UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST. First Round For all Colorado Students Grades 7-12 November 3, 2007

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UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST First Roud For all Colorado Studets Grades 7- November, 7 The positive itegers are,,, 4, 5, 6, 7, 8, 9,,,,. The Pythagorea Theorem says that a + b = c where a, b, ad c are side legths of a right triagle ad c is the hypoteuse. A isosceles triagle has two sides with equal legth.. Poits A, B, C, ad D are evely spaced o the umber lie as show. What reduced fractio correspods to poit C? Express your aswer as a fractio a b.. For which value(s) of, are, +, - the legths of the three sides of a right triagle?. Determie the umber of triagles of all sizes i the shape show. 4. Triagle ABC has iteger side legths. Oe side has legth ad a secod side is twice the legth of the third side. What is the greatest possible perimeter? 5. [4, 4, 5, ] is a collectio of positive itegers whose sum is 5 ad whose product is 96. [,, 4, 8, 8] is aother collectio whose sum is also 5 but whose product is 56. Amog all collectios of positive itegers whose sum is 5 what is the largest product that ca be foud?

6. Six frieds are sittig aroud a campfire. Each perso i tur aouces the total of the ages of the other five people. If 4, 5, 8, 4, 5, ad 9 gives the six sums of each group of five people, what is the age of the oldest perso? 7. ABCD is a rectagle with AB = 5, EC =, AE = AD. Fid AD. 8. (a) Four people Adrew, Beth, Caroly ad Darcey play a game that requires them to split up ito two teams of two players. I how may ways ca they split up? (b) Now suppose that Euler ad Fiboacci joi these four people. I how may ways ca these six people split up ito three teams of two? 9. Triagle ABC is isosceles with AB=AC. The measure of agle BAD is _ ad AD=AE. Determie the measure of agle EDC.. I the diagram to the right (a) what is the sum of the etries i row J? (b) what is the first (leftmost) umber i row Z? A B C D Y Z 7 5 9 5 M 7 9

Brief Solutios FIRST ROUND NOVEMBER 7. 7 7 7 7 7 7 ; Sice =, each segmet has legth. The C is + = 6 4 6 6 6 ; By the Pythagorea Theorem( ) = ( ) +. = =. +. After simplificatio = or. 6; Ay choice of of the 7 base itersectio poits will determie a large triagle. Similarly for the other two sectios. The total umber of triagles is the 7 = = 6. 4. P = 49; The side legths are, 4, ad. 5. 8748; The sum + + + + + + + + = 5 gives the maximum product. 5 is ot eeded as a summad sice it could be replaced by +, yieldig a larger product 6. Similarly, sice 6 = +, 7 = + +, 8 = + +, 9 = + +, oe of these digits are eeded. 6. Age = 9; The sum 4 + 5 + 8 + 4 + 5 + 9 = 665 is five times the sum of all six ages. Sice all six ages sum to 665 = 5, the oldest is 4 = 9. 7. 9 ; Let AD = x ; the AE = x, = x 4 + BE. By the Pythagorea Theorem = ( x ) 5 x ; the 4 x = 9. 8. (a) ; The teams are AB ad CD; AC ad BD; AD ad BC (b) 5; With AB there are ways to form teams of usig C, D, E, F to pair with AB. Similarly, with each of AC, AD, AE, AF there are three ways to complete the split. The total is 5 = 5. 9. Agle EDC = 5 _ ; Label all agles i the triagle ad use two facts the sum of all three agles i ay of the triagles is 8 _ ad the base agles i each of the two isosceles triagles are equal.. (a) The first umber i row J is 9. The sum 9 + 9 + 95 + 97 + 99 + + + 5 + 7 + 9 is. This ca be foud by subtractig oe arithmetic sum from aother. (b) 65; Row Y has 5 elemets i it. The first 5 rows cotai + + + L + 5 = 5 odd umbers. The 5 th odd umber is ( 5) = 649. Row Z starts with 65.

UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST FINAL ROUND For Colorado Studets Grades 7- February, 8. Determie the umber of by square arrays whose row ad colum sums are equal to, usig,, as etries. Etries may be repeated, ad ot all of,, eed be used as the two examples show.. Let S = { a, b, c, d} be a set of four positive itegers. If pairs of distict elemets of S are added, the followig six sums are obtaied 5,,,, 4, 9. Determie the values of a, b, c, ad d. [Hit there are two possibilities.]. A rectagle is iscribed i a square creatig four isosceles right triagles. If the total area of these four triagles is, what is the legth of the diagoal of the rectagle? 4. I the figure there are 8 lie segmets draw from vertex A to the base BC (ot coutig the segmets AB or AC). (a) Determie the total umber of triagles of all sizes. (b) How may triagles are there if there are lies draw from A to iterior poits o BC? 5. The sum of 4,, 5, 8 ad 5 is 8 ad the product of these five umbers is 46,4,, = 464x. 8 (a) Determie the largest umber which is the product of positive itegers whose sum is 8. (b) Determie the largest umber which is the product of positive itegers whose sum is. OVER

6. Poits A ad B are o the same side of lie L i the plae. A is 5 uits away from L, B is 9 uits away from L. The distace betwee A ad B is. For all poits P o L what is the smallest value of the sum AP + PB of the distaces from A to P ad from P to B? 7. Determie the value of a so that the followig fractio reduces to a quotiet of two liear expressios x x + + ( a ) x x + ( a 6) ( a 6) x x + ( a ) 8. Triagle ABC has iteger side legths. Oe side is twice the legth of a secod side. (a) If the third side has legth 4 what is the greatest possible perimeter? (b) If the third side has legth what is the greatest possible perimeter? (c) Now suppose oe side is three times the legth of a secod side ad the third side has legth of 4. What is the maximum perimeter? (d) Geeralize. 9. Let C = + + + L +. (a) Prove that 9C =. (b) Prove that ( + ) 5 = C. (c) Prove that each term i the followig sequece is a perfect square 5, 5, 5, 5, 5,. Let (, ) f be the umber of ways of splittig people ito groups, each of size. As a example, the 4 people A, B, C, D ca be split ito groups AB CD ; AC BD ; ad AD BC. Hece (, ) = (a) Compute f (, ) ad ( 4, ) (b) Cojecture a formula for (, ) (c) Let f (, ) be the umber of ways of splittig {,,,,} Compute f (, ), f (,) ad cojecture a formula for (, ) K ito subsets of size.

Brief Solutios Fial Roud February, 8. ; There are 6 usig s ad 6 more usig o s, ad 9 usig oe.. {, 4, 9, } or {,, 8, }; Order the itegers as a < b < c < d. The two cases are a + b = 5, a + c =, a + d =, b + c =, b + d = 4 ad c + d = 9 givig c b = 5 by subtractig the first two yieldig {, 4, 9, } or a + b = 5, a + c =, b + c =, a + d =, b + d = 4, c + d = 9 yieldig {,, 8, }.. ; Let a, b be the side legths of the small ad large right triagles, respectively. The a + b = ad the legth of the diagoal is d = a + b = =. There are two alterative solutios compress the rectagle ito the diagoal d or expad the rectagle to become a square whose diagoal is parallel to the side of the origial square. 4. (a) ; Ay choice of of the lies will result i a triagle. (b) + = ( + )( + ) 5. (a) (b) If 668 ; for ay the summads 5, 6, 7, 8, 9, ca always be replaced by s ad s. The maximum umber of s will yield the maximum product. = k, = k +, k is the maximum product. If = k +, k is the maximum product. k is the max product. If 6. 8; Reflect PB about the x axis. The x coordiate of B, the reflectio of B, is 8. The distace betwee A (, 5) ad B ( 8, 9) is 8 + 96 = 8. 7. a = 8 ; If c ad d are egatives of each other the the umerator ad deomiator of x + cx x + d x + dx x + c a = 8. x = 4 = each factor by groupig as ( ) x + c x x + c =. So set = ( a 6) ( x c) x c x a ad 8. (a) 57 ; If the side legths are x, x ad 4 the 4 + x > x or x < 4. The maximum perimeter occurs whe x = 9 ad is 9 + 4 + 78 = 57. (b) 4 ; If the sides are x, x, ad the x + > x implies x < ad the sides are ( ), ( ),. (c) 6 ; x + 4 > x implies x <. The 9 + 57 + 4 =6.

9. (a) (b) = + + + L + ; ow subtract C = ( +) = 5 = 5. C = + + L +. (c) Take = 4 ; (C 4 + ) = 9C 4 + C 4 + 4 = 9C 4 C 4 +C 4 +4 = ( 4 )C 4 + C 4 + 4 = 4 C 4 + C 4 + C 4 + 4 = + + + 4 = 5 The proof for geeral is similar.. (a) f (, ) = 5 ; We ca split A, B, C, D, E, F ito groups of as follows; A ca pair with ay of the other 5 i 5 ways. The other four ca split ito i f (, ) = ways. f (4, ) = 7 _ 5 _ i a similar maer. (b) f (, ) = (-)(-)(-5) = (c) f (, ) = ; A ca pair with ay of B, C, D, E, F. f (, ) = ; A ca pair with ay of the other 8. The remaiig 6 ca be split i f (, ) = ways. f (, ) = =