MATHCOUNTS Practice Week 5, 2010 This week’s mini-lesson: how to improve in MATHCOUNTS and get serious about preparing for competitions. Go to the school website, then teachers, then Mr. Lomas. On his homework calendar, you will see a link to Harvest Park MATHCOUNTS. The site is still in the early construction phases, but among other things that you can find here is MATHCOUNTS Must Knows, each of the MATHCOUNTS Practice Worksheets we have done this year as well as 25 different worksheets sorted by level and topic with solutions. Students that still want more should look to www.artofproblemsolving.com. It is the best math competition website on the internet and there are a number of tremendous free preparation tools there including “Go For the Win” and ALCUMUS, as well as a number of different books and classes that are also available. Your one-hour here each week may spark an interest, but will not make you a champion! Contest Opportunity: The first mathleague.org middle school competition of the year will be on Sunday, November 7th at Diablo Valley College. This is NOT a school sponsored event (and since this is on Sunday, I will not be attending). However, for students that would like to participate, the fee is $10 and all information can be found on the mathleague.org website. If you are going to attend, I encourage you to form your own team of four students – you will be responsible for transportation and fees (this is not a school sponsored event). For those that attend, registration is open until November 3rd. The check-in begins at 1:30 and the competition starts at two with the countdown round and awards presented at 5 p.m. All you need to bring is pencil, calculator and your entrance fee ($10). It is usually a good idea to bring a snack and water as well. The address of Diablo Valley College, school map and room numbers are all available on the mathleague.org website. Challenge questions (answers next week – may work with parents this week!): #1 The arithmetic mean of 12 numbers is what percent of the same 12 numbers? Express your solution as a mixed number. #2 Bag A contains 10 marbles: 5 red, 3 blue, 1 yellow and 1 gold. Bag B contains 8 marbles: 4 red, 2 blue, 1 green and 1 silver. What is the probability that two marbles drawn at random, one from each bag, will be the same color? Express your answer as a common fraction. MATHCOUNTS homework (for the student that wants to improve) – • • • Download the MATHCOUNTS Must Knows Try at least one Level 1 worksheet this week – if you want a greater challenge, try a higher level. Read the MOEMS Must Knows to prepare for the Mathematical Olympiad. Our first test is next month! Solutions to Week 3 challenge questions: #1 14. The final value of the sum is 7 . That implies that the common denominator is a factor 10 of 2 and a factor of 5 since the prime factorization of 10 = 2 x 5. Let’s assume, then, that one of the values is 5 (we will let a = 5). The equation then becomes: then we have: simplify to get: 1 b + 1 b 1 c + = 1 c 7 10 = 1 2 − 1 5 = 1 5 + 1 b + 1 c = 7 10 1 2 or with common denominator, b +c bc cross products gives us: 2(b + c ) = bc or we can write a 2 = 1 = 2 bc b +c Guess and check then allows us to figure out what two different positive integers give us 2 when you divide their product by their sum. Shouldn’t take long to see that b and c must be 6 and 3. Therefore, the sum a + b + c = 5 + 6 + 3 = 14. #2 B 15 degrees. Because AX=XY=BC, then triangles AXY, BXY and BCY are all isosceles triangles. Let the measure of angle XAY = x and then angle AYX = x as well. Let angle BXY = y, then A angle XBY = y also. Finally, let angle BCY = z and of course angle BYC = z too. X Y C Now we can see that m∠AXY = 180 − y , because ∠AXY and ∠BXY are supplementary. Similarly, m∠XYB = 180 − 2y since the other two angles of the isosceles triangle both have measure of y. Finally, m∠CBY = 120 − y , because the m∠ABC = 120. So CBY has angles of 120 – y, z, and z. Thus 120 – y + 2z = 180 or 2z – y = 60. From AXY we get the equation (180 – y) + x + x = 180 or 2x = y. We know that m∠AYX + m∠XYB + m∠BYC = 180 so by substitution we get x + (180-2y) + z = 180 OR x + z – 2y = 0. Substitute 2x for y in the first equation to give 2z – 2x = 60 or z = x + 30. We now have both y and z solved for x (y = 2x and z = x + 30. Substitute these last two equations into third equation and solve for x. Rewrite x + z – 2y = 0 as x + (x + 30) - 2(2x) = 0. Simplify to 2x = 30 and x = 15!
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