cara-jitu-menguasai-olimpiade-matematika-smp

Karena banyaknya data adalah 5, maka median adalah data yang ke-3 setelah diurutkan. Kelima data yang telah diurutkan itu adalah 4, 5, 5 + y , 7, 9 5 Akibatnya 5 ≤ 5 + y ≤ 7  0 ≤ y ≤ 2 55  0  y 10 Jadi, kenaikan produksi pada periode ke-5 berkisar antara 0 % sampai 10 %. 4. Akan dicari semua pasangan bilangan bulat (x, y) yang memenuhi sistem persamaan (1) dan (2). … (1) x(y + 1) = y2 – 1 … (2) y(x + 1) = x2 – 1 Dari persamaan (1) diperoleh: x(y + 1) = y2 – 1  x(y + 1) = (y +1)(y – 1) Agar kedua ruas pada persamaan terakhir sama, maka y = - 1 atau jika y  - 1 mengharuskan x = y – 1  Jika y = - 1 Maka pada uraian persamaan (1) tidak berlaku hukum pencoretan (y + 1) pada kedua ruas karena nilai (y + 1) adalah 0. Oleh karena itu nilai y = - 1 disubstitusikan ke persamaan (2) y(x + 1) = x2 – 1  (- 1)(x + 1) = x2 – 1  - x – 1 = x2 – 1  x2 + x = 0  x(x + 1) = 0  x = 0 atau x = - 1 Jadi, pasangan bilangan bulatnya adalah (- 1, - 1) dan (0, - 1)  Jika y  - 1 x = y – 1  y = x + 1 … (3) Kemudian substitusikan persamaan (3) ke persamaan (2) y(x + 1) = x2 – 1  (x + 1) (x + 1) = x2 – 1  x2 + 2x + 1 = x2 – 1  2x = - 2  x=-1 Substitusikan nilai x = - 1 ke persamaan (3) y=x+1  y=-1+1 =0 Jadi, pasangan bilangan bulatnya adalah (- 1, 0). Jadi, seluruh pasangan bilangan bulat (x, y) yang memenuhi persamaan (1) dan (2) adalah (- 1, - 1), (- 1, 0) dan (0, - 1).

5. Perhatikan bahwa titik E terletak di sembarang tempat di Luar persegi ABCD. Karena ABCD persegi, maka kita cukup menempatkan titik E di titik sebelah luar salah satu sisi persegi, misalnya di sebelah luar sisi DC . E DC F A GB Dari segitiga DEF diperoleh EF2 = DE2 – DF2 … (1) Dari segitiga CEF diperoleh EF2 = CE2 – CF2 … (2) Dari persamaan (1) dan (2) diperoleh DE2 – DF2 = CE2 – CF2  CF2 - DF2 = CE2 – DE2 … (3) Dari segitiga AEG diperoleh EG2 = AE2 – AG2 … (4) Dari segitiga BEG diperoleh EG2 = BE2 – BG2 … (5) Dari persamaan (4) dan (5) diperoleh AE2 – AG2 = BE2 – BG2  AE2 – DF2 = BE2 – CF2  CF2 - DF2 = BE2 – AE2 … (6) Dari persamaan (3) dan (6) diperoleh CE2 – DE2 = BE2 – AE2  AE2 + CE2 = BE2 + DE2 Jadi, terbukti bahwa pada persegi ABCD dengan E sebarang titik di luar persegi tersebut, akan berlaku hubungan AE2 + CE2 = BE2 + DE2. 6. Perhatikan bahwa volume air setelah akuarium dikembalikan ke posisi semula sama dengan volume air pada saat akuarium dalam posisi miring. Volume air pada saat akuarium miring adalah volume sebuah prisma yang tingginya sama dengan lebar akuarium, sedangkan alas prisma berupa segitiga siku-siku yang panjang alasnya sama dengan setengah dari panjang akuarium dan tingginya sama dengan tiggi akuarium. Misalkan: P = Panjang akuarium = 100 cm L = Lebar akuarium = 60 cm

T1 = tinggi air pada akuarium Semula = 40 cm T2 = tinggi air setelah akuarium dikembalikan ke posisi semula V1 = Volume air setelah akuarium dikembalikan ke posisi semula V2 = Volume prisma segitiga siku-siku V1 = V2  P × L × T2 = 1 × ( 1 P) × T1 × L (Kedua ruas dibagi dengan PL) 22  T2 = 1 T1 4  T2 = 1 (40 cm) 4 = 10 cm Jadi, tinggi air setelah akuarium dikembalikan ke posisi semula adalah 10 cm. 7. Perhatikan tabel pada soal! Dari semua kolom, hanya kolom c yang mempunyai keteraturan pada setiap barisnya. Setiap baris pada kolom c merupakan kelipatan 4. Perhatikan bahwa khusus untuk bilangan kelipatan 4 yang juga merupakan kelipatan 8, pergerakan bilangan selanjutnya berarah ke sebelah kiri, sedangkan yang lainnya berarah ke sebelah kanan. Selanjutnya akan dicari suatu bilangan kelipatan 4 terbesar yang lebih kecil atau sama dengan 2007 yaitu 2004. Karena 2004 bukan kelipatan 8, maka 2 bilangan asli selanjutnya setelah 2004 yaitu 2005 dan 2006 bergerak ke arah kanan sehingga masing- masing menempati kolom d dan kolom e. Jadi, bilangan 2007 menempati kolom d. 8. Misalkan A = Persentase banyaknya rumah tangga yang memiliki 2 anak atau lebih = 40 % B = Persentase banyaknya rumah tangga yang memiliki tepat 1 anak C = Persentase banyaknya rumah tangga yang tidak memiliki anak = 10 % T = Persentase banyaknya rumah tangga total di Surabaya =A+B+C = B + 50 % = 100 % Akibatnya B = 50 % Kemudian karena 30 % dari B memiliki anak perempuan, maka persentase rumah tangga yang memiliki tepat satu anak perempuan adalah 70 % dari B yaitu 70 % × 50 % = 35 %. Jadi, persentase rumah tangga yang memiliki tepat satu anak di Surabaya adalah 35 %. 9. Perhatikan gambar di bawah ini!

Pada segitiga ABC, BF tegaklurus AC dan AG tegaklurus BE. A 20 cm D F E G B 20 cm C Perhatikan juring ABE ! Karena AB = BC, maka  BAD =  BCE = 450. Akibatnya juring ABE = juring CBD . Luas juring ABE = 450 πr2 3600 = 1 π(AB)2 8 = 1 π 20cm2 8 = 50 π cm2 . Dari segitiga ABC diperoleh AC2 = AB2 + BC2 = (202 + 202) cm = (400 + 400) cm = 800 cm AC = 800 cm = 20 2 cm AF = 1 AC 2 = 1 ( 20 2 ) cm 2 = 10 2 cm Karena BF tegak-lurus dengan AC maka pada segitiga ABF berlaku  ABF =  BAF = 450. Akibatnya BF = AF = 10 2 cm. Perhatikan bahwa CD = AB =r = 20 cm. DF = CD – CF

= CD – AF = (20 - 10 2 ) cm = 10(2 - 2 ) cm Luas segitiga BDE = 1 (DE)(BF) 2 = 1 (2 × DF)(BF) 2 = (DF)(BF) = 10(2 - 2 )(10 2 ) cm2 = 100 2 (2 - 2 ) cm2 = [200 2 - 200] cm2 = 200( 2 - 1) cm2 Dari segitiga BEF diperoleh BE2 = BF2 + EF2 = [(10 2 )2 + (20 - 10 2 )2] cm = [200 + 400 + 200 - 400 2 ] cm = [800 - 400 2 ] cm = 400(2 - 2 )  BE = 400 2  2 cm = 20 2  2 cm BG = 1 BE 2 = 1 (20 2  2 ) cm 2 = 10 2  2 cm Kemudian dari segitiga ABG diperoleh AG2 = AB2 – BG2 = (20)2 – (10 2  2 )2 cm = [400 – 100(2 - 2 )] cm = (400 – 200 + 100 2 ) cm = (200 + 100 2 ) cm = 100(2 + 2 ) cm  AG = 100 2  2 cm = 10 2  2 cm

Luas segitiga ABE = 1 (BE)(AG) 2 cm) 2 = 1 (20 2  2 cm)( 10 2  2   = 100 2  2 2  2 cm2 = 100 4  2 cm2 = 100 2 cm2 Luas tembereng BE = (Luas juring ABE ) – (Luas segitiga ABE) = (50  - 100 2 ) cm2 = 50(  - 2 2 ) cm2 Luas daerah diarsir pada segitiga ABC = 2 × (Luas diarsir pada juring) = 2 × [(Luas juring ABE ) – 2 × (Luas tembereng BE ) - (Luas segitiga BDE)] = 2[50  - (2)50(  - 2 2 ) - 200( 2 - 1)] cm2 = 2[50  - 100  + 200 2 - 200 2 + 200] cm2 = 2[200 - 50  ] cm2 = 100(4 -  ) cm2 Jadi, luas daerah yang diarsir adalah 100(4 -  ) cm2. Catatan : Untuk mencari luas sebarang segitiga yang diketahui ketiga sisinya dapat digunakan juga rumus L = s(s - a)(s - b)(s - c) Dimana L = Luas segitiga a, b, c = Panjang sisi-sisinya s = Setengah keliling = a+b+c 2

10. Pertama-tama akan dipilih sembarang dua titik dari 7 titik yang diketahui untuk dibuat suatu persamaan garis. Selanjutnya substitusikan ketujuh titik tersebut ke persamaan garis yang telah dibuat sehingga tepat 5 titik memenuhi persamaan garis. Apabila tidak tepat 5 titik memenuhi persamaan garis artinya kita telah salah memilih 2 titik sebarang. Misalkan 2 titik yang dipilih adalah titik A = (x1, y1) = (9, 17) dan B = (x1, y1) = (6, 11). Persamaan garis lurus yang melalui A dan B adalah y - y1 = x - x1  y -17 = x - 9 y2 - y1 x2 - x1 11-17 6 - 9  y -17 = x - 9 -6 -3  y -17 = x - 9 2  y -17 = 2x -18  2x – y – 1 = 0 Titik A, B, C, E dan G memenuhi persamaan terakhir di atas sedangkan titik D dan F tidak memenuhi. Jadi, 2 titik yang tidak terletak pada garis yang ditempati kelima titik lainnya adalah titik D(7, 12) dan titik F(5, 10).

LATIHAN BIDANG MATEMATIKA SOAL URAIAN 1. Bentuk paling sederhana dari 2  2 4  12 = … 2. Ada berapa banyak bilangan bulat positif 3 angka, dimana angka ketiga merupakan jumlah kedua angka di depannya? 3. Diketahui ada bilangan 5 digit dengan ciri sebagai berikut: digit puluhan adalah dua kali lipat digit ribuan, serta apabila digit ratusan dan satuan dipertukarkan maka nilai bilangan tersebut tidak berubah. Terdapat berapa banyakkah bilangan tersebut? 4. Tiga sahabat Ani, Ina, dan Nia sedang membicarakan uang saku bulanan yang mereka terima dari orang tua mereka masing-masing. Ani merasa heran, karena uang sakunya hanya setengah dari uang saku Ina, bahkan Nia mendapatkan lebih banyak lagi, karena selisihnya dengan uang saku Ina adalah dua kali lipat dari uang saku Ani. Diketahui jumlah uang saku ketiganya adalah Rp. 350.000. Berapa besarnya uang saku Ani? 5. Tentukan semua pasangan bilangan asli (x,y) sehingga 1 + 1 = 1 ! xy3 6. Diketahui x = 0, 1234567891011 … 998999. Perhatikan bahwa angka-angka di belakang tanda koma adalah bilangan-bilangan asli kurang dari 1.000 yang disusun berurutan . Berikan angka ke-2007 di belakang tanda koma! 7. Terdapat berapa banyakkah bilangan bulat positif tak lebih dari 2007 yang merupakan kelipatan 3 atau 4 tetapi bukan kelipatan 5? 8. Diberikan pola bilangan sebagai berikut 1, 2, 4, 7, 11, 16, … Tentukan bilangan ke-2007 pada barisan tersebut!

9. Jika diketahui n adalah kuadrat suatu bilangan asli, maka tentukan bilangan kuadrat berikutnya yang dinyatakan dalam n! 10. Lima titik A, B, C, D dan E terletak berurutan pada suatu garis lurus. Diketahui bahwa jarak A ke E adalah 20 cm, A ke D 15 cm, B ke E 10 cm, dan titik C terletak di tengah- tengah antara B dan D. Berapakah jarak B ke D?

KUNCI JAWABAN 1. 1 + 3 2. 45 3. 450 4. Rp. 50.000 5. {(2, 6)} 6. 5 7. 804 8. 2.013.022 9. n + n 2 +1 10. 2,5 cm

DAFTAR PUSTAKA Frank C. , 2000, Australian Mathematics Olympiad for Junior High School, Sydney Press. Gardiner A. ,1997, Discovering mathematics, Clarendon press – Oxford. _ , 2001, Text of Canadian Mathematics Olympiad for Junior High school. _ , 2003, Text of Canadian Mathematics Olympiad for junior High School. _ , 2015, Naskah Olimpiade Matematika Tingkat Kabupaten / Kota, Provinsi dan Nasional, Depdiknas.

2003 Fryer Contest (Grade 9) Wednesday, April 16, 2003 1. (a) The marks of 32 mathematics students on Test 1 Marks on Test 1 are all multiples of 10 and are shown on the bar graph. What was the average (mean) mark of the 10 32 students in the class? 9 8 Number of Students 7 6 5 4 3 2 1 0 10 20 30 40 50 60 70 80 90 100 Marks (out of 100) (b) After his first 6 tests, Paul has an average of 86. What will his average be if he scores 100 on his next test? (c) Later in the year, Mary realizes that she needs a mark of 100 on the next test in order to achieve an average of 90 for all her tests. However, if she gets a mark of 70 on the next test, her average will be 87. After she writes the next test, how many tests will she have written? 2. In a game, Xavier and Yolanda take turns calling out whole numbers. The first number called must be a whole number between and including 1 and 9. Each number called after the first must be a whole number which is 1 to 10 greater than the previous number called. (a) The first time the game is played, the person who calls the number 15 is the winner. Explain why Xavier has a winning strategy if he goes first and calls 4. (b) The second time the game is played, the person who calls the number 50 is the winner. If Xavier goes first, how does he guarantee that he will win? 3. In the diagram, ABCD is a square and the coordinates of A and D y are as shown. D(1, 8) C (a) The point P has coordinates (10, 0). Show that the area of A(1, 4) B triangle PCB is 10. (b) Point E(a, 0) is on the x-axis such that triangle CBE lies entirely outside square ABCD. If the area of the triangle is equal to the area of the square, what is the value of a? (c) Show that there is no point F on the x-axis for which the area of triangle ABF is equal to the area of square ABCD. x P(10, 0) 4. For the set of numbers {1, 10, 100} we can obtain 7 distinct numbers as totals of one or more elements of the set. These totals are 1, 10, 100, 1 + 10 = 11, 1 + 100 = 101, 10 + 100 = 110, and 1 + 10 + 100 = 111. The “power-sum” of this set is the sum of these totals, in this case, 444. (a) How many distinct numbers may be obtained as a sum of one or more different numbers from the set {1, 10, 100, 1000}? Calculate the power-sum for this set. (b) Determine the power-sum of the set {1, 10, 100, 1000, 10 000, 100 000, 1 000 000}. over ...

Extensions (Attempt these only when you have completed as much as possible of the four main problems.) Extension to Problem 1: Mary’s teacher records the final marks of the 32 students. The teacher calculates that, for the entire class, the median mark is 80. The teacher also calculates that the difference between the highest and lowest marks is 40 and calculates that the average mark for the entire class is 58. Show that the teacher has made a calculation error. Extension to Problem 2: In the game described in (b), the target number was 50. For what different values of the target number is it guaranteed that Yolanda will have a winning strategy if Xavier goes first? Extension to Problem 3: G is a point on the line passing through the points M (0, 8) and N(3, 10) such that ∆ DCG lies entirely outside the square. If the area of ∆ DCG is equal to the area of the square, determine the coordinates of G. Extension to Problem 4: Consider the set {1, 2, 3, 6, 12, 24, 48, 96}. How many different totals are now possible if a total is defined as the sum of 1 or more elements of a set?

Canadian Mathematics Competition An activity of The Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 2003 Solutions Fryer Contest (Grade 9) © 2003 Waterloo Mathematics Foundation

2003 Fryer Solutions 2 1. (a) Solution 1 The average (mean) is equal to the sum of all of the marks, divided by the total number of marks. Since we know already that there are 32 students (we can check this by looking at the graph), then the average is 1(10) + 2(30) + 2(40) + 1(50) + 4(60) + 6(70) + 9(80) + 4(90) + 3(100) = 2240 = 70 32 32 Therefore, the average mark was 70. Solution 2 The average (mean) is equal to the sum of all of the marks, divided by the total number of marks. Using the bar graph, we can list out all of the marks: 10, 30, 30, 40, 40, 50, 60, 60, 60, 60, 70, 70, 70, 70, 70, 70, 80, 80, 80, 80, 80, 80, 80, 80, 80, 90, 90, 90, 90, 100, 100, 100. Adding these up using a calculator and dividing by 32, we find that the average mark is 70. (b) After his first 6 tests, since Paul’s average is 86, then has gotten a total of 6(86) = 516 marks. After getting 100 on his seventh test, Paul has gotten a total of 516 + 100 = 616 marks, so his new average is 616 = 88. 7 (c) Solution 1 If Mary gets 100, her average becomes 90. If Mary gets 70, her average becomes 87. So a difference of 30 marks on the test gives a difference of 3 marks in the average. Since her average is her total number of marks divided by her total number of tests, and a difference of 30 in the total number of marks makes a difference of 3 in her average, then she will have written 30 = 10 tests. 3 Solution 2 Suppose that after the next test, Mary has written n tests. If her average after getting 100 on the next test is 90, then Mary has earned 90n marks in total after the first n tests, and so 90n − 100 before she writes the nth test. If her average after getting 70 on the next test is 87, then Mary has gotten 87n marks in total after the nth test, and so she will have earned 87n − 70 marks before the next test. Therefore, since the number of marks before her next test is the same in either case,

2003 Fryer Solutions 3 87n − 70 = 90n − 100 30 = 3n n = 10 So Mary will have written 10 tests. Extension We start by using the given information to try to figure out some more things about the marks of the 32 students. Since the median mark is the “middle mark” in a list of marks which is increasing, then there at least 16 students who have marks that are at least 80. Since the difference between the highest and lowest marks is 40, and there are students who got at least 80, then the lowest mark in the class cannot be lower than 40. Since the average mark in the class is 58, then the total number of marks is 32(58) = 1856. So what does this tell us? Since at least 16 students got at least 80, then this accounts for at least 1280 marks, leaving 1856 − 1280 = 576 marks for the remaining 16 students. But the lowest possible mark in the class was 40, so these remaining 16 students got at least 40 each, and so got at least 16(40) = 640 marks in total! So we have an inconsistency in the data. Thus, the teacher made a calculation error. (There is a variety of different ways of reaching this same conclusion. As before, we know that 16 students will have a mark of at least 80, which accounts for 1280 marks. By the same reasoning, the other 16 students would account for the other 576 marks. The average for these 16 students is thus about 34, which implies that at least one of these lower students must have a mark of 34 or lower. This now contradicts the statement that the range is 40 since 80 − 34 > 40.) 2. (a) Solution 1 If Xavier goes first and calls 4, then on her turn Yolanda can call any number from 5 to 14, since her number has to be from 1 to 10 greater than Xavier’s. But if Yolanda calls a number from 5 to 14, then Xavier can call 15 on his next turn, since 15 is from 1 to 10 bigger than any of the possible numbers that Yolanda can call. So Xavier can call 15 on his second turn no matter what Yolanda calls, and is thus always guaranteed to win.

2003 Fryer Solutions 4 Solution 2 If Xavier goes first and calls 4, then Yolanda will call a number of the form 4 + n where n is a whole number between 1 and 10. On his second turn, Xavier can call 15 (and thus win) if the difference between 15 and 4 + n is between 1 and 10. But 15 − (4 + n) = 11− n and since n is between 1 and 10, then 11− n is also between 1 and 10, so Xavier can call 15. Therefore, Xavier’s winning strategy is to call 15 on his second turn. (b) In (a), we saw that if Xavier calls 4, then he can guarantee that he can call 15. Using the same argument, shifting all of the numbers up, to guarantee that he can call 50, he should call 39 on his previous turn. (In this case, Yolanda can call any whole number from 40 to 49, and in any of these cases Xavier can call 50, since 50 is no more than 10 greater than any of these numbers.) In a similar way, to guarantee that he can call 39, he should call 28 on his previous turn, which he can do for the same reasons as above. To guarantee that he can call 28, he should call 17 on his previous turn. To guarantee that he can call 17, he should call 6 on his previous turn, which could be his first turn. Therefore, Xavier’s winning strategy is to call 6 on his first turn, 17 on his second turn, 28 on his third turn, 39 on his fourth turn, and 50 on his last turn. At each step, we are using the fact that Xavier can guarantee that his number on one turn is 11 greater than his number on his previous turn. This is because Yolanda adds 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10 to his previous number, and he can then correspondingly add 10, 9, 8, 7, 6, 5, 4, 3, 2, or 1 to her number, for a total of 11 in each case. Extension In (b), we discovered that Xavier can always guarantee that the difference between his numbers on two successive turns is 11. In fact, Yolanda can do the same thing, using exactly the same strategy as Xavier did. If the target number is between 1 and 9, then Xavier will win on his very first turn by calling that number. If the target number is then 11 greater than a number between 1 and 9, Xavier will win as in (b). Thus, Xavier wins for 12 through 20. What about 10 and 11? In each of these cases, Yolanda can win by choosing 10 or 11 on her first turn, which she can do for any initial choice of Xavier’s, since he chooses a number between 1 and 9. Therefore, Yolanda will also win for 21 and 22, and so also for 32 and 33, and so on. Since either Yolanda or Xavier can repeat their strategy as many times as they want, then Xavier can ensure that he wins if the target number is a multiple of 11 more than one of 1 through 9.

2003 Fryer Solutions 5 Similarly, Yolanda can ensure that she wins if the target number is a multiple of 11 more than 10 or 11, ie. if the target number is a multiple of 11, or 1 less than a multiple of 11. 3. (a) Solution 1 Since ABCD is a square and AD has side length 4, then each of the sides of ABCD has length 4. We can also conclude that B has coordinates (5,4) and C has coordinates (5,8). (Since AD is parallel to the y-axis, then AB is parallel to the x-axis.) If we turn ∆PBC on its side, then its we see y that its base is BC which has length 4. Also the height of the triangle is the vertical D(1, 8) C(5, 8) distance from the line BC to P, which is 5. (We can see this by extending the line CB to a 4 B(5, 4) point X on the x-axis. Then X has coordinates A(1, 4) (5,0) and PX, which has length 5, is 4 perpendicular to CB.) x Therefore, the area of ∆PBC is X(5, 0) 5 P(10, 0) 1 = 1 (4)(5) = 10 . 2 bh 2 Solution 2 Since ABCD is a square and AD has side length 4, then all of the sides of ABCD have length 4. We can also conclude that B has coordinates (5,4) and C has coordinates (5,8). (Since AD is parallel to the y-axis, then AB is parallel to the x-axis.) Extend CB down to a point X on the x-axis. y Point X has coordinates (5,0). Then the area of ∆PBC is the difference D(1, 8) C(5, 8) between the areas of ∆PCX and ∆PBX . ∆PCX has base PX of length 5 and height CX 4 B(5, 4) of length 8. ∆PBX has base PX of length 5 and height BX A(1, 4) of length 4. 4 Therefore, the area of ∆PBC is x 1 (4)(10) − 1 (4)(5) = X(5, 0) 5 P(10, 0) 2 2 10 as required. (b) Solution 1 Since triangle CBE lies entirely outside square ABCD, then the point E must be “to the right” of the square, ie. a must be at least 5.

2003 Fryer Solutions 6 Also, we need to know the area of the square. Since the side length of the square is 4, its area is 16. Thus if we turn ∆CBE on its side, then its we y see that its base is BC which has length 4. Also the height of the triangle is the vertical D(1, 8) C(5, 8) 4 distance from the line BC to E, which is a − 5 since E has coordinates (a,0) and BC is part of the line x = 5. A(1, 4) B(5, 4) Therefore, since the area of ∆CBE is equal to the area of the square, x 1 (4)(a − 5) = 16 E(a, 0) 2 2a − 10 = 16 2a = 26 a = 13 Thus, a = 13. (It is easy to verify that if a = 13, then the height of the triangle is 8 and its base is 4, giving an area of 16.) Solution 2 Since triangle CBE lies entirely outside square ABCD, then the point E must be “to the right” of the square, ie. a must be at least 5. Also, we need to know the area of the square. Since the side length of the square is 4, its area is 16. Extend CB down to a point X on the x-axis. y Point X has coordinates (5,0). Then the area of ∆CBE is the difference D(1, 8) C(5, 8) between the areas of ∆CXE and ∆BXE . ∆CXE has base EX of length a − 5 and 4 B(5, 4) height CX of length 8. ∆BXE has base EX of length a − 5 and height A(1, 4) BX of length 4. 4 Therefore, since the area of ∆CBE is equal to x X(5, 0) a – 5 E(a, 0) the area of the square, 1 (a − 5)(8) − 1 (a − 5)(4) = 16 2 2 4(a − 5) − 2(a − 5) = 16 a−5 = 8 a = 13

2003 Fryer Solutions 7 Thus, a = 13, as required. (c) Suppose that F has coordinates (b,0). y Then triangle ABF has base AB of length 4. D(1, 8) C(5, 8) A(1, 4) The height of triangle ABF is the vertical distance from F to B(5, 4) the line AB, which is always 4, no matter where F is. Thus, the area of triangle ABF is always 1 bh = 1 (4)(4) = 8 , 2 2 which is not equal to the area of the square. x F(b, 0) Extension Since triangle DCG lies entirely outside the square, then G is “above” the line through D and C, ie. the y-coordinate of G is at least 8. Since the area of triangle DCG is equal to the area of the square, then the area of triangle DCG is 16. Now triangle DCB has base DC, which has length y 4, so 1 bh = 1 (4)h = 16 or h = 8. 2 2 G Since the height of triangle DCG is 8, then G has y-coordinate 16, since both D and C have N(3, 10) M(0, 8) y-coordinate 8. D(1, 8) C(5, 8) So we must find the point on the line through M(0,8) and N (3,10) which has y-coordinate 16. To get from M to N, we go 3 to the right and up 2. To get from N to G, we go up 6, so we must go 9 to the right. A(1, 4) B(5, 4) Therefore, G has coordinates G(12,16). x 4. (a) The best approach here is to list the numbers directly. The possible totals are, from smallest to largest: 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111 There 15 possible totals, and their sum (that is, the power-sum) is 8888. (b) Solution 1 First, we consider the numbers that are sums of 1 or more of the numbers from {1,10,100,1000}. In (a), we saw that the sum of these numbers is 8888.

2003 Fryer Solutions 8 What happens when we consider numbers that are sums of 1 or more of the numbers from {1,10,100,1000,10 000}? When we do this, we obtain all 15 of the numbers from the previous paragraph, we obtain the 15 numbers obtained by adding 10 000 to all of the numbers from the previous paragraph, and also the number 10 000. (Either our sum does not include 10 000 as a term, or it does; if it doesn’t, it must be one of the numbers from (a); if it does, it could be 10 000 on its own, or it could be 10 000 plus one of the numbers from (a).) Therefore, we have 15 + 15 + 1 = 31 numbers in total, whose sum is [ ]8888 + 8888 + 15(10 000) + 10 000 = 2(8888) + 160 000 = 2(88 888) = 177 776 What happens when we consider the numbers that are sums of 1 or more of the numbers from {1,10,100,1000,10 000,100 000}? When we do this, we obtain all 31 of the numbers from (a), we obtain the 31 numbers obtained by adding 100 000 to all of the numbers from (a), and also the number 100 000. Therefore, we have 31 + 31 + 1 = 63 numbers in total, whose sum is [ ]177 776 + 177 776 + 31(100 000) + 100 000 = 2(177 776) + 3 200 000 = 3 555 552 What happens when we add 1 000 000 to the set? We then obtain, as before, 63 + 63 + 1 = 127 numbers in total, whose sum is [ ]3 555 552 + 3 555 552 + 63(1 000 000) + 1 000 000 = 2(3 555 552) + 64 000 000 = 71111104 Therefore, the power sum is 71 111 104. Solution 2 There are seven numbers in the given set. When we are considering sums of one or more numbers from the set, each of the seven numbers in the set is either part of the sum, or not part of the sum. So there are two choices (“in” or “out”) for each of the 7 elements. So we can proceed by first choosing the elements we want to add up, and then adding them up. Since for each of the two possibilities for the “1” (ie. chosen or not chosen), there are two possibilities for the “10”, and there are two possibilities for the “100”, and so on. In total, there will be 2 × 2 × 2 × 2 × 2 × 2 × 2 = 27 = 128 ways of choosing elements. Notice that this includes the possibility of choosing no elements at all (since we could choose not to select each of the seven elements). So there are 128 possible sums (including the sum which doesn’t add up any numbers at all!).

2003 Fryer Solutions 9 In how many of these sums is the “1” chosen? If the “1” is chosen, then there are still 2 possibilities for each of the remaining six elements (either chosen or not chosen), so there are 26 = 64 sums with the 1 included, so the 1 contributes 64 to the power-sum. In how many of these sums is the “10” chosen? Using exactly the same reasoning, there are 64 sums which include the 10, so the 10 contributes 640 to the power-sum. Extending this reasoning, each of the 7 elements will contribute to 64 of the sums. (Note that including the “empty” sum doesn’t affect the power-sum.) Therefore, the power sum is 64(1 + 10 + 100 + 1000 + 10000 + 100000 + 1000000) = 64(1111111) = 71111104 Extension We start looking at small numbers to see if we can see a pattern. Using the numbers 1, 2, 3, we can form the sums 1, 2, 3, 1 + 3 = 4 , 1 + 3 = 4 , 2 + 3 = 5, 1 + 2 + 3 = 6 If we include the 6, we can obtain all of these sums, as well as 6 plus each of these sums. In other words, we obtain each of the numbers 1 through 12 as totals. Then including the 12, we can obtain 13 through 24, so we now have each of 1 to 24 as totals. Including the 24, we obtain all numbers up to 48. Including the 48, we obtain all numbers up to 96. Including the 96, we obtain all numbers up to 192. Therefore, there are 192 different totals possible. (We can check as well that the sum of the elements in the original set is 192.)

Canadian Mathematics Competition An activity of The Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario Pascal Contest (Grade 9) Wednesday, February 24, 1999 C.M.C. Sponsors: C.M.C. Supporters: C.M.C. Contributors: Chartered Accountants IBM The Great-West Canada Ltd. Life Assurance Company Canadian Institute of Actuaries Northern Telecom (Nortel) Sybase Inc. (Waterloo) Manulife Financial Equitable Life of Canada Time: 1 hour © 1999 Waterloo Mathematics Foundation Calculators are permitted, providing they are non-programmable and without graphic displays. Instructions 1. Do not open the contest booklet until you are told to do so. 2. You may use rulers, compasses and paper for rough work. 3. Be sure that you understand the coding system for your response form. If you are not sure, ask your teacher to clarify it. All coding must be done with a pencil, preferably HB. Fill in circles completely. 4. On your response form, print your school name, city/town, and province in the box in the upper right corner. 5. Be certain that you code your name, age, sex, grade, and the contest you are writing on the response form. Only those who do so can be counted as official contestants. 6. This is a multiple-choice test. Each question is followed by five possible answers marked A, B, C, D, and E. Only one of these is correct. When you have decided on your choice, fill in the appropriate circles on the response form. 7. Scoring: Each correct answer is worth 5 credits in Part A, 6 credits in Part B, and 8 credits in Part C. There is no penalty for an incorrect answer. Each unanswered question is worth 2 credits, to a maximum of 20 credits. 8. Diagrams are not drawn to scale. They are intended as aids only. 9. When your supervisor instructs you to begin, you will have sixty minutes of working time.

Scoring: There is no penalty for an incorrect answer. Each unanswered question is worth 2 credits, to a maximum of 20 credits. Part A: Each question is worth 5 credits. 1. The value of 4 × 4 + 4 is 2×2–2 (A) 2 (B) 6 (C) 10 (D) 12 (E) 18 ( )2. If k = 2, then k3 – 8 (k + 1) equals (E) – 6 (A) 0 (B) 3 (C) 6 (D) 8 (E) 18 3. If 4(♥)2 = 144, then a value of ♥ is (E) 11 (E) 3975 (A) 3 (B) 6 (C) 9 (D) 12 110° ( )4. Which of the following numbers divide exactly into 15 + 49 ? a° b° (A) 3 (B) 4 (C) 5 (D) 7 60° 5. If 10% of 400 is decreased by 25, the result is (A) 15 (B) 37.5 (C) 65 (D) 260 6. In the diagram, a + b equals (A) 10 (B) 85 (C) 110 (D) 170 (E) 190 7. If 2x – 1 = 5 and 3y + 2 = 17, then the value of 2x + 3y is (A) 8 (B) 19 (C) 21 (D) 23 (E) 25 8. The average of four test marks was 60. The first three marks were 30, 55 and 65. What was the fourth mark? (A) 40 (B) 55 (C) 60 (D) 70 (E) 90 9. In the diagram, each small square is 1 cm by 1 cm. The area of the shaded region, in square centimetres, is (A) 2.75 (B) 3 (C) 3.25 (D) 4.5 (E) 6

10. 10 + 103 equals (B) 8.0 × 103 (C) 4.0 × 101 (D) 1.0 × 104 (E) 1.01 × 103 (A) 2.0 × 103 Part B: Each question is worth 6 credits. 11. Today is Wednesday. What day of the week will it be 100 days from now? (A) Monday (B) Tuesday (C) Thursday (D) Friday (E) Saturday 12. The time on a digital clock is 5:55. How many minutes will pass before the clock next shows a time with all digits identical? (A) 71 (B) 72 (C) 255 (D) 316 (E) 436 13. In Circle Land, the numbers 207 and 4520 are shown in the following way: 2 4 7 5 207 2 4520 In Circle Land, what number does the following diagram represent? 3 1 5 (A) 30 105 (B) 30 150 (C) 3105 (D) 3015 (E) 315 14. An 8 cm cube has a 4 cm square hole cut through its centre, 4 cm as shown. What is the remaining volume, in cm3? 4 cm (A) 64 (B) 128 (C) 256 (D) 384 (E) 448 8 cm 8 cm 8 cm 15. For how many different values of k is the 4-digit number 7k52 divisible by 12? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 16. In an election, Harold received 60% of the votes and Jacquie received all the rest. If Harold won by 24 votes, how many people voted? (A) 40 (B) 60 (C) 72 (D) 100 (E) 120

17. In the parallelogram, the value of x is x° 150° (A) 30 (B) 50 (C) 70 (D) 80 (E) 150 18. In the diagram, AD < BC. What is the perimeter of ABCD? 80° D A7 5 (A) 23 (B) 26 (C) 27 4 (D) 28 (E) 30 BC 19. The numbers 49, 29, 9, 40, 22, 15, 53, 33, 13, 47 are grouped in pairs so that the sum of each pair is the same. Which number is paired with 15? (A) 33 (B) 40 (C) 47 (D) 49 (E) 53 ( )( )( )20. The units (ones) digit in the product (5 + 1) 53 +1 56 +1 512 +1 is (A) 6 (B) 5 (C) 2 (D) 1 (E) 0 Part C: Each question is worth 8 credits. 21. A number is Beprisque if it is the only natural number between a prime number and a perfect square (e.g. 10 is Beprisque but 12 is not). The number of two-digit Beprisque numbers (including 10) is (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 22. If w = 2129 × 381 × 5128, x = 2127 × 381 × 5128, y = 2126 × 382 × 5128, and z = 2125 × 382 × 5129, then the order from smallest to largest is (A) w, x, y, z (B) x, w, y, z (C) x, y, z, w (D) z, y, x, w (E) x, w, z, y 23. Al and Bert must arrive at a town 22.5 km away. They have one bicycle between them and must arrive at the same time. Bert sets out riding at 8 km/h, leaves the bicycle and then walks at 5 km/h. Al walks at 4 km/h, reaches the bicycle and rides at 10 km/h. For how many minutes was the bicycle not in motion? (A) 60 (B) 75 (C) 84 (D) 94 (E) 109 24. A number is formed using the digits 1, 2, ..., 9. Any digit can be used more than once, but adjacent digits cannot be the same. Once a pair of adjacent digits has occurred, that pair, in that order, cannot be used again. How many digits are in the largest such number? (A) 72 (B) 73 (C) 144 (D) 145 (E) 91 continued ...

25. Two circles C1 and C2 touch each other externally and the m line l is a common tangent. The line m is parallel to l and C3 C1 touches the two circles C1 and C3. The three circles are mutually tangent. If the radius of C2 is 9 and the radius of C2 C3 is 4, what is the radius of C1? l (A) 10.4 (B) 11 (C) 8 2 (D) 12 (E) 7 3

Canadian Mathematics Competition An activity of The Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 1999 Solutions Pascal Contest(Grade 9) for the Awards © 1999 Waterloo Mathematics Foundation

1999 Pascal Solutions 2 Part A (E) 18 ANSWER: (C) 1. The value of 4 × 4 + 4 is 2×2–2 (E) – 6 (A) 2 (B) 6 (C) 10 (D) 12 ANSWER: (A) (E) 18 Solution 4 × 4 + 4 = 16 + 4 = 20 = 10 ANSWER: (B) 2×2–2 4–2 2 (E) 11 ( )2. If k = 2, then k3 – 8 (k + 1) equals ANSWER: (E) (E) 3975 (A) 0 (B) 3 (C) 6 (D) 8 ANSWER: (A) Solution ( )For k = 2, k3 – 8 (k +1) ( )= 23 – 8 (2 +1) = 0(3) =0 3. If 4(♥)2 = 144, then a value of ♥ is (A) 3 (B) 6 (C) 9 (D) 12 Solution 4(♥)2 = 144 ♥2 = 36 ♥ = ±6 ( )4. Which of the following numbers divide exactly into 15 + 49 ? (A) 3 (B) 4 (C) 5 (D) 7 Solution 15 + 49 = 15 + 7 = 22 The only integer listed that divides 22 evenly is 11. 5. If 10% of 400 is decreased by 25, the result is (A) 15 (B) 37.5 (C) 65 (D) 260 Solution (10% of 400) – 25 = 40 – 25 = 15.

1999 Pascal Solutions 3 6. In the diagram, a + b equals (A) 10 (B) 85 (C) 110 110° (D) 170 (E) 190 a° b° 60° Solution ANSWER: (E) The number of degrees at the centre of a circle is 360. Thus, a + b + 110 + 60 = 360 (measured in degrees). Therefore a + b = 190. 7. If 2x – 1 = 5 and 3y + 2 = 17, then the value of 2x + 3y is (A) 8 (B) 19 (C) 21 (D) 23 (E) 25 Solution 2x – 1 = 5 , 3y + 2 = 17 2x = 6 3y = 15 Thus, 2x + 3y = 6 + 15 = 21. ANSWER: (C) Note: It is not necessary to solve the equations to find actual values for x and y although this would of course lead to the correct answer. It is, however, a little more efficient to solve for 2x and 3y. 8. The average of four test marks was 60. The first three marks were 30, 55 and 65. What was the fourth mark? (A) 40 (B) 55 (C) 60 (D) 70 (E) 90 Solution The total number of marks scored on the four tests was 4 × 60 or 240. The total number of marks scored on the first three tests was 150. The fourth mark was 240 – 150 = 90. ANSWER: (E) 9. In the diagram, each small square is 1 cm by 1 cm. The area of the shaded region, in square centimetres, is (A) 2.75 (B) 3 (C) 3.25 (D) 4.5 (E) 6 Solution The shaded triangle has a base of 2 cm and a height of 3 cm.

1999 Pascal Solutions 4 Its area is 2 × 3 = 3 (sq. cm). ANSWER: (B) 2 (E) 1.01 × 103 10. 10 + 103 equals (B) 8.0 × 103 (C) 4.0 × 101 (D) 1.0 × 104 ANSWER: (E) (A) 2.0 × 103 Solution 10 + 103 = 10 + 1000 = 1010 = 1.01 × 103 Part B 11. Today is Wednesday. What day of the week will it be 100 days from now? (A) Monday (B) Tuesday (C) Thursday (D) Friday (E) Saturday ANSWER: (D) Solution Since there are 7 days in a week it will be Wednesday in 98 days. In 100 days it will thus be Friday. 12. The time on a digital clock is 5:55. How many minutes will pass before the clock next shows a time with all digits identical? (A) 71 (B) 72 (C) 255 (D) 316 (E) 436 Solution The digits on the clock will next be identical at 11:11. This represents a time difference of 316 minutes. (Notice that times like 6:66, 7:77 etc. are not possible.) ANSWER: (D) 13. In Circle Land, the numbers 207 and 4520 are shown in the following way: 2 4 7 5 207 2 4520 In Circle Land, what number does the following diagram represent? 3 1 5

1999 Pascal Solutions 5 (A) 30 105 (B) 30 150 (C) 3105 (D) 3015 (E) 315 Solution 1 = 3 × 104 = 30 000 3 1 = 1 × 102 = 100 5 = 5 × 100 = 5 The required number is 30 000 + 100 + 5 = 30 105. Solution 2 Since there are four circles around the ‘3’ this corresponds to 3 × 104 = 30 000. The ‘5’ corresponds to a 5 in the units digit which leads to 30 105 as the only correct possibility. ANSWER: (A) 14. An 8 cm cube has a 4 cm square hole cut through its centre, 4 cm as shown. What is the remaining volume, in cm3? 4 cm (A) 64 (B) 128 (C) 256 (D) 384 (E) 448 8 cm 8 cm 8 cm Solution ANSWER: (D) Remaining volume = 8 × 8 × 8 – 8 × 4 × 4 (in cm3) = 8(64 – 16) = 8 × 48 = 384 15. For how many different values of k is the 4-digit number 7k52 divisible by 12? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 Solution Since 12 = 4 × 3 the number 7k52 must be divisible by both 4 and 3. Since 52 is the number formed by the last two digits divisible by 4 then we need only ask, ‘for what values of k is 7k52 divisible by 3?’ If a number is divisible by 3 the sum of its digits must be a multiple of 3. Thus 7 + k + 5 + 2 or 14 + k must be a multiple of 3. The only acceptable values for k are 1, 4 or 7. Thus, are three values. ANSWER: (D)

1999 Pascal Solutions 6 16. In an election, Harold received 60% of the votes and Jacquie received all the rest. If Harold won by 24 votes, how many people voted? (A) 40 (B) 60 (C) 72 (D) 100 (E) 120 Solution If Harold received 60% of the votes this implies that Jacquie received 40% of the total number of votes. The difference between them, 20%, represents 24 votes. Therefore, the total number of votes cast was 5 × 24 = 120 . ANSWER: (E) 17. In the parallelogram, the value of x is x° 150° (A) 30 (B) 50 (C) 70 (D) 80 (E) 150 80° Solution The angle in the parallelogram opposite the angle measuring 80° is also 80°. The angle supplementary to 150° is 30°. In the given triangle we now have, x° + 80° + 30° = 180°. Therefore x = 70. ANSWER: (C) 18. In the diagram, AD < BC. What is the perimeter of ABCD ? A7D (A) 23 (B) 26 (C) 27 4 5 (D) 28 (E) 30 Solution B C From D we draw a line perpendicular to BC that meets A BC at N. Since ADNB is a rectangle and AD BC , 4 7D B 45 DN = 4. We use Pythagoras to find NC = 3. We now know that BC = BN + NC = 7 + 3 = 10. NC The required perimeter is 7 + 5 + 10 + 4 = 26. ANSWER: (B)

1999 Pascal Solutions 7 19. The numbers 49, 29, 9, 40, 22, 15, 53, 33, 13, 47 are grouped in pairs so that the sum of each pair is the same. Which number is paired with 15? (A) 33 (B) 40 (C) 47 (D) 49 (E) 53 Solution If we arrange the numbers in ascending order we would have: 9, 13, 15, 22, 29, 33, 40, 47, 49, 53. If the sum of each pair is equal they would be paired as: 9 ↔ 53, 13 ↔ 49, 15 ↔ 47, 22 ↔ 40 , 29 ↔ 33. ANSWER: (C) ( )( )( )20. The units (ones) digit in the product (5 +1) 53 +1 56 +1 512 +1 is (A) 6 (B) 5 (C) 2 (D) 1 (E) 0 Solution We start by observing that each of 53, 56 and 512 have a units digit of 5. This implies that each of 5 + 1, 53 + 1, 56 + 1 and 512 + 1 will then have a units digit of 6. If we multiply any two numbers having a units digit of 6, their product will also have a units digit of 6. Applying this to the product of four numbers, we see that the final units digit must be a 6. ANSWER: (A) Part C 21. A number is Beprisque if it is the only natural number between a prime number and a perfect square (e.g. 10 is Beprisque but 12 is not). The number of two-digit Beprisque numbers (including 10) is (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 Solution We start with the observation that it is necessary to consider only the odd perfect squares and the integers adjacent to them. It is not necessary to consider the even perfect squares because if we add 2 or subtract 2 from an even number the result is even and it is required by the conditions set out in the question that this number be prime. Considering then the odd perfect squares we have: {9, 10 , 11} , {23, 24 , 25, 26, 27} , {47, 48 , 49, 50, 51} , {79, 80 , 81, 82 , 83} . The Beprisque numbers are those that are circled. ANSWER: (E) 22. If w = 2129 × 381 × 5128, x = 2127 × 381 × 5128, y = 2126 × 382 × 5128, and z = 2125 × 382 × 5129, then the order from smallest to largest is (A) w, x, y, z (B) x, w, y, z (C) x, y, z, w (D) z, y, x, w (E) x, w, z, y Solution We start with the observation that 2125 × 381 × 5128 is a common factor to each of the given numbers.

1999 Pascal Solutions 8 For the basis of comparison, we remove the common factor and write the numbers as follows: ( )w = 24 ⋅ 53 2125 × 381 × 5128 = 2000k ( )x = 22 ⋅ 53 2125 × 381 × 5128 = 500k ( )y = 2 ⋅ 3 ⋅ 53 2125 × 381 × 5128 = 750k ( )z = 3 ⋅ 54 2125 × 381 × 5128 = 1875k , where k = 2125 × 381 × 5128. Thus, x < y < z < w. ANSWER: (C) 23. Al and Bert must arrive at a town 22.5 km away. They have one bicycle between them and must arrive at the same time. Bert sets out riding at 8 km/h, leaves the bicycle and then walks at 5 km/h. Al walks at 4 km/h, reaches the bicycle and rides at 10 km/h. For how many minutes was the bicycle not in motion? (A) 60 (B) 75 (C) 84 (D) 94 (E) 109 Solution Let x represent the distance that Bert rides his bicycle. Therefore, he walks for (22.5 – x) km. Bert’s total time for the trip is  x + 22.5 – x hours and Al’s is  x + 22.5 – x hours.  8 5   4 10  Since their times are equal, x + 22.5 – x = x + 22.5 – x 8 5 4 10 22.5 – x – 22.5 – x = x – x 5 10 4 8 2(22.5 – x) – 22.5 – x = 2x – x 10 10 8 8 22.5 – x = x 10 8 10x = 180 – 8x 18x = 180 x = 10. This means that Bert rode for 1.25 h before he left the bicycle and Al walked for 2.5 h before he picked it up. The bicycle was thus not in motion for 1.25 h or 75 minutes. ANSWER: (B) 24. A number is formed using the digits 1, 2, ..., 9. Any digit can be used more than once, but adjacent digits cannot be the same. Once a pair of adjacent digits has occurred, that pair, in that order, cannot be used again. How many digits are in the largest such number? (A) 72 (B) 73 (C) 144 (D) 145 (E) 91

1999 Pascal Solutions 9 Solution Since there are 9(8) = 72 ordered pairs of consecutive digits, and since the final digit has no successor, we can construct a 73 digit number by adding a 9. The question is, of course, can we actually construct this number? The answer is ‘yes’ and the largest such number is, 98 97 96 95 94 93 92 91 87 86 85 84 83 82 81 76 75 74 73 72 71 65 64 63 62 61 54 53 52 51 43 42 41 32 31 21 9. If we count the numbers in the string we can see that there are actually 73 numbers contained within it. ANSWER: (B) 25. Two circles C1 and C2 touch each other externally and the m C3 line l is a common tangent. The line m is parallel to l and C1 touches the two circles C1 and C3. The three circles are C2 mutually tangent. If the radius of C2 is 9 and the radius of l C3 is 4, what is the radius of C1? (A) 10.4 (B) 11 (C) 8 2 (D) 12 (E) 7 3 Solution m A C3 B We start by joining the centres of the circles to form C1 ∆ C1C2C3. (The lines joining the centres pass through the D C2 corresponding points of tangency.) l Secondly, we construct the rectangle ABC2D as shown in the diagram. If the radius of the circle with centre C1 is r we see that: C1C2 = r + 9, C1C3 = r + 4 and C2C3 = 13. We now label lengths on the rectangle in the way noted in A C3 B the diagram. r –4 r+4 C1 13 2r – 13 r –9 D r+9 C2 To understand this labelling, look for example at C1D. The radius of the large circle is r and the radius of the circle with centre C2 is 9. The length C1D is then r – 9 . This same kind of reasoning can be applied to both C1A and BC2 . Using Pythagoras we can now derive the following: In ∆ AC3C1, C3A2 = (r + 4)2 – (r – 4)2 = 16r . Therefore C3A = 4 r .

1999 Pascal Solutions 10 In ∆ DC1C2, (DC2 )2 = (r + 9)2 – (r – 9)2 ANSWER: (D) = 36r. Therefore DC2 = 6 r . In ∆ BC3C2 , (C3B)2 = 132 – (2r – 13)2 = – 4r2 + 52r . Therefore C3B = – 4r2 + 52r . In a rectangle opposite sides are equal, so: DC2 = C3A + C3B or, 6 r = 4 r + – 4r2 + 52r 2 r = – 4r2 + 52r . Squaring gives, 4r = – 4r2 + 52r 4r2 – 48r = 0 4r(r – 12) = 0 Therefore r = 0 or r = 12. Since r > 0 , r = 12.