The domain of the function. is {x : a x b}. Determine. One factor of. 3x 2 + bx 20. Find b. will each be 39.

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SEPTEMER The poduct of 5 3 8 6 can be epessed as n. Find n. The median of the set {3, 3,, 9, 57, 90} is 6. Detemine the value of. 8 The length of the hypotenuse of a ight tiangle is 5.

1 SEPTEMER The poduct of can be epessed as n. Find n. The median of the set {3, 3,, 9, 57, 90} is 6. Detemine the value of. 8 The length of the hypotenuse of a ight tiangle is 5. If the diffeence between the lengths of the legs is 9, find the sum of the two legs. The thee sides of a tiangle ae all integes. If two of the sides ae 5 and, what is the atio of the numbe of possible obtuse tiangles to the total numbe of possible tiangles? 6 Fom all thee-element subsets of the set of integes fom to 0, inclusive. Then find the poduct of the elements in each subset. How many of these poducts ae divisible by 5? 0 Isosceles tiangle C has vete and altitude D. Etend ay D though point to a point P such that PC = C. Locate a point E, not on PC, such that PC = CE and CE = C. Find m EC. Have you eve noticed that if you cancel the 9s in the faction 9/95, you get the coect eduction, /5? Thee othe factions consisting of two-digit numeatos and denominatos shae this popety. Can you find them? Do not count ecipocals o numeatos and denominatos that shae a facto of. Mas takes Eath days to tavel aound the sun. If the months on Mas ae popotional to the months on Eath, how many Eath seconds will thee be in the month of Septembe on Mas? (n Eath yea is days long.) cicle is inscibed in a quate cicle such that the smalle cicle is tangent to the two adii and to the ac of the quate cicle. Find the atio of the aea of the quate cicle to the aea of the smalle cicle. What is the latest possible month of the yea that the fist Fiday the 3th can occu? What yea and month have the most Fiday the 3ths? (Hint: Repesent the seven weekdays as vetices of a polygon and the passage of time as motion aound the polygon.) Detemine the lagest intege that will divide ( )( ) fo all integal values of. How many seven-digit numbes of the fom a3567b ae divisible by, whee a and b epesent single digits? The unit faction / can be witten as the sum of two unit factions: Find thee unit factions, all diffeent, whose sum is /3. cube of edge 3 in. has -in.-squae holes cut fom thee of its faces though to the opposite face. Find the aea of the suface of the esulting shape, including its inteio. 3 The domain of the function f ( )= is { : a b}. Detemine a+ b One facto of is Find b. 3 + b The sum of the squaes of two numbes is 7, and the poduct of the two numbes is. Find all possible values fo the sum of the two numbes. 7 8 The 5-question MC eam is gaded as follows: 6 points fo a coect answe,.5 points fo leaving the answe blank, and 0 points fo an incoect answe. scoe of 00 o bette is equied to qualify fo the IME. If Sunil gets eactly 3 answes wong, what is the least scoe that he can make to qualify fo the IME? In 7, Chistian Goldbach conjectued that evey even intege geate than can be witten as the sum of two pime integes. Since then, no one has been able to pove o dispove the conjectue. To celebate this fact, find the two pimes that sum to 7 and whose diffeence is the smallest. simple code assigns a numbe to each lette of the alphabet: fo, fo, 3 fo C, and so on, to 6 fo Z. Using these lettes, decode this message: Two fou-digit numbes, abcd, eist such that abcd + a + b + c + d = 03. Find the two numbes. The sum of two positive numbes equals the diffeence of the squaes of the two numbes, which equals the quotient of the lage numbe when divided by the smalle. Find the smalle numbe. The opening enollment at Piedmont High School inceased by 8 students in the fist week of Septembe. In the second week, the total population deceased by 5%. If 6 students wee enolled at the end of the second week, what was the openingday population? Two concentic cicles have adii a and b with a < b. line tangent to the inne cicle at P intesects the bigge cicle at points Q and R. If QR = 0, find the aea of the egion between the two cicles. Solve fo all eal values: 0/ = tiangle is dawn with one side of length 03; the altitude to that side is. If the measue of one angle adjacent to the given side is 30, find the sum of the lengths of the othe two sides of the tiangle. 3 9 Fou identical-looking objects weigh 3, 5, 8, and oz., espectively. Using a balance scale, how can you detemine the heaviest object in just two weighings? Refe to poblem 7. Eplain how you can find the weight of each of the emaining objects in two moe weighings. 8 Place the integes to 5 into the fifteen cicles so that the sum of the five numbes along each side of the tiangle and the sum of the thee cental numbes will each be ltitudes dawn to the shotest two sides of a tiangle have lengths 6 and 0. The length of the thid altitude is geate than h. Find the minimum value of h. 30 National Council of Teaches of Mathematics, 906 ssociation Dive, Reston, V Copyight 03 The National Council of Teaches of Mathematics, Inc. ll ights eseved. This mateial may not be copied o distibuted electonically o in any othe fomat without witten pemission fom NCTM.

2 Solutions to calenda Looking fo moe Calenda poblems? Visit calenda/default.asp?jounal_id= fo a collection of peviously published poblems sotable by topic fom Mathematics Teache. Edited by Magaet Coffey, Magaet.Coffey@fcps.edu, Thomas Jeffeson High School fo Science and Technology, leandia, V 3, and t Kalish, atkalish@ veizon.net, Syosset High School (etied), Syosset, NY 79 The Editoial Panel of Mathematics Teache is consideing sets of poblems submitted by individuals, classes of pospective teaches, and mathematics clubs fo publication in the monthly Calenda. Send poblems to the Calenda editos. Remembe to include a complete solution fo each poblem submitted. Othe souces of poblems in calenda fom available fom NCTM include Calenda Poblems fom the Mathematics Teache (a book featuing moe than 00 poblems, oganized by topic; stock numbe 509, $.95) and the 00 Poblem Poste (stock numbe 307, $9.00). Individual membes eceive a 0 pecent discount off this pice. catalog of educational mateials is available at Eds.. 6/6 = /; 6/65 = /5; and 9/98 = /8 = /. Solving this poblem equies a tial-and-success pocedue, but attempting to solve it using a speadsheet pogam such as Ecel is also a geat challenge. n Ecel speadsheet is available at The speadsheet assumes that the faction (0t + u)/(0u + v) = t/v. If so, then we have the following: 0t+ u + = t 0u v v 0tv + uv = 0ut + tv 9tv + uv = 0ut 0ut v = 9t+ u The speadsheet can be set up so that t and u take on all possible single-digit values, geneating values fo v. When v is integal, a possible solution eists.. 7 in.. Each face is missing a in. squae, so the aea of the emaining potion is 9 = 8 in.. Removing the cube fom a face adds in. of aea. Theefoe, fo each face of the oiginal cube, the aea is in.. Thee ae 6 such faces, so the total suface aea is 7 in and 99. Since abcd + a + b + c + d = 03, 00a + 0b + c + d = 03. Theefoe, a = o a =. When a =, we get 0b + c + d = 0. Thus, b must be 9, since the cay fom the tens column cannot be moe than. Now we have c + d = 03. We know that c must be odd and big enough to cay into the hundeds column, so c = 9. Then d =, and we ae done with that possibility. When a =, 0b + c + d =, so b = 0, c =, and d = 0. We can pove that thee ae no othe such numbes. Each time the units digit inceases by, the sum of the numbe and its digits inceases by. Fo eample, = 05. Each time the tens digit inceases by, the sum of the numbe and its digits inceases by. The maimum sum of fou digits is 36, so the fou-digit numbe must be geate than = 977. Since = 00, we wish to incease the units digit by 6 to incease the total by. Obviously, we cannot do so without affecting the tens digit. Inceasing the tens digit by inceases the sum by, esulting in 0. Inceasing the tens digit by inceases the sum by, esulting in 03. Thus, we then have to educe the units digit by 5 to decease the total by 0 and give us the esult /5. Convet each adicand to a powe of and each adical to a factional eponent. Theefoe, 5 8 = 3/5 and 3 36 = /3. Multiply these numbes by adding the eponents: 3/5 + /3 = 9/5. 5.,875,096 sec. n Eath yea contains days, so days mo. h. 60 min. days h. 60 sec., min. o,875,096, is the numbe of Eath seconds in Septembe on Mas. 6. /. Two estictions need to be satisfied by the domain: and The fist is satisfied by 7/. To satisfy the second, solve Thus, /. The domain of the function must satisfy both sets of conditions. The domain is {: 7/ 9/}, and the aveage of the etemes is /. Mathematics Teache Vol. 07, No. Septembe 03 Copyight 03 The National Council of Teaches of Mathematics, Inc. ll ights eseved. This mateial may not be copied o distibuted electonically o in any othe fomat without witten pemission fom NCTM.

3 7. /. Let a > b > 0. Then a + b = a b = a/b. Then a + b = (a + b)(a b), so a b = and a = + b. Substitute this into the equation: a + b = a/b ( + b) + b = ( + b)/b b( + b) = + b b = b = / = / The set has an even numbe of elements, so the median is the aveage of the two middlemost numbes. In this case, ( + 9)/ = 6 = Let the adius of the quate cicle =, and let the adius of the small cicle =. We can find by applying the Pythagoean theoem o by knowing the atio of the sides in a ight isosceles tiangle. Then, =, so ( ) + = = +. We know that the atio of the aeas of simila figues equals the squae of the atio of coesponding linea measuements. Thus, the atio of the aea of the entie lage cicle to the aea of the smalle cicle equals + ( ) = + = 3+. Since the equested atio involves only the quate cicle, ou final esult is If Sunil gets 3 answes wong, 5 answes coect, and leaves the est of the answes blank, he eceives 5(6) + 3(0) + 7(.5) = = 00.5 points. If he gets only answes coect and leaves 8 blank, his scoe would be 96. One way to appoach this poblem is to assume that Sunil answes the emaining answes coectly, fo a scoe of 3. Fo each question that he omits, he loses.5 points, so if we let epesent the numbe of omitted questions, we solve the inequality ecause must be an intege, the numbe of omitted questions is 7, the numbe of coect answes is 5, and Sunil s scoe is 5(6) + 7(.5). 00 s 80 c o 60 e scoe = numbe coect, 0 c 0 ltenate solution: Find the scoe if questions ae answeed coectly and if answes ae left blank. Plot the odeed pais (c, s) (0, 33) and (, 3) on the coodinate plane and connect them. Obseve values of c whee the segment intesects the line s = 00 and find c, the smallest intege geate than c.. 3. Let be the numbe of students on opening day. Then ( + 8)(0.95) = 6 = Let the legs of the tiangle be and y, so y = 9. Then ( y) = y + y = 9 =,6. y the Pythagoean theoem, + y = 5 =,05. Combining these two equations yields y =,05,6 = 6,86. Since ( + y) = + y + y =,05 + 6,86 = 7,889, then + y = 67. ltenate solution : Recognize that ( + y) = ( y) + y = 9 + (5 9 ) = (5) 9. ltenate solution : In the solutions above, we obtained the equested sum without solving eplicitly fo the two legs. If is the shote leg, then ( + 9) is the longe leg, and we have + ( + 9) = = 0 ( + 3)( ) = 0. The shote leg is, the longe leg is 3, and thei sum is Send student solutions to the Poblem of the Month editos, Séan Madden (smadden@geeleyschools.og) o Ricado Diaz (Ricado.Diaz@unco.edu).. 7. One facto is, so, fo the fist and last tems to be coect, the othe facto must be The middle tem is the sum of (3) + 5() = 7, so b = p. Point P is the midpoint of the tangent segment, so PR = 0. We wish to find the aea of the egion between the two cicles: pb pa = p(b a ). Since OP ^ RQ, OPR is a ight angle, and we can use the Pythagoean theoem to obtain a + 0 = b. Thus, b a = 00, and the aea of the ing is 00p. O a b Q P R 6. /3. ccoding to the tiangle inequality theoem, the only possible lengths fo the thid side ae 8, 9, 0,,, 3,, 5, and 6. When the thid side is 3 o , the tiangle will be a ight tiangle. Theefoe, fo all values geate than 3 o less than 0.9, the tiangle becomes obtuse. Thee ae si such values, so 6/9 = / Facto to see that the stuctue of the epession is five consecutive integes: ( ) ( ) ( + ) ( + ) and an eta. Five consecutive integes ae always divisible by 5, and fou consecutive integes ae always divisible by 8, so we know that the five consecutive integes will be divisible by 0. We can egoup the si integes into two sets of thee consecutive integes: ( ) ( ) and ( + ) ( + ). Thee consecutive integes ae divisible by 3, so the numbe is divisible by 9. Thus, the oiginal epession will be divisible by 0 9 = 360. Vol. 07, No. Septembe 03 Mathematics Teache 3

4 8. ±7. Let the numbes be and y. Then + y = 7, and y =. We wish to find + y. Since ( + y) = + y + y = + y + y = 7 + () = 89, then + y = ±589 = ±7. 9.,,, 5. pply the ules of eponents to ewite the equation as = 0/ + The solution cannot be = 0 given the domain of the oiginal equation. When = ±, the oiginal equation is tue. The emaining solutions can be found by solving the equation 0/ + = 0 + = 3 0 = 0 ( 5)( + ) = 0 = 5 o If the poduct of the elements is divisible by 5, then the subset must contain 5, 0, o both 5 and 0. If 5 is in the subset and 0 is not, 8 C = 8 subsets ae possible. Similaly, if 0 is in the subset and 5 is not, 8 C subsets ae possible. When both 5 and 0 ae in the subset, thee ae 8 possible choices fo the thid element. Thus, = 6 subsets. ltenate solution: Fom the total numbe of subsets, 0 C 3 = 0, subtact the numbe of subsets not containing a 5 o 0: 8 C 3 = 56. The esult is 0 56 = Fo an intege to be divisible by, the sum of the digits in the even positions minus the sum of the digits in the odd positions must be a multiple of. Fo eample, the numbe 5,839 is divisible by since ( ) (5 + 3) =, which is a multiple of. Theefoe, (a b) ( ) = n fo some intege n a + b 5 = n. Since a and b ae single digits, the maimum value fo the left side of the equation is 3, and the minimum value is 5. Thus, when n =, a + b = 6, and we get thee solutions: (9, 7), (8, 8), and (7, 9). When n = 0, a + b = 5, and we get five additional solutions: (5, 0), (, ), (3, ), (, 3), and (, ). Notice that (0, 5) is not a solution because a sevendigit numbe cannot begin with 0... The two pime numbes ae 883 and 859. We can begin by dividing 7. in half to get 87, which is divisible by 3. The pimes suounding 87 ae 863 and 877. Howeve, = 879, which is not pime, and = 865, which is also not pime. Thus, we must look at the net possible pai of pimes namely, 859 and 88. The patne of 88 is 86, which is not pime, but the patne of 859 is 883, which is pime and gives us the esult = Label the tiangle so that the given side,, equals 03 and the altitude is dawn fom C to point D. Then C = 8 since the hypotenuse in a tiangle is twice the side opposite the 30 angle. The length of the side opposite the 60 angle, D, is 3. Theefoe, D = 03 3 = 63. pply the Pythagoean theoem in CD to find the hypotenuse: C = + (63) = = 78 C = 8. The sum of the two sides is = C D ltenate solution: fte obtaining C, use the law of cosines in C to find C The goal will be to pove that CP by SS. We know that PC = CE and PC = CE by constuction. y the efleive popety, C = C, so the tiangles ae conguent. Notice that PC is equilateal since P is equidistant fom and C. The altitude of an equilateal tiangle is also an angle bisecto, so the measue of CPD = 30 = m EC. E P D C 5. /, /3, /56. Notice that so = + n n + nn ( + ), = + = Repeat the application of the theoem fo the faction /: = + = Finally, /3 = / + /3 + /56. This solution is not unique because /3 = /5 + / + /0. Notice that the theoem stated allows us to wite any unit faction as the sum of any finite numbe of othe unit factions. 6. TODY IS THURSDY. No lette is assigned to 0, so the fist numbe must be 0, which is the lette T. The net numbe could be o 5, which is eithe o O. If it is, then the net numbe is 5, the lette E, followed by D. Since TED is not a wod, we know that the second numbe is 5 (O). The thid numbe is, so the thid lette is D. We now have the lettes TOD. The fouth numbe can be () o (L). Reject the L and continue with the net numbe as () o 5 (Y). Continue in this fashion to get to the esult: TODY IS THURSDY. 7. Call the objects a, b, c, and d. Place a and b on one side of the scale and c and d on the othe side. One side must be heavie say, c and d so one of these must be the oz. object. Place c on one side of the scale and d on the othe side. Whicheve side is heavie holds the oz. object. 8. Once you have detemined the heaviest object say, d place a and b back on one side of the scale and d on the othe. Thee esults ae possible: () The objects balance, so a and b add to and must be 3 oz. and 8 oz.; () the side with d is heavie, so a and b ae 3 oz. and 5 oz.; o (3) the side with d is lighte, so a and b ae 5 oz. and 8 oz. Mathematics Teache Vol. 07, No. Septembe 03

The second key is to ealize that the sum of the integes fom to 5, inclusive, is (5)(6)/ = 0 but that the sum of fou 39s is 56.

5 The net weighing will detemine the specific weight of a and b. Then thee will be only one weight emaining fo c. 9. This poblem has many solutions. The fist key is to ealize that the thee middle numbes ae {0,, 5}, {, 3, 5}, and {, 3, }. The second key is to ealize that the sum of the integes fom to 5, inclusive, is (5)(6)/ = 0 but that the sum of fou 39s is 56. Theefoe, the diffeence 56 0 = 36 must be the sum of the thee vetices, which ae each counted twice. Thus, the only possible vetices ae {5,, 7}, {5, 3, 8}, {5,, 9}, {5,, 0}, {, 3, 9}, {,, 0}, and {3,, }. When we select one of the fist two options fo the cente cicles, we have only one coesponding option fo the vetices. The thid option listed fo the cente cicles will povide thee possible sets fo the vetices. Once the thee cente cicles and thee vetices ae filled, the nine emaining numbes become elatively easy to place /. The aea of a tiangle is the poduct of a side times the altitude to that side, so we can say that in any given tiangle a side and its coesponding altitude ae invesely popotional. If we name the shotest two sides a and b and the longest side c, we know by the tiangle inequality theoem that a + b > c. Thus, /6 + /0 > /h /h < 8/30 = /5 h > 5/. Fo an ecel speadsheet fo the solution fo poblem fo the septembe 03 calenda, go to COMING IN NOVEMER MT 03 FOcUs issue eginning algeba: teaching Key concepts Speak at an NCTM Confeence Want to shae you epetise at one of ou 0 Regional Confeences? pply to pesent net yea and join us in Indianapolis, Richmond, o Houston. NCTM confeences help teaches, administatos, and math coaches lean moe about challenges facing schools and how to ovecome them especially in effective mathematics instuction. Submit you online poposal to pesent a session and shae you teaching ideas and pactices. Lean moe and apply to pesent at THE NTION S PREMIER MTH EDUCTION EVENTS 0 REGIONL CONFERENCES & EXPOSITIONS Indianapolis Octobe 9 3 Richmond Novembe Houston Novembe 9 Speakes apply by Septembe 30, 03 Vol. 07, No. septembe 03 MatheMatics teache 5

Online Mathematics Competition Wednesday, November 30, 2016

Math@Mac Online Mathematics Competition Wednesday, Novembe 0, 206 SOLUTIONS. Suppose that a bag contains the nine lettes of the wod OXOMOXO. If you take one lette out of the bag at a time and line them