BRITISH COLUMBIA COLLEGES 2000 SOLUTIONS. 1. Let x be the number of litres of gasoline in the tank prior to lling. Then x +15= 28, or x =6.

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1 BRITISH COLUMBIA COLLEGES HIGH SCHOOL MATHEMATICS CONTEST 000 SOLUTIONS Junior Preliminary. Let x be the number of litres of gasoline in the tank rior to lling. Then x +5= 8, or x = :. The rst gure is comosed of one square of side (consisting of 4 matchsticks) lus squares of side each missing matchstick, for a total of 4 + : 3 = 0 matchsticks. Each subsequent gure consists of the th revious gure lus squares of side each missing matchstick. Thus the n gure in the sequence contains 4 + : 3: n =6n +4 matchsticks. The largest value n for which 500 matchsticks is sucient is thus 8 (which uses u 6: = 496 matchsticks). Now the number of squares in the rst gure is 3 and each subsequent gure contains more squares than the revious one. Therefore the number of th squares in the n gure is n +. For n = 8 this means that 65 squares would bein the largest gure made with 500 matchsticks. Answer is (B) 3. Let ` and w be the length and width (in metres) of the rectangle in question. Since the erimeter is 56 4 metres, we have ` +w =56, or ` + w = 8. We are also told that ` : w =4:3,or` = w. Using this 3 in the rst equation we get 4 w + w =8 3 7 w =8 3 w = which imlies that ` = 6. By the Theorem of Pythagoras the length of the diagonal is + 6 = 400 = 0: Answer is (B) 4. Since there are 3 days in March, there are 3 days between March 3 and Aril 3. That is, the eriod in question is 4 weeks and 3 days. Since Aril 3 is a Tuesday, wemust have March 3 a Saturday, namely 3days earlier in the week. : : 5. The total area of the board is 5x square units. The area of the shaded region is x 4x + x 3x =7x square units. Therefore, the robability ofhitting the shaded area is 7x 7 = =0:8 5x 5 6. Let us comute the sum of the roer divisors of each of the 5 ossible answers in the list: 3 : < 3 6 : ++4+8=5< 6 30 : =4> : ++4++=40< : =43< 50 The only one of these which qualies as an abundant number is The area in question is the area of a traezoid less the area of a semicircle. The area of the semicircle is obviously = cm. The area of the traezoid is 4(4 + 6) = 0 cm.thus the shaded area is 0 0 cm.

2 British Columbia Colleges Junior Preliminary 000 Page 8. Let the coordinates of the oint S be (x; y). Since PS k QR they must have the same sloe: y = = x or 4y 0 3x = Since RS k QP we also have (by the same argument): y = = 0 x or 4y +3x =3 From these two equations in unknowns we easily solve for x = 5 and y =4. Thus x + y =9. 9. The diagram below shows how gures (a), (c), (d), and (e) can be lled with coies of the \T" tile. No matter how one tries gure (b) cannot be lled with coies of it. (a) (c) (d) (e) Answer is (B) 0. Note rst that =( ) = =(3 ) =7 6 =(6 ) =36 Since 7 < 3 < 36, we have 7 < 3 < 36, which means < < 6. Let us rst comute the number of seconds in day, hour, minute, and second, and then multily by 000. Now day lus hour is clearly 5 hours. Then day, hour, lus minute is = 50 minutes. Exressed in seconds this is50 60 = seconds. Thus day, hour, minute, and second is 90;06 seconds. The answer to the roblem is this gure multilied by 000; that is, 80;;000, which to the nearest million is 80;000;000.. If a = b = c, then 00a +0b + c = 00a +0a + a =a. Since a can be any digit, in order for a number to be a factor of the three-digit number, it must be a factor of. The factors of are, 3, 37, and. The only one of these aearing in the list is (x + y) 0 (x 0 y) > 0, x +xy + y 0 x +xy 0 y > 0, 4xy >0, xy > 0 The last condition clearly holds if and only ifx and y have the same sign; that is, both are ositive or both are negative.

3 British Columbia Colleges Senior Preliminary 000 Page 3 4. Let a, b, c be the three sides of the triangle. Let us assume that a b c. Since the erimeter is, we have a + b + c =. Let us now list all ossible sets of integers (a; b; c) satisfying the above conditions: (; ; 0); (; ; 9); (; 3; 8); (; 4; 7); (; 5; 6); (; ; 8); (; 3; 7); (; 4; 6); (; 5; 5); (3; 3; 6); (3; 4; 5); (4; 4; 4) However, it is clear that some of these \triangles" do not actually exist, since in any triangle the sum of the lengths of the two shorter sides must be greater than the length of the longest side. With this additional condition we have only the following triangles (a; b; c): (; 5; 5); (3; 4; 5); (4; 4; 4) We can now examine the 4 statements and conclude that (i), (ii) and (iv) are clearly true. As for (iii), we see that triangle (3; 4; 5) above is right-angled; hence (iii) is false. 5. The area of the original triangle is bh. The new triangle has altitude h + m and base b 0 x. We need to nd x such that thearea of the new triangle is (h + m)(b 0 x). Thus bh = (h + m)(b 0 x) 4 bh = b 0 x (h + m) bh bh + bm 0 bh b(m + h) x = b 0 = = (h + m) (h + m) (h + m) 4 bh. Clearly the area of the new triangle is Senior Preliminary. Antonino averages 5 km/h for the rst 0 km. This means it takes him 0=5 = 4=3 hours to cover the rst 0 km. In order to average 0km/h for a 40 km distance, he must cover the distance in hours. He only has=3 hremaining in which to cover the last 0 km.his seed over this last 0 km then must be (on average) 0=(=3) = 30 km/hr. Answer is (B). Let C be the centre of the circle. Since the oints O and B are equidistant from the centre of the circle and also equidistant from the oint P, and since both O and B lie on the x-axis, we see that P has coordinates (3;y) with y<0. The sloe of OC is =3. Since PO is thetangent line to the circle at O we know that PO? OC. Therefore the sloe of PO is 03=. However, the sloe of PO is comuted to be (y 0 0)=(3 0 0) =y=3. Together these imly that y = 09= ! 3. First of all the number of ossible ways to choose a air ofdistinct students from a set of veis = = 0. From this we need only eliminate those whose age dierence is. Clearly there are!3! exactly 4such, namely (6; 7), (7; 8), (8; 9), and (9; 0). So our robability of success is 6out of 0, or 3=5. 4. The straight sections of the belt are tangent toall 3 ulleys and thus erendicular to the radius of each ulley at the oint of contact. Thus the straight sections of the belt are the same lengths as the distances between the centres of the ulleys, which are 5,, and 5 + = 3; so the straight sections of belt add u to 30 units. The curved sections of belt, when taken together, make uone comlete ulley, or a circle of radius. Thus the curved sections add u to () = 4. So the full length of the belt is units. Answer is (B)

4 British Columbia Colleges Senior Preliminary 000 Page 4 5. Let n be the number of quizzes Mark has already taken. Let x be his total score onall n quizzes. Then we have the following: x +7 =83 n + ) x = 83(n +) 0 7 x +99 =87 n + ) x = 87(n +) 0 99 Solving this system yields n =6. 6. Number the ins as shown in the diagram on the right. There is then large equilateral triangle with 4 ins on a side, namely the one with vertices numbered (; 7; 0). There are also 3 equilateral triangles with 3 ins on a side, namely the ones whose vertices are numbered (; 4; 6), (; 7; 9), and (3; 8; 0). The equilateral triangles with ins on a side come in distinct orientations, one with a single vertex above the horizontal base and one with a single vertex below the horizontal base. For the rst tye we have 6such: (; ; 3), (; 4; 5), (3; 5; 6), (4; 7; 8), (5; 8; 9), and (6; 9; 0). For the second tye we have only 3 such: (; 3; 5), (4; 5; 8), and (5; 6; 9). This gives us a total of 3 equilateral triangles so far. However, there are two others which are skewed somewhat to the edges of the outer triangle: (; 6; 8) and (3; 4; 9), which gives us a total of 5 equilateral triangles The critical idea here is to recognize that when the square covers as much of the triangle as ossible, the triangle will also cover as much of the square as ossible, and that at this oint the amount of triangle covered is the same as the amount of square covered. Let A be the area of the triangle. Then : 0:6A = 36, or A =40cm Let A be the centre of one of the large circles, let B be the oint of contact between the two large circles and let C be the centre of the small circle. Then AB? BC, AC =0+r and BC =0 0 r. From the Theorem of Pythagoras we have (0+r) = 0 + (0 0 r) 00+0r + r = r + r 40r =00 r =:5 Answer is (B) A 0 B 0+r r C 9. See #0 on the Junior aer. 0. Clearly y must be even in order to get integer solutions. The largest ossible value for y is 6 since we must have x>0. When y =6wehave x =,so(x; y) =(; 6) is asolution. Let us consider successively smaller (even) values for y: (x; y) =(4; 4), (7; ), (0; 0), etc. However, the solution (0; 0) and any further ones do not satisfy y>x. So we are left with the solutions (x; y) =(; 6), (4; 4), and 7; ).. Since B and C are ositive integers, we seethat B +=(C +)>, whence its recirocal is smaller than. Therefore, A must reresent theinteger art of 4=5, i.e. A = 4. Then we have 4 5 = or = B C + B + C +

5 British Columbia Colleges Junior Final 000 Page 5 For exactly the same reason as above we see that B must be the integer art of 5=4, i.e. B =. Then =4 ==(C +), which imlies that C =3. Then A +B +3C =4+() + 3(3) = 5: 0 m+n. The number of ways of choosing balls (without relacement) from a box with m + n balls is. The number of ways of choosing white ball ism, and the number of ways of choosing black ball is n. Thus the robability that one ball of each colour is chosen is: mn mn mn mn m + n (m + n)! (m + n)(m + n 0 ) (m + n)(m + n 0 )!(m + n 0 )! = = = 3. Since x builders build z houses in y days, we see that x builders build z=y houses er day, or builder builds z=(xy) houses er day. Hence q builders build qz=(xy) houses er day. Let s be the number of days needed for q builders to build r houses. Then q builders build r=s houses er day. Equating these we get qz=(xy) =r=s, whence s = rxy=(qz) days. 4. Adding the two given equations together we get: Thus x + y =6orx + y = 07. x +xy + y + x + y =4 i.e. (x + y) +(x + y) 0 4 = 0 (x + y 0 6)(x + y +7)=0 5. All the unshaded triangles in the diagram below are right-angled and thus are congruent. By the Theorem of Pythagoras we have x =(70x) +3 =4904x + x +9 4x =58 9 x = 7 87 The area of the shaded arallelogram is 3x = cm. 7 Junior Final 3 cm 7 - x x 3 cm 7 - x x Part A. If we square the digits from 0to9and consider the nal digit of the square we get only the digits 0,, 4, 9, 6, and 5. Since there are no others, we see that 8 is NOT the nal digit of any square.. Each of the 0 straight lines intersects each of the others exactly once. This makes for 90 intersections; however, each of the intersections is counted twice in this aroach, deending uon which of the two lines we consider rst. To get the correct number of intersections we simly divide 90 by to get 45.

6 British Columbia Colleges Junior Final 000 Page 6 3. Let us try successively to make ueach of the given amounts using 6 coins: 9 = = = = Thus each of the rst 4 choices can be made u with 6 coins. In order to make u49wewould need to use 4 ennies. This would require us to make uthetotal of45cents with only coins, which isclearly imossible. 4. First observe that (x 0 y) = x + y 0 xy =80(4) = 0 This means that x 0 y =0,i.e. x = y. In that event we clearly have x 0 y =0. Answer is (B) 5. See #0 on Senior Preliminary above. 6. We are seeking the largest integer n such that n(n +) 500 or n(n +) 000 Since 3 = 04 we see that n<3. Checking n =3we nd 3: 3 = 99. Thus the integer n we seek is 3. The triangular number associated with this value of n is (99) = In order for the roe to be at the lowest ossible oint, that oint must be the middle of the roe. Thus we are faced with solving a right-angled triangle with hyotenuse 40 m and one side of = 4 m. By the Theorem of Pythagoras the third side (x in the diagram at the right) is = 404 = The distance between the two agoles is x = m 50 m 36 m x th 8. Let an be the number of eole who arrive at the n ring of the door bell. Then a n =n0. Let bn th be the number of eole who have arrived after the n b = b = b + a for n n+ n n+ = b +n + n This can be rewritten as b = bn+ 0 b n =n + for n If we write out the rst 0 of these we get a = b 0 b =3 b 0 b =5 3 b 0 b = ring of the door bell. Then we have

7 British Columbia Colleges Junior Final 000 Page 7 When we add all 0 of the above equations together we get b = = 0 :(: + 9: 0 ) = 400 where we have used the well-known formula for the sum of an arithmetic rogression with n terms, having rst term a and common dierence d: n(a +(n0)d). 9. According to the denition of the oeration 3 we have 0 3 (0) = 0 (0) = 0 =0 0. Let a be the length of PQ, QR, and RS. Then the radii of the 3 circles are a, a, and 3a. The area between the inner and middle circles is then (a) 0 a =3a, and the area between the middle and 3 outer circles is (3a) 0 (a) = 5a.Thus the ratio we want is3a :5a =. Part B. (a) For this art of the question, the simlest method is simly to list all the ossible numbers. In increasing order they are: for a total of 0 numbers. ; ; 3; ; 3; 33; ; 3; 33; and 333 (b) Again, most junior students will simly try to list all the ossible integers. In increasing order they are: ; ; 3; 4; ; 3; 4; 33; 34; 44; ; 3; 4; 33; 34; 44; 333; 334; 344; 444; ; 3; 4; 33; 34; 44; 333; 334; 344; 444; 3333; 3334; 3344; 3444; and 4444 for a total of 35 numbers. A more sohisticated aroach (which can be generalized) follows: We rst dene n(k; d) to be the number of k-digit integers ending with the digit d and satisfying the two conditions (i) and (ii) in the roblem statement. Since a k-digit number ending with the digit d consists of aending the digit d to all (k 0 )-digit numbers ending with a digit less than or equal to d, wehave n(k; d) =n(k 0 ; )+n(k 0 ; ) + + n(k 0 ;d) (3) Furthermore, we also have n(;d) = foralldigits d and n(k;) = for all integers k. relationshi (3) allows us to create the following table ofvalues for n(k; d): knd Each entry in the table is the sum of the entries in the revious row u to and including the column containing the given entry (note the resence of Pascal's Triangle in the table). From the answer to arts (a) and (b) are: (a) : n(3; )+n(3; ) + n(3; 3)=+3+6=0 (b) : n(4; )+n(4; ) + n(4; 3) + n(4; 4) = =35 Clearly this table could have been extended to deal with any number k and with any digit d 9. 5 The

8 British Columbia Colleges Senior Final 000 Page 8. Since ABCD is a square the lines AC and BD are erendicular. Since the circle had radius unit, the Theorem of Pythagoras tells usthat AB = BC = CD = DA =. The tangent PC at C is 6 6 erendicular to the diameter AC; thus PCB = 45. Since PA? BC we also have CPB = 45. This makes 4PBC isosceles, which means that PB = BC =. Alying the Theorem of Pythagoras to 4AP D we have PD = AP + AD =( ) + = 8+ = 0 from which we see that PD = Since the rectangle has its sides in the roortion 3 : we will consider rst looking at a 3 rectangle, in which there are vertices which lie on the diagonal. In the original rectangle we need only consider the fteen 3 rectangle which straddle the diagonal in question. The lower left of these has its lower leftmost vertex on the diagonal, and each of these 3 rectangles adds a further vertex to the count for its uer rightmost corner. This gives us a total of + 5 = 6 vertices on the diagonal. 4. (a) We will use the roof in the roblem statement as a model. Consider bc 0ad. Clearly (bc0ad) 0. Exanding gives b c 0 abcd + ad 0 This is easily rearranged to yield bc + ad abcd. (b) We will use art (a) to rove art (b). Since a, b, c, and d are arbitrary real numbers, the inequality in art (a) remains true for any rearrangement of the letters: in articular we have: abcd b c + a d abdc b d + ac adcb d c + a b Recognizing that multilication is commutative for real numbers we can reorganize the roducts in each of the above inequalities and sum the three inequalities to get the desired result. 5. (a) Let us lace (outer) coins next to the original coin sothat they touch each other. Then the centres of the 3 coins form an equilateral triangle with side length equal to twice the radius of a single coin. Therefore the angle between the centres of the (outer) coins measured at the centre of the rst coin is 60. Since 6 such angles make u a full revolution around the inner coin, we can have exactly 6outer coins each touching the original (inner) coin and also touching its other two neighbours. (b) There are 6 non-overlaing saces whose areas we must add; each isfound between 3 coins which simultaneously touch other, and whose centres form the equilateral triangle mentioned in art (a) above. This equilateral triangle has side length, since we are given the radii of the coins as. Our strategy to comute the area of one such sace is to nd the area of the equilateral triangle and subtract the areas of the 3 circular sectors found within the triangle. The altitude of the equilateral triangle with side length can be easily found (Theorem of Pythagoras) as 3. Thus the area of : : : the triangle itself is 3= 3. The area of a single coin is =. The circular sectors within the equilateral triangle are each one-sixth of the area of the coin; there are 3 such sectors which gives us a total area of one-half the area of a single coin to be subtracted from the area of the equilateral triangle. Thus the area of a single sace is 3 0 (=). Since there are 6 such saces, we have atotal area of square units.

9 6 6 British Columbia Colleges Senior Final 000 Page 9 Senior Final Part A The area of the traezoid ABCD is : 4(+0) : = square units. We want tochoose the oint P such that the area of 4CPB is half this area, i.e. square units. Let the base PB have length x. Then the : : area of 4CPB = 4 x =x. Since this must be, wehave x =5.. Let P be the intersection of AC and BD as in the diagram below. The sum of the interior angles of a (regular) entagon is (50 )80 = 540. A So each interior angle in a regular entagon has measure 540 =5 = 08. Since 4ABC is isosceles with vertex angle equal to 08. The base angles, namely BAC and BCA, both have measure 36. Similarly, E B CBD = CDB = 36. Since ABC = 08 and CBD = 36,we see that ABP = = 7. This means that the third angle P in triangle ABP,namely AP B has measure = 7. Thus the angles at P have measure 7 and = 08. D C 3. Let a, b, c be the ages (in integers) of the three children. We may assume that a b c<5. Since the roduct abc = 90, we also know that a>0. The values we seek for a, b, and c are integers which divide evenly into 90 and lie between the values and 4 (inclusive). The only such integers are,, 3, 5, 6,9,and 0. Wewill now look at all ossible roducts which satisfy the above conditions, and for each one we will comute the sum of the ages: a b c sum We are further told that even knowing the sum of the ages would NOT allow us to determine the three ages. Thus we are forced to conclude that the sum must be 6, since there are two distinct sets of ages which sum to 6 in the above table. The only ages found in these two setsare, 3, 5, 9, and 0. We notice that and 6 do not aear. 4. Let V be the volume of a full tub (in litres,say). Then the rate at which the hot water can ll the tub is V=0 litres er minute. Similarly the rate at which the cold water can ll the tub is V=8litres er minute. On the other hand a full tub emties at the rate of V=5litreserminute. If all three are haening at the same time then the rate at which the tub lls is: V V V 4V +5V 0 8V V + 0 = = litres er minute, which means it takes 40 minutes to ll the tub. 5. We rst need to recall that the sum of the rst n integers is given by n(n +)=. The sum we are resented with is now the dierence between the sums of the rst k +9 integers and the rst k integers. Using the above formula we have: (k + 9)(k +0) k(k +) (k +)+(k +)+(k + 9) = 0 k +39k k 0 k = 38k +380 = = 9(k + 0)

10 British Columbia Colleges Senior Final 000 Page 0 Since 9 isrime, in order for 9(k + 0) to be a erfect square, k +0 must contain 9 as a factor. The smallest such value occurs when k +0=9, i.e. when k = 9, and we indeed get a erfect square in this case, namely 9. Answer is (B) 5 6. Let n be the number in question. Then n can be written as 0 + a where a is a number with at most 5digits. Moving the left-most digit (the digit ) to the extreme right roduces a number 0a +. The 5 information in the roblem now tells us that 0a +=3(0 + a) = a, or7a = This yields a = So n = 4857 (and the other number we created is 4857), the sum of whose digits is = Let us organize this solution by considering the size of the largest cube in the subdivision of the original cube. The largest could have a side of size 4 cm, 3cm,cm,orcm. In each of these 4 cases we will determine the minimum number of cubes ossible. In the rst case, when there is a cube of side 4 cm resent, we can only include cubes of side cm to comlete the subdivision, and we would need 6 such 3 3 since the cube of side 4 cm uses u 64 cm of the 5 cm in the original cube. Thus, in this case we have 6 cubes in the subdivision. If we look at the other extreme case, namely when the largest cube in the subdivision has side length of cm, we clearly need 5 cubes for the subdivision. We also note here that we will certainly decrease the number of cubes in a subdivision if we try to relace sets of cubes by larger cubes whenever ossible. Now consider the case when there is a cube of side length 3 cm resent. If we lace it anywhere but in a corner, the subdivision can only be comleted by cubes of side length cm, which gives us + 98 = 99 cubes in total. If we lace it in a corner, we can then lace 4 cubes of side length cm on one side of the larger cube, more such cubeonasecond side and a third such cubeon the third side; this gives us large cube and 7 medium cubes for a total volume 3 of 7 + 7(8) = 83 cm which means we still 4 small cubes, for a grand total of 50 cubes. It is easy to see that if the largest cube has side length cm we can lace at most 8 of them in the original cube and the remainder of the volume must be made u of cubes of side length cm; this gives a total of 8+6=69 cubes. Thus the smallest number of cubes ossible is 50 and in this case there are 7 cubes of side length cm. 8. We need to nd a sequence of all 9 councillors beginning with A and ending with E such that each air of consecutive councillors are `on seaking terms' with each other. When one rst looks at the table rovided, it looks a little daunting. However, a rst observation is that among the councillors other than A and E (who need to aear at the ends of the sequence) councillors F and H are only `on seaking terms' with others, one of which is councillor B. Thus councillor F must receive the rumor from B and ass it toi, orvice versa. Similarly, councillor H must hear the rumor from B and ass it to C, or vice versa. Thus we must have either I 0 F 0 B 0 H 0 C or C 0 H 0 B 0 F 0 I as consecutive councillors in the sequence. Since A and E lie on the ends, and neither of them are `on seaking terms with' either councillors C or I, we see that councillors D and G must be laced one on either end of the above subsequence of 5 councillors. This leaves us with either D 0 I 0 F 0 B 0 H 0 C 0 G or G 0 C 0 H 0 B 0 F 0 I 0 D. Councillors A and E can be laced on the front and rear of either of these sequences to gives the nal sequence as either A 0 D 0 I 0 F 0 B 0 H 0 C 0 G 0 E or A 0 G 0 C 0 H 0 B 0 F 0 I 0 D 0 E. Ineither case the fourth erson after councillor A (who started the rumor) to hear the rumor was councillor B. 9. Let us examine the temerature dierences on the resective airs of thermometers. A dierence of 4 on A corresonds to a dierence of 6 on B; thus they are in the ratio of 3 :. A dierence of on B corresonds to a dierence of 7 on C; thus they are in the ratio of :6.Nowatemerature dro of 8 on A means a dro of on B (using the ratio 3 : ). This results in a temerature dro of7 on C (using the ratio : 6).

11 6 6 British Columbia Colleges Senior Final 000 Page 0. The number of ositive integers less than or equal to n which are multiles of k is the integer art of n=k (that is, erform the division and discard the decimal fraction, if any). This integer is commonly denoted bn=kc. Thus the number of ositive integers between 00 and 000 which are multiles of 6is = = Similarly, the number of ositive integers between 00 and 000 which are multiles of 7 is = = In order to count the number of ositive integers between 00 and 000 which are multiles of 6 or 7 we could add the above numbers. This, however, would count the multiles of both 6 and 7 twice; that is, the multiles of 4 would be counted twice. Thus we need to subtract from this sum the number of ositive integers between 00 and 000 which are multiles of 4. That number is =4704= Therefore, the number of ositive integers between 00 and 000 which are multiles of 6 or 7 is = 54. But we are asked for the number of ositive integers which are multiles of 6 or 7, but NOT BOTH. Thus we need to again subtract the number of multiles of 4 in thisrange, namely 43. The nal answer is = 47. Answer is (B) Part B First draw the radius OD. Let E =. Since DE = r = OD, 4DOE is isosceles. Therefore, DOE =. Since BDO is anexterior angle to 4DOE, it is equal in measure to the sum of the oosite interior angles of the triangle, i.e. BDO =. Now 4BOD is isosceles since two of its sides are radii of the circle. Thus DBO = BDO =. Since BOA is an exterior angle to 4BOE, itis equal in measure to the sum of E = and EBO =. Thus BOA =3, which means that the value k in the roblem is. 3. Since there are 5 regions of equal area which sum to 80 square units, each region has area 36 square units. The dimensions of the inner square are clearly 6 units on a side, and of the outer square are 80 = 6 5 units on a side. Let x and y be the dimensions of one of the four congruent regions, where x<y. Then x + y =6 5and y 0 x =6.Onadding these and dividing by wegety =3( 5+), and then it easily follows that x =3( 5 0 ). 3. (a) Note that 5! = 5: 4: 3: = 0, which endsin the digit 0. Thus n!, where n>5, must also end in the digit 0, since 5! evenly divides n!, for n>5. Thus n! + ends in the digit whenever n 5, which means that 5 can never evenly divide n! + when n 5. We are left to examine the cases n = 4, 3,, and. Since 4! + = 4 + = 5, we seethat5 divides evenly4!+.thus n =4is the largest value of n such that 5evenly divides n!+. (b) Note rst that (x +y)+(y +x) =3x +3y =3(x + y). Thus three evenly divides the sum of the two numbers. This can be rewritten as y +x =3(x + y) 0 (x +y). Suose that 3 evenly divides x+y. This means that x+y =3k for some integer k. Thus y +x =3(x+y) 0 3k =3(x+y 0 k), which means that 3 evenly divides y +x.

12 British Columbia Colleges Senior Final 000 Page 4. (a) Same as Problem #5(b) on the Junior Paer (Part B). (b) Let r be the radius of each of the four larger coins which surround the coin of radius. Then by considering two such neighbouring coins and the coin of radius (as in the diagram below) we have a right-angled triangle when we connect the three centres. The Theorem of Pythagoras then imlies that (r) = (r +) +(r +) r 0 4r 0 =0 4r =r +4r + r 0 r 0 =0 This quadratic has solutions ()(0) 6 8 r = = =6 When we reject the negative root, we are left with r = +. r r r r 5. (a) The slice is in the shae of an isosceles triangle with two sides equal to the altitude of the equilateral triangular faces of side a and the third side of length a. By using the Theorem of Pythagoras on half of one equilateral triangularface we seethat its altitude is given by a 3=. Thus the erimeter of the triangular slice is a +(a 3=) = a( + 3). (b) We must now nd the area of the triangular slice whose sides we found in art (a) above. Consider the altitude h which slitsthe isosceles slice into congruent halves. Each half is a right-angled triangle with hyotenuse a 3= and one side of length a=. The third side is h, which can be found by the Theorem of Pythagoras:! a 3 a 3a a a h = 0 = 0 = 4 4 a a h = = Thus the area of the slice is a a : : a = 4