Part A: Each correct aswer is worth 3 poits. Iteratioal Cotest-Game MATH KANGAROO Caada, 007 Grade ad. Mike is buildig a race track. He wats the cars to start the race i the order preseted o the left, ad to ed the race i the order preseted to the right. Which of the elemets give below should Mike use to replace the elemet X? A) B) C) D) E) Solutio: We eed a piece where the first eterig lie exits third, the secod eterig lie exits first ad the third eterig lie exits secod. Aswer: A si. What is the value of? cos89 A) 0 B) ta C) / ta D) /89 E) Solutio: Usig the properties of the sie ad the cosie, it follows that the umerator ad the deomiator are equal, so the value of the fractio is. Aswer: E 3. O the figure, the triagle ABC is iscribed i a circle with a cetre O. The shaded area is equal to 3. What is the area of the triagle ABC? A) 3 B) C) 5 D) 4 E) 4 3 Solutio: The shaded area is a half of the area of the triagle. Aswer: A 4. At the etrace examiatio to a uiversity, a studet must aswer at least 80% of the questios correctly. So far, Peter has worked o 5 questios. He did ot kow the aswer to 5 of them, but he was sure that he has aswered the other 0 questios correctly. If he does ot guess o the five questios he could ot aswer ad aswers all the remaiig questios i test correctly, he will pass the test at exactly 80%. How may questios are there i the test? A) 0 B) 5 C) 30 D) 35 E) 40 Solutio: Let be the umber of remaiig questios. The, Peter would have aswered (0+) questios correctly, out of (5+) questios i total. As this provides a result of 80%, the equatio to 0 + solve is 0. 8. 5 +
5. A billiard ball is hit from a poit o the vertical board of the table (close to the pocket D), as show i the diagram. It meets the horizotal board at a agle of 45, ad cotiues followig the directio of the arrow. Ito which pocket will the ball fall? A) A B) B C) C D) D E) either of the pockets Solutio: Usig the physics laws of reflectio, it is clear that the ball will approach ad will bouce from all walls at a agle of 45. Therefore, all segmets that trace the path of the ball will pass alog diagoals of the grid squares. The completed path of the ball will be the oe traced o the diagram. 6. Some historias claim that the aciet Egyptias used a strig with kots to costruct a right agle. If the legth of the strig is m ad oe of the kots is at the poit X 3 m from oe of the eds, at what distace from the other ed of the strig should the secod kot be put i order to obtai a right agle at X? A) 3 B) 4 C) 5 D) 6 E) Noe of these Solutio: A right agle at X ca be costructed by foldig the strig ito a right-agled triagle, with the three sectios betwee the kots beig the sides of the triagle. Therefore, the questio is equivalet to fidig two umbers, say, m ad (m<), such that 3, m, ad satisfy the Pythagorea Theorem, that is3 + m, ad 3++m. By oticig that the popular triple 3, 4, 5 has a sum of oe ca quickly obtai the correct aswer. Alteratively, the system of the above equatios may be solved. 7. A islad is ihabited by kights ad liars. Each kight always tells the truth ad each liar always lies. Oce a islader A, whe asked about himself ad aother islader B, claimed that at least oe of A ad B is a liar. Which of the followig seteces is true? A) A is ot able to make the above statemet. B) Both are liars. C) Both are kights. D) A is a liar while B is a kight. E) B is a liar while A is a kight. Solutio: There are four possible combiatios for A ad B, the oes i the aswers B, C, D, ad E. If A is a liar, his statemet would ot be true, so, it meas that oe of the two is a liar, which cotradicts to the assumptio. If A is a kight, the he is tellig the truth, ad it is possible oly if B is the liar. Aswer: E. 8. I the figure, AE is divided ito four equal parts. Three semicircles are costructed takig AE, AD, ad DE as diameters, ad creatig two paths from A to E, as show. What is the ratio of the legth of the curve above AE ad the legth of the curve below AE? A) : B) :3 C) : D) 3: E) :
Solutio: Deote the legth of AB by x. The legth of the curve above AE is ((x) π ) xπ. The legth of the curve below AE is ( (.5 x ) π ) + ( (0.5x) π ) xπ. Aswer: E. Part B: Each correct aswer is worth 4 poits. 9. Give a square ABCD with side, all squares are draw that share at least two vertices with ABCD. What is the area of the regio of all poits covered by at least oe of these squares? A) 5 B) 6 C) 7 D) 8 E) 9 Solutio: The squares that share at least two vertices with ABCD are two groups: () four squares cogruet to it, costructed o each of the sides of ABCD; () four squares with a side equal to the diagoal of ABCD, whose cetres are A,B,C ad D, respectively. All squares that share three or more vertices with ABCD i fact coicide with ABCD. The required regio has the form of a octago ad is composed of five squares with a area of ad four triagles, each of them beig half of the squares. The total area, therefore, is 5+7. 0. Agle β is 5 % less tha agle χ ad 50 % greater tha agle α. Which of the followig is true about agle χ? A) It is 5% greater tha α B) It is 50% greater tha α C) It is 75% greater tha α D) It is 00% greater tha α E) It is 5% greater tha α Solutio: The give iformatio leads to the followig relatioships betwee the agles: β0.75χ; β.5α, which are equivalet to: χ4/3β; α/3β. Therefore, χ is twice as great as α, i.e. it is 00% greater tha α. Aswer: D. x+ x y+ y. Give + 3 3, where x ad y are itegers. What is the value of x? A) 0 B) 3 C) - D) E) Solutio: The give equatio ca be trasformed as follows: x+ x y+ y x y x 3 y + 3 3 ( + ) 3 (3 ) 3. The last equatio is possible oly if x- 3y-0. Hece, x3, y. Aswer: B.. What is the value of cos +cos +cos 3 + +cos 358 +cos 359? A) B) π C) 0 D) 0 E) - Solutio: The sum cotais 359 addeds. Usig the properties of the cosie fuctio, cos + cos 8 cos + cos 8 cos 79 + cos 359 0, ad cos 80 -. Therefore, the value of the give sum is (-). Aswer: A 3. Two semicircles are costructed, as show i the figure. The chord CD is parallel to the diameter AB of the greater semicircle ad touches the smaller semicircle. If the legth of CD is 4, what is the area of the shaded regio? A) π B).5π C) π D) 3π E) more iformatio eeded Solutio: Let O be the cetre of the great semicircle ad OM is the perpedicular from O to CD. It is surprisig that the area of the regio betwee the two semicircles seems to be depedet o the
legth of the chord CD oly, ad idepedet o the radius of ay of the semicircles. The key to justifyig this statemet is the triagle MOD, where MO is the radius, r of the small semicircle, OD is the radius, R, of the great semicircle, ad MD is, as it is the half of CD. By the Pythagorea triagle applied for triagle MOD, R r. O the other had, the area of the shaded regio is ( π R πr ) π ( R r ) π ( ) π. 4. The sum of five cosecutive itegers is equal to the sum of the ext three cosecutive itegers. What is the greatest of these eight umbers? A) 4 B) 8 C) 9 D) E) somethig else Solutio: Deote the greatest of the itegers by x. The, the followig equatio holds: x+(x-)+(x-)(x-3)+(x-4)+(x-5)+(x-6)+(x-7). Its solutio is x. Aswer: D 5. Thomas was bor o his mother s 0 th birthday, ad so they share birthdays. How may times will Thomas s age be a divisor of his mother s age if they both live log lives? A) 4 B) 5 C) 6 D) 7 E) 8 Solutio: If Thomas s age is x, the his mother s age is (x+0). It is required that (x+0) is a multiple of x. This is possible oly if 0 is a multiple of x, which implies that x,, 4, 5, 0, or 0. Therefore, Thomas s age will be a divisor of his mother s age 6 times. 6. Cosider a sphere of radius 3 with cetre at the origi of a Cartesia co-ordiate system. How may poits o the surface of this sphere have iteger co-ordiates? A) 30 B) 4 C) D) 6 E) 3 Solutio: We have to cout the poits (x, y, z), for which the coordiates x, y, z, are iteger umbers ad satisfy the equatio x + y + z 3. There are oly two sums of perfect squares that equal to 9 (99+0+0+4+4). So, the oly groups of iteger umbers that might be coordiates of the poits are ±, ±, ± or ±3, 0, 0. There are 4 poits whose coordiates are the umbers from the first group. I fact, there are 3 possibilities to choose the positio of the (or -). For each of them, there are 8 possible arragemets of the sigs + ad -, so there are 3x84 arragemets for the coordiates. There are 6 poits i the secod group, sice there are three possibilities to choose the positio of the o-zero digit, ad for each of them, there are two choices of the sig. I coclusio, the required umber of poits is 30. Aswer: A. Part C: Each correct aswer is worth 5 poits. 7. Which of the followig umbers caot be writte as x+ x, if x is a iteger umber? A) 870 B) 0 C) 90 D) 60 E) 30 Solutio: Sice the proposed aswers are iteger umbers, it follows that x is a iteger umber, hece, x is a perfect square. O the other had, x+ x x ( x +), which is a product of two cosecutive positive iteger umbers. The umber 60 is the oly oe of the above umbers that caot be represeted i this way. Aswer: D. x 8. If f ( x) ad f ( g( x)) x, what is the equatio of the fuctio g(x)? 3 x + 4
3x + 4 3x x + 4 4x E) Other A) g( x) B) g ( x) C) g( x) D) g( x) x x + 4 4x 3x aswer Solutio: Studets familiar with the cocept of iverse fuctio will otice that g(x) is the iverse fuctio of f(x). To fid its equatio, deote f(x) by y, solve the equatio for x i terms of y ad replace x by g(x) ad y by x: x 4y 4x y 3xy + 4y x x, so g( x). 3x + 4 3y 3x Alteratively, oe ca check which of the equatios for g(x) i the aswers satisfies the equatio f ( g( x)) x. Aswer: D. 9. What is the measure of the acute agles of a rhombus, if its side is the geometrical mea of the diagoals? (Note: The umber C A B is called Geometrical mea of the umbers A ad B). A) 5 B) 30 C) 45 D) 60 E) 75 Solutio: Deote the side of the rhombus ABCD by a, ad the diagoals as follows: AC m, BD. From the give, it follows that a m a m. Applyig the Cosie Law for triagles ABC ad ABD ad the fact that agles A ad B are supplemetary (thus, their cosies are opposite umbers), it follows that m a + a a cos A a 4 a + a m Aswer: B. a (a cos B a ( cos A))(a + a + a cos A ( + cos A)) cos A 4 si 0. The graphic o the right is a piece of the graph of the 3 fuctio f ( x) ax + bx + cx + d. It passes through the poits (-, 0), (0, ) ad (, 0). What is the value of b? A 4 A 30. A) -4 B) - C) 0 D) E) 4 Solutio: Sice the poit (0, ) is o the graph, it follows that f(0)d. Next, f(-) -a+b-c+0, ad f()a+b+c+0. Addig the two last equatios, we get b+40, therefore, b-. Aswer: B. For how may real umbers a does the quadratic equatio x + ax + 007 0 have two iteger roots? A) 3 B) 4 C) 6 D) 8 E) aother aswer Solutio: Let m ad be the roots of the quadratic equatio. Hece, they satisfy the followig two equatios: m+-a, m007 (accordig to the properties of the roots of a polyomial equatio, also kow as Viette s formulas). We are iterested i iteger roots oly; therefore, we must cosider all possible represetatios of 007 as a product of two iteger umbers. I fact, there are exactly 6 such represetatios: 007 007(-) (-007)3 669(-3) (-669)9 3(-9) (-3). The respective values of a for each of them are: a008, -008, 67, -67, 3, -3.
. O a party, five frieds are goig to give each other gifts i such a way that everybody gives oe gift ad receives oe gift (of course, o oe should receive their ow gift). I how may ways is this possible? 3. A) 5 B) 0 C) 44 D) 50 E) 0 Solutio: Deote the frieds by A, B, C, D ad E. The set of possible distributios of gifts ca be split ito two o-overlappig subsets: () Amog the 5 frieds, there are two who exchaged their gifts (i.e. A gets the preset of B ad B gets the preset of A). () Amog the 5 frieds, there is o such pair that exchaged their gifts. We will cout the possibilities for each of the two subsets. () Assume that two of the frieds, say A ad B, exchaged their gifts. It follows that amog the other three frieds, there were t a pair that also exchaged their gifts, because otherwise, the fifth perso will have obody to give his/her gift to, which is ot allowed. For the three frieds, C, D, ad E, there are oly two possibilities to exchage gifts: C gives to D, D gives to E, E gives to C or C gives to E, E gives to D, D gives to C. It follows that the umber of the possibilities i this subset ca be calculated by doublig the possibilities to choose two 5 5 4 frieds out of the five frieds (which is equal to 5 C 0 ). Therefore, there are 0 possible ways to exchage gifts i a way that two frieds would exchage gifts betwee themselves. () Assume that o two frieds exchaged their ow gifts. Without loss of geerality, let A gives a gift to B. The, B caot give the gift to A, so, B gives the gift to somebody of C, D, or E. Assume, B gives the gift to C. Similarly, C must give to either D or E, assume, to D, ad, fially, D gives to to E ad E gives to A, hece, the exchage is oly possible aroud the circle, where the five frieds are arraged i ay order. Therefore, the umber of possibilities i this subset is equal to the umber of differet arragemets of five elemets aroud a circle. Note that two arragemets aroud a circle are differet if ad oly if there is at least oe elemet that has at least oe eighbour that is differet. To cout the umber of there arragemets, first, cosider all possible arragemets of five elemets i a row (there are 5!0 such arragemets). The, ote that, i terms of the defiitio of differet arragemets aroud the circle, there will be groups of arragemets that are ot distict, for istace, ABCDE is equivalet to BCDEA, CDEAB, DEABC, ad EABCD. More specifically, for each permutatio of the five elemets, there will be four more that are equivalet to it. Therefore, the umber of differet arragemets of five elemets aroud a circle is 0/54. I coclusio, the umber of possible ways the five frieds ca exchage gifts is 0+444. 4. What is the sum + + +... +? + 3 + 3 4 3 + 3 4 00 99 + 99 00 A)999/000 B) 99/00 C) 9/0 D) 9 E) Solutio: Each of the fractios i the sum is of the form, where + ( ), 3, 00. We will trasform this fractio as follows:
+ ( ) ( ) ( ( ) ( ) ). ( + ( ) ) ( ( ) ) ( ) ( )( + ) ( ) ( ) ( ). ( )( + ) Next, applyig the result for, 3, 00, the give sum becomes: S + + + +... + + 3 3 4 4 5 98 99 9 S. 00 0 99 00 4. The digits of the sequece 345345345 fill the cells i the table i a spiral-like maer begiig from the marked cell (see the figure). Which digit is writte o the cell placed 00 cells above the marked oe? A) B) C) 3 D) 4 E) 5 Solutio: Suppose the table has bee filled i the specified maer ad the last cell filled is the th cell above the marked oe. Thus, the part of the table that is filled has a shape of a square, whose side is (+) uits log, without the last cells from the upper row, just to the right of the marked cell. The, the total umber of the filled cells i the shape is ( + ), therefore, the required umber is the ( ( + ) )-th term of the sequece 345345345. (it is, i fact, the last umber writte). For 00, ( + ) 4030. To fid the last writte digit, oe must calculate the marked cell is. Aswer: A. Bous : The icreasig sequece, 3, 4, 9, 0,, 3, icludes all the powers of 3 ad all the umbers that ca be writte as the sum of differet powers of 3. What is the 00 th elemet of the sequece? A) 50 B) 98 C) 34 D) 40 E) 3 00 k k k 0 Solutio: First, let us recall that 3 > 3 + 3 +... + 3 + 3 + 3. This ca be easily justified usig the formula for the geometric series for the right-had side or from the idetity k k k 0 x ( x )( x + x +... + x + x + x ) applied for x3. Hece, each of the terms i the above sequece has a uique represetatio as a sum of several powers of 3. Next, for each umber, let us defie a biary sequece of 0s ad s that represets whether a specific power of 3 is icluded i the umber s represetatio as a sum or ot. From right to left, the digits of the biary sequece correspod to the powers 3 0, 3, 3,., etc. For istace, 09+3 +3 0, so the biary sequece defied for 0 is 0. It is clear ow that the biary sequece represetig the 00 th umber of the origial sequece is the umber 00 writte i a Base umerical system, i.e., 0000. It meas that the addeds that compose the 00 th elemet are 3, 3 5, ad 3 6, hece, the 00 th elemet is 9+43+7998. Aswer: B.
Bous : A mathematically skilled spider spis a web ad some of the strigs have legths as show i the picture. If x is a iteger, determie the value of x. A) B)3 C) 5 D) 7 E) 9 Solutio: The key to solvig this problem is the triagle iequality that states that three segmets, a, b, ad c, ca form a triagle if ad oly if each of them is greater tha the differece of the other two ad smaller tha the sum of the other two segmets. Assumig (without loss of geerality) a b c, the iequalities that represet this statemet are: b+c>a>bc; a+c>b>a-c; a+b>c>a-b. I the et, there are two triagles with a side x. Applyig the triagle iequality for each of them, it follows that x<5+94 ad x>7-5. The oly iteger umber that satisfies both coditios is 3, therefore, x3. Aswer: B. Bous 3: A, Belida ad Charles are throwig a die. A wis if she throws a,, or 3; Belida wis if she throws a 4 or 5; Charles wis if he throws a 6. The die rotates from A to Belida to Charles to A, etc., util oe player wis. Calculate the probability that Charles wis. A) /6 B)/8 C) / D) /3 E) It is impossible for Charles to wi Solutio: Deote by A, B, C the evets of wiig the game by A, Belida or Charles, respectively, ad by A, B, C - the evets of ot wiig the game by each of them. The probability of wiig the game by Charles o his first tur is equal to 4 P ( A) P( B) P( C)... If Charles wis the game o his secod tur, it meas that he did ot 6 6 8 wi o his first tur ad A ad Belida have ot wo the game after their secod tur. The the probability 4 5 5 of Charles wiig the game at the secod roud of the game is..... If Charles wis 6 6 6 8 8 the game after turs, it meas that all of them have ot wo after (-) rouds, ad o the last roud, A ad Belida did ot wi while Charles wo. For each of the first (-) rouds, the probability is 4 5 5 4.., ad for the last roud, the probability is... So, the probability that 6 6 8 6 6 8 5 Charles wis the game at the -th roud is.. Sice the umber of rouds i the game is ot 8 8 limited, the overall probability of Charles wiig the game is a ifiite sum: 3 5 5 5 5 P +. +. +. +... +. +.... Usig the formula for the ifiite 8 8 8 8 8 8 8 8 8 geometric series with a ratio less tha, we calculate: P. 8 5 3 8