https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathlover66&feedformat=atom AoPS Wiki - User contributions [en] 2020-10-22T21:01:58Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_21&diff=133895 2015 AMC 10B Problems/Problem 21 2020-09-20T21:15:59Z <p>Mathlover66: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than &lt;math&gt;5&lt;/math&gt; steps left). Suppose the Dash takes &lt;math&gt;19&lt;/math&gt; fewer jumps than Cozy to reach the top of the staircase. Let &lt;math&gt;s&lt;/math&gt; denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of &lt;math&gt;s&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }15&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We can translate this wordy problem into this simple equation:<br /> <br /> &lt;cmath&gt;\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil&lt;/cmath&gt;<br /> <br /> We will proceed to solve this equation via casework.<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: &lt;math&gt;\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}&lt;/math&gt;<br /> <br /> Our equation becomes &lt;math&gt;\frac{s}{2} - 19 = \frac{s}{5} + \frac{j}{5}&lt;/math&gt;, where &lt;math&gt;j \in \{0,1,2,3,4\}&lt;/math&gt; Using the fact that &lt;math&gt;s&lt;/math&gt; is an integer, we quickly find that &lt;math&gt;j=1&lt;/math&gt; and &lt;math&gt;j=4&lt;/math&gt; yield &lt;math&gt;s=64&lt;/math&gt; and &lt;math&gt;s=66&lt;/math&gt;, respectively.<br /> <br /> &lt;br&gt;<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: &lt;math&gt;\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}+\frac{1}{2}&lt;/math&gt;<br /> <br /> Our equation becomes &lt;math&gt;\frac{s}{2} +\frac{1}{2} - 19 = \frac{s}{5} + \frac{j}{5}&lt;/math&gt;, where &lt;math&gt;j \in \{0,1,2,3,4\}&lt;/math&gt; Using the fact that &lt;math&gt;s&lt;/math&gt; is an integer, we quickly find that &lt;math&gt;j=2&lt;/math&gt; yields &lt;math&gt;s=63&lt;/math&gt;. Summing up we get &lt;math&gt;63+64+66=193&lt;/math&gt;. The sum of the digits is &lt;math&gt;\boxed{\textbf{(D)}\; 13}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> We know from the problem that Dash goes &lt;math&gt;3&lt;/math&gt; steps further than Cozy per jump (assuming they aren't within &lt;math&gt;4&lt;/math&gt; steps from the top). That means that if Dash takes &lt;math&gt;19&lt;/math&gt; fewer jumps than Cozy to get to the top of the staircase, the staircase must be at least &lt;math&gt;57&lt;/math&gt; steps high (3*19=57). We then start using guess-and-check:<br /> <br /> &lt;math&gt;57&lt;/math&gt; steps: &lt;math&gt;\left \lceil {57/2} \right \rceil = 29&lt;/math&gt; jumps for Cozy, and &lt;math&gt;\left \lceil {57/5} \right \rceil = 12&lt;/math&gt; jumps for Dash, giving a difference of &lt;math&gt;17&lt;/math&gt; jumps.<br /> <br /> &lt;math&gt;58&lt;/math&gt; steps: &lt;math&gt;\left \lceil {58/2} \right \rceil = 29&lt;/math&gt; jumps for Cozy, and &lt;math&gt;\left \lceil {57/5} \right \rceil = 12&lt;/math&gt; jumps for Dash, giving a difference of &lt;math&gt;17&lt;/math&gt; jumps.<br /> <br /> &lt;math&gt;59&lt;/math&gt; steps: &lt;math&gt;\left \lceil {59/2} \right \rceil = 30&lt;/math&gt; jumps for Cozy, and &lt;math&gt;\left \lceil {59/5} \right \rceil = 12&lt;/math&gt; jumps for Dash, giving a difference of &lt;math&gt;18&lt;/math&gt; jumps.<br /> <br /> &lt;math&gt;\vdots&lt;/math&gt;<br /> <br /> By the time we test &lt;math&gt;61&lt;/math&gt; steps, we notice that when the number of steps exceeds a multiple of &lt;math&gt;2&lt;/math&gt;, the difference in jumps increases. So, we have to find the next number that will increase the difference. &lt;math&gt;62&lt;/math&gt; doesn't because both both Cozy's and Dash's number of jumps increases, but &lt;math&gt;63&lt;/math&gt; does, and &lt;math&gt;64&lt;/math&gt;. &lt;math&gt;65&lt;/math&gt; actually gives a difference of &lt;math&gt;20&lt;/math&gt; jumps, but &lt;math&gt;66&lt;/math&gt; goes back down to &lt;math&gt;19&lt;/math&gt; (because Dash had to take another jump when Cozy didn't). We don't need to go any further because the difference will stay above &lt;math&gt;19&lt;/math&gt; onward.<br /> <br /> Therefore, the possible numbers of steps in the staircase are &lt;math&gt;63&lt;/math&gt;, &lt;math&gt;64&lt;/math&gt;, and &lt;math&gt;66&lt;/math&gt;, giving a sum of &lt;math&gt;193&lt;/math&gt;. The sum of those digits is &lt;math&gt;13&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{D}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> We're looking for natural numbers &lt;math&gt;x&lt;/math&gt; such that &lt;math&gt;\left \lceil{\frac{x}{5}}\right \rceil + 19 = \left \lceil{\frac{x}{2}}\right \rceil&lt;/math&gt;.<br /> <br /> Let's call &lt;math&gt;x = 10a + b&lt;/math&gt;. We now have &lt;math&gt;2a + \left \lceil{\frac{b}{5}}\right \rceil + 19 = 5a + \left \lceil{\frac{b}{2}}\right \rceil&lt;/math&gt;, or<br /> <br /> &lt;math&gt;19 - 3a = \left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil&lt;/math&gt;.<br /> <br /> Obviously, since &lt;math&gt;b \le 10&lt;/math&gt;, this will not work for any value under &lt;math&gt;6&lt;/math&gt;. In addition, since obviously &lt;math&gt;\frac{b}{2} \ge \frac{b}{5}&lt;/math&gt;, this will not work for any value over six, so we have &lt;math&gt;a = 6&lt;/math&gt; and &lt;math&gt;\left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil = 1.&lt;/math&gt;<br /> <br /> This can be achieved when &lt;math&gt;\left \lceil{\frac{b}{5}}\right \rceil = 1&lt;/math&gt; and &lt;math&gt;\left \lceil{\frac{b}{2}}\right \rceil = 2&lt;/math&gt;, or when &lt;math&gt;\left \lceil{\frac{b}{5}}\right \rceil = 2&lt;/math&gt; and &lt;math&gt;\left \lceil{\frac{b}{2}}\right \rceil = 3&lt;/math&gt;.<br /> <br /> Case One:<br /> <br /> We have &lt;math&gt;b \le 5&lt;/math&gt; and &lt;math&gt;3 \le b \le 4&lt;/math&gt;, so &lt;math&gt;b = 3, 4&lt;/math&gt;.<br /> <br /> Case Two:<br /> <br /> We have &lt;math&gt;6 \le b \le 9&lt;/math&gt; and &lt;math&gt;5 \le b \le 6&lt;/math&gt;, so &lt;math&gt;b = 6&lt;/math&gt;.<br /> <br /> We then have &lt;math&gt;63 + 64 + 66 = 193&lt;/math&gt;, which has a digit sum of &lt;math&gt;\boxed{13}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Translate the problem into following equation:<br /> <br /> &lt;math&gt;n = 5D - \{0 \sim 4\} = 2C - \{0 \sim 1\}&lt;/math&gt;<br /> <br /> Since &lt;math&gt;C = D + 19&lt;/math&gt;, we have<br /> <br /> &lt;math&gt;5D - \{0 \sim 4\} = 2D + 38 - \{0 \sim 1\}&lt;/math&gt;<br /> <br /> i.e.,<br /> <br /> &lt;math&gt;3D = 38 + \{0 \sim 4\} - \{0 \sim 1\}&lt;/math&gt;<br /> <br /> We then have &lt;math&gt;D = 13&lt;/math&gt; when &lt;math&gt;\{1\} - \{0\}&lt;/math&gt; or &lt;math&gt;\{2\} - \{1\}&lt;/math&gt; (the dog's last jump has &lt;math&gt;2&lt;/math&gt; steps and the cat's last jump has &lt;math&gt;1&lt;/math&gt; step), which yields &lt;math&gt;n = 64&lt;/math&gt; and &lt;math&gt;n = 63&lt;/math&gt; respectively.<br /> <br /> Another solution is &lt;math&gt;D = 14&lt;/math&gt; when &lt;math&gt;\{4\} - \{0\}&lt;/math&gt;, which yields &lt;math&gt;n = 66&lt;/math&gt;.<br /> <br /> Therefore, with &lt;math&gt;63 + 64 + 66 = 193&lt;/math&gt;, the digit sum is &lt;math&gt;\boxed{13}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/TpHZVbBGmVQ<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_21&diff=133894 2015 AMC 10B Problems/Problem 21 2020-09-20T20:59:04Z <p>Mathlover66: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than &lt;math&gt;5&lt;/math&gt; steps left). Suppose the Dash takes &lt;math&gt;19&lt;/math&gt; fewer jumps than Cozy to reach the top of the staircase. Let &lt;math&gt;s&lt;/math&gt; denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of &lt;math&gt;s&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }15&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We can translate this wordy problem into this simple equation:<br /> <br /> &lt;cmath&gt;\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil&lt;/cmath&gt;<br /> <br /> We will proceed to solve this equation via casework.<br /> <br /> Case &lt;math&gt;1&lt;/math&gt; (&lt;math&gt;s&lt;/math&gt; is an even number): &lt;math&gt;\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}&lt;/math&gt;<br /> <br /> Our equation becomes &lt;math&gt;\frac{s}{2} - 19 = \frac{s}{5} + \frac{j}{5}&lt;/math&gt;, where &lt;math&gt;j \in \{0,1,2,3,4\}&lt;/math&gt; Using the fact that &lt;math&gt;s&lt;/math&gt; is an integer, we quickly find that &lt;math&gt;j=1&lt;/math&gt; and &lt;math&gt;j=4&lt;/math&gt; yield &lt;math&gt;s=64&lt;/math&gt; and &lt;math&gt;s=66&lt;/math&gt;, respectively.<br /> <br /> &lt;br&gt;<br /> <br /> Case &lt;math&gt;2&lt;/math&gt; (&lt;math&gt;s&lt;/math&gt; is an odd number): &lt;math&gt;\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}+\frac{1}{2}&lt;/math&gt;<br /> <br /> Our equation becomes &lt;math&gt;\frac{s}{2} +\frac{1}{2} - 19 = \frac{s}{5} + \frac{j}{5}&lt;/math&gt;, where &lt;math&gt;j \in \{0,1,2,3,4\}&lt;/math&gt; Using the fact that &lt;math&gt;s&lt;/math&gt; is an integer, we quickly find that &lt;math&gt;j=2&lt;/math&gt; yields &lt;math&gt;s=63&lt;/math&gt;. Summing up we get &lt;math&gt;63+64+66=193&lt;/math&gt;. The sum of the digits is &lt;math&gt;\boxed{\textbf{(D)}\; 13}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> We know from the problem that Dash goes &lt;math&gt;3&lt;/math&gt; steps further than Cozy per jump (assuming they aren't within &lt;math&gt;4&lt;/math&gt; steps from the top). That means that if Dash takes &lt;math&gt;19&lt;/math&gt; fewer jumps than Cozy to get to the top of the staircase, the staircase must be at least &lt;math&gt;57&lt;/math&gt; steps high (3*19=57). We then start using guess-and-check:<br /> <br /> &lt;math&gt;57&lt;/math&gt; steps: &lt;math&gt;\left \lceil {57/2} \right \rceil = 29&lt;/math&gt; jumps for Cozy, and &lt;math&gt;\left \lceil {57/5} \right \rceil = 12&lt;/math&gt; jumps for Dash, giving a difference of &lt;math&gt;17&lt;/math&gt; jumps.<br /> <br /> &lt;math&gt;58&lt;/math&gt; steps: &lt;math&gt;\left \lceil {58/2} \right \rceil = 29&lt;/math&gt; jumps for Cozy, and &lt;math&gt;\left \lceil {57/5} \right \rceil = 12&lt;/math&gt; jumps for Dash, giving a difference of &lt;math&gt;17&lt;/math&gt; jumps.<br /> <br /> &lt;math&gt;59&lt;/math&gt; steps: &lt;math&gt;\left \lceil {59/2} \right \rceil = 30&lt;/math&gt; jumps for Cozy, and &lt;math&gt;\left \lceil {59/5} \right \rceil = 12&lt;/math&gt; jumps for Dash, giving a difference of &lt;math&gt;18&lt;/math&gt; jumps.<br /> <br /> &lt;math&gt;\vdots&lt;/math&gt;<br /> <br /> By the time we test &lt;math&gt;61&lt;/math&gt; steps, we notice that when the number of steps exceeds a multiple of &lt;math&gt;2&lt;/math&gt;, the difference in jumps increases. So, we have to find the next number that will increase the difference. &lt;math&gt;62&lt;/math&gt; doesn't because both both Cozy's and Dash's number of jumps increases, but &lt;math&gt;63&lt;/math&gt; does, and &lt;math&gt;64&lt;/math&gt;. &lt;math&gt;65&lt;/math&gt; actually gives a difference of &lt;math&gt;20&lt;/math&gt; jumps, but &lt;math&gt;66&lt;/math&gt; goes back down to &lt;math&gt;19&lt;/math&gt; (because Dash had to take another jump when Cozy didn't). We don't need to go any further because the difference will stay above &lt;math&gt;19&lt;/math&gt; onward.<br /> <br /> Therefore, the possible numbers of steps in the staircase are &lt;math&gt;63&lt;/math&gt;, &lt;math&gt;64&lt;/math&gt;, and &lt;math&gt;66&lt;/math&gt;, giving a sum of &lt;math&gt;193&lt;/math&gt;. The sum of those digits is &lt;math&gt;13&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{D}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> We're looking for natural numbers &lt;math&gt;x&lt;/math&gt; such that &lt;math&gt;\left \lceil{\frac{x}{5}}\right \rceil + 19 = \left \lceil{\frac{x}{2}}\right \rceil&lt;/math&gt;.<br /> <br /> Let's call &lt;math&gt;x = 10a + b&lt;/math&gt;. We now have &lt;math&gt;2a + \left \lceil{\frac{b}{5}}\right \rceil + 19 = 5a + \left \lceil{\frac{b}{2}}\right \rceil&lt;/math&gt;, or<br /> <br /> &lt;math&gt;19 - 3a = \left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil&lt;/math&gt;.<br /> <br /> Obviously, since &lt;math&gt;b \le 10&lt;/math&gt;, this will not work for any value under &lt;math&gt;6&lt;/math&gt;. In addition, since obviously &lt;math&gt;\frac{b}{2} \ge \frac{b}{5}&lt;/math&gt;, this will not work for any value over six, so we have &lt;math&gt;a = 6&lt;/math&gt; and &lt;math&gt;\left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil = 1.&lt;/math&gt;<br /> <br /> This can be achieved when &lt;math&gt;\left \lceil{\frac{b}{5}}\right \rceil = 1&lt;/math&gt; and &lt;math&gt;\left \lceil{\frac{b}{2}}\right \rceil = 2&lt;/math&gt;, or when &lt;math&gt;\left \lceil{\frac{b}{5}}\right \rceil = 2&lt;/math&gt; and &lt;math&gt;\left \lceil{\frac{b}{2}}\right \rceil = 3&lt;/math&gt;.<br /> <br /> Case One:<br /> <br /> We have &lt;math&gt;b \le 5&lt;/math&gt; and &lt;math&gt;3 \le b \le 4&lt;/math&gt;, so &lt;math&gt;b = 3, 4&lt;/math&gt;.<br /> <br /> Case Two:<br /> <br /> We have &lt;math&gt;6 \le b \le 9&lt;/math&gt; and &lt;math&gt;5 \le b \le 6&lt;/math&gt;, so &lt;math&gt;b = 6&lt;/math&gt;.<br /> <br /> We then have &lt;math&gt;63 + 64 + 66 = 193&lt;/math&gt;, which has a digit sum of &lt;math&gt;\boxed{13}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Translate the problem into following equation:<br /> <br /> &lt;math&gt;n = 5D - \{0 \sim 4\} = 2C - \{0 \sim 1\}&lt;/math&gt;<br /> <br /> Since &lt;math&gt;C = D + 19&lt;/math&gt;, we have<br /> <br /> &lt;math&gt;5D - \{0 \sim 4\} = 2D + 38 - \{0 \sim 1\}&lt;/math&gt;<br /> <br /> i.e.,<br /> <br /> &lt;math&gt;3D = 38 + \{0 \sim 4\} - \{0 \sim 1\}&lt;/math&gt;<br /> <br /> We then have &lt;math&gt;D = 13&lt;/math&gt; when &lt;math&gt;\{1\} - \{0\}&lt;/math&gt; or &lt;math&gt;\{2\} - \{1\}&lt;/math&gt; (the dog's last jump has &lt;math&gt;2&lt;/math&gt; steps and the cat's last jump has &lt;math&gt;1&lt;/math&gt; step), which yields &lt;math&gt;n = 64&lt;/math&gt; and &lt;math&gt;n = 63&lt;/math&gt; respectively.<br /> <br /> Another solution is &lt;math&gt;D = 14&lt;/math&gt; when &lt;math&gt;\{4\} - \{0\}&lt;/math&gt;, which yields &lt;math&gt;n = 66&lt;/math&gt;.<br /> <br /> Therefore, with &lt;math&gt;63 + 64 + 66 = 193&lt;/math&gt;, the digit sum is &lt;math&gt;\boxed{13}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/TpHZVbBGmVQ<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_18&diff=133297 2014 AMC 10B Problems/Problem 18 2020-09-07T03:47:10Z <p>Mathlover66: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> A list of &lt;math&gt;11&lt;/math&gt; positive integers has a mean of &lt;math&gt;10&lt;/math&gt;, a median of &lt;math&gt;9&lt;/math&gt;, and a unique mode of &lt;math&gt;8&lt;/math&gt;. What is the largest possible value of an integer in the list?<br /> <br /> &lt;math&gt; \textbf {(A) } 24 \qquad \textbf {(B) } 30 \qquad \textbf {(C) } 31\qquad \textbf {(D) } 33 \qquad \textbf {(E) } 35&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We start off with the fact that the median is &lt;math&gt;9&lt;/math&gt;, so we must have &lt;math&gt;a, b, c, d, e, 9, f, g, h, i, j&lt;/math&gt;, listed in ascending order. Note that the integers do not have to be distinct. <br /> <br /> Since the mode is &lt;math&gt;8&lt;/math&gt;, we have to have at least &lt;math&gt;2&lt;/math&gt; occurrences of &lt;math&gt;8&lt;/math&gt; in the list. If there are &lt;math&gt;2&lt;/math&gt; occurrences of &lt;math&gt;8&lt;/math&gt; in the list, we will have &lt;math&gt;a, b, c, 8, 8, 9, f, g, h, i, j&lt;/math&gt;. In this case, since &lt;math&gt;8&lt;/math&gt; is the unique mode, the rest of the integers have to be distinct. So we minimize &lt;math&gt;a,b,c,f,g,h,i&lt;/math&gt; in order to maximize &lt;math&gt;j&lt;/math&gt;. If we let the list be &lt;math&gt;1,2,3,8,8,9,10,11,12,13,j&lt;/math&gt;, then &lt;math&gt;j = 11 \times 10 - (1+2+3+8+8+9+10+11+12+13) = 33&lt;/math&gt;. <br /> <br /> Next, consider the case where there are &lt;math&gt;3&lt;/math&gt; occurrences of &lt;math&gt;8&lt;/math&gt; in the list. Now, we can have two occurrences of another integer in the list. We try &lt;math&gt;1,1,8,8,8,9,9,10,10,11,j&lt;/math&gt;. Following the same process as above, we get &lt;math&gt;j = 11 \times 10 - (1+1+8+8+8+9+9+10+10+11) = 35&lt;/math&gt;. As this is the highest choice in the list, we know this is our answer. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(E) }35}&lt;/math&gt;<br /> <br /> <br /> ==Solution 2==<br /> Note that x1 + ... + x11 = 110 let x6 = 9 so x1 + ... + x5 + x7+ ... + x11 = 101 now to maximise the value of xi where i ranges from 1 to 11, we let any 7 elements be 1,2,...,7 so x1 + x2 + x3 = 57 now we have to let one of above 3 values = 8 hence x1 + x2 = 49 now let x1 = 35 , x2 =14 hence 35 is the answer.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2014|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_23&diff=133235 2012 AMC 10B Problems/Problem 23 2020-09-06T01:39:57Z <p>Mathlover66: /* Solution 4 */</p> <hr /> <div>[[Category: Introductory Geometry Problems]]<br /> == Problem ==<br /> A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{2 \sqrt{2}}{3}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \frac{2 \sqrt{3}}{3}\qquad\textbf{(E)}\ \sqrt{2} &lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> This tetrahedron has the 4 vertices in these positions: on a corner (lets call this &lt;math&gt;A&lt;/math&gt;) of the cube, and the other three corners (&lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;) adjacent to this corner. We can find the height of the remaining portion of the cube by finding the height of the tetrahedron. We can find the height of this tetrahedron in perspective to the equilateral triangle base (we call this height &lt;math&gt;x&lt;/math&gt;) by finding the volume of the tetrahedron in two ways. &lt;math&gt;\frac{1 \times 1}{2}&lt;/math&gt; is the area of the isosceles base of the tetrahedron. Multiply by the height, &lt;math&gt;1&lt;/math&gt;, and divide by &lt;math&gt;3&lt;/math&gt;, we have the volume of the tetrahedron as &lt;math&gt;\frac{1}{6}&lt;/math&gt;. We set this area equal to one-third the product of our desired height and the area of the equilateral triangle base. First, find the area of the equilateral triangle: &lt;math&gt;[BCD]=\frac{\sqrt{2}^2 \times \sqrt{3}}{4}=\frac{\sqrt{3}}{2}&lt;/math&gt;. So we have: &lt;math&gt;\frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot x=\frac{1}{6}&lt;/math&gt;, and so &lt;math&gt;x=\frac{\sqrt{3}}{3}&lt;/math&gt;. <br /> <br /> Since we know what the height is, we can find the height of the remaining structure. The height of the structure if the tetrahedron was still on would simply be the space diagonal of the cube, &lt;math&gt;\sqrt{3}&lt;/math&gt;, so we just subtract &lt;math&gt;\frac{\sqrt{3}}{3}&lt;/math&gt; from &lt;math&gt;\sqrt{3}&lt;/math&gt; to get &lt;math&gt;\frac{2\sqrt{3}}{3}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can write the volume of a tetrahedron as &lt;math&gt;\frac{Bh}{3}&lt;/math&gt; and revise the formula for the volume of a cube to be &lt;math&gt;Bh&lt;/math&gt;. Also note that the height of the tetrahedron goes through the space diagonal of the cube. So, the remaining part of the cube's space diagonal should be &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the original space diagonal. Hence, the length is &lt;math&gt;\boxed{\textbf{(D)} \: \frac{2\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Plot the cube &lt;math&gt;ABCDEFGH&lt;/math&gt;(let's call it this)on a &lt;math&gt;3-D&lt;/math&gt; cartesian coordinate system so that the two bottom vertices(C and D) of the face at the very front(and on the y-z plane) are on the y-axis. <br /> Cut out the tetrahedron from the cube. Its &lt;math&gt;4&lt;/math&gt; vertices are &lt;math&gt;A&lt;/math&gt;, B, &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E.&lt;/math&gt; Therefore, corner A has now vanished. <br /> The coordinate of corner C is &lt;math&gt;(0,0,0)&lt;/math&gt;, &lt;math&gt;E(1,0,1)&lt;/math&gt;, &lt;math&gt;B(0,1,1)&lt;/math&gt;, and &lt;math&gt;H(1,1,0)&lt;/math&gt;. <br /> Since we are trying to figure out the height of the object, we have to determine which corner is the farthest away from the cut surface. It is corner &lt;math&gt;H&lt;/math&gt;, and the height of the object is the distance between the center of mass of the cut surface and corner &lt;math&gt;H&lt;/math&gt;. <br /> The center of mass of the cut surface is &lt;math&gt;G(1/3, 1/3, 2/3)&lt;/math&gt;. Using the distance between two points formula, the height of the object <br /> &lt;math&gt;\sqrt{(1-\frac{1}{3})^{2}+(1-\frac{1}{3})^{2}+(0-\frac{2}{3})^{2}}=\sqrt{\frac{12}{9}}=\frac{2\sqrt{3}}{3}&lt;/math&gt;. Therefore the answer is &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> &lt;asy&gt;import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); <br /> label(&quot;A&quot;,(0,0,0),S);<br /> label(&quot;B&quot;,(1,0,0),W);<br /> label(&quot;C&quot;,(0,0,1),N);<br /> label(&quot;D&quot;,(1,0,1),NW);<br /> label(&quot;E&quot;,(1,1,0),S);<br /> label(&quot;F&quot;,(0,1,0),E);<br /> label(&quot;G&quot;,(1,1,1),SE);<br /> label(&quot;H&quot;,(0,1,1),NE);<br /> &lt;/asy&gt;<br /> <br /> We can approach this problem similar to the solution above, by attacking it on the Cartesian plane. As mentioned before, it has been proved that the space diagonal intersects the centroid of triangle &lt;math&gt;BGF&lt;/math&gt; (I have defined the tetrahedron as cutting through points &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt;). We can label &lt;math&gt;E&lt;/math&gt; as &lt;math&gt;(0, 0, 0)&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; as &lt;math&gt;(1, 0, 0)&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; as &lt;math&gt;(0, 1, 0)&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; as &lt;math&gt;(0, 0, 1)&lt;/math&gt;. Therefore, the centroid of triangle &lt;math&gt;BGF&lt;/math&gt; would be the average of these three points, or (&lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;). Since &lt;math&gt;C&lt;/math&gt; is defined as &lt;math&gt;(1, 1, 1)&lt;/math&gt;, and the intersection point passes through <br /> (&lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;) (or one-third of the space diagonal), we can conclude that the altitude is &lt;math&gt;\boxed{\textbf{(D)}}&lt;/math&gt;, or &lt;math&gt;\frac{2}{3}&lt;/math&gt; of &lt;math&gt;\sqrt3&lt;/math&gt;.<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/268<br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_23&diff=133234 2012 AMC 10B Problems/Problem 23 2020-09-06T01:39:12Z <p>Mathlover66: /* Solution 4 */</p> <hr /> <div>[[Category: Introductory Geometry Problems]]<br /> == Problem ==<br /> A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{2 \sqrt{2}}{3}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \frac{2 \sqrt{3}}{3}\qquad\textbf{(E)}\ \sqrt{2} &lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> This tetrahedron has the 4 vertices in these positions: on a corner (lets call this &lt;math&gt;A&lt;/math&gt;) of the cube, and the other three corners (&lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;) adjacent to this corner. We can find the height of the remaining portion of the cube by finding the height of the tetrahedron. We can find the height of this tetrahedron in perspective to the equilateral triangle base (we call this height &lt;math&gt;x&lt;/math&gt;) by finding the volume of the tetrahedron in two ways. &lt;math&gt;\frac{1 \times 1}{2}&lt;/math&gt; is the area of the isosceles base of the tetrahedron. Multiply by the height, &lt;math&gt;1&lt;/math&gt;, and divide by &lt;math&gt;3&lt;/math&gt;, we have the volume of the tetrahedron as &lt;math&gt;\frac{1}{6}&lt;/math&gt;. We set this area equal to one-third the product of our desired height and the area of the equilateral triangle base. First, find the area of the equilateral triangle: &lt;math&gt;[BCD]=\frac{\sqrt{2}^2 \times \sqrt{3}}{4}=\frac{\sqrt{3}}{2}&lt;/math&gt;. So we have: &lt;math&gt;\frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot x=\frac{1}{6}&lt;/math&gt;, and so &lt;math&gt;x=\frac{\sqrt{3}}{3}&lt;/math&gt;. <br /> <br /> Since we know what the height is, we can find the height of the remaining structure. The height of the structure if the tetrahedron was still on would simply be the space diagonal of the cube, &lt;math&gt;\sqrt{3}&lt;/math&gt;, so we just subtract &lt;math&gt;\frac{\sqrt{3}}{3}&lt;/math&gt; from &lt;math&gt;\sqrt{3}&lt;/math&gt; to get &lt;math&gt;\frac{2\sqrt{3}}{3}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can write the volume of a tetrahedron as &lt;math&gt;\frac{Bh}{3}&lt;/math&gt; and revise the formula for the volume of a cube to be &lt;math&gt;Bh&lt;/math&gt;. Also note that the height of the tetrahedron goes through the space diagonal of the cube. So, the remaining part of the cube's space diagonal should be &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the original space diagonal. Hence, the length is &lt;math&gt;\boxed{\textbf{(D)} \: \frac{2\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Plot the cube &lt;math&gt;ABCDEFGH&lt;/math&gt;(let's call it this)on a &lt;math&gt;3-D&lt;/math&gt; cartesian coordinate system so that the two bottom vertices(C and D) of the face at the very front(and on the y-z plane) are on the y-axis. <br /> Cut out the tetrahedron from the cube. Its &lt;math&gt;4&lt;/math&gt; vertices are &lt;math&gt;A&lt;/math&gt;, B, &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E.&lt;/math&gt; Therefore, corner A has now vanished. <br /> The coordinate of corner C is &lt;math&gt;(0,0,0)&lt;/math&gt;, &lt;math&gt;E(1,0,1)&lt;/math&gt;, &lt;math&gt;B(0,1,1)&lt;/math&gt;, and &lt;math&gt;H(1,1,0)&lt;/math&gt;. <br /> Since we are trying to figure out the height of the object, we have to determine which corner is the farthest away from the cut surface. It is corner &lt;math&gt;H&lt;/math&gt;, and the height of the object is the distance between the center of mass of the cut surface and corner &lt;math&gt;H&lt;/math&gt;. <br /> The center of mass of the cut surface is &lt;math&gt;G(1/3, 1/3, 2/3)&lt;/math&gt;. Using the distance between two points formula, the height of the object <br /> &lt;math&gt;\sqrt{(1-\frac{1}{3})^{2}+(1-\frac{1}{3})^{2}+(0-\frac{2}{3})^{2}}=\sqrt{\frac{12}{9}}=\frac{2\sqrt{3}}{3}&lt;/math&gt;. Therefore the answer is &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> &lt;asy&gt;import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); <br /> label(&quot;A&quot;,(0,0,0),S);<br /> label(&quot;B&quot;,(1,0,0),W);<br /> label(&quot;C&quot;,(0,0,1),N);<br /> label(&quot;D&quot;,(1,0,1),NW);<br /> label(&quot;E&quot;,(1,1,0),S);<br /> label(&quot;F&quot;,(0,1,0),E);<br /> label(&quot;G&quot;,(1,1,1),SE);<br /> label(&quot;H&quot;,(0,1,1),NE);<br /> &lt;/asy&gt;<br /> <br /> We can approach this problem similar to the solution above, by attacking it on the Cartesian plane. As mentioned before, it has been proved that the space diagonal intersects the centroid of triangle &lt;math&gt;BGF&lt;/math&gt; (I have defined the tetrahedron as cutting through points &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt;). We can label &lt;math&gt;E&lt;/math&gt; as &lt;math&gt;(0, 0, 0)&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; as &lt;math&gt;(1, 0, 0)&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; as &lt;math&gt;(0, 1, 0)&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; as &lt;math&gt;(0, 0, 1)&lt;/math&gt;. Therefore, the centroid of triangle &lt;math&gt;BGF&lt;/math&gt; would be the average of these three points, or (&lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;). Since &lt;math&gt;C&lt;/math&gt; is defined as &lt;math&gt;(1, 1, 1)&lt;/math&gt;, and the intersection point passes through <br /> (&lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;) (or one-third of the space diagonal), we can conclude that the altitude is &lt;math&gt;\boxed{\textbf{(D)}}&lt;/math&gt;, or &lt;math&gt;\frac{2}{3}&lt;/math&gt; of &lt;math&gt;\sqrt(3)&lt;/math&gt;.<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/268<br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_23&diff=133233 2012 AMC 10B Problems/Problem 23 2020-09-06T01:38:55Z <p>Mathlover66: /* Solution 4 */</p> <hr /> <div>[[Category: Introductory Geometry Problems]]<br /> == Problem ==<br /> A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{2 \sqrt{2}}{3}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \frac{2 \sqrt{3}}{3}\qquad\textbf{(E)}\ \sqrt{2} &lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> This tetrahedron has the 4 vertices in these positions: on a corner (lets call this &lt;math&gt;A&lt;/math&gt;) of the cube, and the other three corners (&lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;) adjacent to this corner. We can find the height of the remaining portion of the cube by finding the height of the tetrahedron. We can find the height of this tetrahedron in perspective to the equilateral triangle base (we call this height &lt;math&gt;x&lt;/math&gt;) by finding the volume of the tetrahedron in two ways. &lt;math&gt;\frac{1 \times 1}{2}&lt;/math&gt; is the area of the isosceles base of the tetrahedron. Multiply by the height, &lt;math&gt;1&lt;/math&gt;, and divide by &lt;math&gt;3&lt;/math&gt;, we have the volume of the tetrahedron as &lt;math&gt;\frac{1}{6}&lt;/math&gt;. We set this area equal to one-third the product of our desired height and the area of the equilateral triangle base. First, find the area of the equilateral triangle: &lt;math&gt;[BCD]=\frac{\sqrt{2}^2 \times \sqrt{3}}{4}=\frac{\sqrt{3}}{2}&lt;/math&gt;. So we have: &lt;math&gt;\frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot x=\frac{1}{6}&lt;/math&gt;, and so &lt;math&gt;x=\frac{\sqrt{3}}{3}&lt;/math&gt;. <br /> <br /> Since we know what the height is, we can find the height of the remaining structure. The height of the structure if the tetrahedron was still on would simply be the space diagonal of the cube, &lt;math&gt;\sqrt{3}&lt;/math&gt;, so we just subtract &lt;math&gt;\frac{\sqrt{3}}{3}&lt;/math&gt; from &lt;math&gt;\sqrt{3}&lt;/math&gt; to get &lt;math&gt;\frac{2\sqrt{3}}{3}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can write the volume of a tetrahedron as &lt;math&gt;\frac{Bh}{3}&lt;/math&gt; and revise the formula for the volume of a cube to be &lt;math&gt;Bh&lt;/math&gt;. Also note that the height of the tetrahedron goes through the space diagonal of the cube. So, the remaining part of the cube's space diagonal should be &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the original space diagonal. Hence, the length is &lt;math&gt;\boxed{\textbf{(D)} \: \frac{2\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Plot the cube &lt;math&gt;ABCDEFGH&lt;/math&gt;(let's call it this)on a &lt;math&gt;3-D&lt;/math&gt; cartesian coordinate system so that the two bottom vertices(C and D) of the face at the very front(and on the y-z plane) are on the y-axis. <br /> Cut out the tetrahedron from the cube. Its &lt;math&gt;4&lt;/math&gt; vertices are &lt;math&gt;A&lt;/math&gt;, B, &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E.&lt;/math&gt; Therefore, corner A has now vanished. <br /> The coordinate of corner C is &lt;math&gt;(0,0,0)&lt;/math&gt;, &lt;math&gt;E(1,0,1)&lt;/math&gt;, &lt;math&gt;B(0,1,1)&lt;/math&gt;, and &lt;math&gt;H(1,1,0)&lt;/math&gt;. <br /> Since we are trying to figure out the height of the object, we have to determine which corner is the farthest away from the cut surface. It is corner &lt;math&gt;H&lt;/math&gt;, and the height of the object is the distance between the center of mass of the cut surface and corner &lt;math&gt;H&lt;/math&gt;. <br /> The center of mass of the cut surface is &lt;math&gt;G(1/3, 1/3, 2/3)&lt;/math&gt;. Using the distance between two points formula, the height of the object <br /> &lt;math&gt;\sqrt{(1-\frac{1}{3})^{2}+(1-\frac{1}{3})^{2}+(0-\frac{2}{3})^{2}}=\sqrt{\frac{12}{9}}=\frac{2\sqrt{3}}{3}&lt;/math&gt;. Therefore the answer is &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> &lt;asy&gt;import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); <br /> label(&quot;A&quot;,(0,0,0),S);<br /> label(&quot;B&quot;,(1,0,0),W);<br /> label(&quot;C&quot;,(0,0,1),N);<br /> label(&quot;D&quot;,(1,0,1),NW);<br /> label(&quot;E&quot;,(1,1,0),S);<br /> label(&quot;F&quot;,(0,1,0),E);<br /> label(&quot;G&quot;,(1,1,1),SE);<br /> label(&quot;H&quot;,(0,1,1),NE);<br /> &lt;/asy&gt;<br /> <br /> We can approach this problem similar to the solution above, by attacking it on the Cartesian plane. As mentioned before, it has been proved that the space diagonal intersects the centroid of triangle &lt;math&gt;BGF&lt;/math&gt; (I have defined the tetrahedron as cutting through points &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt;). We can call label &lt;math&gt;E&lt;/math&gt; as &lt;math&gt;(0, 0, 0)&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; as &lt;math&gt;(1, 0, 0)&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; as &lt;math&gt;(0, 1, 0)&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; as &lt;math&gt;(0, 0, 1)&lt;/math&gt;. Therefore, the centroid of triangle &lt;math&gt;BGF&lt;/math&gt; would be the average of these three points, or (&lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;). Since &lt;math&gt;C&lt;/math&gt; is defined as &lt;math&gt;(1, 1, 1)&lt;/math&gt;, and the intersection point passes through <br /> (&lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;) (or one-third of the space diagonal), we can conclude that the altitude is &lt;math&gt;\boxed{\textbf{(D)}}&lt;/math&gt;, or &lt;math&gt;\frac{2}{3}&lt;/math&gt; of &lt;math&gt;\sqrt(3)&lt;/math&gt;.<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/268<br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_23&diff=133232 2012 AMC 10B Problems/Problem 23 2020-09-06T01:38:32Z <p>Mathlover66: /* Solution 4 */</p> <hr /> <div>[[Category: Introductory Geometry Problems]]<br /> == Problem ==<br /> A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{2 \sqrt{2}}{3}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \frac{2 \sqrt{3}}{3}\qquad\textbf{(E)}\ \sqrt{2} &lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> This tetrahedron has the 4 vertices in these positions: on a corner (lets call this &lt;math&gt;A&lt;/math&gt;) of the cube, and the other three corners (&lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;) adjacent to this corner. We can find the height of the remaining portion of the cube by finding the height of the tetrahedron. We can find the height of this tetrahedron in perspective to the equilateral triangle base (we call this height &lt;math&gt;x&lt;/math&gt;) by finding the volume of the tetrahedron in two ways. &lt;math&gt;\frac{1 \times 1}{2}&lt;/math&gt; is the area of the isosceles base of the tetrahedron. Multiply by the height, &lt;math&gt;1&lt;/math&gt;, and divide by &lt;math&gt;3&lt;/math&gt;, we have the volume of the tetrahedron as &lt;math&gt;\frac{1}{6}&lt;/math&gt;. We set this area equal to one-third the product of our desired height and the area of the equilateral triangle base. First, find the area of the equilateral triangle: &lt;math&gt;[BCD]=\frac{\sqrt{2}^2 \times \sqrt{3}}{4}=\frac{\sqrt{3}}{2}&lt;/math&gt;. So we have: &lt;math&gt;\frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot x=\frac{1}{6}&lt;/math&gt;, and so &lt;math&gt;x=\frac{\sqrt{3}}{3}&lt;/math&gt;. <br /> <br /> Since we know what the height is, we can find the height of the remaining structure. The height of the structure if the tetrahedron was still on would simply be the space diagonal of the cube, &lt;math&gt;\sqrt{3}&lt;/math&gt;, so we just subtract &lt;math&gt;\frac{\sqrt{3}}{3}&lt;/math&gt; from &lt;math&gt;\sqrt{3}&lt;/math&gt; to get &lt;math&gt;\frac{2\sqrt{3}}{3}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can write the volume of a tetrahedron as &lt;math&gt;\frac{Bh}{3}&lt;/math&gt; and revise the formula for the volume of a cube to be &lt;math&gt;Bh&lt;/math&gt;. Also note that the height of the tetrahedron goes through the space diagonal of the cube. So, the remaining part of the cube's space diagonal should be &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the original space diagonal. Hence, the length is &lt;math&gt;\boxed{\textbf{(D)} \: \frac{2\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Plot the cube &lt;math&gt;ABCDEFGH&lt;/math&gt;(let's call it this)on a &lt;math&gt;3-D&lt;/math&gt; cartesian coordinate system so that the two bottom vertices(C and D) of the face at the very front(and on the y-z plane) are on the y-axis. <br /> Cut out the tetrahedron from the cube. Its &lt;math&gt;4&lt;/math&gt; vertices are &lt;math&gt;A&lt;/math&gt;, B, &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E.&lt;/math&gt; Therefore, corner A has now vanished. <br /> The coordinate of corner C is &lt;math&gt;(0,0,0)&lt;/math&gt;, &lt;math&gt;E(1,0,1)&lt;/math&gt;, &lt;math&gt;B(0,1,1)&lt;/math&gt;, and &lt;math&gt;H(1,1,0)&lt;/math&gt;. <br /> Since we are trying to figure out the height of the object, we have to determine which corner is the farthest away from the cut surface. It is corner &lt;math&gt;H&lt;/math&gt;, and the height of the object is the distance between the center of mass of the cut surface and corner &lt;math&gt;H&lt;/math&gt;. <br /> The center of mass of the cut surface is &lt;math&gt;G(1/3, 1/3, 2/3)&lt;/math&gt;. Using the distance between two points formula, the height of the object <br /> &lt;math&gt;\sqrt{(1-\frac{1}{3})^{2}+(1-\frac{1}{3})^{2}+(0-\frac{2}{3})^{2}}=\sqrt{\frac{12}{9}}=\frac{2\sqrt{3}}{3}&lt;/math&gt;. Therefore the answer is &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> &lt;asy&gt;import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); <br /> label(&quot;A&quot;,(0,0,0),S);<br /> label(&quot;B&quot;,(1,0,0),W);<br /> label(&quot;C&quot;,(0,0,1),N);<br /> label(&quot;D&quot;,(1,0,1),NW);<br /> label(&quot;E&quot;,(1,1,0),S);<br /> label(&quot;F&quot;,(0,1,0),E);<br /> label(&quot;G&quot;,(1,1,1),SE);<br /> label(&quot;H&quot;,(0,1,1),NE);<br /> &lt;/asy&gt;<br /> <br /> We can approach this problem similar to the solution above, by attacking it on the Cartesian plane. As mentioned before, it has been proved that the space diagonal intersects the centroid of triangle &lt;math&gt;BGF&lt;/math&gt; (I have defined the tetrahedron as cutting through points &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt;). We can call label &lt;math&gt;E&lt;/math&gt; as &lt;math&gt;(0, 0, 0)&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; as &lt;math&gt;(1, 0, 0)&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; as &lt;math&gt;(0, 1, 0)&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; as &lt;math&gt;(0, 0, 1)&lt;/math&gt;. Therefore, the centroid of triangle &lt;math&gt;BGF&lt;/math&gt; would be the average of these three points, or (&lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;). Since &lt;math&gt;C&lt;/math&gt; is defined as &lt;math&gt;(1, 1, 1)&lt;/math&gt;, and the intersection point passes through <br /> (&lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;) (or one-third of the space diagonal), we can conclude that the altitude is &lt;math&gt;\boxed{\textbf{(D)}}&lt;/math&gt;, or &lt;math&gt;\frac{2}{3}&lt;/math&gt; of &lt;math&gt;sqrt(3)&lt;/math&gt;.<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/268<br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_23&diff=133231 2012 AMC 10B Problems/Problem 23 2020-09-06T01:38:08Z <p>Mathlover66: /* Solution 4 */</p> <hr /> <div>[[Category: Introductory Geometry Problems]]<br /> == Problem ==<br /> A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{2 \sqrt{2}}{3}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \frac{2 \sqrt{3}}{3}\qquad\textbf{(E)}\ \sqrt{2} &lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> This tetrahedron has the 4 vertices in these positions: on a corner (lets call this &lt;math&gt;A&lt;/math&gt;) of the cube, and the other three corners (&lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;) adjacent to this corner. We can find the height of the remaining portion of the cube by finding the height of the tetrahedron. We can find the height of this tetrahedron in perspective to the equilateral triangle base (we call this height &lt;math&gt;x&lt;/math&gt;) by finding the volume of the tetrahedron in two ways. &lt;math&gt;\frac{1 \times 1}{2}&lt;/math&gt; is the area of the isosceles base of the tetrahedron. Multiply by the height, &lt;math&gt;1&lt;/math&gt;, and divide by &lt;math&gt;3&lt;/math&gt;, we have the volume of the tetrahedron as &lt;math&gt;\frac{1}{6}&lt;/math&gt;. We set this area equal to one-third the product of our desired height and the area of the equilateral triangle base. First, find the area of the equilateral triangle: &lt;math&gt;[BCD]=\frac{\sqrt{2}^2 \times \sqrt{3}}{4}=\frac{\sqrt{3}}{2}&lt;/math&gt;. So we have: &lt;math&gt;\frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot x=\frac{1}{6}&lt;/math&gt;, and so &lt;math&gt;x=\frac{\sqrt{3}}{3}&lt;/math&gt;. <br /> <br /> Since we know what the height is, we can find the height of the remaining structure. The height of the structure if the tetrahedron was still on would simply be the space diagonal of the cube, &lt;math&gt;\sqrt{3}&lt;/math&gt;, so we just subtract &lt;math&gt;\frac{\sqrt{3}}{3}&lt;/math&gt; from &lt;math&gt;\sqrt{3}&lt;/math&gt; to get &lt;math&gt;\frac{2\sqrt{3}}{3}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can write the volume of a tetrahedron as &lt;math&gt;\frac{Bh}{3}&lt;/math&gt; and revise the formula for the volume of a cube to be &lt;math&gt;Bh&lt;/math&gt;. Also note that the height of the tetrahedron goes through the space diagonal of the cube. So, the remaining part of the cube's space diagonal should be &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the original space diagonal. Hence, the length is &lt;math&gt;\boxed{\textbf{(D)} \: \frac{2\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Plot the cube &lt;math&gt;ABCDEFGH&lt;/math&gt;(let's call it this)on a &lt;math&gt;3-D&lt;/math&gt; cartesian coordinate system so that the two bottom vertices(C and D) of the face at the very front(and on the y-z plane) are on the y-axis. <br /> Cut out the tetrahedron from the cube. Its &lt;math&gt;4&lt;/math&gt; vertices are &lt;math&gt;A&lt;/math&gt;, B, &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E.&lt;/math&gt; Therefore, corner A has now vanished. <br /> The coordinate of corner C is &lt;math&gt;(0,0,0)&lt;/math&gt;, &lt;math&gt;E(1,0,1)&lt;/math&gt;, &lt;math&gt;B(0,1,1)&lt;/math&gt;, and &lt;math&gt;H(1,1,0)&lt;/math&gt;. <br /> Since we are trying to figure out the height of the object, we have to determine which corner is the farthest away from the cut surface. It is corner &lt;math&gt;H&lt;/math&gt;, and the height of the object is the distance between the center of mass of the cut surface and corner &lt;math&gt;H&lt;/math&gt;. <br /> The center of mass of the cut surface is &lt;math&gt;G(1/3, 1/3, 2/3)&lt;/math&gt;. Using the distance between two points formula, the height of the object <br /> &lt;math&gt;\sqrt{(1-\frac{1}{3})^{2}+(1-\frac{1}{3})^{2}+(0-\frac{2}{3})^{2}}=\sqrt{\frac{12}{9}}=\frac{2\sqrt{3}}{3}&lt;/math&gt;. Therefore the answer is &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> &lt;asy&gt;import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); <br /> label(&quot;A&quot;,(0,0,0),S);<br /> label(&quot;B&quot;,(1,0,0),W);<br /> label(&quot;C&quot;,(0,0,1),N);<br /> label(&quot;D&quot;,(1,0,1),NW);<br /> label(&quot;E&quot;,(1,1,0),S);<br /> label(&quot;F&quot;,(0,1,0),E);<br /> label(&quot;G&quot;,(1,1,1),SE);<br /> label(&quot;H&quot;,(0,1,1),NE);<br /> &lt;/asy&gt;<br /> <br /> We can approach this problem similar to the solution above, by attacking it on the Cartesian plane. As mentioned before, it has been proved that the space diagonal intersects the centroid of triangle &lt;math&gt;BGF&lt;/math&gt; (I have defined the tetrahedron as cutting through points &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt;). We can call label &lt;math&gt;E&lt;/math&gt; as &lt;math&gt;(0, 0, 0)&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; as &lt;math&gt;(1, 0, 0)&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; as &lt;math&gt;(0, 1, 0)&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; as &lt;math&gt;(0, 0, 1)&lt;/math&gt;. Therefore, the centroid of triangle &lt;math&gt;BGF&lt;/math&gt; would be the average of these three points, or (&lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;). Since &lt;math&gt;C&lt;/math&gt; is defined as &lt;math&gt;(1, 1, 1)&lt;/math&gt;, and the intersection point passes through <br /> (&lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;) (or one-third of the space diagonal), we can conclude that the altitude is &lt;math&gt;\boxed{\textbf{(D)}}&lt;/math&gt;.<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/268<br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_23&diff=133229 2012 AMC 10B Problems/Problem 23 2020-09-06T01:37:33Z <p>Mathlover66: /* Solution 4 */</p> <hr /> <div>[[Category: Introductory Geometry Problems]]<br /> == Problem ==<br /> A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{2 \sqrt{2}}{3}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \frac{2 \sqrt{3}}{3}\qquad\textbf{(E)}\ \sqrt{2} &lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> This tetrahedron has the 4 vertices in these positions: on a corner (lets call this &lt;math&gt;A&lt;/math&gt;) of the cube, and the other three corners (&lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;) adjacent to this corner. We can find the height of the remaining portion of the cube by finding the height of the tetrahedron. We can find the height of this tetrahedron in perspective to the equilateral triangle base (we call this height &lt;math&gt;x&lt;/math&gt;) by finding the volume of the tetrahedron in two ways. &lt;math&gt;\frac{1 \times 1}{2}&lt;/math&gt; is the area of the isosceles base of the tetrahedron. Multiply by the height, &lt;math&gt;1&lt;/math&gt;, and divide by &lt;math&gt;3&lt;/math&gt;, we have the volume of the tetrahedron as &lt;math&gt;\frac{1}{6}&lt;/math&gt;. We set this area equal to one-third the product of our desired height and the area of the equilateral triangle base. First, find the area of the equilateral triangle: &lt;math&gt;[BCD]=\frac{\sqrt{2}^2 \times \sqrt{3}}{4}=\frac{\sqrt{3}}{2}&lt;/math&gt;. So we have: &lt;math&gt;\frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot x=\frac{1}{6}&lt;/math&gt;, and so &lt;math&gt;x=\frac{\sqrt{3}}{3}&lt;/math&gt;. <br /> <br /> Since we know what the height is, we can find the height of the remaining structure. The height of the structure if the tetrahedron was still on would simply be the space diagonal of the cube, &lt;math&gt;\sqrt{3}&lt;/math&gt;, so we just subtract &lt;math&gt;\frac{\sqrt{3}}{3}&lt;/math&gt; from &lt;math&gt;\sqrt{3}&lt;/math&gt; to get &lt;math&gt;\frac{2\sqrt{3}}{3}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can write the volume of a tetrahedron as &lt;math&gt;\frac{Bh}{3}&lt;/math&gt; and revise the formula for the volume of a cube to be &lt;math&gt;Bh&lt;/math&gt;. Also note that the height of the tetrahedron goes through the space diagonal of the cube. So, the remaining part of the cube's space diagonal should be &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the original space diagonal. Hence, the length is &lt;math&gt;\boxed{\textbf{(D)} \: \frac{2\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Plot the cube &lt;math&gt;ABCDEFGH&lt;/math&gt;(let's call it this)on a &lt;math&gt;3-D&lt;/math&gt; cartesian coordinate system so that the two bottom vertices(C and D) of the face at the very front(and on the y-z plane) are on the y-axis. <br /> Cut out the tetrahedron from the cube. Its &lt;math&gt;4&lt;/math&gt; vertices are &lt;math&gt;A&lt;/math&gt;, B, &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E.&lt;/math&gt; Therefore, corner A has now vanished. <br /> The coordinate of corner C is &lt;math&gt;(0,0,0)&lt;/math&gt;, &lt;math&gt;E(1,0,1)&lt;/math&gt;, &lt;math&gt;B(0,1,1)&lt;/math&gt;, and &lt;math&gt;H(1,1,0)&lt;/math&gt;. <br /> Since we are trying to figure out the height of the object, we have to determine which corner is the farthest away from the cut surface. It is corner &lt;math&gt;H&lt;/math&gt;, and the height of the object is the distance between the center of mass of the cut surface and corner &lt;math&gt;H&lt;/math&gt;. <br /> The center of mass of the cut surface is &lt;math&gt;G(1/3, 1/3, 2/3)&lt;/math&gt;. Using the distance between two points formula, the height of the object <br /> &lt;math&gt;\sqrt{(1-\frac{1}{3})^{2}+(1-\frac{1}{3})^{2}+(0-\frac{2}{3})^{2}}=\sqrt{\frac{12}{9}}=\frac{2\sqrt{3}}{3}&lt;/math&gt;. Therefore the answer is &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> &lt;asy&gt;import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); <br /> label(&quot;A&quot;,(0,0,0),S);<br /> label(&quot;B&quot;,(1,0,0),W);<br /> label(&quot;C&quot;,(0,0,1),N);<br /> label(&quot;D&quot;,(1,0,1),NW);<br /> label(&quot;E&quot;,(1,1,0),S);<br /> label(&quot;F&quot;,(0,1,0),E);<br /> label(&quot;G&quot;,(1,1,1),SE);<br /> label(&quot;H&quot;,(0,1,1),NE);<br /> &lt;/asy&gt;<br /> <br /> We can approach this problem similar to the solution above, by attacking it on the Cartesian plane. As mentioned before, it has been proved that the space diagonal intersects the centroid of triangle &lt;math&gt;BGF&lt;/math&gt; (I have defined the tetrahedron as cutting through points &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;A&lt;/math&gt;). We can call label &lt;math&gt;E&lt;/math&gt; as &lt;math&gt;(0, 0, 0)&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; as &lt;math&gt;(1, 0, 0)&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; as &lt;math&gt;(0, 1, 0)&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; as &lt;math&gt;(0, 0, 1)&lt;/math&gt;. Therefore, the centroid of triangle &lt;math&gt;BGF&lt;/math&gt; would be the average of these three points, or (&lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;). Since &lt;math&gt;C&lt;/math&gt; is defined as &lt;math&gt;(1, 1, 1)&lt;/math&gt;, and the intersection point passes through <br /> (&lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;) (or one-third of the space diagonal), we can conclude that the altitude is $\boxed{\textbf{(D)}}.<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/268<br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_23&diff=133228 2012 AMC 10B Problems/Problem 23 2020-09-06T01:37:05Z <p>Mathlover66: /* Solution 4 */</p> <hr /> <div>[[Category: Introductory Geometry Problems]]<br /> == Problem ==<br /> A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{2 \sqrt{2}}{3}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \frac{2 \sqrt{3}}{3}\qquad\textbf{(E)}\ \sqrt{2} &lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> This tetrahedron has the 4 vertices in these positions: on a corner (lets call this &lt;math&gt;A&lt;/math&gt;) of the cube, and the other three corners (&lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;) adjacent to this corner. We can find the height of the remaining portion of the cube by finding the height of the tetrahedron. We can find the height of this tetrahedron in perspective to the equilateral triangle base (we call this height &lt;math&gt;x&lt;/math&gt;) by finding the volume of the tetrahedron in two ways. &lt;math&gt;\frac{1 \times 1}{2}&lt;/math&gt; is the area of the isosceles base of the tetrahedron. Multiply by the height, &lt;math&gt;1&lt;/math&gt;, and divide by &lt;math&gt;3&lt;/math&gt;, we have the volume of the tetrahedron as &lt;math&gt;\frac{1}{6}&lt;/math&gt;. We set this area equal to one-third the product of our desired height and the area of the equilateral triangle base. First, find the area of the equilateral triangle: &lt;math&gt;[BCD]=\frac{\sqrt{2}^2 \times \sqrt{3}}{4}=\frac{\sqrt{3}}{2}&lt;/math&gt;. So we have: &lt;math&gt;\frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot x=\frac{1}{6}&lt;/math&gt;, and so &lt;math&gt;x=\frac{\sqrt{3}}{3}&lt;/math&gt;. <br /> <br /> Since we know what the height is, we can find the height of the remaining structure. The height of the structure if the tetrahedron was still on would simply be the space diagonal of the cube, &lt;math&gt;\sqrt{3}&lt;/math&gt;, so we just subtract &lt;math&gt;\frac{\sqrt{3}}{3}&lt;/math&gt; from &lt;math&gt;\sqrt{3}&lt;/math&gt; to get &lt;math&gt;\frac{2\sqrt{3}}{3}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can write the volume of a tetrahedron as &lt;math&gt;\frac{Bh}{3}&lt;/math&gt; and revise the formula for the volume of a cube to be &lt;math&gt;Bh&lt;/math&gt;. Also note that the height of the tetrahedron goes through the space diagonal of the cube. So, the remaining part of the cube's space diagonal should be &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the original space diagonal. Hence, the length is &lt;math&gt;\boxed{\textbf{(D)} \: \frac{2\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Plot the cube &lt;math&gt;ABCDEFGH&lt;/math&gt;(let's call it this)on a &lt;math&gt;3-D&lt;/math&gt; cartesian coordinate system so that the two bottom vertices(C and D) of the face at the very front(and on the y-z plane) are on the y-axis. <br /> Cut out the tetrahedron from the cube. Its &lt;math&gt;4&lt;/math&gt; vertices are &lt;math&gt;A&lt;/math&gt;, B, &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E.&lt;/math&gt; Therefore, corner A has now vanished. <br /> The coordinate of corner C is &lt;math&gt;(0,0,0)&lt;/math&gt;, &lt;math&gt;E(1,0,1)&lt;/math&gt;, &lt;math&gt;B(0,1,1)&lt;/math&gt;, and &lt;math&gt;H(1,1,0)&lt;/math&gt;. <br /> Since we are trying to figure out the height of the object, we have to determine which corner is the farthest away from the cut surface. It is corner &lt;math&gt;H&lt;/math&gt;, and the height of the object is the distance between the center of mass of the cut surface and corner &lt;math&gt;H&lt;/math&gt;. <br /> The center of mass of the cut surface is &lt;math&gt;G(1/3, 1/3, 2/3)&lt;/math&gt;. Using the distance between two points formula, the height of the object <br /> &lt;math&gt;\sqrt{(1-\frac{1}{3})^{2}+(1-\frac{1}{3})^{2}+(0-\frac{2}{3})^{2}}=\sqrt{\frac{12}{9}}=\frac{2\sqrt{3}}{3}&lt;/math&gt;. Therefore the answer is &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> &lt;asy&gt;import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); <br /> label(&quot;A&quot;,(0,0,0),S);<br /> label(&quot;B&quot;,(1,0,0),W);<br /> label(&quot;C&quot;,(0,0,1),N);<br /> label(&quot;D&quot;,(1,0,1),NW);<br /> label(&quot;E&quot;,(1,1,0),S);<br /> label(&quot;F&quot;,(0,1,0),E);<br /> label(&quot;G&quot;,(1,1,1),SE);<br /> label(&quot;H&quot;,(0,1,1),NE);<br /> &lt;/asy&gt;<br /> <br /> We can approach this problem similar to the solution above, by attacking it on the Cartesian plane. As mentioned before, it has been proved that the space diagonal intersects the centroid of triangle &lt;math&gt;BGF&lt;/math&gt; (I have defined the tetrahedron as cutting through points &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;A&lt;/math&gt;). We can call label &lt;math&gt;E&lt;/math&gt; as &lt;math&gt;(0, 0, 0)&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; as &lt;math&gt;(1, 0, 0)&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; as &lt;math&gt;(0, 1, 0)&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; as &lt;math&gt;(0, 0, 1)&lt;/math&gt;. Therefore, the centroid of triangle &lt;math&gt;BGF&lt;/math&gt; would be the average of these three points, or (&lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;). Since &lt;math&gt;C&lt;/math&gt; is defined as &lt;math&gt;(1, 1, 1)&lt;/math&gt;, and the intersection point passes through <br /> (&lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\frac{1}{3}&lt;/math&gt;) (or one-third of the space diagonal), we can conclude that the altitude is<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/268<br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_23&diff=133226 2012 AMC 10B Problems/Problem 23 2020-09-06T01:31:44Z <p>Mathlover66: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{2 \sqrt{2}}{3}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \frac{2 \sqrt{3}}{3}\qquad\textbf{(E)}\ \sqrt{2} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> This tetrahedron has the 4 vertices in these positions: on a corner (lets call this &lt;math&gt;A&lt;/math&gt;) of the cube, and the other three corners (&lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;) adjacent to this corner. We can find the height of the remaining portion of the cube by finding the height of the tetrahedron. We can find the height of this tetrahedron in perspective to the equilateral triangle base (we call this height &lt;math&gt;x&lt;/math&gt;) by finding the volume of the tetrahedron in two ways. &lt;math&gt;\frac{1 \times 1}{2}&lt;/math&gt; is the area of the isosceles base of the tetrahedron. Multiply by the height, &lt;math&gt;1&lt;/math&gt;, and divide by &lt;math&gt;3&lt;/math&gt;, we have the volume of the tetrahedron as &lt;math&gt;\frac{1}{6}&lt;/math&gt;. We set this area equal to one-third the product of our desired height and the area of the equilateral triangle base. First, find the area of the equilateral triangle: &lt;math&gt;[BCD]=\frac{\sqrt{2}^2 \times \sqrt{3}}{4}=\frac{\sqrt{3}}{2}&lt;/math&gt;. So we have: &lt;math&gt;\frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot x=\frac{1}{6}&lt;/math&gt;, and so &lt;math&gt;x=\frac{\sqrt{3}}{3}&lt;/math&gt;. <br /> <br /> Since we know what the height is, we can find the height of the remaining structure. The height of the structure if the tetrahedron was still on would simply be the space diagonal of the cube, &lt;math&gt;\sqrt{3}&lt;/math&gt;, so we just subtract &lt;math&gt;\frac{\sqrt{3}}{3}&lt;/math&gt; from &lt;math&gt;\sqrt{3}&lt;/math&gt; to get &lt;math&gt;\frac{2\sqrt{3}}{3}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can write the volume of a tetrahedron as &lt;math&gt;\frac{Bh}{3}&lt;/math&gt; and revise the formula for the volume of a cube to be &lt;math&gt;Bh&lt;/math&gt;. Also note that the height of the tetrahedron goes through the space diagonal of the cube. So, the remaining part of the cube's space diagonal should be &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the original space diagonal. Hence, the length is &lt;math&gt;\boxed{\textbf{(D)} \: \frac{2\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Plot the cube &lt;math&gt;ABCDEFGH&lt;/math&gt;(let's call it this)on a &lt;math&gt;3-D&lt;/math&gt; cartesian coordinate system so that the two bottom vertices(C and D) of the face at the very front(and on the y-z plane) are on the y-axis. <br /> Cut out the tetrahedron from the cube. Its &lt;math&gt;4&lt;/math&gt; vertices are &lt;math&gt;A&lt;/math&gt;, B, &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E.&lt;/math&gt; Therefore, corner A has now vanished. <br /> The coordinate of corner C is &lt;math&gt;(0,0,0)&lt;/math&gt;, &lt;math&gt;E(1,0,1)&lt;/math&gt;, &lt;math&gt;B(0,1,1)&lt;/math&gt;, and &lt;math&gt;H(1,1,0)&lt;/math&gt;. <br /> Since we are trying to figure out the height of the object, we have to determine which corner is the farthest away from the cut surface. It is corner &lt;math&gt;H&lt;/math&gt;, and the height of the object is the distance between the center of mass of the cut surface and corner &lt;math&gt;H&lt;/math&gt;. <br /> The center of mass of the cut surface is &lt;math&gt;G(1/3, 1/3, 2/3)&lt;/math&gt;. Using the distance between two points formula, the height of the object <br /> &lt;math&gt;\sqrt{(1-\frac{1}{3})^{2}+(1-\frac{1}{3})^{2}+(0-\frac{2}{3})^{2}}=\sqrt{\frac{12}{9}}=\frac{2\sqrt{3}}{3}&lt;/math&gt;. Therefore the answer is &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> &lt;asy&gt;import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); <br /> label(&quot;A&quot;,(0,0,0),S);<br /> label(&quot;B&quot;,(1,0,0),W);<br /> label(&quot;C&quot;,(0,0,1),N);<br /> label(&quot;D&quot;,(1,0,1),NW);<br /> label(&quot;E&quot;,(1,1,0),S);<br /> label(&quot;F&quot;,(0,1,0),E);<br /> label(&quot;G&quot;,(1,1,1),SE);<br /> label(&quot;H&quot;,(0,1,1),NE);<br /> &lt;/asy&gt;<br /> <br /> We can approach this problem similar to the solution above, by attacking it on the Cartesian plane. As mentioned before, it has been proved that the space diagonal intersects the centroid of triangle<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/268<br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=22|num-a=24}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_23&diff=133225 2012 AMC 10B Problems/Problem 23 2020-09-06T01:30:34Z <p>Mathlover66: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{2 \sqrt{2}}{3}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \frac{2 \sqrt{3}}{3}\qquad\textbf{(E)}\ \sqrt{2} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> This tetrahedron has the 4 vertices in these positions: on a corner (lets call this &lt;math&gt;A&lt;/math&gt;) of the cube, and the other three corners (&lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;) adjacent to this corner. We can find the height of the remaining portion of the cube by finding the height of the tetrahedron. We can find the height of this tetrahedron in perspective to the equilateral triangle base (we call this height &lt;math&gt;x&lt;/math&gt;) by finding the volume of the tetrahedron in two ways. &lt;math&gt;\frac{1 \times 1}{2}&lt;/math&gt; is the area of the isosceles base of the tetrahedron. Multiply by the height, &lt;math&gt;1&lt;/math&gt;, and divide by &lt;math&gt;3&lt;/math&gt;, we have the volume of the tetrahedron as &lt;math&gt;\frac{1}{6}&lt;/math&gt;. We set this area equal to one-third the product of our desired height and the area of the equilateral triangle base. First, find the area of the equilateral triangle: &lt;math&gt;[BCD]=\frac{\sqrt{2}^2 \times \sqrt{3}}{4}=\frac{\sqrt{3}}{2}&lt;/math&gt;. So we have: &lt;math&gt;\frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot x=\frac{1}{6}&lt;/math&gt;, and so &lt;math&gt;x=\frac{\sqrt{3}}{3}&lt;/math&gt;. <br /> <br /> Since we know what the height is, we can find the height of the remaining structure. The height of the structure if the tetrahedron was still on would simply be the space diagonal of the cube, &lt;math&gt;\sqrt{3}&lt;/math&gt;, so we just subtract &lt;math&gt;\frac{\sqrt{3}}{3}&lt;/math&gt; from &lt;math&gt;\sqrt{3}&lt;/math&gt; to get &lt;math&gt;\frac{2\sqrt{3}}{3}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can write the volume of a tetrahedron as &lt;math&gt;\frac{Bh}{3}&lt;/math&gt; and revise the formula for the volume of a cube to be &lt;math&gt;Bh&lt;/math&gt;. Also note that the height of the tetrahedron goes through the space diagonal of the cube. So, the remaining part of the cube's space diagonal should be &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the original space diagonal. Hence, the length is &lt;math&gt;\boxed{\textbf{(D)} \: \frac{2\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Plot the cube &lt;math&gt;ABCDEFGH&lt;/math&gt;(let's call it this)on a &lt;math&gt;3-D&lt;/math&gt; cartesian coordinate system so that the two bottom vertices(C and D) of the face at the very front(and on the y-z plane) are on the y-axis. <br /> Cut out the tetrahedron from the cube. Its &lt;math&gt;4&lt;/math&gt; vertices are &lt;math&gt;A&lt;/math&gt;, B, &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E.&lt;/math&gt; Therefore, corner A has now vanished. <br /> The coordinate of corner C is &lt;math&gt;(0,0,0)&lt;/math&gt;, &lt;math&gt;E(1,0,1)&lt;/math&gt;, &lt;math&gt;B(0,1,1)&lt;/math&gt;, and &lt;math&gt;H(1,1,0)&lt;/math&gt;. <br /> Since we are trying to figure out the height of the object, we have to determine which corner is the farthest away from the cut surface. It is corner &lt;math&gt;H&lt;/math&gt;, and the height of the object is the distance between the center of mass of the cut surface and corner &lt;math&gt;H&lt;/math&gt;. <br /> The center of mass of the cut surface is &lt;math&gt;G(1/3, 1/3, 2/3)&lt;/math&gt;. Using the distance between two points formula, the height of the object <br /> &lt;math&gt;\sqrt{(1-\frac{1}{3})^{2}+(1-\frac{1}{3})^{2}+(0-\frac{2}{3})^{2}}=\sqrt{\frac{12}{9}}=\frac{2\sqrt{3}}{3}&lt;/math&gt;. Therefore the answer is &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> &lt;asy&gt;import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); <br /> label(&quot;A&quot;,(0,0,0),S);<br /> label(&quot;B&quot;,(1,0,0),W);<br /> label(&quot;C&quot;,(0,0,1),N);<br /> label(&quot;D&quot;,(1,0,1),NW);<br /> label(&quot;E&quot;,(1,1,0),S);<br /> label(&quot;F&quot;,(0,1,0),E);<br /> label(&quot;G&quot;,(1,1,1),SE);<br /> label(&quot;H&quot;,(0,1,1),NE);<br /> &lt;/asy&gt;<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/268<br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=22|num-a=24}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_23&diff=133224 2012 AMC 10B Problems/Problem 23 2020-09-06T01:29:41Z <p>Mathlover66: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{2 \sqrt{2}}{3}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \frac{2 \sqrt{3}}{3}\qquad\textbf{(E)}\ \sqrt{2} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> This tetrahedron has the 4 vertices in these positions: on a corner (lets call this &lt;math&gt;A&lt;/math&gt;) of the cube, and the other three corners (&lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;) adjacent to this corner. We can find the height of the remaining portion of the cube by finding the height of the tetrahedron. We can find the height of this tetrahedron in perspective to the equilateral triangle base (we call this height &lt;math&gt;x&lt;/math&gt;) by finding the volume of the tetrahedron in two ways. &lt;math&gt;\frac{1 \times 1}{2}&lt;/math&gt; is the area of the isosceles base of the tetrahedron. Multiply by the height, &lt;math&gt;1&lt;/math&gt;, and divide by &lt;math&gt;3&lt;/math&gt;, we have the volume of the tetrahedron as &lt;math&gt;\frac{1}{6}&lt;/math&gt;. We set this area equal to one-third the product of our desired height and the area of the equilateral triangle base. First, find the area of the equilateral triangle: &lt;math&gt;[BCD]=\frac{\sqrt{2}^2 \times \sqrt{3}}{4}=\frac{\sqrt{3}}{2}&lt;/math&gt;. So we have: &lt;math&gt;\frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot x=\frac{1}{6}&lt;/math&gt;, and so &lt;math&gt;x=\frac{\sqrt{3}}{3}&lt;/math&gt;. <br /> <br /> Since we know what the height is, we can find the height of the remaining structure. The height of the structure if the tetrahedron was still on would simply be the space diagonal of the cube, &lt;math&gt;\sqrt{3}&lt;/math&gt;, so we just subtract &lt;math&gt;\frac{\sqrt{3}}{3}&lt;/math&gt; from &lt;math&gt;\sqrt{3}&lt;/math&gt; to get &lt;math&gt;\frac{2\sqrt{3}}{3}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can write the volume of a tetrahedron as &lt;math&gt;\frac{Bh}{3}&lt;/math&gt; and revise the formula for the volume of a cube to be &lt;math&gt;Bh&lt;/math&gt;. Also note that the height of the tetrahedron goes through the space diagonal of the cube. So, the remaining part of the cube's space diagonal should be &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the original space diagonal. Hence, the length is &lt;math&gt;\boxed{\textbf{(D)} \: \frac{2\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Plot the cube &lt;math&gt;ABCDEFGH&lt;/math&gt;(let's call it this)on a &lt;math&gt;3-D&lt;/math&gt; cartesian coordinate system so that the two bottom vertices(C and D) of the face at the very front(and on the y-z plane) are on the y-axis. <br /> Cut out the tetrahedron from the cube. Its &lt;math&gt;4&lt;/math&gt; vertices are &lt;math&gt;A&lt;/math&gt;, B, &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E.&lt;/math&gt; Therefore, corner A has now vanished. <br /> The coordinate of corner C is &lt;math&gt;(0,0,0)&lt;/math&gt;, &lt;math&gt;E(1,0,1)&lt;/math&gt;, &lt;math&gt;B(0,1,1)&lt;/math&gt;, and &lt;math&gt;H(1,1,0)&lt;/math&gt;. <br /> Since we are trying to figure out the height of the object, we have to determine which corner is the farthest away from the cut surface. It is corner &lt;math&gt;H&lt;/math&gt;, and the height of the object is the distance between the center of mass of the cut surface and corner &lt;math&gt;H&lt;/math&gt;. <br /> The center of mass of the cut surface is &lt;math&gt;G(1/3, 1/3, 2/3)&lt;/math&gt;. Using the distance between two points formula, the height of the object <br /> &lt;math&gt;\sqrt{(1-\frac{1}{3})^{2}+(1-\frac{1}{3})^{2}+(0-\frac{2}{3})^{2}}=\sqrt{\frac{12}{9}}=\frac{2\sqrt{3}}{3}&lt;/math&gt;. Therefore the answer is &lt;math&gt;\boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> &lt;asy&gt;import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); <br /> &lt;/asy&gt;<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/268<br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=22|num-a=24}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_22&diff=133223 2012 AMC 10B Problems/Problem 22 2020-09-06T00:52:44Z <p>Mathlover66: </p> <hr /> <div>==Problem==<br /> Let (&lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, ... &lt;math&gt;a_{10}&lt;/math&gt;) be a list of the first 10 positive integers such that for each &lt;math&gt;2\le&lt;/math&gt; &lt;math&gt;i&lt;/math&gt; &lt;math&gt;\le10&lt;/math&gt; either &lt;math&gt;a_i + 1&lt;/math&gt; or &lt;math&gt;a_i-1&lt;/math&gt; or both appear somewhere before &lt;math&gt;a_i&lt;/math&gt; in the list. How many such lists are there?<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ \ 362,880&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> If we have 1 as the first number, then the only possible list is &lt;math&gt;(1,2,3,4,5,6,7,8,9,10)&lt;/math&gt;. <br /> <br /> If we have 2 as the first number, then we have 9 ways to choose where the &lt;math&gt;1&lt;/math&gt; goes, and the numbers ascend from the first number, &lt;math&gt;2&lt;/math&gt;, with the exception of the &lt;math&gt;1&lt;/math&gt;.<br /> For example, &lt;math&gt;(2,3,1,4,5,6,7,8,9,10)&lt;/math&gt;, or &lt;math&gt;(2,3,4,1,5,6,7,8,9,10)&lt;/math&gt;. There are &lt;math&gt;\dbinom{9}{1}&lt;/math&gt; ways to do so.<br /> <br /> If we use 3 as the first number, we need to choose 2 spaces to be 2 and 1, respectively. There are &lt;math&gt;\dbinom{9}{2}&lt;/math&gt; ways to do this.<br /> <br /> In the same way, the total number of lists is:<br /> &lt;math&gt;\dbinom{9}{0} +\dbinom{9}{1} + \dbinom{9}{2} + \dbinom{9}{3} + \dbinom{9}{4}.....\dbinom{9}{9}&lt;/math&gt;<br /> <br /> By the binomial theorem, this is &lt;math&gt;2^{9}&lt;/math&gt; = &lt;math&gt;512&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Arrange the spaces and put arrows pointing either up or down between them. Then for each arrangement of arrows there is one and only one list that corresponds to up. For example, all arrows pointing up is &lt;math&gt;(1,2,3,4,5...10)&lt;/math&gt;. There are 9 arrows, so the answer is &lt;math&gt;2^{9}&lt;/math&gt; = &lt;math&gt;512&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;<br /> <br /> NOTE:<br /> Solution cited from: http://www.artofproblemsolving.com/Videos/external.php?video_id=269.<br /> <br /> ==Solution 3==<br /> Notice that the answer to the problem is solely based on the length of the lists, i.e. 10. We can replace 10 with smaller values, such as 2 and 3, and try to find a pattern. If we replace it with 2, we can easily see that there are two possible lists, &lt;math&gt;(1, 2)&lt;/math&gt; and &lt;math&gt;(2, 1)&lt;/math&gt;. If we replace it with 3, there are four lists, &lt;math&gt;(1, 2, 3), (2, 1, 3), (2, 3, 1),&lt;/math&gt; and &lt;math&gt;(3, 2, 1)&lt;/math&gt;. Since 2 and 4 are both powers of 2, it is likely that the number of lists is &lt;math&gt;2^{n-1}&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is the length of the lists. &lt;math&gt;2^{10-1}=512=\boxed{\textbf{(B)}}&lt;/math&gt;<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/269<br /> <br /> ~dolphin7<br /> <br /> ==Video Solution==<br /> https://youtu.be/bXPSv93GVbg<br /> <br /> ~IceMatrix<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_22&diff=133222 2012 AMC 10B Problems/Problem 22 2020-09-06T00:51:45Z <p>Mathlover66: /* Solution 4 */</p> <hr /> <div>==Problem==<br /> Let (&lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, ... &lt;math&gt;a_{10}&lt;/math&gt;) be a list of the first 10 positive integers such that for each &lt;math&gt;2\le&lt;/math&gt; &lt;math&gt;i&lt;/math&gt; &lt;math&gt;\le10&lt;/math&gt; either &lt;math&gt;a_i + 1&lt;/math&gt; or &lt;math&gt;a_i-1&lt;/math&gt; or both appear somewhere before &lt;math&gt;a_i&lt;/math&gt; in the list. How many such lists are there?<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ \ 362,880&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> If we have 1 as the first number, then the only possible list is &lt;math&gt;(1,2,3,4,5,6,7,8,9,10)&lt;/math&gt;. <br /> <br /> If we have 2 as the first number, then we have 9 ways to choose where the &lt;math&gt;1&lt;/math&gt; goes, and the numbers ascend from the first number, &lt;math&gt;2&lt;/math&gt;, with the exception of the &lt;math&gt;1&lt;/math&gt;.<br /> For example, &lt;math&gt;(2,3,1,4,5,6,7,8,9,10)&lt;/math&gt;, or &lt;math&gt;(2,3,4,1,5,6,7,8,9,10)&lt;/math&gt;. There are &lt;math&gt;\dbinom{9}{1}&lt;/math&gt; ways to do so.<br /> <br /> If we use 3 as the first number, we need to choose 2 spaces to be 2 and 1, respectively. There are &lt;math&gt;\dbinom{9}{2}&lt;/math&gt; ways to do this.<br /> <br /> In the same way, the total number of lists is:<br /> &lt;math&gt;\dbinom{9}{0} +\dbinom{9}{1} + \dbinom{9}{2} + \dbinom{9}{3} + \dbinom{9}{4}.....\dbinom{9}{9}&lt;/math&gt;<br /> <br /> By the binomial theorem, this is &lt;math&gt;2^{9}&lt;/math&gt; = &lt;math&gt;512&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Arrange the spaces and put arrows pointing either up or down between them. Then for each arrangement of arrows there is one and only one list that corresponds to up. For example, all arrows pointing up is &lt;math&gt;(1,2,3,4,5...10)&lt;/math&gt;. There are 9 arrows, so the answer is &lt;math&gt;2^{9}&lt;/math&gt; = &lt;math&gt;512&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;<br /> <br /> NOTE:<br /> Solution cited from: http://www.artofproblemsolving.com/Videos/external.php?video_id=269.<br /> <br /> ==Solution 3==<br /> Notice that the answer to the problem is solely based on the length of the lists, i.e. 10. We can replace 10 with smaller values, such as 2 and 3, and try to find a pattern. If we replace it with 2, we can easily see that there are two possible lists, &lt;math&gt;(1, 2)&lt;/math&gt; and &lt;math&gt;(2, 1)&lt;/math&gt;. If we replace it with 3, there are four lists, &lt;math&gt;(1, 2, 3), (2, 1, 3), (2, 3, 1),&lt;/math&gt; and &lt;math&gt;(3, 2, 1)&lt;/math&gt;. Since 2 and 4 are both powers of 2, it is likely that the number of lists is &lt;math&gt;2^{n-1}&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is the length of the lists. &lt;math&gt;2^{10-1}=512=\boxed{\textbf{(B)}}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> We define 'transition' as the move from the &lt;math&gt;n&lt;/math&gt;th index to the &lt;math&gt;n+1&lt;/math&gt;th index. Note that from the transition from the &lt;math&gt;n&lt;/math&gt;th index to the &lt;math&gt;n+1&lt;/math&gt;th index, you can either move to &lt;math&gt;n+1&lt;/math&gt; or &lt;math&gt;n-1&lt;/math&gt;. Therefore, there are &lt;math&gt;2&lt;/math&gt; options for each transition. Since there are 9 total transitions (from the first index to the tenth index), the answer is &lt;math&gt;2^{9}=512=\boxed{\textbf{(B)}}&lt;/math&gt;.<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/269<br /> <br /> ~dolphin7<br /> <br /> ==Video Solution==<br /> https://youtu.be/bXPSv93GVbg<br /> <br /> ~IceMatrix<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_22&diff=133220 2012 AMC 10B Problems/Problem 22 2020-09-06T00:50:17Z <p>Mathlover66: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> Let (&lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, ... &lt;math&gt;a_{10}&lt;/math&gt;) be a list of the first 10 positive integers such that for each &lt;math&gt;2\le&lt;/math&gt; &lt;math&gt;i&lt;/math&gt; &lt;math&gt;\le10&lt;/math&gt; either &lt;math&gt;a_i + 1&lt;/math&gt; or &lt;math&gt;a_i-1&lt;/math&gt; or both appear somewhere before &lt;math&gt;a_i&lt;/math&gt; in the list. How many such lists are there?<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ \ 362,880&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> If we have 1 as the first number, then the only possible list is &lt;math&gt;(1,2,3,4,5,6,7,8,9,10)&lt;/math&gt;. <br /> <br /> If we have 2 as the first number, then we have 9 ways to choose where the &lt;math&gt;1&lt;/math&gt; goes, and the numbers ascend from the first number, &lt;math&gt;2&lt;/math&gt;, with the exception of the &lt;math&gt;1&lt;/math&gt;.<br /> For example, &lt;math&gt;(2,3,1,4,5,6,7,8,9,10)&lt;/math&gt;, or &lt;math&gt;(2,3,4,1,5,6,7,8,9,10)&lt;/math&gt;. There are &lt;math&gt;\dbinom{9}{1}&lt;/math&gt; ways to do so.<br /> <br /> If we use 3 as the first number, we need to choose 2 spaces to be 2 and 1, respectively. There are &lt;math&gt;\dbinom{9}{2}&lt;/math&gt; ways to do this.<br /> <br /> In the same way, the total number of lists is:<br /> &lt;math&gt;\dbinom{9}{0} +\dbinom{9}{1} + \dbinom{9}{2} + \dbinom{9}{3} + \dbinom{9}{4}.....\dbinom{9}{9}&lt;/math&gt;<br /> <br /> By the binomial theorem, this is &lt;math&gt;2^{9}&lt;/math&gt; = &lt;math&gt;512&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Arrange the spaces and put arrows pointing either up or down between them. Then for each arrangement of arrows there is one and only one list that corresponds to up. For example, all arrows pointing up is &lt;math&gt;(1,2,3,4,5...10)&lt;/math&gt;. There are 9 arrows, so the answer is &lt;math&gt;2^{9}&lt;/math&gt; = &lt;math&gt;512&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;<br /> <br /> NOTE:<br /> Solution cited from: http://www.artofproblemsolving.com/Videos/external.php?video_id=269.<br /> <br /> ==Solution 3==<br /> Notice that the answer to the problem is solely based on the length of the lists, i.e. 10. We can replace 10 with smaller values, such as 2 and 3, and try to find a pattern. If we replace it with 2, we can easily see that there are two possible lists, &lt;math&gt;(1, 2)&lt;/math&gt; and &lt;math&gt;(2, 1)&lt;/math&gt;. If we replace it with 3, there are four lists, &lt;math&gt;(1, 2, 3), (2, 1, 3), (2, 3, 1),&lt;/math&gt; and &lt;math&gt;(3, 2, 1)&lt;/math&gt;. Since 2 and 4 are both powers of 2, it is likely that the number of lists is &lt;math&gt;2^{n-1}&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is the length of the lists. &lt;math&gt;2^{10-1}=512=\boxed{\textbf{(B)}}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> Note that from the transition from the &lt;math&gt;n&lt;/math&gt;th index to the &lt;math&gt;n+1&lt;/math&gt;th index, you can either move on to &lt;math&gt;n+1&lt;/math&gt; or &lt;math&gt;n-1&lt;/math&gt;. Since there are 9 total transitions (from the first index to the tenth index), the answer is &lt;math&gt;2^{10-1}=512=\boxed{\textbf{(B)}}&lt;/math&gt;.<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/269<br /> <br /> ~dolphin7<br /> <br /> ==Video Solution==<br /> https://youtu.be/bXPSv93GVbg<br /> <br /> ~IceMatrix<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_16&diff=109558 2012 AMC 12B Problems/Problem 16 2019-08-31T03:36:45Z <p>Mathlover66: /* Solution 4: Another Different Way of Looking at Solution 1 &amp; 3 */</p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #16]] and [[2012 AMC 10B Problems|2012 AMC 10B #24]]}}<br /> <br /> == Problem==<br /> Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 &lt;/math&gt;<br /> <br /> <br /> <br /> ==Solutions==<br /> <br /> === Solution 1===<br /> Let the ordered triple &lt;math&gt;(a,b,c)&lt;/math&gt; denote that &lt;math&gt;a&lt;/math&gt; songs are liked by Amy and Beth, &lt;math&gt;b&lt;/math&gt; songs by Beth and Jo, and &lt;math&gt;c&lt;/math&gt; songs by Jo and Amy. We claim that the only possible triples are &lt;math&gt;(1,1,1), (2,1,1), (1,2,1)(1,1,2)&lt;/math&gt;. <br /> <br /> To show this, observe these are all valid conditions. Second, note that none of &lt;math&gt;a,b,c&lt;/math&gt; can be bigger than 3. Suppose otherwise, that &lt;math&gt;a = 3&lt;/math&gt;. Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to be at least 1. In fact, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to equal 1, otherwise there will be a song liked by all three. Suppose &lt;math&gt;b = 1&lt;/math&gt;. Then we must have &lt;math&gt;c=0&lt;/math&gt; since no song is liked by all three girls, a contradiction.<br /> <br /> '''Case 1''':How many ways are there for &lt;math&gt;(a,b,c)&lt;/math&gt; to equal &lt;math&gt;(1,1,1)&lt;/math&gt;? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So &lt;math&gt;(a,b,c)=(1,1,1)&lt;/math&gt; in &lt;math&gt;4\cdot3\cdot2\cdot4 = 96&lt;/math&gt; ways.<br /> <br /> '''Case 2''':To find the number of ways for &lt;math&gt;(a,b,c) = (2,1,1)&lt;/math&gt;, observe there are &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are &lt;math&gt;6\cdot2\cdot3=36&lt;/math&gt; ways for the girls to like the songs.<br /> <br /> That gives a total of &lt;math&gt;96 + 36 = \boxed{132}&lt;/math&gt; ways for the girls to like the songs, so the answer is &lt;math&gt;(\textrm{\textbf{B}})&lt;/math&gt;.<br /> <br /> === Solution 2 (Answer Choices)===<br /> We begin by noticing that there are four ways to assign a song liked by both Amy and Beth, three ways to assign a song liked by both Amy and Jo (because Jo may not like the song liked by both Amy and Beth), and two ways to assign a song liked by both Beth and Jo (because both Beth and Jo may not like the song liked by the previous pairs. Additionally, there are &lt;math&gt;2\cdot2\cdot2=8&lt;/math&gt; ways to assign song preferences for the fourth song. Multiplying, we obtain an answer of &lt;math&gt;4\cdot3\cdot2\cdot8=192&lt;/math&gt;. However, in doing so, we have committed an egregious error. We have in fact over counted the cases in which the fourth song is liked by two girls but not the third.<br /> <br /> We proceed again by ignoring the cases in which the fourth song is liked by two girls but not the third. There are &lt;math&gt;4\cdot3\cdot2\cdot5=120&lt;/math&gt; of these cases. However, in doing so, we have committed yet another egregious error. The cases in which the fourth song is liked by two girls but not the third have not been accounted for!<br /> <br /> In performing our past two calculations, we have, however, established that the answer has a lower bound of &lt;math&gt;120&lt;/math&gt; and an upper bound of &lt;math&gt;192&lt;/math&gt;. As &lt;math&gt;(\textrm{\textbf{B}})&lt;/math&gt; is the only answer within these bounds, we conclude that the answer must be &lt;math&gt;(\textrm{\textbf{B}})&lt;/math&gt;.<br /> <br /> === Solution 3: A Different Way of Looking at Solution 1===<br /> Let &lt;math&gt;AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt; denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let &lt;math&gt;A, B, J,&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least &lt;math&gt;1\: AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt;, they must be &lt;math&gt;3&lt;/math&gt; songs out of the &lt;math&gt;4&lt;/math&gt; that Amy, Beth, and Jo listened to. The fourth song can be of any type &lt;math&gt;N, A, B, J, AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt; (there is no &lt;math&gt;ABJ&lt;/math&gt; because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange &lt;math&gt;AB, BJ, AJ&lt;/math&gt;, and a song from the set &lt;math&gt;\{N, A, B, J, AB, BJ, AJ\}&lt;/math&gt;.<br /> <br /> Case 1: Fourth song = &lt;math&gt;N, A, B, J &lt;/math&gt;<br /> <br /> Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.<br /> <br /> Number of ways to rearrange = &lt;math&gt;(4!)&lt;/math&gt; rearrangements for each choice &lt;math&gt;*\: 4&lt;/math&gt; choices = &lt;math&gt;96&lt;/math&gt;.<br /> <br /> Case 2: Fourth song = &lt;math&gt;AB, BJ, AJ&lt;/math&gt;<br /> <br /> Note that in Case &lt;math&gt;2&lt;/math&gt;, all three of the choices for the fourth song repeat somewhere in the first three songs.<br /> <br /> Number of ways to rearrange = &lt;math&gt;(4!/2!)&lt;/math&gt; rearrangements for each choice &lt;math&gt;*\: 3&lt;/math&gt; choices = &lt;math&gt;36&lt;/math&gt;.<br /> <br /> &lt;math&gt;96 + 36 = \boxed{\textbf{(B)} \: 132}&lt;/math&gt;.<br /> <br /> <br /> === Solution 4: Another Different Way of Looking at Solution 1 &amp; 3 ===<br /> <br /> There are &lt;math&gt;\binom{4}{3}&lt;/math&gt; ways to choose the three songs that are liked by the three pairs of girls.<br /> <br /> There are &lt;math&gt;3!&lt;/math&gt; ways to determine how the three songs are liked, or which song is liked by which pair of girls.<br /> <br /> In total, there are &lt;math&gt;\binom{4}{3}\cdot3!&lt;/math&gt; possibilities for the first &lt;math&gt;3&lt;/math&gt; songs.<br /> <br /> There are &lt;math&gt;3&lt;/math&gt; cases for the 4th song, call it song D.<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: D is disliked by all &lt;math&gt;3&lt;/math&gt; girls (&quot;_&quot;;) ==&gt; there is only &lt;math&gt;1&lt;/math&gt; possibility.<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: D is liked by exactly &lt;math&gt;1&lt;/math&gt; girl ==&gt; there are &lt;math&gt;3&lt;/math&gt; possibility.<br /> <br /> Case &lt;math&gt;3&lt;/math&gt;: D is liked by exactly &lt;math&gt;2&lt;/math&gt; girls ==&gt; there are &lt;math&gt;3&lt;/math&gt; pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first &lt;math&gt;3&lt;/math&gt; songs liked by the same girls.<br /> <br /> Counting the overlaps, there are &lt;math&gt;3&lt;/math&gt; ways to choose the pair with overlaps and &lt;math&gt;4\cdot3=12&lt;/math&gt; ways to choose what the other &lt;math&gt;2&lt;/math&gt; pairs like independently. In total, there are &lt;math&gt;3\cdot12=36&lt;/math&gt; overlapped possibilities.<br /> <br /> Finally, there are &lt;math&gt;\binom{4}{3}\cdot3!\cdot(3+1+3)-36=132&lt;/math&gt; ways for the songs to be likely by the girls. &lt;math&gt;\boxed{\mathrm{(B)}}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=23|num-a=25}}<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=15|num-a=17}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_16&diff=109557 2012 AMC 12B Problems/Problem 16 2019-08-31T03:34:42Z <p>Mathlover66: /* Solution 4: Another Different Way of Looking at Solution 1 &amp; 3 */</p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #16]] and [[2012 AMC 10B Problems|2012 AMC 10B #24]]}}<br /> <br /> == Problem==<br /> Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 &lt;/math&gt;<br /> <br /> <br /> <br /> ==Solutions==<br /> <br /> === Solution 1===<br /> Let the ordered triple &lt;math&gt;(a,b,c)&lt;/math&gt; denote that &lt;math&gt;a&lt;/math&gt; songs are liked by Amy and Beth, &lt;math&gt;b&lt;/math&gt; songs by Beth and Jo, and &lt;math&gt;c&lt;/math&gt; songs by Jo and Amy. We claim that the only possible triples are &lt;math&gt;(1,1,1), (2,1,1), (1,2,1)(1,1,2)&lt;/math&gt;. <br /> <br /> To show this, observe these are all valid conditions. Second, note that none of &lt;math&gt;a,b,c&lt;/math&gt; can be bigger than 3. Suppose otherwise, that &lt;math&gt;a = 3&lt;/math&gt;. Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to be at least 1. In fact, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to equal 1, otherwise there will be a song liked by all three. Suppose &lt;math&gt;b = 1&lt;/math&gt;. Then we must have &lt;math&gt;c=0&lt;/math&gt; since no song is liked by all three girls, a contradiction.<br /> <br /> '''Case 1''':How many ways are there for &lt;math&gt;(a,b,c)&lt;/math&gt; to equal &lt;math&gt;(1,1,1)&lt;/math&gt;? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So &lt;math&gt;(a,b,c)=(1,1,1)&lt;/math&gt; in &lt;math&gt;4\cdot3\cdot2\cdot4 = 96&lt;/math&gt; ways.<br /> <br /> '''Case 2''':To find the number of ways for &lt;math&gt;(a,b,c) = (2,1,1)&lt;/math&gt;, observe there are &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are &lt;math&gt;6\cdot2\cdot3=36&lt;/math&gt; ways for the girls to like the songs.<br /> <br /> That gives a total of &lt;math&gt;96 + 36 = \boxed{132}&lt;/math&gt; ways for the girls to like the songs, so the answer is &lt;math&gt;(\textrm{\textbf{B}})&lt;/math&gt;.<br /> <br /> === Solution 2 (Answer Choices)===<br /> We begin by noticing that there are four ways to assign a song liked by both Amy and Beth, three ways to assign a song liked by both Amy and Jo (because Jo may not like the song liked by both Amy and Beth), and two ways to assign a song liked by both Beth and Jo (because both Beth and Jo may not like the song liked by the previous pairs. Additionally, there are &lt;math&gt;2\cdot2\cdot2=8&lt;/math&gt; ways to assign song preferences for the fourth song. Multiplying, we obtain an answer of &lt;math&gt;4\cdot3\cdot2\cdot8=192&lt;/math&gt;. However, in doing so, we have committed an egregious error. We have in fact over counted the cases in which the fourth song is liked by two girls but not the third.<br /> <br /> We proceed again by ignoring the cases in which the fourth song is liked by two girls but not the third. There are &lt;math&gt;4\cdot3\cdot2\cdot5=120&lt;/math&gt; of these cases. However, in doing so, we have committed yet another egregious error. The cases in which the fourth song is liked by two girls but not the third have not been accounted for!<br /> <br /> In performing our past two calculations, we have, however, established that the answer has a lower bound of &lt;math&gt;120&lt;/math&gt; and an upper bound of &lt;math&gt;192&lt;/math&gt;. As &lt;math&gt;(\textrm{\textbf{B}})&lt;/math&gt; is the only answer within these bounds, we conclude that the answer must be &lt;math&gt;(\textrm{\textbf{B}})&lt;/math&gt;.<br /> <br /> === Solution 3: A Different Way of Looking at Solution 1===<br /> Let &lt;math&gt;AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt; denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let &lt;math&gt;A, B, J,&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least &lt;math&gt;1\: AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt;, they must be &lt;math&gt;3&lt;/math&gt; songs out of the &lt;math&gt;4&lt;/math&gt; that Amy, Beth, and Jo listened to. The fourth song can be of any type &lt;math&gt;N, A, B, J, AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt; (there is no &lt;math&gt;ABJ&lt;/math&gt; because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange &lt;math&gt;AB, BJ, AJ&lt;/math&gt;, and a song from the set &lt;math&gt;\{N, A, B, J, AB, BJ, AJ\}&lt;/math&gt;.<br /> <br /> Case 1: Fourth song = &lt;math&gt;N, A, B, J &lt;/math&gt;<br /> <br /> Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.<br /> <br /> Number of ways to rearrange = &lt;math&gt;(4!)&lt;/math&gt; rearrangements for each choice &lt;math&gt;*\: 4&lt;/math&gt; choices = &lt;math&gt;96&lt;/math&gt;.<br /> <br /> Case 2: Fourth song = &lt;math&gt;AB, BJ, AJ&lt;/math&gt;<br /> <br /> Note that in Case &lt;math&gt;2&lt;/math&gt;, all three of the choices for the fourth song repeat somewhere in the first three songs.<br /> <br /> Number of ways to rearrange = &lt;math&gt;(4!/2!)&lt;/math&gt; rearrangements for each choice &lt;math&gt;*\: 3&lt;/math&gt; choices = &lt;math&gt;36&lt;/math&gt;.<br /> <br /> &lt;math&gt;96 + 36 = \boxed{\textbf{(B)} \: 132}&lt;/math&gt;.<br /> <br /> <br /> === Solution 4: Another Different Way of Looking at Solution 1 &amp; 3 ===<br /> <br /> There are &lt;math&gt;\binom{4}{3}&lt;/math&gt; ways to choose the three songs that are liked by the three pairs of girls.<br /> <br /> There are &lt;math&gt;3!&lt;/math&gt; ways to determine how the three songs are liked, or which song is liked by which pair of girls.<br /> <br /> In total, there are &lt;math&gt;\binom{4}{3} \cdot 3!&lt;/math&gt; possibilities for the first &lt;math&gt;3&lt;/math&gt; songs.<br /> <br /> There are &lt;math&gt;3&lt;/math&gt; cases for the 4th song, call it song D.<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: D is disliked by all &lt;math&gt;3&lt;/math&gt; girls (&quot;_&quot;;) ==&gt; there is only &lt;math&gt;1&lt;/math&gt; possibility.<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: D is liked by exactly &lt;math&gt;1&lt;/math&gt; girl ==&gt; there are &lt;math&gt;3&lt;/math&gt; possibility.<br /> <br /> Case &lt;math&gt;3&lt;/math&gt;: D is liked by exactly &lt;math&gt;2&lt;/math&gt; girls ==&gt; there are &lt;math&gt;3&lt;/math&gt; pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first &lt;math&gt;3&lt;/math&gt; songs liked by the same girls.<br /> <br /> Counting the overlaps, there are &lt;math&gt;3&lt;/math&gt; ways to choose the pair with overlaps and &lt;math&gt;4\cdot3=12&lt;/math&gt; ways to choose what the other &lt;math&gt;2&lt;/math&gt; pairs like independently. In total, there are &lt;math&gt;3\cdot12=36&lt;/math&gt; overlapped possibilities.<br /> <br /> Finally, there are &lt;math&gt;\binom{4}{3}3!(3+1+3)-36=132&lt;/math&gt; ways for the songs to be likely by the girls. &lt;math&gt;\boxed{\mathrm{(B)}}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=23|num-a=25}}<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=15|num-a=17}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_16&diff=109556 2012 AMC 12B Problems/Problem 16 2019-08-31T03:33:39Z <p>Mathlover66: /* Solution 4: Another Different Way of Looking at Solution 1 &amp; 3 */</p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #16]] and [[2012 AMC 10B Problems|2012 AMC 10B #24]]}}<br /> <br /> == Problem==<br /> Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 &lt;/math&gt;<br /> <br /> <br /> <br /> ==Solutions==<br /> <br /> === Solution 1===<br /> Let the ordered triple &lt;math&gt;(a,b,c)&lt;/math&gt; denote that &lt;math&gt;a&lt;/math&gt; songs are liked by Amy and Beth, &lt;math&gt;b&lt;/math&gt; songs by Beth and Jo, and &lt;math&gt;c&lt;/math&gt; songs by Jo and Amy. We claim that the only possible triples are &lt;math&gt;(1,1,1), (2,1,1), (1,2,1)(1,1,2)&lt;/math&gt;. <br /> <br /> To show this, observe these are all valid conditions. Second, note that none of &lt;math&gt;a,b,c&lt;/math&gt; can be bigger than 3. Suppose otherwise, that &lt;math&gt;a = 3&lt;/math&gt;. Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to be at least 1. In fact, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to equal 1, otherwise there will be a song liked by all three. Suppose &lt;math&gt;b = 1&lt;/math&gt;. Then we must have &lt;math&gt;c=0&lt;/math&gt; since no song is liked by all three girls, a contradiction.<br /> <br /> '''Case 1''':How many ways are there for &lt;math&gt;(a,b,c)&lt;/math&gt; to equal &lt;math&gt;(1,1,1)&lt;/math&gt;? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So &lt;math&gt;(a,b,c)=(1,1,1)&lt;/math&gt; in &lt;math&gt;4\cdot3\cdot2\cdot4 = 96&lt;/math&gt; ways.<br /> <br /> '''Case 2''':To find the number of ways for &lt;math&gt;(a,b,c) = (2,1,1)&lt;/math&gt;, observe there are &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are &lt;math&gt;6\cdot2\cdot3=36&lt;/math&gt; ways for the girls to like the songs.<br /> <br /> That gives a total of &lt;math&gt;96 + 36 = \boxed{132}&lt;/math&gt; ways for the girls to like the songs, so the answer is &lt;math&gt;(\textrm{\textbf{B}})&lt;/math&gt;.<br /> <br /> === Solution 2 (Answer Choices)===<br /> We begin by noticing that there are four ways to assign a song liked by both Amy and Beth, three ways to assign a song liked by both Amy and Jo (because Jo may not like the song liked by both Amy and Beth), and two ways to assign a song liked by both Beth and Jo (because both Beth and Jo may not like the song liked by the previous pairs. Additionally, there are &lt;math&gt;2\cdot2\cdot2=8&lt;/math&gt; ways to assign song preferences for the fourth song. Multiplying, we obtain an answer of &lt;math&gt;4\cdot3\cdot2\cdot8=192&lt;/math&gt;. However, in doing so, we have committed an egregious error. We have in fact over counted the cases in which the fourth song is liked by two girls but not the third.<br /> <br /> We proceed again by ignoring the cases in which the fourth song is liked by two girls but not the third. There are &lt;math&gt;4\cdot3\cdot2\cdot5=120&lt;/math&gt; of these cases. However, in doing so, we have committed yet another egregious error. The cases in which the fourth song is liked by two girls but not the third have not been accounted for!<br /> <br /> In performing our past two calculations, we have, however, established that the answer has a lower bound of &lt;math&gt;120&lt;/math&gt; and an upper bound of &lt;math&gt;192&lt;/math&gt;. As &lt;math&gt;(\textrm{\textbf{B}})&lt;/math&gt; is the only answer within these bounds, we conclude that the answer must be &lt;math&gt;(\textrm{\textbf{B}})&lt;/math&gt;.<br /> <br /> === Solution 3: A Different Way of Looking at Solution 1===<br /> Let &lt;math&gt;AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt; denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let &lt;math&gt;A, B, J,&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least &lt;math&gt;1\: AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt;, they must be &lt;math&gt;3&lt;/math&gt; songs out of the &lt;math&gt;4&lt;/math&gt; that Amy, Beth, and Jo listened to. The fourth song can be of any type &lt;math&gt;N, A, B, J, AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt; (there is no &lt;math&gt;ABJ&lt;/math&gt; because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange &lt;math&gt;AB, BJ, AJ&lt;/math&gt;, and a song from the set &lt;math&gt;\{N, A, B, J, AB, BJ, AJ\}&lt;/math&gt;.<br /> <br /> Case 1: Fourth song = &lt;math&gt;N, A, B, J &lt;/math&gt;<br /> <br /> Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.<br /> <br /> Number of ways to rearrange = &lt;math&gt;(4!)&lt;/math&gt; rearrangements for each choice &lt;math&gt;*\: 4&lt;/math&gt; choices = &lt;math&gt;96&lt;/math&gt;.<br /> <br /> Case 2: Fourth song = &lt;math&gt;AB, BJ, AJ&lt;/math&gt;<br /> <br /> Note that in Case &lt;math&gt;2&lt;/math&gt;, all three of the choices for the fourth song repeat somewhere in the first three songs.<br /> <br /> Number of ways to rearrange = &lt;math&gt;(4!/2!)&lt;/math&gt; rearrangements for each choice &lt;math&gt;*\: 3&lt;/math&gt; choices = &lt;math&gt;36&lt;/math&gt;.<br /> <br /> &lt;math&gt;96 + 36 = \boxed{\textbf{(B)} \: 132}&lt;/math&gt;.<br /> <br /> <br /> === Solution 4: Another Different Way of Looking at Solution 1 &amp; 3 ===<br /> <br /> There are &lt;math&gt;\binom{4}{3}&lt;/math&gt; ways to choose the three songs that are liked by the three pairs of girls.<br /> <br /> There are &lt;math&gt;3!&lt;/math&gt; ways to determine how the three songs are liked, or which song is liked by which pair of girls.<br /> <br /> In total, there are &lt;math&gt;\binom{4}{3} &lt;/math&gt;\cdot&lt;math&gt; 3!&lt;/math&gt; possibilities for the first &lt;math&gt;3&lt;/math&gt; songs.<br /> <br /> There are &lt;math&gt;3&lt;/math&gt; cases for the 4th song, call it song D.<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: D is disliked by all &lt;math&gt;3&lt;/math&gt; girls (&quot;_&quot;;) ==&gt; there is only &lt;math&gt;1&lt;/math&gt; possibility.<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: D is liked by exactly &lt;math&gt;1&lt;/math&gt; girl ==&gt; there are &lt;math&gt;3&lt;/math&gt; possibility.<br /> <br /> Case &lt;math&gt;3&lt;/math&gt;: D is liked by exactly &lt;math&gt;2&lt;/math&gt; girls ==&gt; there are &lt;math&gt;3&lt;/math&gt; pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first &lt;math&gt;3&lt;/math&gt; songs liked by the same girls.<br /> <br /> Counting the overlaps, there are &lt;math&gt;3&lt;/math&gt; ways to choose the pair with overlaps and &lt;math&gt;4\cdot3=12&lt;/math&gt; ways to choose what the other &lt;math&gt;2&lt;/math&gt; pairs like independently. In total, there are &lt;math&gt;3\cdot12=36&lt;/math&gt; overlapped possibilities.<br /> <br /> Finally, there are &lt;math&gt;\binom{4}{3}3!(3+1+3)-36=132&lt;/math&gt; ways for the songs to be likely by the girls. &lt;math&gt;\boxed{\mathrm{(B)}}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=23|num-a=25}}<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=15|num-a=17}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_22&diff=108848 2010 AMC 10B Problems/Problem 22 2019-08-14T02:18:10Z <p>Mathlover66: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1930 \qquad \textbf{(B)}\ 1931 \qquad \textbf{(C)}\ 1932 \qquad \textbf{(D)}\ 1933 \qquad \textbf{(E)}\ 1934&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> We can count the total number of ways to distribute the candies (ignoring the restrictions), and then subtract the overcount to get the answer.<br /> <br /> Each candy has three choices; it can go in any of the three bags.<br /> <br /> Since there are seven candies, that makes the '''total distributions''' &lt;math&gt;3^7=2187&lt;/math&gt;<br /> <br /> <br /> To find the overcount, we calculate the number of invalid distributions: the red or blue bag is empty. <br /> <br /> The number of distributions such that the red bag is empty is equal to &lt;math&gt;2^7&lt;/math&gt;, since it's equivalent to distributing the &lt;math&gt;7&lt;/math&gt; candies into &lt;math&gt;2&lt;/math&gt; bags.<br /> <br /> We know that the number of distributions with the blue bag is empty will be the same number because of the symmetry, so it's also &lt;math&gt;2^7&lt;/math&gt;. <br /> <br /> The case where both the red and the blue bags are empty (all &lt;math&gt;7&lt;/math&gt; candies are in the white bag) are included in both of the above calculations, and this case has only &lt;math&gt;1&lt;/math&gt; distribution.<br /> <br /> '''The total overcount is &lt;math&gt;2^7+2^7-1=2^8-1&lt;/math&gt;'''<br /> <br /> <br /> The final answer will be &lt;math&gt;\text{total}-\text{overcount}=2187-(2^8-1) = 2187-256+1=1931+1=\boxed{\textbf{(C)}\ 1932}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> We can use to our advantage the answer choices &lt;math&gt;\text{AMC}&lt;/math&gt; has given us, and eliminate the obvious wrong answers.<br /> <br /> We can first figure out how many ways there are to take two candies from seven distinct candies to place them into the red/blue bags: &lt;math&gt;7\cdot 6=42&lt;/math&gt;.<br /> <br /> Now we can look at the answer choices to find out which ones are divisible by &lt;math&gt;42&lt;/math&gt;, since the total number of combinations must be &lt;math&gt;42&lt;/math&gt; multiplied by some other number.<br /> <br /> Since answers A, B, D, and E are not divisible by 3 (divisor of 42), the answer must be &lt;math&gt;\boxed{\textbf{(C)}\ 1932}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_22&diff=108847 2010 AMC 10B Problems/Problem 22 2019-08-14T02:17:27Z <p>Mathlover66: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1930 \qquad \textbf{(B)}\ 1931 \qquad \textbf{(C)}\ 1932 \qquad \textbf{(D)}\ 1933 \qquad \textbf{(E)}\ 1934&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> We can count the total number of ways to distribute the candies (ignoring the restrictions), and then subtract the overcount to get the answer.<br /> <br /> Each candy has three choices; it can go in any of the three bags.<br /> <br /> Since there are seven candies, that makes the '''total distributions''' &lt;math&gt;3^7=2187&lt;/math&gt;<br /> <br /> <br /> To find the overcount, we calculate the number of invalid distributions: the red or blue bag is empty. <br /> <br /> The number of distributions such that the red bag is empty is equal to &lt;math&gt;2^7&lt;/math&gt;, since it's equivalent to distributing the &lt;math&gt;7&lt;/math&gt; candies into &lt;math&gt;2&lt;/math&gt; bags.<br /> <br /> We know that the number of distributions with the blue bag is empty will be the same number because of the symmetry, so it's also &lt;math&gt;2^7&lt;/math&gt;. <br /> <br /> The case where both the red and the blue bags are empty (all &lt;math&gt;7&lt;/math&gt; candies are in the white bag) are included in both of the above calculations, and this case has only &lt;math&gt;1&lt;/math&gt; distribution.<br /> <br /> '''The total overcount is &lt;math&gt;2^7+2^7-1=2^8-1&lt;/math&gt;'''<br /> <br /> <br /> The final answer will be &lt;math&gt;\text{total}-\text{overcount}=2187-(2^8-1) = 2187-256+1=1931+1=\boxed{\textbf{(C)}\ 1932}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> We can use to our advantage the answer choices AMC has given us, and eliminate the obvious wrong answers.<br /> <br /> We can first figure out how many ways there are to take two candies from seven distinct candies to place them into the red/blue bags: &lt;math&gt;7\cdot 6=42&lt;/math&gt;.<br /> <br /> Now we can look at the answer choices to find out which ones are divisible by &lt;math&gt;42&lt;/math&gt;, since the total number of combinations must be &lt;math&gt;42&lt;/math&gt; multiplied by some other number.<br /> <br /> Since answers A, B, D, and E are not divisible by 3 (divisor of 42), the answer must be &lt;math&gt;\boxed{\textbf{(C)}\ 1932}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_16&diff=108846 2010 AMC 10B Problems/Problem 16 2019-08-14T02:12:17Z <p>Mathlover66: </p> <hr /> <div>== Problem==<br /> A square of side length &lt;math&gt;1&lt;/math&gt; and a circle of radius &lt;math&gt;\dfrac{\sqrt{3}}{3}&lt;/math&gt; share the same center. What is the area inside the circle, but outside the square?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{\pi}{3}-1 \qquad \textbf{(B)}\ \dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3} \qquad \textbf{(C)}\ \dfrac{\pi}{18} \qquad \textbf{(D)}\ \dfrac{1}{4} \qquad \textbf{(E)}\ \dfrac{2\pi}{9}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The radius of the circle is &lt;math&gt;\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}&lt;/math&gt;. Half the diagonal of the square is &lt;math&gt;\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2} = \sqrt{\frac12}&lt;/math&gt;. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle. Therefore the picture will look something like this:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(5cm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=3;<br /> <br /> real r=sqrt(1/3);<br /> pair O=(0,0);<br /> pair W=(0.5,0.5), X=(0.5,-0.5), Y=(-0.5,-0.5), Z=(-0.5,0.5);<br /> pair A=(-sqrt(1/12),0.5), B=(sqrt(1/12),0.5);<br /> pair V=(0,0.5);<br /> path outer=Circle(O,r);<br /> draw(outer);<br /> draw(W--X--Y--Z--cycle);<br /> draw(O--A);<br /> draw(O--B);<br /> draw(V--O);<br /> <br /> pair[] ps={A,B,V,O};<br /> dot(ps);<br /> <br /> label(&quot;$O$&quot;,O,SW);<br /> label(&quot;$\frac{\sqrt{3}}{3}$&quot;,O--B,SE);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$X$&quot;,V,NW);<br /> label(&quot;$a$&quot;,B--V,S);<br /> label(&quot;$\frac12$&quot;,O--V,W);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Then we proceed to find: 4 &lt;math&gt;\cdot&lt;/math&gt; (area of sector marked off by the two radii - area of the triangle with sides on the square and the two radii).<br /> <br /> First we realize that the radius perpendicular to the side of the square between the two radii marking off the sector splits &lt;math&gt;AB&lt;/math&gt; in half. Let this half-length be &lt;math&gt;a&lt;/math&gt;. Also note that &lt;math&gt;OX=\frac12&lt;/math&gt; because it is half the sidelength of the square. Because this is a right triangle, we can use the [[Pythagorean Theorem]] to solve for &lt;math&gt;a.&lt;/math&gt;<br /> <br /> &lt;cmath&gt;a^2+\left( \frac12 \right) ^2 = \left( \frac{\sqrt{3}}{3} \right) ^2&lt;/cmath&gt;<br /> <br /> Solving, &lt;math&gt;a= \frac{\sqrt{3}}{6}&lt;/math&gt; and &lt;math&gt;2a=\frac{\sqrt{3}}{3}&lt;/math&gt;. Since &lt;math&gt;AB=AO=BO&lt;/math&gt;, &lt;math&gt;\triangle AOB&lt;/math&gt; is an equilateral triangle and the central angle is &lt;math&gt;60^{\circ}&lt;/math&gt;. Therefore the sector has an area &lt;math&gt;\pi \left( \frac{\sqrt{3}}{3} \right) ^2 \left( \frac{60}{360} \right) = \frac{\pi}{18}&lt;/math&gt;.<br /> <br /> Now we turn to the triangle. Since it is equilateral, we can use the formula for the [[area of an equilateral triangle]] which is<br /> <br /> &lt;cmath&gt;\frac{s^2\sqrt{3}}{4} = \frac{\frac13 \sqrt{3}}{4} = \frac{\sqrt{3}}{12}&lt;/cmath&gt;<br /> <br /> Putting it together, we get the answer to be &lt;math&gt;4 \left( \frac{\pi}{18}-\frac{\sqrt{3}}{12} \right)= \boxed{\textbf{(B)}\ \frac{2\pi}{9}-\frac{\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2010|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_16&diff=108845 2010 AMC 10B Problems/Problem 16 2019-08-14T02:11:53Z <p>Mathlover66: /* Solution 2(Intense Sketch) */</p> <hr /> <div>== Problem==<br /> A square of side length &lt;math&gt;1&lt;/math&gt; and a circle of radius &lt;math&gt;\dfrac{\sqrt{3}}{3}&lt;/math&gt; share the same center. What is the area inside the circle, but outside the square?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{\pi}{3}-1 \qquad \textbf{(B)}\ \dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3} \qquad \textbf{(C)}\ \dfrac{\pi}{18} \qquad \textbf{(D)}\ \dfrac{1}{4} \qquad \textbf{(E)}\ \dfrac{2\pi}{9}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The radius of the circle is &lt;math&gt;\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}&lt;/math&gt;. Half the diagonal of the square is &lt;math&gt;\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2} = \sqrt{\frac12}&lt;/math&gt;. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle. Therefore the picture will look something like this:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(5cm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=3;<br /> <br /> real r=sqrt(1/3);<br /> pair O=(0,0);<br /> pair W=(0.5,0.5), X=(0.5,-0.5), Y=(-0.5,-0.5), Z=(-0.5,0.5);<br /> pair A=(-sqrt(1/12),0.5), B=(sqrt(1/12),0.5);<br /> pair V=(0,0.5);<br /> path outer=Circle(O,r);<br /> draw(outer);<br /> draw(W--X--Y--Z--cycle);<br /> draw(O--A);<br /> draw(O--B);<br /> draw(V--O);<br /> <br /> pair[] ps={A,B,V,O};<br /> dot(ps);<br /> <br /> label(&quot;$O$&quot;,O,SW);<br /> label(&quot;$\frac{\sqrt{3}}{3}$&quot;,O--B,SE);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$X$&quot;,V,NW);<br /> label(&quot;$a$&quot;,B--V,S);<br /> label(&quot;$\frac12$&quot;,O--V,W);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Then we proceed to find: 4 &lt;math&gt;\cdot&lt;/math&gt; (area of sector marked off by the two radii - area of the triangle with sides on the square and the two radii).<br /> <br /> First we realize that the radius perpendicular to the side of the square between the two radii marking off the sector splits &lt;math&gt;AB&lt;/math&gt; in half. Let this half-length be &lt;math&gt;a&lt;/math&gt;. Also note that &lt;math&gt;OX=\frac12&lt;/math&gt; because it is half the sidelength of the square. Because this is a right triangle, we can use the [[Pythagorean Theorem]] to solve for &lt;math&gt;a.&lt;/math&gt;<br /> <br /> &lt;cmath&gt;a^2+\left( \frac12 \right) ^2 = \left( \frac{\sqrt{3}}{3} \right) ^2&lt;/cmath&gt;<br /> <br /> Solving, &lt;math&gt;a= \frac{\sqrt{3}}{6}&lt;/math&gt; and &lt;math&gt;2a=\frac{\sqrt{3}}{3}&lt;/math&gt;. Since &lt;math&gt;AB=AO=BO&lt;/math&gt;, &lt;math&gt;\triangle AOB&lt;/math&gt; is an equilateral triangle and the central angle is &lt;math&gt;60^{\circ}&lt;/math&gt;. Therefore the sector has an area &lt;math&gt;\pi \left( \frac{\sqrt{3}}{3} \right) ^2 \left( \frac{60}{360} \right) = \frac{\pi}{18}&lt;/math&gt;.<br /> <br /> Now we turn to the triangle. Since it is equilateral, we can use the formula for the [[area of an equilateral triangle]] which is<br /> <br /> &lt;cmath&gt;\frac{s^2\sqrt{3}}{4} = \frac{\frac13 \sqrt{3}}{4} = \frac{\sqrt{3}}{12}&lt;/cmath&gt;<br /> <br /> Putting it together, we get the answer to be &lt;math&gt;4 \left( \frac{\pi}{18}-\frac{\sqrt{3}}{12} \right)= \boxed{\textbf{(B)}\ \frac{2\pi}{9}-\frac{\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 2 (Intense Sketch)==<br /> After finding the equilateral triangle, you know there is going to be a factor of &lt;math&gt;\sqrt{3}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2010|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_16&diff=108844 2010 AMC 10B Problems/Problem 16 2019-08-14T02:11:29Z <p>Mathlover66: /* Solution 1 */</p> <hr /> <div>== Problem==<br /> A square of side length &lt;math&gt;1&lt;/math&gt; and a circle of radius &lt;math&gt;\dfrac{\sqrt{3}}{3}&lt;/math&gt; share the same center. What is the area inside the circle, but outside the square?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{\pi}{3}-1 \qquad \textbf{(B)}\ \dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3} \qquad \textbf{(C)}\ \dfrac{\pi}{18} \qquad \textbf{(D)}\ \dfrac{1}{4} \qquad \textbf{(E)}\ \dfrac{2\pi}{9}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The radius of the circle is &lt;math&gt;\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}&lt;/math&gt;. Half the diagonal of the square is &lt;math&gt;\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2} = \sqrt{\frac12}&lt;/math&gt;. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle. Therefore the picture will look something like this:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(5cm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=3;<br /> <br /> real r=sqrt(1/3);<br /> pair O=(0,0);<br /> pair W=(0.5,0.5), X=(0.5,-0.5), Y=(-0.5,-0.5), Z=(-0.5,0.5);<br /> pair A=(-sqrt(1/12),0.5), B=(sqrt(1/12),0.5);<br /> pair V=(0,0.5);<br /> path outer=Circle(O,r);<br /> draw(outer);<br /> draw(W--X--Y--Z--cycle);<br /> draw(O--A);<br /> draw(O--B);<br /> draw(V--O);<br /> <br /> pair[] ps={A,B,V,O};<br /> dot(ps);<br /> <br /> label(&quot;$O$&quot;,O,SW);<br /> label(&quot;$\frac{\sqrt{3}}{3}$&quot;,O--B,SE);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$X$&quot;,V,NW);<br /> label(&quot;$a$&quot;,B--V,S);<br /> label(&quot;$\frac12$&quot;,O--V,W);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Then we proceed to find: 4 &lt;math&gt;\cdot&lt;/math&gt; (area of sector marked off by the two radii - area of the triangle with sides on the square and the two radii).<br /> <br /> First we realize that the radius perpendicular to the side of the square between the two radii marking off the sector splits &lt;math&gt;AB&lt;/math&gt; in half. Let this half-length be &lt;math&gt;a&lt;/math&gt;. Also note that &lt;math&gt;OX=\frac12&lt;/math&gt; because it is half the sidelength of the square. Because this is a right triangle, we can use the [[Pythagorean Theorem]] to solve for &lt;math&gt;a.&lt;/math&gt;<br /> <br /> &lt;cmath&gt;a^2+\left( \frac12 \right) ^2 = \left( \frac{\sqrt{3}}{3} \right) ^2&lt;/cmath&gt;<br /> <br /> Solving, &lt;math&gt;a= \frac{\sqrt{3}}{6}&lt;/math&gt; and &lt;math&gt;2a=\frac{\sqrt{3}}{3}&lt;/math&gt;. Since &lt;math&gt;AB=AO=BO&lt;/math&gt;, &lt;math&gt;\triangle AOB&lt;/math&gt; is an equilateral triangle and the central angle is &lt;math&gt;60^{\circ}&lt;/math&gt;. Therefore the sector has an area &lt;math&gt;\pi \left( \frac{\sqrt{3}}{3} \right) ^2 \left( \frac{60}{360} \right) = \frac{\pi}{18}&lt;/math&gt;.<br /> <br /> Now we turn to the triangle. Since it is equilateral, we can use the formula for the [[area of an equilateral triangle]] which is<br /> <br /> &lt;cmath&gt;\frac{s^2\sqrt{3}}{4} = \frac{\frac13 \sqrt{3}}{4} = \frac{\sqrt{3}}{12}&lt;/cmath&gt;<br /> <br /> Putting it together, we get the answer to be &lt;math&gt;4 \left( \frac{\pi}{18}-\frac{\sqrt{3}}{12} \right)= \boxed{\textbf{(B)}\ \frac{2\pi}{9}-\frac{\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 2(Intense Sketch)==<br /> After finding the equilateral triangle, you know there is going to be a factor of &lt;math&gt;\sqrt{3}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;<br /> ==See Also==<br /> {{AMC10 box|year=2010|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_16&diff=108843 2010 AMC 10B Problems/Problem 16 2019-08-14T02:11:00Z <p>Mathlover66: /* Solution 1 */</p> <hr /> <div>== Problem==<br /> A square of side length &lt;math&gt;1&lt;/math&gt; and a circle of radius &lt;math&gt;\dfrac{\sqrt{3}}{3}&lt;/math&gt; share the same center. What is the area inside the circle, but outside the square?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{\pi}{3}-1 \qquad \textbf{(B)}\ \dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3} \qquad \textbf{(C)}\ \dfrac{\pi}{18} \qquad \textbf{(D)}\ \dfrac{1}{4} \qquad \textbf{(E)}\ \dfrac{2\pi}{9}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The radius of the circle is &lt;math&gt;\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}&lt;/math&gt;. Half the diagonal of the square is &lt;math&gt;\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2} = \sqrt{\frac12}&lt;/math&gt;. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle. Therefore the picture will look something like this:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(5cm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=3;<br /> <br /> real r=sqrt(1/3);<br /> pair O=(0,0);<br /> pair W=(0.5,0.5), X=(0.5,-0.5), Y=(-0.5,-0.5), Z=(-0.5,0.5);<br /> pair A=(-sqrt(1/12),0.5), B=(sqrt(1/12),0.5);<br /> pair V=(0,0.5);<br /> path outer=Circle(O,r);<br /> draw(outer);<br /> draw(W--X--Y--Z--cycle);<br /> draw(O--A);<br /> draw(O--B);<br /> draw(V--O);<br /> <br /> pair[] ps={A,B,V,O};<br /> dot(ps);<br /> <br /> label(&quot;$O$&quot;,O,SW);<br /> label(&quot;$\frac{\sqrt{3}}{3}$&quot;,O--B,SE);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$X$&quot;,V,NW);<br /> label(&quot;$a$&quot;,B--V,S);<br /> label(&quot;$\frac12$&quot;,O--V,W);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Then we proceed to find: 4 &lt;math&gt;\cdot (area of sector marked off by the two radii - area of the triangle with sides on the square and the two radii).<br /> <br /> First we realize that the radius perpendicular to the side of the square between the two radii marking off the sector splits &lt;/math&gt;AB&lt;math&gt; in half. Let this half-length be &lt;/math&gt;a&lt;math&gt;. Also note that &lt;/math&gt;OX=\frac12&lt;math&gt; because it is half the sidelength of the square. Because this is a right triangle, we can use the [[Pythagorean Theorem]] to solve for &lt;/math&gt;a.&lt;math&gt;<br /> <br /> &lt;cmath&gt;a^2+\left( \frac12 \right) ^2 = \left( \frac{\sqrt{3}}{3} \right) ^2&lt;/cmath&gt;<br /> <br /> Solving, &lt;/math&gt;a= \frac{\sqrt{3}}{6}&lt;math&gt; and &lt;/math&gt;2a=\frac{\sqrt{3}}{3}&lt;math&gt;. Since &lt;/math&gt;AB=AO=BO&lt;math&gt;, &lt;/math&gt;\triangle AOB&lt;math&gt; is an equilateral triangle and the central angle is &lt;/math&gt;60^{\circ}&lt;math&gt;. Therefore the sector has an area &lt;/math&gt;\pi \left( \frac{\sqrt{3}}{3} \right) ^2 \left( \frac{60}{360} \right) = \frac{\pi}{18}&lt;math&gt;.<br /> <br /> Now we turn to the triangle. Since it is equilateral, we can use the formula for the [[area of an equilateral triangle]] which is<br /> <br /> &lt;cmath&gt;\frac{s^2\sqrt{3}}{4} = \frac{\frac13 \sqrt{3}}{4} = \frac{\sqrt{3}}{12}&lt;/cmath&gt;<br /> <br /> Putting it together, we get the answer to be &lt;/math&gt;4 \left( \frac{\pi}{18}-\frac{\sqrt{3}}{12} \right)= \boxed{\textbf{(B)}\ \frac{2\pi}{9}-\frac{\sqrt{3}}{3}}$<br /> <br /> ==Solution 2(Intense Sketch)==<br /> After finding the equilateral triangle, you know there is going to be a factor of &lt;math&gt;\sqrt{3}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;<br /> ==See Also==<br /> {{AMC10 box|year=2010|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_11&diff=102651 2019 AMC 10B Problems/Problem 11 2019-02-15T00:10:58Z <p>Mathlover66: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar 1 the ratio of blue to green marbles is 9:1, and the ratio of blue to green marbles in Jar 2 is 8:1. There are 95 green marbles in all. How many more blue marbles are in Jar 1 than in Jar 2?<br /> <br /> &lt;math&gt;\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50&lt;/math&gt;<br /> <br /> ==Solution==<br /> Call the amount of marbles in each jar &lt;math&gt;x&lt;/math&gt;, because they are equivalent. Thus, &lt;math&gt;x/10&lt;/math&gt; is the amount of green marbles in &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;x/9&lt;/math&gt; is the amount of green marbles in &lt;math&gt;2&lt;/math&gt;. &lt;math&gt;x/9+x/10=19x/90&lt;/math&gt;, &lt;math&gt;19x/90=95&lt;/math&gt;, and &lt;math&gt;x=450&lt;/math&gt; marbles in each jar. Because the &lt;math&gt;9/10&lt;/math&gt; is the amount of blue marbles in jar &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;8/9&lt;/math&gt; is the amount of blue marbles in jar &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;9x/10-8x/9=x/90&lt;/math&gt;, so there must be &lt;math&gt;5&lt;/math&gt; more marbles in jar &lt;math&gt;1&lt;/math&gt; than jar &lt;math&gt;2&lt;/math&gt;. The answer is &lt;math&gt;(A)&lt;/math&gt;<br /> <br /> (Edited by Lcz)<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Mathlover66 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_11&diff=102650 2019 AMC 10B Problems/Problem 11 2019-02-15T00:10:09Z <p>Mathlover66: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar 1 the ratio of blue to green marbles is 9:1, and the ratio of blue to green marbles in Jar 2 is 8:1. There are 95 green marbles in all. How many more blue marbles are in Jar 1 than in Jar 2?<br /> <br /> &lt;math&gt;\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50&lt;/math&gt;<br /> <br /> ==Solution==<br /> Call the amount of marbles in each jar &lt;math&gt;x&lt;/math&gt;, because they are equivalent. Thus, &lt;math&gt;x/10&lt;/math&gt; is the amount of green marbles in &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;x/9&lt;/math&gt; is the amount of green marbles in &lt;math&gt;2&lt;/math&gt;. &lt;math&gt;x/9+x/10=19x/90&lt;/math&gt;, &lt;math&gt;19x/90=95&lt;/math&gt;, and &lt;math&gt;x=450&lt;/math&gt; marbles in each jar. Because the &lt;math&gt;9/10&lt;/math&gt; is the amount of blue marbles in jar &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;8/9&lt;/math&gt; is the amount of blue marbles in jar &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;9x/10-8x/9=x/90&lt;/math&gt;, so there must be &lt;math&gt;5&lt;/math&gt; more marbles in jar &lt;math&gt;1&lt;/math&gt; than jar &lt;math&gt;2&lt;/math&gt;. The answer is &lt;math&gt;(A)&lt;/math&gt;<br /> <br /> (Edited by math*******)<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Mathlover66