MATH TOURNAMENT 2012 PROBLEMS SOLUTIONS
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1 MATH TOURNAMENT 0 PROBLEMS SOLUTIONS. Consider the eperiment of throwing two 6 sided fair dice, where, the faces are numbered from to 6. What is the probability of the event that the sum of the values of the two dice is 7? (a) (b) 9 (c) 6 (d) 9 (e) There are 6 possible outcomes of which 6 outcomes (,6), (,), (,), (,), (,), (6,) with sum 7. Therefore P 6/6 /6. Find the last digit of the sum 0! +! +! + L + 00! + 0!. (a) (b) (c) (d) 7 (e) 9 After 6!, all of them end with 0. It suffices therefore to just look at the first terms whose sum is Therefore the last digit must be 7.. Squares ABCD and ADEF are in perpendicular planes. If AB, find FC. (a) (b) (c) (d) (e) This problem is basically asking for the length of the main diagonal of a cube of side length. Pythagorean Theorem yields: FC FA + AC FA + AB + BC which implies FC. The probability distribution of your winnings at a casino's card game is shown below.
2 X $0 $ $0 $ P(X) How much should you epect to win if you play the game once? (a) $ (b) $7. (c) $0 (d) $. (e) $ XP (X ) $. Working alone, Al can wash a hotel restaurant s dishes in three hours. Jenny can do the same job in si hours. If they worked together, how long would it take them to complete this task? (a) 9 hours (b) 9 hours (c) hours (d) 9 hours (e) 9 hours hours 6. What is the sum of all positive integer divisors of 0? (a) 0 (b) 7 (c) 7 (d) 8 (e) How many roots of the equation lie in the 0,? interval [ ] (a) 0 (b) (c) (d) (e) The graph of the function which lie in the interval [0, ] f ( ) has -intercepts, all of
3 8. Polynomials P and Q satisfy the equation P ( ) ( + ) Q( ) for all real numbers. When P () is divided by ( ) the remainder when Q () is divided by ( )., the remainder is 0. Determine (a) 0 (b) (c) (d) (e) The remainder when P () is divided by ( ) is P ( ) 0. The remainder when Q () is divided by ( ) is Q (). Setting in the original equation, we get: P ( ) P() ( + ) Q( ) Simplifying yields 0 6 Q () 6 Solving for Q () we get Q ( ). 9. The following is a rectangle made of three congruent squares of side length 8 units. Determine m A + m B + m C. (a) 67. (b) 78.7 (c) 90 (d) 0. (e). Angle C is one of the base angles of a right isosceles triangle. Therefore m C. To determine m A + m B, we use the tangent of the sum formula: tan A + tan B tan( A + B) 6 6, which implies that tan A tan B m A + m B as well. Therefore m A + m B + m C The rational function f ( ) is defined for all real numbers ecept for ±. Given that it has a vertical asymptote a and a slant asymptote y m + b, determine the sum a + m + b. (a) (b) (c) 0 (d) (e)
4 8 The rational function f ( ) has a vertical asymptote and a slant asymptote y. Therefore, the sum a + m + b n+ n+ n+. If N + + 8, what is the smallest positive integer such that the product N is a perfect square for all positive integers n? (a) (b) (c) (d) 7 (e) N 6n+ + n+ + 8 n+ 6n+ + 6n+ + 6n+ 6n+ 6n+ n+ ( + + ) 7 7 ( ) Therefore, the smallest positive integer such that the product N is a perfect square for all positive integers n must be 7.. Let A {,, } and B {,,,,,6,7 }. How many subsets of B contain A? (a) 6 (b) (c) 8 (d) 96 (e) B A {,,6,7} n( B A) Therefore, the number of subsets of set B A is 6. Because ( B A) A B, if we add A to these 6 subsets, we conclude that the 6 subsets of B contain A.. Determine the value of the product ( )( tan )( tan ) ( tan 88 )( tan 89 ) tan K. (a) 0 (b) (c) π (d) π (e) 6 π Using the fact that tan cot(90 ), we see that all terms cancel ecept the tan(90 ) term in the middle, tan Determine the solution set of the equation log ( 9 ) (a) {,} (b) { 0,} (c) { 0,} (d) {,0} (e) {,}.
5 Converting to the eponential form, we obtain ( ) Case : < 0: Solve + Case : 0: Solve + 9. Quadrilateral ABCD is inscribed in a circle in such a way that eactly one pair of opposite sides intercept congruent arcs. Being as specific as possible, what type of quadrilateral is ABCD? (a) Square (b) Rectangle (c) Kite (d) Trapezoid (e) Parallelogram By parallel chords intercepting congruent arcs theorem, the figure must be a trapezoid. (a) 6. Let Evaluate the epression ( ) (b) ( ) 0 (c) (d) 6. 0 (e) Therefore ( ) 0 ( ) ( ) In space, which of the following propositions is false? (a) A line parallel to one of two parallel lines is also parallel to the other line. (b) Three parallel lines may not be coplanar. (c) A line intersecting one of two parallel lines must intersect the other line as well. (d) Through a point not on a given plane, there is eactly one plane parallel to the given plane. (e) Through two distinct points in a plane, there is eactly one plane perpendicular to that
6 plane. All propositions are true ecept (c). 8. In the eperiment of selecting marbles without replacement from a bag containing 6 yellow and red marbles, what is the probability of getting yellow and red marbles? (a) (b) 0 (c) (d) (e) What is the product of all integers n such that n 7 is an integer? (a) - (b) - (c) -6 (d) - (e) -76 The epression n 7 must be a perfect square. Let n 7 k, with k integer. Then n k 7 ( n k)( n + k) 7, which will give the following systems: n k 7 n + k n k 7 n + k n k n + k 7 n k n + k 7 which will happen only if n ± 9. Therefore the product is How many real solutions does the equation ( ) + have? (a) 0 (b) (c) (d) (e) Case : ± Case : , Case : + is even. + is even. ± +, is odd, namely false. Therefore, there are only real solutions. 6
7 n n. Let En k 0 k with n a natural number. Evaluate the difference E n+ En. n (a) (b) n (c) n (d) + n n (e) + n+ n n + n n+ n n En+ En k 0 k k 0 k. Evaluate the epression ( n) 0 8 n m mn. (a) 0 (b) 006 (c) 0 (d) 770 (e) n m 0 8 ( mn n) n( m ) n ( m ) ( ) n m 0 8 n m If a projectile is fired with velocity v at an angle θ, then its vertical displacement y as a function of its horizontal displacement is modeled by the parabola y + ( tanθ ). Find an epression for the range R (the maimum v cos θ horizontal displacement) of this projectile. (a) v sin θ 0 (b) v sinθ (c) v sinθ cosθ 0 v cos θ (d) 0 v sin θ (e) 0 Setting y 0 and solving for, we get: v cos θ tanθ + tanθ 0 0, v cos θ sinθ v cos θ v cos θ tanθ Simplifying, the range is cosθ v cosθ sinθ v sin θ 0. When a camera flash goes off, the batteries immediately begin to recharge the ( ) t / Q t Q e.(the flash's capacitor, which stores electric charge given by ( ) 0 7
8 maimum charge capacity is Q 0 and t is measured in seconds). How long (in seconds) does it take to recharge the capacitor to within 0% of its capacity? (a) ln (b) 9 ln (c) ln0 (d) 0 ln (e) 0 ln 9 t / Solving the equation 0.9Q Q ( e ) for t, we get 0 0 t ln 0. ABC is an isosceles right triangle. BC is etended such that B, C, and D are collinear, and O is the midpoint of AC. If BD AC, find OD. (a) (b) (c) (d) (e) Because ABC is a right isosceles triangle and AC, the Pythagorean Theorem gives AB BC. Moreover, m BAC m BCA. AO OC. DB DC. 8
9 Applying Cosine Law for triangle DOC yields: 6. A regular polygon of side length is inscribed in a unit circle. What is the number of sides of this regular polygon? (a) 0 (b) 8 (c) (d) (e) Consider the regular polygon inscribed in the unit circle with center O: Let OA OB OC. We also have AB BC. Let m AOB m BOC Writing Cosine Law for triangle AOB: AB AO + BO AO BO cos + cos cos / 0 Therefore, the number of sides of the regular polygon is 60/0. 7. A by 0 chess board is colored in grey and white patterns as shown. How many grey squares are there of the total of 0 squares of this chess board? 9
10 (a) (b) 0 (c) (d) (e) For every by 6 rectangle, there are 6 grey squares. Dividing 0 by 6 yields a quotient of 8 and a remainder of. Therefore, there will be grey squares. 8. ABCD is an isosceles trapezoid with perpendicular diagonals and base lengths DC and AB 6. Determine the area of this trapezoid. (a) (b) 6 (c) 8 (d) 0 (e) Both ABE For ABE and DEC are isosceles right triangles. : The altitude drawn to side AB must be half length of side AB. For DEC : The altitude drawn to side DC must be half length of side DC. Therefore, the altitude of the trapezoid is of length The area of the trapezoid 6 9. In a circle centered at O, radius OR is perpendicular to chord AB at point C. If OC and CR, find AB. (a) 8 (b) 0 (c) (d) (e) 6 0
11 ACO is a right triangle with hypotenuse length AO OC + CR 7. Using the Pythagorean Theorem: AC + CO AO AC + 7 AC 8. AB AC 6, because, in a circle, the perpendicular drawn from the center to any chord also bisects that chord. 0. ABC is an equilateral triangle with side length 6. The point O lies on line segment AB. Ray CT is tangent at the point T to the circle of radius 7 centered at O. Determine the length of the line segment CT. (a) 0 (b) (c) (d) (e) Create point H, intersection of the circle with segment BC. The triangle that forms, DBH, is a triangle, with m DHB 90. Therefore BH 7 as well. BC 6, therefore, HC Using the power of point theorem yields: CT CH CB 9 6 CT. The Greek mathematician Archimedes liked the design below so much that he wanted it on his tombstone.
12 When each of the figures is revolved about the vertical line of symmetry, it generates a solid of revolution a cylinder, a sphere, or a cone. Find the ratio of volumes V : V : V. CYLINDER SPHERE CONE (a) : : π (c) : π : (b) : : π (d) : : (e) π : π : Assume the square is a unit square. Using the volume formulas, we get: V π r h π π CYLINDER VSPHERE π r π π VCONE π r h π Therefore, the ratio of volumes is : :.. Find the area of a square with the same perimeter as a regular heagon of area square units. (a) 9 (b) 6 (c) 8 (d) (e) Let a be the side length of the regular heagon. Then solving the equation 6a for a yields a 6. The perimeter of the heagon 6. The square has the same perimeter; therefore, the side length of the square is 9 units. The area of the square is then 8 square units.. In how many ways can Team USA s season of 0 soccer games result in wins, draws, and losses in the 0 World Cup qualification?
13 (a) 60 (b) 600 (c) 0 (d) 909 (e) !!!! 0. The Golden Ratio ϕ is a positive number that satisfies the quadratic equation 0. By substitution, all positive integer powers of ϕ can be epressed in the form a ϕ + b. Determine the value of a + b for ϕ. (a) (b) (c) 8 (d) (e) ϕ ϕ ϕ ( ϕ + )( ϕ + ) ϕ + ϕ + ϕ + + ϕ + ϕ + Therefore a + b. On square ABCD with side length 0, points E, F, G, and H on each side are chosen in such a way that AF BG CH DE 09. Determine the ratio of the area of the square ABCD to the area of the quadrilateral IJKL. (a) (b) (c) (d) (e) AF FB BG CH DE GC HD EA 09 0 By symmetry, the four triangles ABJ, BCK, CDL, DAI are congruent. Moreover, all these four triangles are right triangles. For instance, if m DAI β 90 α. m BAJ α, then
14 Hence all the angles at I, J, K, L are right angles. If CL a, CK b, then LK CL CK a b. Similarly each of the other sides of IJKL equals a b, hence it is a square. AF FB implies AI IJ Compute the area of square ABCD in two different ways: AJ JB AB + JK AJ IJ JK JB JK Substituting these in the above equation we get: JK JK AB + JK JK AB Namely JK 6. A semicircle of radius with center M (0,) and a quarter circle of radius centered at the origin intersect at a point A (, y) in the first quadrant. Find the value of.
15 (a) (b) (c) (d) (e) By symmetry, the midpoint of OM has the same y-coordinate as point A. Because point M has coordinates ( 0,), the value of y must be /. To determine the value of, we can use the equation of either circle. + y y + + ± Because A is in the first quadrant,. 7. A locus is the set of points that satisfy a given condition. Determine an equation of the locus of comple numbers z + iy satisfying the equation z + i 0. (a) ( ) + ( y ) 00 (b) ( ) + ( y ) 6 (c) ( + ) + ( y ) 00 (d) ( ) + ( y ) 6 (e) ( ) + ( y ) 00 ( + ) + ( ) 00 z + i 0 ( + ) + i( y ) 0 y 8. Consider the eponential function f ( ) a where a is a positive number different from. For what value of a does the slope of the tangent line to the graph of f at its y-intercept equal? (a) 0. (b) (c) (d) e (e) π This is the definition of natural eponential function, therefore, (a) a e. 9. Find the coordinates of the point on the circle + y that has the minimum distance from the line + y , Let P(, ) (b) 8 6, 6 (c), 6 (d), ± be a point on the circle. Using point-line distance formula, the distance from P to the line + y 0 is given by: (e), 6
16 ± ± d + Applying the sign test on the epression in the numerator yields ± m + In other words, sign test yields that the epression in absolute value is always negative. It will suffice to minimize this epression. Let f ( ) m + f ( ) ± Solving f () 0 ± ±. Plugging these values in d above, we get: 6 6 ± 6 d 0. or ± 6 d.8 or. 6 Namely d is a minimum for. The corresponding y value is If cos sin, what is the value of cos if is in the interval 0, π? (a) 7 (b) (c) (d) (e) cos sin sin / ( cos sin ) cos sin sin cos / sin cos 7 /6 7 / / 6
17 . Compute i n k k where i and n is the largest perfect cube less than 00. (a) 0 (b) (c) (d) (e) k i k i i k Therefore, i 0 as well. k. Given that the real-valued function f satisfies the functional equation f ( ) f for all > 0, find the minimum value of f (). (a) (b) 7 (c) (d) (e) f ( ) f f ( ) f f ( ) + f f ( ) f 6 f ( ) f ( ) Now and / are two numbers with geometric mean, and arithmetic mean. Solving the inequality G. M. A. M. for f () yields / as the minimum value of f (). 0. Find the sum k k k + + ( k + ). k (a) 0 (b) 0 0 (c) 0 (d) 0 0 (e) The summand can be written as 0 k k k + 0. k k
18 . If a >, what is the value of the following limit? lim 0 a + a a a (a) a (b) a + (c) (d) e (e) a We have the indeterminate form 0/0 as goes to 0. Applying L Hospital s Rule: lim 0 a + a a a lim 0 / a / a + / a ( + a) ( a) (/ a) a( + a) + (/ a) a( a) lim 0 + / a. Determine the value of the following limit. + lim + (a) (b) (c) e (d) e (e) e 6. What is the value of the largest possible real number δ such that + < 0. whenever < δ? (a) /8 (b) /8 (c) 7/8 (d) /6 (e) /6 + < 0. / < < / / < + < 7 / / < + < 9 / /6 < < /6 /6 < < /6 < /6 7. Let a, b, c be positive integers. How many ordered triples ( a, b, c) satisfy the equation a + b + c 0? (a) 0 (b) 8 (c) (d) (e) 0 8
19 n m 8. Let X + m, n {0,,, L,00}. Find the sum of the elements of X. (a) 9 (b) 9 (c) 9 (d) 9 (e) 9 n m n + m y +. Let y n + m. such that y { 0,,, L,700}. But y cannot 0 0 assume all these values. For instance, it is impossible to epress in the form n + m. Similarly, it is impossible to epress, 699, and 697 in this form. Therefore, the sum of 700 k ( ) k the elements of X is ABC is a triangle with BC. A line crosses lines BC, AC and AB at points L, N NA and M respectively, such that LM, MN, LB. Determine the ratio. NC 9
20 (a) /7 (b) 7/ (c) 6/7 (d) / (e) / Using Menelaus Theorem: 0. The figure below shows a tessellation of the plane with equilateral triangles of side length. In the eperiment of tossing a coin of diameter, what is the probability that the coin does not overlap with any triangle verte? (a) 8 π 7 (b) 8 π 8 (c) 8 π 9 (d) 8 π 0 (e) 8 π Consider one of the equilateral triangles: 0
21 The only way the coin will not overlap with any triangle verte will happen if the coin lands in the yellow region shown in the diagram above. Therefore, it suffices to determine the area of this yellow region: The three sectors are each of 60 degree central angle, summing to a 80 degree central angle, i.e., a semicircle. Therefore, the area of the yellow region equals the area of the equilateral triangle of side length the area of a semicircle of radius /. π (/ ) 8 π Dividing this value by the area of the equilateral triangle will give us the desired probability: P π 8 π 8 8 π 8
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