***** O CALCULATORS I THIS ROU D *****

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1 O TST OVMR 009 ROU 1 OMPLX UMRS ( o Trig) ***** O LULTORS I THIS ROU ***** SWRS ) ) ) Note: i= 1 ) Simplif completel: 3 1+ i+ 3i + 4i 3 1 i+ 3i 4i 40 n ) Given: (3+ 3 i) = r, where r and n are both integers etermine the smallest possible value of the sum r + n. ) If 40 9i = + i, compute.

2 O TST - OVMR 009 ROU LGR 1: YTHI G ***** O LULTORS I THIS ROU ***** SWRS ) ) ) ) Find 4 consecutive odd integers whose sum is 13 more than the largest of these integers. ) If + = 3 and = 10, find the largest possible value of. ) train travels 150 miles in w hours. If the rate of the train were increased b mph, the train would arrive at its destination in less hours. Find in terms of w.

3 O TST - OVMR 009 ROU 3 PL GOMTRY: RS OF RTILI R FIGURS ***** O LULTORS I THIS ROU ***** SWRS ) units ) units ) units P ) The area of PQR is 45 square units. The areas of PQS and PSR are unequal. etermine the smaller of the two areas. h Q(-4, 0) S(4, 0) R(11, 0) ) Rectangle has an area of 500 square units. and F are midpoints of two adjacent sides. etermine the area of the larger of the two regions inside created b F. I II F ) Given: quadrilateral with perpendicular diagonals and = 13, = 15, = 5, = 14 To the nearest integer, what is the perimeter of?

4 O TST - OVMR 009 ROU 4 LG 1: FTORI G ITS PPLITIO S ***** O LULTORS I THIS ROU ***** SWRS ) ) ) ) Solve for : = ) Solve for : = ) The polnomial can be written as the product of binomial factors of the form ( a ± b), where a and b are positive integers. etermine the maimum value of.

5 O TST - OVMR 009 ROU 5 TRIG: FU TIO S OF SPIL GLS ***** O LULTORS I THIS ROU ***** SWRS ) (,, ) ) ) ) In simplest form, (tan 40 + tan405 ) 3 = +. etermine the ordered triple (,, ). ) For the purpose of this question, suppose special angles denote angles belonging to the k. famil, 45 famil, 60 famil or the quadrantal famil ( ) ompute tan() given that tan() = 3cot() 1 and is not a special angle. ) In, m = m = 90, = 4, m = 30 and =. Find in simplified radical form.

6 O TST OVMR 009 ROU 6 PL GOMTRY: GLS, TRI GLS PRLLLS ***** O LULTORS I THIS ROU ***** SWRS ) ) ) ) The measures of the verte and base angles of an isosceles triangle are in a 4 : 3 ratio. If the verte angle is the larger of these two angles, compute the measure of an eterior angle at the base. ) regular polgon has 740 diagonals. How man degrees in an eterior angle of this polgon? ) F and is a transversal intersecting and F in points M and respectivel. P is a point between the parallel lines such that m MP= 3m PM m M P= 4m P F If m M= (7 40) and m M F = (5 ), find m P... M.. N F.. P

7 O TST - OVMR 009 ROU 7 TM QUSTIO S ***** LULTORS R PRMITT I THIS ROU ***** SWRS ) Given: z = 1 3i ) ) ) ) ) F) ompute: z z z ) brick mason can do a job in 6 less hours than his apprentice. He and his apprentice work together for 4 hours. fter the fourth hour, the apprentice works alone and finishes the remainder of the job in three hours. If the brick mason had been able to hire k apprentices, each of whom worked at the same rate as his original apprentice, and the all worked together with him from the start, the job would have been finished in a time of one hour or less. What is the minimum possible value of k? ) Point P is located in the interior of rectangle. (No diagram given.) = 5, P = 56, P = 5 and P = 33. ompute. ) Factor completel over the integers. a 4a + a + 6a 4 3 P Q ) PQRS is a rectangle. Let and denote the areas of PQT and SRT respectivel. If m PTS = 60 and PT = tan( TPS) = 1, compute. T F) is a pentagon, = = and =. For integers d and k, m = m = m = m = d and m is 5k. ompute the largest possible value of m that is less than 1. m S R

8 O TST - OVMR 009 SWRS Round 1 lgebra : omple umbers ( o Trig) ) i ) 38 ) Round lgebra 1: nthing 1 81 ) 69, 71, 73, 75 ) (or 0.4) ) w( w ) or equivalent Round 3 Plane Geometr: rea of Rectilinear Figures ) 1 ) or 875 ) 85 Round 4 lgebra 1: Factoring and its pplications ), 8 ) ±, 5 ) 9 Round 5 Trig: Functions of Special ngles [ on-alculator Round] ) (10, 6, 3) ) 3 ) 6+ Round 6 Plane Geometr: ngles, Triangles and Parallels ) 16 ) 9 ) 40 Team Round ) 4 ) a ( a )( a + 1)( a 3) ) 13 ) + 3 ) F) 10 13

9 O TST OVMR 009 SOLUTIO KY Round i+ 3i + 4i 1+ i 3 4i i 1+ i 1+ i 1+ i + i i ) = = = 3 = = = i 1 i+ 3i 4i 1 i 3+ 4i + i 1 i 1+ i 1 i ) = 3 (1 + i) = Note: ( ) ( i ) = i = 3 (1) = = 18 r + n = = 18 = 34. Such equivalent epressions produce larger values of r + n. ) If z = + i, then z = ( ) 40 9i= + i = + i. ut, z = + and z = ( 40) + ( 9) = 41 = 41 ( is a Pthagorean Triple.) (1) = 40 Since z = z, we have + = 41. () = 9 quating the real parts, the imaginar parts and the absolute values, we have these three conditions:. (3) + = 41 () and have opposite signs. 1 (1) + (3) = 1, (3) (1) 1 9 = 81 and (, ) = ±, = = Proof of the fact that for an comple number, z = z. z = ( + i) = + i+ i = + ( ) i Let z = + i. Then ( ) ( ) z = + = + Round = ( ) + ( ) = ( ) z = ( ) + = = + the transitive propert, ) Let the 4 numbers be, +, + 4 and + 6. Then: = = 07 = 69 69, 71, 73, 75 z = z. ) solving (3 ) = 10 or judicious guess and check, (, ) = (5, ) or ( 5, ). The possible values of are.5 or 0.4. The larger value is 0.4. ) Let R denote the new rate and R 1 the original rate. Since R T =, we have R = 150 and R 1 = and = + w w w w learing fractions, 150w= 150( w ) + ( w)( w ) 150w= 150w 300+ w w 300 anceling, we have 300= w w= ( w w) = w w or 300. w( w )

10 O TST OVMR 009 SOLUTIO KY Round 3 P II h I 5 9 Q(-4, 0) S(4, 0) R(11, 0) ) QR = 15, QS = 8 and SR = 7 1 (15) h= 45 h = 6 PSR has the smaller area, = 1 ) rea(i) = 1, rea(ii) = = Thus, regardless of the dimensions of the rectangle, region II has an area 7/8 that of the rectangle 7 (500) = or ) Noting special right triangles , 3( ) and , the problem is almost done. = 165= is onl slightl bigger than the perfect square = < < 40.5 Thus, to the nearest integer, the perimeter of is 85.

11 O TST - OVMR 009 SOLUTIO KY Round ) ross multipling, = = = 4( ) = 4( )( 8) = 0 =, ) = ( 5 ) + ( 4 0) = ( 5 ) 4(5 ) = 0 ( ) = ±, 5 4 (5 ) = 0 ) = ( ) ( 8 56) =( 16 1)( 8 56) = 8 ( )( )( )( 16) = ( 8 )( 4 4 )( )( 4 )( )( 4) ( 8 1)( 4 1)( 1 )( 1)( 1) ( 4 16)( 4)( )( ) = = 9 Round 5 ) = 3 ( 3+ 1) = ( 3+ 1) ) tan() = 3cot() 1 tan () = 3 tan() ( 3+ 1) = ( 3+ 1) ( 3+ 4) = (,, ) = (10, 6, 3) tan () + tan() 3 = (tan + 3)(tan 1) tan() = 3 (1 is etraneous since would be special, i.e. it would belong to the 45 famil.) ) m = 60, =, = 3 m = m = m = 45 =, = 3 and = 3 6 = = 6 Thus, = + = + ( 6 ) = lternate solution In right, cos( ) =. m = = 15 Thus, = 4cos(15 ) = 4cos(45 30 ) = 4( cos45 cos30 + sin 45 sin30 ) 3 1 = 4 + = 6 +

12 O TST - OVMR 009 SOLUTIO KY Round 6 ) 4 + (3) = 180 = 18 base angle: 54 eterior angle: 16 ) n(n 3)/ = 740 n(n 3) = 1480 Rather than tring to factor this quadratic b trial and error, guess at a value for n. If the result is too low, tr a larger value of n; if the result is too high, tr a smaller value of n. n = 35 35(3) = 110 (too low) n = 45 45(4) = 1890 (too high) Since 110 is closer to 1480, we ll start at 40 and step down until we find n. n = 40 40(37) = 1480 ingo! 360 regular polgon with 40 sides has eterior angle with 40 = 9 ) Since M and M F are corresponding angles of parallel lines, we have 7 40 = 5 = 0 m M= 100 a + b = 80 and c + d = 100 m MP= 3m PM b = 3a Thus, (a, b) = (0, 60) m M P= 4m P F c = 4d Thus, (c, d) = (80, 0) Finall, = 180 (b + c) = = 40.. M b c. N d. a. P F

13 Team Round O TST - OVMR 009 SOLUTIO KY ) z z z = z z z = z = z = z z = 1 3i 3= 3i z = 4+ 1 = 4 ) Let 1 denote the rate at which the brick mason works, i.e. the fraction of the job he does in 4 (4+ 3) one hour. Then + = 1 4( + 6) + 7 = = ( 8)( + 3) = = 8 Thus, the mason and apprentice take 8 hours and 14 hours respectivel to complete the job. ssume a minimum of apprentices are needed (1) (1) = 1.5 min = [ 1 apprentices and 1 brick mason take T hours to finish 1 1 T 6T T+ 1 T = 1 + = 1 7T + 48T = P 5- T = 56/55 > 1 ] ) 5 + P = P = 3600 P = 60 ( Refer to note on ontest 1 Round.) Using Heron s formula, rea( P) = 55(30)()(3) = = rea( P) = 84(8)(3)(4) = 3 7 = 1344 Let = = rea(rectangle) = 5 = (5 ) = = 1674 = = ) Hoping to take advantage of a binomial of the form ( a c) thinking of Pascal s triangle: 4 3 a 4a + a + 6a = 4 3 a 4 a + (6 5) a + (10 4) a + (1 5+ 4) 56 +, where c is a constant and Regrouping, we have ( a 4a + 6a 4a + 1) 5( a + a + 1) = ( a 1) 5( a 1) + 4= (( 1) )(( ) ) a 1 a 1 4 = ( 1 1)( 1 1)( 1 )( a + a a + a 1 ) = a ( a )( a + 1)( a 3)

14 Team Round continued O TST - OVMR 009 SOLUTIO KY ) From the diagram at the right, we see that = sin 45 and sin 75 sin 45 sin 45 PTS: = ST = TRS: sin15 1 ST sin 75 ST = = sin15 sin 45 sin15 = sin 45 = tan15sin 45 sin 75 cos15 1 PQ( ) sin = = 1 ( ) = tan15sin 45 = SR tan15 = 3 = + 3 F) 4d + 5k = 540 d = (540 5k)/4 = (135 k) k/4 k must be a multiple of 4. k = 4t m = 0t d = m = m = m = m = (540 0 t) / 4 = 135 5t m = (180 0t)/ = 90 10t m = (135 5t) (90 10t) = t m 90 10t 18 t = = m 45+ 5t 9+ t Tring values of t we get the following ordered pairs: (1, 8/5) (, 14/11) (3, 1). This is a decreasing sequence and t = 4 d d d lternate solution #1: 5d + 5k = 540 d = k = k. 4 4 Since d must be an integer, k must be divisible b 4. Since m = 5k < 180, k < 36. Therefore k = 4, 8, 1, 16,, 3. The chart below indicates as k increases the required ratio decreases and the largest value less than 1 is highlighted. k d ( m ) m m 1 ( ) m ( ) Ratio / / /13 S P k Q T R d lternate solution #: **** 4d + 5k = 540 d must be a multiple of 5. For to be a triangle, 0 < 5k < 180 or 0 < k < 36. Substituting in ****, we have 90 < d < 135. trial and error (and the fact that d is a multiple of 5), we test 130, 15, 10,. These results are contained in the table above, referencing the second column as the ke field.