2001 Solutions Euclid Contest (Grade 12)

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1 Canadian Mathematics Competition An activity of The Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 001 s Euclid Contest (Grade 1) for The CENTRE for EDUCATION in MATHEMATICS and COMUTING Awards 001 Waterloo Mathematics Foundation

2 001 EUCLID SOLUTIONS ( ) =? 1. (a) What are the values of x such that x 9 1 ( ) = x 9 ( ) = x 9 0 x ( x )= 0 [by difference of squares] ( ) + ( x )( x)= 0 so x = or x = 0. ( x ) = 9 x = or x = Therefore x = or x = 0. 4x 1x+ 9= 9 4x 1x = 0 4xx ( )= 0 Therefore x = 0 or x =. (b) If f( x)= x x 5, what are the values of k such that f( k)= k? If f( k)= k, then k k 5= k k 4k 5= 0 ( k 5) ( k+ 1)= 0 so k = 5 or k = 1. (c) Determine all x, y ( ) such that x y + = 5 and x y= 1. 1 (Algebraic) Since x y= 1, then x = y+1 (or y= x 1). So since x + y = 5, then ( y+ 1) + y = 5 or x + ( x 1) = 5

3 001 EUCLID SOLUTIONS y + y+ 1+ y = 5 y + y 4 = 0 x + x x+ 1= 5 x x 4 = 0 y + y 1 = 0 x x 1 = 0 ( y+ 4)( y )= 0 y = 4, ( x 4) ( x+ )= 0 x = 4, and using x = y+1, and using y= x 1, we get x = 4., we get y =, 4. So the solutions are x, y, 4, 4,. ( )= ( ) ( ) (Graphical) lacing each of x + y = 5 and x y= 1 on a grid we have the diagram at the right. (0, 5) y x y = 1 (4, ) (0, 1) (1, 0) (5, 0) x (, 4) x + y = 5 ( )= ( ) Therefore, the solutions are x, y, 4, (4, ).. (a) The vertex of the parabola y= ( x b) + b+ h has coordinates( 5, ). What is the value of h? Since the x-coordinate of the vertex is, then b =. Since the y-coordinate of the vertex is 5, then b+ h=5. Since b =, then h =. (b) In the isosceles triangle AC, A = AC and AC = 40. oint is on AC such that is the bisector of AC. Similarly, Q is on A such that CQ bisects AC. What is the size of A, in degrees? Q A C

4 001 EUCLID SOLUTIONS 4 Let AC = x. Since AC is isosceles, then AC = x. Since bisects AC, A = C = x. Similarly, ACQ = CQ = x. The angles in AC add to 180, so 40 + x + x = 180 x = 5. In A, the angles add to 180, so A = 180 A = 105. Q A 40 x x x x C (c) In the diagram, A = 00, Q = 0, and QR = 100. Also, QR is parallel to AC. Determine the length of C, to the nearest integer. A Q R C 1 Since QR AC, QR = AC = α (alternating angles). From RQ, tan α= C In AC, since tan α= =, let C 5 AC = x and AC = 5 x. (This argument could also be made by just using the fact that RQ and AC are similar.) y ythagoras, x + 5x = 00, x = = Therefore C = 59 m to the nearest metre. 5 A 0 α Q 100 α R C Since QR AC, QR = AC (alternating angles). This means AC ~ RQ (two equal angles). y ythagoras, R = Q + QR A R = = Since AC ~ RQ, 0 Q 100 R C

5 001 EUCLID SOLUTIONS 5 C Q = A R A Q C = R 00 0 = = C is 59 m (to the nearest metre).. (a) In an increasing sequence of numbers with an odd number of terms, the difference between any two consecutive terms is a constant d, and the middle term is 0. When the last 4 terms are removed from the sequence, the middle term of the resulting sequence is 9. What is the value of d? 1 Let the number of terms in the sequence be k + 1. We label the terms a 1, a,..., a k +. 1 The middle term here is a k+ 1 = 0. Since the difference between any two consecutive terms in this increasing sequence is d, am+ 1 am = d for m= 1,,..., k. When the last 4 terms are removed, the last term is now a k so the middle term is then a k 1 = 9. (When four terms are removed from the end, the middle term shifts two terms to the left.) Now = ak+ 1 ak 1 =( ak+ 1 ak)+ ( ak ak 1)= d+ d = d. Therefore d =. If the last four terms are removed from the sequence this results in 0 shifting terms to the left in the new sequence meaning that 0 9 = d, d =. (b) There are two increasing sequences of five consecutive integers, each of which have the property that the sum of the squares of the first three integers in the sequence equals the sum of the squares of the last two. Determine these two sequences. Let n be the smallest integer in one of these sequences. So we want to solve the equation n + ( n+1) + ( n+ ) = ( n+ ) + n+ 4 given problem into an equation). Thus n + n + n+ 1+ n + 4n+ 4= n + n+ 9+ n + 8n+ 1 ( ) (translating the

6 001 EUCLID SOLUTIONS n 8n 0= 0 ( n 10) ( n+ )= 0. So n = 10 or n =. Therefore, the sequences are 10, 11, 1, 1, 14 and, 1, 0, 1,. Verification ( ) + ( ) + = + = and = = 5 4. (a) If f()= t sin πt minimum value? π, what is the smallest positive value of t at which f() t attains its 1 π π Since t > 0, πt >. So sin πt π π πt = t =. ( ) first attains its minimum value when π Rewriting f t Thus f t at the right. Thus f t 1 ()= sin ( ) () as, f t π t () has a period [ ]. π = and appears in the diagram π () attains its minimum at t =. Note that f() t attains a minimum value at t = 0 but since t > 0, the required answer is t =. f (t) t

7 001 EUCLID SOLUTIONS 7 (b) In the diagram, AF = 41, CF = 59, DE is parallel to F, and EF = 5. If AE = EC, determine the length of AE, to decimal places. A D F 5 E C Let the length of AE = EC be x. Then AF = x 5. In, CF, x + 5 = tan ( 59 ). F A x 5 In AF, x 5 = tan ( 41 ). F Solving for F in these two equations and equating, x+ 5 x F = = 5 tan 59 tan 41 so ( tan 41 )( x+ 5)= ( tan 59 )( x 5) 5( tan 59 + tan 41 )= x( tan 59 tan 41 ) 5( tan 59 + tan 41 ) x = tan 59 tan 41 x = Therefore the length of AE is D F 5 E x C 5. (a) Determine all integer values of x such that ( x ) ( x + 5)< 0. Since x 0 for all x, x + 5> 0. Since ( x ) ( x + 5)< 0, x < 0, so x < or < x <. Thus x = 101.,, (b) At present, the sum of the ages of a husband and wife,, is six times the sum of the ages of their children, C. Two years ago, the sum of the ages of the husband and wife was ten times

8 001 EUCLID SOLUTIONS 8 the sum of the ages of the same children. Six years from now, it will be three times the sum of the ages of the same children. Determine the number of children. Let n be the number of children. At the present, = C, where and C are as given. (1) Two years ago, the sum of the ages of the husband and wife was 4, since they were each two years younger. Similarly, the sum of the ages of the children was C n( ) (n is the number of children). So two years ago, 4= 10( C n) (), from the given condition. Similarly, six years from now, + 1 = ( C+ n) (), from the given condition. We want to solve for n. Substituting (1) into each of () and (), C 4 = 10( C n) or 0n 4C= 4 or 5n C= 1 C+ 1= ( C+ n) or 18n+ C= 1 or n+ C= 4 Adding these two equations, n =, so n =. Therefore, there were three children.. (a) Four teams, A,, C, and D, competed in a field hockey tournament. Three Medal Gold Silver ronze coaches predicted who would win the Gold, Silver and ronze medals: Team Coach 1 predicted Gold for A, Silver for, and ronze for C, Coach predicted Gold for, Silver for C, and ronze for D, Coach predicted Gold for C, Silver for A, and ronze for D. Each coach predicted exactly one medal winner correctly. Complete the table in the answer booklet to show which team won which medal. If A wins gold, then Coach 1 has one right. For Coach to get one right, D must win bronze, since A cannot win silver. Since D wins bronze, Coach gets one right. So C can t win silver, so does which means Coach 1 has two right, which can t happen. So A doesn t win gold. If wins gold, then Coach has one right. For Coach 1 to get one right, C wins bronze, as can t win silver. For Coach to get one right, A wins silver. So Gold to, Silver to A and ronze to C satisfies the conditions.

9 001 EUCLID SOLUTIONS 9 (b) In triangle AC, A = C =5 and AC = 0. The circle with diameter C intersects A at X and AC at Y. Determine the length of XY. X A Y C 1 Join Y. Since C is a diameter, then YC = 90. Since A = C, AC is isosceles and Y is an altitude in AC, then AY = YC =15. Let AC = θ. Since AC is isosceles, CA = θ. Since CYX is cyclic, XY = 180 θ and so AXY = θ. Thus AXY is isosceles and so XY = AY =15. Therefore XY = 15. θ θ X 5 15 A 15 Y C θ Join Y. YC = 90 since it is inscribed in a semicircle. AC is isosceles, so altitude Y bisects the base. Therefore Y = 5 15 = 0. Join CX. CX = 90 since it is also inscribed in a semicircle. The area of AC is 1 ( AC) ( Y) = 1 ( A) ( CX) 1 ( 0) ( 0) = 1 ( 5) ( CX) X A 15 Y C CX = 00 = 4. 5 Y 0 4 From AY we conclude that cos AY = = = A 5 5. In XY, applying the Law of Cosines we get ( XY) = ( X) + ( Y) ( X)( Y) cos XY. Now (by ythagoras XC),

10 001 EUCLID SOLUTIONS 10 X = C CX = 5 4 = 49 X = 7. Therefore XY = ( 7)( 0) 4 5 = = 5. Therefore XY = (a) What is the value of x such that log( log ( x ) )=? log ( log ( x ) )= log ( x )= x = x = ( ) 4 x = 1 x = 18 x = 9 (b) () ()= kx Let f( x)= + 9, where k is a real number. If f : f 1:, determine the value of f 9 f. ( ) ( ) From the given condition, k f () 9 1 k f ( ) = = k k ( + 9)= + 9 k k = ( ) k We treat this as a quadratic equation in the variable x =, so 0= x x 18 0= ( x ) ( x+ ). Therefore, k = or k =. Since a > 0 for any a, then k. So k =. We could solve for k here, but this is unnecessary.

11 001 EUCLID SOLUTIONS 11 We calculate f Therefore f ( 9k ) ( k + ) 9k k ( ) ( )= + 9 f 9 9 ( ) ( )= = k = ( ) 9 f 10. = = 10. k 8. (a) On the grid provided in the answer booklet, sketch y= x 4 and y= x. y y = x x y = x 4 4 (b) Determine, with justification, all values of k for which y= x 4 and y= x + k do not intersect. Since each of these two graphs is symmetric about the y-axis (i.e. both are even functions), then we only need to find k so that there are no points of intersection with x 0. So let x 0 and consider the intersection between y= x+ k and y= x 4. Equating, we have, x+ k = x 4. Rearranging, we want x x ( k+ 4)= 0 to have no solutions.

12 001 EUCLID SOLUTIONS 1 For no solutions, the discriminant is negative, i.e k < 0 4k < 0 k < 5. So y= x 4 and y= x + k have no intersection points when k < 5. (c) State the values of k for which y= x 4 and y= x + k intersect in exactly two points. (Justification is not required.) Analysing Graphs For k < 5, there are no points of intersection. When k = 5, the graph with equation y= x + k is tangent to the graph with equation y= x 4 for both x 0 and x 0. So k = 5 is one possibility for two intersection points. y 4 y = x 4 y = x x 4

13 001 EUCLID SOLUTIONS 1 For 5< k < 4 a typical graph appears on the right. i.e. for 5< k < 4, there will be 4 points of intersection. y x 4 When k = 4, a typical graph appears on the right. y x 4

14 001 EUCLID SOLUTIONS 14 So when k > 4, there will only be two points of intersection, as the contact point at the cusp of y= x 4 will be eliminated. An example where k = is shown. y 4 y = x + k, k > x y = x 4 4 So the possibility for exactly two distinct points of intersection are k = 5, k > Triangle AC is right-angled at and has side lengths which are integers. A second triangle, QR, is located inside AC as shown, such that its sides are parallel to the sides of AC and the distance between parallel lines is. Determine the side lengths of all possible triangles AC, such that the area of AC is 9 times that of QR. A Q R C 1 Let the sides of AC be A = c, C = a, AC = b, a, b, c are all integers. Since the sides of QR are all parallel to the sides of AC, then AC is similar to QR. Now the ratio of areas of AC to QR is 9= to 1, so the ratio of side lengths will be to 1. So the sides of QR are Q = c, QR = a, R = b.

15 001 EUCLID SOLUTIONS 15 So we can label the diagram as indicated. We join the corresponding vertices of the two triangles as Area of trapezoid QRC Area of trapezoid CRA Area of trapezoid AQ + Area of QR Area of AC. Doing so gives, a ac ac b c = 7 Or upon simplifying ac = a + b + c (Note that this relationship can be derived in a variety of ways.) ac = c + b + a ac c a = b ac c a a c ac + 9c + 9a ac ac + 18ac= 9 a + c c A c b Q a R a C ) = + (since b= a + c (squaring both sides) ( ) ac( ac c a + 18)= 0 ac c a + 18 = 0 (as ac 0 ) ca ( )= a 18 a 18 c = a 18 c = + a. Since a is a side of a triangle, a > 0. We are now looking for positive integer values such that 18 is also an integer. a The only possible values for a are, 7, 8, 9, 1, 15 and 4. Tabulating the possibilities and calculating values for b and c gives, a c b Thus the only possibilities for the triangle are ( 745,, ), 8157,, The two triangles are similar with areas in the ratio 1:9. Therefore the sides are in the ratio 1:. Let a= C, b= CA, c= A. ( ) and 9115,, ( ). b

16 001 EUCLID SOLUTIONS 1 Then a = Q, b = QR, c = R. Locate points K, L on C; M, N on CA; and T, S on A as shown. C = K + KL + LC a= K + a + a K L T Q R c S Therefore K = a In a similar way, AN Now K. = b. T and ANR Therefore T = K = a Now, A = AS + ST + T c c= b + + a c= b+ a 4 c= b+ a b= c+( a). y ythagoras, a + b = c ASR, both by HL. and AS = AN = b a + [c + ( a)] = c a + c + c( a) + ( a) = c a + ( a) = c( a) a 1a + = c(a ) a a + 18 = c(a ). C M N b A a a+ 18 c = a aa c = ( )+ 18 a 18 c= a+ a. Since a and c are integers, a is a divisor of 18. Also since b< c and b= c+( a), we conclude that a < 0 so a > 0. Thus a can be 1,,,, 9, 18. The values of a are: 7, 8, 9, 1, 15, 4. Matching values for c: 5, 17, 15, 15, 17, 5 Matching values for b: 4, 15, 1, 9, 8, 7

17 001 EUCLID SOLUTIONS 17 The distinct triangles are (7, 4, 5), (8, 15, 17) and (9, 1, 15). 10. oints and Q are located inside the square ACD such that D is parallel to Q and D = Q = Q. Determine the minimum possible value of AD. A Q D C 1 lacing the information on the coordinate axes, the diagram is indicated to the right. We note that has coordinates ( ab, ). y symmetry (or congruency) we can label lengths a and b as shown. Thus Q has coordinates ( a, b). Since D = Q, a + b =( a) + ( b) or, a + b 8a 8b+ 8= ( a b ) + ( ) = 9 is on a circle with centre O 4 4 (, ) with r =. The minimum angle for θ occurs when D is tangent to the circle. A(0, ) (, ) a(a, b) b θ a θ b Q( a, b) D(0, 0) C(, 0) So we have the diagram noted to the right. Since OD makes an angle of 45 with the x -axis then DO = 45 θ and OD = 4. 1 Therefore sin 45 ( θ)= = which means 45 θ = 0 or θ= 15. Thus the minimum value for θ is θ D 45 θ O 4, 4 ( ) Let A = C = CD = DA =1. Join D to. Let AD = θ. Therefore, D = 45 θ. Let D = a and = b and Q = a.

18 001 EUCLID SOLUTIONS 18 We now establish a relationship between a and b. In D, b a a cos 45 θ = + ( )( ) ( ) a or, cos ( 45 θ)= b + a (1) A θ a 45 θ a R b Q In DR, a a + or, cos 45 θ ( )= a Comparing (1) and () gives, a Simplifying this, a or, = a + a 45 cos ( θ) b Now cos ( 45 θ)= b = a a =. () b + a + = 4. a a a + a Now considering a +, we know a a or, a +. a 1 Thus, cos 45 4 ( θ) ( )= cos ( 45 θ). cos 45 θ = a + a. a ( ) has a minimum value for θ = or θ= 15. Join D. Let D meet Q at M. Let AD = θ. y interior alternate angles, = Q and DM = QM. Thus DM QM by A.S.A., so M = QM and DM = M. So M is the midpoint of D and the centre of the square. D C

19 001 EUCLID SOLUTIONS 19 Without loss of generality, let M = 1. Then D =. Since θ+ α= 45 (see diagram), θ will be minimized when α is maximized. A 1 M θ Q α D C Consider MD. Using the sine law, sin α sin ( MD ) =. 1 To maximize α, we maximize sin α. sin ( MD) ut sin α=, so it is maximized when sin ( MD)= 1. In this case, sin α= 1, so α= 0. Therefore, θ= 45 0 = 15, and so the minimum value of θ is We place the diagram on a coordinate grid, with D( 00, ), C( 10, ), ( 01, ), A( 11, ). Let D = Q = Q = a, and AD = θ. Drop a perpendicular from to AD, meeting AD at X. Then X = a sin θ, DX = a cos θ. Therefore the coordinates of are ( asin θ, acos θ). Since D Q, then QC = θ. So by a similar argument (or by using the fact that Q are symmetric through the centre of the square), the coordinates of Q are ( 1 asin θ, 1+ acos θ). A(0, 1) (1, 1) X a a a Q D(0, 0) C(1, 0) ( ) = ( ) + ( ) = + sin θ+ cos θ ( sin θ+ cos θ) = Now Q a, so 1 asin θ 1 acos θ a 4a 4a 4a a

20 001 EUCLID SOLUTIONS 0 Now a a + 4a a = 4a sin θ+ cos θ 0 a + 0 a a + 0 a + a a + 4 a + a 4a + a 4 a + a 4 a and equality occurs when a =. ( ) = sin θ+ cos θ 1 1 = sin θ+ cos θ = cos ( 45 ) sin θ+ sin ( 45 ) cos θ ( ) = sin θ + 45 So sin ( θ+ 45 ) and thus since 0 90 θ, then θ or θ 15. Therefore the minimum possible value of AD is 15.

2009 Euclid Contest. Solutions

2009 Euclid Contest. Solutions Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 009 Euclid Contest Tuesday, April 7, 009 Solutions c 009