<p>Hey. I’m in the eighth grade with mathcounts state competition coming up saturday, and I was wondering if you guys could help me with the following practice problems i couldn’t get:</p>
<li><p>A point (x, y) is randomly selected from the rectangular region bound by (0,0), (0,4), (8,4), (8,0). What is the probability that x^2 + y^2 < or equal to 4?</p></li>
<li><p>In a single elimination tournament, the better player always won. The winner of round 3 was the champion, and the loser of round 3 was the runner up. Eight players were randomly assigned slots in round 1. The winner of the 4 games of round 1 went on to round 2. What is the probability that the runner up was not the second best player in the tournament? </p></li>
</ol>
<p>Round 1: Round 2: Round 3: Champion:
-A - Winner of A and B -Winner of A/B and C/D -Winner of A/B-C/D and E/F-G/H.
-B</p>
<p>-C - Winner of C and D
-D</p>
<p>-E -Winner of E and F -Winner of E/F and G/H.
-F</p>
<p>-G -Winner of G and H
-H</p>
<li><p>Point P is 9 units from the center of a circle of radius 15. How many different chords of the circle contain P and have integer lengths?</p></li>
<li><p>Two numbers are called mirror numbers if the digits of the two numbers are in reverse order. For instance, 421 and 124 are mirror numbers. What is the sum of the two mirror numbers whose product is 110080?</p></li>
</ol>
<p>These are all questions from 2001 State Target with ~3 minutes per question.</p>
<ol>
<li><p>Assuming x and y don't need to be integers.... The area you're interested in is the region bounded by the x-axis, y-axis, and x^2 + y^2 = 4. That equaton is of a circle centered at origin with radius 2. So area of circle = 4pi, area in quadrant I (I think that's quadrant I, maybe II, but you understand) is pi. So answer is pi/32.</p></li>
<li><p>If you look at the number 110080, you see that the last number has to be 5, since 110080 ends in 0, and you can't start a number with zero. You also know the mirror number has to be three digits (500 * 500 = 250000--six digits) and the first digit of the second number has to be 2 or less, otherwise the product would be over 110080. Also, the last digit has to be even (since it 110080 ends in 0) so it must be 2. Now, you know the number is 5n2 and 2n5. You find n by noticing that n2 * n5 = 80. So (10<em>n + 2) * 5 + 20</em>n = 80. And n = 1. So sum = 512 + 215 = 727.</p></li>
</ol>
<ol>
<li>The longest any line on the circle could be is 30 (the diameter) so the chord is < 30. The shortest a chord could be is if it's perpendicular to the radius that intersects P. This chord would be length 12 (isosceles right triangle and side length 6). Therefore, 12<=x<30, so there's 18 integers. (I hope this is right, lol)</li>
</ol>
<p>I'll say it anyway I guess. I rank the guys in skill from 1-8. The runner up must have skill of 4<=x<=6. If his skill is 6, there is a 5/8 chance each guy in his half of the bracket is dumber than him. If his skill is 5, there's a 1/2 chance, if his skill is 4, there's a 3/8 chance. So (1/8)<em>(5/8)^3 + (1/8)</em>(1/2)^3+(1/8)*(3/8)^3 = 27/512.</p>
<p>Hint: Probabilitiy of runner up not being 2nd best = Probability of 2nd best being in the same bracket as the 1st best so that it would have been knocked out in an earlier round.</p>
<p>For problem 3, remember that for any point in a circle, the product of any two pieces of a chord that runs through that point is always the same. So we have 6 * 24 = 144</p>
<p>Soooo the shortest lengthed chord would be 2 * root(144) = 24
longest would be 30.
There are 2 of each chord with lengths 25, 26, 27, 28, 29, and 1 of 24 and 30. Which comes out to be 12 different chords.</p>
<p>(Also, so you can check your answer: the answer to the tournament question is 3/7)</p>