2014 Stark County High School Mathematics Challenge ANSWER KEY

Transcription

1 2014 Stark County High School Mathematics Challenge ANSWER KEY April 26, 2014

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3 TEST 1 Instructions You have ONE HOUR to answer each of the following ten questions to the best of your ability. Each question is worth ten points. Please show all of your work. Partial credit will be awarded for sound reasoning and partial solutions. There will be no talking during the test. If you have any questions as to the wording of a problem, please raise your hand and the proctor will assist you. No other questions will be answered. All tests will be collected promptly at the end of the allotted time. If you finish early, please close your test booklet and quietly exit the room. No one will be re-admitted to the testing room once they have left, i.e., no bathroom breaks will be allowed. Anyone that is deemed by the proctor to be talking, behaving disruptively, or cheating will be asked to leave immediately and will receive a score of zero for the entire competition. You have a short snack break after the test is completed. Good Luck! Do your best! 3

4 4 Problem 1 Find a so that the graph of y = log a x passes through the point (e, 2). We have 2 = log a e. Switch to exponential notation to get a 2 = e. Therefore, a = e or a = e 1/2 (recall that the logarithm base cannot be negative this is why a = e is discarded). Partial Credit: 7 points if a = e is kept as a solution.

5 Problem 2 Napier s logarithm, denoted as Nap.log, is defined as follows: if x is a real number such that x = 10 7 ( ) m for some other real number m, then Nap.log x = m. Prove that Nap.log xy = Nap.log x + Nap.log y Nap.log 1. [Hint: set Nap.log x = m, Nap.log y = n and Nap.log 1 = k, and use the definition of Nap.log.] 5 We have x = 10 ( ) m, ( y = ) n, ( 1 = ) k, implying that xy = 10 ( ) m+n 10 7 = 10 ( ) m+n k (since 10 7 = ( ) k ). Therefore, Nap.log xy = m + n k and the assertion follows. NPC (= No Partial Credit)

6 6 Problem 3 Simplify ( ) log Substitition 1 2 = 8 1/3 gives x = log 8 3 = 8 log 8 3 1/3 = 3 1/3 = 1 3 1/3 NPC

7 7 Problem 4 If a 2 + b 2 = 7ab, prove that log a + b 3 = 1 (log a + log b ). 2 The assertion is equivalent to that is, to log a + b 3 = log ab, a + b = 3 ab, (a + b) 2 = 9 ab. This is indeed what we can get from the assumption: a 2 + b 2 + 2ab = 9ab, (a + b) 2 = 9ab, (a + b) 2 = 9 ab (since the left-hand side is non-negative). Partial Credit: 5 points for getting (a + b) 2 = 9 ab.

8 8 Problem 5 There are 7 white balls and 3 red balls in a bag. On each turn we remove a ball randomly from the bag, and then place a new white ball back in the bag. Find the probability that it takes exactly 4 turns to remove all of the red balls from the bag. On turn 4 we must pull the third red ball. List all the possible sequences of drawing balls in which we pull out the third red ball on the fourth turn. The possible sequences are WRRR, RWRR, and RRWR. Since we replace every ball that is pulled out with a white ball, the probability of the first sequence happening is Similarly the 10 4 probabilities of the remaining sequences happening are , and , respectively. By adding all these probabilities together we have = Partial Credit: 3 points for getting the probability of the sequence WRRR right.

9 9 Problem 6 Find the simplified solutions of the equation ax 2 + bx + c = 0 when a + b + c = 0. 1 Obviously, x = 1 is a solution. Then, (x 1) is a factor of ax 2 + bx + c and the other factor must be (ax c), which gives x = c/a for the other solution. 2 Use b = a c to get ax 2 ax cx + c = 0 and factor by grouping: Then, x = c/a or x = 1. ax(x 1) c(x 1) = 0, (ax c)(x 1) = 0. Partial Credit: 2 points for getting x = 1.

10 10 Problem 7 Find a, b and c so that the graph of the quadratic function f(x) = ax 2 + bx + c has a vertex at ( 2, 1) and passes through the point (0, 3). From the given point we have that c = 3. Also, the x-coordinate of the vertex is b/(2a), so we get the equation b = 4a. Therefore, f(x) = ax 2 + 4ax 3 and, since f( 2) = 1, Then b = 4. NPC 4a 8a 3 = 1, a = 1.

11 11 Problem 8 If a 2 b 2 = 8 and ab = 2, find a 4 + b 4. We have that (a 2 b 2 ) 2 = 64, that is, a 4 2a 2 b 2 + b 4 = 64. Since 2a 2 b 2 = 2(ab) 2 = 8, we get a 4 + b 4 = = 72. NPC

12 12 Problem 9 Prove that 3 2n 1 is divisible by 2 n+2 for any positive integer n. The proof is by mathematical induction. If n = 1, 3 2n 1 = 8, which is divisible by 2 n+2 = 8. Now assume that 3 2k 1 is divisible by 2 k+2 and prove that 3 2k+1 1 is divisible by 2 k+3. We have that 3 2k+1 1 = 3 2 2k 1 = (3 2k) 2 ( ) ( ) 1 = 3 2k 1 3 2k + 1. The first factor is divisible by 2 k+2 by the inductive assumption and the second factor is even, which means that it contains at least one factor of 2. Therefore, the whole expression is divisible by 2 k+2 2 = 2 k+3. Partial Credit: 3 points for getting to 3 2k+1 1.

13 Problem 10 In the sketch below, CD is parallel to AB and the measure of angle t is 90. Find the area of the circle in terms of x. 13 Let E be the point of intersection of AC, BD, and the circle. Since the angle AEB is right, AB is the diameter of the circle. The length of the diameter is 2r, where r is the radius of the circle. The triangles ABE and CDE are similar and we have the proportion and therefore 2r x = 3 5 r = 3 10 x. The area of the circle is πr 2 = (9π/100)x 2 or 0.09πx 2. NPC

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15 TEST 2 Instructions You have THIRTY MINUTES total to answer the following four questions (Step 1 Step 4). Each Step will be scored for correctness. Please show all of your work on the paper. We will look at your work if we find this necessary. The answers to Step 1 Step 4 should be written clearly on the index card provided. Note that each of the steps are connected. That is, if you have a wrong answer to Step 1, it is very likely that you may have incorrect answers to subsequent Steps. If you have any questions as to the wording of a problem, please raise your hand and a proctor will assist you. No other questions will be answered. You must leave the paper with the index card on the table. No bathroom breaks are allowed. Anyone that is deemed by the proctor to be talking, behaving disruptively, or cheating will be asked to leave immediately and will receive a score of zero. 15

16 16 Step 1. Find the midpoint M of the line segment OA with endpoints O(0, 0) and A(4, 2). Show your work on this paper and copy final answer to index card. M((0 + 4)/2, (0 + 2)/2) = M(2, 1) Step 2. Find the slope-intercept form of the equation of the line l perpendicular to OA at M. Show your work on this paper and copy final answer to index card. The slope of OA is 2/4 = 1/2, so the slope of the perpendicular line is 2. This line is y 1 = 2(x 2), or, in the slope-intercept form, y = 2x + 5.

17 Step 3. Find the points of intersection, B and C, of line l and the circle centered at O with radius 10. Show your work on this paper and copy final answer to index card. The equation of the circle is x 2 + y 2 = 10. The system consisting of this equation and the equation y = 2x + 5 is solved by substitution: x 2 + ( 2x + 5) 2 = 10, 5x 2 20x + 25 = 10, 5x 2 20x + 15 = 0, 5(x 2 4x + 3) = 0, 5(x 3)(x 1) = 0, so x = 1 or x = 3. The points are B(1, 3) and C(3, 1). 17 Step 4. Find the equations of the two circles of radius 10, one centered at B and another at C. Show your work on this paper and copy final answer to index card. The circle centered at B is (x 1) 2 + (y 3) 2 centered at C is (x 3) 2 + (y + 1) 2 = 10. = 10 and the circle Remark The above four steps are the steps needed to solve the following problem: Find equations of circles with radius 10 which pass through the points O(0, 0) and A(4, 2).