST Q = 40. Since RST is a straigt line segment, ten RSP + P SQ + T SQ = 180. Terefore, 60 + x + 40 = 180 or x = = 80. Answer: (D) Answer: (C) 6. If te sum of tree consecutive integers is 7, ten te numbers must be 8, 9 and 10. (We could see tis algebraically by calling te integers x, x + 1 and x + and solving te equation x + (x + 1) + (x + ) = 7.) Teir product is = 70. Answer: (C) 7. Te number alfway between two numbers is teir average. Terefore, te number alfway between 1 and 1 is 1( ) = 1( ) = 1( ) = Answer: (D) 8. Since te angle in te sector representing cookies is 90, ten tis sector represents 1 of te 4 total circle. Terefore, 5% of te students cose cookies as teir favourite food. Tus, te percentage of students wo cose sandwices was 100% 30% 5% 35% = 10%. Since tere are 00 students in total, ten = 0 students said tat teir favourite 100 food was sandwices. Answer: (B) 9. Te set S contains 5 multiples of (tat is, even numbers). Wen tese are removed, te set S is left wit only te odd integers from 1 to 49. At tis point, tere are 50 5 = 5 integers in S. We still need to remove te multiples of 3 from S. Since S only contains odd integers at tis point, ten we must remove te odd multiples of 3 between 1 and 49. Tese are 3, 9, 15, 1, 7, 33, 39, 45, of wic tere are 8. Terefore, te number of integers remaining in te set S is 5 8 = 17. Answer: (D)

3 011 Fermat Contest Solutions Page Solution 1 Since P QRS is a square and QR = + 9 = 11, ten P Q = QR = SR = P S = 11. Te eigt of te saded rectangle equals te eigt of te top left rectangle minus te eigt of te top rigt rectangle, or 6 = 4. Te widt of te saded rectangle equals te widt of te top rigt rectangle minus te widt of te bottom rigt rectangle. Since SR = 11, ten te widt of te bottom rigt rectangle is = 1. Terefore, te widt of te saded rectangle is 8 1 = 7. Tus, te area of te saded rectangle is 4 7 = 8. Solution Since P QRS is a square and QR = + 9 = 11, ten P Q = QR = SR = P S = 11. Since te side lengt of te square is 11, ten its area is 11 = 11. Since P Q = 11, ten te widt of te top left rectangle is 11 8 = 3, and so its area is 3 6 = 18. Since P S = 11, ten te eigt of te bottom left rectangle is 11 6 = 5, and so its area is 5 10 = 50. Since SR = 11, ten te widt of te bottom rigt rectangle is = 1, and so its area is 1 9 = 9. Te area of te top rigt rectangle is 8 = 16. Tus, te area of te saded rectangle equals te area of square P QRS minus te combined areas of te four unsaded rectangles, or = 8. Answer: (B) 11. It is possible tat after buying 7 gumballs, Wally as received red, blue, 1 wite, and green gumballs. Tis is te largest number of eac colour tat e could receive witout aving tree gumballs of any one colour. If Wally buys anoter gumball, e will receive a blue or a green or a red gumball. In eac of tese cases, e will ave at least 3 gumballs of one colour. In summary, if Wally buys 7 gumballs, e is not guaranteed to ave 3 of any one colour; if Wally buys 8 gumballs, e is guaranteed to ave 3 of at least one colour. Terefore, te least number tat e must buy to guarantee receiving 3 of te same colour is 8. Answer: (E) 1. Solution 1 A parabola is symmetric about its axis of symmetry. Since te x-intercepts of te given parabola are x = 1 and x = 4, ten te axis of symmetry of te parabola is x = 1+4 = 3. Since te point (3, w) is 3 units to te rigt of te axis of symmetry, ten its y-coordinate (namely w) equals te y-coordinate of te point 3 units to te left of te axis of symmetry, wic is te point wit x = 0. Wen x = 0, we know tat y = 8. Terefore, w = 8. (We could also note tat x = 3 is 1 unit to te left of te rigtmost x-intercept so its y- coordinate is equal to tat of te point 1 unit to te rigt of te leftmost x-intercept.)

4 011 Fermat Contest Solutions Page 4 Solution Since te parabola as x-intercepts of 1 and 4, ten its equation is of te form y = a(x + 1)(x 4) for some value of a. Since te point (0, 8) lies on te parabola, ten 8 = a(1)( 4) or a =. Terefore, te parabola as equation y = (x + 1)(x 4). Since te point (3, w) lies on te parabola, ten w = (4)( 1) = 8. Answer: (E) 13. Since Xavier, Yolanda and Zixuan ave $50 in total, and te ratio of te amount tat Xavier as to te amount tat te oter two ave is 3 :, ten Xavier as 3 of te total, or 3 $50 = $ Terefore, Yolanda and Zixuan togeter ave $50 $30 = $0. We know tat Yolanda as $4 more tan Zixuan, so we must break $0 into two parts, one of wic is $4 larger tan te oter. If Yolanda as $1 and Zixuan as $8, tis satisfies te requirements. Terefore, Zixuan as $8. Answer: (B) 14. Te average of two multiples of 4 must be even, since we can write tese multiples of 4 as 4m and 4n for some integers m and n, wic means tat teir average is 1 (4m + 4n) wic equals m + n or (m + n), wic is a multiple of, and so is even. Eac of te oter four coices may be an odd integer in some cases. Here is an example for eac: (A) Te average of and 4 is 3, wic is not even (B) Te average of 3 and 7 is 5, wic is not even (C) Te average of 1 and 9 is 5, wic is not even (E) Te average of, 3 and 4 is 3, wic is not even Terefore, te correct answer is (D). Answer: (D) 15. Since m and n are consecutive positive integers wit n m > 0, ten n is greater tan m. Terefore, we can write n = m + 1. Since n m > 0, ten (m + 1) m > 0 or m + m + 1 m > 0 or m > 19 or m > 19. Since m is a positive integer, ten m 10. Tus, we want to find te minimum value of n + m = (m + 1) + m = m + m + 1 wen m 10. Tis minimum will occur wen m = 10 (since m + m + 1 increases wit m wen m is a positive integer). Terefore, te minimum possible value is (10 ) + (10) + 1 = 1. Answer: (E) 16. Solution 1 We label te bottom left corner as R and label various side lengts as and w: P X R w Y w Z w Q

5 011 Fermat Contest Solutions Page 5 Since te diagram is made up of rectangles, ten XY is parallel to P R, wic tells us tat Y XZ = RP Q. Also, Y Z is parallel to RQ, wic tells us tat XZY = P QR. Terefore, P RQ is similar to XY Z. Tus, RQ P R = Y Z XY. But Y Z = XY, RQ = 3w and P R = 4. Tis tells us tat 3w 4 = XY XY or 3 4 w = or w = 3 8. Solution As in Solution 1, we label te bottom left corner as R and label various side lengts as and w. Since Y Z = XY, ten te slope of line segment XZ is XY Y Z = XY XY = 1. Since P R = 4 and RQ = 3w, ten te slope of line segment P Q is P R RQ = 4 3w. Since line segment XZ is a portion of line segment P Q, ten te slopes of tese two line segments are equal, so 4 3w = 1 and so w = = 3 8. Answer: (C) 17. Since 3 x = 64 and 3 x = (3 x ), ten (3 x ) = 64 and so 3 x = ±8. Since 3 x > 0, ten 3 x = 8. Tus, 3 x = 1 3 x = 1 8. Answer: (E) 18. We label te stages in tis process as Stage 0 (a square), Stage 1 ( triangles), Stage (4 triangles), Stage 3 (8 triangles), and Stage 4 (16 triangles). We want to determine te lengt of te longest edge of one of te 16 triangles in Stage 4. At Stage 1, we ave two rigt-angled isosceles triangles wit legs of lengt 4. Consider a general rigt-angled isosceles triangle ABC wit legs AB and BC of lengt a. Since tis is a triangle, its ypotenuse AC as lengt a. We split te triangle into two equal pieces by bisecting te rigt-angle at B: B a a A M C Since ABC is isosceles, ten tis bisecting line is bot an altitude and a median. In oter words, it is perpendicular to AC at M and M is te midpoint of AC. Terefore, te two triangular pieces AMB and CMB are identical triangles. Te longest edges of tese triangles (AB and CB) are te legs of te original triangle, and so ave lengt a. Since te longest edge of te original triangle was a, ten te longest edge as been reduced by a factor of. Since we ave sown tat tis is te case for an arbitrary isosceles rigt-angled triangle, we can ten apply tis property to our problem. In Stage 1, te longest edge as lengt 4.

6 011 Fermat Contest Solutions Page 6 Since te longest edge in Stage 1 as lengt 4, ten te longest edge in Stage as lengt 4 = 4. Since te longest edge in Stage as lengt 4, ten te longest edge in Stage 3 as lengt 4 = =. Since te longest edge in Stage 3 as lengt, ten te longest edge in Stage 4 as lengt =. Answer: (B) 19. Suppose tat te radius of te larger circle is r. Join O to P. Ten OP = OS = r. Since Q is te midpoint of P R, and P R = 1, ten P Q = 1 P R = 6. Since OS = r and QS = 4, ten OQ = OS QS = r 4. Since OP Q is rigt-angled at Q, ten by te Pytagorean Teorem, OQ + P Q = OP (r 4) + 6 = r r 8r = r 5 = 8r r = 5 8 Terefore, te radius of te larger circle is 5, or Answer: (C) 0. Since b = ar, c = ar, and te product of a, b and c is 46656, ten a(ar)(ar ) = or a 3 r 3 = or (ar) 3 = or ar = = 36. Terefore, b = ar = 36. Since te sum of a, b and c is 114, ten a + c = 114 b = = 78. Answer: (A) 1. In te given pattern, te rt row contains r integers. Terefore, after n rows, te total number of integers appearing in te pattern is (n ) + (n 1) + n Tis expression is always equal to 1 n(n + 1). (If you ave never seen tis formula before, try to prove it!) Putting tis anoter way, te largest number in te nt row is 1 n(n + 1). To determine wic row te number 400 is in, we want to determine te smallest value of n for wic 1 n(n + 1) 400 or n(n + 1) 800. If n = 7, ten n(n + 1) = 756. If n = 8, ten n(n + 1) = 81. Terefore, 400 appears in te 8t row. Also, te largest integer in te 8t row is 406 and te largest integer in te 7t row is 378. Tus, we want to determine te sum of te integers from 379 (te first integer in te 8t row) to 406, inclusive. We can do tis by calculating te sum of te integers from 1 to 406 and subtracting te sum of te integers from 1 to 378. Since te sum of te integers from 1 to m equals 1 m(m + 1), ten te sum of te integers from 379 to 406 is equal to 1(406)(407) 1(378)(379) = Answer: (A)

7 011 Fermat Contest Solutions Page 7. Since p + p + q 1 p 1 + q = 17, ten 1 q = 17 or 1 p + q pq + 1 q 1 + pq p = 17 or p(pq + 1) q(pq + 1) = 17. Since p and q are positive integers, ten pq + 1 > 0, so we can divide out te common factor in te numerator and denominator to obtain p = 17 or p = 17q. q Since p and q are positive integers, ten q 1. Since p + q 100, ten 17q + q 100 or 18q 100 or q 100 = Since q is a positive integer, ten q 5. Terefore, te combined restriction is 1 q 5, and so tere are five pairs. (We can ceck tat tese pairs are (p, q) = (17, 1), (34, ), (51, 3), (68, 4), (85, 5).) Answer: (E) 3. First, we note tat te tree people are intercangeable in tis problem, so it does not matter wo rides and wo walks at any given moment. We abbreviate te tree people as D, M and P. We call teir starting point A and teir ending point B. Here is a strategy were all tree people are moving at all times and all tree arrive at B at te same time: D and M get on te motorcycle wile P walks. D and M ride te motorcycle to a point Y before B. D drops off M and rides back wile P and M walk toward B. D meets P at point X. D picks up P and tey drive back to B meeting M at B. Point Y is cosen so tat D, M and P arrive at B at te same time. Suppose tat te distance from A to X is a km, from X to Y is d km, and te distance from Y to B is b km. A X Y a d b B In te time tat it takes P to walk from A to X at 6 km/, D rides from A to Y and back to X at 90 km/. Te distance from A to X is a km. Te distance from A to Y and back to X is a + d + d = a + d km. Since te time taken by P and by D is equal, ten a 6 = a + d or 15a = a + d or 7a = d. 90 In te time tat it takes M to walk from Y to B at 6 km/, D rides from Y to X and back to B at 90 km/. Te distance from Y to B is b km, and te distance from Y to X and back to B is d+d+b = b+d km. Since te time taken by M and by D is equal, ten b 6 = b + d or 15b = b + d or 7b = d. 90 Terefore, d = 7a = 7b, and so we can write d = 7a and b = a. Tus, te total distance from A to B is a + d + b = a + 7a + a = 9a km. However, we know tat tis total distance is 135 km, so 9a = 135 or a = 15. Finally, D rides from A to Y to X to B, a total distance of (a + 7a) + 7a + (7a + a) = 3a km.

8 011 Fermat Contest Solutions Page 8 Since a = 15 km and D rides at 90 km/, ten te total time taken for tis strategy is 3 15 = Since we ave a strategy tat takes 3.83, ten te smallest possible time is no more tan Can you explain wy tis is actually te smallest possible time? If we didn t tink of tis strategy, anoter strategy tat we migt try would be: D and M get on te motorcycle wile P walks. D and M ride te motorcycle to B. D drops off M at B and rides back to meet P, wo is still walking. D picks up P and tey drive back to B. (M rests at B.) Tis strategy actually takes 4.15, wic is longer tan te strategy sown above, since M is actually sitting still for some of te time. Answer: (A) 4. Te six possible sums are w + x, w + y, w + z, x + y, x + z, and y + z. Since x < y, ten w + x < w + y. Since w < x, ten w + y < x + y. Since y < z, ten x + y < x + z. Since x < y, ten x + z < y + z. Terefore, we ave w + x < w + y < x + y < x + z < y + z. Tis list includes all of te sums except w + z. Since y < z and w < x, ten w + y < w + z < x + z, but we cannot say for sure weter x + y or w + z is larger. Tus, we know tat w + x is always te smallest sum and tat w + y is always te second smallest sum. Also, we know tat te tird and fourt smallest sums are w + z and x + y in some order. We can conclude tat w + x = 1 and w + y =, and w + z and x + y equal 3 and 4 in some order. From te first and second equations, (w + y) (w + x) = 1 or y x = 1. Case 1: w + z = 3 and x + y = 4 Since y x = 1 and x + y = 4, we add tese to obtain y = 5 or y = 5. Since w + y =, ten w = y = 5 = 1. Since w + z = 3, ten z = 3 w = 3 ( 1 ) = 7. Since x + y = 4, ten x = 4 y = 4 5 = 3. Terefore, we ave w = 1, x = 3, y = 5, and z = 7. We can ceck tat te six sums are 1,, 3, 4, 5, 6, wic are all different as required. Case : w + z = 4 and x + y = 3 Since y x = 1 and x + y = 3, we add tese to obtain y = 4 or y =. Since w + y =, ten w = y = = 0. Since w + z = 4, ten z = 4 w = 4 0 = 4. Since x + y = 3, ten x = 3 y = 3 = 1. Terefore, we ave w = 0, x = 1, y =, and z = 4. We can ceck tat te six sums are 1,, 3, 4, 5, 6, wic are all different as required. Terefore, te two possible values of z are 4 and 7. Te sum of tese values is = 15. Answer: (D)

9 011 Fermat Contest Solutions Page 9 5. Te smallest possible eigt of te pyramid will occur wen te four side faces are just toucing te circumference of te end faces of te cylinder. To see tis, consider starting wit te top vertex of te pyramid muc iger tan in its position wit minimum eigt. In tis iger position, none of te lateral faces of te pyramid touc te cylinder. We gradually lower tis top vertex towards te centre of te square base. Eventually, eac of te lateral faces of te pyramid will touc te rim of one of te circular ends of te cylinder. We cannot lower te top vertex any furter since oterwise part of te cylinder would be outside of te pyramid. Tis is our minimum eigt position. We calculate te eigt of tis pyramid. We label te square base of te pyramid as ABCD, and te top vertex of te pyramid as T. Join AC and BD, te diagonals of te base. Label teir point of intersection, wic is also te centre of te base, as M. Since te square base as side lengt 0, ten AC = BD = 0. Since te diagonals bisect eac oter, ten AM = BM = CM = DM = 10. Note tat T lies directly above M. Let t be te eigt of te pyramid; tat is, let t = T M. We want to calculate t. Suppose tat te central axis of te cylinder lies above AC. Since te midpoint of te central axis lies directly above M, ten te central axis extends a distance of 5 to eiter side of M. Label te points in contact wit AC at te two ends of te cylinder as E and F. Since EM = F M = 5, ten eac end of te cylinder lies a distance of 10 5 from te corner of te base, as measured along te diagonal (tat is, AE = 10 5). B C G M F E A H D From above, te cylinder s footprint on te base of te pyramid is actually a square, since its diameter becomes its widt. Consider a vertical cross section of te pyramid and cylinder troug te end of te cylinder closest to A. L J K O G E H

10 011 Fermat Contest Solutions Page 10 Let L be te point were te cross-section intersects AT and G and H be te points were te cross-section intersects AB and AD, respectively. Tese points G and H are te same points sown in te first diagram. Since BAM = 45 and te cylinder s face is perpendicular to te diagonal of te square base, ten GEA is isosceles and rigt-angled (as is HEA) so GE = HE = AE = Let O be te centre of te end face of te cylinder. Note tat GL and HL are lines tat lie along te faces ABT and ADT of te pyramid. Since te faces ABT and ADT of te pyramid just touc te cylinder s rim, ten GL and HL are tangent to te circular face, say at J and K, respectively. Join O to points G, H and L. Also, join O to points J, K and E. Eac of tese segments is a radius of te circular face, so eac as lengt 5. Since te circle is tangent to faces of te pyramid (including te bottom face) at tese points, ten eac of tese segments is perpendicular to te corresponding side of GHL. Our goal will be to calculate te lengt of LE. Since GE and GJ are tangents to te circle from a common point, ten GJ = GE = Let LE =. Ten LO = 5. Also, let LJ = x. Note tat LJO is similar to LEG, since tey ave a common angle at L and eac is rigtangled. Since tese triangles are similar, ten LJ JO = LE EG. Terefore, x 5 = 10 5 or x = = 1. Also, from te similarity, LG GE = LO OJ. Terefore, x + (10 5) 10 = 5 or x + (10 5) = ( 1)( 5). 5 5 Substituting x =, we obtain (10 5) = ( 1)( 5) Finally, to calculate t, we extract AMT. + 5( 1) = ( 1) ( 5) + 5( 1) = (9 4 ) 5( 1) 10( 1) = (8 4 ) = 10( 1) 8 4 T L A E M

11 011 Fermat Contest Solutions Page 11 Note tat E lies on AM and L lies on AT. Also, T M is perpendicular to AM and LE is perpendicular to AE, wic means tat AEL is similar to AMT. Terefore, T M AM = LE AE, or t 10 = 10, and so 5 t = 10 5( 1) 10( 1) 8 4 = 0 ( 1) 8 4 = 5 ( 1).07 Of te given answers, te smallest possible eigt is closest to.1. Answer: (B)