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1 CO TEST 4 J URY 00 ROU D LYTIC GEOMETRY: YTHI G SWERS ) B) C) (, ) ***** O CLCULTORS I THIS ROU D ***** ) Determine the coordinates (x, y) of all possible intersection points with the x- and y-axes of x (y ). B) Compute the diameter of a circle concentric with tangent to x + 4y x 5y 5x + + and C) The points P(, 5), Q(, 7) and R lie on a parabola whose vertex is at V(, ). The axis of symmetry is parallel to one of the coordinate axes. The focus of the parabola lies on QR. Compute the (x, y) coordinates of the point R.

2 CO TEST 4 - J URY 00 ROU D LG : FCTORI G D/OR EQUTIO S I VOLVI G FCTORI G SWERS ) B) C) ***** O CLCULTORS I THIS ROU D ***** ) Simplify completely. x 5x 0+ 5x B) Compute the value of the integer constant for which a solution of the following equation is. x + x 45 4x 9 4x 7 C) Solve for x. 4 + x x

3 CO TEST 4 - J URY 00 ROU D TRIG: EQUTIO S WITH RESO BLE UMBER OF SOLUTIO S SWERS ) B) C) ***** O CLCULTORS I THIS ROU D ***** ) Let x denote a real number (or an angle measure in radians). Compute the smallest positive value of x for which ( sin ) x x B) If x denotes the unique solution to cot(x) tan(x) 0 between π and π, compute cos(x). C) Solve for θ, where 0 < θ < 0 : sinθ + cosθ

4 CO TEST 4 - J URY 00 ROU D 4 LG : QUDRTIC EQUTIO S / THEORY OF QUDRTICS SWERS ) B) C) ***** O CLCULTORS I THIS ROU D ***** ) Consider the following quadratic equation: x + x + M 0 If M a, the constant term is greater than the coefficient of x. If M b, the equation has equal roots. If M c, the product of the roots is 0. Compute the product abc. B) Find all values of the constant k for which the roots of the quadratic equation y + k y 5ky + y + 7 are numerically equal, but opposite in sign. C) The line x y intersects y x + Bx + C at x and x 7. The low point V has coordinates (, ) Compute the value of C. Q P V

5 CO TEST 4 - J URY 00 ROU D 5 GEOMETRY: SIMILRITY OF POLYGO S SWERS ) units B) : C) : : ***** O CLCULTORS I THIS ROU D ***** ) In trapezoid BCD, D BC, E 4, BE 5, CE 0 and DE. If the area of BEC is 50 units, what is the area of DE? E D B C B) The ratio of the length of the longest diagonal in a regular hexagon to the length of shortest diagonal in regular hexagon B is 4 :. Compute the ratio of the length of shortest diagonal of hexagon to the length of the longest diagonal of hexagon B. FG C) Given: BC is isosceles, DE BC, G and DEC ' B ' is a square Express area( FD) : area( DEC ' B ') : area( DECB) as a simplified ratio. D F E B B' G C' C

6 CO TEST 4 - J URY 00 ROU D LG : YTHI G SWERS ) B) C) minutes seconds ***** O CLCULTORS I THIS ROU D ***** ) Compute: 8 B) Write as a single simplified fraction with no negative exponents. a+ b a b ( a b ) Note: For any real number x 0, x is equivalent to x. C) On the interstate Mario traveling 00 mph passed a state trooper with a radar gun parked beside the road. The trooper immediately decided to give chase. 48 seconds after Mario passed the state trooper s parked car, the trooper had gone / of a mile and had reached his top speed of mph, which he maintained until he overtook Mario. How long after Mario passed the trooper was he apprehended? Express your answer in minutes and seconds, accurate to the nearest second.

7 CO TEST 4 - J URY 00 ROU D 7 TEM QUESTIO S SWERS ) D) B) (,, ) E) : C) F) yards ***** CLCULTORS RE PERMITTED I THIS ROU D ***** ) The maximum height of a parabola above the x-axis is twice the distance between the vertex of the parabola and its focus. If the parabola contains the point (5, 5/) and has an axis of symmetry on the y-axis, compute the distance between its zeros. B), B, C and D are positive integers. x+ y+ Bx+ Cy+ D C x + C xy+ C y + C x+ C y+ C ( )( ) 4 5 ll coefficients in this expansion are either or prime. C, C, C Determine all possible ordered triples ( ) 4 5 C) Given: ( 4 x 4 x)( x) cos 4 sin 4 sin 0, where 0 < x < π. Let (p, q) denote the smallest and eighth largest solutions respectively over the specified interval. Compute q p. (5, 5 ) D) rectangular pane of stamps would contain 48 fewer stamps if it consisted of three more rows each containing 5 fewer stamps. There are stamps on the original pane. Compute the smallest possible integer value of. E) Given: parallelogram BCD, K : KF : 7, DE : EC : area( FDE) Compute the ratio area( BCD ). D F) treasure is located at a point along a straight road with landmarks, B, C and D located (in the given order) as indicated on the map below:.... B C D Note: Relative distances are rarely accurate on these old pirate maps. The following instructions were included: () Start at and go / of the distance to C () Then go / of the distance to D () Then go /4 of the distance to B and dig! If B 0 yards and BC 80 yards and the treasure is buried midway between and D, compute the distance from B to D. K E F C B

8 Round nalytic Geometry: nything CO TEST 4 - J URY 00 SWERS ) ( ±,0) B) 5 C) 9, 9 Round lg: Factoring ) x 5 B) 5 C), Round Trig: Equations ) π B) C) 40 Round 4 lg : Quadratic Equations ) 45 4 or.5 B), C) 7 Round 5 Geometry: Similarity ) 8 B) : C) : 8 : Round lg : nything ) B) a( a+ b) C) 4 minutes 8 seconds Team Round ) 0 D) 4 B) (5, 7, ) only E) : 4 C) F) 00 yards

9 CO TEST 4 - J URY 00 SOLUTIO KEY Round ) Y-intercepts (x 0): (y ) no Y-intercepts ± ( ) X-intercepts (y 0): x x ±,0 B) Completing the square, 5 x+ + 5y P(,5) x + x+ + y x 5y 5x 9 ( /, 0) 4 The distance from this point to x + 4y + 0 can be computed by the point to line ( / ) + 4(0) + 0 distance formula, r d 5 Q(,7) 8 4 P(,5) Q(,7) 4 V(,) V(,) R F(,) C) The points P(, 5) and Q(, 7) lie on the same side of the axis of symmetry. The parabola must open up or to the right. The slope of VP is and the slope of PQ is / Since the slope is decreasing as we move from left to right, the parabola must open to the y 4 p( x ). Substituting P(, 5), we have ( 5 ) 4 p( ) p. Thus, the focus is at (, ) and the slope of QR 7 is 4 and the equation of QR 4 y+ 5 is x 4y 5 or x. Substituting, ( ) 4 4 y+ y 5 ( y ) y 4 y y+ 7 0 ( y )( y 7) 0 y / (x, y) 9, 9 right and, therefore has the form ( )

10 CO TEST 4 - J URY 00 SOLUTIO KEY Round ) x 5x 0+ 5x ( )( ) x x+ 5 (+ x) x 5 x + x 45 4x B) Given: 9 4x 7 and x The right hand side is 8. 7 The left hand side is Equating, 5. (x )(x+ 5) (x+ 5) (+ x)( x) (+ x) 7 7 for x. C) Verify that to avoid division by zero, we require that x 0,, or x 8 5 ( + x) 8 x 8x x x x x x 8x Cross multiplying, x 4x 8x 4x 8x x x 4+ x x x ( ) Multiplying through by (x ), 9( x ) x( x ) x. 9x 8 x + x x x 9x + 8 (x )(x ) 0 x, 4 x 8x 4+ x

11 CO TEST 4 - J URY 00 SOLUTIO KEY Round ) x 0 ( x 0) and () x both produce 0. By inspection, x π/ solves the equation. But is it the smallest? Taking the log of both sides, we have xlog(sin x ) log() 0 Since x > 0, log(sin x) 0 sin x x π/ +nπ and π is the smallest solution. B) tan x tan( x) ( tan x) tan x tan x 0 tan x tan x 0 tan x tan x (x lies in quadrant ) cos(x) (-, ). C) Potential extraneous answers: (cosθ ) θ n sinθ ( + cos θ ) sin θ ( + cosθ + cos θ ) cos θ + cos θ + cos θ 4cos θ + cosθ + 0 cos θ + cosθ + (cos θ + )(cos θ + ) 0 cos θ / θ 0, 40 or cos θ θ 80 (extraneous) Checking: θ 0 : / + ( / ) / / (extraneous) θ 40 : / + ( / ) / / (ok) sin x x lternate solution: Using the identity tan + cos x, Round 4 sinθ sinθ tan( θ / ) θ tan + cosθ + cos ( θ) θ 0 + 0n 00 θ 40 only ) a + a Equal root discriminant 0 4()(b) 0 b 9/8 c 0 c 5 9 Thus, abc or.5 4

12 Round 4 - continued CO TEST 4 - J URY 00 SOLUTIO KEY B) Rewrite equation as y + (k 5k )y 7 0 To have roots that are numerically equal and opposite in sign B must be 0. [Then the roots of x B± B 4C ± 4C + Bx + C 0 would be ] Therefore, k 5k (k + )(k ) 0 k, C) Vertex at (, ) and a vertical axis of symmetry equation of the parabola must be of the y+ a x form ( ) ( ) Substituting in the equation of the line (x y + 7 0), x y. Substituting in the equation of the parabola, 9a a / Expanding, ( ) y + x C + 7 lternate solution (longer, but more straightforward): P(, ), Q(7, ) and V(, ) () 49+ 7B+ C Substituting in the quadratic y x + Bx + C, () 4 B+ C () + B+ C () () B B () () B B 4 dding, 4 (, B), Substituting in (), 4 C + 7 D Round 5 ) Let K denote the area of DE. DE ~ CBE E 4 CE 0 5 area( DE) area( CBE) 5 B E 4 K K C

13 Round 5 - continued B) CO TEST 4 - J URY 00 SOLUTIO KEY Study the diagrams at the right: t t short : Blong t : : FG C) G F G area( DE) area( BC) 9 area( DECB) 8 area( BC ) 9 area( DEC ' B ') 4x 4 area( BC) x 4x 9 Round ) 8 : 4 : ( ) (long diagonal) 4t t t 0 t 0 t t area( FD) area( BC) 8 : 8 : B (short diagonal) t 0 t B 0 t D B' F x x x G x E C' x C B) ( a b ) a+ b a b ( a b) ( a b) ( )( ) + a b a b + a b b b a ( a b)( a b) + ab + b b a ( a+ b)( b a) ab a( a + b) Speeder 00 mph. Trooper mph.. C) mi/min 4 00 mi/hour 00 mi/hr 4 mi. / 4/ - / 7/ T headstart Let t denote the fraction of an hour required to catch the speeder (after the trooper reached his cruising speed.) 4 t 00t+ 7 t 8 hour 0 0 minute minutes 0 sec. Therefore, it took exactly 4 minutes 8 seconds to overtake Mario. S

14 Team Round ) x 4p(y p), where p > 0. CO TEST 4 - J URY 00 SOLUTIO KEY Substituting (5, 5/), 5 4p(5/) + 8p 8p 0p 5 (4p + 5)(p 5) p 5/ Thus, x 0(y 5) x 50 span 0 V(0,p) F(0,p) (0,0) (5, 5 ) B) Since C B and C is prime, B must be. Likewise C C C. ( )( ) ( )( ) 5 ( ) ( ) x+ y+ Bx+ Cy+ D x+ y+ x+ y+ D x + xy+ y + D+ x+ D+ y+ D Since D is prime, we must examine two cases: ) and D is prime ) D and is prime The first case requires (C 4, C 5 ) (D +, D + ). D fails, but D (5, 7) ny other prime values of D will be odd and this forces C 5 to be an even composite number. The second case requires (C 4, C 5 ) ( +, + ). Both and fail. Likewise, any other prime values of will be odd and this forces C 5 to be an even composite number. Therefore, the only ordered triple is (5, 7, ) C) ( cos 4 4x sin 4 4x)( sin x) ( cos 4x+ sin 4x)( cos 4x sin 4x)( sin x) Since the first factor is always equal to, it can be ignored and the original equation simplifies to (cos 8x)(cos x) 0 π (n+ ) π 8x + nπ x π , π, π, π, π, π, π, π π (n+ ) π x + nπ x π, π π π q Thus, p and q 4 p π 4 π

15 Team Round - continued CO TEST 4 - J URY 00 SOLUTIO KEY D) Suppose the original pane had R rows and C columns. Then: 5R R RC (R + )(C 5) C 5R + C R + The smallest value of R that returns a positive integer value for C is 9. [ (R, C) (9, 4) ] Using slope, we create a table of values until the product RC is a perfect square. R C Using this lookup table, the last ordered pair gives us RC ()(49) ( 7) 4 4 The values satisfying this relationship get quite large very quickly. The next three values satisfying RC may be determined with a calculator or spreadsheet. They are: (4)(59) 8 (44) (05)(4) (0) and (988)(4) 4 8 (5) E) Since BK, BKF and BFC have a common altitude (from B), their areas are in the same ratio as their bases, namely K : KF : FC. K Let K x, KF 7x and FC y. Let DE a and CE a. F Since BF ~ CEF, their areas are in a 9 : ratio and CF CE y a D E y x. F B 9x a Let area( BK), area( BKF) 7 and area( BFC). Then area( CEF) and area ( DC). area( FDE) Thus, area( BCD ) 4 4 C B

16 Team Round - continued F) ssign coordinates to the 4 points. CO TEST 4 - J URY 00 SOLUTIO KEY...., 0 B, 0 C, 00 D, 00 + x () start at 0, end at 00 (/ between 0 and 00) () start at 00, end at x (/ of the way to 00 + x) 00 ((00 ) 00) x 00( ) + (00 + x)( ) + + x or by weighted average + () start at x 50+ x, end at (/4 of the way to 0) x 400+ x 400+ x 40+ x 00+ 4x 40 x 50+ x 50+ x x 0( ) + ( ) 50 x By weighted average Midway between and D 50 + x 00 + x 50 + x x x 0 4 BD yards

17 ddendum to the original contest 5C prime omitted question adjusted after the contest so answer as intended ( : 8 : ) area( FD) : area( DEC ' B) : area( DECB) nswer to original question: : : Round changed after contest and note added to B Negative exponents should be avoided in algebra contest at this time of year. question actually asked 8 ns: / B - added to the original question: Note: For any real number x 0, x is equivalent to x.

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

* O CALCULATORS I THIS ROU D *