https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Frestho&feedformat=atom AoPS Wiki - User contributions [en] 2020-11-28T23:35:34Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_17&diff=134897 2013 AMC 12A Problems/Problem 17 2020-10-11T05:36:19Z <p>Frestho: /* Solution 3 (Work backwards) */</p> <hr /> <div>== Problem 17 ==<br /> <br /> A group of &lt;math&gt; 12 &lt;/math&gt; pirates agree to divide a treasure chest of gold coins among themselves as follows. The &lt;math&gt; k^\text{th} &lt;/math&gt; pirate to take a share takes &lt;math&gt; \frac{k}{12} &lt;/math&gt; of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the &lt;math&gt; 12^{\text{th}} &lt;/math&gt; pirate receive?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 720 \qquad \textbf{(B)} \ 1296 \qquad \textbf{(C)} \ 1728 \qquad \textbf{(D)} \ 1925 \qquad \textbf{(E)} \ 3850 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> The first pirate takes &lt;math&gt;\frac{1}{12}&lt;/math&gt; of the &lt;math&gt;x&lt;/math&gt; coins, leaving &lt;math&gt;\frac{11}{12} x&lt;/math&gt;.<br /> <br /> The second pirate takes &lt;math&gt;\frac{2}{12}&lt;/math&gt; of the remaining coins, leaving &lt;math&gt;\frac{10}{12}\cdot <br /> \frac{11}{12}*x&lt;/math&gt;.<br /> <br /> Note that<br /> <br /> &lt;math&gt;12^{11} = (2^2 \cdot 3)^{11} = 2^{22} \cdot 3^{11}&lt;/math&gt;<br /> <br /> &lt;math&gt;11! = 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2&lt;/math&gt;<br /> <br /> <br /> All the &lt;math&gt;2&lt;/math&gt;s and &lt;math&gt;3&lt;/math&gt;s cancel out of &lt;math&gt;11!&lt;/math&gt;, leaving<br /> <br /> &lt;math&gt;11 \cdot 5 \cdot 7 \cdot 5 = 1925&lt;/math&gt;<br /> <br /> in the numerator.<br /> <br /> <br /> We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, &lt;math&gt;x&lt;/math&gt; is the denominator, leaving &lt;math&gt;\boxed{\textbf{(D) }1925}&lt;/math&gt; coins for the twelfth pirate.<br /> <br /> ===Solution 2===<br /> The answer cannot be an even number. Here is why:<br /> <br /> Consider the highest power of 2 that divides the starting number of coins, and consider how this value changes as each pirate takes their share. At each step, the size of the pile is multiplied by some &lt;math&gt;\frac{n}{12}&lt;/math&gt;. This means the highest power of 2 that divides the number of coins is continually decreasing or staying the same (except once, briefly, when we multiply by &lt;math&gt;\frac{2}{3}&lt;/math&gt; for pirate 4, but it immediately drops again in the next step).<br /> <br /> Therefore, if the 12th pirate's coin total were even, then it can't be the smallest possible value, because we can safely cut the initial pot (and all the intermediate totals) in half. We could continue halving the result until the 12th pirate's total is finally odd.<br /> <br /> Only one of the choices given is odd, &lt;math&gt;\boxed{\textbf{(D) }1925}&lt;/math&gt;.<br /> <br /> Note that if there had been multiple odd answer choices, we could repeat the reasoning above to see that the value also must not be divisible by 3.<br /> <br /> ==Solution 3 (Work backwards)==<br /> Let &lt;math&gt;x&lt;/math&gt; be the number of coins the &lt;math&gt;12&lt;/math&gt;th pirate takes. Then the number of coins the &lt;math&gt;k&lt;12&lt;/math&gt;th pirate takes is &lt;math&gt;\frac{12}{1} \cdot \frac{12}{2} \cdots \frac{12}{12-k} x&lt;/math&gt;. For all these to be an integer, we need the denominators to divide into the numerators. Looking at prime factors, obviously there are sufficient &lt;math&gt;2&lt;/math&gt;s and &lt;math&gt;3&lt;/math&gt;s, so we just need &lt;math&gt;5^2 \cdot 7 \cdot 11 = \boxed{\textbf{(D) }1925}&lt;/math&gt; to divide into &lt;math&gt;x&lt;/math&gt;. -Frestho<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2013amc12a/356<br /> <br /> ~dolphin7<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_17&diff=134896 2013 AMC 12A Problems/Problem 17 2020-10-11T05:36:08Z <p>Frestho: /* Solution 3 (Work backwards) */</p> <hr /> <div>== Problem 17 ==<br /> <br /> A group of &lt;math&gt; 12 &lt;/math&gt; pirates agree to divide a treasure chest of gold coins among themselves as follows. The &lt;math&gt; k^\text{th} &lt;/math&gt; pirate to take a share takes &lt;math&gt; \frac{k}{12} &lt;/math&gt; of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the &lt;math&gt; 12^{\text{th}} &lt;/math&gt; pirate receive?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 720 \qquad \textbf{(B)} \ 1296 \qquad \textbf{(C)} \ 1728 \qquad \textbf{(D)} \ 1925 \qquad \textbf{(E)} \ 3850 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> The first pirate takes &lt;math&gt;\frac{1}{12}&lt;/math&gt; of the &lt;math&gt;x&lt;/math&gt; coins, leaving &lt;math&gt;\frac{11}{12} x&lt;/math&gt;.<br /> <br /> The second pirate takes &lt;math&gt;\frac{2}{12}&lt;/math&gt; of the remaining coins, leaving &lt;math&gt;\frac{10}{12}\cdot <br /> \frac{11}{12}*x&lt;/math&gt;.<br /> <br /> Note that<br /> <br /> &lt;math&gt;12^{11} = (2^2 \cdot 3)^{11} = 2^{22} \cdot 3^{11}&lt;/math&gt;<br /> <br /> &lt;math&gt;11! = 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2&lt;/math&gt;<br /> <br /> <br /> All the &lt;math&gt;2&lt;/math&gt;s and &lt;math&gt;3&lt;/math&gt;s cancel out of &lt;math&gt;11!&lt;/math&gt;, leaving<br /> <br /> &lt;math&gt;11 \cdot 5 \cdot 7 \cdot 5 = 1925&lt;/math&gt;<br /> <br /> in the numerator.<br /> <br /> <br /> We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, &lt;math&gt;x&lt;/math&gt; is the denominator, leaving &lt;math&gt;\boxed{\textbf{(D) }1925}&lt;/math&gt; coins for the twelfth pirate.<br /> <br /> ===Solution 2===<br /> The answer cannot be an even number. Here is why:<br /> <br /> Consider the highest power of 2 that divides the starting number of coins, and consider how this value changes as each pirate takes their share. At each step, the size of the pile is multiplied by some &lt;math&gt;\frac{n}{12}&lt;/math&gt;. This means the highest power of 2 that divides the number of coins is continually decreasing or staying the same (except once, briefly, when we multiply by &lt;math&gt;\frac{2}{3}&lt;/math&gt; for pirate 4, but it immediately drops again in the next step).<br /> <br /> Therefore, if the 12th pirate's coin total were even, then it can't be the smallest possible value, because we can safely cut the initial pot (and all the intermediate totals) in half. We could continue halving the result until the 12th pirate's total is finally odd.<br /> <br /> Only one of the choices given is odd, &lt;math&gt;\boxed{\textbf{(D) }1925}&lt;/math&gt;.<br /> <br /> Note that if there had been multiple odd answer choices, we could repeat the reasoning above to see that the value also must not be divisible by 3.<br /> <br /> ==Solution 3 (Work backwards)==<br /> Let &lt;math&gt;x&lt;/math&gt; be the number of coins the &lt;math&gt;12&lt;/math&gt;th pirate takes. Then the number of coins the &lt;math&gt;k&lt;12&lt;/math&gt;th pirate takes is &lt;math&gt;\frac{12}{1} \cdot \frac{12}{2} \cdots \frac{12}{12-k} x&lt;/math&gt;. For all these to be an integer, we need the denominators to divide into the numerators. Looking at prime factors, obviously there are sufficient &lt;math&gt;2&lt;/math&gt;s and &lt;math&gt;3&lt;/math&gt;s, so we just need &lt;math&gt;5^2 \cdot 7 \cdot 11 = \boxed{\textbf{(D) }1925}&lt;/math&gt; to divide into &lt;math&gt;x&lt;/math&gt;.<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2013amc12a/356<br /> <br /> ~dolphin7<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_17&diff=134895 2013 AMC 12A Problems/Problem 17 2020-10-11T05:35:19Z <p>Frestho: /* Solution 3 (Work backwards) */</p> <hr /> <div>== Problem 17 ==<br /> <br /> A group of &lt;math&gt; 12 &lt;/math&gt; pirates agree to divide a treasure chest of gold coins among themselves as follows. The &lt;math&gt; k^\text{th} &lt;/math&gt; pirate to take a share takes &lt;math&gt; \frac{k}{12} &lt;/math&gt; of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the &lt;math&gt; 12^{\text{th}} &lt;/math&gt; pirate receive?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 720 \qquad \textbf{(B)} \ 1296 \qquad \textbf{(C)} \ 1728 \qquad \textbf{(D)} \ 1925 \qquad \textbf{(E)} \ 3850 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> The first pirate takes &lt;math&gt;\frac{1}{12}&lt;/math&gt; of the &lt;math&gt;x&lt;/math&gt; coins, leaving &lt;math&gt;\frac{11}{12} x&lt;/math&gt;.<br /> <br /> The second pirate takes &lt;math&gt;\frac{2}{12}&lt;/math&gt; of the remaining coins, leaving &lt;math&gt;\frac{10}{12}\cdot <br /> \frac{11}{12}*x&lt;/math&gt;.<br /> <br /> Note that<br /> <br /> &lt;math&gt;12^{11} = (2^2 \cdot 3)^{11} = 2^{22} \cdot 3^{11}&lt;/math&gt;<br /> <br /> &lt;math&gt;11! = 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2&lt;/math&gt;<br /> <br /> <br /> All the &lt;math&gt;2&lt;/math&gt;s and &lt;math&gt;3&lt;/math&gt;s cancel out of &lt;math&gt;11!&lt;/math&gt;, leaving<br /> <br /> &lt;math&gt;11 \cdot 5 \cdot 7 \cdot 5 = 1925&lt;/math&gt;<br /> <br /> in the numerator.<br /> <br /> <br /> We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, &lt;math&gt;x&lt;/math&gt; is the denominator, leaving &lt;math&gt;\boxed{\textbf{(D) }1925}&lt;/math&gt; coins for the twelfth pirate.<br /> <br /> ===Solution 2===<br /> The answer cannot be an even number. Here is why:<br /> <br /> Consider the highest power of 2 that divides the starting number of coins, and consider how this value changes as each pirate takes their share. At each step, the size of the pile is multiplied by some &lt;math&gt;\frac{n}{12}&lt;/math&gt;. This means the highest power of 2 that divides the number of coins is continually decreasing or staying the same (except once, briefly, when we multiply by &lt;math&gt;\frac{2}{3}&lt;/math&gt; for pirate 4, but it immediately drops again in the next step).<br /> <br /> Therefore, if the 12th pirate's coin total were even, then it can't be the smallest possible value, because we can safely cut the initial pot (and all the intermediate totals) in half. We could continue halving the result until the 12th pirate's total is finally odd.<br /> <br /> Only one of the choices given is odd, &lt;math&gt;\boxed{\textbf{(D) }1925}&lt;/math&gt;.<br /> <br /> Note that if there had been multiple odd answer choices, we could repeat the reasoning above to see that the value also must not be divisible by 3.<br /> <br /> ==Solution 3 (Work backwards)==<br /> Let &lt;math&gt;x&lt;/math&gt; be the number of coins the &lt;math&gt;12&lt;/math&gt;th pirate takes. Then the number of coins the &lt;math&gt;k&lt;12&lt;/math&gt;th pirate takes is &lt;math&gt;\frac{12}{1} \cdot \frac{12}{2} \cdots \frac{12}{12-k} x&lt;/math&gt;. For all these to be an integer, we need the denominators to divide into the numerators. Looking at prime factors in the denominators, obviously there are sufficient amounts of &lt;math&gt;2&lt;/math&gt;s and &lt;math&gt;3&lt;/math&gt;s, so we just need &lt;math&gt;5^2 \cdot 7 \cdot 11 = \boxed{\textbf{(D) }1925}&lt;/math&gt; to divide into &lt;math&gt;x&lt;/math&gt;.<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2013amc12a/356<br /> <br /> ~dolphin7<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_17&diff=134894 2013 AMC 12A Problems/Problem 17 2020-10-11T05:32:40Z <p>Frestho: /* Solution */</p> <hr /> <div>== Problem 17 ==<br /> <br /> A group of &lt;math&gt; 12 &lt;/math&gt; pirates agree to divide a treasure chest of gold coins among themselves as follows. The &lt;math&gt; k^\text{th} &lt;/math&gt; pirate to take a share takes &lt;math&gt; \frac{k}{12} &lt;/math&gt; of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the &lt;math&gt; 12^{\text{th}} &lt;/math&gt; pirate receive?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 720 \qquad \textbf{(B)} \ 1296 \qquad \textbf{(C)} \ 1728 \qquad \textbf{(D)} \ 1925 \qquad \textbf{(E)} \ 3850 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> The first pirate takes &lt;math&gt;\frac{1}{12}&lt;/math&gt; of the &lt;math&gt;x&lt;/math&gt; coins, leaving &lt;math&gt;\frac{11}{12} x&lt;/math&gt;.<br /> <br /> The second pirate takes &lt;math&gt;\frac{2}{12}&lt;/math&gt; of the remaining coins, leaving &lt;math&gt;\frac{10}{12}\cdot <br /> \frac{11}{12}*x&lt;/math&gt;.<br /> <br /> Note that<br /> <br /> &lt;math&gt;12^{11} = (2^2 \cdot 3)^{11} = 2^{22} \cdot 3^{11}&lt;/math&gt;<br /> <br /> &lt;math&gt;11! = 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2&lt;/math&gt;<br /> <br /> <br /> All the &lt;math&gt;2&lt;/math&gt;s and &lt;math&gt;3&lt;/math&gt;s cancel out of &lt;math&gt;11!&lt;/math&gt;, leaving<br /> <br /> &lt;math&gt;11 \cdot 5 \cdot 7 \cdot 5 = 1925&lt;/math&gt;<br /> <br /> in the numerator.<br /> <br /> <br /> We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, &lt;math&gt;x&lt;/math&gt; is the denominator, leaving &lt;math&gt;\boxed{\textbf{(D) }1925}&lt;/math&gt; coins for the twelfth pirate.<br /> <br /> ===Solution 2===<br /> The answer cannot be an even number. Here is why:<br /> <br /> Consider the highest power of 2 that divides the starting number of coins, and consider how this value changes as each pirate takes their share. At each step, the size of the pile is multiplied by some &lt;math&gt;\frac{n}{12}&lt;/math&gt;. This means the highest power of 2 that divides the number of coins is continually decreasing or staying the same (except once, briefly, when we multiply by &lt;math&gt;\frac{2}{3}&lt;/math&gt; for pirate 4, but it immediately drops again in the next step).<br /> <br /> Therefore, if the 12th pirate's coin total were even, then it can't be the smallest possible value, because we can safely cut the initial pot (and all the intermediate totals) in half. We could continue halving the result until the 12th pirate's total is finally odd.<br /> <br /> Only one of the choices given is odd, &lt;math&gt;\boxed{\textbf{(D) }1925}&lt;/math&gt;.<br /> <br /> Note that if there had been multiple odd answer choices, we could repeat the reasoning above to see that the value also must not be divisible by 3.<br /> <br /> ==Solution 3 (Work backwards)==<br /> Let &lt;math&gt;x&lt;/math&gt; be the number of coins the &lt;math&gt;12&lt;/math&gt;th pirate takes. Then the number of coins the &lt;math&gt;k&lt;12&lt;/math&gt;th pirate takes is &lt;math&gt;\frac{12}{1} \cdot \frac{12}{2} \cdots \frac{12}{12-k} x&lt;/math&gt;. For all these to be an integer, we need the denominators to divide into the numerators. Obviously there are sufficient amounts of prime factors of &lt;math&gt;2&lt;/math&gt;s and &lt;math&gt;3&lt;/math&gt;s, so we just need &lt;math&gt;5^2 \cdot 7 \cdot 11 = \boxed{\textbf{(D) }1925}&lt;/math&gt; to divide into &lt;math&gt;x&lt;/math&gt;.<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2013amc12a/356<br /> <br /> ~dolphin7<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_16&diff=134893 2013 AMC 12A Problems/Problem 16 2020-10-11T05:14:54Z <p>Frestho: /* Solution 1 */</p> <hr /> <div>== Problem==<br /> <br /> &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; are three piles of rocks. The mean weight of the rocks in &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;40&lt;/math&gt; pounds, the mean weight of the rocks in &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;50&lt;/math&gt; pounds, the mean weight of the rocks in the combined piles &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;43&lt;/math&gt; pounds, and the mean weight of the rocks in the combined piles &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; is &lt;math&gt;44&lt;/math&gt; pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> Let pile &lt;math&gt;A&lt;/math&gt; have &lt;math&gt;A&lt;/math&gt; rocks, and so on.<br /> <br /> The total weight of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; can be expressed as &lt;math&gt;44(A + C)&lt;/math&gt;.<br /> <br /> To get the total weight of &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, we add the weight of &lt;math&gt;B&lt;/math&gt; and subtract the weight of &lt;math&gt;A&lt;/math&gt;: &lt;math&gt;44(A + C) + 50B - 40A = 4A + 44C + 50B&lt;/math&gt;<br /> <br /> Therefore, the mean of &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; is &lt;math&gt;\frac{4A + 44C + 50B}{B + C}&lt;/math&gt;, which is simplified to &lt;math&gt;44 + \frac{4A + 6B}{B + C}&lt;/math&gt;.<br /> <br /> We now need to eliminate &lt;math&gt;A&lt;/math&gt; in the numerator. <br /> Since we know that &lt;math&gt;40A + 50B = 43(A + B)&lt;/math&gt;, we have &lt;math&gt;A = \frac{7}{3}B&lt;/math&gt;<br /> <br /> Substituting,<br /> <br /> &lt;math&gt;44 + \frac{4(\frac{7}{3}B) + 6B}{B + C}=44 + \frac{46}{3}*\frac{B}{B + C}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{B}{B + C} &lt; 1&lt;/math&gt;, so the maximum value occurs when &lt;math&gt;C = 1&lt;/math&gt;. Since &lt;math&gt;\frac{46}{3}&lt;/math&gt; must cancel to give an integer, and the only fraction that satisfies both conditions is &lt;math&gt;\frac{45}{46}&lt;/math&gt;<br /> <br /> Plugging in, we get<br /> <br /> &lt;math&gt;44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59&lt;/math&gt;, which is choice E<br /> <br /> ===Solution 2===<br /> Suppose there are &lt;math&gt;A,B,C&lt;/math&gt; rocks in the three piles, and that the mean of pile C is &lt;math&gt;x&lt;/math&gt;, and that the mean of the combination of &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; is &lt;math&gt;y&lt;/math&gt;. We are going to maximize &lt;math&gt;y&lt;/math&gt;, subject to the following conditions:<br /> <br /> &lt;cmath&gt;40A+50B=43(A+B)&lt;/cmath&gt;<br /> &lt;cmath&gt;40A+xC=44(A+C)&lt;/cmath&gt;<br /> &lt;cmath&gt;50B+xC=y(B+C)&lt;/cmath&gt;<br /> <br /> which can be rearranged as:<br /> <br /> &lt;cmath&gt;7B=3A&lt;/cmath&gt;<br /> &lt;cmath&gt;(x-44)C=4A&lt;/cmath&gt;<br /> &lt;cmath&gt;(x-y)C=(y-50)B&lt;/cmath&gt;<br /> <br /> Let us test &lt;math&gt;y=59&lt;/math&gt; is possible. If so, it is already the answer. If not, there will be some contradiction. So the third equation becomes <br /> <br /> &lt;cmath&gt;(x-59)C=9B.&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;15C = (x-44)C - (x-59)C = 4A - 9B&lt;/math&gt;, &lt;math&gt;45C=4(3A)-27B=28B-27B&lt;/math&gt;, &lt;math&gt;105C=28A-9(7B)=A&lt;/math&gt;, therefore,<br /> <br /> &lt;math&gt;A=105C, B=45C, x=4(105)+44=464&lt;/math&gt;, which gives us a consistent solution. Therefore &lt;math&gt;y=59&lt;/math&gt; is the answer.<br /> <br /> (Note: To further illustrate the idea, let us look at &lt;math&gt;y=60&lt;/math&gt; and see what happens. We then get &lt;math&gt;7\cdot 16C = 4A-30A&lt;0&lt;/math&gt;, which is a contradiction!)<br /> <br /> ===Solution 3===<br /> Obtain the 3 equations as in '''solution 2'''.<br /> <br /> &lt;cmath&gt;7B=3A&lt;/cmath&gt;<br /> &lt;cmath&gt;(x-44)C=4A&lt;/cmath&gt;<br /> &lt;cmath&gt;(x-y)C=(y-50)B&lt;/cmath&gt;<br /> <br /> Our goal is to try to isolate &lt;math&gt;y&lt;/math&gt; into an inequality. <br /> The first equation gives &lt;math&gt;A=\frac{7}{3}B&lt;/math&gt;, which we plug into the second equation to get<br /> <br /> &lt;cmath&gt;(x-44)C=\frac{28}{3}B&lt;/cmath&gt;<br /> <br /> To eliminate &lt;math&gt;x&lt;/math&gt;, subtract equation 3 from equation 2:<br /> <br /> &lt;cmath&gt;(x-44)C-(x-y)C=\frac{28}{3}B-(y-50)B&lt;/cmath&gt;<br /> &lt;cmath&gt;(y-44)C=(\frac{178}{3}-y)B&lt;/cmath&gt;<br /> <br /> In order for the coefficients to be positive, &lt;cmath&gt;44&lt;y&lt;\frac{178}{3}&lt;/cmath&gt;<br /> <br /> Thus, the greatest integer value is &lt;math&gt;y=59&lt;/math&gt;, choice &lt;math&gt;(E)&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=15|num-a=17}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_9&diff=126626 2014 AIME II Problems/Problem 9 2020-06-26T18:21:40Z <p>Frestho: /* Solution 3 (Complementary Counting) */</p> <hr /> <div>==Problem==<br /> Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs.<br /> <br /> ==Solution 1 (Casework)==<br /> We know that a subset with less than &lt;math&gt;3&lt;/math&gt; chairs cannot contain &lt;math&gt;3&lt;/math&gt; adjacent chairs. There are only &lt;math&gt;10&lt;/math&gt; sets of &lt;math&gt;3&lt;/math&gt; chairs so that they are all &lt;math&gt;3&lt;/math&gt; adjacent. There are &lt;math&gt;10&lt;/math&gt; subsets of &lt;math&gt;4&lt;/math&gt; chairs where all &lt;math&gt;4&lt;/math&gt; are adjacent, and &lt;math&gt;10 \cdot 5&lt;/math&gt; or &lt;math&gt;50&lt;/math&gt; where there are only &lt;math&gt;3.&lt;/math&gt; If there are &lt;math&gt;5&lt;/math&gt; chairs, &lt;math&gt;10&lt;/math&gt; have all &lt;math&gt;5&lt;/math&gt; adjacent, &lt;math&gt;10 \cdot 4&lt;/math&gt; or &lt;math&gt;40&lt;/math&gt; have &lt;math&gt;4&lt;/math&gt; adjacent, and &lt;math&gt;10 \cdot {5\choose 2}&lt;/math&gt; or &lt;math&gt;100&lt;/math&gt; have &lt;math&gt;3&lt;/math&gt; adjacent. With &lt;math&gt;6&lt;/math&gt; chairs in the subset, &lt;math&gt;10&lt;/math&gt; have all &lt;math&gt;6&lt;/math&gt; adjacent, &lt;math&gt;10(3)&lt;/math&gt; or &lt;math&gt;30&lt;/math&gt; have &lt;math&gt;5&lt;/math&gt; adjacent, &lt;math&gt;10 \cdot {4\choose2}&lt;/math&gt; or &lt;math&gt;60&lt;/math&gt; have &lt;math&gt;4&lt;/math&gt; adjacent, &lt;math&gt;\frac{10 \cdot 3}{2}&lt;/math&gt; or &lt;math&gt;15&lt;/math&gt; have &lt;math&gt;2&lt;/math&gt; groups of &lt;math&gt;3&lt;/math&gt; adjacent chairs, and &lt;math&gt;10 \cdot \left({5\choose2} - 3\right)&lt;/math&gt; or &lt;math&gt;70&lt;/math&gt; have &lt;math&gt;1&lt;/math&gt; group of &lt;math&gt;3&lt;/math&gt; adjacent chairs. All possible subsets with more than &lt;math&gt;6&lt;/math&gt; chairs have at least &lt;math&gt;1&lt;/math&gt; group of &lt;math&gt;3&lt;/math&gt; adjacent chairs, so we add &lt;math&gt;{10\choose7}&lt;/math&gt; or &lt;math&gt;120&lt;/math&gt;, &lt;math&gt;{10\choose8}&lt;/math&gt; or &lt;math&gt;45&lt;/math&gt;, &lt;math&gt;{10\choose9}&lt;/math&gt; or &lt;math&gt;10&lt;/math&gt;, and &lt;math&gt;{10\choose10}&lt;/math&gt; or &lt;math&gt;1.&lt;/math&gt; Adding, we get &lt;math&gt;10 + 10 + 50 + 10 + 40 + 100 + 10 + 30 + 60 + 15 + 70 + 120 + 45 + 10 + 1 = \boxed{581}.&lt;/math&gt;<br /> <br /> ==Solution 2 (PIE)==<br /> Starting with small cases, we see that four chairs give &lt;math&gt;4 + 1 = 5&lt;/math&gt;, five chairs give &lt;math&gt;5 + 5 + 1 = 11&lt;/math&gt;, and six chairs give &lt;math&gt;6 + 6 + 6 + 6 + 1 = 25.&lt;/math&gt; Thus, n chairs should give &lt;math&gt;n 2^{n-4} + 1&lt;/math&gt;, as confirmed above. This claim can be verified by the principle of inclusion-exclusion: there are &lt;math&gt;n 2^{n-3}&lt;/math&gt; ways to arrange &lt;math&gt;3&lt;/math&gt; adjacent chairs, but then we subtract &lt;math&gt;n 2^{n-4}&lt;/math&gt; ways to arrange &lt;math&gt;4.&lt;/math&gt; Finally, we add &lt;math&gt;1&lt;/math&gt; to account for the full subset of chairs. Thus, for &lt;math&gt;n = 10&lt;/math&gt; we get a first count of &lt;math&gt;641.&lt;/math&gt;<br /> <br /> However, we overcount cases in which there are two distinct groups of three or more chairs. Time to casework: we have &lt;math&gt;5&lt;/math&gt; cases for two groups of &lt;math&gt;3&lt;/math&gt; directly opposite each other, &lt;math&gt;5&lt;/math&gt; for two groups of four, &lt;math&gt;20&lt;/math&gt; for two groups of &lt;math&gt;3&lt;/math&gt; not symmetrically opposite, &lt;math&gt;20&lt;/math&gt; for a group of &lt;math&gt;3&lt;/math&gt; and a group of &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;10&lt;/math&gt; for a group of &lt;math&gt;3&lt;/math&gt; and a group of &lt;math&gt;5.&lt;/math&gt; Thus, we have &lt;math&gt;641 - 60 = \boxed{581}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Complementary Counting and Recursion)==<br /> It is possible to use recursion to count the complement. Number the chairs &lt;math&gt;1, 2, 3, ..., 10.&lt;/math&gt; If chair &lt;math&gt;1&lt;/math&gt; is not occupied, then we have a line of &lt;math&gt;9&lt;/math&gt; chairs such that there is no consecutive group of three. If chair &lt;math&gt;1&lt;/math&gt; is occupied, then we split into more cases. If chairs &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;10&lt;/math&gt; are empty, then we have a line of &lt;math&gt;7.&lt;/math&gt; If chair &lt;math&gt;2&lt;/math&gt; is empty but chair &lt;math&gt;10&lt;/math&gt; is occupied, then we have a line of &lt;math&gt;6&lt;/math&gt; chairs (because chair &lt;math&gt;9&lt;/math&gt; cannot be occupied); this is similar to when chair &lt;math&gt;2&lt;/math&gt; is occupied and chair &lt;math&gt;10&lt;/math&gt; is empty. Finally, chairs &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;10&lt;/math&gt; cannot be simultaneously occupied. Thus, we have reduced the problem down to computing &lt;math&gt;T_9 + T_7 + 2T_6&lt;/math&gt;, where &lt;math&gt;T_n&lt;/math&gt; counts the ways to select a subset of chairs &lt;math&gt;\textit{in a line}&lt;/math&gt; from a group of n chairs such that there is no group of &lt;math&gt;3&lt;/math&gt; chairs in a row.<br /> <br /> Now, we notice that &lt;math&gt;T_n = T_{n-1} + T_{n-2} + T_{n-3}&lt;/math&gt; (representing the cases when the first, second, and/or third chair is unoccupied). Also, &lt;math&gt;T_0 = 1, T_1 = 2, T_2 = 4, T_3 = 7&lt;/math&gt;, and hence &lt;math&gt;T_4 = 13, T_5 = 24, T_6 = 44, T_7 = 81, T_8 = 149, T_9 = 274&lt;/math&gt;. Now we know the complement is &lt;math&gt;274 + 81 + 88 = 443&lt;/math&gt;, and subtracting from &lt;math&gt;2^{10} = 1024&lt;/math&gt; gives &lt;math&gt;1024 - 443 = \boxed{581}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=II|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_15&diff=124542 2000 AIME I Problems/Problem 15 2020-06-09T01:54:30Z <p>Frestho: /* Solution 3 (Recursion) */</p> <hr /> <div>== Problem ==<br /> A stack of &lt;math&gt;2000&lt;/math&gt; cards is labelled with the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;2000,&lt;/math&gt; with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: &lt;math&gt;1,2,3,\ldots,1999,2000.&lt;/math&gt; In the original stack of cards, how many cards were above the card labeled &lt;math&gt;1999&lt;/math&gt;?<br /> <br /> == Solution ==<br /> We try to work backwards from when there are 2 cards left, since this is when the 1999 card is laid onto the table. When there are 2 cards left, the 1999 card is on the top of the deck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there are over 1000 cards remaining, some cards have not even made one trip through yet, 2(1024 - 1000) = 48, to be exact. Once these cards go through, 1999 will be the &lt;math&gt;512 - 48 = 464^\text{th}&lt;/math&gt; card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position &lt;math&gt;464 \times 2 = 928&lt;/math&gt;, meaning that there were &lt;math&gt;\boxed{927}&lt;/math&gt; cards are above the one labeled &lt;math&gt;1999&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> To simplify matters, we want a power of &lt;math&gt;2&lt;/math&gt;. Hence, we will add &lt;math&gt;48&lt;/math&gt; 'fake' cards which we must discard in our actual count. Using similar logic as Solution 1, we find that 1999 has position &lt;math&gt;1024&lt;/math&gt; in a &lt;math&gt;2048&lt;/math&gt; card stack, where the fake cards towards the front.<br /> <br /> Let the fake cards have positions &lt;math&gt;1, 3, 5, \cdots, 95&lt;/math&gt;. Then, we know that the real cards filling the gaps between the fake cards must be the cards such that they go to their correct starting positions in the &lt;math&gt;2000&lt;/math&gt; card case, where all of them are below &lt;math&gt;1999&lt;/math&gt;. From this, we know that the cards from positions &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;96&lt;/math&gt; alternate in fake-real-fake-real, where we have the correct order of cards once the first &lt;math&gt;96&lt;/math&gt; have moved and we can start putting real cards on the table. Hence, &lt;math&gt;1999&lt;/math&gt; is in position &lt;math&gt;1024 - 96 = 928&lt;/math&gt;, so &lt;math&gt;\boxed{927}&lt;/math&gt; cards are above it. - Spacesam<br /> <br /> ==Solution 3 (Recursion)==<br /> Consider the general problem: with a stack of &lt;math&gt;n&lt;/math&gt; cards such that they will be laid out &lt;math&gt;1, 2, 3, ..., n&lt;/math&gt; from left to right, how many cards are above the card labeled &lt;math&gt;n-1&lt;/math&gt;?<br /> <br /> Let &lt;math&gt;a_n&lt;/math&gt; be the answer to the above problem.<br /> <br /> As a base case, consider &lt;math&gt;n=2&lt;/math&gt;. Clearly, the stack, from top to bottom, must be &lt;math&gt;(1, 2)&lt;/math&gt;, so &lt;math&gt;a_2=0&lt;/math&gt;. <br /> <br /> Next, let's think about how we can construct a stack of &lt;math&gt;n+1&lt;/math&gt; cards from a stack of &lt;math&gt;n&lt;/math&gt; cards. First, let us renumber the current stack of &lt;math&gt;n&lt;/math&gt; by adding &lt;math&gt;1&lt;/math&gt; to the label of each of the cards. Then we must add a card labeled &quot;&lt;math&gt;1&lt;/math&gt;&quot; somewhere in the new deck.<br /> <br /> Working backwards, we find that we must move the bottom card to the top, then add &quot;&lt;math&gt;1&lt;/math&gt;&quot; to the top of the deck. This is because after one move, the number &quot;&lt;math&gt;1&lt;/math&gt;&quot; will be laid out and the top card will be moved to the bottom, so the deck becomes the same as what we had before, but with everything renumbered correctly such that &lt;math&gt;2, 3, 4, ...&lt;/math&gt; will then be laid out in order.<br /> <br /> Therefore, if &lt;math&gt;a_n \ne n-1&lt;/math&gt; (meaning the card &lt;math&gt;n-1&lt;/math&gt; is not at the bottom of the deck and so it won't be moved to the top), then &lt;math&gt;a_{n+1} = a_n + 2&lt;/math&gt;, since a card from the bottom is moved to be above the &lt;math&gt;n-1&lt;/math&gt; card, and the new card &quot;&lt;math&gt;1&lt;/math&gt;&quot; is added to the top. If &lt;math&gt;a_n = n-1&lt;/math&gt; (meaning the card &lt;math&gt;n-1&lt;/math&gt; is the bottom card), then &lt;math&gt;a_{n+1}=1&lt;/math&gt; because the &lt;math&gt;n-1&lt;/math&gt; card will move to the top and the card &quot;&lt;math&gt;1&lt;/math&gt;&quot; will be added on top of it.<br /> <br /> With these recursions and the base case we found earlier, we calculate &lt;math&gt;a_{2000} = \boxed{927}&lt;/math&gt;. To calculate this by hand, a helpful trick is finding that if &lt;math&gt;a_n=1&lt;/math&gt;, then &lt;math&gt;a_{2n-1}=1&lt;/math&gt; as well. Once we find &lt;math&gt;a_{1537}=1&lt;/math&gt;, the answer is just &lt;math&gt;1+(2000-1537)\cdot2&lt;/math&gt;.<br /> - Frestho<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_3&diff=123948 2015 AIME II Problems/Problem 3 2020-06-05T22:56:11Z <p>Frestho: /* Solution 2 (Shortcut) */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;m&lt;/math&gt; be the least positive integer divisible by &lt;math&gt;17&lt;/math&gt; whose digits sum to &lt;math&gt;17&lt;/math&gt;. Find &lt;math&gt;m&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> The three-digit integers divisible by &lt;math&gt;17&lt;/math&gt;, and their digit sum: &lt;cmath&gt;<br /> \begin{array}{c|c}<br /> m &amp; s(m)\\ \hline<br /> 102 &amp; 3 \\<br /> 119 &amp; 11\\<br /> 136 &amp; 10\\<br /> 153 &amp; 9\\<br /> 170 &amp; 8\\<br /> 187 &amp; 16\\<br /> 204 &amp; 6\\<br /> 221 &amp; 5\\<br /> 238 &amp; 13\\<br /> 255 &amp; 12\\<br /> 272 &amp; 11\\<br /> 289 &amp; 19\\<br /> 306 &amp; 9\\<br /> 323 &amp; 8\\<br /> 340 &amp; 7\\<br /> 357 &amp; 15\\<br /> 374 &amp; 14\\<br /> 391 &amp; 13\\<br /> 408 &amp; 12\\<br /> 425 &amp; 11\\<br /> 442 &amp; 10\\<br /> 459 &amp; 18\\<br /> 476 &amp; 17<br /> \end{array}<br /> &lt;/cmath&gt;<br /> <br /> Thus the answer is &lt;math&gt;\boxed{476}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Shortcut)==<br /> <br /> We can do the same thing as solution 1, except note the following fact: &lt;math&gt;102&lt;/math&gt; is a multiple of &lt;math&gt;17&lt;/math&gt; and its digits sum to &lt;math&gt;3&lt;/math&gt;. <br /> <br /> Therefore, we can add it onto an existing multiple of &lt;math&gt;17&lt;/math&gt; that we know of to have &lt;math&gt;s(m) = 14&lt;/math&gt;, shown in the right-hand column, provided that its units digit is less than &lt;math&gt;8&lt;/math&gt; and its hundreds digit is less than &lt;math&gt;9&lt;/math&gt;. Unfortunately, &lt;math&gt;68&lt;/math&gt; does not fit the criteria, but &lt;math&gt;374&lt;/math&gt; does, meaning that, instead of continually adding multiples of &lt;math&gt;17&lt;/math&gt;, we can stop here and simply add &lt;math&gt;102&lt;/math&gt; to reach our final answer of &lt;math&gt;\boxed{476}&lt;/math&gt;.<br /> <br /> ~Tiblis<br /> <br /> (Comment from another person: Actually, this doesn't work because you can't be sure there are no numbers between 374 and 476 that work. This solution just lucks out.)<br /> <br /> ==Solution 3==<br /> <br /> The digit sum of a base &lt;math&gt;10&lt;/math&gt; integer &lt;math&gt;m&lt;/math&gt; is just &lt;math&gt;m\pmod{9}&lt;/math&gt;. In this problem, we know &lt;math&gt;17\mid m&lt;/math&gt;, or &lt;math&gt;m=17k&lt;/math&gt; for a positive integer &lt;math&gt;k&lt;/math&gt;.<br /> <br /> Also, we know that &lt;math&gt;m\equiv 17\equiv -1\pmod{9}&lt;/math&gt;, or &lt;math&gt;17k\equiv -k\equiv -1\pmod{9}&lt;/math&gt;.<br /> <br /> Obviously &lt;math&gt;k=1&lt;/math&gt; is a solution. This means in general, &lt;math&gt;k=9x+1&lt;/math&gt; is a solution for non-negative integer &lt;math&gt;x&lt;/math&gt;.<br /> <br /> Checking the first few possible solutions, we find that &lt;math&gt;m=\boxed{476}&lt;/math&gt; is the first solution that has &lt;math&gt;s(m)=17&lt;/math&gt;, and we're done.<br /> <br /> ==Solution 4==<br /> <br /> Since the sum of the digits in the base-10 representation of &lt;math&gt;m&lt;/math&gt; is &lt;math&gt;17&lt;/math&gt;, we must have &lt;math&gt;m\equiv 17 \pmod{9}&lt;/math&gt; or &lt;math&gt;m\equiv -1\pmod{9}&lt;/math&gt;.<br /> We also know that since &lt;math&gt;m&lt;/math&gt; is divisible by 17, &lt;math&gt;m\equiv 0 \pmod{17}&lt;/math&gt;.<br /> <br /> To solve this system of linear congruences, we can use the Chinese Remainder Theorem. If we set &lt;math&gt;m\equiv (-1)(17)(8)\pmod {153}&lt;/math&gt;, we find that &lt;math&gt;m\equiv 0\pmod{17}&lt;/math&gt; and &lt;math&gt;m\equiv -1\pmod{9}&lt;/math&gt;, because &lt;math&gt;17\cdot 8\equiv 136 \equiv 1\pmod{9}&lt;/math&gt;. The trick to getting here was to find the number &lt;math&gt;x&lt;/math&gt; such that &lt;math&gt;17x\equiv 1\pmod{9}&lt;/math&gt;, so that when we take things &lt;math&gt;\pmod{9}&lt;/math&gt;, the &lt;math&gt;17&lt;/math&gt; goes away. We can do this using the Extended Euclidean Algorithm or by guess and check to find that &lt;math&gt;x\equiv 8\pmod{9}&lt;/math&gt;.<br /> <br /> Finally, since &lt;math&gt;m\equiv 17\pmod{153}&lt;/math&gt;, we repeatedly add multiples of &lt;math&gt;153&lt;/math&gt; until we get a number in which its digits sum to 17, which first happens when &lt;math&gt;m=\boxed{476}&lt;/math&gt;.<br /> <br /> == See also == <br /> {{AIME box|year=2015|n=II|num-b=2|num-a=4}} {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_23&diff=122873 2017 AMC 12A Problems/Problem 23 2020-05-23T22:02:08Z <p>Frestho: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> For certain real numbers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;, the polynomial &lt;cmath&gt;g(x) = x^3 + ax^2 + x + 10&lt;/cmath&gt;has three distinct roots, and each root of &lt;math&gt;g(x)&lt;/math&gt; is also a root of the polynomial &lt;cmath&gt;f(x) = x^4 + x^3 + bx^2 + 100x + c.&lt;/cmath&gt;What is &lt;math&gt;f(1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;r_1,r_2,&lt;/math&gt; and &lt;math&gt;r_3&lt;/math&gt; be the roots of &lt;math&gt;g(x)&lt;/math&gt;. Let &lt;math&gt;r_4&lt;/math&gt; be the additional root of &lt;math&gt;f(x)&lt;/math&gt;. Then from Vieta's formulas on the quadratic term of &lt;math&gt;g(x)&lt;/math&gt; and the cubic term of &lt;math&gt;f(x)&lt;/math&gt;, we obtain the following:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> r_1+r_2+r_3&amp;=-a \\ <br /> r_1+r_2+r_3+r_4&amp;=-1<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;r_4=a-1&lt;/math&gt;.<br /> <br /> Now applying Vieta's formulas on the constant term of &lt;math&gt;g(x)&lt;/math&gt;, the linear term of &lt;math&gt;g(x)&lt;/math&gt;, and the linear term of &lt;math&gt;f(x)&lt;/math&gt;, we obtain:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> r_1r_2r_3 &amp; = -10\\<br /> r_1r_2+r_2r_3+r_3r_1 &amp;= 1\\ <br /> r_1r_2r_3+r_2r_3r_4+r_3r_4r_1+r_4r_1r_2 &amp; = -100\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Substituting for &lt;math&gt;r_1r_2r_3&lt;/math&gt; in the bottom equation and factoring the remainder of the expression, we obtain:<br /> <br /> &lt;cmath&gt;-10+(r_1r_2+r_2r_3+r_3r_1)r_4=-10+r_4=-100&lt;/cmath&gt;<br /> <br /> It follows that &lt;math&gt;r_4=-90&lt;/math&gt;. But &lt;math&gt;r_4=a-1&lt;/math&gt; so &lt;math&gt;a=-89&lt;/math&gt;<br /> <br /> Now we can factor &lt;math&gt;f(x)&lt;/math&gt; in terms of &lt;math&gt;g(x)&lt;/math&gt; as<br /> <br /> &lt;cmath&gt;f(x)=(x-r_4)g(x)=(x+90)g(x)&lt;/cmath&gt;<br /> <br /> Then &lt;math&gt;f(1)=91g(1)&lt;/math&gt; and<br /> <br /> &lt;cmath&gt;g(1)=1^3-89\cdot 1^2+1+10=-77&lt;/cmath&gt;<br /> <br /> Hence &lt;math&gt;f(1)=91\cdot(-77)=\boxed{\textbf{(C)}\,-7007}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Since all of the roots of &lt;math&gt;g(x)&lt;/math&gt; are distinct and are roots of &lt;math&gt;f(x)&lt;/math&gt;, and the degree of &lt;math&gt;f&lt;/math&gt; is one more than the degree of &lt;math&gt;g&lt;/math&gt;, we have that <br /> <br /> &lt;cmath&gt;f(x) = C(x-k)g(x)&lt;/cmath&gt;<br /> <br /> for some number &lt;math&gt;k&lt;/math&gt;. By comparing &lt;math&gt;x^4&lt;/math&gt; coefficients, we see that &lt;math&gt;C=1&lt;/math&gt;. Thus,<br /> <br /> &lt;cmath&gt;x^4+x^3+bx^2+100x+c=(x-k)(x^3+ax^2+x+10)&lt;/cmath&gt;<br /> <br /> Expanding and equating coefficients we get that<br /> <br /> &lt;cmath&gt;a-k=1,1-ak=b,10-k=100,-10k=c&lt;/cmath&gt;<br /> <br /> The third equation yields &lt;math&gt;k=-90&lt;/math&gt;, and the first equation yields &lt;math&gt;a=-89&lt;/math&gt;. So we have that<br /> <br /> &lt;math&gt;f(1)=(1+90)g(1)=91(1-89+1+10)=(91)(-77)=\boxed{\textbf{(C)}\,-7007}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Let the roots of &lt;math&gt;g(x)&lt;/math&gt; be &lt;math&gt;p,q,r&lt;/math&gt; and the roots of &lt;math&gt;f(x)&lt;/math&gt; be &lt;math&gt;p,q,r,s&lt;/math&gt;. Then by Vietas, &lt;cmath&gt;-100=pqr+pqs+prs+qrs = -10+ s(pq+pr+rs) = -10 + s,&lt;/cmath&gt;so &lt;math&gt;s = -90&lt;/math&gt;. Again by Vietas, &lt;math&gt;p+q+r+s = -a + s = -1 \implies a = -89&lt;/math&gt;. Finally, &lt;math&gt;f(1) = (1-(-90))g(1) = \textbf{(C)}\,-7007&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/MBIiz0mroqk<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=23|num-a=25}}<br /> {{AMC12 box|year=2017|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_6&diff=118940 2019 AIME I Problems/Problem 6 2020-03-09T05:26:26Z <p>Frestho: /* Solution 6 (Alternative PoP) */</p> <hr /> <div>==Problem 6==<br /> In convex quadrilateral &lt;math&gt;KLMN&lt;/math&gt; side &lt;math&gt;\overline{MN}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{KM}&lt;/math&gt;, side &lt;math&gt;\overline{KL}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{LN}&lt;/math&gt;, &lt;math&gt;MN = 65&lt;/math&gt;, and &lt;math&gt;KL = 28&lt;/math&gt;. The line through &lt;math&gt;L&lt;/math&gt; perpendicular to side &lt;math&gt;\overline{KN}&lt;/math&gt; intersects diagonal &lt;math&gt;\overline{KM}&lt;/math&gt; at &lt;math&gt;O&lt;/math&gt; with &lt;math&gt;KO = 8&lt;/math&gt;. Find &lt;math&gt;MO&lt;/math&gt;.<br /> <br /> ==Solution 1 (Trig)==<br /> Let &lt;math&gt;\angle MKN=\alpha&lt;/math&gt; and &lt;math&gt;\angle LNK=\beta&lt;/math&gt;. Note &lt;math&gt;\angle KLP=\beta&lt;/math&gt;. <br /> <br /> Then, &lt;math&gt;KP=28\sin\beta=8\cos\alpha&lt;/math&gt;.<br /> Furthermore, &lt;math&gt;KN=\frac{65}{\sin\alpha}=\frac{28}{\sin\beta} \Rightarrow 65\sin\beta=28\sin\alpha&lt;/math&gt;.<br /> <br /> Dividing the equations gives<br /> &lt;cmath&gt;\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}&lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;MK=\frac{MN}{\tan\alpha}=98&lt;/math&gt;, so &lt;math&gt;MO=MK-KO=\boxed{090}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Similar triangles)==<br /> &lt;asy&gt;<br /> size(250);<br /> real h = sqrt(98^2+65^2);<br /> real l = sqrt(h^2-28^2);<br /> pair K = (0,0);<br /> pair N = (h, 0);<br /> pair M = ((98^2)/h, (98*65)/h);<br /> pair L = ((28^2)/h, (28*l)/h);<br /> pair P = ((28^2)/h, 0);<br /> pair O = ((28^2)/h, (8*65)/h);<br /> draw(K--L--N);<br /> draw(K--M--N--cycle);<br /> draw(L--M);<br /> label(&quot;K&quot;, K, SW);<br /> label(&quot;L&quot;, L, NW);<br /> label(&quot;M&quot;, M, NE);<br /> label(&quot;N&quot;, N, SE);<br /> draw(L--P);<br /> label(&quot;P&quot;, P, S);<br /> dot(O);<br /> label(&quot;O&quot;, shift((1,1))*O, NNE);<br /> label(&quot;28&quot;, scale(1/2)*L, W);<br /> label(&quot;65&quot;, ((M.x+N.x)/2, (M.y+N.y)/2), NE);<br /> &lt;/asy&gt;<br /> <br /> First, let &lt;math&gt;P&lt;/math&gt; be the intersection of &lt;math&gt;LO&lt;/math&gt; and &lt;math&gt;KN&lt;/math&gt; as shown above. Note that &lt;math&gt;m\angle KPL = 90^{\circ}&lt;/math&gt; as given in the problem. Since &lt;math&gt;\angle KPL \cong \angle KLN&lt;/math&gt; and &lt;math&gt;\angle PKL \cong \angle LKN&lt;/math&gt;, &lt;math&gt;\triangle PKL \sim \triangle LKN&lt;/math&gt; by AA similarity. Similarly, &lt;math&gt;\triangle KMN \sim \triangle KPO&lt;/math&gt;. Using these similarities we see that<br /> &lt;cmath&gt;\frac{KP}{KL} = \frac{KL}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}&lt;/cmath&gt;<br /> and<br /> &lt;cmath&gt;\frac{KP}{KO} = \frac{KM}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}&lt;/cmath&gt;<br /> Combining the two equations, we get<br /> &lt;cmath&gt;\frac{8\cdot KM}{KN} = \frac{784}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;8 \cdot KM = 28^2&lt;/cmath&gt;<br /> &lt;cmath&gt;KM = 98&lt;/cmath&gt;<br /> Since &lt;math&gt;KM = KO + MO&lt;/math&gt;, we get &lt;math&gt;MO = 98 -8 = \boxed{090}&lt;/math&gt;.<br /> <br /> Solution by vedadehhc<br /> <br /> ==Solution 3 (Similar triangles, orthocenters)==<br /> Extend &lt;math&gt;KL&lt;/math&gt; and &lt;math&gt;NM&lt;/math&gt; past &lt;math&gt;L&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt; respectively to meet at &lt;math&gt;P&lt;/math&gt;. Let &lt;math&gt;H&lt;/math&gt; be the intersection of diagonals &lt;math&gt;KM&lt;/math&gt; and &lt;math&gt;LN&lt;/math&gt; (this is the orthocenter of &lt;math&gt;\triangle KNP&lt;/math&gt;).<br /> <br /> As &lt;math&gt;\triangle KOL \sim \triangle KHP&lt;/math&gt; (as &lt;math&gt;LO \parallel PH&lt;/math&gt;, using the fact that &lt;math&gt;H&lt;/math&gt; is the orthocenter), we may let &lt;math&gt;OH = 8k&lt;/math&gt; and &lt;math&gt;LP = 28k&lt;/math&gt;.<br /> <br /> Then using similarity with triangles &lt;math&gt;\triangle KLH&lt;/math&gt; and &lt;math&gt;\triangle KMP&lt;/math&gt; we have<br /> <br /> &lt;cmath&gt;\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}&lt;/cmath&gt;<br /> <br /> Cross-multiplying and dividing by &lt;math&gt;4+4k&lt;/math&gt; gives &lt;math&gt;2(8+8k+HM) = 28 \cdot 7 = 196&lt;/math&gt; so &lt;math&gt;MO = 8k + HM = \frac{196}{2} - 8 = \boxed{090}&lt;/math&gt;. (Solution by scrabbler94)<br /> <br /> ==Solution 4 (Algebraic Bashing)==<br /> First, let &lt;math&gt;P&lt;/math&gt; be the intersection of &lt;math&gt;LO&lt;/math&gt; and &lt;math&gt;KN&lt;/math&gt;. We can use the right triangles in the problem to create equations. Let &lt;math&gt;a=NP, b=PK, c=NO, d=OM, e=OP, f=PC,&lt;/math&gt; and &lt;math&gt;g=NC.&lt;/math&gt; We are trying to find &lt;math&gt;d.&lt;/math&gt; We can find &lt;math&gt;7&lt;/math&gt; equations. They are<br /> &lt;cmath&gt;4225+d^2=c^2,&lt;/cmath&gt;<br /> &lt;cmath&gt;4225+d^2+16d+64=a^2+2ab+b^2,&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2+e^2=c^2,&lt;/cmath&gt;<br /> &lt;cmath&gt;b^2+e^2=64,&lt;/cmath&gt;<br /> &lt;cmath&gt;b^2+e^2+2ef+f^2=784,&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2+e^2+2ef+f^2=g^2,&lt;/cmath&gt;<br /> and &lt;cmath&gt;g^2+784=a^2+2ab+b^2.&lt;/cmath&gt;<br /> We can subtract the fifth equation from the sixth equation to get &lt;math&gt;a^2-b^2=g^2-784.&lt;/math&gt; We can subtract the fourth equation from the third equation to get &lt;math&gt;a^2-b^2=c^2-64.&lt;/math&gt; Combining these equations gives &lt;math&gt;c^2-64=g^2-784&lt;/math&gt; so &lt;math&gt;g^2=c^2+720.&lt;/math&gt; Substituting this into the seventh equation gives &lt;math&gt;c^2+1504=a^2+2ab+b^2.&lt;/math&gt; Substituting this into the second equation gives &lt;math&gt;4225+d^2+16d+64=c^2+1504&lt;/math&gt;. Subtracting the first equation from this gives &lt;math&gt;16d+64=1504.&lt;/math&gt; Solving this equation, we find that &lt;math&gt;d=\boxed{090}.&lt;/math&gt;<br /> (Solution by DottedCaculator)<br /> <br /> ==Solution 5 (5-second PoP)==<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair K, L, M, NN, X, O;<br /> K=(-sqrt(98^2+65^2)/2, 0);<br /> NN=(sqrt(98^2+65^2)/2, 0);<br /> L=sqrt(98^2+65^2)/2*dir(180-2*aSin(28/sqrt(98^2+65^2)));<br /> M=sqrt(98^2+65^2)/2*dir(2*aSin(65/sqrt(98^2+65^2)));<br /> X=foot(L, K, NN);<br /> O=extension(L, X, K, M);<br /> draw(K -- L -- M -- NN -- K -- M); draw(L -- NN); draw(arc((K+NN)/2, NN, K));<br /> draw(L -- X, dashed); draw(arc((O+NN)/2, NN, X), dashed);<br /> <br /> draw(rightanglemark(K, L, NN, 100));<br /> draw(rightanglemark(K, M, NN, 100));<br /> draw(rightanglemark(L, X, NN, 100));<br /> dot(&quot;$K$&quot;, K, SW);<br /> dot(&quot;$L$&quot;, L, unit(L));<br /> dot(&quot;$M$&quot;, M, unit(M));<br /> dot(&quot;$N$&quot;, NN, SE);<br /> dot(&quot;$X$&quot;, X, S);<br /> &lt;/asy&gt;<br /> Notice that &lt;math&gt;KLMN&lt;/math&gt; is inscribed in the circle with diameter &lt;math&gt;\overline{KN}&lt;/math&gt; and &lt;math&gt;XOMN&lt;/math&gt; is inscribed in the circle with diameter &lt;math&gt;\overline{ON}&lt;/math&gt;. Furthermore, &lt;math&gt;(XLN)&lt;/math&gt; is tangent to &lt;math&gt;\overline{KL}&lt;/math&gt;. Then, &lt;cmath&gt;KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,&lt;/cmath&gt;and &lt;math&gt;MO=KM-KO=\boxed{090}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 6 (Alternative PoP)==<br /> <br /> &lt;asy&gt;<br /> size(250);<br /> real h = sqrt(98^2+65^2);<br /> real l = sqrt(h^2-28^2);<br /> pair K = (0,0);<br /> pair N = (h, 0);<br /> pair M = ((98^2)/h, (98*65)/h);<br /> pair L = ((28^2)/h, (28*l)/h);<br /> pair P = ((28^2)/h, 0);<br /> pair O = ((28^2)/h, (8*65)/h);<br /> draw(K--L--N);<br /> draw(K--M--N--cycle);<br /> draw(L--M);<br /> label(&quot;K&quot;, K, SW);<br /> label(&quot;L&quot;, L, NW);<br /> label(&quot;M&quot;, M, NE);<br /> label(&quot;N&quot;, N, SE);<br /> draw(L--P);<br /> label(&quot;P&quot;, P, S);<br /> dot(O);<br /> label(&quot;O&quot;, shift((1,1))*O, NNE);<br /> label(&quot;28&quot;, scale(1/2)*L, W);<br /> label(&quot;65&quot;, ((M.x+N.x)/2, (M.y+N.y)/2), NE);<br /> &lt;/asy&gt;<br /> <br /> (Diagram by vedadehhc)<br /> <br /> Call the base of the altitude from &lt;math&gt;L&lt;/math&gt; to &lt;math&gt;NK&lt;/math&gt; point &lt;math&gt;P&lt;/math&gt;. Let &lt;math&gt;PO=x&lt;/math&gt;. Now, we have that &lt;math&gt;KP=\sqrt{64-x^2}&lt;/math&gt; by the Pythagorean Theorem. Once again by Pythagorean, &lt;math&gt;LO=\sqrt{720+x^2}-x&lt;/math&gt;. Using Power of a Point, we have <br /> <br /> &lt;cmath&gt;(KO)(OM)=(LO)(OQ)&lt;/cmath&gt; (&lt;math&gt;Q&lt;/math&gt; is the intersection of &lt;math&gt;OL&lt;/math&gt; with the circle &lt;math&gt;\neq L&lt;/math&gt;)<br /> <br /> &lt;cmath&gt;8(MO)=(\sqrt{720+x^2}+x)(\sqrt{720+x^2}-x)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;8(MO)=720&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;MO=\boxed{090}&lt;/cmath&gt;.<br /> <br /> (Solution by RootThreeOverTwo)<br /> ==Solution 7 (just one pair of similar triangles)==<br /> &lt;asy&gt;<br /> size(250);<br /> real h = sqrt(98^2+65^2);<br /> real l = sqrt(h^2-28^2);<br /> pair K = (0,0);<br /> pair N = (h, 0);<br /> pair M = ((98^2)/h, (98*65)/h);<br /> pair L = ((28^2)/h, (28*l)/h);<br /> pair P = ((28^2)/h, 0);<br /> pair O = ((28^2)/h, (8*65)/h);<br /> draw(K--L--N);<br /> draw(K--M--N--cycle);<br /> draw(L--M);<br /> label(&quot;K&quot;, K, SW);<br /> label(&quot;L&quot;, L, NW);<br /> label(&quot;M&quot;, M, NE);<br /> label(&quot;N&quot;, N, SE);<br /> draw(L--P);<br /> label(&quot;P&quot;, P, S);<br /> dot(O);<br /> label(&quot;O&quot;, shift((1,1))*O, NNE);<br /> label(&quot;28&quot;, scale(1/2)*L, W);<br /> label(&quot;65&quot;, ((M.x+N.x)/2, (M.y+N.y)/2), NE);<br /> &lt;/asy&gt;<br /> Note that since &lt;math&gt;\angle KLN = \angle KMN&lt;/math&gt;, quadrilateral &lt;math&gt;KLMN&lt;/math&gt; is cyclic. Therefore, we have &lt;cmath&gt;\angle LMK = \angle LNK = 90^{\circ} - \angle LKN = \angle KLP,&lt;/cmath&gt;so &lt;math&gt;\triangle KLO \sim \triangle KML&lt;/math&gt;, giving &lt;cmath&gt;\frac{KM}{28} = \frac{28}{8} \implies KM = 98.&lt;/cmath&gt; Therefore, &lt;math&gt;OM = 98-8 = \boxed{90}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> Video Solution:<br /> https://www.youtube.com/watch?v=0AXF-5SsLc8<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_4&diff=118601 2011 AIME I Problems/Problem 4 2020-03-01T06:10:08Z <p>Frestho: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AB=125&lt;/math&gt;, &lt;math&gt;AC=117&lt;/math&gt; and &lt;math&gt;BC=120&lt;/math&gt;. The angle bisector of angle &lt;math&gt;A&lt;/math&gt; intersects &lt;math&gt; \overline{BC} &lt;/math&gt; at point &lt;math&gt;L&lt;/math&gt;, and the angle bisector of angle &lt;math&gt;B&lt;/math&gt; intersects &lt;math&gt; \overline{AC} &lt;/math&gt; at point &lt;math&gt;K&lt;/math&gt;. Let &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt; \overline{BK}&lt;/math&gt; and &lt;math&gt; \overline{AL}&lt;/math&gt;, respectively. Find &lt;math&gt;MN&lt;/math&gt;.<br /> <br /> == Solution 1 == <br /> Extend &lt;math&gt;{CM}&lt;/math&gt; and &lt;math&gt;{CN}&lt;/math&gt; such that they intersect line &lt;math&gt;{AB}&lt;/math&gt; at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, respectively. <br /> Since &lt;math&gt;{BM}&lt;/math&gt; is the angle bisector of angle &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;{CM}&lt;/math&gt; is perpendicular to &lt;math&gt;{BM}&lt;/math&gt;, so &lt;math&gt;BP=BC=120&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;{CP}&lt;/math&gt;. For the same reason, &lt;math&gt;AQ=AC=117&lt;/math&gt;, and &lt;math&gt;N&lt;/math&gt; is the midpoint of &lt;math&gt;{CQ}&lt;/math&gt;.<br /> Hence &lt;math&gt;MN=\frac{PQ}{2}&lt;/math&gt;. But &lt;math&gt;PQ=BP+AQ-AB=120+117-125=112&lt;/math&gt;, so &lt;math&gt;MN=\boxed{056}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;I&lt;/math&gt; be the incenter of &lt;math&gt;ABC&lt;/math&gt;. Now, since &lt;math&gt;IM \perp MC&lt;/math&gt; and &lt;math&gt;IN \perp NC&lt;/math&gt;, we have &lt;math&gt;CMIN&lt;/math&gt; is a cyclic quadrilateral. Consequently, &lt;math&gt;\frac{MN}{\sin \angle MIN} = 2R = CI&lt;/math&gt;. Since &lt;math&gt;\sin \angle MIN = \sin (90 - \frac{\angle BAC}{2}) = \cos \angle IAK&lt;/math&gt;, we have that &lt;math&gt;MN = AI \cdot \cos \angle IAK&lt;/math&gt;. Letting &lt;math&gt;X&lt;/math&gt; be the point of contact of the incircle of &lt;math&gt;ABC&lt;/math&gt; with side &lt;math&gt;AC&lt;/math&gt;, we have &lt;math&gt;AX=MN&lt;/math&gt; thus &lt;math&gt;MN=\frac{117+120-125}{2}=\boxed{056}&lt;/math&gt;<br /> <br /> == Solution 3 (Bash) == <br /> Project &lt;math&gt;I&lt;/math&gt; onto &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; as &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. &lt;math&gt;ID&lt;/math&gt; and &lt;math&gt;IE&lt;/math&gt; are both in-radii of &lt;math&gt;\triangle ABC&lt;/math&gt; so we get right triangles with legs &lt;math&gt;r&lt;/math&gt; (the in-radius length) and &lt;math&gt;s - c = 56&lt;/math&gt;. Since &lt;math&gt;IC&lt;/math&gt; is the hypotenuse for the 4 triangles (&lt;math&gt;\triangle INC, \triangle IMC, \triangle IDC,&lt;/math&gt; and &lt;math&gt;\triangle IEC&lt;/math&gt;), &lt;math&gt;C, D, M, I, N, E&lt;/math&gt; are con-cyclic on a circle we shall denote as &lt;math&gt;\omega&lt;/math&gt; which is also the circumcircle of &lt;math&gt;\triangle CMN&lt;/math&gt; and &lt;math&gt;\triangle CDE&lt;/math&gt;. To find &lt;math&gt;MN&lt;/math&gt;, we can use the Law of Cosines on &lt;math&gt;\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})&lt;/math&gt; where &lt;math&gt;O&lt;/math&gt; is the center of &lt;math&gt;\omega&lt;/math&gt;. Now, the circumradius &lt;math&gt;R&lt;/math&gt; can be found with Pythagorean Theorem with &lt;math&gt;\triangle CDI&lt;/math&gt; or &lt;math&gt;\triangle CEI&lt;/math&gt;: &lt;math&gt;r^2 + 56^2 = (2R)^2&lt;/math&gt;. To find &lt;math&gt;r&lt;/math&gt;, we can use the formula &lt;math&gt;rs = [ABC]&lt;/math&gt; and by Heron's, &lt;math&gt;[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}&lt;/math&gt;. To find &lt;math&gt;\angle MCN&lt;/math&gt;, we can find &lt;math&gt;\angle MIN&lt;/math&gt; since &lt;math&gt;\angle MCN = 180 - \angle MIN&lt;/math&gt;. &lt;math&gt;\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}&lt;/math&gt;. Thus, &lt;math&gt;\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}&lt;/math&gt; and since &lt;math&gt;\angle A + \angle B + \angle C = 180&lt;/math&gt;, we have &lt;math&gt;\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}&lt;/math&gt;. Plugging this into our Law of Cosines formula gives &lt;math&gt;MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})&lt;/math&gt;. To find &lt;math&gt;\cos{\angle C}&lt;/math&gt;, we use LoC on &lt;math&gt;\triangle ABC \implies cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}&lt;/math&gt;. Our formula now becomes &lt;math&gt;MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}&lt;/math&gt;. After simplifying, we get &lt;math&gt;MN^2 = 3136 \implies MN = \boxed{056}&lt;/math&gt;.<br /> <br /> --lucasxia01<br /> <br /> == Solution 4==<br /> <br /> Because &lt;math&gt;\angle CMI = \angle CNI = 90&lt;/math&gt;, &lt;math&gt;CMIN&lt;/math&gt; is cyclic. <br /> <br /> Ptolemy on CMIN:<br /> <br /> &lt;math&gt;CN*MI+CM*IN=CI*MN&lt;/math&gt;<br /> <br /> &lt;math&gt;CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI * MN&lt;/math&gt;<br /> <br /> &lt;math&gt;MN = CI \sin \angle MCN&lt;/math&gt; by angle addition formula.<br /> <br /> &lt;math&gt;\angle MCN = 180 - \angle MIN = 90 - \angle BCI&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;H&lt;/math&gt; be where the incircle touches &lt;math&gt;BC&lt;/math&gt;, then &lt;math&gt;CI \cos \angle BCI = CH = \frac{a+b-c}{2}&lt;/math&gt;.<br /> &lt;math&gt;a=120, b=117, c=125&lt;/math&gt;, for a final answer of &lt;math&gt;\boxed{056}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_13&diff=117547 2019 AIME I Problems/Problem 13 2020-02-09T06:58:52Z <p>Frestho: /* Solution 3 */</p> <hr /> <div>==Problem 13==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has side lengths &lt;math&gt;AB=4&lt;/math&gt;, &lt;math&gt;BC=5&lt;/math&gt;, and &lt;math&gt;CA=6&lt;/math&gt;. Points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; are on ray &lt;math&gt;AB&lt;/math&gt; with &lt;math&gt;AB&lt;AD&lt;AE&lt;/math&gt;. The point &lt;math&gt;F \neq C&lt;/math&gt; is a point of intersection of the circumcircles of &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\triangle EBC&lt;/math&gt; satisfying &lt;math&gt;DF=2&lt;/math&gt; and &lt;math&gt;EF=7&lt;/math&gt;. Then &lt;math&gt;BE&lt;/math&gt; can be expressed as &lt;math&gt;\tfrac{a+b\sqrt{c}}{d}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; are positive integers such that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are relatively prime, and &lt;math&gt;c&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;a+b+c+d&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> &lt;asy&gt;<br /> size(10cm);<br /> pair A, B, C, D, EE, F, X;<br /> B=dir(270-aCos(9/16));<br /> C=dir(270+aCos(9/16));<br /> A=intersectionpoint(circle((0, 0), 1), (B+0.01*(1, 3sqrt(7))) -- (B+100*(1, 3sqrt(7))));<br /> D=B-5/16*(sqrt(2)+1)*(A-B);<br /> EE=B-(5+21*sqrt(2))/16*(A-B);<br /> F=intersectionpoints(circumcircle(A, C, D), circumcircle(B, C, EE));<br /> X=extension(A, B, C, F);<br /> <br /> draw(B -- C -- A -- EE -- F -- C); draw(D -- F);<br /> draw(circumcircle(A, C, D)); draw(circumcircle(C, EE, F));<br /> <br /> dot(&quot;$A$&quot;, A, N);<br /> dot(&quot;$B$&quot;, B, NW);<br /> dot(&quot;$C$&quot;, C, E);<br /> dot(&quot;$D$&quot;, D, SW);<br /> dot(&quot;$E$&quot;, EE, SW);<br /> dot(&quot;$F$&quot;, F, W);<br /> &lt;/asy&gt;<br /> Notice that &lt;cmath&gt;\angle DFE=\angle CFE-\angle CFD=\angle CBE-\angle CAD=180-B-A=C.&lt;/cmath&gt;By the Law of Cosines, &lt;cmath&gt;\cos C=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=\frac34.&lt;/cmath&gt;Then, &lt;cmath&gt;DE^2=DF^2+EF^2-2\cdot DF\cdot EF\cos C=32\implies DE=4\sqrt2.&lt;/cmath&gt;Let &lt;math&gt;X=\overline{AB}\cap\overline{CF}&lt;/math&gt;, &lt;math&gt;a=XB&lt;/math&gt;, and &lt;math&gt;b=XD&lt;/math&gt;. Then, &lt;cmath&gt;XA\cdot XD=XC\cdot XF=XB\cdot XE\implies b(a+4)=a(b+4\sqrt2)\implies b=a\sqrt2.&lt;/cmath&gt;However, since &lt;math&gt;\triangle XFD\sim\triangle XAC&lt;/math&gt;, &lt;math&gt;XF=\tfrac{4+a}3&lt;/math&gt;, but since &lt;math&gt;\triangle XFE\sim\triangle XBC&lt;/math&gt;, &lt;cmath&gt;\frac75=\frac{4+a}{3a}\implies a=\frac54\implies BE=a+a\sqrt2+4\sqrt2=\frac{5+21\sqrt2}4,&lt;/cmath&gt;and the requested sum is &lt;math&gt;5+21+2+4=\boxed{032}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 2==<br /> <br /> Define &lt;math&gt;\omega_1&lt;/math&gt; to be the circumcircle of &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; to be the circumcircle of &lt;math&gt;\triangle EBC&lt;/math&gt;.<br /> <br /> Because of exterior angles, <br /> <br /> &lt;math&gt;\angle ACB = \angle CBE - \angle CAD&lt;/math&gt;<br /> <br /> But &lt;math&gt;\angle CBE = \angle CFE&lt;/math&gt; because &lt;math&gt;CBFE&lt;/math&gt; is cyclic. In addition, &lt;math&gt;\angle CAD = \angle CFD&lt;/math&gt; because &lt;math&gt;CAFD&lt;/math&gt; is cyclic. Therefore, &lt;math&gt;\angle ACB = \angle CFE - \angle CFD&lt;/math&gt;. But &lt;math&gt;\angle CFE - \angle CFD = \angle DFE&lt;/math&gt;, so &lt;math&gt;\angle ACB = \angle DFE&lt;/math&gt;. Using Law of Cosines on &lt;math&gt;\triangle ABC&lt;/math&gt;, we can figure out that &lt;math&gt;\cos(\angle ACB) = \frac{3}{4}&lt;/math&gt;. Since &lt;math&gt;\angle ACB = \angle DFE&lt;/math&gt;, &lt;math&gt;\cos(\angle DFE) = \frac{3}{4}&lt;/math&gt;. We are given that &lt;math&gt;DF = 2&lt;/math&gt; and &lt;math&gt;FE = 7&lt;/math&gt;, so we can use Law of Cosines on &lt;math&gt;\triangle DEF&lt;/math&gt; to find that &lt;math&gt;DE = 4\sqrt{2}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;G&lt;/math&gt; be the intersection of segment &lt;math&gt;\overline{AE}&lt;/math&gt; and &lt;math&gt;\overline{CF}&lt;/math&gt;. Using Power of a Point with respect to &lt;math&gt;G&lt;/math&gt; within &lt;math&gt;\omega_1&lt;/math&gt;, we find that &lt;math&gt;AG \cdot GD = CG \cdot GF&lt;/math&gt;. We can also apply Power of a Point with respect to &lt;math&gt;G&lt;/math&gt; within &lt;math&gt;\omega_2&lt;/math&gt; to find that &lt;math&gt;CG \cdot GF = BG \cdot GE&lt;/math&gt;. Therefore, &lt;math&gt;AG \cdot GD = BG \cdot GE&lt;/math&gt;.<br /> <br /> &lt;math&gt;AG \cdot GD = BG \cdot GE&lt;/math&gt;<br /> <br /> &lt;math&gt;(AB + BG) \cdot GD = BG \cdot (GD + DE)&lt;/math&gt;<br /> <br /> &lt;math&gt;AB \cdot GD + BG \cdot GD = BG \cdot GD + BG \cdot DE&lt;/math&gt;<br /> <br /> &lt;math&gt;AB \cdot GD = BG \cdot DE&lt;/math&gt;<br /> <br /> &lt;math&gt;4 \cdot GD = BG \cdot 4\sqrt{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;GD = BG \cdot \sqrt{2}&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;\triangle GAC&lt;/math&gt; is similar to &lt;math&gt;\triangle GFD&lt;/math&gt;. &lt;math&gt;GF = \frac{BG + 4}{3}&lt;/math&gt;. Also note that &lt;math&gt;\triangle GBC&lt;/math&gt; is similar to &lt;math&gt;\triangle GFE&lt;/math&gt;, which gives us &lt;math&gt;GF = \frac{7 \cdot BG}{5}&lt;/math&gt;. Solving this system of linear equations, we get &lt;math&gt;BG = \frac{5}{4}&lt;/math&gt;. Now, we can solve for &lt;math&gt;BE&lt;/math&gt;, which is equal to &lt;math&gt;BG(\sqrt{2} + 1) + 4\sqrt{2}&lt;/math&gt;. This simplifies to &lt;math&gt;\frac{5 + 21\sqrt{2}}{4}&lt;/math&gt;, which means our answer is &lt;math&gt;\boxed{032}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Construct &lt;math&gt;FC&lt;/math&gt; and let &lt;math&gt;FC\cap AE=K&lt;/math&gt;. Let &lt;math&gt;FK=x&lt;/math&gt;. Using &lt;math&gt;\triangle FKE\sim \triangle BKC&lt;/math&gt;, &lt;cmath&gt;BK=\frac{5}{7}x&lt;/cmath&gt; Using &lt;math&gt;\triangle FDK\sim ACK&lt;/math&gt;, it can be found that &lt;cmath&gt;3x=AK=4+\frac{5}{7}x\to x=\frac{7}{4}&lt;/cmath&gt; This also means that &lt;math&gt;BK=\frac{21}{4}-4=\frac{5}{4}&lt;/math&gt;. It suffices to find &lt;math&gt;KE&lt;/math&gt;. It is easy to see the following: &lt;cmath&gt;180-\angle ABC=\angle KBC=\angle KFE&lt;/cmath&gt; Using reverse Law of Cosines on &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;\cos{\angle ABC}=\frac{1}{8}\to \cos{180-\angle ABC}=\frac{-1}{8}&lt;/math&gt;. Using Law of Cosines on &lt;math&gt;\triangle EFK&lt;/math&gt; gives &lt;math&gt;KE=\frac{21\sqrt 2}{4}&lt;/math&gt;, so &lt;math&gt;BE=\frac{5+21\sqrt 2}{4}\to \textbf{032}&lt;/math&gt;.<br /> -franchester<br /> ==Solution 4 (No &lt;C = &lt;DFE, no LoC)==<br /> Let &lt;math&gt;P=AE\cap CF&lt;/math&gt;. Let &lt;math&gt;CP=5x&lt;/math&gt; and &lt;math&gt;BP=5y&lt;/math&gt;; from &lt;math&gt;\triangle{CBP}\sim\triangle{EFP}&lt;/math&gt; we have &lt;math&gt;EP=7x&lt;/math&gt; and &lt;math&gt;FP=7y&lt;/math&gt;. From &lt;math&gt;\triangle{CAP}\sim\triangle{DFP}&lt;/math&gt; we have &lt;math&gt;\frac{6}{4+5y}=\frac{2}{7y}&lt;/math&gt; giving &lt;math&gt;y=\frac{1}{4}&lt;/math&gt;. So &lt;math&gt;BP=\frac{5}{4}&lt;/math&gt; and &lt;math&gt;FP=\frac{7}{4}&lt;/math&gt;. These similar triangles also gives us &lt;math&gt;DP=\frac{5}{3}x&lt;/math&gt; so &lt;math&gt;DE=\frac{16}{3}x&lt;/math&gt;. Now, Stewart's Theorem on &lt;math&gt;\triangle{FEP}&lt;/math&gt; and cevian &lt;math&gt;FD&lt;/math&gt; tells us that &lt;cmath&gt;\frac{560}{9}x^3+28x=\frac{49}{3}x+\frac{245}{3}x,&lt;/cmath&gt;so &lt;math&gt;x=\frac{3\sqrt{2}}{4}&lt;/math&gt;. Then &lt;math&gt;BE=\frac{5}{4}+7x=\frac{5+21\sqrt{2}}{4}&lt;/math&gt; so the answer is &lt;math&gt;\boxed{032}&lt;/math&gt; as desired. (Solution by Trumpeter, but not added to the Wiki by Trumpeter)<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems&diff=116714 2018 AMC 12B Problems 2020-02-03T06:09:40Z <p>Frestho: /* Problem 23 */</p> <hr /> <div>{{AMC12 Problems|year=2018|ab=B}}<br /> <br /> == Problem 1 ==<br /> <br /> Kate bakes 20-inch by 18-inch pan of cornbread. The cornbread is cut into pieces that measure 2 inches by 2 inches. How many pieces of cornbread does the pan contain?<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 180 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 360&lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> Sam drove 96 miles in 90 minutes. His average speed during the first 30 minutes was 60 mph (miles per hour), and his average speed during the second 30 minutes was 65 mph. What was his average speed, in mph, during the last 30 minutes?<br /> <br /> &lt;math&gt;\textbf{(A) } 64 \qquad \textbf{(B) } 65 \qquad \textbf{(C) } 66 \qquad \textbf{(D) } 67 \qquad \textbf{(E) } 68&lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> A line with slope 2 intersects a line with slope 6 at the point &lt;math&gt;(40,30)&lt;/math&gt;. What is the distance between the &lt;math&gt;x&lt;/math&gt;-intercepts of these two lines? <br /> <br /> &lt;math&gt;\textbf{(A) } 5 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 50&lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> A circle has a chord of length &lt;math&gt;10&lt;/math&gt;, and the distance from the center of the circle to the chord is &lt;math&gt;5&lt;/math&gt;. What is the area of the circle?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }25\pi \qquad<br /> \textbf{(B) }50\pi \qquad<br /> \textbf{(C) }75\pi \qquad<br /> \textbf{(D) }100\pi \qquad<br /> \textbf{(E) }125\pi \qquad<br /> &lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> How many subsets of &lt;math&gt;\{2,3,4,5,6,7,8,9\}&lt;/math&gt; contain at least one prime number?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 128} \qquad \textbf{(B)} \text{ 192} \qquad \textbf{(C)} \text{ 224} \qquad \textbf{(D)} \text{ 240} \qquad \textbf{(E)} \text{ 256}&lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> Suppose &lt;math&gt;S&lt;/math&gt; cans of soda can be purchased from a vending machine for &lt;math&gt;Q&lt;/math&gt; quarters. Which of the following expressions describes the number of cans of soda that can be purchased for &lt;math&gt;D&lt;/math&gt; dollars, where 1 dollar is worth 4 quarters?<br /> <br /> &lt;math&gt;\textbf{(A)} \frac{4DQ}{S} \qquad \textbf{(B)} \frac{4DS}{Q} \qquad \textbf{(C)} \frac{4Q}{DS} \qquad \textbf{(D)} \frac{DQ}{4S} \qquad \textbf{(E)} \frac{DS}{4Q}&lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> <br /> What is the value of<br /> &lt;cmath&gt; \log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27? &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10 &lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> <br /> Line segment &lt;math&gt;\overline{AB}&lt;/math&gt; is a diameter of a circle with &lt;math&gt;AB = 24&lt;/math&gt;. Point &lt;math&gt;C&lt;/math&gt;, not equal to &lt;math&gt;A&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt;, lies on the circle. As point &lt;math&gt;C&lt;/math&gt; moves around the circle, the centroid (center of mass) of &lt;math&gt;\triangle ABC&lt;/math&gt; traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?<br /> <br /> &lt;math&gt;\textbf{(A) } 25 \qquad \textbf{(B) } 38 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 75 &lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> <br /> What is<br /> &lt;cmath&gt; \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ? &lt;/cmath&gt;<br /> <br /> &lt;math&gt; \textbf{(A) }100,100 \qquad<br /> \textbf{(B) }500,500\qquad<br /> \textbf{(C) }505,000 \qquad<br /> \textbf{(D) }1,001,000 \qquad<br /> \textbf{(E) }1,010,000 \qquad &lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> <br /> A list of &lt;math&gt;2018&lt;/math&gt; positive integers has a unique mode, which occurs exactly &lt;math&gt;10&lt;/math&gt; times. What is the least number of distinct values that can occur in the list?<br /> <br /> &lt;math&gt; \textbf{(A) }202 \qquad<br /> \textbf{(B) }223 \qquad<br /> \textbf{(C) }224 \qquad<br /> \textbf{(D) }225 \qquad<br /> \textbf{(E) }234 \qquad &lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> <br /> A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point &lt;math&gt;A&lt;/math&gt; in the figure on the right. The box has base length &lt;math&gt;w&lt;/math&gt; and height &lt;math&gt;h&lt;/math&gt;. What is the area of the sheet of wrapping paper?<br /> <br /> &lt;asy&gt; size(270pt);<br /> defaultpen(fontsize(10pt));<br /> filldraw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey);<br /> dot((-3,3));<br /> label(&quot;$A$&quot;,(-3,3),NW);<br /> draw((1,3)--(-3,-1),dashed+linewidth(.5));<br /> draw((-1,3)--(3,-1),dashed+linewidth(.5));<br /> draw((-1,-3)--(3,1),dashed+linewidth(.5));<br /> draw((1,-3)--(-3,1),dashed+linewidth(.5));<br /> draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5));<br /> draw((0,3)--(0,-3),linetype(&quot;2.5 2.5&quot;)+linewidth(.5));<br /> draw((3,0)--(-3,0),linetype(&quot;2.5 2.5&quot;)+linewidth(.5));<br /> label('$w$',(-1,-1),SW);<br /> label('$w$',(1,-1),SE);<br /> draw((4.5,0)--(6.5,2)--(8.5,0)--(6.5,-2)--cycle);<br /> draw((4.5,0)--(8.5,0));<br /> draw((6.5,2)--(6.5,-2));<br /> label(&quot;$A$&quot;,(6.5,0),NW);<br /> dot((6.5,0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \frac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h &lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> <br /> Side &lt;math&gt;\overline{AB}&lt;/math&gt; of &lt;math&gt;\triangle ABC&lt;/math&gt; has length &lt;math&gt;10&lt;/math&gt;. The bisector of angle &lt;math&gt;A&lt;/math&gt; meets &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, and &lt;math&gt;CD = 3&lt;/math&gt;. The set of all possible values of &lt;math&gt;AC&lt;/math&gt; is an open interval &lt;math&gt;(m,n)&lt;/math&gt;. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }16 \qquad<br /> \textbf{(B) }17 \qquad<br /> \textbf{(C) }18 \qquad<br /> \textbf{(D) }19 \qquad<br /> \textbf{(E) }20 \qquad&lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> <br /> Square &lt;math&gt;ABCD&lt;/math&gt; has side length &lt;math&gt;30&lt;/math&gt;. Point &lt;math&gt;P&lt;/math&gt; lies inside the square so that &lt;math&gt;AP = 12&lt;/math&gt; and &lt;math&gt;BP = 26&lt;/math&gt;. The centroids of &lt;math&gt;\triangle{ABP}&lt;/math&gt;, &lt;math&gt;\triangle{BCP}&lt;/math&gt;, &lt;math&gt;\triangle{CDP}&lt;/math&gt;, and &lt;math&gt;\triangle{DAP}&lt;/math&gt; are the vertices of a convex quadrilateral. What is the area of that quadrilateral? <br /> <br /> &lt;asy&gt;<br /> unitsize(120);<br /> pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3);<br /> draw(A--B--C--D--cycle);<br /> dot(P);<br /> defaultpen(fontsize(10pt));<br /> draw(A--P--B);<br /> draw(C--P--D);<br /> label(&quot;$A$&quot;, A, W);<br /> label(&quot;$B$&quot;, B, W);<br /> label(&quot;$C$&quot;, C, E);<br /> label(&quot;$D$&quot;, D, E);<br /> label(&quot;$P$&quot;, P, N*1.5+E*0.5);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }100\sqrt{2}\qquad\textbf{(B) }100\sqrt{3}\qquad\textbf{(C) }200\qquad\textbf{(D) }200\sqrt{2}\qquad\textbf{(E) }200\sqrt{3}&lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> <br /> Joey, Chloe, and their daughter Zoe all have the same birthday. Joey is 1 year older than Chloe, and Zoe is exactly 1 year old today. Today is the first of the 9 birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?<br /> <br /> &lt;math&gt;\textbf{(A) } 7 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 11 &lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> <br /> How many odd positive 3-digit integers are divisible by 3 but do not contain the digit 3?<br /> <br /> &lt;math&gt;\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120 &lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> <br /> The solutions to the equation &lt;math&gt;(z+6)^8=81&lt;/math&gt; are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled &lt;math&gt;A,B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;. What is the least possible area of &lt;math&gt;\triangle ABC?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{6}\sqrt{6} \qquad \textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2} \qquad \textbf{(C) } 2\sqrt3-3\sqrt2 \qquad \textbf{(D) } \frac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1&lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> <br /> Let &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; be positive integers such that &lt;cmath&gt;\frac{5}{9} &lt; \frac{p}{q} &lt; \frac{4}{7}&lt;/cmath&gt;and &lt;math&gt;q&lt;/math&gt; is as small as possible. What is &lt;math&gt;q-p&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19 &lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> A function &lt;math&gt;f&lt;/math&gt; is defined recursively by &lt;math&gt;f(1)=f(2)=1&lt;/math&gt; and &lt;cmath&gt;f(n)=f(n-1)-f(n-2)+n&lt;/cmath&gt;for all integers &lt;math&gt;n \geq 3&lt;/math&gt;. What is &lt;math&gt;f(2018)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 2016} \qquad \textbf{(B)} \text{ 2017} \qquad \textbf{(C)} \text{ 2018} \qquad \textbf{(D)} \text{ 2019} \qquad \textbf{(E)} \text{ 2020}&lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> Mary chose an even &lt;math&gt;4&lt;/math&gt;-digit number &lt;math&gt;n&lt;/math&gt;. She wrote down all the divisors of &lt;math&gt;n&lt;/math&gt; in increasing order from left to right: &lt;math&gt;1,2,...,\dfrac{n}{2},n&lt;/math&gt;. At some moment Mary wrote &lt;math&gt;323&lt;/math&gt; as a divisor of &lt;math&gt;n&lt;/math&gt;. What is the smallest possible value of the next divisor written to the right of &lt;math&gt;323&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646&lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> <br /> Let &lt;math&gt;ABCDEF&lt;/math&gt; be a regular hexagon with side length &lt;math&gt;1&lt;/math&gt;. Denote by &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; the midpoints of sides &lt;math&gt;\overline {AB}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, and &lt;math&gt;\overline{EF}&lt;/math&gt;, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of &lt;math&gt;\triangle ACE&lt;/math&gt; and &lt;math&gt;\triangle XYZ&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad &lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> <br /> In &lt;math&gt;\triangle{ABC}&lt;/math&gt; with side lengths &lt;math&gt;AB = 13&lt;/math&gt;, &lt;math&gt;AC = 12&lt;/math&gt;, and &lt;math&gt;BC = 5&lt;/math&gt;, let &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt; denote the circumcenter and incenter, respectively. A circle with center &lt;math&gt;M&lt;/math&gt; is tangent to the legs &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; and to the circumcircle of &lt;math&gt;\triangle{ABC}&lt;/math&gt;. What is the area of &lt;math&gt;\triangle{MOI}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 5/2\qquad\textbf{(B)}\ 11/4\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 13/4\qquad\textbf{(E)}\ 7/2&lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> <br /> Consider polynomials &lt;math&gt;P(x)&lt;/math&gt; of degree at most &lt;math&gt;3&lt;/math&gt;, each of whose coefficients is an element of &lt;math&gt;\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}&lt;/math&gt;. How many such polynomials satisfy &lt;math&gt;P(-1) = -9&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286 &lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> <br /> Ajay is standing at point &lt;math&gt;A&lt;/math&gt; near Pontianak, Indonesia, &lt;math&gt;0^\circ&lt;/math&gt; latitude and &lt;math&gt;110^\circ \text{ E}&lt;/math&gt; longitude. Billy is standing at point &lt;math&gt;B&lt;/math&gt; near Big Baldy Mountain, Idaho, USA, &lt;math&gt;45^\circ \text{ N}&lt;/math&gt; latitude and &lt;math&gt;115^\circ \text{ W}&lt;/math&gt; longitude. Assume that Earth is a perfect sphere with center &lt;math&gt;C&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle ACB&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }105 \qquad<br /> \textbf{(B) }112\frac{1}{2} \qquad<br /> \textbf{(C) }120 \qquad<br /> \textbf{(D) }135 \qquad<br /> \textbf{(E) }150 \qquad&lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> Let &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; denote the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;. How many real numbers &lt;math&gt;x&lt;/math&gt; satisfy the equation &lt;math&gt;x^2 + 10,000\lfloor x \rfloor = 10,000x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201&lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> <br /> Circles &lt;math&gt;\omega_1&lt;/math&gt;, &lt;math&gt;\omega_2&lt;/math&gt;, and &lt;math&gt;\omega_3&lt;/math&gt; each have radius &lt;math&gt;4&lt;/math&gt; and are placed in the plane so that each circle is externally tangent to the other two. Points &lt;math&gt;P_1&lt;/math&gt;, &lt;math&gt;P_2&lt;/math&gt;, and &lt;math&gt;P_3&lt;/math&gt; lie on &lt;math&gt;\omega_1&lt;/math&gt;, &lt;math&gt;\omega_2&lt;/math&gt;, and &lt;math&gt;\omega_3&lt;/math&gt; respectively such that &lt;math&gt;P_1P_2=P_2P_3=P_3P_1&lt;/math&gt; and line &lt;math&gt;P_iP_{i+1}&lt;/math&gt; is tangent to &lt;math&gt;\omega_i&lt;/math&gt; for each &lt;math&gt;i=1,2,3&lt;/math&gt;, where &lt;math&gt;P_4 = P_1&lt;/math&gt;. See the figure below. The area of &lt;math&gt;\triangle P_1P_2P_3&lt;/math&gt; can be written in the form &lt;math&gt;\sqrt{a}+\sqrt{b}&lt;/math&gt; for positive integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. What is &lt;math&gt;a+b&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(12);<br /> pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A;<br /> real theta = 41.5;<br /> pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1;<br /> filldraw(P1--P2--P3--cycle, gray(0.9));<br /> draw(Circle(A, 4));<br /> draw(Circle(B, 4));<br /> draw(Circle(C, 4));<br /> dot(P1);<br /> dot(P2);<br /> dot(P3);<br /> defaultpen(fontsize(10pt));<br /> label(&quot;$P_1$&quot;, P1, E*1.5);<br /> label(&quot;$P_2$&quot;, P2, SW*1.5);<br /> label(&quot;$P_3$&quot;, P3, N);<br /> label(&quot;$\omega_1$&quot;, A, W*17);<br /> label(&quot;$\omega_2$&quot;, B, E*17);<br /> label(&quot;$\omega_3$&quot;, C, W*17);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554&lt;/math&gt;<br /> <br /> [[2018 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC12 box|year=2018|ab=B|before=[[2018 AMC 12A Problems]]|after=[[2019 AMC 12A Problems]]}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_24&diff=116711 2008 AMC 10B Problems/Problem 24 2020-02-03T05:20:08Z <p>Frestho: /* Solution 5 */</p> <hr /> <div>==Problem==<br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB = BC = CD&lt;/math&gt;, angle &lt;math&gt;ABC = 70&lt;/math&gt; and angle &lt;math&gt;BCD = 170&lt;/math&gt;. What is the measure of angle &lt;math&gt;BAD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1 ===<br /> Draw the angle bisectors of the angles &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;BCD&lt;/math&gt;. These two bisectors obviously intersect. Let their intersection be &lt;math&gt;P&lt;/math&gt;. <br /> We will now prove that &lt;math&gt;P&lt;/math&gt; lies on the segment &lt;math&gt;AD&lt;/math&gt;. <br /> <br /> Note that the triangles &lt;math&gt;ABP&lt;/math&gt; and &lt;math&gt;CBP&lt;/math&gt; are congruent, as they share the side &lt;math&gt;BP&lt;/math&gt;, and we have &lt;math&gt;AB=BC&lt;/math&gt; and &lt;math&gt;\angle ABP = \angle CBP&lt;/math&gt;.<br /> <br /> Also note that for similar reasons the triangles &lt;math&gt;CBP&lt;/math&gt; and &lt;math&gt;CDP&lt;/math&gt; are congruent.<br /> <br /> Now we can compute their inner angles. &lt;math&gt;BP&lt;/math&gt; is the bisector of the angle &lt;math&gt;ABC&lt;/math&gt;, hence &lt;math&gt;\angle ABP = \angle CBP = 35^\circ&lt;/math&gt;, and thus also &lt;math&gt;\angle CDP = 35^\circ&lt;/math&gt;. (Faster Solution picks up here) &lt;math&gt;CP&lt;/math&gt; is the bisector of the angle &lt;math&gt;BCD&lt;/math&gt;, hence &lt;math&gt;\angle BCP = \angle DCP = 85^\circ&lt;/math&gt;, and thus also &lt;math&gt;\angle BAP = 85^\circ&lt;/math&gt;.<br /> <br /> It follows that &lt;math&gt;\angle APB = \angle BPC = \angle CPD = 180^\circ - 35^\circ - 85^\circ = 60^\circ&lt;/math&gt;. Thus the angle &lt;math&gt;APD&lt;/math&gt; has &lt;math&gt;180^\circ&lt;/math&gt;, and hence &lt;math&gt;P&lt;/math&gt; does indeed lie on &lt;math&gt;AD&lt;/math&gt;. Then obviously &lt;math&gt;\angle BAD = \angle BAP = \boxed{ 85^\circ }&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(.8);<br /> real a=4;<br /> pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120);<br /> draw(A--B--C--D--cycle);<br /> pair P1=B+3*a*dir(145), P2=C+3*a*dir(205);<br /> pair P=intersectionpoint(B--P1,C--P2);<br /> draw(B--P--C);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,N);<br /> label(&quot;$P$&quot;,P,W);<br /> <br /> label(&quot;$35^\circ$&quot;,B + dir(180-17.5));<br /> label(&quot;$35^\circ$&quot;,B + dir(180-35-17.5));<br /> <br /> label(&quot;$85^\circ$&quot;,C + .5*dir(120+42.5));<br /> label(&quot;$85^\circ$&quot;,C + .5*dir(120+85+42.5));<br /> &lt;/asy&gt;<br /> <br /> Faster Solution: Because we now know three angles, we can subtract to get &lt;math&gt;360 - 35 - 85 - 85 - 35 - 35&lt;/math&gt;, or &lt;math&gt;\boxed{85}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Draw the diagonals &lt;math&gt;\overline{BD}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt;, and suppose that they intersect at &lt;math&gt;E&lt;/math&gt;. Then, &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle BCD&lt;/math&gt; are both isosceles, so by angle-chasing, we find that &lt;math&gt;\angle BAC = 55^{\circ}&lt;/math&gt;, &lt;math&gt;\angle CBD = 5^{\circ}&lt;/math&gt;, and &lt;math&gt;\angle BEA = 180 - \angle EBA - \angle BAE = 60^{\circ}&lt;/math&gt;. Draw &lt;math&gt;E'&lt;/math&gt; such that &lt;math&gt;EE'B = 60^{\circ}&lt;/math&gt; and so that &lt;math&gt;E'&lt;/math&gt; is on &lt;math&gt;\overline{AE}&lt;/math&gt;, and draw &lt;math&gt;E''&lt;/math&gt; such that &lt;math&gt;\angle EE''C = 60^{\circ}&lt;/math&gt; and &lt;math&gt;E''&lt;/math&gt; is on &lt;math&gt;\overline{DE}&lt;/math&gt;. It follows that &lt;math&gt;\triangle BEE'&lt;/math&gt; and &lt;math&gt;\triangle CEE''&lt;/math&gt; are both equilateral. Also, it is easy to see that &lt;math&gt;\triangle BEC \cong \triangle DE''C&lt;/math&gt; and &lt;math&gt;\triangle BCE \cong \triangle BAE'&lt;/math&gt; by construction, so that &lt;math&gt;DE'' = BE = EE'&lt;/math&gt; and &lt;math&gt;EE'' = CE = E'A&lt;/math&gt;. Thus, &lt;math&gt;AE = AE' + E'E = EE'' + DE'' = DE&lt;/math&gt;, so &lt;math&gt;\triangle ADE&lt;/math&gt; is isosceles. Since &lt;math&gt;\angle AED = 120^{\circ}&lt;/math&gt;, then &lt;math&gt;\angle DAC = \frac{180 - 120}{2} = 30^{\circ}&lt;/math&gt;, and &lt;math&gt;\angle BAD = 30 + 55 = 85^{\circ}&lt;/math&gt;. <br /> &lt;asy&gt;<br /> import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; <br /> pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); <br /> filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); <br /> dot((0,0),ds); label(&quot;$A$&quot;,(-0.096,0.005),NE*lsf); dot((1,0),ds); label(&quot;$B$&quot;,(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(&quot;$C$&quot;,(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(&quot;$D$&quot;,(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(&quot;$E$&quot;,(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(&quot;$E'$&quot;,(0.1,0.23),NE*lsf); label(&quot;$60^\circ$&quot;,(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(&quot;$E''$&quot;,(0.423,0.957),NE*lsf); label(&quot;$60^\circ$&quot;,(0.761,0.886),NE*lsf,qqwuqq); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> &lt;/asy&gt;<br /> <br /> ===Solution 3 ===<br /> Again, draw the diagonals &lt;math&gt;\overline{BD}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt;, and suppose that they intersect at &lt;math&gt;E&lt;/math&gt;. We find by angle chasing the same way as in solution 2 that &lt;math&gt;m\angle ABE = 65^\circ&lt;/math&gt; and &lt;math&gt;m\angle DCE = 115^\circ&lt;/math&gt;. Applying the Law of Sines to &lt;math&gt;\triangle AEB&lt;/math&gt; and &lt;math&gt;\triangle EDC&lt;/math&gt;, it follows that &lt;math&gt;DE = \frac{2\sin 115^\circ}{\sin \angle DEC} = \frac{2\sin 65^\circ}{\sin \angle AEB} = EA&lt;/math&gt;, so &lt;math&gt;\triangle AED&lt;/math&gt; is isosceles. We finish as we did in solution 2.<br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(.8);<br /> real a=4;<br /> pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120);<br /> draw(A--B--C--D--cycle);<br /> pair P=intersectionpoint(B--D,C--A);<br /> draw(A--C); draw(B--D);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,N);<br /> label(&quot;$E$&quot;,P,W);<br /> &lt;/asy&gt;<br /> <br /> === Solution 4 ===<br /> Start off with the same diagram as solution 1. Now draw &lt;math&gt;\overline{CA}&lt;/math&gt; which creates isosceles &lt;math&gt;\triangle CAB&lt;/math&gt;. We know that the angle bisector of an isosceles triangle splits it in half, we can extrapolate this further to see that it's is &lt;math&gt;\boxed{85}.&lt;/math&gt;<br /> <br /> === Solution 5 ===<br /> Just draw a very accurate diagram with a ruler and protractor and boom.<br /> <br /> ==See also==<br /> {{AMC10 box|year=2008|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_24&diff=116710 2008 AMC 10B Problems/Problem 24 2020-02-03T05:19:47Z <p>Frestho: /* Solution 4 */</p> <hr /> <div>==Problem==<br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB = BC = CD&lt;/math&gt;, angle &lt;math&gt;ABC = 70&lt;/math&gt; and angle &lt;math&gt;BCD = 170&lt;/math&gt;. What is the measure of angle &lt;math&gt;BAD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1 ===<br /> Draw the angle bisectors of the angles &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;BCD&lt;/math&gt;. These two bisectors obviously intersect. Let their intersection be &lt;math&gt;P&lt;/math&gt;. <br /> We will now prove that &lt;math&gt;P&lt;/math&gt; lies on the segment &lt;math&gt;AD&lt;/math&gt;. <br /> <br /> Note that the triangles &lt;math&gt;ABP&lt;/math&gt; and &lt;math&gt;CBP&lt;/math&gt; are congruent, as they share the side &lt;math&gt;BP&lt;/math&gt;, and we have &lt;math&gt;AB=BC&lt;/math&gt; and &lt;math&gt;\angle ABP = \angle CBP&lt;/math&gt;.<br /> <br /> Also note that for similar reasons the triangles &lt;math&gt;CBP&lt;/math&gt; and &lt;math&gt;CDP&lt;/math&gt; are congruent.<br /> <br /> Now we can compute their inner angles. &lt;math&gt;BP&lt;/math&gt; is the bisector of the angle &lt;math&gt;ABC&lt;/math&gt;, hence &lt;math&gt;\angle ABP = \angle CBP = 35^\circ&lt;/math&gt;, and thus also &lt;math&gt;\angle CDP = 35^\circ&lt;/math&gt;. (Faster Solution picks up here) &lt;math&gt;CP&lt;/math&gt; is the bisector of the angle &lt;math&gt;BCD&lt;/math&gt;, hence &lt;math&gt;\angle BCP = \angle DCP = 85^\circ&lt;/math&gt;, and thus also &lt;math&gt;\angle BAP = 85^\circ&lt;/math&gt;.<br /> <br /> It follows that &lt;math&gt;\angle APB = \angle BPC = \angle CPD = 180^\circ - 35^\circ - 85^\circ = 60^\circ&lt;/math&gt;. Thus the angle &lt;math&gt;APD&lt;/math&gt; has &lt;math&gt;180^\circ&lt;/math&gt;, and hence &lt;math&gt;P&lt;/math&gt; does indeed lie on &lt;math&gt;AD&lt;/math&gt;. Then obviously &lt;math&gt;\angle BAD = \angle BAP = \boxed{ 85^\circ }&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(.8);<br /> real a=4;<br /> pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120);<br /> draw(A--B--C--D--cycle);<br /> pair P1=B+3*a*dir(145), P2=C+3*a*dir(205);<br /> pair P=intersectionpoint(B--P1,C--P2);<br /> draw(B--P--C);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,N);<br /> label(&quot;$P$&quot;,P,W);<br /> <br /> label(&quot;$35^\circ$&quot;,B + dir(180-17.5));<br /> label(&quot;$35^\circ$&quot;,B + dir(180-35-17.5));<br /> <br /> label(&quot;$85^\circ$&quot;,C + .5*dir(120+42.5));<br /> label(&quot;$85^\circ$&quot;,C + .5*dir(120+85+42.5));<br /> &lt;/asy&gt;<br /> <br /> Faster Solution: Because we now know three angles, we can subtract to get &lt;math&gt;360 - 35 - 85 - 85 - 35 - 35&lt;/math&gt;, or &lt;math&gt;\boxed{85}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Draw the diagonals &lt;math&gt;\overline{BD}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt;, and suppose that they intersect at &lt;math&gt;E&lt;/math&gt;. Then, &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle BCD&lt;/math&gt; are both isosceles, so by angle-chasing, we find that &lt;math&gt;\angle BAC = 55^{\circ}&lt;/math&gt;, &lt;math&gt;\angle CBD = 5^{\circ}&lt;/math&gt;, and &lt;math&gt;\angle BEA = 180 - \angle EBA - \angle BAE = 60^{\circ}&lt;/math&gt;. Draw &lt;math&gt;E'&lt;/math&gt; such that &lt;math&gt;EE'B = 60^{\circ}&lt;/math&gt; and so that &lt;math&gt;E'&lt;/math&gt; is on &lt;math&gt;\overline{AE}&lt;/math&gt;, and draw &lt;math&gt;E''&lt;/math&gt; such that &lt;math&gt;\angle EE''C = 60^{\circ}&lt;/math&gt; and &lt;math&gt;E''&lt;/math&gt; is on &lt;math&gt;\overline{DE}&lt;/math&gt;. It follows that &lt;math&gt;\triangle BEE'&lt;/math&gt; and &lt;math&gt;\triangle CEE''&lt;/math&gt; are both equilateral. Also, it is easy to see that &lt;math&gt;\triangle BEC \cong \triangle DE''C&lt;/math&gt; and &lt;math&gt;\triangle BCE \cong \triangle BAE'&lt;/math&gt; by construction, so that &lt;math&gt;DE'' = BE = EE'&lt;/math&gt; and &lt;math&gt;EE'' = CE = E'A&lt;/math&gt;. Thus, &lt;math&gt;AE = AE' + E'E = EE'' + DE'' = DE&lt;/math&gt;, so &lt;math&gt;\triangle ADE&lt;/math&gt; is isosceles. Since &lt;math&gt;\angle AED = 120^{\circ}&lt;/math&gt;, then &lt;math&gt;\angle DAC = \frac{180 - 120}{2} = 30^{\circ}&lt;/math&gt;, and &lt;math&gt;\angle BAD = 30 + 55 = 85^{\circ}&lt;/math&gt;. <br /> &lt;asy&gt;<br /> import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; <br /> pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); <br /> filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); <br /> dot((0,0),ds); label(&quot;$A$&quot;,(-0.096,0.005),NE*lsf); dot((1,0),ds); label(&quot;$B$&quot;,(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(&quot;$C$&quot;,(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(&quot;$D$&quot;,(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(&quot;$E$&quot;,(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(&quot;$E'$&quot;,(0.1,0.23),NE*lsf); label(&quot;$60^\circ$&quot;,(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(&quot;$E''$&quot;,(0.423,0.957),NE*lsf); label(&quot;$60^\circ$&quot;,(0.761,0.886),NE*lsf,qqwuqq); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> &lt;/asy&gt;<br /> <br /> ===Solution 3 ===<br /> Again, draw the diagonals &lt;math&gt;\overline{BD}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt;, and suppose that they intersect at &lt;math&gt;E&lt;/math&gt;. We find by angle chasing the same way as in solution 2 that &lt;math&gt;m\angle ABE = 65^\circ&lt;/math&gt; and &lt;math&gt;m\angle DCE = 115^\circ&lt;/math&gt;. Applying the Law of Sines to &lt;math&gt;\triangle AEB&lt;/math&gt; and &lt;math&gt;\triangle EDC&lt;/math&gt;, it follows that &lt;math&gt;DE = \frac{2\sin 115^\circ}{\sin \angle DEC} = \frac{2\sin 65^\circ}{\sin \angle AEB} = EA&lt;/math&gt;, so &lt;math&gt;\triangle AED&lt;/math&gt; is isosceles. We finish as we did in solution 2.<br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(.8);<br /> real a=4;<br /> pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120);<br /> draw(A--B--C--D--cycle);<br /> pair P=intersectionpoint(B--D,C--A);<br /> draw(A--C); draw(B--D);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,N);<br /> label(&quot;$E$&quot;,P,W);<br /> &lt;/asy&gt;<br /> <br /> === Solution 4 ===<br /> Start off with the same diagram as solution 1. Now draw &lt;math&gt;\overline{CA}&lt;/math&gt; which creates isosceles &lt;math&gt;\triangle CAB&lt;/math&gt;. We know that the angle bisector of an isosceles triangle splits it in half, we can extrapolate this further to see that it's is &lt;math&gt;\boxed{85}.&lt;/math&gt;<br /> <br /> == Solution 5 ==<br /> Just draw a very accurate diagram with a ruler and protractor and boom.<br /> <br /> ==See also==<br /> {{AMC10 box|year=2008|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_21&diff=116656 2020 AMC 10A Problems/Problem 21 2020-02-02T22:47:24Z <p>Frestho: /* Solution 4 */</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #19]] and [[2020 AMC 10A Problems|2020 AMC 10A #21]]}}<br /> <br /> ==Problem==<br /> <br /> There exists a unique strictly increasing sequence of nonnegative integers &lt;math&gt;a_1 &lt; a_2 &lt; … &lt; a_k&lt;/math&gt; such that&lt;cmath&gt;\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.&lt;/cmath&gt;What is &lt;math&gt;k?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> First, substitute &lt;math&gt;2^{17}&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt;. <br /> Then, the given equation becomes &lt;math&gt;\frac{a^{17}+1}{a+1}=a^{16}-a^{15}+a^{14}...-a^1+a^0&lt;/math&gt;.<br /> Now consider only &lt;math&gt;a^{16}-a^{15}&lt;/math&gt;. This equals &lt;math&gt;a^{15}(a-1)=a^{15}*(2^{17}-1)&lt;/math&gt;.<br /> Note that &lt;math&gt;2^{17}-1&lt;/math&gt; equals &lt;math&gt;2^{16}+2^{15}+...+1&lt;/math&gt;, since the sum of a geometric sequence is &lt;math&gt;\frac{a^n-1}{a-1}&lt;/math&gt;.<br /> Thus, we can see that &lt;math&gt;a^{16}-a^{15}&lt;/math&gt; forms the sum of 17 different powers of 2. <br /> Applying the same method to each of &lt;math&gt;a^{14}-a^{13}&lt;/math&gt;, &lt;math&gt;a^{12}-a^{11}&lt;/math&gt;, ... , &lt;math&gt;a^{2}-a^{1}&lt;/math&gt;, we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us &lt;math&gt;17*8=136&lt;/math&gt;.<br /> But we must count also the &lt;math&gt;a^0&lt;/math&gt; term. <br /> Thus, Our answer is &lt;math&gt;136+1=\boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ~seanyoon777<br /> <br /> == Solution 2 ==<br /> (This is similar to solution 1)<br /> Let &lt;math&gt;x = 2^{17}&lt;/math&gt;. Then, &lt;math&gt;2^{289} = x^{17}&lt;/math&gt;.<br /> The LHS can be rewritten as &lt;math&gt;\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+\cdots+x^2-x+1=(x-1)(x^{15}+x^{13}+\cdots+x^{1})+1&lt;/math&gt;. Plugging &lt;math&gt;2^{17}&lt;/math&gt; back in for &lt;math&gt;x&lt;/math&gt;, we have &lt;math&gt;(2^{17}-1)(2^{15\cdot{17}}+2^{13\cdot{17}}+\cdots+2^{1\cdot{17}})+1=(2^{16}+2^{15}+\cdots+2^{0})(2^{15\cdot17}+2^{13\cdot17}+\cdots+2^{1\cdot17})+1&lt;/math&gt;. When expanded, this will have &lt;math&gt;17\cdot8+1=137&lt;/math&gt; terms. Therefore, our answer is &lt;math&gt;\boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Intuitive)==<br /> Multiply both sides by &lt;math&gt;2^{17}+1&lt;/math&gt; to get &lt;cmath&gt;2^{289}+1=2^{a_1} + 2^{a_2} + … + 2^{a_k} + 2^{a_1+17} + 2^{a_2+17} + … + 2^{a_k+17}.&lt;/cmath&gt;<br /> <br /> Notice that &lt;math&gt;a_1 = 0&lt;/math&gt;, since there is a &lt;math&gt;1&lt;/math&gt; on the LHS. However, now we have an extra term of &lt;math&gt;2^{18}&lt;/math&gt; on the right from &lt;math&gt;2^{a_1+17}&lt;/math&gt;. To cancel it, we let &lt;math&gt;a_2 = 18&lt;/math&gt;. The two &lt;math&gt;2^{18}&lt;/math&gt;'s now combine into a term of &lt;math&gt;2^{19}&lt;/math&gt;, so we let &lt;math&gt;a_3 = 19&lt;/math&gt;. And so on, until we get to &lt;math&gt;a_{18} = 34&lt;/math&gt;. Now everything we don't want telescopes into &lt;math&gt;2^{35}&lt;/math&gt;. We already have that term since we let &lt;math&gt;a_2 = 18 \implies a_2+17 = 35&lt;/math&gt;. Everything from now on will automatically telescope to &lt;math&gt;2^{52}&lt;/math&gt;. So we let &lt;math&gt;a_{19}&lt;/math&gt; be &lt;math&gt;52&lt;/math&gt;.<br /> <br /> As you can see, we will have to add &lt;math&gt;17&lt;/math&gt; &lt;math&gt;a_n&lt;/math&gt;'s at a time, then &quot;wait&quot; for the sum to automatically telescope for the next &lt;math&gt;17&lt;/math&gt; numbers, etc, until we get to &lt;math&gt;2^{289}&lt;/math&gt;. We only need to add &lt;math&gt;a_n&lt;/math&gt;'s between odd multiples of &lt;math&gt;17&lt;/math&gt; and even multiples. The largest even multiple of &lt;math&gt;17&lt;/math&gt; below &lt;math&gt;289&lt;/math&gt; is &lt;math&gt;17\cdot16&lt;/math&gt;, so we will have to add a total of &lt;math&gt;17\cdot 8&lt;/math&gt; &lt;math&gt;a_n&lt;/math&gt;'s. However, we must not forget we let &lt;math&gt;a_1=0&lt;/math&gt; at the beginning, so our answer is &lt;math&gt;17\cdot8+1 = \boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Note that the expression is equal to something slightly lower than &lt;math&gt;2^{272}&lt;/math&gt;. Clearly, answer choices &lt;math&gt;(D)&lt;/math&gt; and &lt;math&gt;(E)&lt;/math&gt; make no sense because the lowest sum for &lt;math&gt;273&lt;/math&gt; terms is &lt;math&gt;2^{273}-1&lt;/math&gt;. &lt;math&gt;(A)&lt;/math&gt; just makes no sense. &lt;math&gt;(B)&lt;/math&gt; and &lt;math&gt;(C)&lt;/math&gt; are 1 apart, but because the expression is odd, it will have to contain &lt;math&gt;2^0=1&lt;/math&gt;, and because &lt;math&gt;(C)&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; bigger, the answer is &lt;math&gt;\boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ~Lcz<br /> <br /> == Solution 5 ==<br /> In order to shorten expressions, &lt;math&gt;\#&lt;/math&gt; will represent &lt;math&gt;16&lt;/math&gt; consecutive &lt;math&gt;0&lt;/math&gt;s when expressing numbers. &lt;br&gt;<br /> &lt;br&gt;<br /> Think of the problem in binary. We have &lt;br&gt;<br /> &lt;math&gt;\frac{1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2}{1\#1_2}&lt;/math&gt; &lt;br&gt;<br /> Note that &lt;br&gt;<br /> &lt;math&gt;(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) = 2^0(2^{17} + 1) + 2^{34}(2^{17} + 1) + 2^{68}(2^{17} + 1) + \cdots 2^{272}(2^{17} + 1)&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2&lt;/math&gt; &lt;br&gt;<br /> and &lt;br&gt;<br /> &lt;math&gt;(2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{17}(2^{17} + 1) + 2^{51}(2^{17} + 1) + 2^{85}(2^{17} + 1) + \cdots 2^{255}(2^{17} + 1)&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Since &lt;br&gt;<br /> &lt;math&gt;\phantom{=\ } 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;-\ \phantom{1\#} 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2&lt;/math&gt; &lt;br&gt;<br /> this means that&lt;br&gt;<br /> &lt;math&gt;(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{289}&lt;/math&gt; &lt;br&gt;<br /> so &lt;br&gt;<br /> &lt;math&gt;\frac{2^{289}+1}{2^{17}+1} = (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255})&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 2^0 + (2^{34} - 2^{17}) + (2^{68} - 2^{51}) + \cdots + (2^{272} - 2^{255})&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Expressing each of the pairs of the form &lt;math&gt;2^{n + 17} - 2^n&lt;/math&gt; in binary, we have &lt;br&gt;<br /> &lt;math&gt;\phantom{=\ } 1000000000000000000 \cdots 0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;-\ \phantom{10000000000000000} 10 \cdots 0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= \phantom{1} 111111111111111110 \cdots 0_2&lt;/math&gt; &lt;br&gt;<br /> or &lt;br&gt;<br /> &lt;math&gt;2^{n + 17} - 2^n = 2^{n + 16} + 2^{n + 15} + 2^{n + 14} + \cdots + 2^{n}&lt;/math&gt; &lt;br&gt;<br /> This means that each pair has &lt;math&gt;17&lt;/math&gt; terms of the form &lt;math&gt;2^n&lt;/math&gt;. &lt;br&gt;<br /> &lt;br&gt;<br /> Since there are &lt;math&gt;8&lt;/math&gt; of these pairs, there are a total of &lt;math&gt;8 \cdot 17 = 136&lt;/math&gt; terms. Accounting for the &lt;math&gt;2^0&lt;/math&gt; term, which was not in the pair, we have a total of &lt;math&gt;136 + 1 = \boxed{\textbf{(C) } 137}&lt;/math&gt; terms. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Video Solution==<br /> https://youtu.be/Ozp3k2464u4<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2020|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_21&diff=116655 2020 AMC 10A Problems/Problem 21 2020-02-02T22:47:16Z <p>Frestho: /* Solution 3 */</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #19]] and [[2020 AMC 10A Problems|2020 AMC 10A #21]]}}<br /> <br /> ==Problem==<br /> <br /> There exists a unique strictly increasing sequence of nonnegative integers &lt;math&gt;a_1 &lt; a_2 &lt; … &lt; a_k&lt;/math&gt; such that&lt;cmath&gt;\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.&lt;/cmath&gt;What is &lt;math&gt;k?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> First, substitute &lt;math&gt;2^{17}&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt;. <br /> Then, the given equation becomes &lt;math&gt;\frac{a^{17}+1}{a+1}=a^{16}-a^{15}+a^{14}...-a^1+a^0&lt;/math&gt;.<br /> Now consider only &lt;math&gt;a^{16}-a^{15}&lt;/math&gt;. This equals &lt;math&gt;a^{15}(a-1)=a^{15}*(2^{17}-1)&lt;/math&gt;.<br /> Note that &lt;math&gt;2^{17}-1&lt;/math&gt; equals &lt;math&gt;2^{16}+2^{15}+...+1&lt;/math&gt;, since the sum of a geometric sequence is &lt;math&gt;\frac{a^n-1}{a-1}&lt;/math&gt;.<br /> Thus, we can see that &lt;math&gt;a^{16}-a^{15}&lt;/math&gt; forms the sum of 17 different powers of 2. <br /> Applying the same method to each of &lt;math&gt;a^{14}-a^{13}&lt;/math&gt;, &lt;math&gt;a^{12}-a^{11}&lt;/math&gt;, ... , &lt;math&gt;a^{2}-a^{1}&lt;/math&gt;, we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us &lt;math&gt;17*8=136&lt;/math&gt;.<br /> But we must count also the &lt;math&gt;a^0&lt;/math&gt; term. <br /> Thus, Our answer is &lt;math&gt;136+1=\boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ~seanyoon777<br /> <br /> == Solution 2 ==<br /> (This is similar to solution 1)<br /> Let &lt;math&gt;x = 2^{17}&lt;/math&gt;. Then, &lt;math&gt;2^{289} = x^{17}&lt;/math&gt;.<br /> The LHS can be rewritten as &lt;math&gt;\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+\cdots+x^2-x+1=(x-1)(x^{15}+x^{13}+\cdots+x^{1})+1&lt;/math&gt;. Plugging &lt;math&gt;2^{17}&lt;/math&gt; back in for &lt;math&gt;x&lt;/math&gt;, we have &lt;math&gt;(2^{17}-1)(2^{15\cdot{17}}+2^{13\cdot{17}}+\cdots+2^{1\cdot{17}})+1=(2^{16}+2^{15}+\cdots+2^{0})(2^{15\cdot17}+2^{13\cdot17}+\cdots+2^{1\cdot17})+1&lt;/math&gt;. When expanded, this will have &lt;math&gt;17\cdot8+1=137&lt;/math&gt; terms. Therefore, our answer is &lt;math&gt;\boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Intuitive)==<br /> Multiply both sides by &lt;math&gt;2^{17}+1&lt;/math&gt; to get &lt;cmath&gt;2^{289}+1=2^{a_1} + 2^{a_2} + … + 2^{a_k} + 2^{a_1+17} + 2^{a_2+17} + … + 2^{a_k+17}.&lt;/cmath&gt;<br /> <br /> Notice that &lt;math&gt;a_1 = 0&lt;/math&gt;, since there is a &lt;math&gt;1&lt;/math&gt; on the LHS. However, now we have an extra term of &lt;math&gt;2^{18}&lt;/math&gt; on the right from &lt;math&gt;2^{a_1+17}&lt;/math&gt;. To cancel it, we let &lt;math&gt;a_2 = 18&lt;/math&gt;. The two &lt;math&gt;2^{18}&lt;/math&gt;'s now combine into a term of &lt;math&gt;2^{19}&lt;/math&gt;, so we let &lt;math&gt;a_3 = 19&lt;/math&gt;. And so on, until we get to &lt;math&gt;a_{18} = 34&lt;/math&gt;. Now everything we don't want telescopes into &lt;math&gt;2^{35}&lt;/math&gt;. We already have that term since we let &lt;math&gt;a_2 = 18 \implies a_2+17 = 35&lt;/math&gt;. Everything from now on will automatically telescope to &lt;math&gt;2^{52}&lt;/math&gt;. So we let &lt;math&gt;a_{19}&lt;/math&gt; be &lt;math&gt;52&lt;/math&gt;.<br /> <br /> As you can see, we will have to add &lt;math&gt;17&lt;/math&gt; &lt;math&gt;a_n&lt;/math&gt;'s at a time, then &quot;wait&quot; for the sum to automatically telescope for the next &lt;math&gt;17&lt;/math&gt; numbers, etc, until we get to &lt;math&gt;2^{289}&lt;/math&gt;. We only need to add &lt;math&gt;a_n&lt;/math&gt;'s between odd multiples of &lt;math&gt;17&lt;/math&gt; and even multiples. The largest even multiple of &lt;math&gt;17&lt;/math&gt; below &lt;math&gt;289&lt;/math&gt; is &lt;math&gt;17\cdot16&lt;/math&gt;, so we will have to add a total of &lt;math&gt;17\cdot 8&lt;/math&gt; &lt;math&gt;a_n&lt;/math&gt;'s. However, we must not forget we let &lt;math&gt;a_1=0&lt;/math&gt; at the beginning, so our answer is &lt;math&gt;17\cdot8+1 = \boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Note that the expression is equal to something slightly lower than &lt;math&gt;2^{272}&lt;/math&gt;. Clearly, answer choices &lt;math&gt;(D)&lt;/math&gt; and &lt;math&gt;(E)&lt;/math&gt; make no sense because the lowest sum for &lt;math&gt;273&lt;/math&gt; terms is &lt;math&gt;2^{273}-1&lt;/math&gt;. &lt;math&gt;(A)&lt;/math&gt; just makes no sense. &lt;math&gt;(B)&lt;/math&gt; and &lt;math&gt;(C)&lt;/math&gt; are 1 apart, but because the expression is odd, it will have to contain &lt;math&gt;2^0=1&lt;/math&gt;, and because &lt;math&gt;(C)&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; bigger, the answer is &lt;math&gt;\boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ~Lcz<br /> <br /> == Solution 4 ==<br /> In order to shorten expressions, &lt;math&gt;\#&lt;/math&gt; will represent &lt;math&gt;16&lt;/math&gt; consecutive &lt;math&gt;0&lt;/math&gt;s when expressing numbers. &lt;br&gt;<br /> &lt;br&gt;<br /> Think of the problem in binary. We have &lt;br&gt;<br /> &lt;math&gt;\frac{1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2}{1\#1_2}&lt;/math&gt; &lt;br&gt;<br /> Note that &lt;br&gt;<br /> &lt;math&gt;(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) = 2^0(2^{17} + 1) + 2^{34}(2^{17} + 1) + 2^{68}(2^{17} + 1) + \cdots 2^{272}(2^{17} + 1)&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2&lt;/math&gt; &lt;br&gt;<br /> and &lt;br&gt;<br /> &lt;math&gt;(2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{17}(2^{17} + 1) + 2^{51}(2^{17} + 1) + 2^{85}(2^{17} + 1) + \cdots 2^{255}(2^{17} + 1)&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Since &lt;br&gt;<br /> &lt;math&gt;\phantom{=\ } 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;-\ \phantom{1\#} 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2&lt;/math&gt; &lt;br&gt;<br /> this means that&lt;br&gt;<br /> &lt;math&gt;(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{289}&lt;/math&gt; &lt;br&gt;<br /> so &lt;br&gt;<br /> &lt;math&gt;\frac{2^{289}+1}{2^{17}+1} = (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255})&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 2^0 + (2^{34} - 2^{17}) + (2^{68} - 2^{51}) + \cdots + (2^{272} - 2^{255})&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Expressing each of the pairs of the form &lt;math&gt;2^{n + 17} - 2^n&lt;/math&gt; in binary, we have &lt;br&gt;<br /> &lt;math&gt;\phantom{=\ } 1000000000000000000 \cdots 0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;-\ \phantom{10000000000000000} 10 \cdots 0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= \phantom{1} 111111111111111110 \cdots 0_2&lt;/math&gt; &lt;br&gt;<br /> or &lt;br&gt;<br /> &lt;math&gt;2^{n + 17} - 2^n = 2^{n + 16} + 2^{n + 15} + 2^{n + 14} + \cdots + 2^{n}&lt;/math&gt; &lt;br&gt;<br /> This means that each pair has &lt;math&gt;17&lt;/math&gt; terms of the form &lt;math&gt;2^n&lt;/math&gt;. &lt;br&gt;<br /> &lt;br&gt;<br /> Since there are &lt;math&gt;8&lt;/math&gt; of these pairs, there are a total of &lt;math&gt;8 \cdot 17 = 136&lt;/math&gt; terms. Accounting for the &lt;math&gt;2^0&lt;/math&gt; term, which was not in the pair, we have a total of &lt;math&gt;136 + 1 = \boxed{\textbf{(C) } 137}&lt;/math&gt; terms. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Video Solution==<br /> https://youtu.be/Ozp3k2464u4<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2020|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_23&diff=116636 2018 AMC 10B Problems/Problem 23 2020-02-02T18:48:51Z <p>Frestho: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> How many ordered pairs &lt;math&gt;(a, b)&lt;/math&gt; of positive integers satisfy the equation <br /> &lt;cmath&gt;a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),&lt;/cmath&gt;<br /> where &lt;math&gt;\text{gcd}(a,b)&lt;/math&gt; denotes the greatest common divisor of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;\text{lcm}(a,b)&lt;/math&gt; denotes their least common multiple?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> Let &lt;math&gt;x = lcm(a, b)&lt;/math&gt;, and &lt;math&gt;y = gcd(a, b)&lt;/math&gt;. Therefore, &lt;math&gt;a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y&lt;/math&gt;. Thus, the equation becomes<br /> <br /> &lt;cmath&gt;x\cdot y + 63 = 20x + 12y&lt;/cmath&gt;<br /> &lt;cmath&gt;x\cdot y - 20x - 12y + 63 = 0&lt;/cmath&gt;<br /> <br /> Using [[Simon's Favorite Factoring Trick]], we rewrite this equation as<br /> <br /> &lt;cmath&gt;(x - 12)(y - 20) - 240 + 63 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;(x - 12)(y - 20) = 177&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;177 = 3\cdot 59&lt;/math&gt; and &lt;math&gt;x &gt; y&lt;/math&gt;, we have &lt;math&gt;x - 12 = 59&lt;/math&gt; and &lt;math&gt;y - 20 = 3&lt;/math&gt;, or &lt;math&gt;x - 12 = 177&lt;/math&gt; and &lt;math&gt;y - 20 = 1&lt;/math&gt;. This gives us the solutions &lt;math&gt;(71, 23)&lt;/math&gt; and &lt;math&gt;(189, 21)&lt;/math&gt;. Since the &lt;math&gt;\text{GCD}&lt;/math&gt; must be a divisor of the &lt;math&gt;\text{LCM}&lt;/math&gt;, the first pair does not work. Assume &lt;math&gt;a&gt;b&lt;/math&gt;. We must have &lt;math&gt;a = 21 \cdot 9&lt;/math&gt; and &lt;math&gt;b = 21&lt;/math&gt;, and we could then have &lt;math&gt;a&lt;b&lt;/math&gt;, so there are &lt;math&gt;\boxed{2}&lt;/math&gt; solutions.<br /> (awesomeag)<br /> <br /> Edited by IronicNinja, Firebolt360, and mprincess0229~<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=JWGHYUeOx-k<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_23&diff=114951 2016 AMC 10B Problems/Problem 23 2020-01-18T21:05:38Z <p>Frestho: /* Solution 2 (cheap) */</p> <hr /> <div>==Problem==<br /> <br /> In regular hexagon &lt;math&gt;ABCDEF&lt;/math&gt;, points &lt;math&gt;W&lt;/math&gt;, &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are chosen on sides &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, &lt;math&gt;\overline{EF}&lt;/math&gt;, and &lt;math&gt;\overline{FA}&lt;/math&gt; respectively, so lines &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;ZW&lt;/math&gt;, &lt;math&gt;YX&lt;/math&gt;, and &lt;math&gt;ED&lt;/math&gt; are parallel and equally spaced. What is the ratio of the area of hexagon &lt;math&gt;WCXYFZ&lt;/math&gt; to the area of hexagon &lt;math&gt;ABCDEF&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> We draw a diagram to make our work easier:<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z;<br /> A=(0,0);<br /> B=(1,0);<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,sqrt(3));<br /> E=(0,sqrt(3));<br /> F=(-1/2,sqrt(3)/2);<br /> W=(4/3,2sqrt(3)/3);<br /> X=(4/3,sqrt(3)/3);<br /> Y=(-1/3,sqrt(3)/3);<br /> Z=(-1/3,2sqrt(3)/3);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(W--Z);<br /> draw(X--Y);<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$W$&quot;,W,ENE);<br /> label(&quot;$X$&quot;,X,ESE);<br /> label(&quot;$Y$&quot;,Y,WSW);<br /> label(&quot;$Z$&quot;,Z,WNW);<br /> &lt;/asy&gt;<br /> <br /> Assume that &lt;math&gt;AB&lt;/math&gt; is of length &lt;math&gt;1&lt;/math&gt;. Therefore, the area of &lt;math&gt;ABCDEF&lt;/math&gt; is &lt;math&gt;\frac{3\sqrt 3}2&lt;/math&gt;. To find the area of &lt;math&gt;WCXYFZ&lt;/math&gt;, we draw &lt;math&gt;\overline{CF}&lt;/math&gt;, and find the area of the trapezoids &lt;math&gt;WCFZ&lt;/math&gt; and &lt;math&gt;CXYF&lt;/math&gt;. <br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z;<br /> A=(0,0);<br /> B=(1,0);<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,sqrt(3));<br /> E=(0,sqrt(3));<br /> F=(-1/2,sqrt(3)/2);<br /> W=(4/3,2sqrt(3)/3);<br /> X=(4/3,sqrt(3)/3);<br /> Y=(-1/3,sqrt(3)/3);<br /> Z=(-1/3,2sqrt(3)/3);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(W--Z);<br /> draw(X--Y);<br /> draw(F--C--B--E--D--A);<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$W$&quot;,W,ENE);<br /> label(&quot;$X$&quot;,X,ESE);<br /> label(&quot;$Y$&quot;,Y,WSW);<br /> label(&quot;$Z$&quot;,Z,WNW);<br /> &lt;/asy&gt;<br /> <br /> From this, we know that &lt;math&gt;CF=2&lt;/math&gt;. We also know that the combined heights of the trapezoids is &lt;math&gt;\frac{\sqrt 3}3&lt;/math&gt;, since &lt;math&gt;\overline{ZW}&lt;/math&gt; and &lt;math&gt;\overline{YX}&lt;/math&gt; are equally spaced, and the height of each of the trapezoids is &lt;math&gt;\frac{\sqrt 3}6&lt;/math&gt;. From this, we know &lt;math&gt;\overline{ZW}&lt;/math&gt; and &lt;math&gt;\overline{YX}&lt;/math&gt; are each &lt;math&gt;\frac 13&lt;/math&gt; of the way from &lt;math&gt;\overline{CF}&lt;/math&gt; to &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. We know that these are both equal to &lt;math&gt;\frac 53&lt;/math&gt;.<br /> <br /> We find the area of each of the trapezoids, which both happen to be &lt;math&gt;\frac{11}6 \cdot \frac{\sqrt 3}6=\frac{11\sqrt 3}{36}&lt;/math&gt;, and the combined area is &lt;math&gt;\frac{11\sqrt 3}{18}&lt;/math&gt;.<br /> <br /> We find that &lt;math&gt;\dfrac{\frac{11\sqrt 3}{18}}{\frac{3\sqrt 3}2}&lt;/math&gt; is equal to &lt;math&gt;\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}&lt;/math&gt;.<br /> <br /> ==Solution 2 (cheap)==<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z;<br /> A=(0,0);<br /> B=(1,0);<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,sqrt(3));<br /> E=(0,sqrt(3));<br /> F=(-1/2,sqrt(3)/2);<br /> W=(4/3,2sqrt(3)/3);<br /> X=(4/3,sqrt(3)/3);<br /> Y=(-1/3,sqrt(3)/3);<br /> Z=(-1/3,2sqrt(3)/3);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(W--Z);<br /> draw(X--Y);<br /> draw(F--C--B--E--D--A);<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$W$&quot;,W,ENE);<br /> label(&quot;$X$&quot;,X,ESE);<br /> label(&quot;$Y$&quot;,Y,WSW);<br /> label(&quot;$Z$&quot;,Z,WNW);<br /> &lt;/asy&gt;<br /> <br /> First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triangles. There are &lt;math&gt;22&lt;/math&gt; small triangles in hexagon &lt;math&gt;ZWCXYF&lt;/math&gt;, and &lt;math&gt;9 \cdot 6 = 54&lt;/math&gt; small triangles in the whole hexagon. <br /> <br /> Thus, the answer is &lt;math&gt;\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}&lt;/math&gt;.<br /> ==Solution 3 (Similar Triangles)==<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z;<br /> A=(0,0);<br /> B=(1,0);<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,sqrt(3));<br /> E=(0,sqrt(3));<br /> F=(-1/2,sqrt(3)/2);<br /> W=(4/3,2sqrt(3)/3);<br /> X=(4/3,sqrt(3)/3);<br /> Y=(-1/3,sqrt(3)/3);<br /> Z=(-1/3,2sqrt(3)/3);<br /> pair G = (0.5, sqrt(3)*3/2);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(W--Z);<br /> draw(X--Y);<br /> draw(E--G--D);<br /> draw(F--C);<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$W$&quot;,W,ENE);<br /> label(&quot;$X$&quot;,X,ESE);<br /> label(&quot;$Y$&quot;,Y,WSW);<br /> label(&quot;$Z$&quot;,Z,WNW);<br /> label(&quot;$G$&quot;,G,N);<br /> &lt;/asy&gt;<br /> Extend &lt;math&gt;\overline{EF}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt; to meet at point &lt;math&gt;G&lt;/math&gt;, as shown in the diagram. Then &lt;math&gt;\triangle GZW \sim \triangle GFC&lt;/math&gt;. Then &lt;math&gt;[GZW] = \left(\dfrac53\right)^2[GED]&lt;/math&gt; and &lt;math&gt;[GFC] = 2^2[GED]&lt;/math&gt;. Subtracting &lt;math&gt;[GED]&lt;/math&gt;, we find that &lt;math&gt;[EDWZ] = \dfrac{16}{9}[GED]&lt;/math&gt; and &lt;math&gt;[EDCF] = 3[GED]&lt;/math&gt;. Subtracting again, we find that &lt;cmath&gt;[ZWCF] = [EDCF] - [EDWZ] = \dfrac{11}{9}[GED].&lt;/cmath&gt;Finally, &lt;cmath&gt;\dfrac{[WCXYFZ]}{[ABCDEF]} = \dfrac{[ZWCF]}{[EDCF]} = \dfrac{\dfrac{11}{9}[GED]}{3[GED]} = \textbf{(C) } \dfrac{11}{27}.&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_8&diff=114624 2019 AIME I Problems/Problem 8 2020-01-12T04:53:25Z <p>Frestho: /* Solution 3 (Newton Sums) */</p> <hr /> <div>==Problem 8==<br /> Let &lt;math&gt;x&lt;/math&gt; be a real number such that &lt;math&gt;\sin^{10}x+\cos^{10} x = \tfrac{11}{36}&lt;/math&gt;. Then &lt;math&gt;\sin^{12}x+\cos^{12} x = \tfrac{m}{n}&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> We can substitute &lt;math&gt;y = \sin^2{x}&lt;/math&gt;. Since we know that &lt;math&gt;\cos^2{x}=1-\sin^2{x}&lt;/math&gt;, we can do some simplification. <br /> <br /> This yields &lt;math&gt;y^5+(1-y)^5=\frac{11}{36}&lt;/math&gt;. From this, we can substitute again to get some cancellation through binomials. If we let &lt;math&gt;z=1/2-y&lt;/math&gt;, we can simplify the equation to &lt;math&gt;(1/2+z)^5+(1/2-z)^5=\frac{11}{36}&lt;/math&gt;. After using binomial theorem, this simplifies to &lt;math&gt;\frac{1}{16}(80z^4+40z^2+1)=11/36&lt;/math&gt;. If we use the quadratic formula, we obtain the that &lt;math&gt;z^2=\frac{1}{12}&lt;/math&gt;, so &lt;math&gt;z=\pm\frac{1}{2\sqrt{3}}&lt;/math&gt;. By plugging z into &lt;math&gt;(1/2-z)^6+(1/2+z)^6&lt;/math&gt; (which is equal to &lt;math&gt;\sin^{12}{x}+\cos^{12}{x}&lt;/math&gt;, we can either use binomial theorem or sum of cubes to simplify, and we end up with &lt;math&gt;\frac{13}{54}&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{067}&lt;/math&gt;.<br /> <br /> eric2020, inspired by Tommy2002<br /> <br /> ==Solution 2==<br /> <br /> First, for simplicity, let &lt;math&gt;a=\sin{x}&lt;/math&gt; and &lt;math&gt;b=\cos{x}&lt;/math&gt;. Note that &lt;math&gt;a^2+b^2=1&lt;/math&gt;. We then bash the rest of the problem out. Take the tenth power of this expression and get &lt;math&gt;a^{10}+b^{10}+5a^2b^2(a^6+b^6)+10a^4b^4(a^2+b^2)=\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1&lt;/math&gt;. Note that we also have &lt;math&gt;\frac{11}{36}=a^{10}+b^{10}=(a^{10}+b^{10})(a^2+b^2)=a^{12}+b^{12}+a^2b^2(a^8+b^8)&lt;/math&gt;. So, it suffices to compute &lt;math&gt;a^2b^2(a^8+b^8)&lt;/math&gt;. Let &lt;math&gt;y=a^2b^2&lt;/math&gt;. We have from cubing &lt;math&gt;a^2+b^2=1&lt;/math&gt; that &lt;math&gt;a^6+b^6+3a^2b^2(a^2+b^2)=1&lt;/math&gt; or &lt;math&gt;a^6+b^6=1-3y&lt;/math&gt;. Next, using &lt;math&gt;\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1&lt;/math&gt;, we get &lt;math&gt;a^2b^2(a^6+b^6)+2a^4b^4=\frac{5}{36}&lt;/math&gt; or &lt;math&gt;y(1-3y)+2y^2=y-y^2=\frac{5}{36}&lt;/math&gt;. Solving gives &lt;math&gt;y=\frac{5}{6}&lt;/math&gt; or &lt;math&gt;y=\frac{1}{6}&lt;/math&gt;. Clearly &lt;math&gt;y=\frac{5}{6}&lt;/math&gt; is extraneous, so &lt;math&gt;y=\frac{1}{6}&lt;/math&gt;. Now note that &lt;math&gt;a^4+b^4=(a^2+b^2)^2-2a^2b^2=\frac{2}{3}&lt;/math&gt;, and &lt;math&gt;a^8+b^8=(a^4+b^4)^2-2a^4b^4=\frac{4}{9}-\frac{1}{18}=\frac{7}{18}&lt;/math&gt;. Thus we finally get &lt;math&gt;a^{12}+b^{12}=\frac{11}{36}-\frac{7}{18}*\frac{1}{6}=\frac{13}{54}&lt;/math&gt;, giving &lt;math&gt;\boxed{067}&lt;/math&gt;.<br /> <br /> '''- Emathmaster'''<br /> <br /> ==Solution 3 (Newton Sums)==<br /> Newton sums is basically constructing the powers of the roots of the polynomials instead of deconstructing them which was done in Solution &lt;math&gt;2&lt;/math&gt;. Let &lt;math&gt;\sin^2x&lt;/math&gt; and &lt;math&gt;\cos^2x&lt;/math&gt; be the roots of some polynomial &lt;math&gt;F(a)&lt;/math&gt;. Then, by Vieta, &lt;math&gt;F(a)=a^2-a+b&lt;/math&gt; for some &lt;math&gt;b=\sin^2x\cdot\cos^2x&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;S_k=\left(\sin^2x\right)^k+\left(\cos^2x\right)^k&lt;/math&gt;. We want to find &lt;math&gt;S_6&lt;/math&gt;. Clearly &lt;math&gt;S_1=1&lt;/math&gt; and &lt;math&gt;S_2=1-2b&lt;/math&gt;. Newton sums tells us that &lt;math&gt;S_k-S_{k-1}+bS_{k-2}=0\Rightarrow S_k=S_{k-1}-bS_{k-2}&lt;/math&gt; where &lt;math&gt;k\ge 3&lt;/math&gt; for our polynomial &lt;math&gt;F(a)&lt;/math&gt;. <br /> <br /> Bashing, we have<br /> &lt;cmath&gt;S_3=S_2-bS_1\Rightarrow S_3=(1-2b)-b(1)=1-3b&lt;/cmath&gt;<br /> &lt;cmath&gt;S_4=S_3-bS_2\Rightarrow S_4=(1-3b)-b(1-2b)=2b^2-4b+1&lt;/cmath&gt;<br /> &lt;cmath&gt;S_5=S_4-bS_3\Rightarrow S_5=(2b^2-4b+1)-b(1-3b)=5b^2-5b+1=\frac{11}{36}&lt;/cmath&gt;<br /> <br /> Thus &lt;cmath&gt;5b^2-5b+1=\frac{11}{36}\Rightarrow 5b^2-5b+\frac{25}{36}=0, 36b^2-36b+5=0, (6b-1)(6b-5)=0&lt;/cmath&gt;<br /> &lt;math&gt;b=\frac{1}{6} \text{ or } \frac{5}{6}&lt;/math&gt;. Clearly, &lt;math&gt;\sin^2x\cdot\cos^2x\not=\frac{5}{6}&lt;/math&gt; so &lt;math&gt;\sin^2x\cdot\cos^2x=b=\frac{1}{6}&lt;/math&gt;.<br /> <br /> Note &lt;math&gt;S_4=\frac{7}{18}&lt;/math&gt;. Solving for &lt;math&gt;S_6&lt;/math&gt;, we get &lt;math&gt;S_6=S_5-\frac{1}{6}S_4=\frac{13}{54}&lt;/math&gt;. Finally, &lt;math&gt;13+54=\boxed{067}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Factor the first equation. &lt;cmath&gt;\sin^{10}x + \cos^{10}x = (\sin^2x+\cos^2x)(\sin^8x-\sin^6x\cos^2x+\sin^4x\cos^4x-\sin^2x\cos^6x+\cos^8x)&lt;/cmath&gt;<br /> First of all, &lt;math&gt;\sin^4x+\cos^4x = 1-2\sin^2x\cos^2x&lt;/math&gt; because &lt;math&gt;\sin^4x+\cos^4x=(\sin^2x + \cos^2x)^2 -2\sin^2x\cos^2x = 1 - 2\sin^2x\cos^2x&lt;/math&gt;<br /> We group the first, third, and fifth term and second and fourth term. The first group: &lt;cmath&gt; \sin^8+\sin^4x\cos^4x+\cos^8x = (\sin^4x+\cos^4x)^2-\sin^4x\cos^4x)= (1 - 2\sin^2x\cos^2x)^2-\sin^4x\cos^4x)<br /> =1+4\sin^4x\cos^4x-4\sin^2x\cos^2x&lt;/cmath&gt; The second group: &lt;cmath&gt;-\sin^6x\cos^2x-\sin^2x\cos^6x = -\sin^2x\cos^2x(\sin^4x+\cos^4x)=-\sin^2x\cos^2x(1-2\sin^2x\cos^2x) = -\sin^2x\cos^2x+2\sin^4x\cos^4x&lt;/cmath&gt; Add the two together to make &lt;cmath&gt;1+4\sin^4x\cos^4x-4\sin^2x\cos^2x-\sin^2x\cos^2x+2\sin^4x\cos^4x = 1 - 5\sin^2x\cos^2x+5\sin^4x\cos^4x&lt;/cmath&gt; Because this equals &lt;math&gt;\frac{11}{36}&lt;/math&gt;, we have &lt;cmath&gt;5\sin^4x\cos^4x- 5\sin^2x\cos^2x+\frac{25}{36}=0&lt;/cmath&gt; Let &lt;math&gt;\sin^2x\cos^2x = a&lt;/math&gt; so we get &lt;cmath&gt;5a^2- 5a+\frac{25}{36}=0 \Rightarrow a^2-a+\frac{5}{36}&lt;/cmath&gt; Solving the quadratic gives us &lt;cmath&gt;a = \frac{1 \pm \frac{2}{3}}{2}&lt;/cmath&gt; Because &lt;math&gt;\sin^2x\cos^2x \le \frac{1}{4}&lt;/math&gt;, we finally get &lt;math&gt;a = \frac{1 - \frac{2}{3}}{2} = \frac{1}{6}&lt;/math&gt;. <br /> <br /> Now from the second equation, &lt;cmath&gt;\sin^{12}x + \cos^{12}x = (\sin^4x+\cos^4x)(\sin^8x-\sin^4x\cos^4x+\cos^8x)=(1-2\sin^2x\cos^2x)((\sin^4x+\cos^4x)^2-3\sin^4x\cos^4x)=(1-2\sin^2x\cos^2x)((1-2\sin^2x\cos^2x)^2-3\sin^4x\cos^4x)&lt;/cmath&gt; Plug in &lt;math&gt;\sin^2x\cos^2x = \frac{1}{6}&lt;/math&gt; to get &lt;cmath&gt;(1-2(\frac{1}{6}))(1-2(\frac{1}{6})^2-3(\frac{1}{6})^2) = \frac{13}{54}&lt;/cmath&gt;<br /> which yields the answer &lt;math&gt;\boxed{067}&lt;/math&gt;<br /> <br /> ~ZericHang<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_15&diff=114463 2005 AIME II Problems/Problem 15 2020-01-08T01:56:39Z <p>Frestho: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt; w_1 &lt;/math&gt; and &lt;math&gt; w_2 &lt;/math&gt; denote the [[circle]]s &lt;math&gt; x^2+y^2+10x-24y-87=0 &lt;/math&gt; and &lt;math&gt; x^2 +y^2-10x-24y+153=0, &lt;/math&gt; respectively. Let &lt;math&gt; m &lt;/math&gt; be the smallest positive value of &lt;math&gt; a &lt;/math&gt; for which the line &lt;math&gt; y=ax &lt;/math&gt; contains the center of a circle that is externally [[tangent (geometry)|tangent]] to &lt;math&gt; w_2 &lt;/math&gt; and internally tangent to &lt;math&gt; w_1. &lt;/math&gt; Given that &lt;math&gt; m^2=\frac pq, &lt;/math&gt; where &lt;math&gt; p &lt;/math&gt; and &lt;math&gt; q &lt;/math&gt; are relatively prime integers, find &lt;math&gt; p+q. &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Rewrite the given equations as &lt;math&gt;(x+5)^2 + (y-12)^2 = 256&lt;/math&gt; and &lt;math&gt;(x-5)^2 + (y-12)^2 = 16&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;w_3&lt;/math&gt; have center &lt;math&gt;(x,y)&lt;/math&gt; and radius &lt;math&gt;r&lt;/math&gt;. Now, if two circles with radii &lt;math&gt;r_1&lt;/math&gt; and &lt;math&gt;r_2&lt;/math&gt; are externally tangent, then the distance between their centers is &lt;math&gt;r_1 + r_2&lt;/math&gt;, and if they are internally tangent, it is &lt;math&gt;|r_1 - r_2|&lt;/math&gt;. So we have<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> r + 4 &amp;= \sqrt{(x-5)^2 + (y-12)^2} \\<br /> 16 - r &amp;= \sqrt{(x+5)^2 + (y-12)^2} \end{align*} &lt;/cmath&gt;<br /> <br /> Solving for &lt;math&gt;r&lt;/math&gt; in both equations and setting them equal, then simplifying, yields<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> 20 - \sqrt{(x+5)^2 + (y-12)^2} &amp;= \sqrt{(x-5)^2 + (y-12)^2} \\<br /> 20+x &amp;= 2\sqrt{(x+5)^2 + (y-12)^2}<br /> \end{align*} &lt;/cmath&gt;<br /> <br /> Squaring again and canceling yields &lt;math&gt;1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.&lt;/math&gt;<br /> <br /> So the locus of points that can be the center of the circle with the desired properties is an [[ellipse]].<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> size(220); pointpen = black; pen d = linewidth(0.7); pathpen = d; <br /> pair A = (-5, 12), B = (5, 12), C = (0, 0);<br /> D(CR(A,16));D(CR(B,4));D(shift((0,12)) * yscale(3^.5 / 2) * CR(C,10), linetype(&quot;2 2&quot;) + d + red);<br /> D((0,30)--(0,-10),Arrows(4));D((15,0)--(-25,0),Arrows(4));D((0,0)--MP(&quot;y=ax&quot;,(14,14 * (69/100)^.5),E),EndArrow(4));<br /> <br /> void bluecirc (real x) {<br /> pair P = (x, (3 * (25 - x^2 / 4))^.5 + 12); dot(P, blue);<br /> D(CR(P, ((P.x - 5)^2 + (P.y - 12)^2)^.5 - 4) , blue + d + linetype(&quot;4 4&quot;));<br /> }<br /> <br /> bluecirc(-9.2); bluecirc(-4); bluecirc(3);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Since the center lies on the line &lt;math&gt;y = ax&lt;/math&gt;, we substitute for &lt;math&gt;y&lt;/math&gt; and expand: <br /> &lt;cmath&gt;1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75} \Longrightarrow (3+4a^2)x^2 - 96ax + 276 = 0.&lt;/cmath&gt;<br /> <br /> We want the value of &lt;math&gt;a&lt;/math&gt; that makes the line &lt;math&gt;y=ax&lt;/math&gt; tangent to the ellipse, which will mean that for that choice of &lt;math&gt;a&lt;/math&gt; there is only one solution to the most recent equation. But a quadratic has one solution [[iff]] its discriminant is &lt;math&gt;0&lt;/math&gt;, so &lt;math&gt;(-96a)^2 - 4(3+4a^2)(276) = 0&lt;/math&gt;.<br /> <br /> Solving yields &lt;math&gt;a^2 = \frac{69}{100}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{169}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> As above, we rewrite the equations as &lt;math&gt;(x+5)^2 + (y-12)^2 = 256&lt;/math&gt; and &lt;math&gt;(x-5)^2 + (y-12)^2 = 16&lt;/math&gt;. Let &lt;math&gt;F_1=(-5,12)&lt;/math&gt; and &lt;math&gt;F_2=(5,12)&lt;/math&gt;. If a circle with center &lt;math&gt;C=(a,b)&lt;/math&gt; and radius &lt;math&gt;r&lt;/math&gt; is externally tangent to &lt;math&gt;w_2&lt;/math&gt; and internally tangent to &lt;math&gt;w_1&lt;/math&gt;, then &lt;math&gt;CF_1=16-r&lt;/math&gt; and &lt;math&gt;CF_2=4+r&lt;/math&gt;. Therefore, &lt;math&gt;CF_1+CF_2=20&lt;/math&gt;. In particular, the locus of points &lt;math&gt;C&lt;/math&gt; that can be centers of circles must be an ellipse with foci &lt;math&gt;F_1&lt;/math&gt; and &lt;math&gt;F_2&lt;/math&gt; and major axis &lt;math&gt;20&lt;/math&gt;.<br /> <br /> Clearly, the minimum value of the slope &lt;math&gt;a&lt;/math&gt; will occur when the line &lt;math&gt;y=ax&lt;/math&gt; is tangent to this ellipse. Suppose that this point of tangency is denoted by &lt;math&gt;T&lt;/math&gt;, and the line &lt;math&gt;y=ax&lt;/math&gt; is denoted by &lt;math&gt;\ell&lt;/math&gt;. Then we reflect the ellipse over &lt;math&gt;\ell&lt;/math&gt; to a new ellipse with foci &lt;math&gt;F_1'&lt;/math&gt; and &lt;math&gt;F_2'&lt;/math&gt; as shown below.<br /> &lt;center&gt;&lt;asy&gt;<br /> size(220); <br /> pair F1 = (-5, 12), F2 = (5, 12),C=(0,12);<br /> draw(circle(F1,16));<br /> draw(circle(F2,4));<br /> draw(ellipse(C,10,5*sqrt(3)));<br /> xaxis(&quot;$x$&quot;,Arrows);<br /> yaxis(&quot;$y$&quot;,Arrows);<br /> dot(F1^^F2^^C);<br /> <br /> real l(real x) {return sqrt(69)*x/10;}<br /> path g=graph(l,-7,14);<br /> draw(g);<br /> draw(reflect((0,0),(10,l(10)))*ellipse(C,10,5*sqrt(3)));<br /> pair T=intersectionpoint(ellipse(C,10,5*sqrt(3)),(0,0)--(10,l(10)));<br /> dot(T);<br /> pair F1P=reflect((0,0),(10,l(10)))*F1;<br /> pair F2P=reflect((0,0),(10,l(10)))*F2;<br /> dot(F1P^^F2P);<br /> dot((0,0));<br /> label(&quot;$F_1$&quot;,F1,N,fontsize(9));<br /> label(&quot;$F_2$&quot;,F2,N,fontsize(9));<br /> label(&quot;$F_1'$&quot;,F1P,SE,fontsize(9));<br /> label(&quot;$F_2'$&quot;,F2P,SE,fontsize(9));<br /> label(&quot;$O$&quot;,(0,0),NW,fontsize(9));<br /> label(&quot;$\ell$&quot;,(13,l(13)),SE,fontsize(9));<br /> label(&quot;$T$&quot;,T,NW,fontsize(9));<br /> draw((0,0)--F1--F2--F2P--F1P--cycle);<br /> draw(F1--F2P^^F2--F1P);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> By the reflection property of ellipses (i.e., the angle of incidence to a tangent line is equal to the angle of reflection for any path that travels between the foci), we know that &lt;math&gt;F_1&lt;/math&gt;, &lt;math&gt;T&lt;/math&gt;, and &lt;math&gt;F_2'&lt;/math&gt; are collinear, and similarly, &lt;math&gt;F_2&lt;/math&gt;, &lt;math&gt;T&lt;/math&gt; and &lt;math&gt;F_1'&lt;/math&gt; are collinear. Therefore, &lt;math&gt;OF_1F_2F_2'F_1'&lt;/math&gt; is a pentagon with &lt;math&gt;OF_1=OF_2=OF_1'=OF_2'=13&lt;/math&gt;, &lt;math&gt;F_1F_2=F_1'F_2'=10&lt;/math&gt;, and &lt;math&gt;F_1F_2'=F_1'F_2=20&lt;/math&gt;. Note that &lt;math&gt;\ell&lt;/math&gt; bisects &lt;math&gt;\angle F_1'OF_1&lt;/math&gt;. We can bisect this angle by bisecting &lt;math&gt;\angle F_1'OF_2&lt;/math&gt; and &lt;math&gt;F_2OF_1&lt;/math&gt; separately.<br /> <br /> We proceed using complex numbers. Triangle &lt;math&gt;F_2OF_1'&lt;/math&gt; is isosceles with side lengths &lt;math&gt;13,13,20&lt;/math&gt;. The height of this from the base of &lt;math&gt;20&lt;/math&gt; is &lt;math&gt;\sqrt{69}&lt;/math&gt;. Therefore, the complex number &lt;math&gt;\sqrt{69}+10i&lt;/math&gt; represents the bisection of &lt;math&gt;\angle F_1'OF_2&lt;/math&gt;.<br /> <br /> Similarly, using the 5-12-13 triangles, we easily see that &lt;math&gt;12+5i&lt;/math&gt; represents the bisection of the angle &lt;math&gt;F_2OF_1&lt;/math&gt;. Therefore, we can add these two angles together by multiplying the complex numbers, finding<br /> &lt;cmath&gt;\text{arg}\left((\sqrt{69}+10i)(12+5i)\right)=\frac{1}{2}\angle F_1'OF_1.&lt;/cmath&gt;<br /> Now the point &lt;math&gt;F_1&lt;/math&gt; is given by the complex number &lt;math&gt;-5+12i&lt;/math&gt;. Therefore, to find a point on line &lt;math&gt;\ell&lt;/math&gt;, we simply subtract &lt;math&gt;\frac{1}{2}\angle F_1'OF_1&lt;/math&gt;, which is the same as multiplying &lt;math&gt;-5+12i&lt;/math&gt; by the conjugate of &lt;math&gt;(\sqrt{69}+10i)(12+5i)&lt;/math&gt;. We find<br /> &lt;cmath&gt;(-5+12i)(\sqrt{69}-10i)(12-5i)=169(10+i\sqrt{69}).&lt;/cmath&gt;<br /> In particular, note that the tangent of the argument of this complex number is &lt;math&gt;\sqrt{69}/10&lt;/math&gt;, which must be the slope of the tangent line. Hence &lt;math&gt;a^2=69/100&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{169}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> We use the same reflection as in Solution 2. As &lt;math&gt;OF_1'=OF_2=13&lt;/math&gt;, we know that &lt;math&gt;\triangle OF_1'F_2&lt;/math&gt; is isosceles. Hence &lt;math&gt;\angle F_2F_1'O=\angle F_1'F_2O&lt;/math&gt;. But by symmetry, we also know that &lt;math&gt;\angle OF_1T=\angle F_2F_1'O&lt;/math&gt;. Hence &lt;math&gt;\angle OF_1T=\angle F_1'F_2O&lt;/math&gt;. In particular, as &lt;math&gt;\angle OF_1T=\angle OF_2T&lt;/math&gt;, this implies that &lt;math&gt;O, F_1, F_2&lt;/math&gt;, and &lt;math&gt;T&lt;/math&gt; are concyclic.<br /> <br /> Let &lt;math&gt;X&lt;/math&gt; be the intersection of &lt;math&gt;F_2F_1'&lt;/math&gt; with the &lt;math&gt;x&lt;/math&gt;-axis. As &lt;math&gt;F_1F_2&lt;/math&gt; is parallel to the &lt;math&gt;x&lt;/math&gt;-axis, we know that &lt;cmath&gt;\angle TXO=180-\angle F_1F_2T.\tag{1}&lt;/cmath&gt; But &lt;cmath&gt;180-\angle F_1F_2T=\angle F_2F_1T+\angle F_1TF_2.\tag{2}&lt;/cmath&gt; By the fact that &lt;math&gt;OF_1F_2T&lt;/math&gt; is cyclic, &lt;cmath&gt;\angle F_2F_1T=\angle F_2OT\qquad\text{and}\qquad \angle F_1TF_2=\angle F_1OF_2.\tag{3}&lt;/cmath&gt; Therefore, combining (1), (2), and (3), we find that<br /> &lt;cmath&gt;\angle TXO=\angle F_2OT+\angle F_1OF_2=\angle F_1OT.\tag{4}&lt;/cmath&gt;<br /> <br /> By symmetry, we also know that<br /> &lt;cmath&gt;\angle F_1TO=\angle OTF_1'.\tag{5}&lt;/cmath&gt;<br /> Therefore, (4) and (5) show by AA similarity that &lt;math&gt;\triangle F_1OT\sim \triangle OXT&lt;/math&gt;. Therefore, &lt;math&gt;\angle XOT=\angle OF_1T&lt;/math&gt;.<br /> <br /> Now as &lt;math&gt;OF_1=OF_2'=13&lt;/math&gt;, we know that &lt;math&gt;\triangle OF_1F_2'&lt;/math&gt; is isosceles, and as &lt;math&gt;F_1F_2'=20&lt;/math&gt;, we can drop an altitude to &lt;math&gt;F_1F_2'&lt;/math&gt; to easily find that &lt;math&gt;\tan \angle OF_1T=\sqrt{69}/10&lt;/math&gt;. Therefore, &lt;math&gt;\tan\angle XOT&lt;/math&gt;, which is the desired slope, must also be &lt;math&gt;\sqrt{69}/10&lt;/math&gt;. As before, we conclude that the answer is &lt;math&gt;\boxed{169}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> [[Image:2005_AIME_II_-15.png||center|800px]]<br /> First, rewrite the equations for the circles as &lt;math&gt;(x+5)^2+(y-12)^2=16^2&lt;/math&gt; and &lt;math&gt;(x-5)^2+(y-12)^2=4^2&lt;/math&gt;. <br /> Then, choose a point &lt;math&gt;(a,b)&lt;/math&gt; that is a distance of &lt;math&gt;x&lt;/math&gt; from both circles. Use the distance formula between &lt;math&gt;(a,b)&lt;/math&gt; and each of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; (in the diagram above). The distances, as can be seen in the diagram above are &lt;math&gt;16-x&lt;/math&gt; and &lt;math&gt;4+x&lt;/math&gt;, respectively.<br /> &lt;cmath&gt;(a-5)^2+(b-12)^2=(4+x)^2&lt;/cmath&gt;<br /> &lt;cmath&gt;(a+5)^2+(b-12)^2=(16-x)^2&lt;/cmath&gt;<br /> Subtracting the first equation from the second gives &lt;cmath&gt;20a=240-40x\rightarrow a=12-2x\rightarrow x=6-\frac a2&lt;/cmath&gt;<br /> Substituting this into the first equation gives<br /> &lt;cmath&gt;a^2-10a+25+b^2-24b+144=100-10a+\frac{a^2}4&lt;/cmath&gt;<br /> &lt;cmath&gt;b^2-24b+69+\frac{3a^2}4=0&lt;/cmath&gt;<br /> Now, instead of converting this to the equation of an eclipse, solve for &lt;math&gt;b&lt;/math&gt; and then divide by &lt;math&gt;a&lt;/math&gt;.<br /> &lt;cmath&gt;b=\frac{24\pm\sqrt{300-3a^2}}{2}&lt;/cmath&gt;<br /> We take the smaller root to minimize &lt;math&gt;\frac b a&lt;/math&gt;.<br /> &lt;cmath&gt;\frac b a=\frac{24-\sqrt{300-3a^2}}{2a}=\frac{24-\sqrt3\cdot\sqrt{100-a^2}}{2a}=\frac{12}a-\frac{\sqrt3}{2a}\sqrt{100-a^2}&lt;/cmath&gt;<br /> Now, let &lt;math&gt;10\cos\theta=a&lt;/math&gt;. This way, &lt;math&gt;\sqrt{100-a^2}=10\sin\theta&lt;/math&gt;.<br /> Substitute this in. &lt;math&gt;\frac{b}{a}=\frac{12}{10\cos\theta}-\frac{\sqrt3\cdot\sin\theta}{2\cos\theta}=\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta&lt;/math&gt;<br /> Then, take the derivative of this and set it to 0 to find the minimum value.<br /> &lt;math&gt;\frac{6}{5}\sec\theta\tan\theta-\frac{\sqrt3}{2}\sec^2\theta=0\rightarrow\frac{6}{5}\sin\theta-frac{\sqrt3}{2}=0\rightarrow\sin\theta=\frac{5\sqrt3}{12}&lt;/math&gt;<br /> Then, use this value of &lt;math&gt;\sin\theta&lt;/math&gt; to find the minimum of &lt;math&gt;\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta&lt;/math&gt; to get &lt;math&gt;\frac{\sqrt{69}}{10}\rightarrow\left(\frac{\sqrt{69}}{10}\right)^2=\frac{69}{100}\rightarrow69+100=\boxed{169}&lt;/math&gt;<br /> <br /> ==Solution 5 (probably fastest)==<br /> Like before, notice that the distances from the centers of the given circles to the desired center are &lt;math&gt;4+r&lt;/math&gt; and &lt;math&gt;16-r&lt;/math&gt;, which add up to &lt;math&gt;20&lt;/math&gt;. This means that the possible centers of the third circle lie on an ellipse with foci &lt;math&gt;(-5, 12)&lt;/math&gt; and &lt;math&gt;(5, 12)&lt;/math&gt;. Using the fact that the sum of the distances from the foci is &lt;math&gt;20&lt;/math&gt;, we find that the semi-major axis has length &lt;math&gt;10&lt;/math&gt; and the semi-minor axis has length &lt;math&gt;5\sqrt{3}&lt;/math&gt;. Therefore, the equation of the ellipse is &lt;cmath&gt;\dfrac{x}{100}+\dfrac{(y-12)}{75} = 1,&lt;/cmath&gt; where the numbers &lt;math&gt;100&lt;/math&gt; and &lt;math&gt;75&lt;/math&gt; come from &lt;math&gt;10^2&lt;/math&gt; and &lt;math&gt;(5\sqrt{3})^2&lt;/math&gt; respectively. <br /> <br /> <br /> We proceed to find &lt;math&gt;m&lt;/math&gt; using the same method as Solution 1.<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_20&diff=113299 2004 AMC 10B Problems/Problem 20 2019-12-23T23:11:27Z <p>Frestho: /* Solution 4 (Menelaus) */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; lie on &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;, respectively. If &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BE&lt;/math&gt; intersect at &lt;math&gt;T&lt;/math&gt; so that &lt;math&gt;\frac{AT}{DT}=3&lt;/math&gt; and &lt;math&gt;\frac{BT}{ET}=4&lt;/math&gt;, what is &lt;math&gt;\frac{CD}{BD}&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{8} \qquad \mathrm{(B) \ } \frac{2}{9} \qquad \mathrm{(C) \ } \frac{3}{10} \qquad \mathrm{(D) \ } \frac{4}{11} \qquad \mathrm{(E) \ } \frac{5}{12} &lt;/math&gt;<br /> <br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=5*dir(60), C=5*(1,0), D=B + (11/15)*(C-B), E = A + (11/16)*(C-A);<br /> draw(A--B--C--cycle);<br /> draw(A--D);<br /> draw(B--E);<br /> pair T=intersectionpoint(A--D,B--E);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,N);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$T$&quot;,T,2*WNW);<br /> &lt;/asy&gt;<br /> <br /> == Solution 1 (Triangle Areas) ==<br /> <br /> We use the square bracket notation &lt;math&gt;[\cdot]&lt;/math&gt; to denote area.<br /> <br /> WLOG, we can assume &lt;math&gt;[\triangle BTD] = 1&lt;/math&gt;. Then &lt;math&gt;[\triangle BTA] = 3&lt;/math&gt;, and &lt;math&gt;[\triangle ATE] = 3/4&lt;/math&gt;. We have &lt;math&gt;CD/BD = [\triangle ACD]/[\triangle ABD]&lt;/math&gt;, so we need to find the area of quadrilateral &lt;math&gt;TDCE&lt;/math&gt;.<br /> <br /> Draw the line segment &lt;math&gt;TC&lt;/math&gt; to form the two triangles &lt;math&gt;\triangle TDC&lt;/math&gt; and &lt;math&gt;\triangle TEC&lt;/math&gt;. Let &lt;math&gt;x = [\triangle TDC]&lt;/math&gt;, and &lt;math&gt;y = [\triangle TEC]&lt;/math&gt;. By considering triangles &lt;math&gt;\triangle BTC&lt;/math&gt; and &lt;math&gt;\triangle ETC&lt;/math&gt;, we obtain &lt;math&gt;(1+x)/y=4&lt;/math&gt;, and by considering triangles &lt;math&gt;\triangle ATC&lt;/math&gt; and &lt;math&gt;\triangle DTC&lt;/math&gt;, we obtain &lt;math&gt;(3/4+y)/x=3&lt;/math&gt;. Solving, we get &lt;math&gt;x=4/11&lt;/math&gt;, &lt;math&gt;y=15/44&lt;/math&gt;, so the area of quadrilateral &lt;math&gt;TDEC&lt;/math&gt; is &lt;math&gt;x+y=31/44&lt;/math&gt;.<br /> <br /> Therefore &lt;math&gt;\frac{CD}{BD}=\frac{\frac{3}{4}+\frac{31}{44}}{3+1}=\boxed{\textbf{(D)} \frac{4}{11}}&lt;/math&gt;<br /> <br /> == Solution 1.5 (Triangle Areas, Alternate Approach) ==<br /> <br /> We observe that &lt;math&gt;\frac{BC}{CD} = \frac{[BAE]}{[DAE]}&lt;/math&gt;. The proof is that if &lt;math&gt;B_H&lt;/math&gt; and &lt;math&gt;D_H&lt;/math&gt; are the feet of the altitudes from &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, respectively, to &lt;math&gt;AC&lt;/math&gt;, then both sides of that equation are equal to &lt;math&gt;\frac{BB_H}{DD_H}&lt;/math&gt;.<br /> <br /> From there, we can easily finish:<br /> &lt;cmath&gt;\frac{BC}{CD} = \frac{[BAE]}{[DAE]} = \frac{[BAT]\cdot\frac{5}{4}}{[EAT]\cdot \frac{4}{3}} = 4\cdot\frac{5}{4}\cdot\frac{3}{4} = \frac{15}{4}&lt;/cmath&gt;<br /> and thus &lt;math&gt;\frac{CD}{BD} = \boxed{\textbf{(D)}\frac 4{11}}&lt;/math&gt;.<br /> <br /> == Solution 2 (Mass points) ==<br /> <br /> The presence of only ratios in the problem essentially cries out for mass points.<br /> <br /> As per the problem, we assign a mass of &lt;math&gt;1&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;, and a mass of &lt;math&gt;3&lt;/math&gt; to &lt;math&gt;D&lt;/math&gt;. Then, to balance &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; on &lt;math&gt;T&lt;/math&gt;, &lt;math&gt;T&lt;/math&gt; has a mass of &lt;math&gt;4&lt;/math&gt;.<br /> <br /> Now, were we to assign a mass of &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; and a mass of &lt;math&gt;4&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;, we'd have &lt;math&gt;5T&lt;/math&gt;. Scaling this down by &lt;math&gt;4/5&lt;/math&gt; (to get &lt;math&gt;4T&lt;/math&gt;, which puts &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; in terms of the masses of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;), we assign a mass of &lt;math&gt;\frac{4}{5}&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; and a mass of &lt;math&gt;\frac{16}{5}&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;.<br /> <br /> Now, to balance &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; on &lt;math&gt;E&lt;/math&gt;, we must give &lt;math&gt;C&lt;/math&gt; a mass of &lt;math&gt;\frac{16}{5}-1=\frac{11}{5}&lt;/math&gt;.<br /> <br /> Finally, the ratio of &lt;math&gt;CD&lt;/math&gt; to &lt;math&gt;BD&lt;/math&gt; is given by the ratio of the mass of &lt;math&gt;B&lt;/math&gt; to the mass of &lt;math&gt;C&lt;/math&gt;, which is &lt;math&gt;\frac{4}{5}\cdot\frac{5}{11}=\boxed{\textbf{(D)}\ \frac{4}{11}}&lt;/math&gt;.<br /> <br /> == Solution 3 (Coordinates) ==<br /> <br /> Affine transformations preserve ratios of distances, and for any pair of triangles, there is an affine transformation that maps the first one onto the second one. This is why the answer is the same for any &lt;math&gt;\triangle ABC&lt;/math&gt;, and we just need to compute it for any single triangle.<br /> <br /> We can choose the points &lt;math&gt;A=(-3,0)&lt;/math&gt;, &lt;math&gt;B=(0,4)&lt;/math&gt;, and &lt;math&gt;D=(1,0)&lt;/math&gt;. This way we will have &lt;math&gt;T=(0,0)&lt;/math&gt;, and &lt;math&gt;E=(0,-1)&lt;/math&gt;. The situation is <br /> shown in the picture below:<br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(0.8);<br /> pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1);<br /> draw(A--B--C--cycle);<br /> draw(A--D);<br /> draw(B--E);<br /> pair T=intersectionpoint(A--D,B--E);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,N);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$T$&quot;,T,NW);<br /> label(&quot;$3$&quot;,A--T,N);<br /> label(&quot;$4$&quot;,B--T,W);<br /> label(&quot;$1$&quot;,D--T,N);<br /> label(&quot;$1$&quot;,E--T,W);<br /> <br /> &lt;/asy&gt;<br /> <br /> The point &lt;math&gt;C&lt;/math&gt; is the intersection of the lines &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt;. The points on the first line have the form &lt;math&gt;(t,4-4t)&lt;/math&gt;, the points on the second line have the form &lt;math&gt;(t,-1-t/3)&lt;/math&gt;. Solving for &lt;math&gt;t&lt;/math&gt; we get &lt;math&gt;t=15/11&lt;/math&gt;, hence &lt;math&gt;C=(15/11,-16/11)&lt;/math&gt;.<br /> <br /> The ratio &lt;math&gt;CD/BD&lt;/math&gt; can now be computed simply by observing the &lt;math&gt;x&lt;/math&gt; coordinates of &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;<br /> \frac{CD}{BD} = \frac{15/11 - 1}{1 - 0} = \boxed{\textbf{(D)}\frac 4{11}}<br /> &lt;/cmath&gt;<br /> == Solution 4 (Menelaus) ==<br /> By Menelaus on triangle &lt;math&gt;BDT&lt;/math&gt;, we have that &lt;cmath&gt;\begin{align*}1&amp;=\dfrac{DA}{TA}\cdot\dfrac{TE}{BE}\cdot\dfrac{BC}{DC} \\ &amp;= \dfrac43 \cdot \dfrac 15 \cdot \dfrac {BC}{DC},\end{align*}&lt;/cmath&gt;giving &lt;math&gt;\dfrac{BC}{DC} = \dfrac{15}{4}&lt;/math&gt;. Therefore, &lt;math&gt;\dfrac{CD}{BD} = \boxed{\textbf{(D)}\dfrac{4}{11}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> <br /> {{AMC10 box|year=2004|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_20&diff=113298 2004 AMC 10B Problems/Problem 20 2019-12-23T23:11:00Z <p>Frestho: /* Solution 3 (Coordinates) */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; lie on &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;, respectively. If &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BE&lt;/math&gt; intersect at &lt;math&gt;T&lt;/math&gt; so that &lt;math&gt;\frac{AT}{DT}=3&lt;/math&gt; and &lt;math&gt;\frac{BT}{ET}=4&lt;/math&gt;, what is &lt;math&gt;\frac{CD}{BD}&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{8} \qquad \mathrm{(B) \ } \frac{2}{9} \qquad \mathrm{(C) \ } \frac{3}{10} \qquad \mathrm{(D) \ } \frac{4}{11} \qquad \mathrm{(E) \ } \frac{5}{12} &lt;/math&gt;<br /> <br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=5*dir(60), C=5*(1,0), D=B + (11/15)*(C-B), E = A + (11/16)*(C-A);<br /> draw(A--B--C--cycle);<br /> draw(A--D);<br /> draw(B--E);<br /> pair T=intersectionpoint(A--D,B--E);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,N);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$T$&quot;,T,2*WNW);<br /> &lt;/asy&gt;<br /> <br /> == Solution 1 (Triangle Areas) ==<br /> <br /> We use the square bracket notation &lt;math&gt;[\cdot]&lt;/math&gt; to denote area.<br /> <br /> WLOG, we can assume &lt;math&gt;[\triangle BTD] = 1&lt;/math&gt;. Then &lt;math&gt;[\triangle BTA] = 3&lt;/math&gt;, and &lt;math&gt;[\triangle ATE] = 3/4&lt;/math&gt;. We have &lt;math&gt;CD/BD = [\triangle ACD]/[\triangle ABD]&lt;/math&gt;, so we need to find the area of quadrilateral &lt;math&gt;TDCE&lt;/math&gt;.<br /> <br /> Draw the line segment &lt;math&gt;TC&lt;/math&gt; to form the two triangles &lt;math&gt;\triangle TDC&lt;/math&gt; and &lt;math&gt;\triangle TEC&lt;/math&gt;. Let &lt;math&gt;x = [\triangle TDC]&lt;/math&gt;, and &lt;math&gt;y = [\triangle TEC]&lt;/math&gt;. By considering triangles &lt;math&gt;\triangle BTC&lt;/math&gt; and &lt;math&gt;\triangle ETC&lt;/math&gt;, we obtain &lt;math&gt;(1+x)/y=4&lt;/math&gt;, and by considering triangles &lt;math&gt;\triangle ATC&lt;/math&gt; and &lt;math&gt;\triangle DTC&lt;/math&gt;, we obtain &lt;math&gt;(3/4+y)/x=3&lt;/math&gt;. Solving, we get &lt;math&gt;x=4/11&lt;/math&gt;, &lt;math&gt;y=15/44&lt;/math&gt;, so the area of quadrilateral &lt;math&gt;TDEC&lt;/math&gt; is &lt;math&gt;x+y=31/44&lt;/math&gt;.<br /> <br /> Therefore &lt;math&gt;\frac{CD}{BD}=\frac{\frac{3}{4}+\frac{31}{44}}{3+1}=\boxed{\textbf{(D)} \frac{4}{11}}&lt;/math&gt;<br /> <br /> == Solution 1.5 (Triangle Areas, Alternate Approach) ==<br /> <br /> We observe that &lt;math&gt;\frac{BC}{CD} = \frac{[BAE]}{[DAE]}&lt;/math&gt;. The proof is that if &lt;math&gt;B_H&lt;/math&gt; and &lt;math&gt;D_H&lt;/math&gt; are the feet of the altitudes from &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, respectively, to &lt;math&gt;AC&lt;/math&gt;, then both sides of that equation are equal to &lt;math&gt;\frac{BB_H}{DD_H}&lt;/math&gt;.<br /> <br /> From there, we can easily finish:<br /> &lt;cmath&gt;\frac{BC}{CD} = \frac{[BAE]}{[DAE]} = \frac{[BAT]\cdot\frac{5}{4}}{[EAT]\cdot \frac{4}{3}} = 4\cdot\frac{5}{4}\cdot\frac{3}{4} = \frac{15}{4}&lt;/cmath&gt;<br /> and thus &lt;math&gt;\frac{CD}{BD} = \boxed{\textbf{(D)}\frac 4{11}}&lt;/math&gt;.<br /> <br /> == Solution 2 (Mass points) ==<br /> <br /> The presence of only ratios in the problem essentially cries out for mass points.<br /> <br /> As per the problem, we assign a mass of &lt;math&gt;1&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;, and a mass of &lt;math&gt;3&lt;/math&gt; to &lt;math&gt;D&lt;/math&gt;. Then, to balance &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; on &lt;math&gt;T&lt;/math&gt;, &lt;math&gt;T&lt;/math&gt; has a mass of &lt;math&gt;4&lt;/math&gt;.<br /> <br /> Now, were we to assign a mass of &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; and a mass of &lt;math&gt;4&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;, we'd have &lt;math&gt;5T&lt;/math&gt;. Scaling this down by &lt;math&gt;4/5&lt;/math&gt; (to get &lt;math&gt;4T&lt;/math&gt;, which puts &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; in terms of the masses of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;), we assign a mass of &lt;math&gt;\frac{4}{5}&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; and a mass of &lt;math&gt;\frac{16}{5}&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;.<br /> <br /> Now, to balance &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; on &lt;math&gt;E&lt;/math&gt;, we must give &lt;math&gt;C&lt;/math&gt; a mass of &lt;math&gt;\frac{16}{5}-1=\frac{11}{5}&lt;/math&gt;.<br /> <br /> Finally, the ratio of &lt;math&gt;CD&lt;/math&gt; to &lt;math&gt;BD&lt;/math&gt; is given by the ratio of the mass of &lt;math&gt;B&lt;/math&gt; to the mass of &lt;math&gt;C&lt;/math&gt;, which is &lt;math&gt;\frac{4}{5}\cdot\frac{5}{11}=\boxed{\textbf{(D)}\ \frac{4}{11}}&lt;/math&gt;.<br /> <br /> == Solution 3 (Coordinates) ==<br /> <br /> Affine transformations preserve ratios of distances, and for any pair of triangles, there is an affine transformation that maps the first one onto the second one. This is why the answer is the same for any &lt;math&gt;\triangle ABC&lt;/math&gt;, and we just need to compute it for any single triangle.<br /> <br /> We can choose the points &lt;math&gt;A=(-3,0)&lt;/math&gt;, &lt;math&gt;B=(0,4)&lt;/math&gt;, and &lt;math&gt;D=(1,0)&lt;/math&gt;. This way we will have &lt;math&gt;T=(0,0)&lt;/math&gt;, and &lt;math&gt;E=(0,-1)&lt;/math&gt;. The situation is <br /> shown in the picture below:<br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(0.8);<br /> pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1);<br /> draw(A--B--C--cycle);<br /> draw(A--D);<br /> draw(B--E);<br /> pair T=intersectionpoint(A--D,B--E);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,N);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$T$&quot;,T,NW);<br /> label(&quot;$3$&quot;,A--T,N);<br /> label(&quot;$4$&quot;,B--T,W);<br /> label(&quot;$1$&quot;,D--T,N);<br /> label(&quot;$1$&quot;,E--T,W);<br /> <br /> &lt;/asy&gt;<br /> <br /> The point &lt;math&gt;C&lt;/math&gt; is the intersection of the lines &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt;. The points on the first line have the form &lt;math&gt;(t,4-4t)&lt;/math&gt;, the points on the second line have the form &lt;math&gt;(t,-1-t/3)&lt;/math&gt;. Solving for &lt;math&gt;t&lt;/math&gt; we get &lt;math&gt;t=15/11&lt;/math&gt;, hence &lt;math&gt;C=(15/11,-16/11)&lt;/math&gt;.<br /> <br /> The ratio &lt;math&gt;CD/BD&lt;/math&gt; can now be computed simply by observing the &lt;math&gt;x&lt;/math&gt; coordinates of &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;<br /> \frac{CD}{BD} = \frac{15/11 - 1}{1 - 0} = \boxed{\textbf{(D)}\frac 4{11}}<br /> &lt;/cmath&gt;<br /> == Solution 4 (Menelaus) ==<br /> By Menelaus on triangle &lt;math&gt;BDT&lt;/math&gt;, we have that &lt;cmath&gt;\begin{align*}1&amp;=\dfrac{DA}{TA}\cdot\dfrac{TE}{BE}\cdot\dfrac{BC}{DC} \\ &amp;= \dfrac43 \cdot \dfrac 15 \cdot \dfrac {BC}{DC},\end{align*}&lt;/cmath&gt;giving &lt;math&gt;\dfrac{BC}{DC} = \dfrac{15}{4}&lt;/math&gt;. Therefore, &lt;math&gt;\dfrac{CD}{BD} = \boxed{\dfrac{4}{11}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> <br /> {{AMC10 box|year=2004|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_15&diff=113114 2000 AIME I Problems/Problem 15 2019-12-21T01:26:58Z <p>Frestho: /* Solution 3 (Recursion) */</p> <hr /> <div>== Problem ==<br /> A stack of &lt;math&gt;2000&lt;/math&gt; cards is labelled with the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;2000,&lt;/math&gt; with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: &lt;math&gt;1,2,3,\ldots,1999,2000.&lt;/math&gt; In the original stack of cards, how many cards were above the card labeled &lt;math&gt;1999&lt;/math&gt;?<br /> <br /> == Solution ==<br /> We try to work backwards from when there are 2 cards left, since this is when the 1999 card is laid onto the table. When there are 2 cards left, the 1999 card is on the top of the deck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there are over 1000 cards remaining, some cards have not even made one trip through yet, 2(1024 - 1000) = 48, to be exact. Once these cards go through, 1999 will be the &lt;math&gt;512 - 48 = 464^\text{th}&lt;/math&gt; card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position &lt;math&gt;464 \times 2 = 928&lt;/math&gt;, meaning that there were &lt;math&gt;\boxed{927}&lt;/math&gt; cards are above the one labeled &lt;math&gt;1999&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> To simplify matters, we want a power of &lt;math&gt;2&lt;/math&gt;. Hence, we will add &lt;math&gt;48&lt;/math&gt; 'fake' cards which we must discard in our actual count. Using similar logic as Solution 1, we find that 1999 has position &lt;math&gt;1024&lt;/math&gt; in a &lt;math&gt;2048&lt;/math&gt; card stack, where the fake cards towards the front.<br /> <br /> Let the fake cards have positions &lt;math&gt;1, 3, 5, \cdots, 95&lt;/math&gt;. Then, we know that the real cards filling the gaps between the fake cards must be the cards such that they go to their correct starting positions in the &lt;math&gt;2000&lt;/math&gt; card case, where all of them are below &lt;math&gt;1999&lt;/math&gt;. From this, we know that the cards from positions &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;96&lt;/math&gt; alternate in fake-real-fake-real, where we have the correct order of cards once the first &lt;math&gt;96&lt;/math&gt; have moved and we can start putting real cards on the table. Hence, &lt;math&gt;1999&lt;/math&gt; is in position &lt;math&gt;1024 - 96 = 928&lt;/math&gt;, so &lt;math&gt;\boxed{927}&lt;/math&gt; cards are above it. - Spacesam<br /> <br /> ==Solution 3 (Recursion)==<br /> Consider the general problem: with a stack of &lt;math&gt;n&lt;/math&gt; cards such that they will be laid out &lt;math&gt;1, 2, 3, ..., n&lt;/math&gt; from left to right, how many cards are above the card labeled &lt;math&gt;n-1&lt;/math&gt;?<br /> <br /> Let &lt;math&gt;a_n&lt;/math&gt; be the answer to the above problem.<br /> <br /> As a base case, consider &lt;math&gt;n=2&lt;/math&gt;. Clearly, the stack, from top to bottom, must be &lt;math&gt;(1, 2)&lt;/math&gt;, so &lt;math&gt;a_n=0&lt;/math&gt;. <br /> <br /> Next, let's think about how we can construct a stack of &lt;math&gt;n+1&lt;/math&gt; cards from a stack of &lt;math&gt;n&lt;/math&gt; cards. First, let us renumber the current stack of &lt;math&gt;n&lt;/math&gt; by adding &lt;math&gt;1&lt;/math&gt; to the label of each of the cards. Then we must add a card labeled &quot;&lt;math&gt;1&lt;/math&gt;&quot;. <br /> <br /> Working backwards, we find that we must move the bottom card to the top, then add &quot;&lt;math&gt;1&lt;/math&gt;&quot; to the top of the deck. <br /> <br /> Therefore, if &lt;math&gt;a_n \ne n-1&lt;/math&gt; (meaning the card &lt;math&gt;n-1&lt;/math&gt; is not at the bottom of the deck and so it won't be moved to the top), then &lt;math&gt;a_{n+1} = a_n + 2&lt;/math&gt;, since a card from the bottom is moved to be above the &lt;math&gt;n-1&lt;/math&gt; card, and the new card &quot;&lt;math&gt;1&lt;/math&gt;&quot; is added to the top. If &lt;math&gt;a_n = n-1&lt;/math&gt; (meaning the card &lt;math&gt;n-1&lt;/math&gt; is the bottom card), then &lt;math&gt;a_{n+1}=1&lt;/math&gt; because it will move to the top and the card &quot;&lt;math&gt;1&lt;/math&gt;&quot; will be added on top of it.<br /> <br /> With these recursions and the base case we found earlier, we calculate &lt;math&gt;a_{2000} = \boxed{927}&lt;/math&gt;. To calculate this by hand, a helpful trick is finding that if &lt;math&gt;a_n=1&lt;/math&gt;, then &lt;math&gt;a_{2n-1}=1&lt;/math&gt; as well. Once we find &lt;math&gt;a_{1537}=1&lt;/math&gt;, the answer is just &lt;math&gt;1+(2000-1537)\cdot2&lt;/math&gt;.<br /> - Frestho<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_15&diff=113094 2000 AIME I Problems/Problem 15 2019-12-20T17:05:12Z <p>Frestho: /* Solution 3 (Recursion) */</p> <hr /> <div>== Problem ==<br /> A stack of &lt;math&gt;2000&lt;/math&gt; cards is labelled with the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;2000,&lt;/math&gt; with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: &lt;math&gt;1,2,3,\ldots,1999,2000.&lt;/math&gt; In the original stack of cards, how many cards were above the card labeled &lt;math&gt;1999&lt;/math&gt;?<br /> <br /> == Solution ==<br /> We try to work backwards from when there are 2 cards left, since this is when the 1999 card is laid onto the table. When there are 2 cards left, the 1999 card is on the top of the deck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there are over 1000 cards remaining, some cards have not even made one trip through yet, 2(1024 - 1000) = 48, to be exact. Once these cards go through, 1999 will be the &lt;math&gt;512 - 48 = 464^\text{th}&lt;/math&gt; card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position &lt;math&gt;464 \times 2 = 928&lt;/math&gt;, meaning that there were &lt;math&gt;\boxed{927}&lt;/math&gt; cards are above the one labeled &lt;math&gt;1999&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> To simplify matters, we want a power of &lt;math&gt;2&lt;/math&gt;. Hence, we will add &lt;math&gt;48&lt;/math&gt; 'fake' cards which we must discard in our actual count. Using similar logic as Solution 1, we find that 1999 has position &lt;math&gt;1024&lt;/math&gt; in a &lt;math&gt;2048&lt;/math&gt; card stack, where the fake cards towards the front.<br /> <br /> Let the fake cards have positions &lt;math&gt;1, 3, 5, \cdots, 95&lt;/math&gt;. Then, we know that the real cards filling the gaps between the fake cards must be the cards such that they go to their correct starting positions in the &lt;math&gt;2000&lt;/math&gt; card case, where all of them are below &lt;math&gt;1999&lt;/math&gt;. From this, we know that the cards from positions &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;96&lt;/math&gt; alternate in fake-real-fake-real, where we have the correct order of cards once the first &lt;math&gt;96&lt;/math&gt; have moved and we can start putting real cards on the table. Hence, &lt;math&gt;1999&lt;/math&gt; is in position &lt;math&gt;1024 - 96 = 928&lt;/math&gt;, so &lt;math&gt;\boxed{927}&lt;/math&gt; cards are above it. - Spacesam<br /> <br /> ==Solution 3 (Recursion)==<br /> Consider the general problem: with a stack of &lt;math&gt;n&lt;/math&gt; cards such that they will be laid out &lt;math&gt;1, 2, 3, ..., n&lt;/math&gt; from left to right, how many cards are above the card labeled &lt;math&gt;n-1&lt;/math&gt;?<br /> <br /> Let &lt;math&gt;a_n&lt;/math&gt; be the answer to the above problem.<br /> <br /> As a base case, consider &lt;math&gt;n=2&lt;/math&gt;. Clearly, the stack must be (top to bottom) &lt;math&gt;(1, 2)&lt;/math&gt;, so &lt;math&gt;a_n=0&lt;/math&gt;. <br /> <br /> Next, let's think about how we can construct a stack of &lt;math&gt;n+1&lt;/math&gt; cards from a stack of &lt;math&gt;n&lt;/math&gt; cards. First, let us renumber the current stack of &lt;math&gt;n&lt;/math&gt; by adding &lt;math&gt;1&lt;/math&gt; to the label of each of the cards. Then we must add a card labeled &quot;&lt;math&gt;1&lt;/math&gt;&quot;. <br /> <br /> Working backwards, we find that we must move the bottom card to the top, then add &quot;&lt;math&gt;1&lt;/math&gt;&quot; to the top of the deck. <br /> <br /> Therefore, if &lt;math&gt;a_n \ne n-1&lt;/math&gt; (meaning the card &lt;math&gt;n-1&lt;/math&gt; is not at the bottom of the deck and so it won't be moved to the top), then &lt;math&gt;a_{n+1} = a_n + 2&lt;/math&gt;, since a card from the bottom is moved to be above the &lt;math&gt;n-1&lt;/math&gt; card, and the new card &quot;&lt;math&gt;1&lt;/math&gt;&quot; is added to the top. If &lt;math&gt;a_n = n-1&lt;/math&gt; (meaning the card &lt;math&gt;n-1&lt;/math&gt; is the bottom card), then &lt;math&gt;a_{n+1}=1&lt;/math&gt; because it will move to the top and the card &quot;&lt;math&gt;1&lt;/math&gt;&quot; will be added on top of it.<br /> <br /> With these recursions and the base case we found earlier, we calculate &lt;math&gt;a_{2000} = \boxed{927}&lt;/math&gt;. To calculate this by hand, a helpful trick is finding that if &lt;math&gt;a_n=1&lt;/math&gt;, then &lt;math&gt;a_{2n-1}=1&lt;/math&gt; as well. Once we find &lt;math&gt;a_{1537}=1&lt;/math&gt;, the answer is just &lt;math&gt;1+(2000-1537)\cdot2&lt;/math&gt;.<br /> - Frestho<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_15&diff=113087 2000 AIME I Problems/Problem 15 2019-12-20T06:06:11Z <p>Frestho: /* Solution 3 (Recursion) */</p> <hr /> <div>== Problem ==<br /> A stack of &lt;math&gt;2000&lt;/math&gt; cards is labelled with the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;2000,&lt;/math&gt; with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: &lt;math&gt;1,2,3,\ldots,1999,2000.&lt;/math&gt; In the original stack of cards, how many cards were above the card labeled &lt;math&gt;1999&lt;/math&gt;?<br /> <br /> == Solution ==<br /> We try to work backwards from when there are 2 cards left, since this is when the 1999 card is laid onto the table. When there are 2 cards left, the 1999 card is on the top of the deck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there are over 1000 cards remaining, some cards have not even made one trip through yet, 2(1024 - 1000) = 48, to be exact. Once these cards go through, 1999 will be the &lt;math&gt;512 - 48 = 464^\text{th}&lt;/math&gt; card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position &lt;math&gt;464 \times 2 = 928&lt;/math&gt;, meaning that there were &lt;math&gt;\boxed{927}&lt;/math&gt; cards are above the one labeled &lt;math&gt;1999&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> To simplify matters, we want a power of &lt;math&gt;2&lt;/math&gt;. Hence, we will add &lt;math&gt;48&lt;/math&gt; 'fake' cards which we must discard in our actual count. Using similar logic as Solution 1, we find that 1999 has position &lt;math&gt;1024&lt;/math&gt; in a &lt;math&gt;2048&lt;/math&gt; card stack, where the fake cards towards the front.<br /> <br /> Let the fake cards have positions &lt;math&gt;1, 3, 5, \cdots, 95&lt;/math&gt;. Then, we know that the real cards filling the gaps between the fake cards must be the cards such that they go to their correct starting positions in the &lt;math&gt;2000&lt;/math&gt; card case, where all of them are below &lt;math&gt;1999&lt;/math&gt;. From this, we know that the cards from positions &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;96&lt;/math&gt; alternate in fake-real-fake-real, where we have the correct order of cards once the first &lt;math&gt;96&lt;/math&gt; have moved and we can start putting real cards on the table. Hence, &lt;math&gt;1999&lt;/math&gt; is in position &lt;math&gt;1024 - 96 = 928&lt;/math&gt;, so &lt;math&gt;\boxed{927}&lt;/math&gt; cards are above it. - Spacesam<br /> <br /> ==Solution 3 (Recursion)==<br /> Consider the general problem: with a stack of &lt;math&gt;n&lt;/math&gt; cards such that they will be laid out &lt;math&gt;1, 2, 3, ..., n&lt;/math&gt; from left to right, how many cards are above the card labeled &lt;math&gt;n-1&lt;/math&gt;?<br /> <br /> Let &lt;math&gt;a_n&lt;/math&gt; be the answer to the above problem.<br /> <br /> As a base case, consider &lt;math&gt;n=2&lt;/math&gt;. Clearly, the stack must be (top to bottom) &lt;math&gt;(1, 2)&lt;/math&gt;, so &lt;math&gt;a_n=0&lt;/math&gt;. <br /> <br /> Next, let's think about how we can construct a stack of &lt;math&gt;n+1&lt;/math&gt; cards from a stack of &lt;math&gt;n&lt;/math&gt; cards. First, let us renumber the current stack of &lt;math&gt;n&lt;/math&gt; by adding &lt;math&gt;1&lt;/math&gt; to the label of each of the cards. Then we must add a card labeled &quot;&lt;math&gt;1&lt;/math&gt;&quot;. <br /> <br /> Working backwards, we find that we must move the bottom card to the top, then add &quot;&lt;math&gt;1&lt;/math&gt;&quot; to the top of the deck. <br /> <br /> Therefore, if &lt;math&gt;a_n \ne n-1&lt;/math&gt; (meaning the card &lt;math&gt;n-1&lt;/math&gt; is not at the bottom of the deck and so it won't be moved to the top), then &lt;math&gt;a_{n+1} = a_n + 2&lt;/math&gt;, since a card from the bottom is moved to be above the &lt;math&gt;n-1&lt;/math&gt; card, and the new card &quot;&lt;math&gt;1&lt;/math&gt;&quot; is added to the top. If &lt;math&gt;a_n = n-1&lt;/math&gt; (meaning the card &lt;math&gt;n-1&lt;/math&gt; is the bottom card), then &lt;math&gt;a_{n+1}=1&lt;/math&gt; because it will move to the top and the card &quot;&lt;math&gt;1&lt;/math&gt;&quot; will be added on top of it.<br /> <br /> With these recursions and the base case we found earlier, we calculate &lt;math&gt;a_{2000} = \boxed{927}&lt;/math&gt;. To calculate this by hand, note that a helpful trick is finding that if &lt;math&gt;a_n=1&lt;/math&gt;, then &lt;math&gt;a_{2n-1}=1&lt;/math&gt; as well. Once we find &lt;math&gt;a_{1537}=1&lt;/math&gt;, the answer is just &lt;math&gt;1+(2000-1537)\cdot2&lt;/math&gt;.<br /> - Frestho<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_15&diff=113086 2000 AIME I Problems/Problem 15 2019-12-20T06:01:59Z <p>Frestho: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A stack of &lt;math&gt;2000&lt;/math&gt; cards is labelled with the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;2000,&lt;/math&gt; with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: &lt;math&gt;1,2,3,\ldots,1999,2000.&lt;/math&gt; In the original stack of cards, how many cards were above the card labeled &lt;math&gt;1999&lt;/math&gt;?<br /> <br /> == Solution ==<br /> We try to work backwards from when there are 2 cards left, since this is when the 1999 card is laid onto the table. When there are 2 cards left, the 1999 card is on the top of the deck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there are over 1000 cards remaining, some cards have not even made one trip through yet, 2(1024 - 1000) = 48, to be exact. Once these cards go through, 1999 will be the &lt;math&gt;512 - 48 = 464^\text{th}&lt;/math&gt; card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position &lt;math&gt;464 \times 2 = 928&lt;/math&gt;, meaning that there were &lt;math&gt;\boxed{927}&lt;/math&gt; cards are above the one labeled &lt;math&gt;1999&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> To simplify matters, we want a power of &lt;math&gt;2&lt;/math&gt;. Hence, we will add &lt;math&gt;48&lt;/math&gt; 'fake' cards which we must discard in our actual count. Using similar logic as Solution 1, we find that 1999 has position &lt;math&gt;1024&lt;/math&gt; in a &lt;math&gt;2048&lt;/math&gt; card stack, where the fake cards towards the front.<br /> <br /> Let the fake cards have positions &lt;math&gt;1, 3, 5, \cdots, 95&lt;/math&gt;. Then, we know that the real cards filling the gaps between the fake cards must be the cards such that they go to their correct starting positions in the &lt;math&gt;2000&lt;/math&gt; card case, where all of them are below &lt;math&gt;1999&lt;/math&gt;. From this, we know that the cards from positions &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;96&lt;/math&gt; alternate in fake-real-fake-real, where we have the correct order of cards once the first &lt;math&gt;96&lt;/math&gt; have moved and we can start putting real cards on the table. Hence, &lt;math&gt;1999&lt;/math&gt; is in position &lt;math&gt;1024 - 96 = 928&lt;/math&gt;, so &lt;math&gt;\boxed{927}&lt;/math&gt; cards are above it. - Spacesam<br /> <br /> ==Solution 3 (Recursion)==<br /> Consider the general problem: with a stack of &lt;math&gt;n&lt;/math&gt; cards such that they will be laid out &lt;math&gt;1, 2, 3, ..., n&lt;/math&gt; from left to right, how many cards are above the card labeled &lt;math&gt;n-1&lt;/math&gt;?<br /> <br /> Let &lt;math&gt;a_n&lt;/math&gt; be the answer to the above problem.<br /> <br /> As a base case, consider &lt;math&gt;n=2&lt;/math&gt;. Clearly, the stack must be (top to bottom) &lt;math&gt;(1, 2)&lt;/math&gt;, so &lt;math&gt;a_n=0&lt;/math&gt;. <br /> <br /> Next, let's think about how we can construct a stack of &lt;math&gt;n+1&lt;/math&gt; cards from a stack of &lt;math&gt;n&lt;/math&gt; cards. First, let us renumber the current stack of &lt;math&gt;n&lt;/math&gt; by adding &lt;math&gt;1&lt;/math&gt; to the label of each of the cards. Then we must add a card labeled &quot;&lt;math&gt;1&lt;/math&gt;&quot;. <br /> <br /> Working backwards, we find that we must move the bottom card to the top, then add &quot;&lt;math&gt;1&lt;/math&gt;&quot; to the top of the deck. <br /> <br /> Therefore, if &lt;math&gt;a_n \ne n-1&lt;/math&gt; (meaning the card &lt;math&gt;n-1&lt;/math&gt; is not at the bottom of the deck and so it won't be moved to the top), then &lt;math&gt;a_{n+1} = a_n + 2&lt;/math&gt;, since a card from the bottom is moved to be above the &lt;math&gt;n-1&lt;/math&gt; card, and the new card &quot;&lt;math&gt;1&lt;/math&gt;&quot; is added to the top. If &lt;math&gt;a_n = n-1&lt;/math&gt; (meaning the card &lt;math&gt;n-1&lt;/math&gt; is the bottom card), then &lt;math&gt;a_{n+1}=1&lt;/math&gt; because it will move to the top and the card &quot;&lt;math&gt;1&lt;/math&gt;&quot; will be added on top of it.<br /> <br /> With these recursions and the base case we found earlier, we calculate &lt;math&gt;a_{2000} = \boxed{927}&lt;/math&gt;. To calculate this by hand, note that a helpful trick is finding that if &lt;math&gt;a_n=1&lt;/math&gt;, then &lt;math&gt;a_{2n-1}=1&lt;/math&gt; as well. Once we find &lt;math&gt;a_{1537}=1&lt;/math&gt;, the answer is just &lt;math&gt;1+(2000-1537)\cdot2&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=113078 2019 AMC 8 Problems/Problem 24 2019-12-20T01:31:29Z <p>Frestho: /* Solution 9 */</p> <hr /> <div><br /> <br /> <br /> ==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; s that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and left &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Draw &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{AF}&lt;/math&gt; such that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE = ED&lt;/math&gt;. &lt;math&gt;FC=3XD&lt;/math&gt; so &lt;math&gt;BC=4BF&lt;/math&gt;. Since &lt;math&gt;AF=3EF&lt;/math&gt; (&lt;math&gt;XE=EF&lt;/math&gt; and &lt;math&gt;AX=\frac13 AF&lt;/math&gt;, so &lt;math&gt;XE=EF=\frac13 AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{3}&lt;/math&gt; of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, so the area of &lt;math&gt;BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{(B) 30}&lt;/math&gt; ~[[User:heeeeeeeheeeee|heeeeeeeheeeee]]<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of 3. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of 3 also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}(30)&lt;/math&gt;<br /> -Brudder<br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{B}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{(B) 30}&lt;/math&gt;. ~~SmileKat32<br /> <br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Since &lt;math&gt;CD = 2AD&lt;/math&gt;, triangle &lt;math&gt;CDB&lt;/math&gt; has four times the area of triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we get &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the congruent. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and therefore four times the area of triangle &lt;math&gt;BEF&lt;/math&gt;, giving &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> <br /> ==Solution 5 (Area Ratios)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> As before we figure out the areas labeled in the diagram. Then we note that &lt;cmath&gt;\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.&lt;/cmath&gt;Solving gives &lt;math&gt;x = \boxed{\textbf{(B) }30}&lt;/math&gt;. <br /> (Credit to scrabbler94 for the idea)<br /> <br /> ==Solution 6(Coordinate Bashing)==<br /> Let &lt;math&gt;ADB&lt;/math&gt; be a right triangle, and &lt;math&gt;BD=CD&lt;/math&gt;<br /> <br /> Let &lt;math&gt;A=(-2\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;B=(0, 4\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;C=(4\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;D=(0, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;E=(0, 2\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;F=(\sqrt{30}, 3\sqrt{30})&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{AE}&lt;/math&gt; can be described with the equation &lt;math&gt;y=x-2\sqrt{30}&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{BC}&lt;/math&gt; can be described with &lt;math&gt;x+y=4\sqrt{30}&lt;/math&gt;<br /> <br /> Solving, we get &lt;math&gt;x=3\sqrt{30}&lt;/math&gt; and &lt;math&gt;y=\sqrt{30}&lt;/math&gt;<br /> <br /> Now we can find &lt;math&gt;EF=BF=2\sqrt{15}&lt;/math&gt;<br /> <br /> &lt;math&gt;[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{(B) 30}\blacksquare&lt;/math&gt;<br /> <br /> -Trex4days<br /> <br /> == Solution 7 ==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(15cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */<br /> <br /> /* draw figures */<br /> draw(circle((0,0), 5), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> draw((-4,-3)--(0,5), linewidth(2)); <br /> draw((0,5)--(4,3), linewidth(2)); <br /> draw((12,-1)--(-4,-3), linewidth(2)); <br /> draw((0,5)--(0,-5), linewidth(2)); <br /> draw((-4,-3)--(0,-5), linewidth(2)); <br /> draw((4,3)--(0,2.48), linewidth(2)); <br /> draw((4,3)--(12,-1), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> /* dots and labels */<br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((-5,0),dotstyle); <br /> dot((-4,-3),dotstyle); <br /> label(&quot;B&quot;, (-4.45,-3.38), NE * labelscalefactor); <br /> dot((4,3),dotstyle); <br /> label(&quot;$D$&quot;, (4.15,3.2), NE * labelscalefactor); <br /> dot((0,5),dotstyle); <br /> label(&quot;A&quot;, (-0.09,5.26), NE * labelscalefactor); <br /> dot((12,-1),dotstyle); <br /> label(&quot;C&quot;, (12.23,-1.24), NE * labelscalefactor); <br /> dot((0,-5),dotstyle); <br /> label(&quot;$G$&quot;, (0.19,-4.82), NE * labelscalefactor); <br /> dot((0,2.48),dotstyle); <br /> label(&quot;I&quot;, (-0.33,2.2), NE * labelscalefactor); <br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((0,-2.5),dotstyle); <br /> label(&quot;F&quot;, (0.23,-2.2), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;A[\Delta XYZ]&lt;/math&gt; = &lt;math&gt;Area&lt;/math&gt; &lt;math&gt;of&lt;/math&gt; &lt;math&gt;Triangle&lt;/math&gt; &lt;math&gt;XYZ&lt;/math&gt; <br /> <br /> <br /> &lt;math&gt;A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;A[\Delta ABE] = A[\Delta AED] = 60&lt;/math&gt; (the median divides the area of the triangle into two equal parts)<br /> <br /> <br /> Construction: Draw a circumcircle around &lt;math&gt;\Delta ABD&lt;/math&gt; with &lt;math&gt;BD&lt;/math&gt; as is diameter. Extend &lt;math&gt;AF&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that it meets the circle at &lt;math&gt;G&lt;/math&gt;. Draw line &lt;math&gt;BG&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;A[\Delta ABD] = A[\Delta ABG] = 120&lt;/math&gt; (Since &lt;math&gt;\square ABGD&lt;/math&gt; is cyclic)<br /> <br /> <br /> But &lt;math&gt;A[\Delta ABE]&lt;/math&gt; is common in both with an area of 60. So, &lt;math&gt;A[\Delta AED] = A[\Delta BEG]&lt;/math&gt;.<br /> <br /> \therefore &lt;math&gt;A[\Delta AED] \cong A[\Delta BEG]&lt;/math&gt; (SAS Congruency Theorem).<br /> <br /> In &lt;math&gt;\Delta AED&lt;/math&gt;, let &lt;math&gt;DI&lt;/math&gt; be the median of &lt;math&gt;\Delta AED&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta AID] = 30 = A[\Delta EID]&lt;/math&gt;<br /> <br /> <br /> Rotate &lt;math&gt;\Delta DEA&lt;/math&gt; to meet &lt;math&gt;D&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; at &lt;math&gt;G&lt;/math&gt;. &lt;math&gt;DE&lt;/math&gt; will fit exactly in &lt;math&gt;BE&lt;/math&gt; (both are radii of the circle). From the above solutions, &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;.<br /> <br /> &lt;math&gt;AE&lt;/math&gt; is a radius and &lt;math&gt;EF&lt;/math&gt; is half of it implies &lt;math&gt;EF&lt;/math&gt; = &lt;math&gt;\frac{radius}{2}&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta BEF] \cong A[\Delta DEI]&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;A[\Delta BEF] = 30&lt;/math&gt;<br /> <br /> <br /> ~phoenixfire &amp; flamewavelight<br /> <br /> == Solution 8 ==<br /> The diagram is very inaccurate. <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF, M;<br /> B = (0,0); C = (3,0); M = (1.45,0);<br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> draw(EE--M);<br /> label(&quot;$M$&quot;,M,S);<br /> label(&quot;$1$&quot;,A--DD,N);<br /> label(&quot;$2$&quot;,DD--C,N);<br /> label(&quot;$1$&quot;,EE--M,N);<br /> &lt;/asy&gt;<br /> Note: All numbers above &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{EM}&lt;/math&gt; are ratios, not actual side lengths. If somebody could edit this to make the diagram more accurate it would be greatly appreicated. &lt;br /&gt;<br /> Using the ratio of &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, we find the area of &lt;math&gt;\triangle ADB&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt; and the area of &lt;math&gt;\triangle BDC&lt;/math&gt; is &lt;math&gt;240&lt;/math&gt;. Also using the fact that &lt;math&gt;E&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt;, we know &lt;math&gt;\triangle ADE = \triangle ABE = 60&lt;/math&gt;.<br /> Let &lt;math&gt;M&lt;/math&gt; be a point such &lt;math&gt;\overline{EM}&lt;/math&gt; is parellel to &lt;math&gt;\overline{CD}&lt;/math&gt;. We immediatley know that &lt;math&gt;\triangle BEM \sim BDC&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt;. Using that we can conclude &lt;math&gt;EM&lt;/math&gt; has ratio &lt;math&gt;1&lt;/math&gt;. Using &lt;math&gt;\triangle EFM \sim \triangle AFC&lt;/math&gt;, we get &lt;math&gt;EF:AE = 1:2&lt;/math&gt;. Therefore using the fact that &lt;math&gt;\triangle EBF&lt;/math&gt; is in &lt;math&gt;\triangle ABF&lt;/math&gt;, the area has ratio &lt;math&gt;\triangle BEF : \triangle ABE=1:2&lt;/math&gt; and we know &lt;math&gt;\triangle ABE&lt;/math&gt; has area &lt;math&gt;60&lt;/math&gt; so &lt;math&gt;\triangle BEF&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{B} \, 30}&lt;/math&gt;. - fath2012<br /> <br /> ==Solution 9==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(F--D);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$x$&quot;, (D+E+F)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120+2x$&quot;, (D+F+C)/3);<br /> &lt;/asy&gt;<br /> Labeling the areas in the diagram, we have:&lt;br /&gt;<br /> &lt;math&gt;[DBC]=240=[BFE]+[FED]+[FDC]=x+x+120+2x=120+4x&lt;/math&gt; so &lt;math&gt;240=120+4x, 120=4x, 30=x&lt;/math&gt;.&lt;br /&gt;<br /> So our answer is &lt;math&gt;\boxed{\textbf{(B)} 30}&lt;/math&gt;. ~~RWhite<br /> ==Solution 10 (Menelaus's Theorem)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> By Menelaus's Theorem on triangle &lt;math&gt;BCD&lt;/math&gt;, we have &lt;cmath&gt;\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.&lt;/cmath&gt; Therefore, &lt;cmath&gt;[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_8&diff=112459 2012 AMC 10A Problems/Problem 8 2019-12-02T04:00:13Z <p>Frestho: /* Solution 2 (Faster) */</p> <hr /> <div>{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #6]] and [[2012 AMC 10A Problems|2012 AMC 10A #8]]}}<br /> <br /> == Problem ==<br /> The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Let the three numbers be equal to &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;. We can now write three equations:<br /> <br /> &lt;math&gt;a+b=12&lt;/math&gt;<br /> <br /> &lt;math&gt;b+c=17&lt;/math&gt;<br /> <br /> &lt;math&gt;a+c=19&lt;/math&gt;<br /> <br /> Adding these equations together, we get that<br /> <br /> &lt;math&gt;2(a+b+c)=48&lt;/math&gt; and<br /> <br /> &lt;math&gt;a+b+c=24&lt;/math&gt;<br /> <br /> Substituting the original equations into this one, we find<br /> <br /> &lt;math&gt;c+12=24&lt;/math&gt;<br /> <br /> &lt;math&gt;a+17=24&lt;/math&gt;<br /> <br /> &lt;math&gt;b+19=24&lt;/math&gt;<br /> <br /> Therefore, our numbers are 12, 7, and 5. The middle number is &lt;math&gt;\boxed{\textbf{(D)}\ 7}&lt;/math&gt;<br /> <br /> ==Solution 2 (Faster)==<br /> <br /> Let the three numbers be &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;a&lt;b&lt;c&lt;/math&gt;. We get the three equations: <br /> <br /> &lt;math&gt;a+b=12&lt;/math&gt;<br /> <br /> &lt;math&gt;a+c=17&lt;/math&gt;<br /> <br /> &lt;math&gt;b+c=19&lt;/math&gt;<br /> <br /> We add the first and last equations and then subtract the second one.<br /> <br /> &lt;math&gt;(a+b)+(b+c)-(a+c) = 12+19-17 \Rightarrow 2b=14 \Rightarrow b = 7&lt;/math&gt;<br /> <br /> Because &lt;math&gt;b&lt;/math&gt; is the middle number, the middle number is &lt;math&gt;\boxed{\textbf{(D)}\ 7}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Just guess and check. You can do it in your head in about 45 seconds and easily check your answer.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2012|ab=A|num-b=7|num-a=9}}<br /> {{AMC12 box|year=2012|ab=A|num-b=5|num-a=7}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_7&diff=112218 2019 AIME I Problems/Problem 7 2019-11-27T04:15:43Z <p>Frestho: /* Solution 2 (Crappier Solution) */</p> <hr /> <div>==Problem 7==<br /> There are positive integers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; that satisfy the system of equations &lt;cmath&gt;<br /> \begin{align*}<br /> \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &amp;= 60\\<br /> \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &amp;= 570.<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> Let &lt;math&gt;m&lt;/math&gt; be the number of (not necessarily distinct) prime factors in the prime factorization of &lt;math&gt;x&lt;/math&gt;, and let &lt;math&gt;n&lt;/math&gt; be the number of (not necessarily distinct) prime factors in the prime factorization of &lt;math&gt;y&lt;/math&gt;. Find &lt;math&gt;3m+2n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Add the two equations to get that &lt;math&gt;\log x+\log y+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630&lt;/math&gt;.<br /> Then, we use the theorem &lt;math&gt;\log a+\log b=\log ab&lt;/math&gt; to get the equation, &lt;math&gt;\log (xy)+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630&lt;/math&gt;.<br /> Using the theorem that &lt;math&gt;\gcd(x,y) \cdot \text{lcm}(x,y)=x\cdot y&lt;/math&gt;, along with the previously mentioned theorem, we can get the equation &lt;math&gt;3\log(xy)=630&lt;/math&gt;.<br /> This can easily be simplified to &lt;math&gt;\log(xy)=210&lt;/math&gt;, or &lt;math&gt;xy = 10^{210}&lt;/math&gt;.<br /> <br /> &lt;math&gt;10^{210}&lt;/math&gt; can be factored into &lt;math&gt;2^{210} \cdot 5^{210}&lt;/math&gt;, and &lt;math&gt;m+n&lt;/math&gt; equals to the sum of the exponents of 2 and 5, which is &lt;math&gt;210+210 = 420&lt;/math&gt;.<br /> Multiply by two to get &lt;math&gt;2m +2n&lt;/math&gt;, which is &lt;math&gt;840&lt;/math&gt;.<br /> Then, use the first equation (&lt;math&gt;\log x + 2\log(\gcd(x,y)) = 60&lt;/math&gt;) to show that x has to have lower degrees of 2 and 5 than y. Therefore, making the &lt;math&gt;gcd x&lt;/math&gt;. Then, turn the equation into &lt;math&gt;3\log x = 60&lt;/math&gt;, which yields &lt;math&gt;\log x = 20&lt;/math&gt;, or &lt;math&gt;x = 10^{20}&lt;/math&gt;.<br /> Factor this into &lt;math&gt;2^{20} \cdot 5^{20}&lt;/math&gt;, and add the two 20's, resulting in m, which is 40.<br /> Add &lt;math&gt;m&lt;/math&gt; to &lt;math&gt;2m + 2n&lt;/math&gt; (which is 840) to get &lt;math&gt;40+840 = \boxed{880}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Crappier Solution)==<br /> <br /> First simplifying the first and second equations, we get that <br /> <br /> &lt;cmath&gt;\log_{10}(x\cdot\text{gcd}(x,y)^2)=60&lt;/cmath&gt;<br /> &lt;cmath&gt;\log_{10}(y\cdot\text{lcm}(x,y)^2)=570&lt;/cmath&gt;<br /> <br /> <br /> Thus, when the two equations are added, we have that<br /> &lt;cmath&gt;\log_{10}(x\cdot y\cdot\text{gcd}^2\cdot\text{lcm}^2)=630&lt;/cmath&gt;<br /> When simplified, this equals <br /> &lt;cmath&gt;\log_{10}(x^3y^3)=630&lt;/cmath&gt;<br /> so this means that<br /> &lt;cmath&gt;x^3y^3=10^{630}&lt;/cmath&gt;<br /> so<br /> &lt;cmath&gt;xy=10^{210}.&lt;/cmath&gt;<br /> <br /> Now, the following cannot be done on a proof contest but let's (intuitively) assume that &lt;math&gt;x&lt;y&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both powers of &lt;math&gt;10&lt;/math&gt;. This means the first equation would simplify to &lt;cmath&gt;x^3=10^{60}&lt;/cmath&gt; and &lt;cmath&gt;y^3=10^{570}.&lt;/cmath&gt; Therefore, &lt;math&gt;x=10^{20}&lt;/math&gt; and &lt;math&gt;y=10^{190}&lt;/math&gt; and if we plug these values back, it works! &lt;math&gt;10^{20}&lt;/math&gt; has &lt;math&gt;20\cdot2=40&lt;/math&gt; total factors and &lt;math&gt;10^{190}&lt;/math&gt; has &lt;math&gt;190\cdot2=380&lt;/math&gt; so &lt;cmath&gt;3\cdot 40 + 2\cdot 380 = \boxed{880}.&lt;/cmath&gt;<br /> <br /> Please remember that you should only assume on these math contests because they are timed; this would technically not be a valid solution.<br /> <br /> ==Solution 3 (Easy Solution)==<br /> Let &lt;math&gt;x=10^a&lt;/math&gt; and &lt;math&gt;y=10^b&lt;/math&gt; and &lt;math&gt;a&lt;b&lt;/math&gt;. Then the given equations become &lt;math&gt;3a=60&lt;/math&gt; and &lt;math&gt;3b=570&lt;/math&gt;. Therefore, &lt;math&gt;x=10^{20}=2^{20}\cdot5^{20}&lt;/math&gt; and &lt;math&gt;y=10^{190}=2^{190}\cdot5^{190}&lt;/math&gt;. Our answer is &lt;math&gt;3(20+20)+2(190+190)=\boxed{880}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_22&diff=111959 2018 AMC 8 Problems/Problem 22 2019-11-21T17:28:29Z <p>Frestho: /* Solution 3 */</p> <hr /> <div>==Problem 22==<br /> Point &lt;math&gt;E&lt;/math&gt; is the midpoint of side &lt;math&gt;\overline{CD}&lt;/math&gt; in square &lt;math&gt;ABCD,&lt;/math&gt; and &lt;math&gt;\overline{BE}&lt;/math&gt; meets diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; at &lt;math&gt;F.&lt;/math&gt; The area of quadrilateral &lt;math&gt;AFED&lt;/math&gt; is &lt;math&gt;45.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> draw((0,0)--(6,0)--(6,6)--(0,6)--cycle);<br /> draw((0,6)--(6,0)); draw((3,0)--(6,6));<br /> label(&quot;$A$&quot;,(0,6),NW);<br /> label(&quot;$B$&quot;,(6,6),NE);<br /> label(&quot;$C$&quot;,(6,0),SE);<br /> label(&quot;$D$&quot;,(0,0),SW);<br /> label(&quot;$E$&quot;,(3,0),S);<br /> label(&quot;$F$&quot;,(4,2),E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the area of &lt;math&gt;\triangle CEF&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt;. Thus, the area of triangle &lt;math&gt;\triangle ACD&lt;/math&gt; is &lt;math&gt;45+x&lt;/math&gt; and the area of the square is &lt;math&gt;2(45+x) = 90+2x&lt;/math&gt;.<br /> <br /> By AA similarity, &lt;math&gt;\triangle CEF \sim \triangle ABF&lt;/math&gt; with a 1:2 ratio, so the area of triangle &lt;math&gt;\triangle ABF&lt;/math&gt; is &lt;math&gt;4x&lt;/math&gt;. Now consider trapezoid &lt;math&gt;ABED&lt;/math&gt;. Its area is &lt;math&gt;45+4x&lt;/math&gt;, which is three-fourths the area of the square. We set up an equation in &lt;math&gt;x&lt;/math&gt;:<br /> <br /> &lt;cmath&gt; 45+4x = \frac{3}{4}\left(90+2x\right) &lt;/cmath&gt;<br /> Solving, we get &lt;math&gt;x = 9&lt;/math&gt;. The area of square &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B)} 108}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can use analytic geometry for this problem.<br /> <br /> Let us start by giving &lt;math&gt;D&lt;/math&gt; the coordinate &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt; the coordinate &lt;math&gt;(0,1)&lt;/math&gt;, and so forth. &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{EB}&lt;/math&gt; can be represented by the equations &lt;math&gt;y=-x+1&lt;/math&gt; and &lt;math&gt;y=2x-1&lt;/math&gt;, respectively. Solving for their intersection gives point &lt;math&gt;F&lt;/math&gt; coordinates &lt;math&gt;\left(\frac{2}{3},\frac{1}{3}\right)&lt;/math&gt;. <br /> <br /> Now, &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;EFC&lt;/math&gt;’s area is simply &lt;math&gt;\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}&lt;/math&gt; or &lt;math&gt;\frac{1}{12}&lt;/math&gt;. This means that pentagon &lt;math&gt;ABCEF&lt;/math&gt;’s area is &lt;math&gt;\frac{1}{2}+\frac{1}{12}=\frac{7}{12}&lt;/math&gt; of the entire square, and it follows that quadrilateral &lt;math&gt;AFED&lt;/math&gt;’s area is &lt;math&gt;\frac{5}{12}&lt;/math&gt; of the square. <br /> <br /> The area of the square is then &lt;math&gt;\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}&lt;/math&gt;.<br /> ==Solution 3==<br /> Note that triangle &lt;math&gt;ABC&lt;/math&gt; has half the area of the square and triangle &lt;math&gt;FEC&lt;/math&gt; has &lt;math&gt;\dfrac1{12}&lt;/math&gt;th. Thus the area of the quadrilateral is &lt;math&gt;1-1/2-1/12=5/12&lt;/math&gt;th the area of the square. The area of the square is then &lt;math&gt;45\cdot\dfrac{12}{5}=\boxed{\textbf{(B)}108}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AMC8 box|year=2018|num-b=21|num-a=23}}<br /> Set s to be the bottom left triangle.<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_22&diff=111958 2018 AMC 8 Problems/Problem 22 2019-11-21T17:27:49Z <p>Frestho: /* Solution 2 */</p> <hr /> <div>==Problem 22==<br /> Point &lt;math&gt;E&lt;/math&gt; is the midpoint of side &lt;math&gt;\overline{CD}&lt;/math&gt; in square &lt;math&gt;ABCD,&lt;/math&gt; and &lt;math&gt;\overline{BE}&lt;/math&gt; meets diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; at &lt;math&gt;F.&lt;/math&gt; The area of quadrilateral &lt;math&gt;AFED&lt;/math&gt; is &lt;math&gt;45.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> draw((0,0)--(6,0)--(6,6)--(0,6)--cycle);<br /> draw((0,6)--(6,0)); draw((3,0)--(6,6));<br /> label(&quot;$A$&quot;,(0,6),NW);<br /> label(&quot;$B$&quot;,(6,6),NE);<br /> label(&quot;$C$&quot;,(6,0),SE);<br /> label(&quot;$D$&quot;,(0,0),SW);<br /> label(&quot;$E$&quot;,(3,0),S);<br /> label(&quot;$F$&quot;,(4,2),E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the area of &lt;math&gt;\triangle CEF&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt;. Thus, the area of triangle &lt;math&gt;\triangle ACD&lt;/math&gt; is &lt;math&gt;45+x&lt;/math&gt; and the area of the square is &lt;math&gt;2(45+x) = 90+2x&lt;/math&gt;.<br /> <br /> By AA similarity, &lt;math&gt;\triangle CEF \sim \triangle ABF&lt;/math&gt; with a 1:2 ratio, so the area of triangle &lt;math&gt;\triangle ABF&lt;/math&gt; is &lt;math&gt;4x&lt;/math&gt;. Now consider trapezoid &lt;math&gt;ABED&lt;/math&gt;. Its area is &lt;math&gt;45+4x&lt;/math&gt;, which is three-fourths the area of the square. We set up an equation in &lt;math&gt;x&lt;/math&gt;:<br /> <br /> &lt;cmath&gt; 45+4x = \frac{3}{4}\left(90+2x\right) &lt;/cmath&gt;<br /> Solving, we get &lt;math&gt;x = 9&lt;/math&gt;. The area of square &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B)} 108}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can use analytic geometry for this problem.<br /> <br /> Let us start by giving &lt;math&gt;D&lt;/math&gt; the coordinate &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt; the coordinate &lt;math&gt;(0,1)&lt;/math&gt;, and so forth. &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{EB}&lt;/math&gt; can be represented by the equations &lt;math&gt;y=-x+1&lt;/math&gt; and &lt;math&gt;y=2x-1&lt;/math&gt;, respectively. Solving for their intersection gives point &lt;math&gt;F&lt;/math&gt; coordinates &lt;math&gt;\left(\frac{2}{3},\frac{1}{3}\right)&lt;/math&gt;. <br /> <br /> Now, &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;EFC&lt;/math&gt;’s area is simply &lt;math&gt;\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}&lt;/math&gt; or &lt;math&gt;\frac{1}{12}&lt;/math&gt;. This means that pentagon &lt;math&gt;ABCEF&lt;/math&gt;’s area is &lt;math&gt;\frac{1}{2}+\frac{1}{12}=\frac{7}{12}&lt;/math&gt; of the entire square, and it follows that quadrilateral &lt;math&gt;AFED&lt;/math&gt;’s area is &lt;math&gt;\frac{5}{12}&lt;/math&gt; of the square. <br /> <br /> The area of the square is then &lt;math&gt;\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}&lt;/math&gt;.<br /> ==Solution 3==<br /> Note that triangle &lt;math&gt;ABC&lt;/math&gt; has half the area of the square and triangle &lt;math&gt;FEC&lt;/math&gt; has &lt;math&gt;\dfrac1{12}&lt;/math&gt;th. This the area of the quadrilateral is &lt;math&gt;1-1/2-1/12=5/12&lt;/math&gt;th the area of the square. The area of the square is then &lt;math&gt;45\cdot\dfrac{12}{5}=\boxed{\textbf{(B)}108}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AMC8 box|year=2018|num-b=21|num-a=23}}<br /> Set s to be the bottom left triangle.<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=111885 2019 AMC 8 Problems/Problem 24 2019-11-21T04:57:56Z <p>Frestho: /* Solution 5 (Area Ratios) */</p> <hr /> <div><br /> <br /> <br /> ==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; s that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and left &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Draw &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{AF}&lt;/math&gt; such that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE&lt;/math&gt;=&lt;math&gt;ED&lt;/math&gt;. &lt;math&gt;FC&lt;/math&gt;=3&lt;math&gt;XD&lt;/math&gt; so &lt;math&gt;BC&lt;/math&gt;=4&lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;AF&lt;/math&gt;=3&lt;math&gt;EF&lt;/math&gt;( &lt;math&gt;XE&lt;/math&gt;=&lt;math&gt;EF&lt;/math&gt; and &lt;math&gt;AX&lt;/math&gt;=1/3&lt;math&gt;AF&lt;/math&gt;, so &lt;math&gt;XE&lt;/math&gt;=&lt;math&gt;EF&lt;/math&gt;=1/3&lt;math&gt;AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to 1/3 of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is 360, so the area of &lt;math&gt;BEF&lt;/math&gt;=1/3*1/4*360=&lt;math&gt;\boxed{(B) 30}&lt;/math&gt;~heeeeeeeheeeee<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of 3. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of 3 also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}(30)&lt;/math&gt;<br /> -Brudder<br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{B}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{The area of triangle ABE}{The area of triangle ACE}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{(B) 30}&lt;/math&gt;. ~~SmileKat32<br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Since &lt;math&gt;AD = 2CD&lt;/math&gt;, triangle &lt;math&gt;CDB&lt;/math&gt; has four times the area of triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we get &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the congruent. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and therefore four times the area of triangle &lt;math&gt;BEF&lt;/math&gt;, giving &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> ==Solution 5 (Area Ratios)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> As before we figure out the areas labeled in the diagram. Then we note that &lt;cmath&gt;\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.&lt;/cmath&gt;Solving gives &lt;math&gt;x = \boxed{\textbf{(B) }30}&lt;/math&gt;. <br /> (Credit to scrabbler94 for the idea)<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=111884 2019 AMC 8 Problems/Problem 24 2019-11-21T04:57:35Z <p>Frestho: /* Solution 4 (Similar Triangles) */</p> <hr /> <div><br /> <br /> <br /> ==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; s that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and left &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Draw &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{AF}&lt;/math&gt; such that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE&lt;/math&gt;=&lt;math&gt;ED&lt;/math&gt;. &lt;math&gt;FC&lt;/math&gt;=3&lt;math&gt;XD&lt;/math&gt; so &lt;math&gt;BC&lt;/math&gt;=4&lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;AF&lt;/math&gt;=3&lt;math&gt;EF&lt;/math&gt;( &lt;math&gt;XE&lt;/math&gt;=&lt;math&gt;EF&lt;/math&gt; and &lt;math&gt;AX&lt;/math&gt;=1/3&lt;math&gt;AF&lt;/math&gt;, so &lt;math&gt;XE&lt;/math&gt;=&lt;math&gt;EF&lt;/math&gt;=1/3&lt;math&gt;AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to 1/3 of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is 360, so the area of &lt;math&gt;BEF&lt;/math&gt;=1/3*1/4*360=&lt;math&gt;\boxed{(B) 30}&lt;/math&gt;~heeeeeeeheeeee<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of 3. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of 3 also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}(30)&lt;/math&gt;<br /> -Brudder<br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{B}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{The area of triangle ABE}{The area of triangle ACE}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{(B) 30}&lt;/math&gt;. ~~SmileKat32<br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Since &lt;math&gt;AD = 2CD&lt;/math&gt;, triangle &lt;math&gt;CDB&lt;/math&gt; has four times the area of triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we get &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the congruent. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and therefore four times the area of triangle &lt;math&gt;BEF&lt;/math&gt;, giving &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> ==Solution 5 (Area Ratios)==<br /> [asy]<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;&lt;math&gt;A&lt;/math&gt;&quot;, A, N);<br /> label(&quot;&lt;math&gt;B&lt;/math&gt;&quot;, B, WSW);<br /> label(&quot;&lt;math&gt;C&lt;/math&gt;&quot;, C, ESE);<br /> label(&quot;&lt;math&gt;D&lt;/math&gt;&quot;, D, dir(0)*1.5);<br /> label(&quot;&lt;math&gt;E&lt;/math&gt;&quot;, E, SSE);<br /> label(&quot;&lt;math&gt;F&lt;/math&gt;&quot;, F, S);<br /> label(&quot;&lt;math&gt;60&lt;/math&gt;&quot;, (A+E+D)/3);<br /> label(&quot;&lt;math&gt;60&lt;/math&gt;&quot;, (A+E+B)/3);<br /> label(&quot;&lt;math&gt;120&lt;/math&gt;&quot;, (D+E+C)/3);<br /> label(&quot;&lt;math&gt;x&lt;/math&gt;&quot;, (B+E+F)/3);<br /> label(&quot;&lt;math&gt;120-x&lt;/math&gt;&quot;, (F+E+C)/3);<br /> [/asy]<br /> As before we figure out the areas labeled in the diagram. Then we note that &lt;cmath&gt;\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.&lt;/cmath&gt;Solving gives &lt;math&gt;x = \boxed{\textbf{(B) }30}&lt;/math&gt;. (Credit to scrabbler94 for the idea)<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=111883 2019 AMC 8 Problems/Problem 24 2019-11-21T04:43:22Z <p>Frestho: /* Solution 1 */</p> <hr /> <div><br /> <br /> <br /> ==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; s that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and left &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Draw &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{AF}&lt;/math&gt; such that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE&lt;/math&gt;=&lt;math&gt;ED&lt;/math&gt;. &lt;math&gt;FC&lt;/math&gt;=3&lt;math&gt;XD&lt;/math&gt; so &lt;math&gt;BC&lt;/math&gt;=4&lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;AF&lt;/math&gt;=3&lt;math&gt;EF&lt;/math&gt;( &lt;math&gt;XE&lt;/math&gt;=&lt;math&gt;EF&lt;/math&gt; and &lt;math&gt;AX&lt;/math&gt;=1/3&lt;math&gt;AF&lt;/math&gt;, so &lt;math&gt;XE&lt;/math&gt;=&lt;math&gt;EF&lt;/math&gt;=1/3&lt;math&gt;AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to 1/3 of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is 360, so the area of &lt;math&gt;BEF&lt;/math&gt;=1/3*1/4*360=&lt;math&gt;\boxed{(B) 30}&lt;/math&gt;~heeeeeeeheeeee<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of 3. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of 3 also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}(30)&lt;/math&gt;<br /> -Brudder<br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{B}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{The area of triangle ABE}{The area of triangle ACE}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{(B) 30}&lt;/math&gt;. ~~SmileKat32<br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Since &lt;math&gt;AD = 2CD&lt;/math&gt;, triangle &lt;math&gt;CDB&lt;/math&gt; has four times the area of triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we get &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the congruent. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and therefore four times the area of triangle &lt;math&gt;BEF&lt;/math&gt;, giving &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=111882 2019 AMC 8 Problems/Problem 24 2019-11-21T04:42:02Z <p>Frestho: /* Solution 4 (Similar Triangles) */</p> <hr /> <div><br /> <br /> <br /> ==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; s that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and left &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Draw &lt;math&gt;X&lt;/math&gt; so that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE&lt;/math&gt;=&lt;math&gt;ED&lt;/math&gt;. &lt;math&gt;FC&lt;/math&gt;=3&lt;math&gt;XD&lt;/math&gt; so &lt;math&gt;BC&lt;/math&gt;=4&lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;AF&lt;/math&gt;=3&lt;math&gt;EF&lt;/math&gt;( &lt;math&gt;XE&lt;/math&gt;=&lt;math&gt;EF&lt;/math&gt; and &lt;math&gt;AX&lt;/math&gt;=1/3&lt;math&gt;AF&lt;/math&gt;, so &lt;math&gt;XE&lt;/math&gt;=&lt;math&gt;EF&lt;/math&gt;=1/3&lt;math&gt;AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to 1/3 of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is 360, so the area of &lt;math&gt;BEF&lt;/math&gt;=1/3*1/4*360=&lt;math&gt;\boxed{(B) 30}&lt;/math&gt;~heeeeeeeheeeee<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of 3. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of 3 also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}(30)&lt;/math&gt;<br /> -Brudder<br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{B}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{The area of triangle ABE}{The area of triangle ACE}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{(B) 30}&lt;/math&gt;. ~~SmileKat32<br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Since &lt;math&gt;AD = 2CD&lt;/math&gt;, triangle &lt;math&gt;CDB&lt;/math&gt; has four times the area of triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we get &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the congruent. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and therefore four times the area of triangle &lt;math&gt;BEF&lt;/math&gt;, giving &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=111881 2019 AMC 8 Problems/Problem 24 2019-11-21T04:37:19Z <p>Frestho: /* Solution 4 (Similar Triangles) */</p> <hr /> <div><br /> <br /> <br /> ==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; s that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and left &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Draw &lt;math&gt;X&lt;/math&gt; so that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE&lt;/math&gt;=&lt;math&gt;ED&lt;/math&gt;. &lt;math&gt;FC&lt;/math&gt;=3&lt;math&gt;XD&lt;/math&gt; so &lt;math&gt;BC&lt;/math&gt;=4&lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;AF&lt;/math&gt;=3&lt;math&gt;EF&lt;/math&gt;( &lt;math&gt;XE&lt;/math&gt;=&lt;math&gt;EF&lt;/math&gt; and &lt;math&gt;AX&lt;/math&gt;=1/3&lt;math&gt;AF&lt;/math&gt;, so &lt;math&gt;XE&lt;/math&gt;=&lt;math&gt;EF&lt;/math&gt;=1/3&lt;math&gt;AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to 1/3 of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is 360, so the area of &lt;math&gt;BEF&lt;/math&gt;=1/3*1/4*360=&lt;math&gt;\boxed{(B) 30}&lt;/math&gt;~heeeeeeeheeeee<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of 3. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of 3 also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}(30)&lt;/math&gt;<br /> -Brudder<br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{B}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{The area of triangle ABE}{The area of triangle ACE}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{(B) 30}&lt;/math&gt;. ~~SmileKat32<br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Note that &lt;math&gt;AD = 2CD&lt;/math&gt;, so triangle &lt;math&gt;CDB&lt;/math&gt; has an area four times as large as triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we have &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the same. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and four times the area of triangle &lt;math&gt;BEF&lt;/math&gt; so &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=111880 2019 AMC 8 Problems/Problem 24 2019-11-21T04:34:57Z <p>Frestho: /* Solution 4 (Similar Triangles) */</p> <hr /> <div><br /> <br /> <br /> ==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; s that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and left &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Draw &lt;math&gt;X&lt;/math&gt; so that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE&lt;/math&gt;=&lt;math&gt;ED&lt;/math&gt;. &lt;math&gt;FC&lt;/math&gt;=3&lt;math&gt;XD&lt;/math&gt; so &lt;math&gt;BC&lt;/math&gt;=4&lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;AF&lt;/math&gt;=3&lt;math&gt;EF&lt;/math&gt;( &lt;math&gt;XE&lt;/math&gt;=&lt;math&gt;EF&lt;/math&gt; and &lt;math&gt;AX&lt;/math&gt;=1/3&lt;math&gt;AF&lt;/math&gt;, so &lt;math&gt;XE&lt;/math&gt;=&lt;math&gt;EF&lt;/math&gt;=1/3&lt;math&gt;AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to 1/3 of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is 360, so the area of &lt;math&gt;BEF&lt;/math&gt;=1/3*1/4*360=&lt;math&gt;\boxed{(B) 30}&lt;/math&gt;~heeeeeeeheeeee<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of 3. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of 3 also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}(30)&lt;/math&gt;<br /> -Brudder<br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{B}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{The area of triangle ABE}{The area of triangle ACE}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{(B) 30}&lt;/math&gt;. ~~SmileKat32<br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Note that triangle &lt;math&gt;CDB&lt;/math&gt; has sides twice as long as triangle &lt;math&gt;ADG&lt;/math&gt;, so triangle &lt;math&gt;CDB&lt;/math&gt; has an area four times as large. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we have &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have that &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas,so their bases must be the same. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and four times the area of triangle &lt;math&gt;BEF&lt;/math&gt; so &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=111879 2019 AMC 8 Problems/Problem 24 2019-11-21T04:34:01Z <p>Frestho: /* Solution 3 */</p> <hr /> <div><br /> <br /> <br /> ==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; s that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and left &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Draw &lt;math&gt;X&lt;/math&gt; so that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE&lt;/math&gt;=&lt;math&gt;ED&lt;/math&gt;. &lt;math&gt;FC&lt;/math&gt;=3&lt;math&gt;XD&lt;/math&gt; so &lt;math&gt;BC&lt;/math&gt;=4&lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;AF&lt;/math&gt;=3&lt;math&gt;EF&lt;/math&gt;( &lt;math&gt;XE&lt;/math&gt;=&lt;math&gt;EF&lt;/math&gt; and &lt;math&gt;AX&lt;/math&gt;=1/3&lt;math&gt;AF&lt;/math&gt;, so &lt;math&gt;XE&lt;/math&gt;=&lt;math&gt;EF&lt;/math&gt;=1/3&lt;math&gt;AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to 1/3 of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is 360, so the area of &lt;math&gt;BEF&lt;/math&gt;=1/3*1/4*360=&lt;math&gt;\boxed{(B) 30}&lt;/math&gt;~heeeeeeeheeeee<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of 3. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of 3 also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}(30)&lt;/math&gt;<br /> -Brudder<br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{B}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{The area of triangle ABE}{The area of triangle ACE}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{(B) 30}&lt;/math&gt;. ~~SmileKat32<br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> [asy]<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;&lt;math&gt;A&lt;/math&gt;&quot;, A, N);<br /> label(&quot;&lt;math&gt;B&lt;/math&gt;&quot;, B, WSW);<br /> label(&quot;&lt;math&gt;C&lt;/math&gt;&quot;, C, ESE);<br /> label(&quot;&lt;math&gt;D&lt;/math&gt;&quot;, D, dir(0)*1.5);<br /> label(&quot;&lt;math&gt;E&lt;/math&gt;&quot;, E, SE);<br /> label(&quot;&lt;math&gt;F&lt;/math&gt;&quot;, F, S);<br /> label(&quot;&lt;math&gt;G&lt;/math&gt;&quot;, G, ENE);<br /> [/asy]<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Note that triangle &lt;math&gt;CDB&lt;/math&gt; has sides twice as long as triangle &lt;math&gt;ADG&lt;/math&gt;, so triangle &lt;math&gt;CDB&lt;/math&gt; has an area four times as large. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we have &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have that &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas,so their bases must be the same. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and four times the area of triangle &lt;math&gt;BEF&lt;/math&gt; so &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> [asy]<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;&lt;math&gt;A&lt;/math&gt;&quot;, A, N);<br /> label(&quot;&lt;math&gt;B&lt;/math&gt;&quot;, B, WSW);<br /> label(&quot;&lt;math&gt;C&lt;/math&gt;&quot;, C, ESE);<br /> label(&quot;&lt;math&gt;D&lt;/math&gt;&quot;, D, dir(0)*1.5);<br /> label(&quot;&lt;math&gt;E&lt;/math&gt;&quot;, E, SE);<br /> label(&quot;&lt;math&gt;F&lt;/math&gt;&quot;, F, S);<br /> label(&quot;&lt;math&gt;G&lt;/math&gt;&quot;, G, ENE);<br /> label(&quot;&lt;math&gt;60&lt;/math&gt;&quot;, (A+E+D)/3);<br /> label(&quot;&lt;math&gt;60&lt;/math&gt;&quot;, (A+E+B)/3);<br /> label(&quot;&lt;math&gt;60&lt;/math&gt;&quot;, (A+G+D)/3);<br /> label(&quot;&lt;math&gt;30&lt;/math&gt;&quot;, (B+E+F)/3);<br /> [/asy]<br /> (Credit to MP8148 for the idea)<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_14&diff=111365 2006 AMC 12A Problems/Problem 14 2019-11-13T02:07:43Z <p>Frestho: /* Solution 3 */</p> <hr /> <div>{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #14]] and [[2006 AMC 10A Problems/Problem 22|2006 AMC 10A #22]]}}<br /> <br /> == Problem ==<br /> <br /> Two farmers agree that pigs are worth &lt;math&gt;300&lt;/math&gt; dollars and that goats are worth &lt;math&gt;210&lt;/math&gt; dollars. When one farmer owes the other money, he pays the debt in pigs or goats, with &quot;change&quot; received in the form of goats or pigs as necessary. (For example, a &lt;math&gt;390&lt;/math&gt; dollar debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 10\qquad \mathrm{(C) \ } 30\qquad \mathrm{(D) \ } 90\qquad \mathrm{(E) \ } 210&lt;/math&gt;<br /> <br /> == Solutions ==<br /> <br /> ===Solution 1===<br /> <br /> The problem can be restated as an equation of the form &lt;math&gt;300p + 210g = x&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; is the number of pigs, &lt;math&gt;g&lt;/math&gt; is the number of goats, and &lt;math&gt;x&lt;/math&gt; is the positive debt. The problem asks us to find the lowest ''x'' possible. ''&lt;math&gt;p&lt;/math&gt;'' and ''&lt;math&gt;g&lt;/math&gt;'' must be [[integer]]s, which makes the equation a [[Diophantine equation]]. Bezout’s Identity tells us that the smallest &lt;math&gt;c&lt;/math&gt; for the Diophantine equation &lt;math&gt;am + bn = c&lt;/math&gt; to have solutions is when &lt;math&gt;c&lt;/math&gt; is the [[greatest common divisor]] of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. Therefore, the answer is &lt;math&gt;gcd(300,210)&lt;/math&gt;, which is &lt;math&gt;30&lt;/math&gt;, &lt;math&gt;\mathrm{(C) \ }&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> Alternatively, note that &lt;math&gt;300p + 210g = 30(10p + 7g)&lt;/math&gt; is divisible by 30 no matter what &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;g&lt;/math&gt; are, so our answer must be divisible by 30. In addition, three goats minus two pigs gives us &lt;math&gt;630 - 600 = 30&lt;/math&gt; exactly. Since our theoretical best can be achieved, it must really be the best, and the answer is &lt;math&gt;\mathrm{(C) \ }&lt;/math&gt;.<br /> debt that can be resolved.<br /> <br /> ===Solution 3===<br /> <br /> Let us simplify this problem. Dividing by &lt;math&gt;30&lt;/math&gt;, we get a pig to be: &lt;math&gt;\frac{300}{30} = 10&lt;/math&gt;, and a goat to be &lt;math&gt;\frac{210}{30}= 7&lt;/math&gt;. <br /> It becomes evident that if you exchange &lt;math&gt;5&lt;/math&gt; pigs for &lt;math&gt;7&lt;/math&gt; goats, we get the smallest positive difference - &lt;math&gt;5\cdot 10 - 7\cdot 7 = 50-49 = 1&lt;/math&gt;, since we can't made a non-integer with a linear combination of integers. <br /> Since we originally divided by &lt;math&gt;30&lt;/math&gt;, we need to multiply again, thus getting the answer: &lt;math&gt;1\cdot 30 = \mathrm{(C) 30}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2006|ab=A|num-b=13|num-a=15}}<br /> {{AMC10 box|year=2006|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_17&diff=110915 2007 AMC 12B Problems/Problem 17 2019-11-05T16:42:48Z <p>Frestho: /* Cheap Solution that most people probably used */</p> <hr /> <div>==Problem 17==<br /> If &lt;math&gt;a&lt;/math&gt; is a nonzero integer and &lt;math&gt;b&lt;/math&gt; is a positive number such that &lt;math&gt;ab^2=\log_{10}b&lt;/math&gt;, what is the median of the set &lt;math&gt;\{0,1,a,b,1/b\}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ a \qquad \mathrm{(D)}\ b \qquad \mathrm{(E)}\ \frac{1}{b}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Note that if &lt;math&gt;a&lt;/math&gt; is positive, then, the equation will have no solutions for &lt;math&gt;b&lt;/math&gt;. This becomes more obvious by noting that at &lt;math&gt;b=1&lt;/math&gt;, &lt;math&gt;ab^2 &gt; \log_{10} b&lt;/math&gt;. The LHS quadratic function will increase faster than the RHS logarithmic function, so they will never intersect.<br /> <br /> This puts &lt;math&gt;a&lt;/math&gt; as the smallest in the set since it must be negative.<br /> <br /> Checking the new equation: &lt;math&gt;-b^2 = \log_{10}b&lt;/math&gt;<br /> <br /> Near &lt;math&gt;b=0&lt;/math&gt;, &lt;math&gt;-b^2 &gt; \log_{10} b&lt;/math&gt; but at &lt;math&gt;b=1&lt;/math&gt;, &lt;math&gt;-b^2 &lt; \log_{10} b&lt;/math&gt;<br /> <br /> This implies that the solution occurs somewhere in between: &lt;math&gt;0 &lt; b &lt; 1&lt;/math&gt;<br /> <br /> This also implies that &lt;math&gt;\frac{1}{b} &gt; 1&lt;/math&gt;<br /> <br /> This makes our set (ordered)<br /> &lt;math&gt;\{a,0,b,1,1/b\}&lt;/math&gt;<br /> <br /> The median is &lt;math&gt;b \Rightarrow \mathrm {(D)}&lt;/math&gt;<br /> ==Cheap Solution that most people probably used==<br /> Led &lt;math&gt;b=0.1&lt;/math&gt;. Then &lt;math&gt;a\cdot0.01 = -1,&lt;/math&gt; giving &lt;math&gt;a=-100&lt;/math&gt;. Then the ordered set is &lt;math&gt;\{-100, 0, 0.1, 1, 10\}&lt;/math&gt; and the median is &lt;math&gt;0.1=b,&lt;/math&gt; so the answer is &lt;math&gt;\mathrm {(D)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2007|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_17&diff=110914 2007 AMC 12B Problems/Problem 17 2019-11-05T16:42:18Z <p>Frestho: /* Cheap Solution that most people probably used */</p> <hr /> <div>==Problem 17==<br /> If &lt;math&gt;a&lt;/math&gt; is a nonzero integer and &lt;math&gt;b&lt;/math&gt; is a positive number such that &lt;math&gt;ab^2=\log_{10}b&lt;/math&gt;, what is the median of the set &lt;math&gt;\{0,1,a,b,1/b\}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ a \qquad \mathrm{(D)}\ b \qquad \mathrm{(E)}\ \frac{1}{b}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Note that if &lt;math&gt;a&lt;/math&gt; is positive, then, the equation will have no solutions for &lt;math&gt;b&lt;/math&gt;. This becomes more obvious by noting that at &lt;math&gt;b=1&lt;/math&gt;, &lt;math&gt;ab^2 &gt; \log_{10} b&lt;/math&gt;. The LHS quadratic function will increase faster than the RHS logarithmic function, so they will never intersect.<br /> <br /> This puts &lt;math&gt;a&lt;/math&gt; as the smallest in the set since it must be negative.<br /> <br /> Checking the new equation: &lt;math&gt;-b^2 = \log_{10}b&lt;/math&gt;<br /> <br /> Near &lt;math&gt;b=0&lt;/math&gt;, &lt;math&gt;-b^2 &gt; \log_{10} b&lt;/math&gt; but at &lt;math&gt;b=1&lt;/math&gt;, &lt;math&gt;-b^2 &lt; \log_{10} b&lt;/math&gt;<br /> <br /> This implies that the solution occurs somewhere in between: &lt;math&gt;0 &lt; b &lt; 1&lt;/math&gt;<br /> <br /> This also implies that &lt;math&gt;\frac{1}{b} &gt; 1&lt;/math&gt;<br /> <br /> This makes our set (ordered)<br /> &lt;math&gt;\{a,0,b,1,1/b\}&lt;/math&gt;<br /> <br /> The median is &lt;math&gt;b \Rightarrow \mathrm {(D)}&lt;/math&gt;<br /> ==Cheap Solution that most people probably used==<br /> Led &lt;math&gt;b=0.1&lt;/math&gt;. Then &lt;math&gt;a*0.01 = -1,&lt;/math&gt; giving &lt;math&gt;a=-100&lt;/math&gt;. Then the ordered set is &lt;math&gt;\{-100, 0, 0.1, 1, 10\}&lt;/math&gt; and the median is &lt;math&gt;0.1=b,&lt;/math&gt; so the answer is &lt;math&gt;\mathrm {(D)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2007|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_17&diff=110913 2007 AMC 12B Problems/Problem 17 2019-11-05T16:40:37Z <p>Frestho: /* Solution */</p> <hr /> <div>==Problem 17==<br /> If &lt;math&gt;a&lt;/math&gt; is a nonzero integer and &lt;math&gt;b&lt;/math&gt; is a positive number such that &lt;math&gt;ab^2=\log_{10}b&lt;/math&gt;, what is the median of the set &lt;math&gt;\{0,1,a,b,1/b\}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ a \qquad \mathrm{(D)}\ b \qquad \mathrm{(E)}\ \frac{1}{b}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Note that if &lt;math&gt;a&lt;/math&gt; is positive, then, the equation will have no solutions for &lt;math&gt;b&lt;/math&gt;. This becomes more obvious by noting that at &lt;math&gt;b=1&lt;/math&gt;, &lt;math&gt;ab^2 &gt; \log_{10} b&lt;/math&gt;. The LHS quadratic function will increase faster than the RHS logarithmic function, so they will never intersect.<br /> <br /> This puts &lt;math&gt;a&lt;/math&gt; as the smallest in the set since it must be negative.<br /> <br /> Checking the new equation: &lt;math&gt;-b^2 = \log_{10}b&lt;/math&gt;<br /> <br /> Near &lt;math&gt;b=0&lt;/math&gt;, &lt;math&gt;-b^2 &gt; \log_{10} b&lt;/math&gt; but at &lt;math&gt;b=1&lt;/math&gt;, &lt;math&gt;-b^2 &lt; \log_{10} b&lt;/math&gt;<br /> <br /> This implies that the solution occurs somewhere in between: &lt;math&gt;0 &lt; b &lt; 1&lt;/math&gt;<br /> <br /> This also implies that &lt;math&gt;\frac{1}{b} &gt; 1&lt;/math&gt;<br /> <br /> This makes our set (ordered)<br /> &lt;math&gt;\{a,0,b,1,1/b\}&lt;/math&gt;<br /> <br /> The median is &lt;math&gt;b \Rightarrow \mathrm {(D)}&lt;/math&gt;<br /> ==Cheap Solution that most people probably used==<br /> Led &lt;math&gt;b=0.1&lt;/math&gt;. Then &lt;math&gt;a/100 = -1,&lt;/math&gt; giving &lt;math&gt;a=-100&lt;/math&gt;. Then the ordered set is &lt;math&gt;\{-100, 0, 0.1, 1, 10\}&lt;/math&gt; and the median is &lt;math&gt;0.1=b,&lt;/math&gt; so the answer is &lt;math&gt;\mathrm {(D)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2007|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_3&diff=110153 2019 AIME I Problems/Problem 3 2019-10-06T20:21:05Z <p>Frestho: /* Solution 3 */</p> <hr /> <div><br /> ==Problem 3==<br /> In &lt;math&gt;\triangle PQR&lt;/math&gt;, &lt;math&gt;PR=15&lt;/math&gt;, &lt;math&gt;QR=20&lt;/math&gt;, and &lt;math&gt;PQ=25&lt;/math&gt;. Points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; lie on &lt;math&gt;\overline{PQ}&lt;/math&gt;, points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; lie on &lt;math&gt;\overline{QR}&lt;/math&gt;, and points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; lie on &lt;math&gt;\overline{PR}&lt;/math&gt;, with &lt;math&gt;PA=QB=QC=RD=RE=PF=5&lt;/math&gt;. Find the area of hexagon &lt;math&gt;ABCDEF&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> We know the area of the hexagon &lt;math&gt;ABCDEF&lt;/math&gt; to be &lt;math&gt;\triangle PQR- \triangle PAF- \triangle BCQ- \triangle RED &lt;/math&gt;. Since &lt;math&gt;PR^2+RQ^2=PQ^2&lt;/math&gt;, we know that &lt;math&gt;\triangle PRQ&lt;/math&gt; is a right triangle. Thus the area of &lt;math&gt;\triangle PQR&lt;/math&gt; is &lt;math&gt;150&lt;/math&gt;. Another way to compute the area is &lt;cmath&gt;\frac12 \cdot PQ\cdot RQ \sin \angle PQR = \frac12 \cdot 500 \cdot \sin \angle PQR=150 \implies \sin \angle PQR = \frac35.&lt;/cmath&gt; Then the area of &lt;math&gt;\triangle BCQ = \frac12 \cdot BQ \cdot CQ \cdot \sin \angle PQR= \frac{25}{2}\cdot \frac{3}{5}=\frac{15}{2}&lt;/math&gt;. Preceding in a similar fashion for &lt;math&gt;\triangle PAF&lt;/math&gt;, the area of &lt;math&gt;\triangle PAF&lt;/math&gt; is &lt;math&gt;10&lt;/math&gt;. Since &lt;math&gt;\angle ERD = 90^{\circ}&lt;/math&gt;, the area of &lt;math&gt;\triangle RED=\frac{25}{2}&lt;/math&gt;. Thus our desired answer is &lt;math&gt;150-\frac{15}{2}-10-\frac{25}{2}=\boxed{120}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;R&lt;/math&gt; be the origin. Noticing that the triangle is a 3-4-5 right triangle, we can see that &lt;math&gt;A=(4,12), B=(16,3), C=(15,0), D=(5,0), E=(0,5)&lt;/math&gt;, and &lt;math&gt;F=(0,10)&lt;/math&gt;. Using the shoelace theorem, the area is &lt;math&gt;\boxed{120}&lt;/math&gt;.<br /> Shoelace theorem:Suppose the polygon &lt;math&gt;P&lt;/math&gt; has vertices &lt;math&gt;(a_1, b_1)&lt;/math&gt;, &lt;math&gt;(a_2, b_2)&lt;/math&gt;, ... , &lt;math&gt;(a_n, b_n)&lt;/math&gt;, listed in clockwise order. Then the area of &lt;math&gt;P&lt;/math&gt; is<br /> <br /> &lt;cmath&gt;\dfrac{1}{2} |(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1)|&lt;/cmath&gt;<br /> You can also go counterclockwise order, as long as you find the absolute value of the answer.<br /> <br /> .<br /> <br /> ==Solution 3 (Easiest, uses only basic geometry too)==<br /> Note that &lt;math&gt;\triangle{PQR}&lt;/math&gt; has area &lt;math&gt;150&lt;/math&gt; and is a 3-4-5 right triangle. Then, by similar triangles, the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;QC&lt;/math&gt; has length 3 and the altitude from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;FP&lt;/math&gt; has length 4, so &lt;math&gt;[QBC]+[DRE]+[AFP]=\frac{15}{2}+\frac{25}{2}+\frac{20}{2}=30&lt;/math&gt;, meaning that &lt;math&gt;[ABCDEF]=\boxed{120}&lt;/math&gt;. <br /> -Stormersyle<br /> <br /> ==Solution 4==<br /> Knowing that &lt;math&gt;\triangle{PQR}&lt;/math&gt; has area 150 and is a 3-4-5 triangle, we can find the area of the smaller triangles &lt;math&gt;\triangle{DRE}&lt;/math&gt;, &lt;math&gt;\triangle{APF}&lt;/math&gt;, and &lt;math&gt;\triangle{CQB}&lt;/math&gt; and subtract them from &lt;math&gt;\triangle{PQR}&lt;/math&gt; to obtain our answer. First off, we know &lt;math&gt;\triangle{DRE}&lt;/math&gt; has area &lt;math&gt;12.5&lt;/math&gt; since it is a right triangle. To the find the areas of &lt;math&gt;\triangle{APF}&lt;/math&gt; and &lt;math&gt;\triangle{CQB}&lt;/math&gt; , we can use Law of Cosines (&lt;math&gt;c^2 = a^2 + b^2 - 2ab\cos C&lt;/math&gt;) to find the lengths of &lt;math&gt;AF&lt;/math&gt; and &lt;math&gt;CB&lt;/math&gt;, respectively. Computing gives &lt;math&gt;AF = \sqrt{20}&lt;/math&gt; and &lt;math&gt;CB = \sqrt{10}&lt;/math&gt;. Now, using Heron's Formula, we find &lt;math&gt;\triangle{APF} = 10&lt;/math&gt; and &lt;math&gt;\triangle{CQB} = 7.5&lt;/math&gt;. Adding these and subtracting from &lt;math&gt;\triangle{PQR}&lt;/math&gt;, we get &lt;math&gt;150 - (10 + 7.5 + 12.5) = \boxed{120}&lt;/math&gt; -Starsher<br /> <br /> ==Video Solution==<br /> <br /> https://www.youtube.com/watch?v=4jOfXNiQ6WM<br /> <br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2011_IMO_Problems/Problem_2&diff=108508 2011 IMO Problems/Problem 2 2019-08-09T00:43:03Z <p>Frestho: /* Weblinks */</p> <hr /> <div>Let &lt;math&gt;\mathcal{S}&lt;/math&gt; be a finite set of at least two points in the plane. Assume that no three points of &lt;math&gt;\mathcal S&lt;/math&gt; are collinear. A ''windmill'' is a process that starts with a line &lt;math&gt;\ell&lt;/math&gt; going through a single point &lt;math&gt;P \in \mathcal S&lt;/math&gt;. The line rotates clockwise about the ''pivot'' &lt;math&gt;P&lt;/math&gt; until the first time that the line meets some other point belonging to &lt;math&gt;\mathcal S&lt;/math&gt;. This point, &lt;math&gt;Q&lt;/math&gt;, takes over as the new pivot, and the line now rotates clockwise about &lt;math&gt;Q&lt;/math&gt;, until it next meets a point of &lt;math&gt;\mathcal S&lt;/math&gt;. This process continues indefinitely. <br /> Show that we can choose a point &lt;math&gt;P&lt;/math&gt; in &lt;math&gt;\mathcal S&lt;/math&gt; and a line &lt;math&gt;\ell&lt;/math&gt; going through &lt;math&gt;P&lt;/math&gt; such that the resulting windmill uses each point of &lt;math&gt;\mathcal S&lt;/math&gt; as a pivot infinitely many times.<br /> <br /> ==Solution==<br /> Choose a coordinate system so that all points in &lt;math&gt;\mathcal S&lt;/math&gt; have distinct x-coordinates. Number the points &lt;math&gt;P_i=(x_i,y_i)&lt;/math&gt; of &lt;math&gt;\mathcal S&lt;/math&gt; by increasing x-coordinates: &lt;math&gt;x_1&lt;x_2&lt;\ldots&lt;x_N&lt;/math&gt;.<br /> <br /> In order to divide the set &lt;math&gt;\mathcal S&lt;/math&gt; into two halves, define &lt;math&gt;n&lt;/math&gt; so that &lt;math&gt;N=2n+1+d&lt;/math&gt; where &lt;math&gt;d=0&lt;/math&gt; for an odd number of points and &lt;math&gt;d=1&lt;/math&gt; for an even number of points.<br /> <br /> Start the &quot;windmill&quot; process with the line &lt;math&gt;\ell&lt;/math&gt; going vertically through the point &lt;math&gt;P_{n+1}&lt;/math&gt;. Attach a down-up direction to this line so that we can color all points as follows: Points to the left of &lt;math&gt;\ell&lt;/math&gt; (with lower x-coordinates) are blue, the pivot point on &lt;math&gt;\ell&lt;/math&gt; is white and point to the right of &lt;math&gt;\ell&lt;/math&gt; (with higher x-coordinates) are red. We have now &lt;math&gt;n&lt;/math&gt; blue points, one white point and &lt;math&gt;n+d&lt;/math&gt; red points.<br /> <br /> After processing the &quot;windmill&quot; by 180 degrees, the line &lt;math&gt;\ell&lt;/math&gt; goes vertically up-down. Now, points with lower x-coordinates are to the right of &lt;math&gt;\ell&lt;/math&gt; and colored red; points with higher x-coordinates are to the left of &lt;math&gt;\ell&lt;/math&gt; and colored blue.<br /> <br /> Note that at each pivot exchange, the old pivot point enters the same side of &lt;math&gt;\ell&lt;/math&gt; where the new pivot point came from. This means that throughout the &quot;windmill&quot; process, the number of blue points and the number of red points stay constant, respectively: We still have &lt;math&gt;n&lt;/math&gt; blue points, one white point and &lt;math&gt;n+d&lt;/math&gt; red points. This means that the current pivot point is &lt;math&gt;P_{n+d}&lt;/math&gt;. <br /> <br /> Note that all blue and all red points changed their color from the start of the &quot;windmill&quot; process. This implies that every point was a pivot at some stage of the rotation.<br /> <br /> For every 180 degrees of &quot;windmill&quot; rotation, the same argument applies: all colored points must change their color and hence be a pivot at some stage. Infinitely many rotations imply infinitely many color changes. This completes the proof.<br /> <br /> ==See Also==<br /> *[[IMO Problems and Solutions]]<br /> <br /> ==Weblinks==<br /> * [http://bernikr.github.io/windmill/ A interactive visualization of the Problem]<br /> <br /> * [https://www.youtube.com/watch?v=M64HUIJFTZM/ A video solution]</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_7&diff=104491 2019 AIME I Problems/Problem 7 2019-03-15T15:35:29Z <p>Frestho: /* Solution 2 (Crappier Solution) */</p> <hr /> <div>==Problem 7==<br /> There are positive integers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; that satisfy the system of equations &lt;cmath&gt;\log_{10} x + 2 \log_{10} (\gcd(x,y)) = 60&lt;/cmath&gt;&lt;cmath&gt;\log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) = 570.&lt;/cmath&gt; Let &lt;math&gt;m&lt;/math&gt; be the number of (not necessarily distinct) prime factors in the prime factorization of &lt;math&gt;x&lt;/math&gt;, and let &lt;math&gt;n&lt;/math&gt; be the number of (not necessarily distinct) prime factors in the prime factorization of &lt;math&gt;y&lt;/math&gt;. Find &lt;math&gt;3m+2n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Add the two equations to get that &lt;math&gt;\log x+\log y+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630&lt;/math&gt;.<br /> Then, we use the theorem &lt;math&gt;\log a+\log b=\log ab&lt;/math&gt; to get the equation, &lt;math&gt;\log (xy)+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630&lt;/math&gt;.<br /> Using the theorem that &lt;math&gt;\gcd(x,y) \cdot \text{lcm}(x,y)=x\cdot y&lt;/math&gt;, along with the previously mentioned theorem, we can get the equation &lt;math&gt;3\log(xy)=630&lt;/math&gt;.<br /> This can easily be simplified to &lt;math&gt;\log(xy)=210&lt;/math&gt;, or &lt;math&gt;xy = 10^{210}&lt;/math&gt;.<br /> <br /> &lt;math&gt;10^{210}&lt;/math&gt; can be factored into &lt;math&gt;2^{210} \cdot 5^{210}&lt;/math&gt;, and &lt;math&gt;m+n&lt;/math&gt; equals to the sum of the exponents of 2 and 5, which is &lt;math&gt;210+210 = 420&lt;/math&gt;.<br /> Multiply by two to get &lt;math&gt;2m +2n&lt;/math&gt;, which is &lt;math&gt;840&lt;/math&gt;.<br /> Then, use the first equation (&lt;math&gt;\log x + 2\log(\gcd(x,y)) = 60&lt;/math&gt;) to show that x has to have lower degrees of 2 and 5 than y. Therefore, making the &lt;math&gt;gcd x&lt;/math&gt;. Then, turn the equation into &lt;math&gt;3\log x = 60&lt;/math&gt;, which yields &lt;math&gt;\log x = 20&lt;/math&gt;, or &lt;math&gt;x = 10^20&lt;/math&gt;.<br /> Factor this into &lt;math&gt;2^{20} \cdot 5^{20}&lt;/math&gt;, and add the two 20's, resulting in m, which is 40.<br /> Add &lt;math&gt;m&lt;/math&gt; to &lt;math&gt;2m + 2n&lt;/math&gt; (which is 840) to get &lt;math&gt;40+840 = \boxed{880}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Crappier Solution)==<br /> <br /> First simplifying the first and second equations, we get that <br /> <br /> &lt;cmath&gt;\log_{10}(x\cdot\text{gcd}(x,y)^2)=60&lt;/cmath&gt;<br /> &lt;cmath&gt;\log_{10}(y\cdot\text{lcm}(x,y)^2)=570&lt;/cmath&gt;<br /> <br /> <br /> Thus, when the two equations are added, we have that<br /> &lt;cmath&gt;\log_{10}(x\cdot y\cdot\text{gcd}\cdot\text{lcm}^2)=630&lt;/cmath&gt;<br /> When simplified, this equals <br /> &lt;cmath&gt;\log_{10}(x^3y^3)=630&lt;/cmath&gt;<br /> so this means that<br /> &lt;cmath&gt;x^3y^3=10^{630}&lt;/cmath&gt;<br /> so<br /> &lt;cmath&gt;xy=10^{210}.&lt;/cmath&gt;<br /> <br /> Now, the following cannot be done on a proof contest but let's (intuitively) assume that &lt;math&gt;x&lt;y&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both powers of &lt;math&gt;10&lt;/math&gt;. This means the first equation would simplify to &lt;cmath&gt;x^3=10^{60}&lt;/cmath&gt; and &lt;cmath&gt;y^3=10^{570}.&lt;/cmath&gt; Therefore, &lt;math&gt;x=10^{20}&lt;/math&gt; and &lt;math&gt;y=10^{190}&lt;/math&gt; and if we plug these values back, it works! &lt;math&gt;10^{20}&lt;/math&gt; has &lt;math&gt;20\cdot2=40&lt;/math&gt; total factors and &lt;math&gt;10^{190}&lt;/math&gt; has &lt;math&gt;190\cdot2=380&lt;/math&gt; so &lt;cmath&gt;3\cdot 40 + 2\cdot 380 = \boxed{880}.&lt;/cmath&gt;<br /> <br /> Please remember that you should only assume on these math contests because they are timed; this would technically not be a valid solution.<br /> <br /> ==Solution 3 (Easy Solution)==<br /> Let &lt;math&gt;x=10^a&lt;/math&gt; and &lt;math&gt;y=10^b&lt;/math&gt; and &lt;math&gt;a&lt;b&lt;/math&gt;. Then the given equations become &lt;math&gt;3a=60&lt;/math&gt; and &lt;math&gt;3b=570&lt;/math&gt;. Therefore, &lt;math&gt;x=10^{20}=2^{20}\cdot5^{20}&lt;/math&gt; and &lt;math&gt;y=10^{190}=2^{190}\cdot5^{190}&lt;/math&gt;. Our answer is &lt;math&gt;3(20+20)+2(190+190)=\boxed{880}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_23&diff=101971 2011 AMC 10A Problems/Problem 23 2019-02-13T02:24:02Z <p>Frestho: /* Solution */</p> <hr /> <div>== Problem ==<br /> Seven students count from 1 to 1000 as follows:<br /> <br /> •Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br /> <br /> •Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br /> <br /> •Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br /> <br /> •Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br /> <br /> •Finally, George says the only number that no one else says.<br /> <br /> What number does George say?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 &lt;/math&gt;<br /> <br /> == Solution ==<br /> First look at the numbers Alice says. &lt;math&gt;1, 3, 4, 6, 7, 9 \cdots&lt;/math&gt; skipping every number that is congruent to &lt;math&gt;2 \pmod 3&lt;/math&gt;. Thus, Barbara says those numbers EXCEPT every second - being &lt;math&gt;2 + 3^1 \equiv 5 \pmod{3^2=9}&lt;/math&gt;. So Barbara skips every number congruent to &lt;math&gt;5 \pmod 9&lt;/math&gt;. We continue and see: <br /> <br /> Alice skips &lt;math&gt;2 \pmod 3&lt;/math&gt;, Barbara skips &lt;math&gt;5 \pmod 9&lt;/math&gt;, Candice skips &lt;math&gt;14 \pmod {27}&lt;/math&gt;, Debbie skips &lt;math&gt;41 \pmod {81}&lt;/math&gt;, Eliza skips &lt;math&gt;122 \pmod {243}&lt;/math&gt;, and Fatima skips &lt;math&gt;365 \pmod {729}&lt;/math&gt;.<br /> <br /> Since the only number congruent to &lt;math&gt;365 \pmod {729}&lt;/math&gt; and less than &lt;math&gt;1,000&lt;/math&gt; is &lt;math&gt;365&lt;/math&gt;, the correct answer is &lt;math&gt; \boxed{365\ \mathbf{(C)}} &lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> After Alice says all her numbers, the numbers not mentioned yet are &lt;cmath&gt;\text{Alice: } 2,5,8,11,14,17,\cdots,998.&lt;/cmath&gt;After Barbara says all her numbers, the numbers that haven't been said yet are &lt;cmath&gt;\text{Barbara: } 5,14,23,32,41,50,\cdots,995.&lt;/cmath&gt;After Candice, the list is &lt;cmath&gt;\text{Candice: } 14,41,68,\cdots,986.&lt;/cmath&gt;Notice how each list is an arithmetic sequence where the common ratio is thrice the common ratio of the previous list and the first term is the second term of the previous list. Now that the pattern is clear, we construct the rest of the lists: &lt;cmath&gt;\text{Debbie: } 41,122,203,\cdots,959&lt;/cmath&gt;&lt;cmath&gt;\text{Eliza: } 122,365,608,878&lt;/cmath&gt;&lt;cmath&gt;\text{Fatima: } \boxed{\textbf{(C) } 365}&lt;/cmath&gt;<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_19&diff=101968 2011 AMC 10A Problems/Problem 19 2019-02-13T02:12:16Z <p>Frestho: /* Solution 3 */</p> <hr /> <div>== Problem 19 ==<br /> <br /> In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let the population of the town in &lt;math&gt;1991&lt;/math&gt; be &lt;math&gt;p^2&lt;/math&gt;. Let the population in &lt;math&gt;2001&lt;/math&gt; be &lt;math&gt;q^2+9&lt;/math&gt;. It follows that &lt;math&gt;p^2+150=q^2+9&lt;/math&gt;. Rearrange this equation to get &lt;math&gt;141=q^2-p^2=(q-p)(q+p)&lt;/math&gt;. Since &lt;math&gt;q&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are both positive integers with &lt;math&gt;q&gt;p&lt;/math&gt;, &lt;math&gt;(q-p)&lt;/math&gt; and &lt;math&gt;(q+p)&lt;/math&gt; also must be, and thus, they are both factors of &lt;math&gt;141&lt;/math&gt;. We have two choices for pairs of factors of &lt;math&gt;141&lt;/math&gt;: &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;141&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;47&lt;/math&gt;. Assuming the former pair, since &lt;math&gt;(q-p)&lt;/math&gt; must be less than &lt;math&gt;(q+p)&lt;/math&gt;, &lt;math&gt;q-p=1&lt;/math&gt; and &lt;math&gt;q+p=141&lt;/math&gt;. Solve to get &lt;math&gt;p=70, q=71&lt;/math&gt;. Since &lt;math&gt;p^2+300&lt;/math&gt; is not a perfect square, this is not the correct pair. Solve for the other pair to get &lt;math&gt;p=22, q=25&lt;/math&gt;. This time, &lt;math&gt;p^2+300=22^2+300=784=28^2&lt;/math&gt;. This is the correct pair. Now, we find the percent increase from &lt;math&gt;22^2=484&lt;/math&gt; to &lt;math&gt;28^2=784&lt;/math&gt;. Since the increase is &lt;math&gt;300&lt;/math&gt;, the percent increase is &lt;math&gt;\frac{300}{484}\times100\%\approx\boxed{\textbf{(E)}\ 62\%}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Proceed through the difference of squares for &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;:<br /> &lt;math&gt;141=q^2-p^2=(q-p)(q+p)&lt;/math&gt;<br /> <br /> However, instead of testing both pairs of factors we take a more certain approach. Here &lt;math&gt;r^2&lt;/math&gt; is the population of the town in 2011.<br /> &lt;math&gt;r^2-p^2=300&lt;/math&gt;<br /> &lt;math&gt;(r-p)(r+p)=300&lt;/math&gt;<br /> Test through pairs of &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; that makes sure &lt;math&gt;p=22&lt;/math&gt; or &lt;math&gt;p=70&lt;/math&gt;. <br /> Then go through the same routine as demonstrated above to finish this problem. <br /> <br /> Note that this approach might take more testing if one is not familiar with finding factors.<br /> <br /> == Solution 3 ==<br /> Since all the answer choices are around &lt;math&gt;50\%&lt;/math&gt;, we know the town's starting population must be around &lt;math&gt;600&lt;/math&gt;. We list perfect squares from &lt;math&gt;400&lt;/math&gt; to &lt;math&gt;1000&lt;/math&gt;.<br /> &lt;cmath&gt;441, 484, 529,576,625,676,729,784,841,900,961&lt;/cmath&gt;We see that &lt;math&gt;484&lt;/math&gt; and &lt;math&gt;784&lt;/math&gt; differ by &lt;math&gt;300&lt;/math&gt;, and can we can confirm that &lt;math&gt;484&lt;/math&gt; is the correct starting number by noting that &lt;math&gt;484+150=634=25^2+9&lt;/math&gt;. Thus, the answer is &lt;math&gt;784/484-1\approx \boxed{\textbf{(E) } 62\%}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_19&diff=101967 2011 AMC 10A Problems/Problem 19 2019-02-13T02:11:55Z <p>Frestho: /* Solution 3 */</p> <hr /> <div>== Problem 19 ==<br /> <br /> In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let the population of the town in &lt;math&gt;1991&lt;/math&gt; be &lt;math&gt;p^2&lt;/math&gt;. Let the population in &lt;math&gt;2001&lt;/math&gt; be &lt;math&gt;q^2+9&lt;/math&gt;. It follows that &lt;math&gt;p^2+150=q^2+9&lt;/math&gt;. Rearrange this equation to get &lt;math&gt;141=q^2-p^2=(q-p)(q+p)&lt;/math&gt;. Since &lt;math&gt;q&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are both positive integers with &lt;math&gt;q&gt;p&lt;/math&gt;, &lt;math&gt;(q-p)&lt;/math&gt; and &lt;math&gt;(q+p)&lt;/math&gt; also must be, and thus, they are both factors of &lt;math&gt;141&lt;/math&gt;. We have two choices for pairs of factors of &lt;math&gt;141&lt;/math&gt;: &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;141&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;47&lt;/math&gt;. Assuming the former pair, since &lt;math&gt;(q-p)&lt;/math&gt; must be less than &lt;math&gt;(q+p)&lt;/math&gt;, &lt;math&gt;q-p=1&lt;/math&gt; and &lt;math&gt;q+p=141&lt;/math&gt;. Solve to get &lt;math&gt;p=70, q=71&lt;/math&gt;. Since &lt;math&gt;p^2+300&lt;/math&gt; is not a perfect square, this is not the correct pair. Solve for the other pair to get &lt;math&gt;p=22, q=25&lt;/math&gt;. This time, &lt;math&gt;p^2+300=22^2+300=784=28^2&lt;/math&gt;. This is the correct pair. Now, we find the percent increase from &lt;math&gt;22^2=484&lt;/math&gt; to &lt;math&gt;28^2=784&lt;/math&gt;. Since the increase is &lt;math&gt;300&lt;/math&gt;, the percent increase is &lt;math&gt;\frac{300}{484}\times100\%\approx\boxed{\textbf{(E)}\ 62\%}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Proceed through the difference of squares for &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;:<br /> &lt;math&gt;141=q^2-p^2=(q-p)(q+p)&lt;/math&gt;<br /> <br /> However, instead of testing both pairs of factors we take a more certain approach. Here &lt;math&gt;r^2&lt;/math&gt; is the population of the town in 2011.<br /> &lt;math&gt;r^2-p^2=300&lt;/math&gt;<br /> &lt;math&gt;(r-p)(r+p)=300&lt;/math&gt;<br /> Test through pairs of &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; that makes sure &lt;math&gt;p=22&lt;/math&gt; or &lt;math&gt;p=70&lt;/math&gt;. <br /> Then go through the same routine as demonstrated above to finish this problem. <br /> <br /> Note that this approach might take more testing if one is not familiar with finding factors.<br /> <br /> == Solution 3 ==<br /> Since all the answer choices are around &lt;math&gt;50\%&lt;/math&gt;, we know the town's starting population must be around &lt;math&gt;600&lt;/math&gt;. We list perfect squares from &lt;math&gt;400-1000&lt;/math&gt;.<br /> &lt;cmath&gt;441, 484, 529,576,625,676,729,784,841,900,961&lt;/cmath&gt;We see that &lt;math&gt;484&lt;/math&gt; and &lt;math&gt;784&lt;/math&gt; differ by &lt;math&gt;300&lt;/math&gt;, and can we can confirm that &lt;math&gt;484&lt;/math&gt; is the correct starting number by noting that &lt;math&gt;484+150=634=25^2+9&lt;/math&gt;. Thus, the answer is &lt;math&gt;784/484-1\approx \boxed{\textbf{(E) } 62\%}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Frestho https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_19&diff=101966 2011 AMC 10A Problems/Problem 19 2019-02-13T02:11:37Z <p>Frestho: /* Solution 3 */</p> <hr /> <div>== Problem 19 ==<br /> <br /> In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let the population of the town in &lt;math&gt;1991&lt;/math&gt; be &lt;math&gt;p^2&lt;/math&gt;. Let the population in &lt;math&gt;2001&lt;/math&gt; be &lt;math&gt;q^2+9&lt;/math&gt;. It follows that &lt;math&gt;p^2+150=q^2+9&lt;/math&gt;. Rearrange this equation to get &lt;math&gt;141=q^2-p^2=(q-p)(q+p)&lt;/math&gt;. Since &lt;math&gt;q&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are both positive integers with &lt;math&gt;q&gt;p&lt;/math&gt;, &lt;math&gt;(q-p)&lt;/math&gt; and &lt;math&gt;(q+p)&lt;/math&gt; also must be, and thus, they are both factors of &lt;math&gt;141&lt;/math&gt;. We have two choices for pairs of factors of &lt;math&gt;141&lt;/math&gt;: &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;141&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;47&lt;/math&gt;. Assuming the former pair, since &lt;math&gt;(q-p)&lt;/math&gt; must be less than &lt;math&gt;(q+p)&lt;/math&gt;, &lt;math&gt;q-p=1&lt;/math&gt; and &lt;math&gt;q+p=141&lt;/math&gt;. Solve to get &lt;math&gt;p=70, q=71&lt;/math&gt;. Since &lt;math&gt;p^2+300&lt;/math&gt; is not a perfect square, this is not the correct pair. Solve for the other pair to get &lt;math&gt;p=22, q=25&lt;/math&gt;. This time, &lt;math&gt;p^2+300=22^2+300=784=28^2&lt;/math&gt;. This is the correct pair. Now, we find the percent increase from &lt;math&gt;22^2=484&lt;/math&gt; to &lt;math&gt;28^2=784&lt;/math&gt;. Since the increase is &lt;math&gt;300&lt;/math&gt;, the percent increase is &lt;math&gt;\frac{300}{484}\times100\%\approx\boxed{\textbf{(E)}\ 62\%}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Proceed through the difference of squares for &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;:<br /> &lt;math&gt;141=q^2-p^2=(q-p)(q+p)&lt;/math&gt;<br /> <br /> However, instead of testing both pairs of factors we take a more certain approach. Here &lt;math&gt;r^2&lt;/math&gt; is the population of the town in 2011.<br /> &lt;math&gt;r^2-p^2=300&lt;/math&gt;<br /> &lt;math&gt;(r-p)(r+p)=300&lt;/math&gt;<br /> Test through pairs of &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; that makes sure &lt;math&gt;p=22&lt;/math&gt; or &lt;math&gt;p=70&lt;/math&gt;. <br /> Then go through the same routine as demonstrated above to finish this problem. <br /> <br /> Note that this approach might take more testing if one is not familiar with finding factors.<br /> <br /> == Solution 3 ==<br /> Since all the answer choices are around &lt;math&gt;50\%&lt;/math&gt;, we know the town's starting population must be around &lt;math&gt;600&lt;/math&gt;. We list perfect squares from &lt;math&gt;400-1000&lt;/math&gt;.<br /> &lt;cmath&gt;441, 484, 529,576,625,676,729,784,841,900,961&lt;/cmath&gt;We see that &lt;math&gt;484&lt;/math&gt; and &lt;math&gt;784&lt;/math&gt; differ by &lt;math&gt;300&lt;/math&gt;, and can we can confirm that &lt;math&gt;484&lt;/math&gt; is the correct starting number by noting that &lt;math&gt;484+150=634=25^2+9&lt;/math&gt;. Thus, the answer is &lt;math&gt;784/484-1\approx \boxed{\text{ (E)} 62\%}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Frestho