https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Kkwang&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-11T10:56:52Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_17&diff=74754 2010 AMC 10B Problems/Problem 17 2016-01-22T03:33:59Z <p>Kkwang: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Every high school in the city of Euclid sent a team of &lt;math&gt;3&lt;/math&gt; students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed &lt;math&gt;37&lt;/math&gt;th and &lt;math&gt;64&lt;/math&gt;th, respectively. How many schools are in the city?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26&lt;/math&gt;<br /> <br /> == Solution ==<br /> Let the &lt;math&gt;n&lt;/math&gt; be the number of schools, &lt;math&gt;3n&lt;/math&gt; be the number of contestants, and &lt;math&gt;x&lt;/math&gt; be Andrea's place. Since the number of participants divided by three is the number of schools, &lt;math&gt;n\geq\frac{64}3=21\frac13&lt;/math&gt;. Andrea received a higher score than her teammates, so &lt;math&gt;x\leq36&lt;/math&gt;. Since &lt;math&gt;36&lt;/math&gt; is the maximum possible median, then &lt;math&gt;2*36-1=71&lt;/math&gt; is the maximum possible number of participants. Therefore, &lt;math&gt;3n\leq71\Rightarrow n\leq\frac{71}3=23\frac23&lt;/math&gt;. This yields the compound inequality: &lt;math&gt;21\frac13\leq n\leq<br /> 23\frac23&lt;/math&gt;. Since a set with an even number of elements has a median that is the average of the two middle terms, an occurrence that cannot happen in this situation, &lt;math&gt;n&lt;/math&gt; cannot be even. &lt;math&gt;\boxed{\textbf{(B)}\ 23}&lt;/math&gt; is the only other option.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;n&lt;/math&gt; = the number of schools that participated in the contest.<br /> <br /> Then &lt;math&gt;3n&lt;/math&gt; students participated in the contest.<br /> <br /> Since Andrea's score was the median score of all the students, the number of students in the contest must be odd - we know that no two students can have the same score (from the problem), and we know that if the number of students was even, the middle two scores would have to be the same to get a whole number for the median.<br /> <br /> So, we can now write an expression for the median score (Andrea's score) in terms of the number of students who participated in the contest (i.e. &lt;math&gt;3n&lt;/math&gt;):<br /> <br /> &lt;math&gt;\frac{3n+1}{2}&lt;/math&gt;<br /> <br /> Also, since we know that this expression must be smaller than &lt;math&gt;37&lt;/math&gt; (Andrea, whose score was the median, got a better score than Beth and Carla), we can write an inequality and solve it for &lt;math&gt;n&lt;/math&gt;:<br /> <br /> &lt;math&gt;\frac{3n+1}{2}&lt;37&lt;/math&gt;<br /> <br /> Solving this inequality for &lt;math&gt;n&lt;/math&gt;, we get that:<br /> <br /> &lt;math&gt;n&lt;\frac{37}{3} \mathrm{(which \quad is \quad a \quad little\quad over \quad 24)}&lt;/math&gt;<br /> <br /> So this eliminates answer choices &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. <br /> <br /> But we already know that &lt;math&gt;3n&lt;/math&gt; must be odd, implying that &lt;math&gt;n&lt;/math&gt; must also be odd! So at this point, the only odd answer choice is &lt;math&gt;\boxed{\textbf{(B)}\ 23}&lt;/math&gt;, and we are done.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2010|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_17&diff=74753 2010 AMC 10B Problems/Problem 17 2016-01-22T03:31:11Z <p>Kkwang: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Every high school in the city of Euclid sent a team of &lt;math&gt;3&lt;/math&gt; students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed &lt;math&gt;37&lt;/math&gt;th and &lt;math&gt;64&lt;/math&gt;th, respectively. How many schools are in the city?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26&lt;/math&gt;<br /> <br /> == Solution ==<br /> Let the &lt;math&gt;n&lt;/math&gt; be the number of schools, &lt;math&gt;3n&lt;/math&gt; be the number of contestants, and &lt;math&gt;x&lt;/math&gt; be Andrea's place. Since the number of participants divided by three is the number of schools, &lt;math&gt;n\geq\frac{64}3=21\frac13&lt;/math&gt;. Andrea received a higher score than her teammates, so &lt;math&gt;x\leq36&lt;/math&gt;. Since &lt;math&gt;36&lt;/math&gt; is the maximum possible median, then &lt;math&gt;2*36-1=71&lt;/math&gt; is the maximum possible number of participants. Therefore, &lt;math&gt;3n\leq71\Rightarrow n\leq\frac{71}3=23\frac23&lt;/math&gt;. This yields the compound inequality: &lt;math&gt;21\frac13\leq n\leq<br /> 23\frac23&lt;/math&gt;. Since a set with an even number of elements has a median that is the average of the two middle terms, an occurrence that cannot happen in this situation, &lt;math&gt;n&lt;/math&gt; cannot be even. &lt;math&gt;\boxed{\textbf{(B)}\ 23}&lt;/math&gt; is the only other option.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;n&lt;/math&gt; = the number of schools that participated in the contest.<br /> <br /> Then &lt;math&gt;3n&lt;/math&gt; students participated in the contest.<br /> <br /> Since Andrea's score was the median score of all the students, the number of students in the contest must be odd - we know that no two students can have the same score (from the problem), and we know that if the number of students was even, the middle two scores would have to be the same to get a whole number for the median.<br /> <br /> So, we can now write an expression for the median score (Andrea's score) in terms of the number of students who participated in the contest (i.e. &lt;math&gt;3n&lt;/math&gt;):<br /> <br /> &lt;math&gt;\frac{3n+1}{2}&lt;/math&gt;<br /> <br /> Also, since we know that this expression must be smaller than &lt;math&gt;37&lt;/math&gt; (Andrea, whose score was the median, got a better score than Beth and Carla), we can write an inequality and solve it for &lt;math&gt;n&lt;/math&gt;:<br /> <br /> &lt;math&gt;\frac{3n+1}{2}&lt;37&lt;/math&gt;<br /> <br /> Solving this inequality for &lt;math&gt;n&lt;/math&gt;, we get that:<br /> <br /> &lt;math&gt;n&lt;\frac{37}{3} \mathrm{(which &amp;nbsp;is &amp;nbsp;a &amp;nbsp;little &amp;nbsp;over &amp;nbsp;24)}&lt;/math&gt;<br /> <br /> So this eliminates answer choices &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. <br /> <br /> But we already know that &lt;math&gt;3n&lt;/math&gt; must be odd, implying that &lt;math&gt;n&lt;/math&gt; must also be odd! So at this point, the only odd answer choice is &lt;math&gt;\boxed{\textbf{(B)}\ 23}&lt;/math&gt;, and we are done.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2010|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_17&diff=74752 2010 AMC 10B Problems/Problem 17 2016-01-22T03:29:56Z <p>Kkwang: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Every high school in the city of Euclid sent a team of &lt;math&gt;3&lt;/math&gt; students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed &lt;math&gt;37&lt;/math&gt;th and &lt;math&gt;64&lt;/math&gt;th, respectively. How many schools are in the city?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26&lt;/math&gt;<br /> <br /> == Solution ==<br /> Let the &lt;math&gt;n&lt;/math&gt; be the number of schools, &lt;math&gt;3n&lt;/math&gt; be the number of contestants, and &lt;math&gt;x&lt;/math&gt; be Andrea's place. Since the number of participants divided by three is the number of schools, &lt;math&gt;n\geq\frac{64}3=21\frac13&lt;/math&gt;. Andrea received a higher score than her teammates, so &lt;math&gt;x\leq36&lt;/math&gt;. Since &lt;math&gt;36&lt;/math&gt; is the maximum possible median, then &lt;math&gt;2*36-1=71&lt;/math&gt; is the maximum possible number of participants. Therefore, &lt;math&gt;3n\leq71\Rightarrow n\leq\frac{71}3=23\frac23&lt;/math&gt;. This yields the compound inequality: &lt;math&gt;21\frac13\leq n\leq<br /> 23\frac23&lt;/math&gt;. Since a set with an even number of elements has a median that is the average of the two middle terms, an occurrence that cannot happen in this situation, &lt;math&gt;n&lt;/math&gt; cannot be even. &lt;math&gt;\boxed{\textbf{(B)}\ 23}&lt;/math&gt; is the only other option.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;n&lt;/math&gt; = the number of schools that participated in the contest.<br /> <br /> Then &lt;math&gt;3n&lt;/math&gt; students participated in the contest.<br /> <br /> Since Andrea's score was the median score of all the students, the number of students in the contest must be odd - we know that no two students can have the same score (from the problem), and we know that if the number of students was even, the middle two scores would have to be the same to get a whole number for the median.<br /> <br /> So, we can now write an expression for the median score (Andrea's score) in terms of the number of students who participated in the contest (i.e. &lt;math&gt;3n&lt;/math&gt;):<br /> <br /> &lt;math&gt;\frac{3n+1}{2}&lt;/math&gt;<br /> <br /> Also, since we know that this expression must be smaller than &lt;math&gt;37&lt;/math&gt; (Andrea, whose score was the median, got a better score than Beth and Carla), we can write an inequality and solve it for &lt;math&gt;n&lt;/math&gt;:<br /> <br /> &lt;math&gt;\frac{3n+1}{2}&lt;37&lt;/math&gt;<br /> <br /> Solving this inequality for &lt;math&gt;n&lt;/math&gt;, we get that:<br /> <br /> &lt;math&gt;n&lt;\frac{37}{3} \mathrm{(which is a little over 24)}&lt;/math&gt;<br /> <br /> So this eliminates answer choices &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. <br /> <br /> But we already know that &lt;math&gt;3n&lt;/math&gt; must be odd, implying that &lt;math&gt;n&lt;/math&gt; must also be odd! So at this point, the only odd answer choice is &lt;math&gt;\boxed{\textbf{(B)}\ 23}&lt;/math&gt;, and we are done.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2010|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_21&diff=74702 2011 AMC 10B Problems/Problem 21 2016-01-19T01:40:48Z <p>Kkwang: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Brian writes down four integers &lt;math&gt;w &gt; x &gt; y &gt; z&lt;/math&gt; whose sum is &lt;math&gt;44&lt;/math&gt;. The pairwise positive differences of these numbers are &lt;math&gt;1, 3, 4, 5, 6,&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt;. What is the sum of the possible values for &lt;math&gt;w&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> The largest difference, &lt;math&gt;9,&lt;/math&gt; must be between &lt;math&gt;w&lt;/math&gt; and &lt;math&gt;z.&lt;/math&gt;<br /> <br /> The smallest difference, &lt;math&gt;1,&lt;/math&gt; must be directly between two integers. This also means the differences directly between the other two should add up to &lt;math&gt;8.&lt;/math&gt; The only remaining differences that would make this possible are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;5.&lt;/math&gt; However, those two differences can't be right next to each other because they would make a difference of &lt;math&gt;8.&lt;/math&gt; This means &lt;math&gt;1&lt;/math&gt; must be the difference between &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;x.&lt;/math&gt; We can express the possible configurations as the lines.<br /> <br /> <br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(14mm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=4;<br /> <br /> pair Z1=(0,1), Y1=(1,1), X1=(2,1), W1=(3,1);<br /> pair Z4=(4,1), Y4=(5,1), X4=(6,1), W4=(7,1);<br /> <br /> draw(Z1--W1); draw(Z4--W4);<br /> <br /> pair[] ps={W1,W4,X1,X4,Y1,Y4,Z1,Z4};<br /> dot(ps);<br /> label(&quot;$z$&quot;,Z1,N); label(&quot;$y$&quot;,Y1,N); label(&quot;$x$&quot;,X1,N); label(&quot;$w$&quot;,W1,N);<br /> label(&quot;$z$&quot;,Z4,N); label(&quot;$y$&quot;,Y4,N); label(&quot;$x$&quot;,X4,N); label(&quot;$w$&quot;,W4,N);<br /> <br /> label(&quot;$1$&quot;,(X1--Y1),N); label(&quot;$1$&quot;,(X4--Y4),N);<br /> label(&quot;$3$&quot;,(Y1--Z1),N); label(&quot;$3$&quot;,(W4--X4),N); label(&quot;$5$&quot;,(X1--W1),N); label(&quot;$5$&quot;,(Y4--Z4),N);<br /> <br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> If we look at the first number line, you can express &lt;math&gt;x&lt;/math&gt; as &lt;math&gt;w-5,&lt;/math&gt; &lt;math&gt;y&lt;/math&gt; as &lt;math&gt;w-6,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; as &lt;math&gt;w-9.&lt;/math&gt; Since the sum of all these integers equal &lt;math&gt;44&lt;/math&gt;,<br /> &lt;cmath&gt;\begin{align*}<br /> w+w-5+w-6+w-9&amp;=44\\<br /> 4w&amp;=64\\<br /> w&amp;=16 \end{align*}&lt;/cmath&gt;<br /> You can do something similar to this with the second number line to find the other possible value of &lt;math&gt;w.&lt;/math&gt;<br /> &lt;cmath&gt;\begin{align*}<br /> w+w-3+w-4+w-9&amp;=44\\<br /> 4w&amp;=60\\<br /> w&amp;=15 \end{align*}&lt;/cmath&gt;<br /> The sum of the possible values of &lt;math&gt;w&lt;/math&gt; is &lt;math&gt;16+15 = \boxed{\textbf{(B) }31}&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> <br /> First, like Solution 1, we know that &lt;math&gt;w-z=9 \ \text{(1)}&lt;/math&gt;, because no sum could be smaller. Next, we find the sum of all the differences; since &lt;math&gt;w&lt;/math&gt; is in the positive part of a difference 3 times, and has no differences where it contributes as the negative part, the sum of the differences includes &lt;math&gt;3w&lt;/math&gt;. Continuing in this way, we find that &lt;cmath&gt;3w+x-y-3z=28 \ \text{(2)}&lt;/cmath&gt;. Now, we can subtract &lt;math&gt;3w-3z=27&lt;/math&gt; from (2) to get &lt;math&gt;x-y=1 \ \text{(3)}&lt;/math&gt;. Also, adding (2) with &lt;math&gt;w+x+y+z=44&lt;/math&gt; gives &lt;math&gt;4w+2x-2z=72&lt;/math&gt;, or &lt;math&gt;2w+x-z=36&lt;/math&gt;. Subtracting (1) from this gives &lt;math&gt;w+x=27&lt;/math&gt;. Since we know &lt;math&gt;w-z&lt;/math&gt; and &lt;math&gt;x-y&lt;/math&gt;, we find that &lt;cmath&gt;(w-z)+(x-y)=(w-y)+(x-z)=9+1=10&lt;/cmath&gt;. This means that &lt;math&gt;w-y&lt;/math&gt; and &lt;math&gt;x-z&lt;/math&gt; must be 4 and 6, in some order. If &lt;math&gt;w-y=6&lt;/math&gt;, then subtracting this from (3) gives &lt;math&gt;(w-y)-(x-y)=6-1=5&lt;/math&gt;, so &lt;math&gt;w-x=5&lt;/math&gt;. This means that &lt;math&gt;(w-x)+(w+x)=2w=27+5=32&lt;/math&gt;, so &lt;math&gt;w=16&lt;/math&gt;. Similarly, &lt;math&gt;w&lt;/math&gt; can also equal &lt;math&gt;15&lt;/math&gt;.<br /> <br /> Now if you are in a rush, you would have just answered &lt;math&gt;16+15=\boxed{\textbf{(B) }31}&lt;/math&gt;. But we do have to check if these work. In fact, they do, giving solutions &lt;math&gt;\boxed{16, 11, 10, 7}&lt;/math&gt; and &lt;math&gt;\boxed{15, 12, 11, 6}&lt;/math&gt;.<br /> <br /> == See Also==<br /> <br /> {{AMC10 box|year=2011|ab=B|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_21&diff=74701 2011 AMC 10B Problems/Problem 21 2016-01-19T01:40:18Z <p>Kkwang: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> Brian writes down four integers &lt;math&gt;w &gt; x &gt; y &gt; z&lt;/math&gt; whose sum is &lt;math&gt;44&lt;/math&gt;. The pairwise positive differences of these numbers are &lt;math&gt;1, 3, 4, 5, 6,&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt;. What is the sum of the possible values for &lt;math&gt;w&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93&lt;/math&gt;<br /> <br /> == Solution ==<br /> The largest difference, &lt;math&gt;9,&lt;/math&gt; must be between &lt;math&gt;w&lt;/math&gt; and &lt;math&gt;z.&lt;/math&gt;<br /> <br /> The smallest difference, &lt;math&gt;1,&lt;/math&gt; must be directly between two integers. This also means the differences directly between the other two should add up to &lt;math&gt;8.&lt;/math&gt; The only remaining differences that would make this possible are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;5.&lt;/math&gt; However, those two differences can't be right next to each other because they would make a difference of &lt;math&gt;8.&lt;/math&gt; This means &lt;math&gt;1&lt;/math&gt; must be the difference between &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;x.&lt;/math&gt; We can express the possible configurations as the lines.<br /> <br /> <br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(14mm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=4;<br /> <br /> pair Z1=(0,1), Y1=(1,1), X1=(2,1), W1=(3,1);<br /> pair Z4=(4,1), Y4=(5,1), X4=(6,1), W4=(7,1);<br /> <br /> draw(Z1--W1); draw(Z4--W4);<br /> <br /> pair[] ps={W1,W4,X1,X4,Y1,Y4,Z1,Z4};<br /> dot(ps);<br /> label(&quot;$z$&quot;,Z1,N); label(&quot;$y$&quot;,Y1,N); label(&quot;$x$&quot;,X1,N); label(&quot;$w$&quot;,W1,N);<br /> label(&quot;$z$&quot;,Z4,N); label(&quot;$y$&quot;,Y4,N); label(&quot;$x$&quot;,X4,N); label(&quot;$w$&quot;,W4,N);<br /> <br /> label(&quot;$1$&quot;,(X1--Y1),N); label(&quot;$1$&quot;,(X4--Y4),N);<br /> label(&quot;$3$&quot;,(Y1--Z1),N); label(&quot;$3$&quot;,(W4--X4),N); label(&quot;$5$&quot;,(X1--W1),N); label(&quot;$5$&quot;,(Y4--Z4),N);<br /> <br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> If we look at the first number line, you can express &lt;math&gt;x&lt;/math&gt; as &lt;math&gt;w-5,&lt;/math&gt; &lt;math&gt;y&lt;/math&gt; as &lt;math&gt;w-6,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; as &lt;math&gt;w-9.&lt;/math&gt; Since the sum of all these integers equal &lt;math&gt;44&lt;/math&gt;,<br /> &lt;cmath&gt;\begin{align*}<br /> w+w-5+w-6+w-9&amp;=44\\<br /> 4w&amp;=64\\<br /> w&amp;=16 \end{align*}&lt;/cmath&gt;<br /> You can do something similar to this with the second number line to find the other possible value of &lt;math&gt;w.&lt;/math&gt;<br /> &lt;cmath&gt;\begin{align*}<br /> w+w-3+w-4+w-9&amp;=44\\<br /> 4w&amp;=60\\<br /> w&amp;=15 \end{align*}&lt;/cmath&gt;<br /> The sum of the possible values of &lt;math&gt;w&lt;/math&gt; is &lt;math&gt;16+15 = \boxed{\textbf{(B) }31}&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> <br /> First, like Solution 1, we know that &lt;math&gt;w-z=9 \ \text{(1)}&lt;/math&gt;, because no sum could be smaller. Next, we find the sum of all the differences; since &lt;math&gt;w&lt;/math&gt; is in the positive part of a difference 3 times, and has no differences where it contributes as the negative part, the sum of the differences includes &lt;math&gt;3w&lt;/math&gt;. Continuing in this way, we find that &lt;cmath&gt;3w+x-y-3z=28 \ \text{(2)}&lt;/cmath&gt;. Now, we can subtract &lt;math&gt;3w-3z=27&lt;/math&gt; from (2) to get &lt;math&gt;x-y=1 \ \text{(3)}&lt;/math&gt;. Also, adding (2) with &lt;math&gt;w+x+y+z=44&lt;/math&gt; gives &lt;math&gt;4w+2x-2z=72&lt;/math&gt;, or &lt;math&gt;2w+x-z=36&lt;/math&gt;. Subtracting (1) from this gives &lt;math&gt;w+x=27&lt;/math&gt;. Since we know &lt;math&gt;w-z&lt;/math&gt; and &lt;math&gt;x-y&lt;/math&gt;, we find that &lt;cmath&gt;(w-z)+(x-y)=(w-y)+(x-z)=9+1=10&lt;/cmath&gt;. This means that &lt;math&gt;w-y&lt;/math&gt; and &lt;math&gt;x-z&lt;/math&gt; must be 4 and 6, in some order. If &lt;math&gt;w-y=6&lt;/math&gt;, then subtracting this from (3) gives &lt;math&gt;(w-y)-(x-y)=6-1=5&lt;/math&gt;, so &lt;math&gt;w-x=5&lt;/math&gt;. This means that &lt;math&gt;(w-x)+(w+x)=2w=27+5=32&lt;/math&gt;, so &lt;math&gt;w=16&lt;/math&gt;. Similarly, &lt;math&gt;w&lt;/math&gt; can also equal &lt;math&gt;15&lt;/math&gt;.<br /> <br /> Now if you are in a rush, you would have just answered &lt;math&gt;16+15=\boxed{\textbf{(B) }31}&lt;/math&gt;. But we do have to check if these work. In fact, they do, giving solutions &lt;math&gt;\boxed{16, 11, 10, 7}&lt;/math&gt; and &lt;math&gt;\boxed{15, 12, 11, 6}&lt;/math&gt;.<br /> <br /> == See Also==<br /> <br /> {{AMC10 box|year=2011|ab=B|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_20&diff=74592 2009 AMC 10A Problems/Problem 20 2016-01-17T01:57:53Z <p>Kkwang: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Andrea and Lauren are &lt;math&gt;20&lt;/math&gt; kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of &lt;math&gt;1&lt;/math&gt; kilometer per minute. After &lt;math&gt;5&lt;/math&gt; minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 20<br /> \qquad<br /> \mathrm{(B)}\ 30<br /> \qquad<br /> \mathrm{(C)}\ 55<br /> \qquad<br /> \mathrm{(D)}\ 65<br /> \qquad<br /> \mathrm{(E)}\ 80<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let their speeds in kilometers per hour be &lt;math&gt;v_A&lt;/math&gt; and &lt;math&gt;v_L&lt;/math&gt;. We know that &lt;math&gt;v_A=3v_L&lt;/math&gt; and that &lt;math&gt;v_A+v_L=60&lt;/math&gt;. (The second equation follows from the fact that &lt;math&gt;1\mathrm km/min = 60\mathrm km/h&lt;/math&gt;.) This solves to &lt;math&gt;v_A=45&lt;/math&gt; and &lt;math&gt;v_L=15&lt;/math&gt;. <br /> <br /> As the distance decreases at a rate of &lt;math&gt;1&lt;/math&gt; kilometer per minute, after &lt;math&gt;5&lt;/math&gt; minutes the distance between them will be &lt;math&gt;20-5=15&lt;/math&gt; kilometers.<br /> <br /> From this point on, only Lauren will be riding her bike. As there are &lt;math&gt;15&lt;/math&gt; kilometers remaining and &lt;math&gt;v_L=15&lt;/math&gt;, she will need exactly an hour to get to Andrea. Therefore the total time in minutes is &lt;math&gt;5+60 = \boxed{65}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_20&diff=74591 2009 AMC 10A Problems/Problem 20 2016-01-17T01:56:59Z <p>Kkwang: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Andrea and Lauren are &lt;math&gt;20&lt;/math&gt; kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of &lt;math&gt;1&lt;/math&gt; kilometer per minute. After &lt;math&gt;5&lt;/math&gt; minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 20<br /> \qquad<br /> \mathrm{(B)}\ 30<br /> \qquad<br /> \mathrm{(C)}\ 55<br /> \qquad<br /> \mathrm{(D)}\ 65<br /> \qquad<br /> \mathrm{(E)}\ 80<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let their speeds in kilometers per hour be &lt;math&gt;v_A&lt;/math&gt; and &lt;math&gt;v_L&lt;/math&gt;. We know that &lt;math&gt;v_A=3v_L&lt;/math&gt; and that &lt;math&gt;v_A+v_L=60&lt;/math&gt;. (The second equation follows from the fact that &lt;math&gt;1,\unit{\mathrm km/min} = 60,\unit{\mathrm km/h}&lt;/math&gt;.) This solves to &lt;math&gt;v_A=45&lt;/math&gt; and &lt;math&gt;v_L=15&lt;/math&gt;. <br /> <br /> As the distance decreases at a rate of &lt;math&gt;1&lt;/math&gt; kilometer per minute, after &lt;math&gt;5&lt;/math&gt; minutes the distance between them will be &lt;math&gt;20-5=15&lt;/math&gt; kilometers.<br /> <br /> From this point on, only Lauren will be riding her bike. As there are &lt;math&gt;15&lt;/math&gt; kilometers remaining and &lt;math&gt;v_L=15&lt;/math&gt;, she will need exactly an hour to get to Andrea. Therefore the total time in minutes is &lt;math&gt;5+60 = \boxed{65}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_20&diff=74590 2009 AMC 10A Problems/Problem 20 2016-01-17T01:56:39Z <p>Kkwang: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Andrea and Lauren are &lt;math&gt;20&lt;/math&gt; kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of &lt;math&gt;1&lt;/math&gt; kilometer per minute. After &lt;math&gt;5&lt;/math&gt; minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 20<br /> \qquad<br /> \mathrm{(B)}\ 30<br /> \qquad<br /> \mathrm{(C)}\ 55<br /> \qquad<br /> \mathrm{(D)}\ 65<br /> \qquad<br /> \mathrm{(E)}\ 80<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let their speeds in kilometers per hour be &lt;math&gt;v_A&lt;/math&gt; and &lt;math&gt;v_L&lt;/math&gt;. We know that &lt;math&gt;v_A=3v_L&lt;/math&gt; and that &lt;math&gt;v_A+v_L=60&lt;/math&gt;. (The second equation follows from the fact that &lt;math&gt;1\,\unit{\mathrm km/min} = 60\,\unit{\mathrm km/h}&lt;/math&gt;.) This solves to &lt;math&gt;v_A=45&lt;/math&gt; and &lt;math&gt;v_L=15&lt;/math&gt;. <br /> <br /> As the distance decreases at a rate of &lt;math&gt;1&lt;/math&gt; kilometer per minute, after &lt;math&gt;5&lt;/math&gt; minutes the distance between them will be &lt;math&gt;20-5=15&lt;/math&gt; kilometers.<br /> <br /> From this point on, only Lauren will be riding her bike. As there are &lt;math&gt;15&lt;/math&gt; kilometers remaining and &lt;math&gt;v_L=15&lt;/math&gt;, she will need exactly an hour to get to Andrea. Therefore the total time in minutes is &lt;math&gt;5+60 = \boxed{65}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_19&diff=74589 2009 AMC 10A Problems/Problem 19 2016-01-17T01:55:05Z <p>Kkwang: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Circle &lt;math&gt;A&lt;/math&gt; has radius &lt;math&gt;100&lt;/math&gt;. Circle &lt;math&gt;B&lt;/math&gt; has an integer radius &lt;math&gt;r&lt;100&lt;/math&gt; and remains internally tangent to circle &lt;math&gt;A&lt;/math&gt; as it rolls once around the circumference of circle &lt;math&gt;A&lt;/math&gt;. The two circles have the same points of tangency at the beginning and end of cirle &lt;math&gt;B&lt;/math&gt;'s trip. How many possible values can &lt;math&gt;r&lt;/math&gt; have?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 4\<br /> \qquad<br /> \mathrm{(B)}\ 8\<br /> \qquad<br /> \mathrm{(C)}\ 9\<br /> \qquad<br /> \mathrm{(D)}\ 50\<br /> \qquad<br /> \mathrm{(E)}\ 90\<br /> \qquad<br /> &lt;/math&gt;<br /> <br /> [[Category: Introductory Geometry Problems]]<br /> <br /> == Solution ==<br /> <br /> The circumference of circle A is &lt;math&gt;200\pi&lt;/math&gt;, and the circumference of circle B with radius &lt;math&gt;r&lt;/math&gt; is &lt;math&gt;2r\pi&lt;/math&gt;. Since circle B makes a complete revolution and ''ends up on the same point'', the circumference of A must be a perfect factor of the circumference of B, therefore the quotient must be an integer.<br /> <br /> &lt;math&gt;So\qquad\frac{200\pi}{2\pi \cdot r} = \frac{100}{r}&lt;/math&gt;<br /> <br /> R must then be a factor of 100, excluding 100 (because then circle B would be the same size as circle A). &lt;math&gt;100\: =\: 2^2\; \cdot \; 5^2&lt;/math&gt;. Therefore 100 has &lt;math&gt;(2+1)\; \cdot \; (2+1)\;&lt;/math&gt; factors*. But you need to subtract 1 from 9, in order to exclude 100. Therefore the answer is &lt;math&gt;\boxed{8}&lt;/math&gt;.<br /> <br /> *The number of factors of &lt;math&gt;a^x\: \cdot \: b^y\: \cdot \: c^z\;...&lt;/math&gt; and so on, where &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are prime numbers, is &lt;math&gt;(x+1)(y+1)(z+1)...&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_11&diff=74338 2005 AMC 12B Problems/Problem 11 2016-01-06T04:49:31Z <p>Kkwang: /* Solution */</p> <hr /> <div>{{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #11]] and [[2005 AMC 10B Problems|2005 AMC 10B #15]]}}<br /> == Problem ==<br /> An envelope contains eight bills: &lt;math&gt;2&lt;/math&gt; ones, &lt;math&gt;2&lt;/math&gt; fives, &lt;math&gt;2&lt;/math&gt; tens, and &lt;math&gt;2&lt;/math&gt; twenties. Two bills are drawn at random without replacement. What is the probability that their sum is &amp;#36;&lt;math&gt;20&lt;/math&gt; or more?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ {{{\frac{1}{4}}}} \qquad \mathrm{(B)}\ {{{\frac{2}{5}}}} \qquad \mathrm{(C)}\ {{{\frac{3}{7}}}} \qquad \mathrm{(D)}\ {{{\frac{1}{2}}}} \qquad \mathrm{(E)}\ {{{\frac{2}{3}}}}&lt;/math&gt;<br /> <br /> == Solution 1==<br /> The only way to get a total of &amp;#36;&lt;math&gt;20&lt;/math&gt; or more is if you pick a twenty and another bill, or if you pick both tens. There are a total of &lt;math&gt;\dbinom{8}{2}=\dfrac{8\times7}{2\times1}=28&lt;/math&gt; ways to choose &lt;math&gt;2&lt;/math&gt; bills out of &lt;math&gt;8&lt;/math&gt;. There are &lt;math&gt;12&lt;/math&gt; ways to choose a twenty and some other non-twenty bill. There is &lt;math&gt;1&lt;/math&gt; way to choose both twenties, and also &lt;math&gt;1&lt;/math&gt; way to choose both tens. Adding these up, we find that there are a total of &lt;math&gt;14&lt;/math&gt; ways to attain a sum of &lt;math&gt;20&lt;/math&gt; or greater, so there is a total probability of &lt;math&gt;\dfrac{14}{28}=\boxed{\mathrm{(D)}\ \dfrac{1}{2}}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> TBE<br /> <br /> == See also ==<br /> {{AMC10 box|year=2005|ab=B|num-b=14|num-a=16}}<br /> {{AMC12 box|year=2005|ab=B|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_14&diff=74305 2005 AMC 10A Problems/Problem 14 2016-01-04T03:55:15Z <p>Kkwang: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 &lt;/math&gt;<br /> <br /> ==Solution==<br /> If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits. <br /> <br /> Doing some [[casework]]: <br /> <br /> If the middle digit is &lt;math&gt;1&lt;/math&gt;, possible numbers range from &lt;math&gt;111&lt;/math&gt; to &lt;math&gt;210&lt;/math&gt;. So there are &lt;math&gt;2&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;2&lt;/math&gt;, possible numbers range from &lt;math&gt;123&lt;/math&gt; to &lt;math&gt;420&lt;/math&gt;. So there are &lt;math&gt;4&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;3&lt;/math&gt;, possible numbers range from &lt;math&gt;135&lt;/math&gt; to &lt;math&gt;630&lt;/math&gt;. So there are &lt;math&gt;6&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;4&lt;/math&gt;, possible numbers range from &lt;math&gt;147&lt;/math&gt; to &lt;math&gt;840&lt;/math&gt;. So there are &lt;math&gt;8&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;5&lt;/math&gt;, possible numbers range from &lt;math&gt;159&lt;/math&gt; to &lt;math&gt;951&lt;/math&gt;. So there are &lt;math&gt;9&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;6&lt;/math&gt;, possible numbers range from &lt;math&gt;369&lt;/math&gt; to &lt;math&gt;963&lt;/math&gt;. So there are &lt;math&gt;7&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;7&lt;/math&gt;, possible numbers range from &lt;math&gt;579&lt;/math&gt; to &lt;math&gt;975&lt;/math&gt;. So there are &lt;math&gt;5&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;8&lt;/math&gt;, possible numbers range from &lt;math&gt;789&lt;/math&gt; to &lt;math&gt;987&lt;/math&gt;. So there are &lt;math&gt;3&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;9&lt;/math&gt;, the only possible number is &lt;math&gt;999&lt;/math&gt;. So there is &lt;math&gt;1&lt;/math&gt; number in this case. <br /> <br /> So the total number of three-digit numbers that satisfy the property is &lt;math&gt;2+4+6+8+9+7+5+3+1=45\Rightarrow E&lt;/math&gt;<br /> <br /> == Solution 2 (much faster and slicker)==<br /> Alternatively, we could note that the middle digit is uniquely defined by the first and third digits, since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:<br /> <br /> If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are &lt;math&gt;5\cdot5=25&lt;/math&gt; numbers in this case.<br /> <br /> If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are &lt;math&gt;4\cdot5=20&lt;/math&gt; numbers here.<br /> <br /> The total number, then, is &lt;math&gt;20+25=45\Rightarrow E&lt;/math&gt;<br /> <br /> ==See Also==<br /> *[[2005 AMC 10A Problems]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 13|Previous Problem]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 15|Next Problem]]<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_14&diff=74304 2005 AMC 10A Problems/Problem 14 2016-01-04T03:54:38Z <p>Kkwang: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 &lt;/math&gt;<br /> <br /> ==Solution==<br /> If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits. <br /> <br /> Doing some [[casework]]: <br /> <br /> If the middle digit is &lt;math&gt;1&lt;/math&gt;, possible numbers range from &lt;math&gt;111&lt;/math&gt; to &lt;math&gt;210&lt;/math&gt;. So there are &lt;math&gt;2&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;2&lt;/math&gt;, possible numbers range from &lt;math&gt;123&lt;/math&gt; to &lt;math&gt;420&lt;/math&gt;. So there are &lt;math&gt;4&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;3&lt;/math&gt;, possible numbers range from &lt;math&gt;135&lt;/math&gt; to &lt;math&gt;630&lt;/math&gt;. So there are &lt;math&gt;6&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;4&lt;/math&gt;, possible numbers range from &lt;math&gt;147&lt;/math&gt; to &lt;math&gt;840&lt;/math&gt;. So there are &lt;math&gt;8&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;5&lt;/math&gt;, possible numbers range from &lt;math&gt;159&lt;/math&gt; to &lt;math&gt;951&lt;/math&gt;. So there are &lt;math&gt;9&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;6&lt;/math&gt;, possible numbers range from &lt;math&gt;369&lt;/math&gt; to &lt;math&gt;963&lt;/math&gt;. So there are &lt;math&gt;7&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;7&lt;/math&gt;, possible numbers range from &lt;math&gt;579&lt;/math&gt; to &lt;math&gt;975&lt;/math&gt;. So there are &lt;math&gt;5&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;8&lt;/math&gt;, possible numbers range from &lt;math&gt;789&lt;/math&gt; to &lt;math&gt;987&lt;/math&gt;. So there are &lt;math&gt;3&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;9&lt;/math&gt;, the only possible number is &lt;math&gt;999&lt;/math&gt;. So there is &lt;math&gt;1&lt;/math&gt; number in this case. <br /> <br /> So the total number of three-digit numbers that satisfy the property is &lt;math&gt;2+4+6+8+9+7+5+3+1=45\Rightarrow E&lt;/math&gt;<br /> <br /> == Solution 2 (much faster and slicker)==<br /> Alternatively, we could note that the middle digit is uniquely defined by the first and third digits, since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:<br /> <br /> If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are &lt;math&gt;5\cdot5=25&lt;/math&gt; numbers in this case.<br /> <br /> If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are &lt;math&gt;4\cdot5=20&lt;/math&gt; numbers here.<br /> <br /> The total number, then, is &lt;math&gt;20+25=45\Rightarrow E&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> By listing the first two hundred numbers you'll see a pattern. The &lt;math&gt;(-)&lt;/math&gt; is where the middle number should go.<br /> <br /> &lt;math&gt;1-0&lt;/math&gt; Can't do &lt;math&gt; \frac{1+0}{2)=\frac{1}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;1-1&lt;/math&gt; Can do &lt;math&gt;(1+1)/2=1&lt;/math&gt;<br /> <br /> &lt;math&gt;1-2&lt;/math&gt; Can't do &lt;math&gt;(1+2)=3/2&lt;/math&gt;<br /> <br /> &lt;math&gt;1-3&lt;/math&gt; Can do &lt;math&gt;(1+3)/2=2&lt;/math&gt;<br /> <br /> &lt;math&gt;1-4&lt;/math&gt; Can't do &lt;math&gt;(1+4)=5/2&lt;/math&gt;<br /> <br /> &lt;math&gt;1-5&lt;/math&gt; Can do &lt;math&gt;(1+5)/2=3&lt;/math&gt;<br /> <br /> &lt;math&gt;1-6&lt;/math&gt; Can't do &lt;math&gt;(1+6)=7/2&lt;/math&gt;<br /> <br /> &lt;math&gt;1-7&lt;/math&gt; Can do &lt;math&gt;(1+7)/2=2&lt;/math&gt;<br /> <br /> &lt;math&gt;1-8&lt;/math&gt; Can't do &lt;math&gt;(1+8)=9/2&lt;/math&gt;<br /> <br /> &lt;math&gt;1-9&lt;/math&gt; Can do &lt;math&gt;(1+9)/2=2&lt;/math&gt;<br /> <br /> Total can do: &lt;math&gt;5&lt;/math&gt;<br /> <br /> &lt;math&gt;2-0&lt;/math&gt; Can't do &lt;math&gt;(2+0)/2=0.5&lt;/math&gt;<br /> <br /> &lt;math&gt;2-1&lt;/math&gt; Can do &lt;math&gt;(2+1)/2=1&lt;/math&gt;<br /> <br /> &lt;math&gt;2-2&lt;/math&gt; Can't do &lt;math&gt;(2+2)=3/2&lt;/math&gt;<br /> <br /> &lt;math&gt;2-3&lt;/math&gt; Can do &lt;math&gt;(2+3)/2=2&lt;/math&gt;<br /> <br /> &lt;math&gt;2-4&lt;/math&gt; Can't do &lt;math&gt;(2+4)=5/2&lt;/math&gt;<br /> <br /> &lt;math&gt;2-5&lt;/math&gt; Can do &lt;math&gt;(2+5)/2=2&lt;/math&gt;<br /> <br /> &lt;math&gt;2-6&lt;/math&gt; Can't do &lt;math&gt;(2+6)=7/2&lt;/math&gt;<br /> <br /> &lt;math&gt;2-7&lt;/math&gt; Can do &lt;math&gt;(2+7)/2=2&lt;/math&gt;<br /> <br /> &lt;math&gt;2-8&lt;/math&gt; Can't do &lt;math&gt;(2+8)=9/2&lt;/math&gt;<br /> <br /> &lt;math&gt;2-9&lt;/math&gt; Can do &lt;math&gt;(2+9)/2=2&lt;/math&gt;<br /> <br /> Total can do: &lt;math&gt;5&lt;/math&gt;<br /> <br /> SUMMARY<br /> <br /> You can now see the pattern. There are &lt;math&gt;5&lt;/math&gt; can do's each hundred. So then &lt;math&gt;9*5= \boxed{45} \Rightarrow \mathrm{(E)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> *[[2005 AMC 10A Problems]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 13|Previous Problem]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 15|Next Problem]]<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_14&diff=74303 2005 AMC 10A Problems/Problem 14 2016-01-04T03:53:47Z <p>Kkwang: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 &lt;/math&gt;<br /> <br /> ==Solution==<br /> If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits. <br /> <br /> Doing some [[casework]]: <br /> <br /> If the middle digit is &lt;math&gt;1&lt;/math&gt;, possible numbers range from &lt;math&gt;111&lt;/math&gt; to &lt;math&gt;210&lt;/math&gt;. So there are &lt;math&gt;2&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;2&lt;/math&gt;, possible numbers range from &lt;math&gt;123&lt;/math&gt; to &lt;math&gt;420&lt;/math&gt;. So there are &lt;math&gt;4&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;3&lt;/math&gt;, possible numbers range from &lt;math&gt;135&lt;/math&gt; to &lt;math&gt;630&lt;/math&gt;. So there are &lt;math&gt;6&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;4&lt;/math&gt;, possible numbers range from &lt;math&gt;147&lt;/math&gt; to &lt;math&gt;840&lt;/math&gt;. So there are &lt;math&gt;8&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;5&lt;/math&gt;, possible numbers range from &lt;math&gt;159&lt;/math&gt; to &lt;math&gt;951&lt;/math&gt;. So there are &lt;math&gt;9&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;6&lt;/math&gt;, possible numbers range from &lt;math&gt;369&lt;/math&gt; to &lt;math&gt;963&lt;/math&gt;. So there are &lt;math&gt;7&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;7&lt;/math&gt;, possible numbers range from &lt;math&gt;579&lt;/math&gt; to &lt;math&gt;975&lt;/math&gt;. So there are &lt;math&gt;5&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;8&lt;/math&gt;, possible numbers range from &lt;math&gt;789&lt;/math&gt; to &lt;math&gt;987&lt;/math&gt;. So there are &lt;math&gt;3&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;9&lt;/math&gt;, the only possible number is &lt;math&gt;999&lt;/math&gt;. So there is &lt;math&gt;1&lt;/math&gt; number in this case. <br /> <br /> So the total number of three-digit numbers that satisfy the property is &lt;math&gt;2+4+6+8+9+7+5+3+1=45\Rightarrow E&lt;/math&gt;<br /> <br /> == Solution 2 (much faster and slicker)==<br /> Alternatively, we could note that the middle digit is uniquely defined by the first and third digits, since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:<br /> <br /> If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are &lt;math&gt;5\cdot5=25&lt;/math&gt; numbers in this case.<br /> <br /> If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are &lt;math&gt;4\cdot5=20&lt;/math&gt; numbers here.<br /> <br /> The total number, then, is &lt;math&gt;20+25=45\Rightarrow E&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> By listing the first two hundred numbers you'll see a pattern. The &lt;math&gt;(-)&lt;/math&gt; is where the middle number should go.<br /> <br /> &lt;math&gt;1-0&lt;/math&gt; Can't do &lt;math&gt;(1+0)/2=0.5&lt;/math&gt;<br /> <br /> &lt;math&gt;1-1&lt;/math&gt; Can do &lt;math&gt;(1+1)/2=1&lt;/math&gt;<br /> <br /> &lt;math&gt;1-2&lt;/math&gt; Can't do &lt;math&gt;(1+2)=3/2&lt;/math&gt;<br /> <br /> &lt;math&gt;1-3&lt;/math&gt; Can do &lt;math&gt;(1+3)/2=2&lt;/math&gt;<br /> <br /> &lt;math&gt;1-4&lt;/math&gt; Can't do &lt;math&gt;(1+4)=5/2&lt;/math&gt;<br /> <br /> &lt;math&gt;1-5&lt;/math&gt; Can do &lt;math&gt;(1+5)/2=3&lt;/math&gt;<br /> <br /> &lt;math&gt;1-6&lt;/math&gt; Can't do &lt;math&gt;(1+6)=7/2&lt;/math&gt;<br /> <br /> &lt;math&gt;1-7&lt;/math&gt; Can do &lt;math&gt;(1+7)/2=2&lt;/math&gt;<br /> <br /> &lt;math&gt;1-8&lt;/math&gt; Can't do &lt;math&gt;(1+8)=9/2&lt;/math&gt;<br /> <br /> &lt;math&gt;1-9&lt;/math&gt; Can do &lt;math&gt;(1+9)/2=2&lt;/math&gt;<br /> <br /> Total can do: &lt;math&gt;5&lt;/math&gt;<br /> <br /> &lt;math&gt;2-0&lt;/math&gt; Can't do &lt;math&gt;(2+0)/2=0.5&lt;/math&gt;<br /> <br /> &lt;math&gt;2-1&lt;/math&gt; Can do &lt;math&gt;(2+1)/2=1&lt;/math&gt;<br /> <br /> &lt;math&gt;2-2&lt;/math&gt; Can't do &lt;math&gt;(2+2)=3/2&lt;/math&gt;<br /> <br /> &lt;math&gt;2-3&lt;/math&gt; Can do &lt;math&gt;(2+3)/2=2&lt;/math&gt;<br /> <br /> &lt;math&gt;2-4&lt;/math&gt; Can't do &lt;math&gt;(2+4)=5/2&lt;/math&gt;<br /> <br /> &lt;math&gt;2-5&lt;/math&gt; Can do &lt;math&gt;(2+5)/2=2&lt;/math&gt;<br /> <br /> &lt;math&gt;2-6&lt;/math&gt; Can't do &lt;math&gt;(2+6)=7/2&lt;/math&gt;<br /> <br /> &lt;math&gt;2-7&lt;/math&gt; Can do &lt;math&gt;(2+7)/2=2&lt;/math&gt;<br /> <br /> &lt;math&gt;2-8&lt;/math&gt; Can't do &lt;math&gt;(2+8)=9/2&lt;/math&gt;<br /> <br /> &lt;math&gt;2-9&lt;/math&gt; Can do &lt;math&gt;(2+9)/2=2&lt;/math&gt;<br /> <br /> Total can do: &lt;math&gt;5&lt;/math&gt;<br /> <br /> SUMMARY<br /> <br /> You can now see the pattern. There are &lt;math&gt;5&lt;/math&gt; can do's each hundred. So then &lt;math&gt;9*5= \boxed{45} \Rightarrow \mathrm{(E)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> *[[2005 AMC 10A Problems]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 13|Previous Problem]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 15|Next Problem]]<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_9&diff=74301 2005 AMC 10A Problems/Problem 9 2016-01-04T02:42:44Z <p>Kkwang: /* Solution */</p> <hr /> <div>==Problem==<br /> Three tiles are marked &lt;math&gt;X&lt;/math&gt; and two other tiles are marked &lt;math&gt;O&lt;/math&gt;. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads &lt;math&gt;XOXOX&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{12}\qquad \mathrm{(B) \ } \frac{1}{10}\qquad \mathrm{(C) \ } \frac{1}{6}\qquad \mathrm{(D) \ } \frac{1}{4}\qquad \mathrm{(E) \ } \frac{1}{3} &lt;/math&gt;<br /> <br /> ==Solution==<br /> There are &lt;math&gt;\frac{5!}{2!3!}=10&lt;/math&gt; distinct arrangements of three &lt;math&gt;X&lt;/math&gt;'s and two &lt;math&gt;O&lt;/math&gt;'s. <br /> <br /> There is only &lt;math&gt;1&lt;/math&gt; distinct arrangement that reads &lt;math&gt;XOXOX&lt;/math&gt;<br /> <br /> Therefore the desired [[probability]] is &lt;math&gt;\boxed{\frac{1}{10}} \Rightarrow \mathrm{(B)}&lt;/math&gt;<br /> <br /> ==See Also==<br /> *[[2005 AMC 10A Problems]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 8|Previous Problem]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 10|Next Problem]]<br /> <br /> *[[Combination]]<br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_8&diff=74300 2005 AMC 10A Problems/Problem 8 2016-01-04T02:37:57Z <p>Kkwang: /* Solution */</p> <hr /> <div>== Problem ==<br /> In the figure, the length of side &lt;math&gt;AB&lt;/math&gt; of square &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;\sqrt{50}&lt;/math&gt; and &lt;math&gt;BE&lt;/math&gt;=1. What is the area of the inner square &lt;math&gt;EFGH&lt;/math&gt;?<br /> <br /> [[File:AMC102005Aq.png]]<br /> <br /> &lt;math&gt; \textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42 &lt;/math&gt;<br /> <br /> ==Solution==<br /> We see that side &lt;math&gt;BE&lt;/math&gt;, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I'll not prove this). So, &lt;math&gt;AH = 1&lt;/math&gt;. Then &lt;math&gt;HB = HE + BE = HE + 1&lt;/math&gt;, and &lt;math&gt;HE&lt;/math&gt; is one of the sides of the square whose area we want to find. So:<br /> <br /> &lt;cmath&gt;1^2 + (HE+1)^2=\sqrt{50}^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1 + (HE+1)^2=50&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(HE+1)^2=49&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;HE+1=7&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;HE=6&lt;/cmath&gt; <br /> So, the area of the square is &lt;math&gt;6^2=\boxed{36} \Rightarrow \mathrm{(C)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_8&diff=74299 2005 AMC 10A Problems/Problem 8 2016-01-04T02:36:29Z <p>Kkwang: /* Solution */</p> <hr /> <div>== Problem ==<br /> In the figure, the length of side &lt;math&gt;AB&lt;/math&gt; of square &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;\sqrt{50}&lt;/math&gt; and &lt;math&gt;BE&lt;/math&gt;=1. What is the area of the inner square &lt;math&gt;EFGH&lt;/math&gt;?<br /> <br /> [[File:AMC102005Aq.png]]<br /> <br /> &lt;math&gt; \textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42 &lt;/math&gt;<br /> <br /> ==Solution==<br /> We see that side &lt;math&gt;BE&lt;/math&gt;, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I'll not prove this). So, &lt;math&gt;AH = 1&lt;/math&gt;. Then &lt;math&gt;HB = HE + BE = HE + 1&lt;/math&gt;, and &lt;math&gt;HE&lt;/math&gt; is one of the sides of the square whose area we want to find. So:<br /> <br /> &lt;math&gt;1^2 + (HE+1)^2=\sqrt{50}^2&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + (HE+1)^2=50&lt;/math&gt;<br /> <br /> &lt;math&gt;(HE+1)^2=49&lt;/math&gt;<br /> <br /> &lt;math&gt;HE+1=7&lt;/math&gt;<br /> <br /> &lt;math&gt;HE=6&lt;/math&gt; So, the area of the square is &lt;math&gt;6^2=\boxed{36} \Rightarrow (C)&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_8&diff=74298 2005 AMC 10A Problems/Problem 8 2016-01-04T02:35:45Z <p>Kkwang: /* Solution */</p> <hr /> <div>== Problem ==<br /> In the figure, the length of side &lt;math&gt;AB&lt;/math&gt; of square &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;\sqrt{50}&lt;/math&gt; and &lt;math&gt;BE&lt;/math&gt;=1. What is the area of the inner square &lt;math&gt;EFGH&lt;/math&gt;?<br /> <br /> [[File:AMC102005Aq.png]]<br /> <br /> &lt;math&gt; \textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42 &lt;/math&gt;<br /> <br /> ==Solution==<br /> We see that side &lt;math&gt;BE&lt;/math&gt;, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I'll not prove this). So, &lt;math&gt;AH = 1&lt;/math&gt;. Then &lt;math&gt;HB = HE + BE = HE + 1&lt;/math&gt;, and &lt;math&gt;HE&lt;/math&gt; is one of the sides of the square whose area we want to find. So:<br /> <br /> &lt;math&gt;1^2 + (HE+1)^2=\sqrt{50}^2&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + (HE+1)^2=50&lt;/math&gt;<br /> <br /> &lt;math&gt;(HE+1)^2=49&lt;/math&gt;<br /> <br /> &lt;math&gt;HE+1=7&lt;/math&gt;<br /> <br /> &lt;math&gt;HE=6&lt;/math&gt; So, the area of the square is &lt;math&gt;6^2=\boxed{36} \rightarrow (C)&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_23&diff=74297 2005 AMC 10A Problems/Problem 23 2016-01-04T02:17:11Z <p>Kkwang: /* Problem */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;AB&lt;/math&gt; be a diameter of a circle and let &lt;math&gt;C&lt;/math&gt; be a point on &lt;math&gt;AB&lt;/math&gt; with &lt;math&gt;2\cdot AC=BC&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be points on the circle such that &lt;math&gt;DC \perp AB&lt;/math&gt; and &lt;math&gt;DE&lt;/math&gt; is a second diameter. What is the ratio of the area of &lt;math&gt;\triangle DCE&lt;/math&gt; to the area of &lt;math&gt;\triangle ABD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2.5cm);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=3;<br /> pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);<br /> pair D=dir(aCos(C.x)), E=(-D.x,-D.y);<br /> draw(A--B--D--cycle);<br /> draw(D--E--C);<br /> draw(unitcircle,white);<br /> drawline(D,C);<br /> dot(O);<br /> clip(unitcircle);<br /> draw(unitcircle);<br /> label(&quot;$E$&quot;,E,SSE);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$A$&quot;,A,W);<br /> label(&quot;$D$&quot;,D,NNW);<br /> label(&quot;$C$&quot;,C,SW);<br /> draw(rightanglemark(D,C,B,2));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> [[File:Circlenc1.png]]<br /> <br /> Let us assume that the diameter is of length &lt;math&gt;1&lt;/math&gt;. <br /> <br /> &lt;math&gt;AC&lt;/math&gt; is &lt;math&gt;\frac{1}{3}&lt;/math&gt; of diameter and &lt;math&gt;CO&lt;/math&gt; is &lt;math&gt;\frac{1}{2}-\frac{1}{3} = \frac{1}{6}&lt;/math&gt;. <br /> <br /> &lt;math&gt;OD&lt;/math&gt; is the radius of the circle, so using the Pythagorean theorem height &lt;math&gt;CD&lt;/math&gt; of &lt;math&gt;\triangle AOC&lt;/math&gt; is &lt;math&gt;\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2} = \frac{\sqrt{2}}{3}&lt;/math&gt;. This is also the height of the &lt;math&gt;\triangle ABD&lt;/math&gt;.<br /> <br /> Area of the &lt;math&gt;\triangle DCO&lt;/math&gt; is &lt;math&gt;\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}&lt;/math&gt; = &lt;math&gt;\frac{\sqrt{2}}{36}&lt;/math&gt;.<br /> <br /> The height of &lt;math&gt;\triangle DCE&lt;/math&gt; can be found using the area of &lt;math&gt;\triangle DCO&lt;/math&gt; and &lt;math&gt;DO&lt;/math&gt; as base. <br /> <br /> Hence the height of &lt;math&gt;\triangle DCE&lt;/math&gt; is &lt;math&gt;\frac{\frac{\sqrt{2}}{36}}{\frac{1}{2}\cdot\frac{1}{2}}&lt;/math&gt; = &lt;math&gt;\frac{\sqrt{2}}{9}&lt;/math&gt;. <br /> <br /> The diameter is the base for both the triangles &lt;math&gt;\triangle DCE&lt;/math&gt; and &lt;math&gt;\triangle ABD&lt;/math&gt;. <br /> <br /> Hence, the ratio of the area of &lt;math&gt;\triangle DCE&lt;/math&gt; to the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is<br /> &lt;math&gt;\frac{\frac{\sqrt{2}}{9}}{\frac{\sqrt{2}}{3}}&lt;/math&gt; = &lt;math&gt;\frac{1}{3} \Rightarrow C&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Since &lt;math&gt;\triangle DCE&lt;/math&gt; and &lt;math&gt;\triangle ABD&lt;/math&gt; share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;DE&lt;/math&gt;. <br /> <br /> &lt;asy&gt;<br /> import graph;<br /> import olympiad;<br /> pair O,A,B,C,D,E,F;<br /> O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774);<br /> draw(Circle((0,0),15)); <br /> draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);draw(A--D);draw(D--B);<br /> label(&quot;A&quot;,A,NE);label(&quot;B&quot;,B,W);label(&quot;C&quot;,C,SE);label(&quot;D&quot;,D,NE);label(&quot;E&quot;,E,SW);label(&quot;O&quot;,O,SW);label(&quot;F&quot;,F,NW);<br /> markscalefactor=0.2;<br /> draw(anglemark(C,F,D),blue);draw(anglemark(D,C,B),blue);<br /> &lt;/asy&gt;<br /> &lt;math&gt;OD=r, OC=\frac{1}{3}r&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;m\angle DCO=m\angle DFC=90^\circ&lt;/math&gt;, then &lt;math&gt;\triangle DCO\cong \triangle DFC&lt;/math&gt;. So the ratio of the two altitudes is &lt;math&gt;\frac{CF}{DC}=\frac{OC}{DO}=\frac{1}{3}\Rightarrow \text{(C)}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Say the center of the circle is point &lt;math&gt;O&lt;/math&gt;;<br /> Without loss of generality, assume &lt;math&gt;AC=2&lt;/math&gt;, so &lt;math&gt;CB=4&lt;/math&gt; and the diameter and radius are &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;, respectively. Therefore, &lt;math&gt;CO=1&lt;/math&gt;, and &lt;math&gt;DO=3&lt;/math&gt;.<br /> The area of &lt;math&gt;\triangle DCE&lt;/math&gt; can be expressed as &lt;math&gt;\frac{1}{2}(CD)(6)\text{sin }(CDE).&lt;/math&gt; &lt;math&gt;\frac{1}{2}(CD)(6)&lt;/math&gt; happens to be the area of &lt;math&gt;\triangle ABD&lt;/math&gt;. Furthermore, &lt;math&gt;\text{sin } CDE = \frac{CO}{DO},&lt;/math&gt; or &lt;math&gt;\frac{1}{3}.&lt;/math&gt; Therefore, the ratio is &lt;math&gt;\frac{1}{3}.&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Ratio Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_14&diff=74295 2005 AMC 10A Problems/Problem 14 2016-01-04T02:11:17Z <p>Kkwang: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 &lt;/math&gt;<br /> <br /> ==Solution==<br /> If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits. <br /> <br /> Doing some [[casework]]: <br /> <br /> If the middle digit is &lt;math&gt;1&lt;/math&gt;, possible numbers range from &lt;math&gt;111&lt;/math&gt; to &lt;math&gt;210&lt;/math&gt;. So there are &lt;math&gt;2&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;2&lt;/math&gt;, possible numbers range from &lt;math&gt;123&lt;/math&gt; to &lt;math&gt;420&lt;/math&gt;. So there are &lt;math&gt;4&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;3&lt;/math&gt;, possible numbers range from &lt;math&gt;135&lt;/math&gt; to &lt;math&gt;630&lt;/math&gt;. So there are &lt;math&gt;6&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;4&lt;/math&gt;, possible numbers range from &lt;math&gt;147&lt;/math&gt; to &lt;math&gt;840&lt;/math&gt;. So there are &lt;math&gt;8&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;5&lt;/math&gt;, possible numbers range from &lt;math&gt;159&lt;/math&gt; to &lt;math&gt;951&lt;/math&gt;. So there are &lt;math&gt;9&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;6&lt;/math&gt;, possible numbers range from &lt;math&gt;369&lt;/math&gt; to &lt;math&gt;963&lt;/math&gt;. So there are &lt;math&gt;7&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;7&lt;/math&gt;, possible numbers range from &lt;math&gt;579&lt;/math&gt; to &lt;math&gt;975&lt;/math&gt;. So there are &lt;math&gt;5&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;8&lt;/math&gt;, possible numbers range from &lt;math&gt;789&lt;/math&gt; to &lt;math&gt;987&lt;/math&gt;. So there are &lt;math&gt;3&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;9&lt;/math&gt;, the only possible number is &lt;math&gt;999&lt;/math&gt;. So there is &lt;math&gt;1&lt;/math&gt; number in this case. <br /> <br /> So the total number of three-digit numbers that satisfy the property is &lt;math&gt;2+4+6+8+9+7+5+3+1=45\Rightarrow E&lt;/math&gt;<br /> <br /> == Solution 2 (much faster and slicker)==<br /> Alternatively, we could note that the middle digit is uniquely defined by the first and third digits, since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:<br /> <br /> If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are &lt;math&gt;5\cdot5=25&lt;/math&gt; numbers in this case.<br /> <br /> If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are &lt;math&gt;4\cdot5=20&lt;/math&gt; numbers here.<br /> <br /> The total number, then, is &lt;math&gt;20+25=45\Rightarrow E&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> By listing the first two hundred numbers you'll see a pattern. The (-) is where the middle number should go.<br /> <br /> 1-0 Can't do (1+0)/2=0.5<br /> <br /> 1-1 Can do (1+1)/2=1<br /> <br /> 1-2 Can't do (1+2)=3/2<br /> <br /> 1-3 Can do (1+3)/2=2<br /> <br /> 1-4 Can't do (1+4)=5/2<br /> <br /> 1-5 Can do (1+5)/2=2<br /> <br /> 1-6 Can't do (1+6)=7/2<br /> <br /> 1-7 Can do (1+7)/2=2<br /> <br /> 1-8 Can't do (1+8)=9/2<br /> <br /> 1-9 Can do (1+9)/2=2<br /> <br /> Total can do: 5<br /> <br /> 2-0 Can't do (2+0)/2=0.5<br /> <br /> 2-1 Can do (2+1)/2=1<br /> <br /> 2-2 Can't do (2+2)=3/2<br /> <br /> 2-3 Can do (2+3)/2=2<br /> <br /> 2-4 Can't do (2+4)=5/2<br /> <br /> 2-5 Can do (2+5)/2=2<br /> <br /> 2-6 Can't do (2+6)=7/2<br /> <br /> 2-7 Can do (2+7)/2=2<br /> <br /> 2-8 Can't do (2+8)=9/2<br /> <br /> 2-9 Can do (2+9)/2=2<br /> <br /> Total can do: 5<br /> <br /> SUMMARY<br /> <br /> You can now see the pattern. There are 5 can do's each hundred. So then 9*5=&lt;math&gt;45&lt;/math&gt;. <br /> ==See Also==<br /> *[[2005 AMC 10A Problems]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 13|Previous Problem]]<br /> <br /> *[[2005 AMC 10A Problems/Problem 15|Next Problem]]<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_22&diff=74257 2013 AMC 10B Problems/Problem 22 2016-01-03T01:19:42Z <p>Kkwang: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> The regular octagon &lt;math&gt;ABCDEFGH&lt;/math&gt; has its center at &lt;math&gt;J&lt;/math&gt;. Each of the vertices and the center are to be associated with one of the digits &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;9&lt;/math&gt;, with each digit used once, in such a way that the sums of the numbers on the lines &lt;math&gt;AJE&lt;/math&gt;, &lt;math&gt;BJF&lt;/math&gt;, &lt;math&gt;CJG&lt;/math&gt;, and &lt;math&gt;DJH&lt;/math&gt; are all equal. In how many ways can this be done?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 384 \qquad\textbf{(B)}\ 576 \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456 &lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G,H,J;<br /> A=(20,20(2+sqrt(2)));<br /> B=(20(1+sqrt(2)),20(2+sqrt(2)));<br /> C=(20(2+sqrt(2)),20(1+sqrt(2)));<br /> D=(20(2+sqrt(2)),20);<br /> E=(20(1+sqrt(2)),0);<br /> F=(20,0);<br /> G=(0,20);<br /> H=(0,20(1+sqrt(2)));<br /> J=(10(2+sqrt(2)),10(2+sqrt(2)));<br /> draw(A--B);<br /> draw(B--C);<br /> draw(C--D);<br /> draw(D--E);<br /> draw(E--F);<br /> draw(F--G);<br /> draw(G--H);<br /> draw(H--A);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(E);<br /> dot(F);<br /> dot(G);<br /> dot(H);<br /> dot(J);<br /> label(&quot;A&quot;,A,NNW);<br /> label(&quot;B&quot;,B,NNE);<br /> label(&quot;C&quot;,C,ENE);<br /> label(&quot;D&quot;,D,ESE);<br /> label(&quot;E&quot;,E,SSE);<br /> label(&quot;F&quot;,F,SSW);<br /> label(&quot;G&quot;,G,WSW);<br /> label(&quot;H&quot;,H,WNW);<br /> label(&quot;J&quot;,J,SE);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> <br /> First of all, note that &lt;math&gt;J&lt;/math&gt; must be &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, or &lt;math&gt;9&lt;/math&gt; to preserve symmetry. We also notice that &lt;math&gt;A+E = B+F = C+G = D+H&lt;/math&gt;. <br /> <br /> WLOG assume that &lt;math&gt;J = 1&lt;/math&gt;. Thus the pairs of vertices must be &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;. There are &lt;math&gt;4! = 24&lt;/math&gt; ways to assign these to the vertices. Furthermore, there are &lt;math&gt;2^{4} = 16&lt;/math&gt; ways to switch them (i.e. do &lt;math&gt;2&lt;/math&gt; &lt;math&gt;9&lt;/math&gt; instead of &lt;math&gt;9&lt;/math&gt; &lt;math&gt;2&lt;/math&gt;). <br /> <br /> Thus, there are &lt;math&gt;16(24) = 384&lt;/math&gt; ways for each possible J value. There are &lt;math&gt;3&lt;/math&gt; possible J values that still preserve symmetry: &lt;math&gt;384(3) = \boxed{\textbf{(C) }1152}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> As in solution 1, &lt;math&gt;J&lt;/math&gt; must be &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, or &lt;math&gt;9&lt;/math&gt; giving us 3 choices. Additionally &lt;math&gt;A+E = B+F = C+G = D+H&lt;/math&gt;. This means once we choose &lt;math&gt;J&lt;/math&gt; there are &lt;math&gt;8&lt;/math&gt; remaining choices. Going clockwise from &lt;math&gt;A&lt;/math&gt; we count, &lt;math&gt;8&lt;/math&gt; possibilities for &lt;math&gt;A&lt;/math&gt;. Choosing &lt;math&gt;A&lt;/math&gt; also determines &lt;math&gt;E&lt;/math&gt; which leaves &lt;math&gt;6&lt;/math&gt; choices for &lt;math&gt;B&lt;/math&gt;, once &lt;math&gt;B&lt;/math&gt; is chosen it also determines &lt;math&gt;F&lt;/math&gt; leaving &lt;math&gt;4&lt;/math&gt; choices for &lt;math&gt;C&lt;/math&gt;. Once &lt;math&gt;C&lt;/math&gt; is chosen it determines &lt;math&gt;G&lt;/math&gt; leaving &lt;math&gt;2&lt;/math&gt; choices for &lt;math&gt;D&lt;/math&gt;. Choosing &lt;math&gt;D&lt;/math&gt; determines &lt;math&gt;H&lt;/math&gt;, exhausting the numbers. To get the answer we multiply &lt;math&gt;2*4*6*8*3=\boxed{\textbf{(C) }1152}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2013|ab=B|num-b=21|num-a=23}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_17&diff=74240 2013 AMC 10A Problems/Problem 17 2016-01-02T05:40:06Z <p>Kkwang: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next 365-day period will exactly two friends visit her?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The 365-day time period can be split up into &lt;math&gt;6&lt;/math&gt; 60-day time periods, because after &lt;math&gt;60&lt;/math&gt; days, all three of them visit again (Least common multiple of &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt;).<br /> You can find how many times each pair of visitors can meet by finding the LCM of their visiting days and dividing that number by 60.<br /> Remember to subtract 1, because you do not wish to count the 60th day, when all three visit.<br /> <br /> A and B visit &lt;math&gt;\frac{60}{3 \cdot 4}-1 = 4&lt;/math&gt; times.<br /> B and C visit &lt;math&gt;\frac{60}{4 \cdot 5}-1 = 2&lt;/math&gt; times.<br /> C and A visit &lt;math&gt;\frac{60}{3 \cdot 5}-1 = 3&lt;/math&gt; times.<br /> <br /> This is a total of &lt;math&gt;9&lt;/math&gt; visits per &lt;math&gt;60&lt;/math&gt; day period.<br /> Therefore, the total number of 2-person visits is &lt;math&gt;9 \cdot 6 = \boxed{\textbf{(B) }54}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_24&diff=74091 2003 AMC 10B Problems/Problem 24 2015-12-30T04:16:13Z <p>Kkwang: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> The ﬁrst four terms in an arithmetic sequence are &lt;math&gt;x+y&lt;/math&gt;,&lt;math&gt;x-y&lt;/math&gt; ,&lt;math&gt;xy&lt;/math&gt; , and &lt;math&gt;\frac{x}{y}&lt;/math&gt;, in that order. What is the ﬁfth term? <br /> <br /> &lt;math&gt; \textbf{(A)}\ -\frac{15}{8}\qquad\textbf{(B)}\ -\frac{6}{5}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \frac{27}{20}\qquad\textbf{(E)}\ \frac{123}{40} &lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> <br /> The difference between consecutive terms is &lt;math&gt;(x-y)-(x+y)=-2y.&lt;/math&gt; Therefore we can also express the third and fourth terms as &lt;math&gt;x-3y&lt;/math&gt; and &lt;math&gt;x-5y.&lt;/math&gt; Then we can set them equal to &lt;math&gt;xy&lt;/math&gt; and &lt;math&gt;\frac{x}{y}&lt;/math&gt; because they are the same thing.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> xy&amp;=x-3y\\<br /> xy-x&amp;=-3y\\<br /> x(y-1)&amp;=-3y\\<br /> x&amp;=\frac{-3y}{y-1}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Substitute into our other equation.<br /> <br /> &lt;cmath&gt;<br /> \frac{x}{y}=x-5y&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{-3}{y-1}=\frac{-3y}{y-1}-5y&lt;/cmath&gt;<br /> &lt;cmath&gt;-3=-3y-5y(y-1)&lt;/cmath&gt;<br /> &lt;cmath&gt;0=5y^2-2y-3&lt;/cmath&gt;<br /> &lt;cmath&gt;0=(5y+3)(y-1)&lt;/cmath&gt;<br /> &lt;cmath&gt;y=-\frac35, 1&lt;/cmath&gt;<br /> <br /> But &lt;math&gt;y&lt;/math&gt; cannot be &lt;math&gt;1&lt;/math&gt; because then every term would be equal to &lt;math&gt;x.&lt;/math&gt; Therefore &lt;math&gt;y=-\frac35.&lt;/math&gt; Substituting the value for &lt;math&gt;y&lt;/math&gt; into any of the equations, we get &lt;math&gt;x=-\frac98.&lt;/math&gt; Finally,<br /> <br /> &lt;cmath&gt; \frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\boxed{\textbf{(E)}\ \frac{123}{40}}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2003|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_22&diff=74089 2003 AMC 10B Problems/Problem 22 2015-12-30T04:09:09Z <p>Kkwang: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A clock chimes once at &lt;math&gt;30&lt;/math&gt; minutes past each hour and chimes on the hour according to the hour. For example, at &lt;math&gt;1 \text{PM}&lt;/math&gt; there is one chime and at noon and midnight there are twelve chimes. Starting at &lt;math&gt;11:15 \text{AM}&lt;/math&gt; on &lt;math&gt;\text{February 26, 2003},&lt;/math&gt; on what date will the &lt;math&gt;2003^{\text{rd}}&lt;/math&gt; chime occur?<br /> <br /> &lt;math&gt;\textbf{(A) } \text{March 8} \qquad\textbf{(B) } \text{March 9} \qquad\textbf{(C) } \text{March 10} \qquad\textbf{(D) } \text{March 20} \qquad\textbf{(E) } \text{March 21}&lt;/math&gt;<br /> <br /> ==Solution==<br /> First, find how many chimes will have already happened before midnight (the beginning of the day) of &lt;math&gt;\text{February 27, 2003}.&lt;/math&gt; &lt;math&gt;13&lt;/math&gt; half-hours have passed, and the number of chimes according to the hour is &lt;math&gt;1+2+3+\cdots+12.&lt;/math&gt; The total number of chimes is &lt;math&gt;13+78=91&lt;/math&gt;<br /> <br /> Every day, there will be &lt;math&gt;24&lt;/math&gt; half-hours and &lt;math&gt;2(1+2+3+\cdots+12)&lt;/math&gt; chimes according to the arrow, resulting in &lt;math&gt;24+156=180&lt;/math&gt; total chimes.<br /> <br /> On &lt;math&gt;\text{February 27},&lt;/math&gt; the number of chimes that still need to occur is &lt;math&gt;2003-91=1912.&lt;/math&gt; &lt;math&gt;1912 \div 180=10 \text{R}112.&lt;/math&gt; Round up, and it is &lt;math&gt;11&lt;/math&gt; days past &lt;math&gt;\text{February 27},&lt;/math&gt; which is &lt;math&gt;\boxed{\textbf{(B) \ } \text{March 9}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2003|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_22&diff=74088 2003 AMC 10B Problems/Problem 22 2015-12-30T04:07:49Z <p>Kkwang: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A clock chimes once at &lt;math&gt;30&lt;/math&gt; minutes past each hour and chimes on the hour according to the hour. For example, at &lt;math&gt;1 \text{PM}&lt;/math&gt; there is one chime and at noon and midnight there are twelve chimes. Starting at &lt;math&gt;11:15 \text{AM}&lt;/math&gt; on &lt;math&gt;\text{February 26, 2003},&lt;/math&gt; on what date will the &lt;math&gt;2003^{\text{rd}}&lt;/math&gt; chime occur?<br /> <br /> &lt;math&gt;\textbf{(A) } \text{March 8} \qquad\textbf{(B) } \text{March 9} \qquad\textbf{(C) } \text{March 10} \qquad\textbf{(D) } \text{March 20} \qquad\textbf{(E) } \text{March 21}&lt;/math&gt;<br /> <br /> ==Solution==<br /> First, find how many chimes will have already happened before midnight (the beginning of the day) of &lt;math&gt;\text{February 27, 2003}.&lt;/math&gt; &lt;math&gt;13&lt;/math&gt; half-hours have passed, and the number of chimes according to the hour is &lt;math&gt;1+2+3+\cdots+12.&lt;/math&gt; The total number of chimes is &lt;math&gt;13+78=91&lt;/math&gt;<br /> <br /> Every day, there will be &lt;math&gt;24&lt;/math&gt; half-hours and &lt;math&gt;2(1+2+3+\cdots+12)&lt;/math&gt; chimes according to the arrow, resulting in &lt;math&gt;24+156=180&lt;/math&gt; total chimes.<br /> <br /> On &lt;math&gt;\text{February 27},&lt;/math&gt; the number of chimes that still need to occur is &lt;math&gt;2003-91=1912.&lt;/math&gt; &lt;math&gt;1912/180=10\text{r}112.&lt;/math&gt; Round up, and it is &lt;math&gt;11&lt;/math&gt; days past &lt;math&gt;\text{February 27},&lt;/math&gt; which is &lt;math&gt;\boxed{\textbf{(B) \ } \text{March 9}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2003|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_22&diff=74087 2003 AMC 10B Problems/Problem 22 2015-12-30T04:07:25Z <p>Kkwang: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A clock chimes once at &lt;math&gt;30&lt;/math&gt; minutes past each hour and chimes on the hour according to the hour. For example, at &lt;math&gt;1 \text{PM}&lt;/math&gt; there is one chime and at noon and midnight there are twelve chimes. Starting at &lt;math&gt;11:15 \text{AM}&lt;/math&gt; on &lt;math&gt;\text{February 26, 2003},&lt;/math&gt; on what date will the &lt;math&gt;2003^{\text{rd}}&lt;/math&gt; chime occur?<br /> <br /> &lt;math&gt;\textbf{(A) } \text{March 8} \qquad\textbf{(B) } \text{March 9} \qquad\textbf{(C) } \text{March 10} \qquad\textbf{(D) } \text{March 20} \qquad\textbf{(E) } \text{March 21}&lt;/math&gt;<br /> <br /> ==Solution==<br /> First, find how many chimes will have already happened before midnight (the beginning of the day) of &lt;math&gt;\text{February 27, 2003}.&lt;/math&gt; &lt;math&gt;13&lt;/math&gt; half-hours have passed, and the number of chimes according to the hour is &lt;math&gt;1+2+3+\cdots+12.&lt;/math&gt; The total number of chimes is &lt;math&gt;13+78=91&lt;/math&gt;<br /> <br /> Every day, there will be &lt;math&gt;24&lt;/math&gt; half-hours and &lt;math&gt;2(1+2+3+\cdots+12)&lt;/math&gt; chimes according to the arrow, resulting in &lt;math&gt;24+156=180&lt;/math&gt; total chimes.<br /> <br /> On &lt;math&gt;\text{February 27},&lt;/math&gt; the number of chimes that still need to occur is &lt;math&gt;&lt;/math&gt;2003-91=1912.&lt;math&gt; &lt;/math&gt;1912/180=10\text{r}112.&lt;math&gt;&lt;/math&gt; Round up, and it is &lt;math&gt;11&lt;/math&gt; days past &lt;math&gt;\text{February 27},&lt;/math&gt; which is &lt;math&gt;\boxed{\textbf{(B) \ } \text{March 9}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2003|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_20&diff=74084 2003 AMC 10B Problems/Problem 20 2015-12-30T04:00:50Z <p>Kkwang: /* Solution */</p> <hr /> <div>{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #14]] and [[2003 AMC 10B Problems|2003 AMC 10B #20]]}}<br /> <br /> ==Problem==<br /> In rectangle &lt;math&gt;ABCD, AB=5&lt;/math&gt; and &lt;math&gt;BC=3&lt;/math&gt;. Points &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; are on &lt;math&gt;\overline{CD}&lt;/math&gt; so that &lt;math&gt;DF=1&lt;/math&gt; and &lt;math&gt;GC=2&lt;/math&gt;. Lines &lt;math&gt;AF&lt;/math&gt; and &lt;math&gt;BG&lt;/math&gt; intersect at &lt;math&gt;E&lt;/math&gt;. Find the area of &lt;math&gt;\triangle AEB&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(A) } 10 \qquad\textbf{(B) } \frac{21}{2} \qquad\textbf{(C) } 12 \qquad\textbf{(D) } \frac{25}{2} \qquad\textbf{(E) } 15&lt;/math&gt;<br /> <br /> ==Solution==<br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(8mm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=4;<br /> <br /> pair A=(0,0), B=(5,0), C=(5,3), D=(0,3);<br /> pair F=(1,3), G=(3,3);<br /> pair E=(5/3,5);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(A--E);<br /> draw(B--E);<br /> <br /> pair[] ps={A,B,C,D,E,F,G}; dot(ps);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$E$&quot;,E,N);<br /> label(&quot;$F$&quot;,F,SE);<br /> label(&quot;$G$&quot;,G,SW);<br /> label(&quot;$1$&quot;,midpoint(D--F),N);<br /> label(&quot;$2$&quot;,midpoint(G--C),N);<br /> label(&quot;$5$&quot;,midpoint(A--B),S);<br /> label(&quot;$3$&quot;,midpoint(A--D),W);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> &lt;math&gt;\triangle EFG \sim \triangle EAB&lt;/math&gt; because &lt;math&gt;FG \parallel AB.&lt;/math&gt; The ratio of &lt;math&gt;\triangle EFG&lt;/math&gt; to &lt;math&gt;\triangle EAB&lt;/math&gt; is &lt;math&gt;2:5&lt;/math&gt; since &lt;math&gt;AB=5&lt;/math&gt; and &lt;math&gt;FG=2&lt;/math&gt; from subtraction. If we let &lt;math&gt;h&lt;/math&gt; be the height of &lt;math&gt;\triangle EAB,&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\frac{2}{5} = \frac{h-3}{h}&lt;/cmath&gt;<br /> &lt;cmath&gt;2h = 5h-15&lt;/cmath&gt;<br /> &lt;cmath&gt;3h = 15&lt;/cmath&gt;<br /> &lt;cmath&gt;h = 5&lt;/cmath&gt;<br /> <br /> The height is &lt;math&gt;5&lt;/math&gt; so the area of &lt;math&gt;\triangle EAB&lt;/math&gt; is &lt;math&gt;\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2003|ab=B|num-b=13|num-a=15}}<br /> {{AMC10 box|year=2003|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_20&diff=74054 2002 AMC 10A Problems/Problem 20 2015-12-29T01:38:52Z <p>Kkwang: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Points &lt;math&gt;A,B,C,D,E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; lie, in that order, on &lt;math&gt;\overline{AF}&lt;/math&gt;, dividing it into five segments, each of length 1. Point &lt;math&gt;G&lt;/math&gt; is not on line &lt;math&gt;AF&lt;/math&gt;. Point &lt;math&gt;H&lt;/math&gt; lies on &lt;math&gt;\overline{GD}&lt;/math&gt;, and point &lt;math&gt;J&lt;/math&gt; lies on &lt;math&gt;\overline{GF}&lt;/math&gt;. The line segments &lt;math&gt;\overline{HC}, \overline{JE},&lt;/math&gt; and &lt;math&gt;\overline{AG}&lt;/math&gt; are parallel. Find &lt;math&gt;HC/JE&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{CH}&lt;/math&gt; are parallel, triangles &lt;math&gt;\triangle GAD&lt;/math&gt; and &lt;math&gt;\triangle HCD&lt;/math&gt; are similar. Hence, &lt;math&gt;\frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{JE}&lt;/math&gt; are parallel, triangles &lt;math&gt;\triangle GAF&lt;/math&gt; and &lt;math&gt;\triangle JEF&lt;/math&gt; are similar. Hence, &lt;math&gt;\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{CH}{EJ} = (\frac{CH}{AG})\div(\frac{EJ}{AG}) = (\frac{1}{3})\div(\frac{1}{5}) = \boxed{\frac{5}{3}}&lt;/math&gt;. The answer is &lt;math&gt;\boxed{(D) 5/3}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> As &lt;math&gt;\overline{JE}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AG}&lt;/math&gt;, angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. From AA similarity, &lt;math&gt;\triangle AGF \sim \triangle EJF&lt;/math&gt;; hence &lt;math&gt;\frac {AG}{JE} =5&lt;/math&gt;. Similarly, &lt;math&gt;\frac {AG}{HC} = 3&lt;/math&gt;. Thus, &lt;math&gt;\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2002|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_20&diff=74053 2002 AMC 10A Problems/Problem 20 2015-12-29T01:36:05Z <p>Kkwang: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Points &lt;math&gt;A,B,C,D,E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; lie, in that order, on &lt;math&gt;\overline{AF}&lt;/math&gt;, dividing it into five segments, each of length 1. Point &lt;math&gt;G&lt;/math&gt; is not on line &lt;math&gt;AF&lt;/math&gt;. Point &lt;math&gt;H&lt;/math&gt; lies on &lt;math&gt;\overline{GD}&lt;/math&gt;, and point &lt;math&gt;J&lt;/math&gt; lies on &lt;math&gt;\overline{GF}&lt;/math&gt;. The line segments &lt;math&gt;\overline{HC}, \overline{JE},&lt;/math&gt; and &lt;math&gt;\overline{AG}&lt;/math&gt; are parallel. Find &lt;math&gt;HC/JE&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{CH}&lt;/math&gt; are parallel, triangles &lt;math&gt;GAD&lt;/math&gt; and &lt;math&gt;HCD&lt;/math&gt; are similar. Hence, &lt;math&gt;\frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{JE}&lt;/math&gt; are parallel, triangles &lt;math&gt;GAF&lt;/math&gt; and &lt;math&gt;JEF&lt;/math&gt; are similar. Hence, &lt;math&gt;\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{CH}{EJ} = (\frac{CH}{AG})\div(\frac{EJ}{AG}) = (\frac{1}{3})\div(\frac{1}{5}) = \boxed{\frac{5}{3}}&lt;/math&gt;. The answer is &lt;math&gt;\boxed{(D)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> As &lt;math&gt;\overline{JE}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AG}&lt;/math&gt;, angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. From AA similarity, &lt;math&gt;\triangle AGF \sim \triangle EJF&lt;/math&gt;; hence &lt;math&gt;\frac {AG}{JE} =5&lt;/math&gt;. Similarly, &lt;math&gt;\frac {AG}{HC} = 3&lt;/math&gt;. Thus, &lt;math&gt;\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2002|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_20&diff=74052 2002 AMC 10A Problems/Problem 20 2015-12-29T01:35:31Z <p>Kkwang: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Points &lt;math&gt;A,B,C,D,E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; lie, in that order, on &lt;math&gt;\overline{AF}&lt;/math&gt;, dividing it into five segments, each of length 1. Point &lt;math&gt;G&lt;/math&gt; is not on line &lt;math&gt;AF&lt;/math&gt;. Point &lt;math&gt;H&lt;/math&gt; lies on &lt;math&gt;\overline{GD}&lt;/math&gt;, and point &lt;math&gt;J&lt;/math&gt; lies on &lt;math&gt;\overline{GF}&lt;/math&gt;. The line segments &lt;math&gt;\overline{HC}, \overline{JE},&lt;/math&gt; and &lt;math&gt;\overline{AG}&lt;/math&gt; are parallel. Find &lt;math&gt;HC/JE&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{CH}&lt;/math&gt; are parallel, from AA similarity, triangles &lt;math&gt;GAD&lt;/math&gt; and &lt;math&gt;HCD&lt;/math&gt; are similar. Hence, &lt;math&gt;\frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{JE}&lt;/math&gt; are parallel, from AA similarity, triangles &lt;math&gt;GAF&lt;/math&gt; and &lt;math&gt;JEF&lt;/math&gt; are similar. Hence, &lt;math&gt;\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{CH}{EJ} = (\frac{CH}{AG})\div(\frac{EJ}{AG}) = (\frac{1}{3})\div(\frac{1}{5}) = \boxed{\frac{5}{3}}&lt;/math&gt;. The answer is &lt;math&gt;\boxed{(D)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> As &lt;math&gt;\overline{JE}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AG}&lt;/math&gt;, angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. From AA similarity, &lt;math&gt;\triangle AGF \sim \triangle EJF&lt;/math&gt;; hence &lt;math&gt;\frac {AG}{JE} =5&lt;/math&gt;. Similarly, &lt;math&gt;\frac {AG}{HC} = 3&lt;/math&gt;. Thus, &lt;math&gt;\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2002|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_20&diff=74051 2002 AMC 10A Problems/Problem 20 2015-12-29T01:30:25Z <p>Kkwang: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Points &lt;math&gt;A,B,C,D,E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; lie, in that order, on &lt;math&gt;\overline{AF}&lt;/math&gt;, dividing it into five segments, each of length 1. Point &lt;math&gt;G&lt;/math&gt; is not on line &lt;math&gt;AF&lt;/math&gt;. Point &lt;math&gt;H&lt;/math&gt; lies on &lt;math&gt;\overline{GD}&lt;/math&gt;, and point &lt;math&gt;J&lt;/math&gt; lies on &lt;math&gt;\overline{GF}&lt;/math&gt;. The line segments &lt;math&gt;\overline{HC}, \overline{JE},&lt;/math&gt; and &lt;math&gt;\overline{AG}&lt;/math&gt; are parallel. Find &lt;math&gt;HC/JE&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{CH}&lt;/math&gt; are parallel, from AA similarity, triangles &lt;math&gt;GAD&lt;/math&gt; and &lt;math&gt;HCD&lt;/math&gt; are similar. Hence, &lt;math&gt;\frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{JE}&lt;/math&gt; are parallel, from AA similarity, triangles &lt;math&gt;GAF&lt;/math&gt; and &lt;math&gt;JEF&lt;/math&gt; are similar. Hence, &lt;math&gt;\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{CH}{EJ} = (\frac{CH}{AG})/(\frac{EJ}{AG}) = (\frac{1}{3})/(\frac{1}{5}) = \boxed{\frac{5}{3}}&lt;/math&gt;. The answer is &lt;math&gt;\boxed{(D)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> As &lt;math&gt;\overline{JE}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AG}&lt;/math&gt;, angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. From AA similarity, &lt;math&gt;\triangle AGF \sim \triangle EJF&lt;/math&gt;; hence &lt;math&gt;\frac {AG}{JE} =5&lt;/math&gt;. Similarly, &lt;math&gt;\frac {AG}{HC} = 3&lt;/math&gt;. Thus, &lt;math&gt;\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2002|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_20&diff=74050 2002 AMC 10A Problems/Problem 20 2015-12-29T01:27:25Z <p>Kkwang: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Points &lt;math&gt;A,B,C,D,E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; lie, in that order, on &lt;math&gt;\overline{AF}&lt;/math&gt;, dividing it into five segments, each of length 1. Point &lt;math&gt;G&lt;/math&gt; is not on line &lt;math&gt;AF&lt;/math&gt;. Point &lt;math&gt;H&lt;/math&gt; lies on &lt;math&gt;\overline{GD}&lt;/math&gt;, and point &lt;math&gt;J&lt;/math&gt; lies on &lt;math&gt;\overline{GF}&lt;/math&gt;. The line segments &lt;math&gt;\overline{HC}, \overline{JE},&lt;/math&gt; and &lt;math&gt;\overline{AG}&lt;/math&gt; are parallel. Find &lt;math&gt;HC/JE&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{CH}&lt;/math&gt; are parallel, from AA similarity, triangles &lt;math&gt;GAD&lt;/math&gt; and &lt;math&gt;HCD&lt;/math&gt; are similar. Hence, &lt;math&gt;\frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{JE}&lt;/math&gt; are parallel, triangles &lt;math&gt;GAF&lt;/math&gt; and &lt;math&gt;JEF&lt;/math&gt; are similar. Hence, &lt;math&gt;\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{CH}{EJ} = (\frac{CH}{AG})/(\frac{EJ}{AG}) = (\frac{1}{3})/(\frac{1}{5}) = \boxed{\frac{5}{3}}&lt;/math&gt;. The answer is &lt;math&gt;\boxed{(D)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> As &lt;math&gt;\overline{JE}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AG}&lt;/math&gt;, angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. From AA similarity, &lt;math&gt;\triangle AGF \sim \triangle EJF&lt;/math&gt;; hence &lt;math&gt;\frac {AG}{JE} =5&lt;/math&gt;. Similarly, &lt;math&gt;\frac {AG}{HC} = 3&lt;/math&gt;. Thus, &lt;math&gt;\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2002|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_20&diff=74049 2002 AMC 10A Problems/Problem 20 2015-12-29T01:26:46Z <p>Kkwang: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Points &lt;math&gt;A,B,C,D,E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; lie, in that order, on &lt;math&gt;\overline{AF}&lt;/math&gt;, dividing it into five segments, each of length 1. Point &lt;math&gt;G&lt;/math&gt; is not on line &lt;math&gt;AF&lt;/math&gt;. Point &lt;math&gt;H&lt;/math&gt; lies on &lt;math&gt;\overline{GD}&lt;/math&gt;, and point &lt;math&gt;J&lt;/math&gt; lies on &lt;math&gt;\overline{GF}&lt;/math&gt;. The line segments &lt;math&gt;\overline{HC}, \overline{JE},&lt;/math&gt; and &lt;math&gt;\overline{AG}&lt;/math&gt; are parallel. Find &lt;math&gt;HC/JE&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{CH}&lt;/math&gt; are parallel,from AA similarity, triangles &lt;math&gt;GAD&lt;/math&gt; and &lt;math&gt;HCD&lt;/math&gt; are similar. Hence, &lt;math&gt;\frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{JE}&lt;/math&gt; are parallel, triangles &lt;math&gt;GAF&lt;/math&gt; and &lt;math&gt;JEF&lt;/math&gt; are similar. Hence, &lt;math&gt;\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{CH}{EJ} = (\frac{CH}{AG})/(\frac{EJ}{AG}) = (\frac{1}{3})/(\frac{1}{5}) = \boxed{\frac{5}{3}}&lt;/math&gt;. The answer is &lt;math&gt;\boxed{(D)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> As &lt;math&gt;\overline{JE}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AG}&lt;/math&gt;, angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. From AA similarity, &lt;math&gt;\triangle AGF \sim \triangle EJF&lt;/math&gt;; hence &lt;math&gt;\frac {AG}{JE} =5&lt;/math&gt;. Similarly, &lt;math&gt;\frac {AG}{HC} = 3&lt;/math&gt;. Thus, &lt;math&gt;\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2002|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_20&diff=74042 2002 AMC 10A Problems/Problem 20 2015-12-28T23:27:21Z <p>Kkwang: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Points &lt;math&gt;A,B,C,D,E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; lie, in that order, on &lt;math&gt;\overline{AF}&lt;/math&gt;, dividing it into five segments, each of length 1. Point &lt;math&gt;G&lt;/math&gt; is not on line &lt;math&gt;AF&lt;/math&gt;. Point &lt;math&gt;H&lt;/math&gt; lies on &lt;math&gt;\overline{GD}&lt;/math&gt;, and point &lt;math&gt;J&lt;/math&gt; lies on &lt;math&gt;\overline{GF}&lt;/math&gt;. The line segments &lt;math&gt;\overline{HC}, \overline{JE},&lt;/math&gt; and &lt;math&gt;\overline{AG}&lt;/math&gt; are parallel. Find &lt;math&gt;HC/JE&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{CH}&lt;/math&gt; are parallel, triangles &lt;math&gt;GAD&lt;/math&gt; and &lt;math&gt;HCD&lt;/math&gt; are similar. Hence, &lt;math&gt;\frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{JE}&lt;/math&gt; are parallel, triangles &lt;math&gt;GAF&lt;/math&gt; and &lt;math&gt;JEF&lt;/math&gt; are similar. Hence, &lt;math&gt;\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{CH}{EJ} = (\frac{CH}{AG})/(\frac{EJ}{AG}) = (\frac{1}{3})/(\frac{1}{5}) = \boxed{\frac{5}{3}}&lt;/math&gt;. The answer is &lt;math&gt;\boxed{(D)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> As &lt;math&gt;\overline{JE}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AG}&lt;/math&gt;, angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. From AA similarity, &lt;math&gt;\triangle AGF \sim \triangle EJF&lt;/math&gt;; hence &lt;math&gt;\frac {AG}{JE} =5&lt;/math&gt;. Similarly, &lt;math&gt;\frac {AG}{HC} = 3&lt;/math&gt;. Thus, &lt;math&gt;\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2002|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Kkwang https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_20&diff=74038 2002 AMC 10A Problems/Problem 20 2015-12-28T23:15:31Z <p>Kkwang: /* Solution */</p> <hr /> <div>==Problem==<br /> Points &lt;math&gt;A,B,C,D,E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; lie, in that order, on &lt;math&gt;\overline{AF}&lt;/math&gt;, dividing it into five segments, each of length 1. Point &lt;math&gt;G&lt;/math&gt; is not on line &lt;math&gt;AF&lt;/math&gt;. Point &lt;math&gt;H&lt;/math&gt; lies on &lt;math&gt;\overline{GD}&lt;/math&gt;, and point &lt;math&gt;J&lt;/math&gt; lies on &lt;math&gt;\overline{GF}&lt;/math&gt;. The line segments &lt;math&gt;\overline{HC}, \overline{JE},&lt;/math&gt; and &lt;math&gt;\overline{AG}&lt;/math&gt; are parallel. Find &lt;math&gt;HC/JE&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since &lt;math&gt;AG&lt;/math&gt; and &lt;math&gt;CH&lt;/math&gt; are parallel, triangles &lt;math&gt;GAD&lt;/math&gt; and &lt;math&gt;HCD&lt;/math&gt; are similar. Hence, &lt;math&gt;CH/AG = CD/AD = 1/3&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;AG&lt;/math&gt; and &lt;math&gt;JE&lt;/math&gt; are parallel, triangles &lt;math&gt;GAF&lt;/math&gt; and &lt;math&gt;JEF&lt;/math&gt; are similar. Hence, &lt;math&gt;EJ/AG = EF/AF = 1/5&lt;/math&gt;. Therefore, &lt;math&gt;CH/EJ = (CH/AG)/(EJ/AG) = (1/3)/(1/5) = \boxed{5/3}&lt;/math&gt;. The answer is (D).<br /> <br /> ==Solution 2==<br /> As &lt;math&gt;\overline{JE}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AG}&lt;/math&gt;, angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. From AA similarity, &lt;math&gt;\triangle AGF \sim \triangle EJF&lt;/math&gt;; hence &lt;math&gt;\frac {AG}{JE} =5&lt;/math&gt;. Similarly, &lt;math&gt;\frac {AG}{HC} = 3&lt;/math&gt;. Thus, &lt;math&gt;\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2002|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Kkwang