Westmont College Mathematics Contest Team Exam Grades 9 – 10, February 9, 2008 SCHOOL: Instructions: Put the name of your school in the space above, and on each solution page. Calculators are not permitted. This is a team exam. There are only five questions, so your team will be graded both on the accuracy of your answer and the clarity with which you express your answer. Turn in only one solution per problem per team. Work each problem out on scratch paper at first, then copy the solution you want graded to the space provided below the corresponding question, making sure your writing is legible and coherent. Be sure you justify your answer rather than merely stating what the answer is. Use the back of the paper if you run out of room. When time is up, put all your answers in the envelope provided (make sure your correct school name is on the envelope), and turn it into the exam proctor. Team Exam Answer Key, Grades 9 – 10 Problem 1 Diophantus was a Greek mathematician who lived around AD 250. He is known for his work in algebra, and on his tomb the following epitaph was found. This tomb holds Diophantus. Ah, what a marvel! And the tomb tells scientifically the measure of his life. God vouchsafed (granted) that he should be a boy for the sixth part of his life; when a twelfth was added, his cheeks grew a beard; He kindled for him the light of marriage after an additional seventh, and in the fifth year after his marriage He granted him a son. Alas! Late gotten and miserable child, when he had reached the measure of half his father’s life, the chill grave took him. After consoling his grief by this science of numbers for four years, he reached the end of his life. How old was Diophantus when he died? Solution: Diophantus was 84 years old when he died Explanation: Let x be the age of Diophantus when he died. Then, according to the epitaph, 1 1 1 1 x = x + x + x + 5 + x + 4, 6 12 7 2 7 12 42 14 x = x + x + x + x + 9, 84 84 84 84 75 x = x + 9, 84 x− 75 x = 9, 84 9 x = 9, 84 x = 84. Westmont College Mathematics Contest, February 9, 2008 Team Exam Answer Key, Grades 9 – 10 Problem 2 Adam has made it to Round 2 of the California chess championship and wants to get to Round 3. To do so he must play three more games and win at least two of them in a row. His opponent for each game is to be either Bill or Cheryl. He can choose whom to play first, but then he must alternate opponents. He knows that each time he plays Bill the probability that he wins is 12 , and each time he plays Cheryl his probability of winning is 41 . Whom should he play first, and why? Solution: He should play Cheryl first Explanation: Let’s compute the probabilities for each of the three mutually exclusive ways that Adam can win at least two games in a row depending on the person he plays first. We present the results in the following tables. First, suppose Adam plays Bill first. Outcome Bill Cheryl Bill Probability Win Win Win 1 2 × 1 4 × 1 2 = 1 16 Win Win Loss 1 2 × 1 4 × 1 2 = 1 16 Loss Win Win 1 2 × 1 4 × 1 2 = 1 16 The events are mutually exclusive, so the probability of winning at least two games in a row when 1 1 1 3 starting with Bill as an opponent is 16 + 16 + 16 = 16 . The next table shows the computations if Adam plays Cheryl first. Outcome Cheryl Bill Cheryl Probability Win Win Win 1 4 × 1 2 × 1 4 = 1 32 Win Win Loss 1 4 × 1 2 × 3 4 = 3 32 Loss Win Win 3 4 × 1 2 × 1 4 = 3 32 Again, the events are mutually exclusive, so the probability of winning at least two games in a row 1 3 3 7 when starting with Cheryl as an opponent is 32 + 32 + 32 = 32 . Thus, the rather surprising result is that Adam should play Cheryl first, because the probability of 7 his winning at least two games in a row with this strategy is 32 . If he plays Bill first, probability of 3 6 his winning at least two games in a row is only 16 = 32 . Westmont College Mathematics Contest, February 9, 2008 Team Exam Answer Key, Grades 9 – 10 Problem 3 Mr. RED, Mr. WHITE, Mr. TAN, and Mr. BLUE entered the Grape Juice Cafe. They ordered a red, a white, a tan, and a blue drink, but none of them ordered the same color as his name. They sat at a square table, and no drink name matched the name of the person at the opposite side of the table. Furthermore, Mr. RED was on Mr. WHITE’s right, and Mr. WHITE was to the right of the white drink. The waiter noted all these things and went to collect the drinks from the kitchen. On the way he realized that it wasn’t clear who had ordered what. He knew that either Mr. TAN and the blue drink or the tan drink and Mr. BLUE went together, and that either Mr. TAN and the white drink or the tan drink and Mr. WHITE should be paired off. Who ordered what drink? Solution: Mr. RED ordered the blue drink Mr. WHITE ordered the red drink Mr. TAN ordered the white drink Mr. BLUE ordered the tan drink Explanation: Figure 1 on the right illustrates the initial setup. Suppose, according to the information in the second paragraph, that Mr. TAN and the blue drink go together. Thus, Mr. TAN cannot be with the white drink. Looking at the options after the and in the second paragraph reveals that the only possible scenario is for Mr. WHITE to go with the tan drink. But then Mr. TAN, with the blue drink, would have to be seated across from Mr. WHITE, with the tan drink. According to the rules in the first paragraph this situation is impossible: Mr. TAN cannot be seated across from the tan drink. Our supposition that Mr. TAN was paired with the blue drink lead to an impossible situation, so it must be that Mr. BLUE is paired with the tan drink. From the initial setup, he must be opposite Mr. WHITE. The only person unaccounted for is Mr. TAN, who must be paired with the white drink, as seen in Figure 2. The only drinks left are red and blue. Mr. RED, of course, cannot be paired with the red drink. Therefore, Mr. RED is paired with the blue drink and Mr. WHITE is paired with the red drink. The other persons have already been paired with their drinks, giving the solution listed above. white RED WHITE Figure 1: Initial setup BLUE tan TAN white RED WHITE Figure 2: Required pairs Westmont College Mathematics Contest, February 9, 2008 Team Exam Answer Key, Grades 9 – 10 Problem 4 The center of a circle and two chords are shown in the figure below. The chords are, respectively, 14 and 10 centimeters long, and they are six centimeters apart. a. What is the area of the circle? b. What is the length of the chord that is parallel to the given chords and midway between them? 14 cm 6 cm 10 cm Solution: 50π cm2 √ 2 46 cm (a) (b) Explanation: Let x be the distance from the center of the circle to the 14 cm chord. Thus, the distance from the center to the 10 cm chord is 6 − x, as shown in the figure below. With r as the radius of the circle the Pythagorean theorem gives r2 = x2 + 49 = (x − 6)2 + 25. Hence, x2 + 49 = x2 − 12x + 36 + 25, or 12x = 61 − 49 = 12. Therefore, x = 1. Using the Pythagorean theorem again gives r2 = 50, so the area of the circle is 50π cm2 . The chord midway between the given chords and√parallel to them must be 2 cm from the center. If its length is 2d, then d2 + 22 = r2 = 50, so d = 46. x 7 cm r 6 cm 6−x r 5 cm Westmont College Mathematics Contest, February 9, 2008 Team Exam Answer Key, Grades 9 – 10 Problem 5 In 2007 the U.S. began producing dollar coins with the image of presidents on one of the sides. Suppose two such coins with the image of George Washington’s head up are placed side by side. The coin on the left is rotated (without slipping) half way along the perimeter of the coin on the right until it is on the opposite side as shown in the figure below. What is the orientation of the rotated dollar (head up or head down) when it gets to its final position? Solution: Head up Explanation: First, consider how far the center of the coin on the left moves. If each coin has radius r, then the distance between the centers of the two coins is 2r. Therefore, the center of the coin on the left moves along the perimeter of half a circle (i.e., π radians) whose radius is 2r, so the distance it travels is (π)(2r) = 2πr. Next, imagine the coin on the left rotating about its center as the center travels along a straight line of length 2πr. The circumference of the coin is also of length 2πr, so the coin will end with the same orientation that it had at the start, which is head up. Of course, there is no difference between this outcome and the outcome where the center travels along the perimeter of the half circle described above. Westmont College Mathematics Contest, February 9, 2008

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