WISCONSIN HIGH SCHOOL STATE MATHEMATICS MEET WISCONSIN MATHEMATICS COUNCIL March 4 8, Solutions

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1 WISCONSIN HIGH SCHOOL STATE MATHEMATICS MEET WISCONSIN MATHEMATICS COUNCIL March 4 8, 2013 Problem set #1 Solutions a b a b 32 a b , , and , so the three ordered pairs are (2, 4), (4, 2), and (16, 1). 2. The center of circle O lies on angle bisector AD 5 3. In triangle AOG, OG 3, so AO 6. Thus the circle lies inside A ABC, and OD AD AO 5 3 6, so (a, b) (5, 6). 3. Since x 2 9x + 3 (x r)(x s) x 2 (r + s)x + rs, we have r + s 9 and rs 3. So r 2 + s 2 (r + s) 2 2rs and r 2 s Therefore, x 2 + bx + c (x r 2 )(x s 2 ), which becomes x 2 (r 2 + s 2 )x + r 2 s 2 x 2 75x + 9, and the ordered pair (b, c) is (-75, 9). B H O D G C Problem set #2 1. Cross multiply and solve: (3+ x)(x + 8) (x + 4)(6 + x) x 2 +11x + 24 x 2 +10x + 24 x 0 2. If 6 and 25 are factors, then so are 2, 3, and 5. Of course, 1 is also a factor, as well as all combinations of multiplying 2, 3, 5, and another factor of 5 together. If you list them out, they are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, and 150. The largest factor is Computing the first few terms of each sequence, we learn that {g i } i!, while {h i } (i + 1)!. When we make a ratio of g 2013, it is 2013! h ! Problem set #3 1. There are a few trigonometric identities needed, all based on definitions of the functions or variations on the most common identity of sin 2 θ + cos 2 θ 1:

2 1 tan 2 θ +1 + cos2 θ csc 2 θ 1 1 sec 2 θ + cos2 θ cot 2 θ cos 2 θ + cos2 θ cos 2 θ sin 2 θ cos 2 θ + sin 2 θ 1 2. The probability of choosing six people with different birth months involves multiplying six fractions together representing the number of months to choose from divided by the number of months available to pick from: To get the probability of 1728 choosing at least 2 people with the same birth month, we subtract this result from 1, to get U 3. Label the inner triangle ABC, and denote the area of that triangle as [ABC], which is pq T. Additionally, we have 2 [ABWV] q 2, [BCSR] p 2, and [ACTU] p 2 + q 2 (from V A the Pythagorean Theorem). Finally, [VAU], [CST], and q [BRW] are all pq W, so [RSTUVW] is the sum of each of B p 2 C these seven regions: 4 pq 2 + p2 + q 2 + ( p 2 + q 2 ), which becomes 2p 2 + 2pq + 2q , or p 2 + pq + q ( p + q) 2 pq, or 961+ pq ( p + q) 2. Since , R S we have p + q > 31. If p + q 32, then pq 63, and no such combination exists. The same occurs for p + q 33 and 34, but for 35, (p, q) (24, 11) works ( ) Therefore, p 24 and q 11. Problem set #4 1. Since log a b is not defined, that means a 1. Therefore, we need the value of log b b There are 120 (5!) different permutations for all of these 5-digit numbers. In one-fifth of these numbers, the ones digit is 1, one-fifth of these numbers has the ones digit of 2, and so on. Therefore, adding the ones place together, we have 24( ) 360. The same thing happens for the tens digits, the hundreds digit, the thousands digit, and the tenthousands digit. The sum for all of these digits in their places, written from left to right, is 3,600, , , , ,999,960.

3 3. Shown here is one of several methods to solve this problem. First, we rewrite x + y + z 0 as x + y -z. We then cube both sides to get x 3 + 3x 2 y + 3y 2 x + y 3 -z 3. Rearranging the equation, we get x 3 + y 3 + z 3-3x 2 y 3y 2 x. We know that x 3 + y 3 + z 3 3, so substituting, we get -3xy(x + y) 3, or xy(x + y) -1. This also tells us that both xy 0 and x + y 0. Now take the fifth power of x + y -z to get x 5 + 5x 4 y +10x 3 y 2 +10x 2 y 3 + 5xy 4 + y 5 -z 5. Rearranging the equation gives us x 5 + y 5 + z 5-5xy( x 3 + 2x 2 y + 2xy 2 + y 3 ). We know that x 5 + y 5 + z 5 0, so we get -5xy x 3 + 2x 2 y + 2xy 2 + y 3 ( ) 0, or xy( x 3 + 2x 2 y + 2xy 2 + y ) 3 0. Also, we know that xy 0, so we get x 3 + 2x 2 y + 2xy 2 + y 3 x 3 + y 3 + 2xy(x + y) 0. We know that xy(x + y) -1, so this simplifies the previous equation to x 3 + y 3 2. Finally, we know that x 3 + y 3 + z 3 3, so z 3 must equal 1. By symmetry, x 3 y 3 1. Since 2013 is divisible by 3, x y z To show that this value is possible, let x 1, y i, and z i, which are the cube roots of 1. Then x + y + z x 5 + y 5 + z 5 0, and x 3 + y 3 + z 3 x y z Team Problem set 1. cos3α cos(2α + α ) (1 2sinα )cosα sin2α sinα cosα(1 2sin 2 α 2sin 2 α ) cosα(1 4sin 2 α ) sin3α sin(2α + α ) sin2α cosα + sinα cos2α 2sinα cos 2 α + sinα(1 2sin 2 α ) sinα(2cos 2 α +1 2sin 2 α ) sinα(2cos 2 α 1+ (2 2sin 2 α )) sinα(2cos 2 α 1+ 2cos 2 α ) sinα(4cos 2 α 1) Therefore, cos3α cosα 1 4sin2 α 1 3. Therefore, sin3α sinα 4cos2 α 1 4(1 sin 2 α ) 1 4 4sin 2 α 1 3 4sin 2 α 2 + (1 4sin 2 α ) Here is another method to solve the problem. Start with the difference between the two fractions sin3α cos3α and sinα cosa :

4 sin3α sinα cos3α sin3α cosα cos3α sinα cosα sinα cosα sin(3α α ) sinα cosα sin2α sinα cosα 2sinα cosα sinα cosα 2 sin3α sinα cos3α cosα sin3α sinα sin3α sinα First of all, a < 200, otherwise c > 1000, which will not be a 3-digit number. Also, a cannot be even, otherwise c 5a ends in a 0, which is also not allowed. Try the first few smallest possibilities for a: a b c digits 3 and 6 are repeated digits 1 and 3 are repeated This works! 3. Draw line m through C parallel to. Draw line n through A parallel to. Call a, b, and c the radii of circles A, B, and C respectively. F B. E. C A. D n m AD a c and AC a + c, so DC BE b c and BC b + c, so CE Therefore, DE 2( ac + bc ). AB a + b and BF a b, so AF (a + c) 2 (a c) 2 2 ac. (b + c) 2 (b c) 2 2 bc. (a + b) 2 (a b) 2 2 ab. Since AF DE, we can write the following: 2 ab 2( ac + bc ) ab ac + bc ab ac + bc + 2 ac ab ac + 2c ab + bc bc

5 ab c( a + 2 ab + b) ab c a + 2 ab + b 4. Compute a few small values of the function to get started: f (1) f (0) f (1) + f (1) 3 f (0) 3+ 3 f (0) 2 f (1) f (1) f (2) + f (0) 3 3 f (2) + 2 f (2) 7 f (2) f (1) f (3) + f (1) 7 3 f (3) + 3 f (3) 18 f (3) f (1) f (4) + f (2) 18 3 f (4) + 7 f (4) 47 f (4) f (3) f (7) + f (1) f (7) + 3 f (7) Two methods of solving this problem are presented here, one is by direct computations, the other purely symbolic. Both involve recognizing a telescoping sum, which when you see how the pattern works, it uses the first and the last terms of the sum, with most of the middle terms cancelling out. The direct computation method will be presented first and so on to the last term in the expression, which is written as The sum becomes The following method is the symbolic method. Let the sum be denoted by S: S 1+ 1 a (a +1) 2 a 4 + 2a 3 + 3a 2 + 2a +1 a 2 (a +1) 2

6 (a 2 + a +1) 2 (a 2 + a) 2 a 2 + a +1 a 2 + a 1 1+ a 2 + a a 1 a +1, a telescoping sum 6. If all the judges gave 1 st place to the same participant, then c 1 9. If exactly 2 people got 1 st place, one of them got at least 5 ones. His score is at most (the worst he can get from the other judges). So c If 3 people got 1 st places, the sum of their scores is at most So c If 4 people got 1 st places, then the sum of their scores is at least , and c Five or more 1 st places is not possible. Example for c 1 24: Each of the top 3 gets the scores 1,1,1,3,3,3,4,4,4 (not in the same order). The next gets 2,2,2,2,2,5,5,5,5, and the next gets 5,5,5,5,5,2,2,2,2. The rest get arbitrary grades with difference 3. This gives c 1 24.

7 2013 Wisconsin Mathematics Council State High School Math Meet Feedback Form Dear teachers and students: We at the Wisconsin Mathematics Council would like to get your feedback on this year s State Math Meet. Please take a moment to fill out this form and return it with the other materials that you need to return to WMC. Your input will help to shape the future of these math meets to make them a more enjoyable and enriching experience. 1. Were there any questions you felt were too easy or too difficult? Which one(s) and why? 2. Were there questions on math subjects you were expecting to see, but did not? If so, what did you expect to see? 3. Were there questions on math subjects you saw but did not expect to see? If so, which ones? 4. Was there enough time to finish the team questions? 5. Did you like the first event, where calculators were not allowed? Were they too easy, too hard, or just the right amount of difficulty? 6. We try to include a wide variety of problems with some interesting ways to solve them. Were there any questions that sparked further discussion among students and teachers as a result of seeing them here? Which one(s)? 7. Any additional comments you would like to make about this year s problem set or the meet in general: Thank you for taking the time to complete this feedback form. Your comments are invaluable to us. We hope to have you compete again with us next year!

Given an arc of length s on a circle of radius r, the radian measure of the central angle subtended by the arc is given by θ = s r :

Given an arc of length s on a circle of radius r, the radian measure of the central angle subtended by the arc is given by θ = s r : To convert from radians (rad) to degrees ( ) and vice versa, use the