2019 Chapter Competition Solutions - Mathcounts
2019 Chapter Competition SolutionsAre you wondering how we could have possibly thought that a Mathlete would be ableto answer a particular Sprint Round problem without a calculator?Are you wondering how we could have possibly thought that a Mathlete would be ableto answer a particular Target Round problem in less 3 minutes?Are you wondering how we could have possibly thought that a particular Team Roundproblem would be solved by a team of only four Mathletes?The following pages provide solutions to the Sprint, Target and Team Rounds of the2019 MATHCOUNTS Chapter Competition. These solutions provide creative andconcise ways of solving the problems from the competition.There are certainly numerous other solutions that also lead to the correct answer,some even more creative and more concise!We encourage you to find a variety of approaches to solving these fun and challengingMATHCOUNTS problems.Special thanks to solutions authorHoward Ludwigfor graciously and voluntarily sharing his solutionswith the MATHCOUNTS community.
2019 Chapter Sprint Round Solutions1. For a nonnegative real number ππ, the square of the square root of ππ yields ππ. Since the square root of ππ is 4, ππmust be the square of 4, which is 4 4 ππππ.π¦π¦2. We start at the origin, where the two double-arrowed axis lines meet, and,7since we want π₯π₯ 2, we proceed horizontally to the right (toward the π₯π₯) to65where we see the 2. We then proceed vertically parallel to the axis marked π¦π¦4until we meet the blue function line, which we see to be 3 units above the3π₯π₯-axis. So the answer is 3.21 1 11 2 3 4 5 6 7π₯π₯3. Line up the values vertically so that the units digit of all the values are in a column as in the figure with theproblem and add the digits column by column:1101011 01010 101101 010112 233.4. 82 62 64 36 ππππ.5. With 63 people owning a dog, the remaining 100 63 37 people must own just a cat, since all 100 peopleown a dog, a cat or both. However, 58 people own a cat, and only 37 cats have been accounted for. Thatmeans 58 37 ππππ cats are owned by the people owning a dog also.6. 6applesbin 4bins bundle2bundlescrate ππππapples.crate7. If all 9 coins were pennies, then they would have a value of 9 . Because the total value is to be 29 , we are20 short. Each time we replace a penny by a nickel, the total value increases by 4 . To make up that 20 by20 substituting nickels for pennies will require 5 such substitutions, thus ending up with 5 nickels.8. There are 60 seconds in a minute, so4 32 heartbeats60 sec 1 min15 sec 32 60 heartbeats15 min 32 4heartbeatsmin ππππππheartbeats.min9. The first five prime numbers in increasing order are 2, 3, 5, 7, 11. The median value is the third prime, 5.10. 90% 9/10, so 108 9π₯π₯10and π₯π₯ 109 108 10 1089 10 12 ππππππ.11. A regular hexagon is composed of 6 congruent equilateral triangles as shown in the figure.The shaded region covers 5 of the 6 congruent triangles, so the shaded area is ππ/ππ of thehexagonβs area.12. 2 2ππ8 ππ4so ππ 4 2 ππ.
13. Here we have three smaller triangles joined in a row such that each pair of adjacent triangles forms atriangle, with there being two such pairs (left with middle and middle with right) and all three takentogether forming a single triangle. That means 3 2 1 6 total triangles. Had ππ distinct line segments beendrawn from the vertex to the base of the original triangle, partitioning the original triangle into ππ 1 smalltriangles, the total number of triangles would have been the triangular numberππππ 1 (ππ 1)(ππ 2)/2.14. In the Fastball row under Min Speed we see the value 80 mi/h. In the Knuckleball row under Max Speed wesee the value 70 mi/h. The absolute difference is 80 70 mi/h 10 mi/h ππππ mi/h.15. The first six positive prime numbers are 2, 3, 5, 7, 11 and 13. We do not need to determine the actualproduct, only how many zeros are at the right end of the integerβin other words how many factors of 10are generated by the product. The only ways to get factors of 10 are by multiplying by a multiple of 10 andby multiplying be a multiple of 2 and by a multiple of 5. In the first six primes there are no multiples of 10,but there is one multiple of 2 and one multiple of 5, having one factor of 2 and one factor of 5, respectively,which combine to yield one factor of 10, so the number of 0s at the right end of the product is 1.16.10 lap10 min 1lap min2.5mi lap60minh 1 2.5 60mihr ππππππmi.h17. The sum of the two solutions of the quadratic equation π΄π΄π₯π₯ 2 π΅π΅π΅π΅ πΆπΆ 0 is given be π΅π΅/π΄π΄ [and, as a sidenote, the product of the solutions is πΆπΆ/π΄π΄]. Because the two solutions are given as 2 and 7, the sum of the twoπ΅π΅ππsolutions is 9. Now, π΄π΄ 1 and π΅π΅ ππ, so 9 ππ, so that ππ ππ.π΄π΄118. The way of selecting the most total days without selecting three consecutive days by starting at either endand alternately selecting two and skipping one. Thus, starting from the left, days 1, 2, 4, 5, and 7 are selectedwithout three in a row anywhere. There are only two days left to choose, #3 and #6. Choosing either onemakes at least three consecutive days. Thus, making 5 selections is not enough to guarantee threeconsecutive days but 6 is enough.19. (3 4) (20 16) 32 42 202 162 5 12 52 122 ππππ.The evaluations of the radicals can be done in any of several ways:Just crunch numbers: 32 42 9 16 25 5; 202 162 400 256 144 12; 52 122 25 144 169 13.Recognize each square root except the last is re-squared, so 32 42 202 162 9 16 400 256 169 13.Note Pythagorean triples: 3-4-5 scaled by 1 and 4 to 3-4-5 and 12-16-20, respectively; and lastly 5-12-13.The operation combines two legs to yield the hypotenuse.The operation combines the hypotenuse and one leg to yield the other leg.20. Jones catches up 800 m in 4 min and, thus at a rate of:800 m4 min 200going 50 km/h, so the car is going 50 β 12 km/h 38 km/h.m1 km60 min 1hmin1000 m 12km.hJones is
21. Each of 20 people on the winning team shook hands with each of 20 people on the losing team for a total ofππ 20 20 400 handshakes. Each of the 20 people on the winning team fist-bumped each of the other19 teammates, but we are double counting because player A fist-bumping player B is the same action asplayer B fist-bumping A. So, there were ππ 20 19/2 10 19 190 fist bumps.ππ ππ 400 190 ππππππ.22. The sections of spinner 1 are the prime numbers less than 10: 2, 3, 5, 7. The sections of spinner 2 are thepositive perfect squares less than 40: 1, 4, 9, 16, 25, 36.The 2 on spinner 1 is not relatively prime to these values on spinner 2: 4, 16, and 36.The 3 on spinner 1 is not relatively prime to these values on spinner 2: 9 and 36.The 5 on spinner 1 is not relatively prime to these values on spinner 2: 25.The 7 on spinner 1 is not relatively prime to these values on spinner 2: β.There are, thus, 6 pairs that qualify; there are 4 6 24 total pairs.The desired probability is, therefore, 6/24 ππ/ππ.23. The sum of the three values is (20 π΄π΄) (30 π΄π΄) (40 π΄π΄) 90 3π΄π΄, but we are told that their sum is100 π΄π΄. For 90 3π΄π΄ 100 π΄π΄, we must have 2π΄π΄ 10, so π΄π΄ ππ.24. There are 4 4 4 64 equally likely outcomes of rolling the die. The only ways to get a sum of 7 are:3!6rolling 4, 2 and 1 in any order, which can occur in 1! 1! 1! 1 1 1 6 distinct, equally likely ways;3!6 3 distinct, equally likely ways;2! 1!2 13!6rolling 3, 2 and 2 in any order, which can occur in 3 distinct, equally likely ways.1! 2!1 26 3 312ππThus, the probability that the sum of the numbers rolled is 7 is .6464ππππrolling 3,3 and 1 in any order, which can occur in25. We are to determine π π , given the information in the figure shown and π π ππ 44 cm.The reason for bisecting the angle formed by the 33 cm and 55 cm sides is that we areinstructed to bisect the maximal acute angle. Now, a 33 cm-44 cm-55 cm triangle is a3-4-5 right triangle with a scaling of 11 cm, so the angle opposite the 55 cm side is aright angle, not acute. The larger acute angle is then the angle opposite the next longest33 cmπππ π 55 cmππ55 cmside, which is the 44 cm side. We can use the triangle angle-bisector theorem, which says that π π 33 cm , soππ π π π π 55 cm 33 cm33 cm 883383 . Because ππ π π 44 cm, we have44 cmπ π 83 , so π π 3 44 cm8 3 112cm ππππcm.ππ26. The only Q is in the center square, so that is where we must start. From there, moving left, right, up, ordown, gets us to a U, so that is 4 ways to make QU. Due to symmetry all four ways behave alike, so we cananalyze just one choice for the U and multiply by 4. Regardless of which U we move to, three out of the fourpossible moves left, right, up, or down get us to an E to make QUE. We must be careful here, though, becausenot all 3 Es yield the same number of options for the next move:(i) We see that 2 of the Es are interior on a diagonal, each of which leads to any of 4 Us. From any ofthese Us, we can move to any of 3 Es to make QUEUE.(ii) But 1 of the Es is along an outer edge in the middle and leads to any of 3 Us. From any of these Us,we can move to any of 3 Es to make QUEUE.Thus, we have a total of 4[(2 4 3) (1 3 3)] 4(24 9) 4(33) ππππππ ways.
27. Here we have a problem relating the average speed over a whole trip to the average speed over each ofseveral legs of equal length. The overall average is the harmonic mean of the speeds for each leg. For two12π£π£ π£π£legs with average speeds of π£π£1 and π£π£2 , respectively, the harmonic mean is 1 1 1 1 2 . Thus, we havethe overall average speed being π£π£a 2π£π£1 π£π£2π£π£1 π£π£2 2 π£π£1 π£π£2π£π£1 π£π£2. We are given the actual value for the difference between π£π£1 andπ£π£2 , but for now letβs simply say that π£π£2 π£π£1 ππ. For solving equations with lots of peculiar numbers andmeasurement units, it is often easier to solve in terms of general variables and substitute actual quantities2π£π£1 π£π£2, andonce we express the variable to be solved in terms of the other variables. We have π£π£a 2π£π£1 (π£π£1 ππ),2π£π£1 πππ£π£a π£π£1 π£π£2but we need to rearrange this to express π£π£1 in terms of π£π£a and ππ. Multiplying both sides by2π£π£1 ππ yields π£π£a (2π£π£1 ππ) 2π£π£1 (π£π£1 ππ), which expands to 2π£π£a π£π£1 πππ£π£a 2π£π£1 2 2πππ£π£1 . Rearranging thisinto conventional form for a quadratic equation in terms of π£π£1 yields 2π£π£1 2 2(ππ π£π£a )π£π£1 πππ£π£a 0. Usingthe quadratic formula, we get π£π£1 2(ππ π£π£a ) [2(ππ π£π£a )]2 4(2)( πππ£π£a )2 2 (π£π£a ππ) (π£π£a ππ)2 2πππ£π£a2. We know that theoverall average speed is given by the total distance 2 180 mi 360 mi divided by the total time 7.5 h. So,we have π£π£a 20mihπ£π£1 360 mi7.5 hand π£π£a ππ 283 120 mi4 120 mi480 mimimi 48 . We are given that π£π£1 π£π£2 20 , so ππ3 2.5 h4 2.5 h10 hhhmimimi48 20 28 . Substituting into our quadratic formula yields:hhh mimi 2mimi 28 2 20 48hhhh2 14mihmi 2hmi 2h 196 480 14mi h 676mih (14 26)mi.hThis yields a choice of 40 mi h or 12 mi h, but a negative result does not make sense for this problem, sothe answer must be π£π£1 ππππ mi h .28. Start with 20 and work up, incorporating new factors that have not yet been covered. Now, we are dealingwith numbers in the 20s (or less due to common factors)βletβs just say an average of 25 102 /22 . With 71014such factors, we would be at most about (102 /22 )7 1014 /214 1.6 104 6 109, so letβs work 20through 27 and see where we stand. [NOTE that billion refers to 109 .]2022 5;21 3 7, both new, so 20 2122 3 5 7;22 2 11, with new 22 but old 2, so 20 21 1122 3 5 7 11;23 is a new prime, so 20 21 11 2322 3 5 7 11 23;24 23 3 with one new 2, two old 2βs and old 3, so 20 21 22 23 23 3 5 7 11 23;23 3 52 7 11 23;25 52 with a new and an old 5, so 100 21 22 2326 2 13 with a new 13 and an old 2, so 100 21 22 23 1323 3 52 7 11 13 23;23 33 52 7 11 13 23;27 33 with two new and one old 3, so 100 21 22 23 117Pause: Letβs assess where we are with 100 21 22 23 117. The product of the first 4 factors isstraightforward to do exactly with only mental power; for the rest an approximation is likely good enough.21 22 23 22 (222 12 ) 223 22 23 113 22 8 1331 22 10 648 22 10 626.The 100 part tacks on two more 0βs for 1 062 600, just over 106 ; 117 is only a little over 102 , for a total of alittle over 108 . We need another factor of 10 or so to get there.28 22 7, all of which is old, so nothing new to contribute;29 is a prime, so new, and provides us with our needed factor of at least 10, making the answer 29.
29. My favorite approach to this type of problem is somewhat unconventionalβbut only in the sense ofcombining properties of similarity and proportionality in such a way as to combine steps unconventionally,but quite validly, to reduce the count and complexity of steps, but all equivalent to a more traditionalapproach. The area of a triangle is 1/2 times the base times the height. For the ratio of areas of two trianglesthe 1/2 cancels out, leaving the product of the ratio of the bases time the ratio of the heights. Letβs regardthe base of ABC to be segment BC and the base of PQR to be segment PQ. It is very convenient they areAPparallel to one another. With AB AQAC1 PQ1must be the same , so5 BC5 ,15the know the ratio of the bases is .Now we need to deal with the ratio of the heights of the two triangles, and this will be a 2-step process: first,the ratio of the height of PQ above BC to the height of A above BC, and, second, the ratio of the height of PQabove R to the height of PQ above BC. The product of these two will be the ratio of the height of PQ above Rto the height of A above BC, which product is the ratio of the heights of the two triangles of concern. We use1similarity to realize that the height of A above PQ is the height of A above BC, so the ratio of the height of514PQ above BC to the height of A above BC is 1 minus the ratio 5, which is 5 as the answer to the first part. Forthe second part, we need to determine the horizontal distance from line BQ to line CP going horizontally atsome height, and left-to-right will be regarded as a positive value and right-to-left as a negative value (aform of what is often referred to a displacement [physics] or directed distance [mathematics]). At the heightof base PQ, we are going from Q to P, which is right-to-left, so negative. Since we care only about ratios andscaling, we can regard the distance as normalized to 1. At the base BC, the point on BQ is B and the pointon CP is C, with C to the right of B, so a positive distance that is 5 times the magnitude of the distance 1from Q to P, with our given scaling. The magnitude of 1 is 1, which multiplied by 5 is 5. The two lines1intersect at R so the distance from R to R is 0. Now, 0 is of the way from 1 to 5, so the height of R below16645PQ is the height of BC below PQ, which is the height of BC below A, making the height of PQR equal to45162the height of ABC. With a ratio of bases1512ππareas is .515ππππ 15being and the ratio of heights being2, the ratio15of the[The explanation is long because there are some atypical details to discuss, but the calculations are shortwhen you are used to the concept. It is very quick to see that the base of RPQ is 1/5 as wide as for ABC,1645and then the height of RPQ is 2that of ABC, so1515the ratio of the areas is 215 ππ.]ππππ30. We can reorder the terms so that ππ ππ ππ. It must be that ππ must be 2 or 3; otherwise, ππ being 1 makes the36sum of the three reciprocals greater than 1, and the maximum sum for ππ 3 is . For the sum to have a47denominator of 7, at least one of ππ, ππ, and ππ must be a multiple of 7, and it is not ππ since ππ is 2 or 3; neithercan it be ππ since then ππ would also have to be at least 7, and the sum of the reciprocals cannot reach the671ππ67131721 3, making ππ ππ (a811615Thus, ππ must be 2, making . Therefore,ππππ7214requisite . Thus, with ππ 3 and ππ 7, then contradiction of specified conditions) if ππ 3.145 ππ 8, so ππ21 14(the first condition based on ππ being an arbitrarily large integer, making 1/ππ arbitrarily close to30, and the second condition based on ππ 7 since ππ is a multiple of 7). Thus, ππ must be 3 or 4. For ππ 3, ππ11would have to be ππ 5 1 , which is the reciprocal of an integer and, therefore, suitable. For 14 342completeness (but not necessary in the competition), for ππ 4, ππ would have to be ππ not the reciprocal of an integer, so is not an acceptable answer.Therefore, we end up with: ππ 2, ππ 3, ππ 42; ππ ππ ππ 2 3 42 ππππ.15 114 4 328, which is
2019 Chapter Target Round Solutions1. The price increase of each orange is 0.69 0.49 0.20. Therefore, the price increase of 6 oranges is6 0.20 ππ. ππππ. [WARNING: That second decimal place is required for monetary amounts.]2. The π¦π¦-intercept is determined by solving for π¦π¦ when π₯π₯ 0. For Chris that is at π¦π¦ 7, while for Sebastian thatis π¦π¦ ππ. For Sebastianβs to be double Chrisβ, ππ 2 7 14.The π₯π₯-intercept is determined by solving for π₯π₯ when y 0. For Chris that is at π₯π₯ 7/3 while for Sebastianππππthat is π₯π₯ ππ/ππ. For Sebastianβs to be double Chrisβ, Therefore, ππ ππ 3 14 ππππ.14ππ73 2 141, so3ππ 1/3 and ππ 3.3. In 12 min, the Cubes have gained 21 18 3 points over the Bisectors, so the rate of gain is3 points12 min 1 point,4 minor 1 point every 4 minutes. The game lasts 2 20 min 40 min. Therefore, the total gain of points by the1 ptCubes over the Bisectors is 4 min 40 min ππππ points.4. Let point X be the point of intersection of the bisector of D and line AB, forming ADX, and point Z be thepoint of intersection of the two dashed lines, forming BXZ. We are given ππ XAD 84 and ππ ADX 32 /2 16 , so ππ DXA 180 (84 16 ) 80 . DXB is the supplement of DXA, and, thus, has ameasure of 100 ; this is the same as BXZ. We are given ππ ZDB 32 /2 16 . Therefore, for XZB, whichis the angle in question, ππ XZB 180 (100 16 ) ππππ .5. As the figure shows, we can draw line segments AW and AY perpendicular to lines QRand RS, respectively. Because A is the center of square PQRS, W and Y are themidpoints of their respective sides of the square. Triangles AWX and AYZ arecongruent. The shaded area is quadrilateral AXRZ. That area is equal to the sum of theareas of quadrilateral AXRY and triangle AYZ. Since triangles AWX and AYZ arecongruent, their areas are equal. Thus, the given shaded area is equal to the sum of theareas of quadrilateral AXRY and triangle AWX, which is the same as the area of square1AWRYβa square that is similar to square PQRS, with linear scale factor of . The ratio21 22ππππof the area AWRY to the area of PQRS is the square of this linear scale factor, thus . Note that thisanswer does not depend in any way on the degree of rotation of the larger square about the center of thesmaller square.6. Let the desired tens digit be ππ, the wrongly input tens digit be π€π€, and the units digit be π’π’. Then the desirednumber to square is 10ππ π’π’ and the wrongly squared number is 10π€π€ π’π’. The difference in their squares is2340 (10π€π€ π’π’)2 (10ππ π’π’)2 (100π€π€ 2 20π€π€π€π€ π’π’2 ) (100ππ2 20ππππ π’π’2 ) 100(π€π€ 2 ππ2 ) 20(π€π€ ππ)π’π’. Dividing both sides by 20 yields: 117 5(π€π€ 2 ππ2 ) (π€π€ ππ)π’π’ [5(π€π€ ππ) π’π’](π€π€ ππ). Weare dealing with tenβs digits ππ and π€π€ of 2-digit numbers, with π€π€ ππ, so we must have 1 ππ π€π€ 9. Thus,1 π€π€ ππ 8 and π€π€ ππ must divide 117 32 13, meaning that π€π€ ππ is 1 or 3. If π€π€ ππ 1, then5(π€π€ ππ) π’π’ 117, but π€π€ 9, ππ 8, and π’π’ 9, so 5(π€π€ ππ) π’π’ 94 and can
The following pages provide solutions to the Sprint, Target and Team Rounds of the 2019 MATHCOUNTS Chapter Competition. ese solutions provide creative and Th concise ways of solving the problems from the competition. There are certainly numerous other solutions that also lead to thecorrect answer , some even more creative and more concise!
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
math team placed fifth out of 32 middle schools competing in the MATHCOUNTS 2006 Oahu Chapter Competition, and are two of the six Oahu math teams that will represent Oahu in the Hawaii State MATHCOUNTS Competition on March 11, 2006. Other public . The first is the Sprint Round with its 30 problems, and then comes the Target
The MATHCOUNTS Competition Program is designed to excite and challenge middle school students. With four levels of competition - school, chapter (local), state and national - the Competition Program provides students with the incentive to prepare throughout the school year to represent their schools at these MATHCOUNTS-hosted* events. MATHCOUNTS
2018 AMC 8 Problem 2 exactly the same as 2012 3 is MathCounts State Sprint Problem 3, and very similar to the following 4 problems: 2017 MathCounts Chapter Countdown #49 2016 MathCounts National Sprint #11 2016 - 2017 MathCounts School
2005 MathCounts National Competition in Detroit, be it known that Detroit will host the 2005 Major League Baseball All-Star Game, the 2006 Super Bowl, and the 2009 NCAA Menβs Basketball Final Four. Obviously, the National MathCounts Competition fits right in with these other world-class c
2019 MATHCOUNTS State Competition. These solutions provide creative and concise ways of solving the problems from the competition. There are certainly numerous other solutions that also lead to the correct answer, some even more creative and more concise! We encourage you to find a variety of approaches to solving thesefun and challenging
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Scrum 1 Agile has become one of the big buzzwords in the software development industry. But what exactly is agile development? Put simply, agile development is a different way of executing software development teams and projects. To understand what is new, let us recap the traditional methods. In conventional software development, the product requirements are finalized before proceeding with .
