2019 Chapter Competition Solutions

Transcription

1 019 Capter Competition Solutions Are you wondering ow we could ave possibly tougt tat a Matlete would be able to answer a particular Sprint Round problem witout a calculator? Are you wondering ow we could ave possibly tougt tat a Matlete would be able to answer a particular Target Round problem in less 3 nutes? Are you wondering ow we could ave possibly tougt tat a particular Team Round problem would be solved by a team of only four Matletes? Te following pages provide solutions to te Sprint, Target and Team Rounds of te 019 MATHCOUNTS Capter Competition. Tese solutions provide creative and concise ways of solving te problems from te competition. Tere are certainly numerous oter solutions tat also lead to te correct answer, some even more creative and more concise! We encourage you to find a variety of approaces to solving tese fun and callenging MATHCOUNTS problems. Special tanks to solutions autor Howard Ludwig for graciously and voluntarily saring is solutions wit te MATHCOUNTS community.

2 019 Capter Sprint Round Solutions 1. For a nonnegative real number nn, te square of te square root of nn yields nn. Since te square root of nn is, nn must be te square of, wic is = yy. We start at te origin, were te two double-arrowed axis lines meet, and, 7 since we want xx =, we proceed orizontally to te rigt (toward te xx) to were we see te. We ten proceed vertically parallel to te axis marked yy until we meet te blue function line, wic we see to be 3 units above te xx-axis. So te answer is Line up te values vertically so tat te units digit of all te values are in a column as in te figure wit te problem and add te digits column by column: = 6 36 =. 5. Wit 63 people owning a dog, te remaining = 37 people must own just a cat, since all 100 people own a dog, a cat or bot. However, 58 people own a cat, and only 37 cats ave been accounted for. Tat means = 11 cats are owned by te people owning a dog also apples bins bundles apples =. bin bundle crate crate 7. If all 9 coins were pennies, ten tey would ave a value of 9. Because te total value is to be 9, we are 0 sort. Eac time we replace a penny by a nickel, te total value increases by. To make up tat 0 by substituting nickels for pennies will require 0 = 5 suc substitutions, tus ending up wit 5 nickels. 8. Tere are 60 seconds in a nute, so 3 eartbeats 15 sec 60 sec 3 60 eartbeats = = 3 eartbeats 1 n 15 n n = eartbeats. n 9. Te first five prime numbers in increasing order are, 3, 5, 7, 11. Te median value is te tird prime, % = 9/10, so 108 = xx and xx = 108 = 10 = 10 1 = A regular exagon is composed of 6 congruent equilateral triangles as sown in te figure. Te saded region covers 5 of te 6 congruent triangles, so te saded area is 55/66 of te exagon s area xx 1. = nn 8 = nn so nn = = 88.

3 13. Here we ave tree smaller triangles joined in a row suc tat eac pair of adjacent triangles forms a triangle, wit tere being two suc pairs (left wit ddle and ddle wit rigt) and all tree taken togeter forng a single triangle. Tat means = 6 total triangles. Had nn distinct line segments been drawn from te vertex to te base of te original triangle, partitioning te original triangle into nn + 1 small triangles, te total number of triangles would ave been te triangular number TT nn+1 = (nn + 1)(nn + )/. 1. In te Fastball row under Min Speed we see te value 80 /. In te Knuckleball row under Max Speed we see te value 70 /. Te absolute difference is / = 10 / = 1111 /. 15. Te first six positive prime numbers are, 3, 5, 7, 11 and 13. We do not need to deterne te actual product, only ow many zeros are at te rigt end of te integer in oter words ow many factors of 10 are generated by te product. Te only ways to get factors of 10 are by multiplying by a multiple of 10 and by multiplying be a multiple of and by a multiple of 5. In te first six primes tere are no multiples of 10, but tere is one multiple of and one multiple of 5, aving one factor of and one factor of 5, respectively, wic combine to yield one factor of 10, so te number of 0s at te rigt end of te product is lap = 1 lap n n n lap = r = Te sum of te two solutions of te quadratic equation AAxx + BBBB + CC = 0 is given be BB/AA [and, as a side note, te product of te solutions is CC/AA]. Because te two solutions are given as and 7, te sum of te two solutions is 9. Now, AA = 1 and BB = aa, so 9 = BB = aa = aa, so tat aa = 99. AA Te way of selecting te most total days witout selecting tree consecutive days by starting at eiter end and alternately selecting two and skipping one. Tus, starting from te left, days 1,,, 5, and 7 are selected witout tree in a row anywere. Tere are only two days left to coose, #3 and #6. Coosing eiter one makes at least tree consecutive days. Tus, making 5 selections is not enoug to guarantee tree consecutive days but 6 is enoug. 19. (3 ) (0 16) = = 5 1 = = Te evaluations of te radicals can be done in any of several ways: Just crunc numbers: 3 + = = 5 = 5; 0 16 = = 1 = 1; = = 169 = 13. Recognize eac square root except te last is re-squared, so = = 169 = 13. Note Pytagorean triples: 3--5 scaled by 1 and to 3--5 and , respectively; and lastly Te operation combines two legs to yield te ypotenuse. Te operation combines te ypotenuse and one leg to yield te oter leg. 0. Jones catces up 800 m in n and, tus at a rate of: 800 m n = 00 m n 1 km 1000 m 60 n 1 = 1 km. Jones is going 50 km/, so te car is going 50 1 km/ = 38 km/.

4 1. Eac of 0 people on te winning team sook ands wit eac of 0 people on te losing team for a total of nn = 0 0 = 00 andsakes. Eac of te 0 people on te winning team fist-bumped eac of te oter 19 teammates, but we are double counting because player A fist-bumping player B is te same action as player B fist-bumping A. So, tere were mm = 0 19/ = = 190 fist bumps. nn + mm = = Te sections of spinner 1 are te prime numbers less tan 10:, 3, 5, 7. Te sections of spinner are te positive perfect squares less tan 0: 1,, 9, 16, 5, 36. Te on spinner 1 is not relatively prime to tese values on spinner :, 16, and 36. Te 3 on spinner 1 is not relatively prime to tese values on spinner : 9 and 36. Te 5 on spinner 1 is not relatively prime to tese values on spinner : 5. Te 7 on spinner 1 is not relatively prime to tese values on spinner :. Tere are, tus, 6 pairs tat qualify; tere are 6 = total pairs. Te desired probability is, terefore, 6/ = 11/. 3. Te sum of te tree values is (0 + AA) + (30 + AA) + (0 + AA) = AA, but we are told tat teir sum is AA. For AA = AA, we must ave AA = 10, so AA = 55.. Tere are = 6 equally likely outcomes of rolling te die. Te only ways to get a sum of 7 are: rolling, and 1 in any order, wic can occur in 3! = 6 1! 1! 1! rolling 3,3 and 1 in any order, wic can occur in 3!! 1! = 6 1 rolling 3, and in any order, wic can occur in 3! 1!! = 6 1 Tus, te probability tat te sum of te numbers rolled is 7 is = 6 distinct, equally likely ways; = 3 distinct, equally likely ways; = 3 distinct, equally likely ways. = 1 6 = We are to deterne ss, given te information in te figure sown and ss + ll = cm. Te reason for bisecting te angle formed by te 33 cm and 55 cm sides is tat we are 33 cm instructed to bisect te maximal acute angle. Now, a 33 cm- cm-55 cm triangle is a 3--5 rigt triangle wit a scaling of 11 cm, so te angle opposite te 55 cm side is a rigt angle, not acute. Te larger acute angle is ten te angle opposite te next longest ss ll side, wic is te cm side. We can use te triangle angle-bisector teorem, wic says tat ll 55 cm = ss 33 cm ll+ss 55 cm+33 cm = = 88 cm. Because ll + ss = cm, we ave = 8 3 cm, so ss = = 3 11 ss 33 cm 33 3 ss 3 8 cm = 3333 cm. 6. Te only Q is in te center square, so tat is were we must start. From tere, moving left, rigt, up, or down, gets us to a U, so tat is ways to make QU. Due to symmetry all four ways beave alike, so we can analyze just one coice for te U and multiply by. Regardless of wic U we move to, tree out of te four possible moves left, rigt, up, or down get us to an E to make QUE. We must be careful ere, toug, because not all 3 Es yield te same number of options for te next move: (i) We see tat of te Es are interior on a diagonal, eac of wic leads to any of Us. From any of tese Us, we can move to any of 3 Es to make QUEUE. (ii) But 1 of te Es is along an outer edge in te ddle and leads to any of 3 Us. From any of tese Us, we can move to any of 3 Es to make QUEUE. Tus, we ave a total of [( 3) + (1 3 3)] = ( + 9) = (33) = ways. 55 cm

5 7. Here we ave a problem relating te average speed over a wole trip to te average speed over eac of several legs of equal lengt. Te overall average is te armonic mean of te speeds for eac leg. For two 1 legs wit average speeds of vv 1 and vv, respectively, te armonic mean is. Tus, we ave vv 1 +vv = vv 1vv 1 1 vv1 + 1 vv te overall average speed being vv a = vv 1vv. We are given te actual value for te difference between vv vv 1 +vv 1 and vv, but for now let s simply say tat vv = vv 1 + dd. For solving equations wit lots of peculiar numbers and measurement units, it is often easier to solve in terms of general variables and substitute actual quantities once we express te variable to be solved in terms of te oter variables. We ave vv a = vv 1vv, and vv 1 +vv vv a = vv 1 (vv 1+dd), but we need to rearrange tis to express vv vv 1 +dd 1 in terms of vv a and dd. Multiplying bot sides by vv 1 + dd yields vv a (vv 1 + dd) = vv 1 (vv 1 + dd), wic expands to vv a vv 1 + ddvv a = vv 1 + ddvv 1. Rearranging tis into conventional form for a quadratic equation in terms of vv 1 yields vv 1 + (dd vv a )vv 1 ddvv a = 0. Using te quadratic formula, we get vv 1 = (dd vv a )± [(dd vv a )] ()( ddvv a ) = (vv a dd)± (vv a dd) +ddvv a. We know tat te overall average speed is given by te total distance 180 = 360 divided by te total time 7.5. So, we ave vv a = = and vv a dd = 8 8 ± 8 0 = = = 8. We are given tat vv 10 1 = vv 0, so dd = = 8. Substituting into our quadratic formula yields: vv 1 = = 1 ± = 1 ± 676 = (1 ± 6). Tis yields a coice of 0 or 1, but a negative result does not make sense for tis problem, so te answer must be vv 1 = Start wit 0 and work up, incorporating new factors tat ave not yet been covered. Now, we are dealing wit numbers in te 0s (or less due to common factors) let s just say an average of 5 = 10 /. Wit 7 suc factors, we would be at most about (10 / ) 7 = 10 1 / , so let s work 0 troug 7 and see were we stand. [NOTE tat billion refers to 10 9.] 0 5; 1 = 3 7, bot new, so ; = 11, wit new but old, so ; 3 is a new prime, so ; = 3 3 wit one new, two old s and old 3, so ; 5 = 5 wit a new and an old 5, so ; 6 = 13 wit a new 13 and an old, so ; 7 = 3 3 wit two new and one old 3, so ; Pause: Let s assess were we are wit Te product of te first factors is straigtforward to do exactly wit only mental power; for te rest an approximation is likely good enoug. 1 3 = ( 1 ) = 3 = = = = Te 100 part tacks on two more 0 s for , just over 10 6 ; 117 is only a little over 10, for a total of a little over We need anoter factor of 10 or so to get tere. 8 = 7, all of wic is old, so noting new to contribute; 9 is a prime, so new, and provides us wit our needed factor of at least 10, making te answer 9.

6 9. My favorite approac to tis type of problem is somewat unconventional but only in te sense of combining properties of silarity and proportionality in suc a way as to combine steps unconventionally, but quite validly, to reduce te count and complexity of steps, but all equivalent to a more traditional approac. Te area of a triangle is 1/ times te base times te eigt. For te ratio of areas of two triangles te 1/ cancels out, leaving te product of te ratio of te bases time te ratio of te eigts. Let s regard te base of ABC to be segment BC and te base of PQR to be segment PQ. It is very convenient tey are parallel to one anoter. Wit AP = AQ = 1, PQ must be te same 1, so te know te ratio of te bases is 1. AB AC 5 BC 5 5 Now we need to deal wit te ratio of te eigts of te two triangles, and tis will be a -step process: first, te ratio of te eigt of PQ above BC to te eigt of A above BC, and, second, te ratio of te eigt of PQ above R to te eigt of PQ above BC. Te product of tese two will be te ratio of te eigt of PQ above R to te eigt of A above BC, wic product is te ratio of te eigts of te two triangles of concern. We use silarity to realize tat te eigt of A above PQ is 1 te eigt of A above BC, so te ratio of te eigt of 5 PQ above BC to te eigt of A above BC is 1 nus te ratio 1, wic is as te answer to te first part. For 5 5 te second part, we need to deterne te orizontal distance from line BQ to line CP going orizontally at some eigt, and left-to-rigt will be regarded as a positive value and rigt-to-left as a negative value (a form of wat is often referred to a displacement [pysics] or directed distance [matematics]). At te eigt of base PQ, we are going from Q to P, wic is rigt-to-left, so negative. Since we care only about ratios and scaling, we can regard te distance as normalized to 1. At te base BC, te point on BQ is B and te point on CP is C, wit C to te rigt of B, so a positive distance tat is 5 times te magnitude of te distance 1 from Q to P, wit our given scaling. Te magnitude of 1 is 1, wic multiplied by 5 is 5. Te two lines intersect at R so te distance from R to R is 0. Now, 0 is 1 of te way from 1 to 5, so te eigt of R below 6 PQ is 1 te eigt of BC below PQ, wic is te eigt of BC below A, making te eigt of PQR equal to = te eigt of ABC. Wit a ratio of bases being 1 and te ratio of eigts being, te ratio of te areas is 1 = [Te explanation is long because tere are some atypical details to discuss, but te calculations are sort wen you are used to te concept. It is very quick to see tat te base of RPQ is 1/5 as wide as for ABC, and ten te eigt of RPQ is 1 = tat of ABC, so te ratio of te areas is 1 =.] We can reorder te terms so tat aa bb cc. It must be tat aa must be or 3; oterwise, aa being 1 makes te sum of te tree reciprocals greater tan 1, and te maximum sum for aa > 3 is 3 < 6. For te sum to ave a 7 denonator of 7, at least one of aa, bb, and cc must be a multiple of 7, and it is not aa since aa is or 3; neiter can it be bb since ten cc would also ave to be at least 7, and te sum of te reciprocals cannot reac te requisite 6. Tus, wit aa = 3 and cc 7, ten bb = 8 1, so bb < 3, making bb < aa (a contradiction of specified conditions) if aa = 3. Tus, aa must be, making = 6 1 = 5. Terefore, bb cc < bb (te first condition based on cc being an arbitrarily large integer, making 1/cc arbitrarily close to 5 3 0, and te second condition based on cc 7 since cc is a multiple of 7). Tus, bb must be 3 or. For bb = 3, cc would ave to be cc = 1, wic is te reciprocal of an integer and, terefore, suitable. For = 1 completeness (but not necessary in te competition), for bb =, cc would ave to be cc = 1 not te reciprocal of an integer, so is not an acceptable answer. Terefore, we end up wit: aa =, bb = 3, cc = ; aa + bb + cc = = = 3 8, wic is

7 019 Capter Target Round Solutions 1. Te price increase of eac orange is $0.69 $0.9 = $0.0. Terefore, te price increase of 6 oranges is 6 $0.0 = $11.. [WARNING: Tat second decimal place is required for monetary amounts.]. Te yy-intercept is deterned by solving for yy wen xx = 0. For Cris tat is at yy = 7, wile for Sebastian tat is yy = bb. For Sebastian s to be double Cris, bb = 7 = 1. Te xx-intercept is deterned by solving for xx wen y = 0. For Cris tat is at xx = 7/3 wile for Sebastian tat is xx = bb/aa. For Sebastian s to be double Cris, bb = 1 = 7 1 =, so 1 = 1/3 and aa = 3. aa aa 3 3 aa Terefore, aa + bb = = In 1 n, te Cubes ave gained 1 18 = 3 points over te Bisectors, so te rate of gain is 3 points = 1 point, 1 n n or 1 point every nutes. Te game lasts 0 n = 0 n. Terefore, te total gain of points by te Cubes over te Bisectors is 1 pt 0 n = 1111 points. n. Let point X be te point of intersection of te bisector of D and line AB, forng ADX, and point Z be te point of intersection of te two dased lines, forng BXZ. We are given mm XAD = 8 and mm ADX = 3 / = 16, so mm DXA = 180 ( ) = 80. DXB is te supplement of DXA, and, tus, as a measure of 100 ; tis is te same as BXZ. We are given mm ZDB = 3 / = 16. Terefore, for XZB, wic is te angle in question, mm XZB = 180 ( ) = As te figure sows, we can draw line segments AW and AY perpendicular to lines QR and RS, respectively. Because A is te center of square PQRS, W and Y are te dpoints of teir respective sides of te square. Triangles AWX and AYZ are congruent. Te saded area is quadrilateral AXRZ. Tat area is equal to te sum of te areas of quadrilateral AXRY and triangle AYZ. Since triangles AWX and AYZ are congruent, teir areas are equal. Tus, te given saded area is equal to te sum of te areas of quadrilateral AXRY and triangle AWX, wic is te same as te area of square AWRY a square tat is silar to square PQRS, wit linear scale factor of 1. Te ratio of te area AWRY to te area of PQRS is te square of tis linear scale factor, tus 1 = 11. Note tat tis answer does not depend in any way on te degree of rotation of te larger square about te center of te smaller square. 6. Let te desired tens digit be dd, te wrongly input tens digit be ww, and te units digit be uu. Ten te desired number to square is 10dd + uu and te wrongly squared number is 10ww + uu. Te difference in teir squares is 30 = (10ww + uu) (10dd + uu) = (100ww + 0wwww + uu ) (100dd + 0dddd + uu ) = 100(ww dd ) + 0(ww dd)uu. Dividing bot sides by 0 yields: 117 = 5(ww dd ) + (ww dd)uu = [5(ww + dd) + uu](ww dd). We are dealing wit ten s digits dd and ww of -digit numbers, wit ww > dd, so we must ave 1 dd < ww 9. Tus, 1 ww dd 8 and ww dd must divide 117 = 3 13, meaning tat ww dd is 1 or 3. If ww dd = 1, ten 5(ww + dd) + uu = 117, but ww 9, dd 8, and uu 9, so 5(ww + dd) + uu 9 and cannot be 117. Terefore, ww dd = 3. Now we ave 117 = [5(ww + dd) + uu](3), so 5(ww + dd) + uu = 39, wic means: ww + dd = 6 and uu = 9, or ww + dd = 7 and uu =. Since ww dd is odd (3), ww + dd must likewise be odd. Tus, ww + dd = 7 and uu = olds. Now, we are to find (10ww + uu) + (10dd + uu) = 10(ww + dd) + uu = = 7777.

8 7. In order to traverse from point A to point B, one must pass troug exactly one of points C, D, E or F sown. Tere is some number of pats from A to eac of te points C, D, E and F, and for eac suc pat tere is some number of pats from tere to B. So, we can deterne te product of tose two values for eac point C, D, E and F and add te four products togeter. Te number of pats between any two lattice points tat is rr steps rigt and dd steps down is te number of combinations of rr + dd tings taken dd at a time. (i) For A C: rr = 0, dd = 5 and 5C 5 = 1 and for C B: rr = 5, dd = 0 and 5C 0 = 1. So, it follows tat tere is 1 1 = 1 pat from A C B. (ii) For A D: rr = 1, dd = and 5C = 5 and for D B: rr =, dd = 1 and 5C 1 = 5. So, it follows tat tere are 5 5 = 5 pats from A D B. (iii) For A E: rr =, dd = and 6C = 15 and for E B: rr = 1, dd = 3 and C 3 =. So, it follows tat tere are 15 = 60 pats from A E B. (iv) For A F: rr = 5, dd = 1 and 6C 1 = 6 and for F B: rr = 0, dd = and C = 1. So, it follows tat tere are 6 1 = 6 pats from A F B. Total number of pats is = 9999 pats. A C D E F B 8. For te product of integers to be 1, eac integer must be +1 or 1, and te count of 1 factors must be even. Tis means tat for a factor tat is a difference of two letter values, it is necessary (but not sufficient) tat te letters be consecutive in terms of assignment to te sequential numbers 1,, 3,, 5, 6. Terefore, MM and AA must be consecutive, so eiter MM, AA (yields M A = 1) or AA, MM (yields M A = +1); TT must be consecutive wit bot HH and EE, so TT must be between HH and EE, as EE, TT, HH or HH, TT, EE; EE must be consecutive wit bot LL and TT, so EE must be between LL and TT, as LL, EE, TT or TT, EE, LL. Putting te last two lines togeter yields LL, EE, TT, HH (yields (T H)(L E)(T E) = ( 1)( 1)(+1) = +1) or HH, TT, EE, LL (yields (T H)(L E)(T E) = (+1)(+1)( 1) = 1). Tus, we need MM, AA to go wit HH, TT, EE, LL, eiter order, and AA, MM to go wit LL, EE, TT, HH, eiter order. Tus, we ave coices tat satisfy te product criterion: MM = 1, AA =, HH = 3, TT =, EE = 5, LL = 6; HH = 1, TT =, EE = 3, LL =, MM = 5, AA = 6; AA = 1, MM =, LL = 3, EE =, TT = 5, HH = 6; LL = 1, EE =, TT = 3, HH =, AA = 5, MM = 6. Tere are 6! = 70 possible arrangements of te letters wit te numeric values. Terefore, te fraction wit te desired result is =

9 019 Capter Team Round Solutions 1. Let s try a greedy algoritm, adding one triangle at a time, maxizing te contribution to te perimeter eac time. If we attac a new triangle onto a side of 5 inces, tat 5-inc side as disappeared, but we now ave 1 inces sowing, for a net increase of 1 5 = 9 inces to te perimeter. If we attac a new triangle onto a side of 7 inces, tat 7-inc side as disappeared, but we now ave 1 inces sowing, for a net increase of 1 7 = 5 inces to te perimeter. Tus, we sould want to attac along a side of 5 inces wenever possible and attac along a side of 7 inces only if suc is te only coice. First, we start wit one triangle and ave perimeter 7 inces + 5 inces = 19 inces, and we ave two 7-inc and one 5-inc sides to attac to. Let s attac te second triangle along te one available 5-inc side, increasing te perimeter to 19 inces + 9 inces = 8 inces, and we ave four 7-inc sides to attac to. Since we ave only 7-inc sides to attac te tird triangle to, doing so increases te perimeter to 8 inces + 5 inces = 33 inces = 5 5 = (5 ) 5 = 5 5, so nn = Let s use LL for te amount Lior as at any given time, CC for te amount Celine as at any given time, and EE for te amount Elliot as at any given time. We start off wit LL =, CC = 0, EE = 0. 3 CC takes 1/ of LL: CC = 1 = 1, LL = 1 = 1, EE = EE takes 1/ of LL: EE = 1 1 = 1, LL = 1 1 = 1, CC = 1. 6 CC and EE take [add to teir own] 1/3 of LL, tus = 1 1, wic is also te amount LL keeps: CC = = 11, wit EE ending up wit 1 and LL wit 1, toug tere is no need to deterne te last two Being te region between two concentric circles, an annulus as an area equal to te difference of te area witin te larger circle nus te area witin te smaller circle, tus of te form π(rr rr ). Te radius RR of te larger circle is easy to deterne te distance between te center point of te rotation and te fartest vertex. (Remember, we really need te square of te distance, so tere is no need to do te square root portion of te Pytagorean formula only to square it again later.) Between (0; 3) and ( ; 0): ( 0) + [0 ( 3)] = + 9 = 13. Between (0; 3) and (10; ): (10 0) + [ ( 3)] = = 15. Between (0; 3) and ( 6; 6): ( 6 0) + [6 ( 3)] = = 117. Terefore, RR = 15. Te radius of te inner circle is not so easy it is not necessarily te distance between te center point of rotation and te nearest vertex, as te sortest distance could be on a side between two vertices. Drawing a quick sketc is likely to convince under time pressure of a competition tat te nearest point on te triangle is a vertex, and tat is good enoug. If you are really nervous and paranoid, toug, te only time issues can arise (and it gt not be serious even ten) is wen te negative reciprocal of te slope of a side is between te value of te slope between te rotation center point and eac of te two end vertices of tat side. Tis does occur for te side between (10; ) and (6; 6), inclusive, for wic te nimum distance squared between te rotation center point and tat side is 11.5, wic is greater tan te distance squared between te rotation center point and vertex (; 0), so we will take 13 as te nimum distance squared, rr. Terefore, te answer is π(rr rr ) = π(15 13) = π units. 5. To average te same overall speed ss, Crystal must spend = 0.3 = 3 of er time at speed 1.1ss and = 0.1 = 1 11 at speed 0.7ss, wic means 3 = 3333 of te distance at speed 1.1ss

10 6. If we number te rows of te table as rr, wit rr = 1 for te top content (non-eader) row to rr = 6 for te bottom row, and te columns as cc, wit cc = 1 for te left content (non-eader) column and cc = 6 for te rigt column, ten te value in te cell at row rr and column cc is 6(rr 1) + cc, wit te first part depending on only te row number and te last part depending on only te column number. Te eader column at te far left tells ow many cells in tat row are used and, terefore, ow many times 6(rr 1) sows up in te sum due to row rr; silarly, te eader row at te top tells ow many cells in tat column are used and, terefore, ow many times cc sows up in te sum due to column cc. By splitting tis way, eac part of te value expression is andled exactly once and is in a meaningful way no double counting, etc. Terefore, te sum of te coin-occupied values is: = = For a product of factors in te range of 1 to 10 to ave a units digit of 0 (be divisible by 10), eiter 10 is one of te factors, or else 5 is one of te factors along wit at least one even number. We must be carefully in assessing te probability because bot conditions may occur togeter. Let s look at two main cases: 1. Of te values, 1 of tem is a 10; we do not care wat te oter 3 values are.. Of te values, 1 of tem is a 5, at least 1 (meaning 1,, or 3) is an even number oter tan 10 (,, 6, 8), and te remaining ones (, 1, or 0) is an odd number oter tan 5. Te two cases were deliberately cosen to be mutually exclusive, so te probabilities of te two can be merely added. We can use te generalized or multivariate ypergeometric probability distribution. For case 1, we ave = = (We will wait until te end to reduce te fractions.) For case, we can decompose into 3 mutually exclusive subcases: 1,, or 3 values cosen from,, 6, and 8, yielding = = 1 10 ( ) = Terefore, te total probability is = = Te rectangle eigt, 1, is te diameter of eac circle, so te radius of te circles is 6. A Let s put te origin of a coordinate system at te center O of te rectangle wit te xx-axis P O Q orizontal and yy-axis vertical. Ten te centers of circles Q and P are at (±3; 0) and te B two circles intersect at A and B, 0; ±3 3. Te area of te saded region is te area of circle P nus te area of overlap between te two circles. Te overlap region can be partitioned into a segment of circle P bounded by AB and arc BQA, and a segment of circle Q bounded by AB and arc BPA; tese two segments are congruent, so te area of te overlap region is double te area of eiter segment. We can find te area of segment AOBQA by subtracting te area of te sector of circle P formed by arc BQA nus te area of PAB. AOP and BOP are. Te ratio of eac of legs AO and BO to leg OP is 3, meaning tat mm APO = mm BPO = 60, so te measure of APB and arc BQA are 60 = 10. Tus, te sector of circle P formed by arc BQA is 1/3 of circle P and as area 1π; APB as base AB of lengt 6 3 and eigt PO of 3, so area of PAB is = 9 3. Te overlap region as area 1π 9 3 = π 18 3, so te saded area is 36π π 18 3 = 1π Te area of te rectangle is 18 1 = 6, so te saded area occupies 1π undredt as 0.3. = π = , wic rounds to te nearest

11 9. Eiter Felix or Oscar goes in te red cage; te oter goes in te blue cage tis yields options. For eac of tose two options, tere are remaining amsters. Any of te may be second to occupy te red cage, and after tat coice is made any of te 3 remaining ones may be tird to occupy te red cage. Now, te last go in te blue cage. Tus, it appears tat tere are 3 = ways of selecting wic amsters will occupy te red cage. However, we could ave swapped wic amster was cosen second versus wic amster was cosen tird, and we would end up wit te same occupancy in eac cage. Terefore, we need to divide by to avoid double counting, so we end up wit / = 1111 ways. We could also ave done it as (for Felix and Oscar) times te number of ways (combinations) to distribute tings (te remaining amsters) taken at a time ( to go in te red box) to get! = = 1111 ways.!! 10. Te result does not depend on te sape nor on te size of te triangle. To make te A intuition a little easier, let s use a rigt triangle: Te area of APQ is 1 AP RQ = 1 AP QR AP AB BC = QR AB BC AB BC 1 P AB BC. R Te rigt end of te equation contains AP QR B times te area of ABC, and expression tat AB BC we are given to be 1 AP te area of ABC. Terefore, QR = 1 PB AP. We are given tat = aa, so = 1 AB BC AP AB PB AB = aa aa+1 aa+1 and, wit only te former being important to us. Let qq = QC QA, wic we wis to be at most 1. By silarity, qq = QC = BC QR = BC QQQQ 1, so = 1. Terefore, 1 1 = 1 and aa + 1 =, so QA QR QR BBBB qq+1 aa+1 qq+1 qq+1 aa = 1 = 1 qq qq+1 1+qq 1 aa ; symmetrically, qq =. Because aa is restricted to 0 < aa < 1, tat means tat qq is 1+aa restricted to 0 < qq < 1. However, we want to know te fraction of values for wic qq 1. Now, we must be extremely careful ere. It gt now appear tat te probability is 1 because te interval 0 < qq 1 is 1 as wide as te interval 0 < qq < 1. However, we do not know tat te various values of qq are equally likely over tese intervals. We are given only tat aa is randomly cosen meaning all values in te specified interval are equally likely; qq is not called out as random. Te region of interest for qq is 0 < qq 1, wic corresponds to 3 aa < 1, wic is = of te allowed interval for te randomly cosen aa, so te probability is. [Because tis value is not 1, it demonstrates tat te various possible values for qq are not 55 equally likely unlike wit aa. Tis is to be expected since te relationsip between aa and qq is non-linear.] Q C