https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Crazyeyemoody907&feedformat=atom AoPS Wiki - User contributions [en] 2020-09-19T23:07:28Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_11&diff=122531 1988 AIME Problems/Problem 11 2020-05-16T19:09:48Z <p>Crazyeyemoody907: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;w_1, w_2, \dots, w_n&lt;/math&gt; be [[complex number]]s. A line &lt;math&gt;L&lt;/math&gt; in the [[complex plane]] is called a mean [[line]] for the [[point]]s &lt;math&gt;w_1, w_2, \dots, w_n&lt;/math&gt; if &lt;math&gt;L&lt;/math&gt; contains points (complex numbers) &lt;math&gt;z_1, z_2, \dots, z_n&lt;/math&gt; such that<br /> &lt;cmath&gt;<br /> \sum_{k = 1}^n (z_k - w_k) = 0.<br /> &lt;/cmath&gt;<br /> For the numbers &lt;math&gt;w_1 = 32 + 170i&lt;/math&gt;, &lt;math&gt;w_2 = - 7 + 64i&lt;/math&gt;, &lt;math&gt;w_3 = - 9 + 200i&lt;/math&gt;, &lt;math&gt;w_4 = 1 + 27i&lt;/math&gt;, and &lt;math&gt;w_5 = - 14 + 43i&lt;/math&gt;, there is a unique mean line with &lt;math&gt;y&lt;/math&gt;-intercept 3. Find the [[slope]] of this mean line.<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> &lt;math&gt;\sum_{k=1}^5 z_k - \sum_{k=1}^5 w_k = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{k=1}^5 z_k = 3 + 504i&lt;/math&gt;<br /> <br /> Each &lt;math&gt;z_k = x_k + y_ki&lt;/math&gt; lies on the complex line &lt;math&gt;y = mx + 3&lt;/math&gt;, so we can rewrite this as <br /> <br /> &lt;math&gt;\sum_{k=1}^5 z_k = \sum_{k=1}^5 x_k + \sum_{k=1}^n y_ki&lt;/math&gt;<br /> <br /> &lt;math&gt;3 + 504i = \sum_{k=1}^5 x_k + i \sum_{k=1}^5 (mx_k + 3)&lt;/math&gt;<br /> <br /> Matching the real parts and the imaginary parts, we get that &lt;math&gt;\sum_{k=1}^5 x_k = 3&lt;/math&gt; and &lt;math&gt;\sum_{k=1}^5 (mx_k + 3) = 504&lt;/math&gt;. Simplifying the second summation, we find that &lt;math&gt;m\sum_{k=1}^5 x_k = 504 - 3 \cdot 5 = 489&lt;/math&gt;, and substituting, the answer is &lt;math&gt;m \cdot 3 = 489 \Longrightarrow m = 163&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> We know that <br /> <br /> &lt;math&gt;\sum_{k=1}^5 w_k = 3 + 504i&lt;/math&gt;<br /> <br /> And because the sum of the 5 &lt;math&gt;z&lt;/math&gt;'s must cancel this out,<br /> <br /> &lt;math&gt;\sum_{k=1}^5 z_k = 3 + 504i&lt;/math&gt;<br /> <br /> We write the numbers in the form &lt;math&gt;a + bi&lt;/math&gt; and we know that <br /> <br /> &lt;math&gt;\sum_{k=1}^5 a_k = 3&lt;/math&gt; and &lt;math&gt;\sum_{k=1}^5 b_k = 504&lt;/math&gt;<br /> <br /> The line is of equation &lt;math&gt;y=mx+3&lt;/math&gt;. Substituting in the polar coordinates, we have &lt;math&gt;b_k = ma_k + 3&lt;/math&gt;.<br /> <br /> Summing all 5 of the equations given for each &lt;math&gt;k&lt;/math&gt;, we get <br /> <br /> &lt;math&gt;504 = 3m + 15&lt;/math&gt;<br /> <br /> Solving for &lt;math&gt;m&lt;/math&gt;, the slope, we get &lt;math&gt;\boxed{163}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_11&diff=122530 1988 AIME Problems/Problem 11 2020-05-16T19:09:37Z <p>Crazyeyemoody907: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;w_1, w_2, \dots, w_n&lt;/math&gt; be [[complex number]]s. A line &lt;math&gt;L&lt;/math&gt; in the [[complex plane]] is called a mean [[line]] for the [[point]]s &lt;math&gt;w_1, w_2, \dots, w_n&lt;/math&gt; if &lt;math&gt;L&lt;/math&gt; contains points (complex numbers) &lt;math&gt;z_1, z_2, \dots, z_n&lt;/math&gt; such that<br /> &lt;cmath&gt;<br /> \sum_{k = 1}^n (z_k - w_k) = 0.<br /> &lt;/cmath&gt;<br /> For the numbers &lt;math&gt;w_1 = 32 + 170i&lt;/math&gt;, &lt;math&gt;w_2 = - 7 + 64i&lt;/math&gt;, &lt;math&gt;w_3 = - 9 + 200i&lt;/math&gt;, &lt;math&gt;w_4 = 1 + 27i&lt;/math&gt;, and &lt;math&gt;w_5 = - 14 + 43i&lt;/math&gt;, there is a unique mean line with &lt;math&gt;y&lt;/math&gt;-intercept 3. Find the [[slope]] of this mean line.<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> &lt;math&gt;\sum_{k=1}^5 z_k - \sum_{k=1}^5 w_k = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{k=1}^5 z_k = 3 + 504i&lt;/math&gt;<br /> <br /> Each &lt;math&gt;z_k = x_k + y_ki&lt;/math&gt; lies on the complex line &lt;math&gt;y = mx + 3&lt;/math&gt;, so we can rewrite this as <br /> <br /> &lt;math&gt;\sum_{k=1}^5 z_k = \sum_{k=1}^5 x_k + \sum_{k=1}^n y_ki&lt;/math&gt;<br /> <br /> &lt;math&gt;3 + 504i = \sum_{k=1}^5 x_k + i \sum_{k=1}^5 (mx_k + 3)&lt;/math&gt;<br /> <br /> Matching the real parts and the imaginary parts, we get that &lt;math&gt;\sum_{k=1}^5 x_k = 3&lt;/math&gt; and &lt;math&gt;\sum_{k=1}^5 (mx_k + 3) = 504&lt;/math&gt;. Simplifying the second summation, we find that &lt;math&gt;m\sum_{k=1}^5 x_k = 504 - 3 \cdot 5 = 489&lt;/math&gt;, and substituting, the answer is &lt;math&gt;m \cdot 3 = 489 \Longrightarrow m = 163&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We know that <br /> <br /> &lt;math&gt;\sum_{k=1}^5 w_k = 3 + 504i&lt;/math&gt;<br /> <br /> And because the sum of the 5 &lt;math&gt;z&lt;/math&gt;'s must cancel this out,<br /> <br /> &lt;math&gt;\sum_{k=1}^5 z_k = 3 + 504i&lt;/math&gt;<br /> <br /> We write the numbers in the form &lt;math&gt;a + bi&lt;/math&gt; and we know that <br /> <br /> &lt;math&gt;\sum_{k=1}^5 a_k = 3&lt;/math&gt; and &lt;math&gt;\sum_{k=1}^5 b_k = 504&lt;/math&gt;<br /> <br /> The line is of equation &lt;math&gt;y=mx+3&lt;/math&gt;. Substituting in the polar coordinates, we have &lt;math&gt;b_k = ma_k + 3&lt;/math&gt;.<br /> <br /> Summing all 5 of the equations given for each &lt;math&gt;k&lt;/math&gt;, we get <br /> <br /> &lt;math&gt;504 = 3m + 15&lt;/math&gt;<br /> <br /> Solving for &lt;math&gt;m&lt;/math&gt;, the slope, we get &lt;math&gt;\boxed{163}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_9&diff=122529 1988 AIME Problems/Problem 9 2020-05-16T19:04:34Z <p>Crazyeyemoody907: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Find the smallest positive integer whose [[perfect cube|cube]] ends in &lt;math&gt;888&lt;/math&gt;.<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of &lt;math&gt;(10k + 2)^3&lt;/math&gt;; using the [[binomial theorem]] gives us &lt;math&gt;1000k^3 + 600k^2 + 120k + 8&lt;/math&gt;. Since we are looking for the tens digit, &lt;math&gt;\mod{100}&lt;/math&gt; we get &lt;math&gt;20k + 8 \equiv 88 \pmod{100}&lt;/math&gt;. This is true if the tens digit is either &lt;math&gt;4&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;. Casework:<br /> *&lt;math&gt;4&lt;/math&gt;: Then our cube must be in the form of &lt;math&gt;(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}&lt;/math&gt;. Hence the lowest possible value for the hundreds digit is &lt;math&gt;4&lt;/math&gt;, and so &lt;math&gt;442&lt;/math&gt; is a valid solution. <br /> *&lt;math&gt;9&lt;/math&gt;: Then our cube is &lt;math&gt;(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}&lt;/math&gt;. The lowest possible value for the hundreds digit is &lt;math&gt;1&lt;/math&gt;, and we get &lt;math&gt;192&lt;/math&gt;. Hence, since &lt;math&gt;192 &lt; 442&lt;/math&gt;, the answer is &lt;math&gt;\fbox{192}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> &lt;math&gt;n^3 \equiv 888 \pmod{1000} \implies n^3 \equiv 0 \pmod 8&lt;/math&gt; and &lt;math&gt;n^3 \equiv 13 \pmod{125}&lt;/math&gt;.<br /> &lt;math&gt;n \equiv 2 \pmod 5&lt;/math&gt; due to the last digit of &lt;math&gt;n^3&lt;/math&gt;. Let &lt;math&gt;n = 5a + 2&lt;/math&gt;. By expanding, &lt;math&gt;125a^3 + 150a^2 + 60a + 8 \equiv 13 \pmod{125} \implies 5a^2 + 12a \equiv 1 \pmod{25}&lt;/math&gt;. <br /> <br /> By looking at the last digit again, we see &lt;math&gt;a \equiv 3 \pmod5&lt;/math&gt;, so we let &lt;math&gt;a = 5a_1 + 3&lt;/math&gt; where &lt;math&gt;a_1 \in \mathbb{Z^+}&lt;/math&gt;. Plugging this in to &lt;math&gt;5a^2 + 12a \equiv 1 \pmod{25}&lt;/math&gt; gives &lt;math&gt;10a_1 + 6 \equiv 1 \pmod{25}&lt;/math&gt;. Obviously, &lt;math&gt;a_1 \equiv 2 \pmod 5&lt;/math&gt;, so we let &lt;math&gt;a_1 = 5a_2 + 2&lt;/math&gt; where &lt;math&gt;a_2&lt;/math&gt; can be any non-negative integer. <br /> <br /> Therefore, &lt;math&gt;n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67&lt;/math&gt;. &lt;math&gt;n&lt;/math&gt; must also be a multiple of &lt;math&gt;8&lt;/math&gt;, so &lt;math&gt;125a_2 + 67 \equiv 5a_2 + 3 \pmod 8 \implies a_2 = 1,9,17 \ldots&lt;/math&gt;. Therefore, the minimum of &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;125 + 67 = \boxed{192}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Let &lt;math&gt;x^3 = 1000a + 888&lt;/math&gt;. We factor an &lt;math&gt;8&lt;/math&gt; out of the right hand side, and we note that &lt;math&gt;x&lt;/math&gt; must be of the form &lt;math&gt;x = 2y&lt;/math&gt;, where &lt;math&gt;y&lt;/math&gt; is a positive integer. Then, this becomes &lt;math&gt;y^3 = 125a + 111&lt;/math&gt;. Taking mod &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;25&lt;/math&gt;, and &lt;math&gt;125&lt;/math&gt;, we get &lt;math&gt;y^3 \equiv 1\pmod 5&lt;/math&gt;, &lt;math&gt;y^3 \equiv 11\pmod{25}&lt;/math&gt;, and &lt;math&gt;y^3 \equiv 111\pmod{125}&lt;/math&gt;. <br /> <br /> We can work our way up, and find that &lt;math&gt;y\equiv 1\pmod 5&lt;/math&gt;, &lt;math&gt;y\equiv 21\pmod{25}&lt;/math&gt;, and finally &lt;math&gt;y\equiv 96\pmod{125}&lt;/math&gt;. This gives us our smallest value, &lt;math&gt;y = 96&lt;/math&gt;, so &lt;math&gt;x = \boxed{192}&lt;/math&gt;, as desired. - Spacesam<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_9&diff=122528 1988 AIME Problems/Problem 9 2020-05-16T19:04:15Z <p>Crazyeyemoody907: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Find the smallest positive integer whose [[perfect cube|cube]] ends in &lt;math&gt;888&lt;/math&gt;.<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of &lt;math&gt;(10k + 2)^3&lt;/math&gt;; using the [[binomial theorem]] gives us &lt;math&gt;1000k^3 + 600k^2 + 120k + 8&lt;/math&gt;. Since we are looking for the tens digit, &lt;math&gt;\mod{100}&lt;/math&gt; we get &lt;math&gt;20k + 8 \equiv 88 \pmod{100}&lt;/math&gt;. This is true if the tens digit is either &lt;math&gt;4&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;. Casework:<br /> *&lt;math&gt;4&lt;/math&gt;: Then our cube must be in the form of &lt;math&gt;(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}&lt;/math&gt;. Hence the lowest possible value for the hundreds digit is &lt;math&gt;4&lt;/math&gt;, and so &lt;math&gt;442&lt;/math&gt; is a valid solution. <br /> *&lt;math&gt;9&lt;/math&gt;: Then our cube is &lt;math&gt;(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}&lt;/math&gt;. The lowest possible value for the hundreds digit is &lt;math&gt;1&lt;/math&gt;, and we get &lt;math&gt;192&lt;/math&gt;. Hence, since &lt;math&gt;192 &lt; 442&lt;/math&gt;, the answer is &lt;math&gt;\fbox{192}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> &lt;math&gt;n^3 \equiv 888 \pmod{1000} \implies n^3 \equiv 0 \pmod 8&lt;/math&gt; and &lt;math&gt;n^3 \equiv 13 \pmod{125}&lt;/math&gt;.<br /> &lt;math&gt;n \equiv 2 \pmod 5&lt;/math&gt; due to the last digit of &lt;math&gt;n^3&lt;/math&gt;. Let &lt;math&gt;n = 5a + 2&lt;/math&gt;. By expanding, &lt;math&gt;125a^3 + 150a^2 + 60a + 8 \equiv 13 \pmod{125} \implies 5a^2 + 12a \equiv 1 \pmod{25}&lt;/math&gt;. <br /> <br /> By looking at the last digit again, we see &lt;math&gt;a \equiv 3 \pmod5&lt;/math&gt;, so we let &lt;math&gt;a = 5a_1 + 3&lt;/math&gt; where &lt;math&gt;a_1 \in \mathbb{Z^+}&lt;/math&gt;. Plugging this in to &lt;math&gt;5a^2 + 12a \equiv 1 \pmod{25}&lt;/math&gt; gives &lt;math&gt;10a_1 + 6 \equiv 1 \pmod{25}&lt;/math&gt;. Obviously, &lt;math&gt;a_1 \equiv 2 \pmod 5&lt;/math&gt;, so we let &lt;math&gt;a_1 = 5a_2 + 2&lt;/math&gt; where &lt;math&gt;a_2&lt;/math&gt; can be any non-negative integer. <br /> <br /> Therefore, &lt;math&gt;n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67&lt;/math&gt;. &lt;math&gt;n&lt;/math&gt; must also be a multiple of &lt;math&gt;8&lt;/math&gt;, so &lt;math&gt;125a_2 + 67 \equiv 5a_2 + 3 \pmod 8 \implies a_2 = 1,9,17 \ldots&lt;/math&gt;. Therefore, the minimum of &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;125 + 67 = \boxed{192}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> Let &lt;math&gt;x^3 = 1000a + 888&lt;/math&gt;. We factor an &lt;math&gt;8&lt;/math&gt; out of the right hand side, and we note that &lt;math&gt;x&lt;/math&gt; must be of the form &lt;math&gt;x = 2y&lt;/math&gt;, where &lt;math&gt;y&lt;/math&gt; is a positive integer. Then, this becomes &lt;math&gt;y^3 = 125a + 111&lt;/math&gt;. Taking mod &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;25&lt;/math&gt;, and &lt;math&gt;125&lt;/math&gt;, we get &lt;math&gt;y^3 \equiv 1\pmod 5&lt;/math&gt;, &lt;math&gt;y^3 \equiv 11\pmod{25}&lt;/math&gt;, and &lt;math&gt;y^3 \equiv 111\pmod{125}&lt;/math&gt;. <br /> <br /> We can work our way up, and find that &lt;math&gt;y\equiv 1\pmod 5&lt;/math&gt;, &lt;math&gt;y\equiv 21\pmod{25}&lt;/math&gt;, and finally &lt;math&gt;y\equiv 96\pmod{125}&lt;/math&gt;. This gives us our smallest value, &lt;math&gt;y = 96&lt;/math&gt;, so &lt;math&gt;x = \boxed{192}&lt;/math&gt;, as desired. - Spacesam<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_9&diff=122527 1988 AIME Problems/Problem 9 2020-05-16T19:03:50Z <p>Crazyeyemoody907: /* Solution */</p> <hr /> <div>== Problem ==<br /> Find the smallest positive integer whose [[perfect cube|cube]] ends in &lt;math&gt;888&lt;/math&gt;.<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of &lt;math&gt;(10k + 2)^3&lt;/math&gt;; using the [[binomial theorem]] gives us &lt;math&gt;1000k^3 + 600k^2 + 120k + 8&lt;/math&gt;. Since we are looking for the tens digit, &lt;math&gt;\mod{100}&lt;/math&gt; we get &lt;math&gt;20k + 8 \equiv 88 \pmod{100}&lt;/math&gt;. This is true if the tens digit is either &lt;math&gt;4&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;. Casework:<br /> *&lt;math&gt;4&lt;/math&gt;: Then our cube must be in the form of &lt;math&gt;(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}&lt;/math&gt;. Hence the lowest possible value for the hundreds digit is &lt;math&gt;4&lt;/math&gt;, and so &lt;math&gt;442&lt;/math&gt; is a valid solution. <br /> *&lt;math&gt;9&lt;/math&gt;: Then our cube is &lt;math&gt;(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}&lt;/math&gt;. The lowest possible value for the hundreds digit is &lt;math&gt;1&lt;/math&gt;, and we get &lt;math&gt;192&lt;/math&gt;. Hence, since &lt;math&gt;192 &lt; 442&lt;/math&gt;, the answer is &lt;math&gt;\fbox{192}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> &lt;math&gt;n^3 \equiv 888 \pmod{1000} \implies n^3 \equiv 0 \pmod 8&lt;/math&gt; and &lt;math&gt;n^3 \equiv 13 \pmod{125}&lt;/math&gt;.<br /> &lt;math&gt;n \equiv 2 \pmod 5&lt;/math&gt; due to the last digit of &lt;math&gt;n^3&lt;/math&gt;. Let &lt;math&gt;n = 5a + 2&lt;/math&gt;. By expanding, &lt;math&gt;125a^3 + 150a^2 + 60a + 8 \equiv 13 \pmod{125} \implies 5a^2 + 12a \equiv 1 \pmod{25}&lt;/math&gt;. <br /> <br /> By looking at the last digit again, we see &lt;math&gt;a \equiv 3 \pmod5&lt;/math&gt;, so we let &lt;math&gt;a = 5a_1 + 3&lt;/math&gt; where &lt;math&gt;a_1 \in \mathbb{Z^+}&lt;/math&gt;. Plugging this in to &lt;math&gt;5a^2 + 12a \equiv 1 \pmod{25}&lt;/math&gt; gives &lt;math&gt;10a_1 + 6 \equiv 1 \pmod{25}&lt;/math&gt;. Obviously, &lt;math&gt;a_1 \equiv 2 \pmod 5&lt;/math&gt;, so we let &lt;math&gt;a_1 = 5a_2 + 2&lt;/math&gt; where &lt;math&gt;a_2&lt;/math&gt; can be any non-negative integer. <br /> <br /> Therefore, &lt;math&gt;n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67&lt;/math&gt;. &lt;math&gt;n&lt;/math&gt; must also be a multiple of &lt;math&gt;8&lt;/math&gt;, so &lt;math&gt;125a_2 + 67 \equiv 5a_2 + 3 \pmod 8 \implies a_2 = 1,9,17 \ldots&lt;/math&gt;. Therefore, the minimum of &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;125 + 67 = \boxed{192}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> Let &lt;math&gt;x^3 = 1000a + 888&lt;/math&gt;. We factor an &lt;math&gt;8&lt;/math&gt; out of the right hand side, and we note that &lt;math&gt;x&lt;/math&gt; must be of the form &lt;math&gt;x = 2y&lt;/math&gt;, where &lt;math&gt;y&lt;/math&gt; is a positive integer. Then, this becomes &lt;math&gt;y^3 = 125a + 111&lt;/math&gt;. Taking mod &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;25&lt;/math&gt;, and &lt;math&gt;125&lt;/math&gt;, we get &lt;math&gt;y^3 \equiv 1\pmod 5&lt;/math&gt;, &lt;math&gt;y^3 \equiv 11\pmod{25}&lt;/math&gt;, and &lt;math&gt;y^3 \equiv 111\pmod{125}&lt;/math&gt;. <br /> <br /> We can work our way up, and find that &lt;math&gt;y\equiv 1\pmod 5&lt;/math&gt;, &lt;math&gt;y\equiv 21\pmod{25}&lt;/math&gt;, and finally &lt;math&gt;y\equiv 96\pmod{125}&lt;/math&gt;. This gives us our smallest value, &lt;math&gt;y = 96&lt;/math&gt;, so &lt;math&gt;x = \boxed{192}&lt;/math&gt;, as desired. - Spacesam<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_6&diff=122526 1988 AIME Problems/Problem 6 2020-05-16T19:00:58Z <p>Crazyeyemoody907: /* Solution */</p> <hr /> <div>== Problem ==<br /> It is possible to place positive integers into the vacant twenty-one squares of the &lt;math&gt;5 \times 5&lt;/math&gt; square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (*).<br /> <br /> [[Image:1988_AIME-6.png]]<br /> <br /> __TOC__<br /> == Solutions ==<br /> === Solution 1 (specific) ===<br /> Let the coordinates of the square at the bottom left be &lt;math&gt;(0,0)&lt;/math&gt;, the square to the right &lt;math&gt;(1,0)&lt;/math&gt;, etc.<br /> <br /> Label the leftmost column (from bottom to top) &lt;math&gt;0, a, 2a, 3a, 4a&lt;/math&gt; and the bottom-most row (from left to right) &lt;math&gt;0, b, 2b, 3b, 4b&lt;/math&gt;. Our method will be to use the given numbers to set up equations to solve for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, and then calculate &lt;math&gt;(*)&lt;/math&gt;.<br /> <br /> &lt;math&gt;\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a &amp; &amp; &amp; * &amp; \\<br /> \hline 3a &amp; 74 &amp; &amp; &amp; \\<br /> \hline 2a &amp; &amp; &amp; &amp; 186 \\<br /> \hline a &amp; &amp; 103 &amp; &amp; \\<br /> \hline 0 &amp; b &amp; 2b &amp; 3b &amp; 4b \\<br /> \hline \end{tabular}&lt;/math&gt;<br /> <br /> We can compute the squares at the intersections of two existing numbers in terms of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;; two such equations will give us the values of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. On the fourth row from the bottom, the common difference is &lt;math&gt;74 - 3a&lt;/math&gt;, so the square at &lt;math&gt;(2,3)&lt;/math&gt; has a value of &lt;math&gt;148 - 3a&lt;/math&gt;. On the third column from the left, the common difference is &lt;math&gt;103 - 2b&lt;/math&gt;, so that square also has a value of &lt;math&gt;2b + 3(103 - 2b) = 309 - 4b&lt;/math&gt;. Equating, we get &lt;math&gt;148 - 3a = 309 - 4b \Longrightarrow 4b - 3a = 161&lt;/math&gt;.<br /> <br /> Now we compute the square &lt;math&gt;(2,2)&lt;/math&gt;. By rows, this value is simply the average of &lt;math&gt;2a&lt;/math&gt; and &lt;math&gt;186&lt;/math&gt;, so it is equal to &lt;math&gt;\frac{2a + 186}{2} = a + 93&lt;/math&gt;. By columns, the common difference is &lt;math&gt;103 - 2b&lt;/math&gt;, so our value is &lt;math&gt;206 - 2b&lt;/math&gt;. Equating, &lt;math&gt;a + 93 = 206 - 2b \Longrightarrow a + 2b = 113&lt;/math&gt;.<br /> <br /> Solving <br /> &lt;cmath&gt;<br /> \begin{align*}4b - 3a &amp;= 161\\<br /> a + 2b &amp;= 113<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> gives &lt;math&gt;a = 13&lt;/math&gt;, &lt;math&gt;b = 50&lt;/math&gt;. Now it is simple to calculate &lt;math&gt;(4,3)&lt;/math&gt;. One way to do it is to see that &lt;math&gt;(2,2)&lt;/math&gt; has &lt;math&gt;206 - 2b = 106&lt;/math&gt; and &lt;math&gt;(4,2)&lt;/math&gt; has &lt;math&gt;186&lt;/math&gt;, so &lt;math&gt;(3,2)&lt;/math&gt; has &lt;math&gt;\frac{106 + 186}{2} = 146&lt;/math&gt;. Now, &lt;math&gt;(3,0)&lt;/math&gt; has &lt;math&gt;3b = 150&lt;/math&gt;, so &lt;math&gt;(3,2) = \frac{(3,0) + (3,4)}{2} \Longrightarrow (3,4) = * = \boxed{142}&lt;/math&gt;.<br /> <br /> === Solution 2 (general) ===<br /> First, let &lt;math&gt;a =&lt;/math&gt; the number to be placed in the first column, fourth row. Let &lt;math&gt;b =&lt;/math&gt; the number to be placed in the second column, fifth row. We can determine the entire first column and fifth row in terms of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;:<br /> <br /> &lt;math&gt;\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a &amp; &amp; &amp; &amp; \\<br /> \hline 3a &amp; &amp; &amp; &amp; \\<br /> \hline 2a &amp; &amp; &amp; &amp; \\<br /> \hline a &amp; &amp; &amp; &amp; \\<br /> \hline 0 &amp; b &amp; 2b &amp; 3b &amp; 4b \\<br /> \hline \end{tabular}&lt;/math&gt;<br /> <br /> Next, let &lt;math&gt;a + b + c =&lt;/math&gt; the number to be placed in the second column, fourth row. We can determine the entire second column and fourth row in terms of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;:<br /> <br /> &lt;math&gt;\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a &amp; 4a + b + 4c &amp; &amp; &amp; \\<br /> \hline 3a &amp; 3a + b + 3c &amp; &amp; &amp; \\<br /> \hline 2a &amp; 2a + b + 2c &amp; &amp; &amp; \\<br /> \hline a &amp; a + b + c &amp; a + 2b + 2c &amp; a + 3b + 3c &amp; a + 4b + 4c \\<br /> \hline 0 &amp; b &amp; 2b &amp; 3b &amp; 4b \\<br /> \hline \end{tabular}&lt;/math&gt;<br /> <br /> We have now determined at least two values in each row and column. We can finish the table without introducing any more variables:<br /> <br /> &lt;math&gt;\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a &amp; 4a + b + 4c &amp; 4a + 2b + 8c &amp; 4a + 3b + 12c &amp; 4a + 4b + 16c \\<br /> \hline 3a &amp; 3a + b + 3c &amp; 3a + 2b + 6c &amp; 3a + 3b + 9c &amp; 3a + 4b + 12c \\<br /> \hline 2a &amp; 2a + b + 2c &amp; 2a + 2b + 4c &amp; 2a + 3b + 6c &amp; 2a + 4b + 8c \\<br /> \hline a &amp; a + b + c &amp; a + 2b + 2c &amp; a + 3b + 3c &amp; a + 4b + 4c \\<br /> \hline 0 &amp; b &amp; 2b &amp; 3b &amp; 4b \\<br /> \hline \end{tabular}&lt;/math&gt;<br /> <br /> We now have a system of equations.<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;3a + b + 3c = 74&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;2a + 4b + 8c = 186&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;a + 2b + 2c = 103&lt;/math&gt;&lt;/div&gt;<br /> <br /> Solving, we find that &lt;math&gt;(a,b,c) = (13,50, - 5)&lt;/math&gt;. The number in the square marked by the asterisk is &lt;math&gt;4a + 3b + 12c = \boxed{142}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_11&diff=122350 1987 AIME Problems/Problem 11 2020-05-13T05:10:16Z <p>Crazyeyemoody907: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Find the largest possible value of &lt;math&gt;k&lt;/math&gt; for which &lt;math&gt;3^{11}&lt;/math&gt; is expressible as the sum of &lt;math&gt;k&lt;/math&gt; consecutive [[positive integer]]s.<br /> ==Solutions==<br /> === Solution 1===<br /> Let us write down one such sum, with &lt;math&gt;m&lt;/math&gt; terms and first term &lt;math&gt;n + 1&lt;/math&gt;:<br /> <br /> &lt;math&gt;3^{11} = (n + 1) + (n + 2) + \ldots + (n + m) = \frac{1}{2} m(2n + m + 1)&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;m(2n + m + 1) = 2 \cdot 3^{11}&lt;/math&gt; so &lt;math&gt;m&lt;/math&gt; is a [[divisor]] of &lt;math&gt;2\cdot 3^{11}&lt;/math&gt;. However, because &lt;math&gt;n \geq 0&lt;/math&gt; we have &lt;math&gt;m^2 &lt; m(m + 1) \leq 2\cdot 3^{11}&lt;/math&gt; so &lt;math&gt;m &lt; \sqrt{2\cdot 3^{11}} &lt; 3^6&lt;/math&gt;. Thus, we are looking for large factors of &lt;math&gt;2\cdot 3^{11}&lt;/math&gt; which are less than &lt;math&gt;3^6&lt;/math&gt;. The largest such factor is clearly &lt;math&gt;2\cdot 3^5 = 486&lt;/math&gt;; for this value of &lt;math&gt;m&lt;/math&gt; we do indeed have the valid [[expression]] &lt;math&gt;3^{11} = 122 + 123 + \ldots + 607&lt;/math&gt;, for which &lt;math&gt;k=\boxed{486}&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> First note that if &lt;math&gt;k&lt;/math&gt; is odd, and &lt;math&gt;n&lt;/math&gt; is the middle term, the sum is equal to kn. If &lt;math&gt;k&lt;/math&gt; is even, then we have the sum equal to &lt;math&gt;kn+k/2&lt;/math&gt; which is going to be even. Since &lt;math&gt;3^{11}&lt;/math&gt; is odd, we see that &lt;math&gt;k&lt;/math&gt; is odd. <br /> <br /> Thus, we have &lt;math&gt;nk=3^{11} \implies n=3^{11}/k&lt;/math&gt;. Also, note &lt;math&gt;n-(k+1)/2=0 \implies n=(k+1)/2.&lt;/math&gt; Subsituting &lt;math&gt;n=3^{11}/k&lt;/math&gt;, we have &lt;math&gt;k^2+k=2*3^{11}&lt;/math&gt;. Proceed as in solution 1.<br /> <br /> == See also ==<br /> {{AIME box|year=1987|num-b=10|num-a=12}}<br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_11&diff=122349 1987 AIME Problems/Problem 11 2020-05-13T05:10:03Z <p>Crazyeyemoody907: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Find the largest possible value of &lt;math&gt;k&lt;/math&gt; for which &lt;math&gt;3^{11}&lt;/math&gt; is expressible as the sum of &lt;math&gt;k&lt;/math&gt; consecutive [[positive integer]]s.<br /> ==Solutions==<br /> === Solution 1===<br /> Let us write down one such sum, with &lt;math&gt;m&lt;/math&gt; terms and first term &lt;math&gt;n + 1&lt;/math&gt;:<br /> <br /> &lt;math&gt;3^{11} = (n + 1) + (n + 2) + \ldots + (n + m) = \frac{1}{2} m(2n + m + 1)&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;m(2n + m + 1) = 2 \cdot 3^{11}&lt;/math&gt; so &lt;math&gt;m&lt;/math&gt; is a [[divisor]] of &lt;math&gt;2\cdot 3^{11}&lt;/math&gt;. However, because &lt;math&gt;n \geq 0&lt;/math&gt; we have &lt;math&gt;m^2 &lt; m(m + 1) \leq 2\cdot 3^{11}&lt;/math&gt; so &lt;math&gt;m &lt; \sqrt{2\cdot 3^{11}} &lt; 3^6&lt;/math&gt;. Thus, we are looking for large factors of &lt;math&gt;2\cdot 3^{11}&lt;/math&gt; which are less than &lt;math&gt;3^6&lt;/math&gt;. The largest such factor is clearly &lt;math&gt;2\cdot 3^5 = 486&lt;/math&gt;; for this value of &lt;math&gt;m&lt;/math&gt; we do indeed have the valid [[expression]] &lt;math&gt;3^{11} = 122 + 123 + \ldots + 607&lt;/math&gt;, for which &lt;math&gt;k=\boxed{486}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> First note that if &lt;math&gt;k&lt;/math&gt; is odd, and &lt;math&gt;n&lt;/math&gt; is the middle term, the sum is equal to kn. If &lt;math&gt;k&lt;/math&gt; is even, then we have the sum equal to &lt;math&gt;kn+k/2&lt;/math&gt; which is going to be even. Since &lt;math&gt;3^{11}&lt;/math&gt; is odd, we see that &lt;math&gt;k&lt;/math&gt; is odd. <br /> <br /> Thus, we have &lt;math&gt;nk=3^{11} \implies n=3^{11}/k&lt;/math&gt;. Also, note &lt;math&gt;n-(k+1)/2=0 \implies n=(k+1)/2.&lt;/math&gt; Subsituting &lt;math&gt;n=3^{11}/k&lt;/math&gt;, we have &lt;math&gt;k^2+k=2*3^{11}&lt;/math&gt;. Proceed as in solution 1.<br /> <br /> == See also ==<br /> {{AIME box|year=1987|num-b=10|num-a=12}}<br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_9&diff=122348 1987 AIME Problems/Problem 9 2020-05-13T04:57:15Z <p>Crazyeyemoody907: /* Note */</p> <hr /> <div>== Problem ==<br /> [[Triangle]] &lt;math&gt;ABC&lt;/math&gt; has [[right angle]] at &lt;math&gt;B&lt;/math&gt;, and contains a [[point]] &lt;math&gt;P&lt;/math&gt; for which &lt;math&gt;PA = 10&lt;/math&gt;, &lt;math&gt;PB = 6&lt;/math&gt;, and &lt;math&gt;\angle APB = \angle BPC = \angle CPA&lt;/math&gt;. Find &lt;math&gt;PC&lt;/math&gt;.<br /> <br /> [[Image:AIME_1987_Problem_9.png]]<br /> == Solution ==<br /> Let &lt;math&gt;PC = x&lt;/math&gt;. Since &lt;math&gt;\angle APB = \angle BPC = \angle CPA&lt;/math&gt;, each of them is equal to &lt;math&gt;120^\circ&lt;/math&gt;. By the [[Law of Cosines]] applied to triangles &lt;math&gt;\triangle APB&lt;/math&gt;, &lt;math&gt;\triangle BPC&lt;/math&gt; and &lt;math&gt;\triangle CPA&lt;/math&gt; at their respective angles &lt;math&gt;P&lt;/math&gt;, remembering that &lt;math&gt;\cos 120^\circ = -\frac12&lt;/math&gt;, we have<br /> <br /> &lt;cmath&gt;AB^2 = 36 + 100 + 60 = 196, BC^2 = 36 + x^2 + 6x, CA^2 = 100 + x^2 + 10x&lt;/cmath&gt;<br /> <br /> Then by the [[Pythagorean Theorem]], &lt;math&gt;AB^2 + BC^2 = CA^2&lt;/math&gt;, so <br /> <br /> &lt;cmath&gt;x^2 + 10x + 100 = x^2 + 6x + 36 + 196&lt;/cmath&gt; <br /> <br /> and<br /> <br /> &lt;cmath&gt;4x = 132 \Longrightarrow x = \boxed{033}.&lt;/cmath&gt;<br /> === Note ===<br /> This is the Fermat point of the triangle.<br /> <br /> == See also ==<br /> {{AIME box|year=1987|num-b=8|num-a=10}}<br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=Ordered_tuple&diff=122347 Ordered tuple 2020-05-13T04:55:02Z <p>Crazyeyemoody907: uwu</p> <hr /> <div>A ordered tuple is a tuple that is ordered.</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_3&diff=122346 1987 AIME Problems/Problem 3 2020-05-13T04:38:23Z <p>Crazyeyemoody907: /* Solution */</p> <hr /> <div>== Problem ==<br /> By a proper [[divisor]] of a [[natural number]] we mean a [[positive]] [[integer|integral]] divisor other than 1 and the number itself. A natural number greater than 1 will be called ''nice'' if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers?<br /> <br /> == Solution ==<br /> Let &lt;math&gt;p(n)&lt;/math&gt; denote the product of the distinct proper divisors of &lt;math&gt;n&lt;/math&gt;. A number &lt;math&gt;n&lt;/math&gt; is ''nice'' in one of two instances: <br /> #It has exactly two distinct [[prime]] divisors. <br /> #:If we let &lt;math&gt;n = pq&lt;/math&gt;, where &lt;math&gt;p,q&lt;/math&gt; are the prime factors, then its proper divisors are &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;, and &lt;math&gt;p(n) = p \cdot q = n&lt;/math&gt;.<br /> #It is the cube of a prime number. <br /> #:If we let &lt;math&gt;n=p^3&lt;/math&gt; with &lt;math&gt;p&lt;/math&gt; prime, then its proper divisors are &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;p^2&lt;/math&gt;, and &lt;math&gt;p(n) = p \cdot p^2 =n&lt;/math&gt;. <br /> <br /> We now show that the above are the only two cases. Suppose that another nice number existed that does not fall into one of these two categories. Then we can either express it in the form &lt;math&gt;n = pqr&lt;/math&gt; (with &lt;math&gt;p,q&lt;/math&gt; prime and &lt;math&gt;r &gt; 1&lt;/math&gt;) or &lt;math&gt;n = p^e&lt;/math&gt; (with &lt;math&gt;e \neq 3&lt;/math&gt;). <br /> In the former case, it suffices to note that &lt;math&gt;p(n) \ge (pr) \cdot (qr) = pqr^2 &gt; pqr = n&lt;/math&gt;. <br /> <br /> In the latter case, then &lt;math&gt;p(n) = p \cdot p^2 \cdots p^{(e-1)} = p^{(e-1)e/2}&lt;/math&gt;.<br /> <br /> For &lt;math&gt;p(n) = n&lt;/math&gt;, we need &lt;math&gt;p^{(e-1)e/2} = p^e \Longrightarrow e^2 - e = 2e \Longrightarrow &lt;/math&gt; &lt;math&gt;e = 0 or e = 3&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;e \neq 3&lt;/math&gt;, in the case &lt;math&gt;e = 0 \Longrightarrow n = 1&lt;/math&gt; does not work. <br /> <br /> Thus, listing out the first ten numbers to fit this form, &lt;math&gt;2 \cdot 3 = 6,\ 2^3 = 8,\ 2 \cdot 5 = 10,&lt;/math&gt; &lt;math&gt;\ 2 \cdot 7 = 14,\ 3 \cdot 5 = 15,\ 3 \cdot 7 = 21,&lt;/math&gt; &lt;math&gt;\ 2 \cdot 11 = 22,\ 2 \cdot 13 = 26,&lt;/math&gt; &lt;math&gt;\ 3^3 = 27,\ 3 \cdot 11 = 33&lt;/math&gt;.<br /> [[Sum]]ming these yields &lt;math&gt;\boxed{182}&lt;/math&gt;.<br /> <br /> Alternatively, we could note that &lt;math&gt;n&lt;/math&gt; is only nice when it only has two divisors, which, when multiplied, clearly yield &lt;math&gt;n&lt;/math&gt;. We know that when the [[prime factorization]] of &lt;math&gt;n = a_1^{b_1} \cdot a_2^{b_2} \cdot a_3^{b_3} . . . \cdot a_m^{b_m}&lt;/math&gt;, the number of factors &lt;math&gt;f(n)&lt;/math&gt; of &lt;math&gt;n&lt;/math&gt; is &lt;cmath&gt;f(n) = (b_1 + 1)(b_2 +1)(b_3 +1) . . . (b_m +1).&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;n&lt;/math&gt; is nice, it may only have &lt;math&gt;4&lt;/math&gt; factors (&lt;math&gt;1&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;p&lt;/math&gt;, and &lt;math&gt;q&lt;/math&gt;). This means that &lt;math&gt;f(n) = 4&lt;/math&gt;. The number &lt;math&gt;4&lt;/math&gt; can only be factored into &lt;math&gt;(2)(2)&lt;/math&gt; or &lt;math&gt;(4)(1)&lt;/math&gt;, which means that either &lt;math&gt;b_1 = 1&lt;/math&gt; and &lt;math&gt;b_2 = 1&lt;/math&gt;, or &lt;math&gt;b_1 = 3&lt;/math&gt;. Therefore the only two cases are &lt;math&gt;n = pq&lt;/math&gt;, or &lt;math&gt;n = p^3&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1987|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_11&diff=122281 1986 AIME Problems/Problem 11 2020-05-11T05:16:23Z <p>Crazyeyemoody907: /* Solution 4(calculus) */</p> <hr /> <div>== Problem ==<br /> The [[polynomial]] &lt;math&gt;1-x+x^2-x^3+\cdots+x^{16}-x^{17}&lt;/math&gt; may be written in the form &lt;math&gt;a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}&lt;/math&gt;, where &lt;math&gt;y=x+1&lt;/math&gt; and the &lt;math&gt;a_i&lt;/math&gt;'s are [[constant]]s. Find the value of &lt;math&gt;a_2&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> Using the [[geometric series]] formula, &lt;math&gt;1 - x + x^2 + \cdots - x^{17} = \frac {1 - x^{18}}{1 + x} = \frac {1-x^{18}}{y}&lt;/math&gt;. Since &lt;math&gt;x = y - 1&lt;/math&gt;, this becomes &lt;math&gt;\frac {1-(y - 1)^{18}}{y}&lt;/math&gt;. We want &lt;math&gt;a_2&lt;/math&gt;, which is the coefficient of the &lt;math&gt;y^3&lt;/math&gt; term in &lt;math&gt;-(y - 1)^{18}&lt;/math&gt; (because the &lt;math&gt;y&lt;/math&gt; in the denominator reduces the degrees in the numerator by &lt;math&gt;1&lt;/math&gt;). By the [[Binomial Theorem]], this is &lt;math&gt;(-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Again, notice &lt;math&gt;x = y - 1&lt;/math&gt;. So<br /> <br /> &lt;cmath&gt;\begin{align*}1 - x + x^2 + \cdots - x^{17} &amp; = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\<br /> &amp; = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}.&lt;/cmath&gt;<br /> <br /> We want the coefficient of the &lt;math&gt;y^2&lt;/math&gt; term of each power of each binomial, which by the binomial theorem is &lt;math&gt;{2\choose 2} + {3\choose 2} + \cdots + {17\choose 2}&lt;/math&gt;. The [[Hockey Stick Identity]] tells us that this quantity is equal to &lt;math&gt;{18\choose 3} = \boxed{816}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Again, notice &lt;math&gt;x=y-1&lt;/math&gt;. Substituting &lt;math&gt;y-1&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;f(x)&lt;/math&gt; gives: <br /> &lt;cmath&gt;\begin{align*}1 - x + x^2 + \cdots - x^{17} &amp; = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\<br /> &amp; = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}.&lt;/cmath&gt;<br /> From binomial theorem, the coefficient of the &lt;math&gt;y^2&lt;/math&gt; term is &lt;math&gt;{2\choose 0} + {3\choose 1} + \cdots + {17\choose 15}&lt;/math&gt;. This is actually the sum of the first 16 triangular numbers, which evaluates to &lt;math&gt;\frac{(16)(17)(18)}{6} = \boxed{816}&lt;/math&gt;.<br /> <br /> === Solution 4(calculus) ===<br /> Let &lt;math&gt;f(x)=1-x+x^2-x^3+\cdots+x^{16}-x^{17}&lt;/math&gt; and &lt;math&gt;g(y)=a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}&lt;/math&gt;.<br /> <br /> Then, since &lt;math&gt;f(x)=g(y)&lt;/math&gt;,<br /> &lt;cmath&gt;\frac{d^2f}{dx^2}=\frac{d^2g}{dy^2}&lt;/cmath&gt;<br /> &lt;math&gt;\frac{d^2f}{dx^2} = 2\cdot 1 - 3\cdot 2x+\cdots-17\cdot 16x^{15}&lt;/math&gt; by the power rule.<br /> <br /> Similarly, &lt;math&gt;\frac{d^2g}{dy^2} = a_2(2\cdot 1) + a_3(3\cdot 2y)+\cdots+a_{17}(17\cdot 16y^{15})&lt;/math&gt;<br /> <br /> Now, notice that if &lt;math&gt;x = -1&lt;/math&gt;, then &lt;math&gt;y = 0&lt;/math&gt;, so &lt;math&gt;f^{''}(-1) = g^{''}(0)&lt;/math&gt;<br /> <br /> &lt;math&gt;g^{''}(0)= 2a_2&lt;/math&gt;, and &lt;math&gt;f^{''}(-1) = 2\cdot 1 + 3\cdot 2 +\cdots + 16\cdot 17&lt;/math&gt;.<br /> <br /> Now, we can use the hockey stick theorem to see that &lt;math&gt;2\cdot 1 + 3\cdot 2 +\cdots + 16\cdot 17 = 2\binom{18}{3}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;2a_2 = 2\binom{18}{3}\rightarrow a_2 = \binom{18}{3}=\boxed{816}&lt;/math&gt;<br /> <br /> -AOPS81619<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_9&diff=122278 1986 AIME Problems/Problem 9 2020-05-11T02:25:23Z <p>Crazyeyemoody907: /* Solution */</p> <hr /> <div>== Problem ==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB= 425&lt;/math&gt;, &lt;math&gt;BC=450&lt;/math&gt;, and &lt;math&gt;AC=510&lt;/math&gt;. An interior [[point]] &lt;math&gt;P&lt;/math&gt; is then drawn, and [[segment]]s are drawn through &lt;math&gt;P&lt;/math&gt; [[parallel]] to the sides of the [[triangle]]. If these three segments are of an equal length &lt;math&gt;d&lt;/math&gt;, find &lt;math&gt;d&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> &lt;center&gt;&lt;asy&gt;<br /> size(200);<br /> pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10);<br /> pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425));<br /> /* construct remaining points */<br /> pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C);<br /> pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); <br /> D(MP(&quot;A&quot;,A,s)--MP(&quot;B&quot;,B,N,s)--MP(&quot;C&quot;,C,s)--cycle);<br /> dot(MP(&quot;D&quot;,D,NE,s));dot(MP(&quot;E&quot;,E,NW,s));dot(MP(&quot;F&quot;,F,s));dot(MP(&quot;D'&quot;,Da,NE,s));dot(MP(&quot;E'&quot;,Ea,NW,s));dot(MP(&quot;F'&quot;,Fa,s));<br /> D(D--Ea);D(Da--F);D(Fa--E);<br /> MP(&quot;450&quot;,(B+C)/2,NW);MP(&quot;425&quot;,(A+B)/2,NE);MP(&quot;510&quot;,(A+C)/2);<br /> /*P copied from above solution*/<br /> pair P = IP(D--Ea,E--Fa); dot(MP(&quot;P&quot;,P,N)); <br /> &lt;/asy&gt;&lt;/center&gt; &lt;!-- Asymptote replacement for Image:1986_AIME-9.png by azjps --&gt;<br /> <br /> Let the points at which the segments hit the triangle be called &lt;math&gt;D, D', E, E', F, F'&lt;/math&gt; as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are [[similar triangles|similar]] (&lt;math&gt;\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF&lt;/math&gt;). The remaining three sections are [[parallelogram]]s.<br /> <br /> By similar triangles, &lt;math&gt;BE'=\frac{d}{510}\cdot450=\frac{15}{17}d&lt;/math&gt; and &lt;math&gt;EC=\frac{d}{425}\cdot450=\frac{18}{17}d&lt;/math&gt;. Since &lt;math&gt;FD'=BC-EE'&lt;/math&gt;, we have &lt;math&gt;900-\frac{33}{17}d=d&lt;/math&gt;, so &lt;math&gt;d=\boxed{306}&lt;/math&gt;.<br /> <br /> ===Solution 2 ===<br /> &lt;asy&gt;<br /> size(200);<br /> pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10);<br /> pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425));<br /> /* construct remaining points */<br /> pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C);<br /> pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); <br /> /* Construct P */<br /> pair P = IP(D--Ea,E--Fa); dot(MP(&quot;P&quot;,P,NE)); <br /> pair X = IP(L(A,P,4), B--C); dot(MP(&quot;X&quot;,X,NW));<br /> pair Y = IP(L(B,P,4), C--A); dot(MP(&quot;Y&quot;,Y,NE));<br /> pair Z = IP(L(C,P,4), A--B); dot(MP(&quot;Z&quot;,Z,N));<br /> <br /> D(A--X); D(B--Y); D(C--Z);<br /> D(MP(&quot;A&quot;,A,s)--MP(&quot;B&quot;,B,N,s)--MP(&quot;C&quot;,C,s)--cycle);<br /> MP(&quot;450&quot;,(B+C)/2,NW);MP(&quot;425&quot;,(A+B)/2,NE);MP(&quot;510&quot;,(A+C)/2);<br /> &lt;/asy&gt;<br /> <br /> Construct cevians &lt;math&gt;AX&lt;/math&gt;, &lt;math&gt;BY&lt;/math&gt; and &lt;math&gt;CZ&lt;/math&gt; through &lt;math&gt;P&lt;/math&gt;. Place masses of &lt;math&gt;x,y,z&lt;/math&gt; on &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; respectively; then &lt;math&gt;P&lt;/math&gt; has mass &lt;math&gt;x+y+z&lt;/math&gt;.<br /> <br /> Notice that &lt;math&gt;Z&lt;/math&gt; has mass &lt;math&gt;x+y&lt;/math&gt;. On the other hand, by similar triangles, &lt;math&gt;\frac{CP}{CZ} = \frac{d}{AB}&lt;/math&gt;. Hence by mass points we find that &lt;cmath&gt; \frac{x+y}{x+y+z} = \frac{d}{AB} &lt;/cmath&gt; Similarly, we obtain &lt;cmath&gt; \frac{y+z}{x+y+z} = \frac{d}{BC} \qquad \text{and} \qquad \frac{z+x}{x+y+z} = \frac{d}{CA} &lt;/cmath&gt; Summing these three equations yields &lt;cmath&gt; \frac{d}{AB} + \frac{d}{BC} + \frac{d}{CA} = \frac{x+y}{x+y+z} + \frac{y+z}{x+y+z} + \frac{z+x}{x+y+z} = \frac{2x+2y+2z}{x+y+z} = 2 &lt;/cmath&gt;<br /> <br /> Hence, &lt;center&gt;&lt;math&gt; d = \frac{2}{\frac{1}{AB} + \frac{1}{BC} + \frac{1}{CA}} = \frac{2}{\frac{1}{510} + \frac{1}{450} + \frac{1}{425}} = \frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}&lt;/math&gt;&lt;math&gt;= \frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}=\frac{10}{\frac{10}{306}} = \boxed{306}&lt;/math&gt;&lt;/center&gt;<br /> <br /> === Solution 3 ===<br /> &lt;center&gt;&lt;asy&gt;<br /> size(200);<br /> pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10);<br /> pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425));<br /> /* construct remaining points */<br /> pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C);<br /> pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); <br /> D(MP(&quot;A&quot;,A,s)--MP(&quot;B&quot;,B,N,s)--MP(&quot;C&quot;,C,s)--cycle);<br /> dot(MP(&quot;D&quot;,D,NE,s));dot(MP(&quot;E&quot;,E,NW,s));dot(MP(&quot;F&quot;,F,s));dot(MP(&quot;D'&quot;,Da,NE,s));dot(MP(&quot;E'&quot;,Ea,NW,s));dot(MP(&quot;F'&quot;,Fa,s));<br /> D(D--Ea);D(Da--F);D(Fa--E);<br /> MP(&quot;450&quot;,(B+C)/2,NW);MP(&quot;425&quot;,(A+B)/2,NE);MP(&quot;510&quot;,(A+C)/2);<br /> /*P copied from above solution*/<br /> pair P = IP(D--Ea,E--Fa); dot(MP(&quot;P&quot;,P,N)); <br /> &lt;/asy&gt;&lt;/center&gt; &lt;!-- Asymptote replacement for Image:1986_AIME-9.png by azjps --&gt;<br /> <br /> Let the points at which the segments hit the triangle be called &lt;math&gt;D, D', E, E', F, F'&lt;/math&gt; as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are [[similar triangles|similar]] (&lt;math&gt;\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF&lt;/math&gt;). The remaining three sections are [[parallelogram]]s.<br /> <br /> Since &lt;math&gt;PDAF'&lt;/math&gt; is a parallelogram, we find &lt;math&gt;PF' = AD&lt;/math&gt;, and similarly &lt;math&gt;PE = BD'&lt;/math&gt;. So &lt;math&gt;d = PF' + PE = AD + BD' = 425 - DD'&lt;/math&gt;. Thus &lt;math&gt;DD' = 425 - d&lt;/math&gt;. By the same logic, &lt;math&gt;EE' = 450 - d&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;\triangle DPD' \sim \triangle ABC&lt;/math&gt;, we have the [[proportion]]:<br /> <br /> &lt;center&gt;&lt;math&gt;\frac{425-d}{425} = \frac{PD}{510} \Longrightarrow PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d&lt;/math&gt;&lt;/center&gt;<br /> <br /> Doing the same with &lt;math&gt;\triangle PEE'&lt;/math&gt;, we find that &lt;math&gt;PE' =510 - \frac{17}{15}d&lt;/math&gt;. Now, &lt;math&gt;d = PD + PE' = 510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = \boxed{306}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Define the points the same as above.<br /> <br /> Let &lt;math&gt;[CE'PF] = a&lt;/math&gt;, &lt;math&gt;[E'EP] = b&lt;/math&gt;, &lt;math&gt;[BEPD'] = c&lt;/math&gt;, &lt;math&gt;[D'PD] = d&lt;/math&gt;, &lt;math&gt;[DAF'P] = e&lt;/math&gt; and &lt;math&gt;[F'D'P] = f&lt;/math&gt;<br /> <br /> The key theorem we apply here is that the ratio of the areas of 2 similar triangles is the ratio of a pair of corresponding sides squared.<br /> <br /> Let the length of the segment be &lt;math&gt;x&lt;/math&gt; and the area of the triangle be &lt;math&gt;A&lt;/math&gt;, using the theorem, we get:<br /> <br /> &lt;math&gt;\frac {d + e + f}{A} = \left(\frac {x}{BC}\right)^2&lt;/math&gt;, &lt;math&gt;\frac {b + c + d}{A}= \left(\frac {x}{AC}\right)^2&lt;/math&gt;, &lt;math&gt;\frac {a + b + f}{A} = \left(\frac {x}{AB}\right)^2&lt;/math&gt;.<br /> Adding all these together and using &lt;math&gt;a + b + c + d + e + f = A&lt;/math&gt; we get<br /> &lt;math&gt;\frac {f + d + b}{A} + 1 = x^2 \cdot \left(\frac {1}{BC^2} + \frac {1}{AC^2} + \frac {1}{AB^2}\right)&lt;/math&gt;<br /> <br /> Using [[corresponding angles]] from parallel lines, it is easy to show that &lt;math&gt;\triangle ABC \sim \triangle F'PF&lt;/math&gt;; since &lt;math&gt;ADPF'&lt;/math&gt; and &lt;math&gt;CFPE'&lt;/math&gt; are parallelograms, it is easy to show that &lt;math&gt;FF' = AC - x&lt;/math&gt;<br /> <br /> Now we have the side length [[ratio]], so we have the area ratio<br /> &lt;math&gt;\frac {f}{A} = \left(\frac {AC - x}{AC}\right)^2 = \left(1 - \frac {x}{AC}\right)^2&lt;/math&gt;. By symmetry, we have<br /> &lt;math&gt;\frac {d}{A} = \left(1 - \frac {x}{AB}\right)^2&lt;/math&gt; and &lt;math&gt;\frac {b}{A} = \left(1 - \frac {x}{BC}\right)^2&lt;/math&gt;<br /> <br /> Substituting these into our initial equation, we have<br /> &lt;math&gt;1 + \sum_{cyc}\left(1 - \frac {x}{AB}\right) - \frac {x^2}{AB^2} = 0&lt;/math&gt;<br /> &lt;math&gt;\implies 1 + \sum_{cyc}1 - 2 \cdot \frac {x}{AB} = 0&lt;/math&gt;<br /> &lt;math&gt;\implies \frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x&lt;/math&gt;<br /> and the answer follows after some hideous computation.<br /> <br /> ===Solution 5===<br /> Refer to the diagram in solution 2; let &lt;math&gt;a^2=[E'EP]&lt;/math&gt;, &lt;math&gt;b^2=[D'DP]&lt;/math&gt;, and &lt;math&gt;c^2=[F'FP]&lt;/math&gt;. Now, note that &lt;math&gt;[E'BD]&lt;/math&gt;, &lt;math&gt;[D'DP]&lt;/math&gt;, and &lt;math&gt;[E'EP]&lt;/math&gt; are similar, so through some similarities we find that &lt;math&gt;\frac{E'P}{PD}=\frac{a}{b}\implies\frac{E'D}{PD}=\frac{a+b}{b}\implies[E'BD]=b^2\left(\frac{a+b}{b}\right)^2=(a+b)^2&lt;/math&gt;. Similarly, we find that &lt;math&gt;[D'AF]=(b+c)^2&lt;/math&gt; and &lt;math&gt;[F'CE]=(c+a)^2&lt;/math&gt;, so &lt;math&gt;[ABC]=(a+b+c)^2&lt;/math&gt;. Now, again from similarity, it follows that &lt;math&gt;\frac{d}{510}=\frac{a+b}{a+b+c}&lt;/math&gt;, &lt;math&gt;\frac{d}{450}=\frac{b+c}{a+b+c}&lt;/math&gt;, and &lt;math&gt;\frac{d}{425}=\frac{c+a}{a+b+c}&lt;/math&gt;, so adding these together, simplifying, and solving gives &lt;math&gt;d=\frac{2}{\frac{1}{425}+\frac{1}{450}+\frac{1}{510}}=\frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}=\frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}&lt;/math&gt;<br /> &lt;math&gt;=\frac{10}{\frac{10}{306}}=\boxed{306}&lt;/math&gt;.<br /> <br /> === Solution 6 ===<br /> Refer to the diagram from Solution 2. Notice that because &lt;math&gt;CE'PF&lt;/math&gt;, &lt;math&gt;AF'PD&lt;/math&gt;, and &lt;math&gt;BD'PE&lt;/math&gt; are parallelograms, &lt;math&gt;\overline{DD'} = 425-d&lt;/math&gt;, &lt;math&gt;\overline{EE'} = 450-d&lt;/math&gt;, and &lt;math&gt;\overline{FF'} = 510-d&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;F'P = x&lt;/math&gt;. Then, because &lt;math&gt;\triangle ABC \sim \triangle F'PF&lt;/math&gt;, &lt;math&gt;\frac{AB}{AC}=\frac{F'P}{F'F}&lt;/math&gt;, so &lt;math&gt;\frac{425}{510}=\frac{x}{510-d}&lt;/math&gt;. Simplifying the LHS and cross-multiplying, we have &lt;math&gt;6x=2550-5d&lt;/math&gt;. From the same triangles, we can find that &lt;math&gt;FP=\frac{18}{17}x&lt;/math&gt;.<br /> <br /> &lt;math&gt;\triangle PEE'&lt;/math&gt; is also similar to &lt;math&gt;\triangle F'PF&lt;/math&gt;. Since &lt;math&gt;EF'=d&lt;/math&gt;, &lt;math&gt;EP=d-x&lt;/math&gt;. We now have &lt;math&gt;\frac{PE}{EE'}=\frac{F'P}{FP}&lt;/math&gt;, and &lt;math&gt;\frac{d-x}{450-d}=\frac{17}{18}&lt;/math&gt;. Cross multiplying, we have &lt;math&gt;18d-18x=450 \cdot 17-17d&lt;/math&gt;. Using the previous equation to substitute for &lt;math&gt;x&lt;/math&gt;, we have: &lt;cmath&gt;18d-3\cdot2550+15d=450\cdot17-17d&lt;/cmath&gt; This is a linear equation in one variable, and we can solve to get &lt;math&gt;d=\boxed{306}&lt;/math&gt;<br /> <br /> *I did not show the multiplication in the last equation because most of it cancels out when solving.<br /> <br /> (Note: I chose &lt;math&gt;F'P&lt;/math&gt; to be &lt;math&gt;x&lt;/math&gt; only because that is what I had written when originally solving. The solution would work with other choices for &lt;math&gt;x&lt;/math&gt;.)<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_7&diff=122275 1986 AIME Problems/Problem 7 2020-05-11T02:18:29Z <p>Crazyeyemoody907: /* Solution */</p> <hr /> <div>== Problem ==<br /> The increasing [[sequence]] &lt;math&gt;1,3,4,9,10,12,13\cdots&lt;/math&gt; consists of all those positive [[integer]]s which are [[exponent|powers]] of 3 or sums of distinct powers of 3. Find the &lt;math&gt;100^{\mbox{th}}&lt;/math&gt; term of this sequence.<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> <br /> Rewrite all of the terms in base 3. Since the numbers are sums of ''distinct'' powers of 3, in base 3 each number is a sequence of 1s and 0s (if there is a 2, then it is no longer the sum of distinct powers of 3). Therefore, we can recast this into base 2 (binary) in order to determine the 100th number. &lt;math&gt;100&lt;/math&gt; is equal to &lt;math&gt;64 + 32 + 4&lt;/math&gt;, so in binary form we get &lt;math&gt;1100100&lt;/math&gt;. However, we must change it back to base 10 for the answer, which is &lt;math&gt;3^6 + 3^5 + 3^2 = 729 + 243 + 9 = \boxed {981}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Notice that the first term of the sequence is &lt;math&gt;1&lt;/math&gt;, the second is &lt;math&gt;3&lt;/math&gt;, the fourth is &lt;math&gt;9&lt;/math&gt;, and so on. Thus the &lt;math&gt;64th&lt;/math&gt; term of the sequence is &lt;math&gt;729&lt;/math&gt;. Now out of &lt;math&gt;64&lt;/math&gt; terms which are of the form &lt;math&gt;729&lt;/math&gt; + &lt;math&gt;'''S'''&lt;/math&gt;, &lt;math&gt;32&lt;/math&gt; of them include &lt;math&gt;243&lt;/math&gt; and &lt;math&gt;32&lt;/math&gt; do not. The smallest term that includes &lt;math&gt;243&lt;/math&gt;, i.e. &lt;math&gt;972&lt;/math&gt;, is greater than the largest term which does not, or &lt;math&gt;854&lt;/math&gt;. So the &lt;math&gt;96&lt;/math&gt;th term will be &lt;math&gt;972&lt;/math&gt;, then &lt;math&gt;973&lt;/math&gt;, then &lt;math&gt;975&lt;/math&gt;, then &lt;math&gt;976&lt;/math&gt;, and finally &lt;math&gt;\boxed{981}&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> After the &lt;math&gt;n&lt;/math&gt;th power of 3 in the sequence, the number of terms after that power but before the &lt;math&gt;(n+1)&lt;/math&gt;th power of 3 is equal to the number of terms before the &lt;math&gt;n&lt;/math&gt;th power, because those terms after the &lt;math&gt;n&lt;/math&gt;th power are just the &lt;math&gt;n&lt;/math&gt;th power plus all the distinct combinations of powers of 3 before it, which is just all the terms before it. Adding the powers of &lt;math&gt;3&lt;/math&gt; and the terms that come after them, we see that the &lt;math&gt;100&lt;/math&gt;th term is after &lt;math&gt;729&lt;/math&gt;, which is the &lt;math&gt;64&lt;/math&gt;th term. Also, note that the &lt;math&gt;k&lt;/math&gt;th term after the &lt;math&gt;n&lt;/math&gt;th power of 3 is equal to the power plus the &lt;math&gt;k&lt;/math&gt;th term in the entire sequence. Thus, the &lt;math&gt;100&lt;/math&gt;th term is &lt;math&gt;729&lt;/math&gt; plus the &lt;math&gt;36&lt;/math&gt;th term. Using the same logic, the &lt;math&gt;36&lt;/math&gt;th term is &lt;math&gt;243&lt;/math&gt; plus the &lt;math&gt;4&lt;/math&gt;th term, &lt;math&gt;9&lt;/math&gt;. We now have &lt;math&gt;729+243+9=\boxed{981}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=6|num-a=8}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=122213 2011 USAJMO Problems/Problem 3 2020-05-09T18:43:36Z <p>Crazyeyemoody907: /* See also */</p> <hr /> <div>== Problem ==<br /> <br /> For a point &lt;math&gt;P = (a, a^2)&lt;/math&gt; in the coordinate plane, let &lt;math&gt;\ell(P)&lt;/math&gt; denote the line passing through &lt;math&gt;P&lt;/math&gt; with slope &lt;math&gt;2a&lt;/math&gt;. Consider the set of triangles with vertices of the form &lt;math&gt;P_1 = (a_1, a_1^2)&lt;/math&gt;, &lt;math&gt;P_2 = (a_2, a_2^2)&lt;/math&gt;, &lt;math&gt;P_3 = (a_3, a_3^2)&lt;/math&gt;, such that the intersections of the lines &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle &lt;math&gt;\triangle&lt;/math&gt;. Find the locus of the center of &lt;math&gt;\triangle&lt;/math&gt; as &lt;math&gt;P_1P_2P_3&lt;/math&gt; ranges over all such triangles.<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> Note that all the points &lt;math&gt;P=(a,a^2)&lt;/math&gt; belong to the parabola &lt;math&gt;y=x^2&lt;/math&gt; which we will denote &lt;math&gt;p&lt;/math&gt;. This parabola has a focus &lt;math&gt;F=\left(0,\frac{1}{4}\right)&lt;/math&gt; and directrix &lt;math&gt;y=-\frac{1}{4}&lt;/math&gt; which we will denote &lt;math&gt;d&lt;/math&gt;. We will prove that the desired locus is &lt;math&gt;d&lt;/math&gt;.<br /> <br /> First note that for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt;, the line &lt;math&gt;\ell(P)&lt;/math&gt; is the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;. This is because &lt;math&gt;\ell(P)&lt;/math&gt; contains &lt;math&gt;P&lt;/math&gt; and because &lt;math&gt;[\frac{d}{dx}] x^2=2x&lt;/math&gt;. If you don't like calculus, you can also verify that &lt;math&gt;\ell(P)&lt;/math&gt; has equation &lt;math&gt;y=2a(x-a)+a^2&lt;/math&gt; and does not intersect &lt;math&gt;y=x^2&lt;/math&gt; at any point besides &lt;math&gt;P&lt;/math&gt;. Now for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt; let &lt;math&gt;P'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;P&lt;/math&gt; onto &lt;math&gt;d&lt;/math&gt;. Then by the definition of parabolas, &lt;math&gt;PP'=PF&lt;/math&gt;. Let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;. Since &lt;math&gt;PP'=PF&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt; passes through &lt;math&gt;P&lt;/math&gt;. Suppose &lt;math&gt;K&lt;/math&gt; is any other point on &lt;math&gt;q&lt;/math&gt; and let &lt;math&gt;K'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;K&lt;/math&gt; to &lt;math&gt;d&lt;/math&gt;. Then in right &lt;math&gt;\Delta KK'P'&lt;/math&gt;, &lt;math&gt;KK'&lt;/math&gt; is a leg and so &lt;math&gt;KK'&lt;KP'=KF&lt;/math&gt;. Therefore &lt;math&gt;K&lt;/math&gt; cannot be on &lt;math&gt;p&lt;/math&gt;. This implies that &lt;math&gt;q&lt;/math&gt; is exactly the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;, that is &lt;math&gt;q=\ell(P)&lt;/math&gt;. So we have proved Lemma 1: If &lt;math&gt;P&lt;/math&gt; is a point on &lt;math&gt;p&lt;/math&gt; then &lt;math&gt;\ell(P)&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;.<br /> <br /> We need another lemma before we proceed. Lemma 2: If &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt; with orthocenter &lt;math&gt;H&lt;/math&gt;, then the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt;, &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, and &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are collinear with &lt;math&gt;H&lt;/math&gt;.<br /> <br /> Proof of Lemma 2: Say the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are &lt;math&gt;C'&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt;, and the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; are &lt;math&gt;A'&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt;. Then we angle chase &lt;math&gt;\angle JYZ=\angle HYZ=\angle HXZ=\angle JXZ=m(JZ)/2&lt;/math&gt; where &lt;math&gt;m(JZ)&lt;/math&gt; is the measure of minor arc &lt;math&gt;JZ&lt;/math&gt; on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. This implies that &lt;math&gt;J&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;, and similarly &lt;math&gt;I&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. Therefore &lt;math&gt;\angle C'HJ=\angle FJH=m(XF)/2&lt;/math&gt;, and &lt;math&gt;\angle A'HX=\angle FIX=m(FX)/2&lt;/math&gt;. So &lt;math&gt;\angle C'HJ = \angle A'HX&lt;/math&gt;. Since &lt;math&gt;J&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear it follows that &lt;math&gt;C'&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt; and &lt;math&gt;A'&lt;/math&gt; are collinear. Similarly, the reflection of &lt;math&gt;F&lt;/math&gt; over &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; also lies on this line, and so the claim is proved.<br /> <br /> Now suppose &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; are three points of &lt;math&gt;p&lt;/math&gt; and let &lt;math&gt;\ell(A)\cap\ell(B)=X&lt;/math&gt;, &lt;math&gt;\ell(A)\cap\ell(C)=Y&lt;/math&gt;, and &lt;math&gt;\ell(B)\cap\ell(C)=Z&lt;/math&gt;. Also let &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; be the midpoints of &lt;math&gt;\overline{A'F}&lt;/math&gt;, &lt;math&gt;\overline{B'F}&lt;/math&gt;, and &lt;math&gt;\overline{C'F}&lt;/math&gt; respectively. Then since &lt;math&gt;\overleftrightarrow{A''B''}\parallel \overline{A'B'}=d&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{B''C''}\parallel \overline{B'C'}=d&lt;/math&gt;, it follows that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are collinear. By Lemma 1, we know that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are the feet of the altitudes from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{XZ}&lt;/math&gt;, and &lt;math&gt;\overline{YZ}&lt;/math&gt;. Therefore by the Simson Line Theorem, &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. If &lt;math&gt;H&lt;/math&gt; is the orthocenter of &lt;math&gt;\Delta XYZ&lt;/math&gt;, then by Lemma 2, it follows that &lt;math&gt;H&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{A'C'}=d&lt;/math&gt;. It follows that the locus described in the problem is a subset of &lt;math&gt;d&lt;/math&gt;.<br /> <br /> Since we claim that the locus described in the problem is &lt;math&gt;d&lt;/math&gt;, we still need to show that for any choice of &lt;math&gt;H&lt;/math&gt; on &lt;math&gt;d&lt;/math&gt; there exists an equilateral triangle with center &lt;math&gt;H&lt;/math&gt; such that the lines containing the sides of the triangle are tangent to &lt;math&gt;p&lt;/math&gt;. So suppose &lt;math&gt;H&lt;/math&gt; is any point on &lt;math&gt;d&lt;/math&gt; and let the circle centered at &lt;math&gt;H&lt;/math&gt; through &lt;math&gt;F&lt;/math&gt; be &lt;math&gt;O&lt;/math&gt;. Then suppose &lt;math&gt;A&lt;/math&gt; is one of the intersections of &lt;math&gt;d&lt;/math&gt; with &lt;math&gt;O&lt;/math&gt;. Let &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, and construct the ray through &lt;math&gt;F&lt;/math&gt; on the same halfplane of &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt; as &lt;math&gt;A&lt;/math&gt; that makes an angle of &lt;math&gt;2\theta&lt;/math&gt; with &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt;. Say this ray intersects &lt;math&gt;O&lt;/math&gt; in a point &lt;math&gt;B&lt;/math&gt; besides &lt;math&gt;F&lt;/math&gt;, and let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{HB}&lt;/math&gt;. Since &lt;math&gt;\angle HFB=2\theta&lt;/math&gt; and &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, we have &lt;math&gt;\angle BFA=\theta&lt;/math&gt;. By the inscribed angles theorem, it follows that &lt;math&gt;\angle AHB=2\theta&lt;/math&gt;. Also since &lt;math&gt;HF&lt;/math&gt; and &lt;math&gt;HB&lt;/math&gt; are both radii, &lt;math&gt;\Delta HFB&lt;/math&gt; is isosceles and &lt;math&gt;\angle HBF=\angle HFB=2\theta&lt;/math&gt;. Let &lt;math&gt;P_1'&lt;/math&gt; be the reflection of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;q&lt;/math&gt;. Then &lt;math&gt;2\theta=\angle FBH=\angle C'HB&lt;/math&gt;, and so &lt;math&gt;\angle C'HB=\angle AHB&lt;/math&gt;. It follows that &lt;math&gt;P_1'&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{AH}=d&lt;/math&gt;, which means &lt;math&gt;q&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{FP_1'}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;q&lt;/math&gt; intersect &lt;math&gt;O&lt;/math&gt; in points &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt; and let &lt;math&gt;X&lt;/math&gt; be the point diametrically opposite to &lt;math&gt;B&lt;/math&gt; on &lt;math&gt;O&lt;/math&gt;. Also let &lt;math&gt;\overline{HB}&lt;/math&gt; intersect &lt;math&gt;q&lt;/math&gt; at &lt;math&gt;M&lt;/math&gt;. Then &lt;math&gt;HM=HB/2=HZ/2&lt;/math&gt;. Therefore &lt;math&gt;\Delta HMZ&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; right triangle and so &lt;math&gt;\angle ZHB=60^{\circ}&lt;/math&gt;. So &lt;math&gt;\angle ZHY=120^{\circ}&lt;/math&gt; and by the inscribed angles theorem, &lt;math&gt;\angle ZXY=60^{\circ}&lt;/math&gt;. Since &lt;math&gt;ZX=ZY&lt;/math&gt; it follows that &lt;math&gt;\Delta ZXY&lt;/math&gt; is and equilateral triangle with center &lt;math&gt;H&lt;/math&gt;.<br /> <br /> By Lemma 2, it follows that the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, call them &lt;math&gt;P_2'&lt;/math&gt; and &lt;math&gt;P_3'&lt;/math&gt;, lie on &lt;math&gt;d&lt;/math&gt;. Let the intersection of &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_1'&lt;/math&gt; be &lt;math&gt;P_1&lt;/math&gt;, the intersection of &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_2'&lt;/math&gt; be &lt;math&gt;P_2&lt;/math&gt;, and the intersection of &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_3'&lt;/math&gt; be &lt;math&gt;P_3&lt;/math&gt;. Then by the definitions of &lt;math&gt;P_1'&lt;/math&gt;, &lt;math&gt;P_2'&lt;/math&gt;, and &lt;math&gt;P_3'&lt;/math&gt; it follows that &lt;math&gt;FP_i=P_iP_i'&lt;/math&gt; for &lt;math&gt;i=1,2,3&lt;/math&gt; and so &lt;math&gt;P_1&lt;/math&gt;, &lt;math&gt;P_2&lt;/math&gt;, and &lt;math&gt;P_3&lt;/math&gt; are on &lt;math&gt;p&lt;/math&gt;. By lemma 1, &lt;math&gt;\ell(P_1)=\overleftrightarrow{YZ}&lt;/math&gt;, &lt;math&gt;\ell(P_2)=\overleftrightarrow{XY}&lt;/math&gt;, and &lt;math&gt;\ell(P_3)=\overleftrightarrow{XZ}&lt;/math&gt;. Therefore the intersections of &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, and &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle with center &lt;math&gt;H&lt;/math&gt;, which finishes the proof.<br /> --Killbilledtoucan<br /> <br /> ===Solution 2===<br /> <br /> Note that the lines &lt;math&gt;l(P_1), l(P_2), l(P_3)&lt;/math&gt; are &lt;cmath&gt;y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,&lt;/cmath&gt; respectively. It is easy to deduce that the three points of intersection are &lt;cmath&gt;\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).&lt;/cmath&gt; The slopes of each side of this equilateral triangle are &lt;cmath&gt;2a_1,2a_2,2a_3,&lt;/cmath&gt; and we want to find the locus of &lt;cmath&gt;\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).&lt;/cmath&gt; We know that &lt;cmath&gt;2a_1=\tan(\theta), 2a_2=\tan (\theta + 120), 2a_3=\tan (\theta-120)&lt;/cmath&gt; for some &lt;math&gt;\theta.&lt;/math&gt; Therefore, we can use the tangent addition formula to deduce &lt;cmath&gt;\frac{a_1+a_2+a_3}{3}=\frac{\tan(\theta)+\tan (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}&lt;/cmath&gt; and &lt;cmath&gt;\begin{align*}<br /> \frac{a_1a_2+a_2a_3+a_3a_1}{3}&amp;=\frac{\tan\theta (\tan(\theta-120)+\tan(\theta+120))+\tan(\theta-120)\tan(\theta+120)}{12}\\<br /> &amp;=\frac{9\tan^2\theta-3}{12(1-3\tan^2\theta)}\\<br /> &amp;=-\frac{1}{4}.\end{align*}&lt;/cmath&gt; Now we show that &lt;math&gt;\frac{a_1+a_2+a_3}{3}&lt;/math&gt; can be any real number. Let's say &lt;cmath&gt;\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}=k&lt;/cmath&gt; for some real number &lt;math&gt;k.&lt;/math&gt; Multiplying both sides by &lt;math&gt;2-\tan^2\theta&lt;/math&gt; and rearranging yields a cubic in &lt;math&gt;\tan\theta.&lt;/math&gt; Clearly this cubic has at least one real solution. As &lt;math&gt;\tan \theta&lt;/math&gt; can take on any real number, all values of &lt;math&gt;k&lt;/math&gt; are possible, and our answer is the line &lt;cmath&gt;\boxed{y=-\frac{1}{4}.}&lt;/cmath&gt; Of course, as the denominator could equal 0, we must check &lt;math&gt;\tan \theta=\pm \frac{1}{\sqrt{3}}.&lt;/math&gt; &lt;cmath&gt;3\tan \theta-\tan^3\theta=k(2-6\tan^2\theta).&lt;/cmath&gt; The left side is nonzero, while the right side is zero, so these values of &lt;math&gt;\theta&lt;/math&gt; do not contribute to any values of &lt;math&gt;k.&lt;/math&gt; So, our answer remains the same. &lt;math&gt;\blacksquare&lt;/math&gt; ~ Benq<br /> == See also ==<br /> {{USAJMO newbox|year=2011|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_6&diff=122212 1985 AIME Problems/Problem 6 2020-05-09T18:18:49Z <p>Crazyeyemoody907: /* Solution */</p> <hr /> <div>== Problem ==<br /> As shown in the figure, [[triangle]] &lt;math&gt;ABC&lt;/math&gt; is divided into six smaller triangles by [[line]]s drawn from the [[vertex | vertices]] through a common interior point. The [[area]]s of four of these triangles are as indicated. Find the area of triangle &lt;math&gt;ABC&lt;/math&gt;. <br /> <br /> [[Image:AIME 1985 Problem 6.png]]<br /> == Solution 1==<br /> Let the interior point be &lt;math&gt;P&lt;/math&gt;, let the points on &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt; be &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;, respectively. Let &lt;math&gt;x&lt;/math&gt; be the area of &lt;math&gt;\triangle APE&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be the area of &lt;math&gt;\triangle CPD&lt;/math&gt;. Note that &lt;math&gt;\triangle APF&lt;/math&gt; and &lt;math&gt;\triangle BPF&lt;/math&gt; share the same [[altitude]] from &lt;math&gt;P&lt;/math&gt;, so the [[ratio]] of their areas is the same as the ratio of their bases. Similarly, &lt;math&gt;\triangle ACF&lt;/math&gt; and &lt;math&gt;\triangle BCF&lt;/math&gt; share the same altitude from &lt;math&gt;C&lt;/math&gt;, so the ratio of their areas is the same as the ratio of their bases. Moreover, the two pairs of bases are actually the same, and thus in the same ratio. As a result, we have:<br /> &lt;math&gt;\frac{40}{30} = \frac{124 + x}{65 + y}&lt;/math&gt; or equivalently &lt;math&gt;372 + 3x = 260 + 4y&lt;/math&gt; and so &lt;math&gt;4y = 3x+ 112&lt;/math&gt;.<br /> <br /> Applying identical reasoning to the triangles with bases &lt;math&gt;\overline{CD}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt;, we get &lt;math&gt;\frac{y}{35} = \frac{x+y+84}{105}&lt;/math&gt; so that &lt;math&gt;3y = x + y + 84&lt;/math&gt; and &lt;math&gt;2y = x + 84&lt;/math&gt;. Substituting from this equation into the previous one gives &lt;math&gt;x = 56&lt;/math&gt;, from which we get &lt;math&gt;y = 70&lt;/math&gt; and so the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;56 + 40 + 30 + 35 + 70 + 84 = \Rightarrow \boxed{315}&lt;/math&gt;.<br /> <br /> == Solution 2 == <br /> This problem can be done using mass points. Assign B a weight of 1 and realize that many of the triangles have the same altitude. After continuously using the formulas that state (The sum of the two weights) = (The middle weight), and (The weight &lt;math&gt;\times&lt;/math&gt; side) = (Other weight) &lt;math&gt;\times&lt;/math&gt; (The other side), the problem yields the answer &lt;math&gt;\boxed{315}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> Let the interior point be &lt;math&gt;P&lt;/math&gt; and let the points on &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt; be &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;, respectively. Also, let &lt;math&gt;[APE]=x,[CPD]=y.&lt;/math&gt; Then notice that by Ceva's, &lt;math&gt;\frac{FB\cdot DC\cdot EA}{DB\cdot CE\cdot AF}=1.&lt;/math&gt; However, we can deduce &lt;math&gt;\frac{FB}{AF}=\frac{3}{4}&lt;/math&gt; from the fact that &lt;math&gt;[AFP]&lt;/math&gt; and &lt;math&gt;[BPF]&lt;/math&gt; share the same height. Similarly, &lt;math&gt;x=\frac{84CE}{EA}&lt;/math&gt; and &lt;math&gt;y=\frac{35DC}{BD}.&lt;/math&gt; Plug and chug and you get &lt;math&gt;xy=84\cdot 35\cdot \frac{3}{4}=2205.&lt;/math&gt; Then notice by the same height reasoning, &lt;math&gt;\frac{84}{x}=\frac{119+y}{x+70}.&lt;/math&gt; Clear the fractions and combine like terms to get &lt;math&gt;35x=5880-xy.&lt;/math&gt; We know &lt;math&gt;xy=2205&lt;/math&gt; so subtraction yields &lt;math&gt;35x=3675,&lt;/math&gt; or &lt;math&gt;x=105.&lt;/math&gt; Plugging this in to our previous ratio statement yields &lt;math&gt;\frac{84}{105}=\frac{4}{5}=\frac{119+y}{175},&lt;/math&gt; so &lt;math&gt;y=21.&lt;/math&gt; Basic addition gives us &lt;math&gt;105+84+21+35+30+40=\boxed{315}.&lt;/math&gt;<br /> <br /> -dchen<br /> <br /> == See also ==<br /> {{AIME box|year=1985|num-b=5|num-a=7}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_14&diff=121991 1993 AIME Problems/Problem 14 2020-05-04T05:16:41Z <p>Crazyeyemoody907: /* Solution */</p> <hr /> <div>== Problem ==<br /> A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form &lt;math&gt;\sqrt{N}\,&lt;/math&gt;, for a positive integer &lt;math&gt;N\,&lt;/math&gt;. Find &lt;math&gt;N\,&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Answer: &lt;math&gt;448&lt;/math&gt;.<br /> <br /> Solution: Put the rectangle on the coordinate plane so its vertices are at &lt;math&gt;(\pm4,\pm3)&lt;/math&gt;, for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, &lt;math&gt;O&lt;/math&gt;.<br /> <br /> Note that such a rectangle is unstuck if its four vertices are in or on the edge of all four quadrants, and it is not the same rectangle as the big one. Let the four vertices of this rectangle be &lt;math&gt;A(4,y)&lt;/math&gt;, &lt;math&gt;B(-x,3)&lt;/math&gt;, &lt;math&gt;C(-4,-y)&lt;/math&gt; and &lt;math&gt;D(x,-3)&lt;/math&gt; for nonnegative &lt;math&gt;x,y&lt;/math&gt;. Then this is a rectangle, so &lt;math&gt;OA=OB&lt;/math&gt;, or &lt;math&gt;16+y^2=9+x^2&lt;/math&gt;, so &lt;math&gt;x^2=y^2+7&lt;/math&gt;.<br /> <br /> [[Image:1993 AIME 14 Diagram.png|center]]<br /> <br /> Reflect &lt;math&gt;D&lt;/math&gt; across the side of the rectangle containing &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;D'(-8-x,-3)&lt;/math&gt;. Then &lt;math&gt;BD'=\sqrt{(-8-x-(-x))^2+(3-(-3))^2}=10&lt;/math&gt; is constant, and the perimeter of the rectangle is equal to &lt;math&gt;2(BC+CD')&lt;/math&gt;. The midpoint of &lt;math&gt;\overline{BD'}&lt;/math&gt; is &lt;math&gt;(-4-x,0)&lt;/math&gt;, and since &lt;math&gt;-4&gt;-4-x&lt;/math&gt; and &lt;math&gt;-y\le0&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; always lies below &lt;math&gt;\overline{BD'}&lt;/math&gt;.<br /> <br /> If &lt;math&gt;y&lt;/math&gt; is positive, it can be decreased to &lt;math&gt;y'&lt;y&lt;/math&gt;. This causes &lt;math&gt;x&lt;/math&gt; to decrease as well, to &lt;math&gt;x'&lt;/math&gt;, where &lt;math&gt;x'^2=y'^2+7&lt;/math&gt; and &lt;math&gt;x'&lt;/math&gt; is still positive. If &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D'&lt;/math&gt; are held in place as everything else moves, then &lt;math&gt;C&lt;/math&gt; moves &lt;math&gt;(y-y')&lt;/math&gt; units up and &lt;math&gt;(x-x')&lt;/math&gt; units left to &lt;math&gt;C'&lt;/math&gt;, which must lie within &lt;math&gt;\triangle BCD'&lt;/math&gt;. Then we must have &lt;math&gt;BC'+C'D'&lt;BC+CD'&lt;/math&gt;, and the perimeter of the rectangle is decreased. Therefore, the minimum perimeter must occur with &lt;math&gt;y=0&lt;/math&gt;, so &lt;math&gt;x=\sqrt7&lt;/math&gt;.<br /> <br /> By the distance formula, this minimum perimeter is &lt;cmath&gt;2\left(\sqrt{(4-\sqrt7)^2+3^2}+\sqrt{(4+\sqrt7)^2+3^2}\right)=4\left(\sqrt{8-2\sqrt7}+\sqrt{8+2\sqrt7}\right)&lt;/cmath&gt; &lt;cmath&gt;=4(\sqrt7-1+\sqrt7+1)=8\sqrt7=\sqrt{448}.&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> Note that the diagonal of the rectangle with minimum perimeter must have the diagonal along the middle segment of length 8 of the rectangle (any other inscribed rectangle can be rotated a bit, then made smaller; this one can't because then the rectangle cannot be inscribed since its longest diagonal is less than 8 in length). Then since a rectangle must have right angles, we draw a circle of radius 4 around the center of the rectangle. Picking the two midpoints on the sides of length 6 and opposite intersection points on the segments of length 8, we form a rectangle. Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be the sides of the rectangle. Then &lt;math&gt;ab = 3(8) = 24&lt;/math&gt; since both are twice the area of the same right triangle, and &lt;math&gt;a^2+b^2 = 64&lt;/math&gt;. So &lt;math&gt;(a+b)^2 = 64+2(24) = 112&lt;/math&gt;, so &lt;math&gt;2(a+b) = \sqrt{\boxed{448}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_14&diff=121989 1993 AIME Problems/Problem 14 2020-05-04T05:16:27Z <p>Crazyeyemoody907: /* Solution */</p> <hr /> <div>== Problem ==<br /> A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form &lt;math&gt;\sqrt{N}\,&lt;/math&gt;, for a positive integer &lt;math&gt;N\,&lt;/math&gt;. Find &lt;math&gt;N\,&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Answer: &lt;math&gt; 448 &lt;/math&gt;.<br /> <br /> Solution: Put the rectangle on the coordinate plane so its vertices are at &lt;math&gt;(\pm4,\pm3)&lt;/math&gt;, for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, &lt;math&gt;O&lt;/math&gt;.<br /> <br /> Note that such a rectangle is unstuck if its four vertices are in or on the edge of all four quadrants, and it is not the same rectangle as the big one. Let the four vertices of this rectangle be &lt;math&gt;A(4,y)&lt;/math&gt;, &lt;math&gt;B(-x,3)&lt;/math&gt;, &lt;math&gt;C(-4,-y)&lt;/math&gt; and &lt;math&gt;D(x,-3)&lt;/math&gt; for nonnegative &lt;math&gt;x,y&lt;/math&gt;. Then this is a rectangle, so &lt;math&gt;OA=OB&lt;/math&gt;, or &lt;math&gt;16+y^2=9+x^2&lt;/math&gt;, so &lt;math&gt;x^2=y^2+7&lt;/math&gt;.<br /> <br /> [[Image:1993 AIME 14 Diagram.png|center]]<br /> <br /> Reflect &lt;math&gt;D&lt;/math&gt; across the side of the rectangle containing &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;D'(-8-x,-3)&lt;/math&gt;. Then &lt;math&gt;BD'=\sqrt{(-8-x-(-x))^2+(3-(-3))^2}=10&lt;/math&gt; is constant, and the perimeter of the rectangle is equal to &lt;math&gt;2(BC+CD')&lt;/math&gt;. The midpoint of &lt;math&gt;\overline{BD'}&lt;/math&gt; is &lt;math&gt;(-4-x,0)&lt;/math&gt;, and since &lt;math&gt;-4&gt;-4-x&lt;/math&gt; and &lt;math&gt;-y\le0&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; always lies below &lt;math&gt;\overline{BD'}&lt;/math&gt;.<br /> <br /> If &lt;math&gt;y&lt;/math&gt; is positive, it can be decreased to &lt;math&gt;y'&lt;y&lt;/math&gt;. This causes &lt;math&gt;x&lt;/math&gt; to decrease as well, to &lt;math&gt;x'&lt;/math&gt;, where &lt;math&gt;x'^2=y'^2+7&lt;/math&gt; and &lt;math&gt;x'&lt;/math&gt; is still positive. If &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D'&lt;/math&gt; are held in place as everything else moves, then &lt;math&gt;C&lt;/math&gt; moves &lt;math&gt;(y-y')&lt;/math&gt; units up and &lt;math&gt;(x-x')&lt;/math&gt; units left to &lt;math&gt;C'&lt;/math&gt;, which must lie within &lt;math&gt;\triangle BCD'&lt;/math&gt;. Then we must have &lt;math&gt;BC'+C'D'&lt;BC+CD'&lt;/math&gt;, and the perimeter of the rectangle is decreased. Therefore, the minimum perimeter must occur with &lt;math&gt;y=0&lt;/math&gt;, so &lt;math&gt;x=\sqrt7&lt;/math&gt;.<br /> <br /> By the distance formula, this minimum perimeter is &lt;cmath&gt;2\left(\sqrt{(4-\sqrt7)^2+3^2}+\sqrt{(4+\sqrt7)^2+3^2}\right)=4\left(\sqrt{8-2\sqrt7}+\sqrt{8+2\sqrt7}\right)&lt;/cmath&gt; &lt;cmath&gt;=4(\sqrt7-1+\sqrt7+1)=8\sqrt7=\sqrt{448}.&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> Note that the diagonal of the rectangle with minimum perimeter must have the diagonal along the middle segment of length 8 of the rectangle (any other inscribed rectangle can be rotated a bit, then made smaller; this one can't because then the rectangle cannot be inscribed since its longest diagonal is less than 8 in length). Then since a rectangle must have right angles, we draw a circle of radius 4 around the center of the rectangle. Picking the two midpoints on the sides of length 6 and opposite intersection points on the segments of length 8, we form a rectangle. Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be the sides of the rectangle. Then &lt;math&gt;ab = 3(8) = 24&lt;/math&gt; since both are twice the area of the same right triangle, and &lt;math&gt;a^2+b^2 = 64&lt;/math&gt;. So &lt;math&gt;(a+b)^2 = 64+2(24) = 112&lt;/math&gt;, so &lt;math&gt;2(a+b) = \sqrt{\boxed{448}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_14&diff=121990 1993 AIME Problems/Problem 14 2020-05-04T05:16:27Z <p>Crazyeyemoody907: /* Solution */</p> <hr /> <div>== Problem ==<br /> A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form &lt;math&gt;\sqrt{N}\,&lt;/math&gt;, for a positive integer &lt;math&gt;N\,&lt;/math&gt;. Find &lt;math&gt;N\,&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Answer: &lt;math&gt; 448 &lt;/math&gt;.<br /> <br /> Solution: Put the rectangle on the coordinate plane so its vertices are at &lt;math&gt;(\pm4,\pm3)&lt;/math&gt;, for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, &lt;math&gt;O&lt;/math&gt;.<br /> <br /> Note that such a rectangle is unstuck if its four vertices are in or on the edge of all four quadrants, and it is not the same rectangle as the big one. Let the four vertices of this rectangle be &lt;math&gt;A(4,y)&lt;/math&gt;, &lt;math&gt;B(-x,3)&lt;/math&gt;, &lt;math&gt;C(-4,-y)&lt;/math&gt; and &lt;math&gt;D(x,-3)&lt;/math&gt; for nonnegative &lt;math&gt;x,y&lt;/math&gt;. Then this is a rectangle, so &lt;math&gt;OA=OB&lt;/math&gt;, or &lt;math&gt;16+y^2=9+x^2&lt;/math&gt;, so &lt;math&gt;x^2=y^2+7&lt;/math&gt;.<br /> <br /> [[Image:1993 AIME 14 Diagram.png|center]]<br /> <br /> Reflect &lt;math&gt;D&lt;/math&gt; across the side of the rectangle containing &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;D'(-8-x,-3)&lt;/math&gt;. Then &lt;math&gt;BD'=\sqrt{(-8-x-(-x))^2+(3-(-3))^2}=10&lt;/math&gt; is constant, and the perimeter of the rectangle is equal to &lt;math&gt;2(BC+CD')&lt;/math&gt;. The midpoint of &lt;math&gt;\overline{BD'}&lt;/math&gt; is &lt;math&gt;(-4-x,0)&lt;/math&gt;, and since &lt;math&gt;-4&gt;-4-x&lt;/math&gt; and &lt;math&gt;-y\le0&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; always lies below &lt;math&gt;\overline{BD'}&lt;/math&gt;.<br /> <br /> If &lt;math&gt;y&lt;/math&gt; is positive, it can be decreased to &lt;math&gt;y'&lt;y&lt;/math&gt;. This causes &lt;math&gt;x&lt;/math&gt; to decrease as well, to &lt;math&gt;x'&lt;/math&gt;, where &lt;math&gt;x'^2=y'^2+7&lt;/math&gt; and &lt;math&gt;x'&lt;/math&gt; is still positive. If &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D'&lt;/math&gt; are held in place as everything else moves, then &lt;math&gt;C&lt;/math&gt; moves &lt;math&gt;(y-y')&lt;/math&gt; units up and &lt;math&gt;(x-x')&lt;/math&gt; units left to &lt;math&gt;C'&lt;/math&gt;, which must lie within &lt;math&gt;\triangle BCD'&lt;/math&gt;. Then we must have &lt;math&gt;BC'+C'D'&lt;BC+CD'&lt;/math&gt;, and the perimeter of the rectangle is decreased. Therefore, the minimum perimeter must occur with &lt;math&gt;y=0&lt;/math&gt;, so &lt;math&gt;x=\sqrt7&lt;/math&gt;.<br /> <br /> By the distance formula, this minimum perimeter is &lt;cmath&gt;2\left(\sqrt{(4-\sqrt7)^2+3^2}+\sqrt{(4+\sqrt7)^2+3^2}\right)=4\left(\sqrt{8-2\sqrt7}+\sqrt{8+2\sqrt7}\right)&lt;/cmath&gt; &lt;cmath&gt;=4(\sqrt7-1+\sqrt7+1)=8\sqrt7=\sqrt{448}.&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> Note that the diagonal of the rectangle with minimum perimeter must have the diagonal along the middle segment of length 8 of the rectangle (any other inscribed rectangle can be rotated a bit, then made smaller; this one can't because then the rectangle cannot be inscribed since its longest diagonal is less than 8 in length). Then since a rectangle must have right angles, we draw a circle of radius 4 around the center of the rectangle. Picking the two midpoints on the sides of length 6 and opposite intersection points on the segments of length 8, we form a rectangle. Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be the sides of the rectangle. Then &lt;math&gt;ab = 3(8) = 24&lt;/math&gt; since both are twice the area of the same right triangle, and &lt;math&gt;a^2+b^2 = 64&lt;/math&gt;. So &lt;math&gt;(a+b)^2 = 64+2(24) = 112&lt;/math&gt;, so &lt;math&gt;2(a+b) = \sqrt{\boxed{448}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_14&diff=121988 1993 AIME Problems/Problem 14 2020-05-04T05:14:04Z <p>Crazyeyemoody907: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form &lt;math&gt;\sqrt{N}\,&lt;/math&gt;, for a positive integer &lt;math&gt;N\,&lt;/math&gt;. Find &lt;math&gt;N\,&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Answer: 448.<br /> <br /> Solution: Put the rectangle on the coordinate plane so its vertices are at &lt;math&gt;(\pm4,\pm3)&lt;/math&gt;, for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, &lt;math&gt;O&lt;/math&gt;.<br /> <br /> Note that such a rectangle is unstuck if its four vertices are in or on the edge of all four quadrants, and it is not the same rectangle as the big one. Let the four vertices of this rectangle be &lt;math&gt;A(4,y)&lt;/math&gt;, &lt;math&gt;B(-x,3)&lt;/math&gt;, &lt;math&gt;C(-4,-y)&lt;/math&gt; and &lt;math&gt;D(x,-3)&lt;/math&gt; for nonnegative &lt;math&gt;x,y&lt;/math&gt;. Then this is a rectangle, so &lt;math&gt;OA=OB&lt;/math&gt;, or &lt;math&gt;16+y^2=9+x^2&lt;/math&gt;, so &lt;math&gt;x^2=y^2+7&lt;/math&gt;.<br /> <br /> [[Image:1993 AIME 14 Diagram.png|center]]<br /> <br /> Reflect &lt;math&gt;D&lt;/math&gt; across the side of the rectangle containing &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;D'(-8-x,-3)&lt;/math&gt;. Then &lt;math&gt;BD'=\sqrt{(-8-x-(-x))^2+(3-(-3))^2}=10&lt;/math&gt; is constant, and the perimeter of the rectangle is equal to &lt;math&gt;2(BC+CD')&lt;/math&gt;. The midpoint of &lt;math&gt;\overline{BD'}&lt;/math&gt; is &lt;math&gt;(-4-x,0)&lt;/math&gt;, and since &lt;math&gt;-4&gt;-4-x&lt;/math&gt; and &lt;math&gt;-y\le0&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; always lies below &lt;math&gt;\overline{BD'}&lt;/math&gt;.<br /> <br /> If &lt;math&gt;y&lt;/math&gt; is positive, it can be decreased to &lt;math&gt;y'&lt;y&lt;/math&gt;. This causes &lt;math&gt;x&lt;/math&gt; to decrease as well, to &lt;math&gt;x'&lt;/math&gt;, where &lt;math&gt;x'^2=y'^2+7&lt;/math&gt; and &lt;math&gt;x'&lt;/math&gt; is still positive. If &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D'&lt;/math&gt; are held in place as everything else moves, then &lt;math&gt;C&lt;/math&gt; moves &lt;math&gt;(y-y')&lt;/math&gt; units up and &lt;math&gt;(x-x')&lt;/math&gt; units left to &lt;math&gt;C'&lt;/math&gt;, which must lie within &lt;math&gt;\triangle BCD'&lt;/math&gt;. Then we must have &lt;math&gt;BC'+C'D'&lt;BC+CD'&lt;/math&gt;, and the perimeter of the rectangle is decreased. Therefore, the minimum perimeter must occur with &lt;math&gt;y=0&lt;/math&gt;, so &lt;math&gt;x=\sqrt7&lt;/math&gt;.<br /> <br /> By the distance formula, this minimum perimeter is &lt;cmath&gt;2\left(\sqrt{(4-\sqrt7)^2+3^2}+\sqrt{(4+\sqrt7)^2+3^2}\right)=4\left(\sqrt{8-2\sqrt7}+\sqrt{8+2\sqrt7}\right)&lt;/cmath&gt; &lt;cmath&gt;=4(\sqrt7-1+\sqrt7+1)=8\sqrt7=\sqrt{448}.&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> Note that the diagonal of the rectangle with minimum perimeter must have the diagonal along the middle segment of length 8 of the rectangle (any other inscribed rectangle can be rotated a bit, then made smaller; this one can't because then the rectangle cannot be inscribed since its longest diagonal is less than 8 in length). Then since a rectangle must have right angles, we draw a circle of radius 4 around the center of the rectangle. Picking the two midpoints on the sides of length 6 and opposite intersection points on the segments of length 8, we form a rectangle. Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be the sides of the rectangle. Then &lt;math&gt;ab = 3(8) = 24&lt;/math&gt; since both are twice the area of the same right triangle, and &lt;math&gt;a^2+b^2 = 64&lt;/math&gt;. So &lt;math&gt;(a+b)^2 = 64+2(24) = 112&lt;/math&gt;, so &lt;math&gt;2(a+b) = \sqrt{\boxed{448}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_13&diff=121987 1993 AIME Problems/Problem 13 2020-05-04T05:08:35Z <p>Crazyeyemoody907: /* Solution 5 */</p> <hr /> <div>== Problem ==<br /> Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let &lt;math&gt;t\,&lt;/math&gt; be the amount of time, in seconds, before Jenny and Kenny can see each other again. If &lt;math&gt;t\,&lt;/math&gt; is written as a fraction in lowest terms, what is the sum of the numerator and denominator?<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1===<br /> Consider the unit cicle of radius 50. Assume that they start at points &lt;math&gt;(-50,100)&lt;/math&gt; and &lt;math&gt;(-50,-100).&lt;/math&gt; Then at time &lt;math&gt;t&lt;/math&gt;, they end up at points &lt;math&gt;(-50+t,100)&lt;/math&gt; and &lt;math&gt;(-50+3t,-100).&lt;/math&gt;<br /> The equation of the line connecting these points and the equation of the circle are &lt;cmath&gt;\begin{align}y&amp;=-\frac{100}{t}x+200-\frac{5000}{t}\\50^2&amp;=x^2+y^2\end{align}.&lt;/cmath&gt; When they see each other again, the line connecting the two points will be tangent to the circle at the point &lt;math&gt;(x,y).&lt;/math&gt; Since the radius is perpendicular to the tangent we get &lt;cmath&gt;-\frac{x}{y}=-\frac{100}{t}&lt;/cmath&gt; or &lt;math&gt;xt=100y.&lt;/math&gt; Now substitute &lt;cmath&gt;y= \frac{xt}{100}&lt;/cmath&gt; into &lt;math&gt;(2)&lt;/math&gt; and get &lt;cmath&gt;x=\frac{5000}{\sqrt{100^2+t^2}}.&lt;/cmath&gt; Now substitute this and &lt;cmath&gt;y=\frac{xt}{100}&lt;/cmath&gt; into &lt;math&gt;(1)&lt;/math&gt; and solve for &lt;math&gt;t&lt;/math&gt; to get &lt;cmath&gt;t=\frac{160}{3}.&lt;/cmath&gt; Finally, the sum of the numerator and denominator is &lt;math&gt;160+3=\boxed{163}.&lt;/math&gt;<br /> <br /> ===Solution 2=== <br /> <br /> Let &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; be Kenny's initial and final points respectively and define &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; similarly for Jenny. Let &lt;math&gt;O&lt;/math&gt; be the center of the building. Also, let &lt;math&gt;X&lt;/math&gt; be the intersection of &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt;. Finaly, let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be the points of tangency of circle &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; respectively. <br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> size(8cm); defaultpen(linewidth(0.7)); <br /> pair A,B,C,D,P,Q,O,X; <br /> A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); X=(300,0); <br /> dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(X);<br /> draw(A--B--X--cycle); <br /> draw(C--D); <br /> draw(P--O--Q); <br /> draw(O--X); <br /> draw(Circle(O,50)); <br /> label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,NNW); label(&quot;$C$&quot;,C,S); label(&quot;$D$&quot;,D,NE); label(&quot;$P$&quot;,P,S); label(&quot;$Q$&quot;,Q,NE); label(&quot;$O$&quot;,O,W); label(&quot;$X$&quot;,X,ESE); <br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> From the problem statement, &lt;math&gt;AB=3t&lt;/math&gt;, and &lt;math&gt;CD=t&lt;/math&gt;. Since &lt;math&gt;\Delta ABX \sim \Delta CDX&lt;/math&gt;, &lt;math&gt;CX=AC\cdot\left(\frac{CD}{AB-CD}\right)=200\cdot\left(\frac{t}{3t-t}\right)=100&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;PC=100&lt;/math&gt;, &lt;math&gt;PX=200&lt;/math&gt;. So, &lt;math&gt;\tan(\angle OXP)=\frac{OP}{PX}=\frac{50}{200}=\frac{1}{4}&lt;/math&gt;. <br /> <br /> Since circle &lt;math&gt;O&lt;/math&gt; is tangent to &lt;math&gt;BX&lt;/math&gt; and &lt;math&gt;AX&lt;/math&gt;, &lt;math&gt;OX&lt;/math&gt; is the [[angle bisector]] of &lt;math&gt;\angle BXA&lt;/math&gt;. <br /> <br /> Thus, &lt;math&gt;\tan(\angle BXA)=\tan(2\angle OXP)=\frac{2\tan(\angle OXP)}{1- \tan^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}=\frac{8}{15}&lt;/math&gt;. <br /> <br /> Therefore, &lt;math&gt;t = CD = CX\cdot\tan(\angle BXA) = 100 \cdot \frac{8}{15} = \frac{160}{3}&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{163}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> &lt;asy&gt;<br /> size(8cm); defaultpen(linewidth(0.7)); <br /> pair A,B,C,D,P,Q,O,R,S; <br /> A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); R=(100,106.667); S=(0,53.333);<br /> dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(R); dot(S);<br /> draw(A--B--D--C--cycle);<br /> draw(P--O); draw(D--S);<br /> draw(O--Q--R--cycle);<br /> draw(Circle(O,50)); <br /> label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,NNW); label(&quot;$C$&quot;,(200,-205),S); label(&quot;$D$&quot;,D,NE); label(&quot;$P$&quot;,(100,-205),S); label(&quot;$Q$&quot;,Q,NE); label(&quot;$O$&quot;,O,SW); label(&quot;$R$&quot;,R,NE); label(&quot;$S$&quot;,S,W);<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;t&lt;/math&gt; be the time they walk. Then &lt;math&gt;CD=t&lt;/math&gt; and &lt;math&gt;AB=3t&lt;/math&gt;.<br /> <br /> <br /> Draw a line from point &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; such that &lt;math&gt;OQ&lt;/math&gt; is perpendicular to &lt;math&gt;BD&lt;/math&gt;. Further, draw a line passing through points &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt;, so &lt;math&gt;OP&lt;/math&gt; is parallel to &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; and is midway between those two lines. Then &lt;math&gt;PR=\dfrac{AB+CD}{2}=\dfrac{3t+t}{2}=2t&lt;/math&gt;. Draw another line passing through point &lt;math&gt;D&lt;/math&gt; and parallel to &lt;math&gt;AC&lt;/math&gt;, and call the point of intersection of this line with &lt;math&gt;AB&lt;/math&gt; as &lt;math&gt;S&lt;/math&gt;. Then &lt;math&gt;SB=AB-AS=3t-t=2t&lt;/math&gt;.<br /> <br /> <br /> We see that &lt;math&gt;m\angle SBD=m\angle ORQ&lt;/math&gt; since they are corresponding angles, and thus by angle-angle similarity, &lt;math&gt;\triangle QOR\sim\triangle SDB&lt;/math&gt;.<br /> <br /> <br /> Then<br /> &lt;cmath&gt;\begin{align*}<br /> \dfrac{OQ}{DS}=\dfrac{RO}{BD}&amp;\implies\dfrac{50}{200}=\dfrac{RO}{\sqrt{200^2+4t^2}}\\<br /> &amp;\implies RO=\dfrac{1}{4}\left(\sqrt{200^2+4t^2}\right)\\<br /> &amp;\implies RO=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right)<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> And we obtain<br /> &lt;cmath&gt;\begin{align*}<br /> PR-OP&amp;=RO\\<br /> 2t-50&amp;=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right)\\<br /> 4t-100&amp;=\sqrt{100^2+t^2}\\<br /> (4t-100)^2&amp;=\left(\sqrt{100^2+t^2}\right)^2\\<br /> 16t^2-800t+100^2&amp;=t^2+100^2\\<br /> 15t^2&amp;=800t\\<br /> t&amp;=\dfrac{800}{15}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> so we have &lt;math&gt;t=\frac{160}{3}&lt;/math&gt;, and our answer is thus &lt;math&gt;160+3=\boxed{163}&lt;/math&gt;.<br /> ===Solution 4===<br /> <br /> We can use areas to find the answer. Since Jenny and Kenny are 200 meters apart, we know that they are side by side, and that the line connecting the two of them is tangent to the circular building. <br /> <br /> We know that the radius of the circle is 50, and that &lt;math&gt;\overline{AJ} = x&lt;/math&gt;, &lt;math&gt;\overline{BK} = 3x&lt;/math&gt;.<br /> <br /> Illustration:<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A = (0,0), B = (200,0), K = (200,160), J = (0,53.3333), O = (100,50);<br /> pair D = tangent(J,O,50,2), F = tangent(A,O,50);<br /> pair[] p = {A,B,K,J,O,D,F};<br /> for (pair point : p) {<br /> dot(point);<br /> }<br /> label(&quot;A&quot;,A,SW);<br /> label(&quot;B&quot;,B,SE);<br /> label(&quot;K&quot;,K,NE);<br /> label(&quot;J&quot;,J,NW);<br /> label(&quot;O&quot;,O,SW);<br /> label(&quot;D&quot;,D,N);<br /> label(&quot;F&quot;,F,S);<br /> draw(Circle(O,50));<br /> draw(A--B--K--J--cycle);<br /> draw(F--O--J);<br /> draw(D--O);<br /> draw(O--K);<br /> label(&quot;$50$&quot;,(100,26.666),0.5E);<br /> label(&quot;$50$&quot;,midpoint(D--O),0.5NE);<br /> label(&quot;$x$&quot;,midpoint(A--J),0.5W);<br /> label(&quot;$3x$&quot;,midpoint(B--K),0.5E);<br /> label(&quot;$200$&quot;,midpoint(A--B),4S);<br /> draw(rightanglemark(J,A,F,150));<br /> draw(rightanglemark(J,D,O,150));<br /> draw(rightanglemark(A,F,O,150));<br /> draw(rightanglemark(F,B,K,150));<br /> &lt;/asy&gt;<br /> <br /> By areas, &lt;math&gt;[OJK] + [AJOF] + [OFBK] = [ABKJ]&lt;/math&gt;. Having right trapezoids, &lt;math&gt;[AFOJ] = \frac{x+50}{2} \cdot 100&lt;/math&gt;. The other areas of right trapezoids can be calculated in the same way. We just need to find &lt;math&gt;[OJK]&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt;.<br /> <br /> If we bring &lt;math&gt;\overline{AB}&lt;/math&gt; up to where the point J is, we have by the Pythagorean Theorem, &lt;math&gt;\overline{JK} = 2\sqrt{x^2+10000} \Rightarrow [OJK] = \frac{1}{2} \overline{JK} \cdot \overline{OD}&lt;/math&gt;.<br /> <br /> Now we have everything to solve for &lt;math&gt;x&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;[OJK] + [AJOF] + [OFBK] = [ABKJ]&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;50 \sqrt{x^2 + 10000} + \frac{x+50}{2} \cdot 100 + \frac{3x+50}{2} \cdot 100 = \frac{x+3x}{2} \cdot 200&lt;/cmath&gt;<br /> <br /> After isolating the radical, dividing by 50, and squaring, we obtain: &lt;math&gt;15x^2 - 800x = 0 \Rightarrow x = \frac{160}{3}&lt;/math&gt;.<br /> <br /> Since Jenny walks &lt;math&gt;\frac{160}{3}&lt;/math&gt; meters at 1 m/s, our answer is &lt;math&gt;160+3 = \fbox{163}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> Basically, draw out a good diagram, and the rest is done.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_13&diff=121986 1993 AIME Problems/Problem 13 2020-05-04T05:08:20Z <p>Crazyeyemoody907: /* Solution 5 */</p> <hr /> <div>== Problem ==<br /> Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let &lt;math&gt;t\,&lt;/math&gt; be the amount of time, in seconds, before Jenny and Kenny can see each other again. If &lt;math&gt;t\,&lt;/math&gt; is written as a fraction in lowest terms, what is the sum of the numerator and denominator?<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1===<br /> Consider the unit cicle of radius 50. Assume that they start at points &lt;math&gt;(-50,100)&lt;/math&gt; and &lt;math&gt;(-50,-100).&lt;/math&gt; Then at time &lt;math&gt;t&lt;/math&gt;, they end up at points &lt;math&gt;(-50+t,100)&lt;/math&gt; and &lt;math&gt;(-50+3t,-100).&lt;/math&gt;<br /> The equation of the line connecting these points and the equation of the circle are &lt;cmath&gt;\begin{align}y&amp;=-\frac{100}{t}x+200-\frac{5000}{t}\\50^2&amp;=x^2+y^2\end{align}.&lt;/cmath&gt; When they see each other again, the line connecting the two points will be tangent to the circle at the point &lt;math&gt;(x,y).&lt;/math&gt; Since the radius is perpendicular to the tangent we get &lt;cmath&gt;-\frac{x}{y}=-\frac{100}{t}&lt;/cmath&gt; or &lt;math&gt;xt=100y.&lt;/math&gt; Now substitute &lt;cmath&gt;y= \frac{xt}{100}&lt;/cmath&gt; into &lt;math&gt;(2)&lt;/math&gt; and get &lt;cmath&gt;x=\frac{5000}{\sqrt{100^2+t^2}}.&lt;/cmath&gt; Now substitute this and &lt;cmath&gt;y=\frac{xt}{100}&lt;/cmath&gt; into &lt;math&gt;(1)&lt;/math&gt; and solve for &lt;math&gt;t&lt;/math&gt; to get &lt;cmath&gt;t=\frac{160}{3}.&lt;/cmath&gt; Finally, the sum of the numerator and denominator is &lt;math&gt;160+3=\boxed{163}.&lt;/math&gt;<br /> <br /> ===Solution 2=== <br /> <br /> Let &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; be Kenny's initial and final points respectively and define &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; similarly for Jenny. Let &lt;math&gt;O&lt;/math&gt; be the center of the building. Also, let &lt;math&gt;X&lt;/math&gt; be the intersection of &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt;. Finaly, let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be the points of tangency of circle &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; respectively. <br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> size(8cm); defaultpen(linewidth(0.7)); <br /> pair A,B,C,D,P,Q,O,X; <br /> A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); X=(300,0); <br /> dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(X);<br /> draw(A--B--X--cycle); <br /> draw(C--D); <br /> draw(P--O--Q); <br /> draw(O--X); <br /> draw(Circle(O,50)); <br /> label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,NNW); label(&quot;$C$&quot;,C,S); label(&quot;$D$&quot;,D,NE); label(&quot;$P$&quot;,P,S); label(&quot;$Q$&quot;,Q,NE); label(&quot;$O$&quot;,O,W); label(&quot;$X$&quot;,X,ESE); <br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> From the problem statement, &lt;math&gt;AB=3t&lt;/math&gt;, and &lt;math&gt;CD=t&lt;/math&gt;. Since &lt;math&gt;\Delta ABX \sim \Delta CDX&lt;/math&gt;, &lt;math&gt;CX=AC\cdot\left(\frac{CD}{AB-CD}\right)=200\cdot\left(\frac{t}{3t-t}\right)=100&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;PC=100&lt;/math&gt;, &lt;math&gt;PX=200&lt;/math&gt;. So, &lt;math&gt;\tan(\angle OXP)=\frac{OP}{PX}=\frac{50}{200}=\frac{1}{4}&lt;/math&gt;. <br /> <br /> Since circle &lt;math&gt;O&lt;/math&gt; is tangent to &lt;math&gt;BX&lt;/math&gt; and &lt;math&gt;AX&lt;/math&gt;, &lt;math&gt;OX&lt;/math&gt; is the [[angle bisector]] of &lt;math&gt;\angle BXA&lt;/math&gt;. <br /> <br /> Thus, &lt;math&gt;\tan(\angle BXA)=\tan(2\angle OXP)=\frac{2\tan(\angle OXP)}{1- \tan^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}=\frac{8}{15}&lt;/math&gt;. <br /> <br /> Therefore, &lt;math&gt;t = CD = CX\cdot\tan(\angle BXA) = 100 \cdot \frac{8}{15} = \frac{160}{3}&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{163}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> &lt;asy&gt;<br /> size(8cm); defaultpen(linewidth(0.7)); <br /> pair A,B,C,D,P,Q,O,R,S; <br /> A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); R=(100,106.667); S=(0,53.333);<br /> dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(R); dot(S);<br /> draw(A--B--D--C--cycle);<br /> draw(P--O); draw(D--S);<br /> draw(O--Q--R--cycle);<br /> draw(Circle(O,50)); <br /> label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,NNW); label(&quot;$C$&quot;,(200,-205),S); label(&quot;$D$&quot;,D,NE); label(&quot;$P$&quot;,(100,-205),S); label(&quot;$Q$&quot;,Q,NE); label(&quot;$O$&quot;,O,SW); label(&quot;$R$&quot;,R,NE); label(&quot;$S$&quot;,S,W);<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;t&lt;/math&gt; be the time they walk. Then &lt;math&gt;CD=t&lt;/math&gt; and &lt;math&gt;AB=3t&lt;/math&gt;.<br /> <br /> <br /> Draw a line from point &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; such that &lt;math&gt;OQ&lt;/math&gt; is perpendicular to &lt;math&gt;BD&lt;/math&gt;. Further, draw a line passing through points &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt;, so &lt;math&gt;OP&lt;/math&gt; is parallel to &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; and is midway between those two lines. Then &lt;math&gt;PR=\dfrac{AB+CD}{2}=\dfrac{3t+t}{2}=2t&lt;/math&gt;. Draw another line passing through point &lt;math&gt;D&lt;/math&gt; and parallel to &lt;math&gt;AC&lt;/math&gt;, and call the point of intersection of this line with &lt;math&gt;AB&lt;/math&gt; as &lt;math&gt;S&lt;/math&gt;. Then &lt;math&gt;SB=AB-AS=3t-t=2t&lt;/math&gt;.<br /> <br /> <br /> We see that &lt;math&gt;m\angle SBD=m\angle ORQ&lt;/math&gt; since they are corresponding angles, and thus by angle-angle similarity, &lt;math&gt;\triangle QOR\sim\triangle SDB&lt;/math&gt;.<br /> <br /> <br /> Then<br /> &lt;cmath&gt;\begin{align*}<br /> \dfrac{OQ}{DS}=\dfrac{RO}{BD}&amp;\implies\dfrac{50}{200}=\dfrac{RO}{\sqrt{200^2+4t^2}}\\<br /> &amp;\implies RO=\dfrac{1}{4}\left(\sqrt{200^2+4t^2}\right)\\<br /> &amp;\implies RO=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right)<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> And we obtain<br /> &lt;cmath&gt;\begin{align*}<br /> PR-OP&amp;=RO\\<br /> 2t-50&amp;=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right)\\<br /> 4t-100&amp;=\sqrt{100^2+t^2}\\<br /> (4t-100)^2&amp;=\left(\sqrt{100^2+t^2}\right)^2\\<br /> 16t^2-800t+100^2&amp;=t^2+100^2\\<br /> 15t^2&amp;=800t\\<br /> t&amp;=\dfrac{800}{15}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> so we have &lt;math&gt;t=\frac{160}{3}&lt;/math&gt;, and our answer is thus &lt;math&gt;160+3=\boxed{163}&lt;/math&gt;.<br /> ===Solution 4===<br /> <br /> We can use areas to find the answer. Since Jenny and Kenny are 200 meters apart, we know that they are side by side, and that the line connecting the two of them is tangent to the circular building. <br /> <br /> We know that the radius of the circle is 50, and that &lt;math&gt;\overline{AJ} = x&lt;/math&gt;, &lt;math&gt;\overline{BK} = 3x&lt;/math&gt;.<br /> <br /> Illustration:<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A = (0,0), B = (200,0), K = (200,160), J = (0,53.3333), O = (100,50);<br /> pair D = tangent(J,O,50,2), F = tangent(A,O,50);<br /> pair[] p = {A,B,K,J,O,D,F};<br /> for (pair point : p) {<br /> dot(point);<br /> }<br /> label(&quot;A&quot;,A,SW);<br /> label(&quot;B&quot;,B,SE);<br /> label(&quot;K&quot;,K,NE);<br /> label(&quot;J&quot;,J,NW);<br /> label(&quot;O&quot;,O,SW);<br /> label(&quot;D&quot;,D,N);<br /> label(&quot;F&quot;,F,S);<br /> draw(Circle(O,50));<br /> draw(A--B--K--J--cycle);<br /> draw(F--O--J);<br /> draw(D--O);<br /> draw(O--K);<br /> label(&quot;$50$&quot;,(100,26.666),0.5E);<br /> label(&quot;$50$&quot;,midpoint(D--O),0.5NE);<br /> label(&quot;$x$&quot;,midpoint(A--J),0.5W);<br /> label(&quot;$3x$&quot;,midpoint(B--K),0.5E);<br /> label(&quot;$200$&quot;,midpoint(A--B),4S);<br /> draw(rightanglemark(J,A,F,150));<br /> draw(rightanglemark(J,D,O,150));<br /> draw(rightanglemark(A,F,O,150));<br /> draw(rightanglemark(F,B,K,150));<br /> &lt;/asy&gt;<br /> <br /> By areas, &lt;math&gt;[OJK] + [AJOF] + [OFBK] = [ABKJ]&lt;/math&gt;. Having right trapezoids, &lt;math&gt;[AFOJ] = \frac{x+50}{2} \cdot 100&lt;/math&gt;. The other areas of right trapezoids can be calculated in the same way. We just need to find &lt;math&gt;[OJK]&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt;.<br /> <br /> If we bring &lt;math&gt;\overline{AB}&lt;/math&gt; up to where the point J is, we have by the Pythagorean Theorem, &lt;math&gt;\overline{JK} = 2\sqrt{x^2+10000} \Rightarrow [OJK] = \frac{1}{2} \overline{JK} \cdot \overline{OD}&lt;/math&gt;.<br /> <br /> Now we have everything to solve for &lt;math&gt;x&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;[OJK] + [AJOF] + [OFBK] = [ABKJ]&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;50 \sqrt{x^2 + 10000} + \frac{x+50}{2} \cdot 100 + \frac{3x+50}{2} \cdot 100 = \frac{x+3x}{2} \cdot 200&lt;/cmath&gt;<br /> <br /> After isolating the radical, dividing by 50, and squaring, we obtain: &lt;math&gt;15x^2 - 800x = 0 \Rightarrow x = \frac{160}{3}&lt;/math&gt;.<br /> <br /> Since Jenny walks &lt;math&gt;\frac{160}{3}&lt;/math&gt; meters at 1 m/s, our answer is &lt;math&gt;160+3 = \fbox{163}&lt;/math&gt;.<br /> <br /> = Solution 5 =<br /> Basically, draw out a good diagram, and the rest is done.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_13&diff=121985 1993 AIME Problems/Problem 13 2020-05-04T05:07:30Z <p>Crazyeyemoody907: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let &lt;math&gt;t\,&lt;/math&gt; be the amount of time, in seconds, before Jenny and Kenny can see each other again. If &lt;math&gt;t\,&lt;/math&gt; is written as a fraction in lowest terms, what is the sum of the numerator and denominator?<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1===<br /> Consider the unit cicle of radius 50. Assume that they start at points &lt;math&gt;(-50,100)&lt;/math&gt; and &lt;math&gt;(-50,-100).&lt;/math&gt; Then at time &lt;math&gt;t&lt;/math&gt;, they end up at points &lt;math&gt;(-50+t,100)&lt;/math&gt; and &lt;math&gt;(-50+3t,-100).&lt;/math&gt;<br /> The equation of the line connecting these points and the equation of the circle are &lt;cmath&gt;\begin{align}y&amp;=-\frac{100}{t}x+200-\frac{5000}{t}\\50^2&amp;=x^2+y^2\end{align}.&lt;/cmath&gt; When they see each other again, the line connecting the two points will be tangent to the circle at the point &lt;math&gt;(x,y).&lt;/math&gt; Since the radius is perpendicular to the tangent we get &lt;cmath&gt;-\frac{x}{y}=-\frac{100}{t}&lt;/cmath&gt; or &lt;math&gt;xt=100y.&lt;/math&gt; Now substitute &lt;cmath&gt;y= \frac{xt}{100}&lt;/cmath&gt; into &lt;math&gt;(2)&lt;/math&gt; and get &lt;cmath&gt;x=\frac{5000}{\sqrt{100^2+t^2}}.&lt;/cmath&gt; Now substitute this and &lt;cmath&gt;y=\frac{xt}{100}&lt;/cmath&gt; into &lt;math&gt;(1)&lt;/math&gt; and solve for &lt;math&gt;t&lt;/math&gt; to get &lt;cmath&gt;t=\frac{160}{3}.&lt;/cmath&gt; Finally, the sum of the numerator and denominator is &lt;math&gt;160+3=\boxed{163}.&lt;/math&gt;<br /> <br /> ===Solution 2=== <br /> <br /> Let &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; be Kenny's initial and final points respectively and define &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; similarly for Jenny. Let &lt;math&gt;O&lt;/math&gt; be the center of the building. Also, let &lt;math&gt;X&lt;/math&gt; be the intersection of &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt;. Finaly, let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be the points of tangency of circle &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; respectively. <br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> size(8cm); defaultpen(linewidth(0.7)); <br /> pair A,B,C,D,P,Q,O,X; <br /> A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); X=(300,0); <br /> dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(X);<br /> draw(A--B--X--cycle); <br /> draw(C--D); <br /> draw(P--O--Q); <br /> draw(O--X); <br /> draw(Circle(O,50)); <br /> label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,NNW); label(&quot;$C$&quot;,C,S); label(&quot;$D$&quot;,D,NE); label(&quot;$P$&quot;,P,S); label(&quot;$Q$&quot;,Q,NE); label(&quot;$O$&quot;,O,W); label(&quot;$X$&quot;,X,ESE); <br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> From the problem statement, &lt;math&gt;AB=3t&lt;/math&gt;, and &lt;math&gt;CD=t&lt;/math&gt;. Since &lt;math&gt;\Delta ABX \sim \Delta CDX&lt;/math&gt;, &lt;math&gt;CX=AC\cdot\left(\frac{CD}{AB-CD}\right)=200\cdot\left(\frac{t}{3t-t}\right)=100&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;PC=100&lt;/math&gt;, &lt;math&gt;PX=200&lt;/math&gt;. So, &lt;math&gt;\tan(\angle OXP)=\frac{OP}{PX}=\frac{50}{200}=\frac{1}{4}&lt;/math&gt;. <br /> <br /> Since circle &lt;math&gt;O&lt;/math&gt; is tangent to &lt;math&gt;BX&lt;/math&gt; and &lt;math&gt;AX&lt;/math&gt;, &lt;math&gt;OX&lt;/math&gt; is the [[angle bisector]] of &lt;math&gt;\angle BXA&lt;/math&gt;. <br /> <br /> Thus, &lt;math&gt;\tan(\angle BXA)=\tan(2\angle OXP)=\frac{2\tan(\angle OXP)}{1- \tan^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}=\frac{8}{15}&lt;/math&gt;. <br /> <br /> Therefore, &lt;math&gt;t = CD = CX\cdot\tan(\angle BXA) = 100 \cdot \frac{8}{15} = \frac{160}{3}&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{163}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> &lt;asy&gt;<br /> size(8cm); defaultpen(linewidth(0.7)); <br /> pair A,B,C,D,P,Q,O,R,S; <br /> A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); R=(100,106.667); S=(0,53.333);<br /> dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(R); dot(S);<br /> draw(A--B--D--C--cycle);<br /> draw(P--O); draw(D--S);<br /> draw(O--Q--R--cycle);<br /> draw(Circle(O,50)); <br /> label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,NNW); label(&quot;$C$&quot;,(200,-205),S); label(&quot;$D$&quot;,D,NE); label(&quot;$P$&quot;,(100,-205),S); label(&quot;$Q$&quot;,Q,NE); label(&quot;$O$&quot;,O,SW); label(&quot;$R$&quot;,R,NE); label(&quot;$S$&quot;,S,W);<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;t&lt;/math&gt; be the time they walk. Then &lt;math&gt;CD=t&lt;/math&gt; and &lt;math&gt;AB=3t&lt;/math&gt;.<br /> <br /> <br /> Draw a line from point &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; such that &lt;math&gt;OQ&lt;/math&gt; is perpendicular to &lt;math&gt;BD&lt;/math&gt;. Further, draw a line passing through points &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt;, so &lt;math&gt;OP&lt;/math&gt; is parallel to &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; and is midway between those two lines. Then &lt;math&gt;PR=\dfrac{AB+CD}{2}=\dfrac{3t+t}{2}=2t&lt;/math&gt;. Draw another line passing through point &lt;math&gt;D&lt;/math&gt; and parallel to &lt;math&gt;AC&lt;/math&gt;, and call the point of intersection of this line with &lt;math&gt;AB&lt;/math&gt; as &lt;math&gt;S&lt;/math&gt;. Then &lt;math&gt;SB=AB-AS=3t-t=2t&lt;/math&gt;.<br /> <br /> <br /> We see that &lt;math&gt;m\angle SBD=m\angle ORQ&lt;/math&gt; since they are corresponding angles, and thus by angle-angle similarity, &lt;math&gt;\triangle QOR\sim\triangle SDB&lt;/math&gt;.<br /> <br /> <br /> Then<br /> &lt;cmath&gt;\begin{align*}<br /> \dfrac{OQ}{DS}=\dfrac{RO}{BD}&amp;\implies\dfrac{50}{200}=\dfrac{RO}{\sqrt{200^2+4t^2}}\\<br /> &amp;\implies RO=\dfrac{1}{4}\left(\sqrt{200^2+4t^2}\right)\\<br /> &amp;\implies RO=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right)<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> And we obtain<br /> &lt;cmath&gt;\begin{align*}<br /> PR-OP&amp;=RO\\<br /> 2t-50&amp;=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right)\\<br /> 4t-100&amp;=\sqrt{100^2+t^2}\\<br /> (4t-100)^2&amp;=\left(\sqrt{100^2+t^2}\right)^2\\<br /> 16t^2-800t+100^2&amp;=t^2+100^2\\<br /> 15t^2&amp;=800t\\<br /> t&amp;=\dfrac{800}{15}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> so we have &lt;math&gt;t=\frac{160}{3}&lt;/math&gt;, and our answer is thus &lt;math&gt;160+3=\boxed{163}&lt;/math&gt;.<br /> ===Solution 4===<br /> <br /> We can use areas to find the answer. Since Jenny and Kenny are 200 meters apart, we know that they are side by side, and that the line connecting the two of them is tangent to the circular building. <br /> <br /> We know that the radius of the circle is 50, and that &lt;math&gt;\overline{AJ} = x&lt;/math&gt;, &lt;math&gt;\overline{BK} = 3x&lt;/math&gt;.<br /> <br /> Illustration:<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A = (0,0), B = (200,0), K = (200,160), J = (0,53.3333), O = (100,50);<br /> pair D = tangent(J,O,50,2), F = tangent(A,O,50);<br /> pair[] p = {A,B,K,J,O,D,F};<br /> for (pair point : p) {<br /> dot(point);<br /> }<br /> label(&quot;A&quot;,A,SW);<br /> label(&quot;B&quot;,B,SE);<br /> label(&quot;K&quot;,K,NE);<br /> label(&quot;J&quot;,J,NW);<br /> label(&quot;O&quot;,O,SW);<br /> label(&quot;D&quot;,D,N);<br /> label(&quot;F&quot;,F,S);<br /> draw(Circle(O,50));<br /> draw(A--B--K--J--cycle);<br /> draw(F--O--J);<br /> draw(D--O);<br /> draw(O--K);<br /> label(&quot;$50$&quot;,(100,26.666),0.5E);<br /> label(&quot;$50$&quot;,midpoint(D--O),0.5NE);<br /> label(&quot;$x$&quot;,midpoint(A--J),0.5W);<br /> label(&quot;$3x$&quot;,midpoint(B--K),0.5E);<br /> label(&quot;$200$&quot;,midpoint(A--B),4S);<br /> draw(rightanglemark(J,A,F,150));<br /> draw(rightanglemark(J,D,O,150));<br /> draw(rightanglemark(A,F,O,150));<br /> draw(rightanglemark(F,B,K,150));<br /> &lt;/asy&gt;<br /> <br /> By areas, &lt;math&gt;[OJK] + [AJOF] + [OFBK] = [ABKJ]&lt;/math&gt;. Having right trapezoids, &lt;math&gt;[AFOJ] = \frac{x+50}{2} \cdot 100&lt;/math&gt;. The other areas of right trapezoids can be calculated in the same way. We just need to find &lt;math&gt;[OJK]&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt;.<br /> <br /> If we bring &lt;math&gt;\overline{AB}&lt;/math&gt; up to where the point J is, we have by the Pythagorean Theorem, &lt;math&gt;\overline{JK} = 2\sqrt{x^2+10000} \Rightarrow [OJK] = \frac{1}{2} \overline{JK} \cdot \overline{OD}&lt;/math&gt;.<br /> <br /> Now we have everything to solve for &lt;math&gt;x&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;[OJK] + [AJOF] + [OFBK] = [ABKJ]&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;50 \sqrt{x^2 + 10000} + \frac{x+50}{2} \cdot 100 + \frac{3x+50}{2} \cdot 100 = \frac{x+3x}{2} \cdot 200&lt;/cmath&gt;<br /> <br /> After isolating the radical, dividing by 50, and squaring, we obtain: &lt;math&gt;15x^2 - 800x = 0 \Rightarrow x = \frac{160}{3}&lt;/math&gt;.<br /> <br /> Since Jenny walks &lt;math&gt;\frac{160}{3}&lt;/math&gt; meters at 1 m/s, our answer is &lt;math&gt;160+3 = \fbox{163}&lt;/math&gt;.<br /> <br /> == Solution 5 ==<br /> Basically, draw out a good diagram, and the rest is done.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=109370 Gmaas 2019-08-25T23:21:14Z <p>Crazyeyemoody907: Replaced content with &quot;lol&quot;</p> <hr /> <div>lol</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=Talk:Gmaas&diff=108765 Talk:Gmaas 2019-08-11T21:57:20Z <p>Crazyeyemoody907: Replaced content with &quot;lol i deleted ur spam!-crazyeyemoody907&quot;</p> <hr /> <div>lol i deleted ur spam!-crazyeyemoody907</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=108764 Gmaas 2019-08-11T21:47:44Z <p>Crazyeyemoody907: </p> <hr /> <div>ha lol i deleted ur spamming!!!-crazyeyemoody907</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=108763 Gmaas 2019-08-11T21:47:23Z <p>Crazyeyemoody907: This was a spam page.</p> <hr /> <div>ha lol i deleted ur spamming!!!</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=Asymptote:_Help&diff=106893 Asymptote: Help 2019-06-23T23:31:21Z <p>Crazyeyemoody907: </p> <hr /> <div>&lt;asy&gt;<br /> size(150);<br /> real ticklen=3;<br /> real tickspace=2;<br /> <br /> real ticklength=0.1cm;<br /> real axisarrowsize=0.14cm;<br /> pen axispen=black+1.3bp;<br /> real vectorarrowsize=0.2cm;<br /> real tickdown=-0.5;<br /> real tickdownlength=-0.15inch;<br /> real tickdownbase=0.3;<br /> real wholetickdown=tickdown;<br /> void rr_cartesian_axes(real xleft, real xright, real ybottom, real ytop, real xstep=1, real ystep=1, bool useticks=false, bool complexplane=false, bool usegrid=true) {<br /> import graph;<br /> real i;<br /> if(complexplane) {<br /> label(&quot;$\textnormal{Re}$&quot;,(xright,0),SE);<br /> label(&quot;$\textnormal{Im}$&quot;,(0,ytop),NW);<br /> } else {<br /> label(&quot;$x$&quot;,(xright+0.4,-0.5));<br /> label(&quot;$y$&quot;,(-0.5,ytop+0.2)); <br /> }<br /> <br /> ylimits(ybottom,ytop);<br /> xlimits( xleft, xright);<br /> real[] TicksArrx,TicksArry;<br /> <br /> for(i=xleft+xstep; i&lt;xright; i+=xstep) {<br /> if(abs(i) &gt;0.1) {<br /> TicksArrx.push(i);<br /> }<br /> }<br /> for(i=ybottom+ystep; i&lt;ytop; i+=ystep) {<br /> if(abs(i) &gt;0.1) {<br /> TicksArry.push(i);<br /> }<br /> }<br /> <br /> if(usegrid) {<br /> xaxis(BottomTop(extend=false), Ticks(&quot;%&quot;, TicksArrx ,pTick=gray(0.22),extend=true),p=invisible);//,above=true);<br /> yaxis(LeftRight(extend=false),Ticks(&quot;%&quot;, TicksArry ,pTick=gray(0.22),extend=true), p=invisible);//,Arrows);<br /> }<br /> if(useticks) {<br /> xequals(0, ymin=ybottom, ymax=ytop, p=axispen, Ticks(&quot;%&quot;,TicksArry , pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize));<br /> yequals(0, xmin=xleft, xmax=xright, p=axispen, Ticks(&quot;%&quot;,TicksArrx , pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize));<br /> <br /> } else {<br /> xequals(0, ymin=ybottom, ymax=ytop, p=axispen, above=true, Arrows(size=axisarrowsize));<br /> yequals(0, xmin=xleft, xmax=xright, p=axispen, above=true, Arrows(size=axisarrowsize));<br /> }<br /> };<br /> rr_cartesian_axes(-5,5,-5,6);<br /> draw((-4,4)--(-1,0)--(0,2)--(4,-4),red);<br /> &lt;/asy&gt;</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=User:Crazyeyemoody907&diff=106150 User:Crazyeyemoody907 2019-06-07T11:50:11Z <p>Crazyeyemoody907: Not writing one you stalkers</p> <hr /> <div>Crazyeyemoody is a person.</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12_Problems/Problem_16&diff=100749 2018 AMC 12 Problems/Problem 16 2019-01-21T23:46:40Z <p>Crazyeyemoody907: Created page with &quot;See 2018 AMC 10 Problems/Problem 21&quot;</p> <hr /> <div>See 2018 AMC 10 Problems/Problem 21</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=Chicken_McNugget_Theorem&diff=100562 Chicken McNugget Theorem 2019-01-17T00:30:35Z <p>Crazyeyemoody907: /* Olympiad */</p> <hr /> <div>The '''Chicken McNugget Theorem''' (or '''Postage Stamp Problem''' or '''Frobenius Coin Problem''') states that for any two [[relatively prime]] [[positive integer]]s &lt;math&gt;m,n&lt;/math&gt;, the greatest integer that cannot be written in the form &lt;math&gt;am + bn&lt;/math&gt; for [[nonnegative]] integers &lt;math&gt;a, b&lt;/math&gt; is &lt;math&gt;mn-m-n&lt;/math&gt;.<br /> <br /> A consequence of the theorem is that there are exactly &lt;math&gt;\frac{(m - 1)(n - 1)}{2}&lt;/math&gt; positive integers which cannot be expressed in the form &lt;math&gt;am + bn&lt;/math&gt;. The proof is based on the fact that in each pair of the form &lt;math&gt;(k, (m - 1)(n - 1) - k+1)&lt;/math&gt;, exactly one element is expressible.<br /> <br /> == Origins ==<br /> There are many stories surrounding the origin of the Chicken McNugget theorem. However, the most popular by far remains that of the Chicken McNugget. Originally, McDonald's sold its nuggets in packs of 9 and 20. Math enthusiasts were curious to find the largest number of nuggets that could not have been bought with these packs, thus creating the Chicken McNugget Theorem (the answer worked out to be 151 nuggets). The Chicken McNugget Theorem has also been called the Frobenius Coin Problem or the Frobenius Problem, after German mathematician Ferdinand Frobenius inquired about the largest amount of currency that could not have been made with certain types of coins.<br /> <br /> <br /> <br /> <br /> <br /> ==Proof 1==<br /> &lt;b&gt;Definition&lt;/b&gt;. An integer &lt;math&gt;N \in \mathbb{Z}&lt;/math&gt; will be called &lt;i&gt;purchasable&lt;/i&gt; if there exist nonnegative integers &lt;math&gt;a,b&lt;/math&gt; such that &lt;math&gt;am+bn = N&lt;/math&gt;.<br /> <br /> We would like to prove that &lt;math&gt;mn-m-n&lt;/math&gt; is the largest non-purchasable integer. We are required to show that (1) &lt;math&gt;mn-m-n&lt;/math&gt; is non-purchasable, and (2) every &lt;math&gt;N &gt; mn-m-n&lt;/math&gt; is purchasable. <br /> Note that all purchasable integers are nonnegative, thus the set of non-purchasable integers is nonempty.<br /> <br /> &lt;b&gt;Lemma&lt;/b&gt;. Let &lt;math&gt;A_{N} \subset \mathbb{Z} \times \mathbb{Z}&lt;/math&gt; be the set of solutions &lt;math&gt;(x,y)&lt;/math&gt; to &lt;math&gt;xm+yn = N&lt;/math&gt;. Then &lt;math&gt;A_{N} = \{(x+kn,y-km) \;:\; k \in \mathbb{Z}\}&lt;/math&gt; for any &lt;math&gt;(x,y) \in A_{N}&lt;/math&gt;.<br /> <br /> &lt;i&gt;Proof&lt;/i&gt;: By [[Bezout's Lemma]], there exist integers &lt;math&gt;x',y'&lt;/math&gt; such that &lt;math&gt;x'm+y'n = 1&lt;/math&gt;. Then &lt;math&gt;(Nx')m+(Ny')n = N&lt;/math&gt;. Hence &lt;math&gt;A_{N}&lt;/math&gt; is nonempty. It is easy to check that &lt;math&gt;(Nx'+kn,Ny'-km) \in A_{N}&lt;/math&gt; for all &lt;math&gt;k \in \mathbb{Z}&lt;/math&gt;. We now prove that there are no others. Suppose &lt;math&gt;(x_{1},y_{1})&lt;/math&gt; and &lt;math&gt;(x_{2},y_{2})&lt;/math&gt; are solutions to &lt;math&gt;xm+yn=N&lt;/math&gt;. Then &lt;math&gt;x_{1}m+y_{1}n = x_{2}m+y_{2}n&lt;/math&gt; implies &lt;math&gt;m(x_{1}-x_{2}) = n(y_{2}-y_{1})&lt;/math&gt;. Since &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are coprime and &lt;math&gt;m&lt;/math&gt; divides &lt;math&gt;n(y_{2}-y_{1})&lt;/math&gt;, &lt;math&gt;m&lt;/math&gt; divides &lt;math&gt;y_{2}-y_{1}&lt;/math&gt; and &lt;math&gt;y_{2} \equiv y_{1} \pmod{m}&lt;/math&gt;. Similarly &lt;math&gt;x_{2} \equiv x_{1} \pmod{n}&lt;/math&gt;. Let &lt;math&gt;k_{1},k_{2}&lt;/math&gt; be integers such that &lt;math&gt;x_{2}-x_{1} = k_{1}n&lt;/math&gt; and &lt;math&gt;y_{2}-y_{1} = k_{2}m&lt;/math&gt;. Then &lt;math&gt;m(-k_{1}n) = n(k_{2}m)&lt;/math&gt; implies &lt;math&gt;k_{1} = -k_{2}.&lt;/math&gt; We have the desired result. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> &lt;b&gt;Lemma&lt;/b&gt;. For any integer &lt;math&gt;N&lt;/math&gt;, there exists unique &lt;math&gt;(a_{N},b_{N}) \in \mathbb{Z} \times \{0,1,\ldots,m-1\}&lt;/math&gt; such that &lt;math&gt;a_{N}m + b_{N}n = N&lt;/math&gt;.<br /> <br /> &lt;i&gt;Proof&lt;/i&gt;: By the division algorithm, there exists &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;0 \le y-km \le m-1&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> &lt;b&gt;Lemma&lt;/b&gt;. &lt;math&gt;N&lt;/math&gt; is purchasable if and only if &lt;math&gt;a_{N} \ge 0&lt;/math&gt;.<br /> <br /> &lt;i&gt;Proof&lt;/i&gt;: If &lt;math&gt;a_{N} \ge 0&lt;/math&gt;, then we may simply pick &lt;math&gt;(a,b) = (a_{N},b_{N})&lt;/math&gt; so &lt;math&gt;N&lt;/math&gt; is purchasable. If &lt;math&gt;a_{N} &lt; 0&lt;/math&gt;, then &lt;math&gt;a_{N}+kn &lt; 0&lt;/math&gt; if &lt;math&gt;k \le 0&lt;/math&gt; and &lt;math&gt;b_{N}-km &lt; 0&lt;/math&gt; if &lt;math&gt;k &gt; 0&lt;/math&gt;, hence at least one coordinate of &lt;math&gt;(a_{N}+kn,b_{N}-km)&lt;/math&gt; is negative for all &lt;math&gt;k \in \mathbb{Z}&lt;/math&gt;. Thus &lt;math&gt;N&lt;/math&gt; is not purchasable. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> Thus the set of non-purchasable integers is &lt;math&gt;\{xm+yn \;:\; x&lt;0,0 \le y \le m-1\}&lt;/math&gt;. We would like to find the maximum of this set. <br /> Since both &lt;math&gt;m,n&lt;/math&gt; are positive, the maximum is achieved when &lt;math&gt;x = -1&lt;/math&gt; and &lt;math&gt;y = m-1&lt;/math&gt; so that &lt;math&gt;xm+yn = (-1)m+(m-1)n = mn-m-n&lt;/math&gt;.<br /> <br /> ==Proof 2==<br /> We start with this statement taken from [[Fermat%27s_Little_Theorem#Proof_2_.28Inverses.29|Proof 2 of Fermat's Little Theorem]]:<br /> <br /> &quot;Let &lt;math&gt;S = \{1,2,3,\cdots, p-1\}&lt;/math&gt;. Then, we claim that the set &lt;math&gt;a \cdot S&lt;/math&gt;, consisting of the product of the elements of &lt;math&gt;S&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt;, taken modulo &lt;math&gt;p&lt;/math&gt;, is simply a permutation of &lt;math&gt;S&lt;/math&gt;. In other words, <br /> <br /> &lt;center&gt;&lt;cmath&gt;S \equiv \{1a, 2a, \cdots, (p-1)a\} \pmod{p}.&lt;/cmath&gt;&lt;/center&gt;&lt;br&gt;<br /> <br /> Clearly none of the &lt;math&gt;ia&lt;/math&gt; for &lt;math&gt;1 \le i \le p-1&lt;/math&gt; are divisible by &lt;math&gt;p&lt;/math&gt;, so it suffices to show that all of the elements in &lt;math&gt;a \cdot S&lt;/math&gt; are distinct. Suppose that &lt;math&gt;ai \equiv aj \pmod{p}&lt;/math&gt; for &lt;math&gt;i \neq j&lt;/math&gt;. Since &lt;math&gt;\text{gcd}\, (a,p) = 1&lt;/math&gt;, by the cancellation rule, that reduces to &lt;math&gt;i \equiv j \pmod{p}&lt;/math&gt;, which is a contradiction.&quot;<br /> <br /> Because &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are coprime, we know that multiplying the residues of &lt;math&gt;m&lt;/math&gt; by &lt;math&gt;n&lt;/math&gt; simply permutes these residues. Each of these permuted residues is purchasable (using the definition from Proof 1), because, in the form &lt;math&gt;am+bn&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; is the original residue. We now prove the following lemma.<br /> <br /> &lt;b&gt;Lemma&lt;/b&gt;: For any nonnegative integer &lt;math&gt;c &lt; m&lt;/math&gt;, &lt;math&gt;cn&lt;/math&gt; is the least purchasable number &lt;math&gt;\equiv cn \bmod m&lt;/math&gt;.<br /> <br /> &lt;i&gt;Proof&lt;/i&gt;: Any number that is less than &lt;math&gt;cn&lt;/math&gt; and congruent to it &lt;math&gt;\bmod m&lt;/math&gt; can be represented in the form &lt;math&gt;cn-dm&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; is a positive integer. If this is purchasable, we can say &lt;math&gt;cn-dm=am+bn&lt;/math&gt; for some nonnegative integers &lt;math&gt;a, b&lt;/math&gt;. This can be rearranged into &lt;math&gt;(a+d)m=(c-b)n&lt;/math&gt;, which implies that &lt;math&gt;(a+d)&lt;/math&gt; is a multiple of &lt;math&gt;n&lt;/math&gt; (since &lt;math&gt;\gcd(m, n)=1&lt;/math&gt;). We can say that &lt;math&gt;(a+d)=gn&lt;/math&gt; for some positive integer &lt;math&gt;g&lt;/math&gt;, and substitute to get &lt;math&gt;gmn=(c-b)n&lt;/math&gt;. Because &lt;math&gt;c &lt; m&lt;/math&gt;, &lt;math&gt;(c-b)n &lt; mn&lt;/math&gt;, and &lt;math&gt;gmn &lt; mn&lt;/math&gt;. We divide by &lt;math&gt;mn&lt;/math&gt; to get &lt;math&gt;g&lt;1&lt;/math&gt;. However, we defined &lt;math&gt;g&lt;/math&gt; to be a positive integer, and all positive integers are greater than or equal to &lt;math&gt;1&lt;/math&gt;. Therefore, we have a contradiction, and &lt;math&gt;cn&lt;/math&gt; is the least purchasable number congruent to &lt;math&gt;cn \bmod m&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> This means that because &lt;math&gt;cn&lt;/math&gt; is purchasable, every number that is greater than &lt;math&gt;cn&lt;/math&gt; and congruent to it &lt;math&gt;\bmod m&lt;/math&gt; is also purchasable (because these numbers are in the form &lt;math&gt;am+bn&lt;/math&gt; where &lt;math&gt;b=c&lt;/math&gt;). Another result of this Lemma is that &lt;math&gt;cn-m&lt;/math&gt; is the greatest number &lt;math&gt;\equiv cn \bmod m&lt;/math&gt; that is not purchasable. &lt;math&gt;c \leq m-1&lt;/math&gt;, so &lt;math&gt;cn-m \leq (m-1)n-m=mn-m-n&lt;/math&gt;, which shows that &lt;math&gt;mn-m-n&lt;/math&gt; is the greatest number in the form &lt;math&gt;cn-m&lt;/math&gt;. Any number greater than this and congruent to some &lt;math&gt;cn \bmod m&lt;/math&gt; is purchasable, because that number is greater than &lt;math&gt;cn&lt;/math&gt;. All numbers are congruent to some &lt;math&gt;cn&lt;/math&gt;, and thus all numbers greater than &lt;math&gt;mn-m-n&lt;/math&gt; are purchasable.<br /> <br /> Putting it all together, we can say that for any coprime &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;mn-m-n&lt;/math&gt; is the greatest number not representable in the form &lt;math&gt;am + bn&lt;/math&gt; for nonnegative integers &lt;math&gt;a, b&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> ==Corollary==<br /> This corollary is based off of Proof 2, so it is necessary to read that proof before this corollary. We prove the following lemma.<br /> <br /> &lt;b&gt;Lemma&lt;/b&gt; For any integer &lt;math&gt;k&lt;/math&gt;, exactly one of the integers &lt;math&gt;k&lt;/math&gt;, &lt;math&gt;mn-m-n-k&lt;/math&gt; is not purchasable.<br /> <br /> &lt;i&gt;Proof&lt;/i&gt;: Because every number is congruent to some residue of &lt;math&gt;m&lt;/math&gt; permuted by &lt;math&gt;n&lt;/math&gt;, we can set &lt;math&gt;k \equiv cn \bmod m&lt;/math&gt; for some &lt;math&gt;c&lt;/math&gt;. We can break this into two cases.<br /> <br /> &lt;i&gt;Case 1&lt;/i&gt;: &lt;math&gt;k \leq cn-m&lt;/math&gt;. This implies that &lt;math&gt;k&lt;/math&gt; is not purchasable, and that &lt;math&gt;mn-m-n-k \geq mn-m-n-(cn-m) = n(m-1-c)&lt;/math&gt;. &lt;math&gt;n(m-1-c)&lt;/math&gt; is a permuted residue, and a result of the lemma in Proof 2 was that a permuted residue is the least number congruent to itself &lt;math&gt;\bmod m&lt;/math&gt; that is purchasable. Therefore, &lt;math&gt;mn-m-n-k \equiv n(m-1-c) \bmod m&lt;/math&gt; and &lt;math&gt;mn-m-n-k \geq n(m-1-c)&lt;/math&gt;, so &lt;math&gt;mn-m-n-k&lt;/math&gt; is purchasable.<br /> <br /> &lt;i&gt;Case 2&lt;/i&gt;: &lt;math&gt;k &gt; cn-m&lt;/math&gt;. This implies that &lt;math&gt;k&lt;/math&gt; is purchasable, and that &lt;math&gt;mn-m-n-k &lt; mn-m-n-(cn-m) = n(m-1-c)&lt;/math&gt;. Again, because &lt;math&gt;n(m-1-c)&lt;/math&gt; is the least number congruent to itself &lt;math&gt;\bmod m&lt;/math&gt; that is purchasable, and because &lt;math&gt;mn-m-n-k \equiv n(m-1-c) \bmod m&lt;/math&gt; and &lt;math&gt;mn-m-n-k &lt; n(m-1-c)&lt;/math&gt;, &lt;math&gt;mn-m-n-k&lt;/math&gt; is not purchasable.<br /> <br /> We now limit the values of &lt;math&gt;k&lt;/math&gt; to all integers &lt;math&gt;0 \leq k \leq \frac{mn-m-n}{2}&lt;/math&gt;, which limits the values of &lt;math&gt;mn-m-n-k&lt;/math&gt; to &lt;math&gt;mn-m-n \geq mn-m-n-k \geq \frac{mn-m-n}{2}&lt;/math&gt;. Because &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are coprime, only one of them can be a multiple of &lt;math&gt;2&lt;/math&gt;. Therefore, &lt;math&gt;mn-m-n \equiv (0)(1)-0-1 \equiv -1 \equiv 1 \bmod 2&lt;/math&gt;, showing that &lt;math&gt;\frac{mn-m-n}{2}&lt;/math&gt; is not an integer and that &lt;math&gt;\frac{mn-m-n-1}{2}&lt;/math&gt; and &lt;math&gt;\frac{mn-m-n+1}{2}&lt;/math&gt; are integers. We can now set limits that are equivalent to the previous on the values of &lt;math&gt;k&lt;/math&gt; and &lt;math&gt;mn-m-n-k&lt;/math&gt; so that they cover all integers form &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;mn-m-n&lt;/math&gt; without overlap: &lt;math&gt;0 \leq k \leq \frac{mn-m-n-1}{2}&lt;/math&gt; and &lt;math&gt;\frac{mn-m-n+1}{2} \leq mn-m-n-k \leq mn-m-n&lt;/math&gt;. There are &lt;math&gt;\frac{mn-m-n-1}{2}+1=\frac{(m-1)(n-1)}{2}&lt;/math&gt; values of &lt;math&gt;k&lt;/math&gt;, and each is paired with a value of &lt;math&gt;mn-m-n-k&lt;/math&gt;, so we can make &lt;math&gt;\frac{(m-1)(n-1)}{2}&lt;/math&gt; different ordered pairs of the form &lt;math&gt;(k, mn-m-n-k)&lt;/math&gt;. The coordinates of these ordered pairs cover all integers from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;mn-m-n&lt;/math&gt; inclusive, and each contains exactly one not-purchasable integer, so that means that there are &lt;math&gt;\frac{(m-1)(n-1)}{2}&lt;/math&gt; different not-purchasable integers from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;mn-m-n&lt;/math&gt;. All integers greater than &lt;math&gt;mn-m-n&lt;/math&gt; are purchasable, so that means there are a total of &lt;math&gt;\frac{(m-1)(n-1)}{2}&lt;/math&gt; integers &lt;math&gt;\geq 0&lt;/math&gt; that are not purchasable.<br /> <br /> In other words, for every pair of coprime integers &lt;math&gt;m, n&lt;/math&gt;, there are exactly &lt;math&gt;\frac{(m-1)(n-1)}{2}&lt;/math&gt; nonnegative integers that cannot be represented in the form &lt;math&gt;am + bn&lt;/math&gt; for nonnegative integers &lt;math&gt;a, b&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> ==Generalization==<br /> If &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are not relatively prime, then we can simply rearrange &lt;math&gt;am+bn&lt;/math&gt; into the form<br /> &lt;cmath&gt;\gcd(m,n) \left( a\frac{m}{\gcd(m,n)}+b\frac{n}{\gcd(m,n)} \right)&lt;/cmath&gt;<br /> &lt;math&gt;\frac{m}{\gcd(m,n)}&lt;/math&gt; and &lt;math&gt;\frac{n}{\gcd(m,n)}&lt;/math&gt; are relatively prime, so we apply Chicken McNugget to find a bound<br /> &lt;cmath&gt;\frac{mn}{\gcd(m,n)^{2}}-\frac{m}{\gcd(m,n)}-\frac{n}{\gcd(m,n)}&lt;/cmath&gt;<br /> We can simply multiply &lt;math&gt;\gcd(m,n)&lt;/math&gt; back into the bound to get<br /> &lt;cmath&gt;\frac{mn}{\gcd(m,n)}-m-n=\textrm{lcm}(m, n)-m-n&lt;/cmath&gt;<br /> Therefore, all multiples of &lt;math&gt;\gcd(m, n)&lt;/math&gt; greater than &lt;math&gt;\textrm{lcm}(m, n)-m-n&lt;/math&gt; are representable in the form &lt;math&gt;am+bn&lt;/math&gt; for some positive integers &lt;math&gt;a, b&lt;/math&gt;.<br /> <br /> =Problems=<br /> <br /> ===Simple===<br /> *Marcy buys paint jars in containers of &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;. What's the largest number of paint jars that Marcy can't obtain? <br /> <br /> Answer: &lt;math&gt;5&lt;/math&gt; containers<br /> <br /> *Bay Area Rapid food sells chicken nuggets. You can buy packages of &lt;math&gt;11&lt;/math&gt; or &lt;math&gt;7&lt;/math&gt;. What is the largest integer &lt;math&gt;n&lt;/math&gt; such that there is no way to buy exactly &lt;math&gt;n&lt;/math&gt; nuggets? Can you Generalize ?(ACOPS) <br /> <br /> Answer: &lt;math&gt;n=59&lt;/math&gt; <br /> <br /> *If a game of American Football has only scores of field goals (3 points) and touchdowns with the extra point (7 points), then what is the greatest score that cannot be the score of a team in this football game (ignoring time constraints)?<br /> <br /> Answer: &lt;math&gt;11&lt;/math&gt; points<br /> <br /> ===Intermediate===<br /> *Ninety-four bricks, each measuring &lt;math&gt;4''\times10''\times19'',&lt;/math&gt; are to stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contributes &lt;math&gt;4''\,&lt;/math&gt; or &lt;math&gt;10''\,&lt;/math&gt; or &lt;math&gt;19''\,&lt;/math&gt; to the total height of the tower. How many different tower heights can be achieved using all ninety-four of the bricks? [[1994 AIME Problems/Problem 11|AIME]]<br /> <br /> ===Olympiad===<br /> *On the real number line, paint red all points that correspond to integers of the form &lt;math&gt;81x+100y&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are positive integers. Paint the remaining integer points blue. Find a point &lt;math&gt;P&lt;/math&gt; on the line such that, for every integer point &lt;math&gt;T&lt;/math&gt;, the reflection of &lt;math&gt;T&lt;/math&gt; with respect to &lt;math&gt;P&lt;/math&gt; is an integer point of a different colour than &lt;math&gt;T&lt;/math&gt;. (India TST)<br /> <br /> *Let &lt;math&gt;S&lt;/math&gt; be a set of integers (not necessarily positive) such that<br /> <br /> (a) there exist &lt;math&gt;a,b \in S&lt;/math&gt; with &lt;math&gt;\gcd(a,b)=\gcd(a-2,b-2)=1&lt;/math&gt;;<br /> <br /> (b) if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are elements of &lt;math&gt;S&lt;/math&gt; (possibly equal), then &lt;math&gt;x^2-y&lt;/math&gt; also belongs to &lt;math&gt;S&lt;/math&gt;. <br /> <br /> Prove that &lt;math&gt;S&lt;/math&gt; is the set of all integers. (USAMO)<br /> <br /> ==See Also==<br /> *[[Theorem]]<br /> *[[Prime]]<br /> <br /> [[Category:Theorems]]<br /> [[Category:Number theory]]</div> Crazyeyemoody907 https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_10&diff=100552 2018 AIME I Problems/Problem 10 2019-01-16T03:50:44Z <p>Crazyeyemoody907: /* Solution 3 (Similar to Solution 2 but with slightly modified wording so it is easier to understand) */</p> <hr /> <div><br /> ==Problem==<br /> The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point &lt;math&gt;A&lt;/math&gt;. At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug only walks in a clockwise direction. For example, the bug could travel along the path &lt;math&gt;AJABCHCHIJA&lt;/math&gt;, which has &lt;math&gt;10&lt;/math&gt; steps. Let &lt;math&gt;n&lt;/math&gt; be the number of paths with &lt;math&gt;15&lt;/math&gt; steps that begin and end at point &lt;math&gt;A&lt;/math&gt;. Find the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;<br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(32);<br /> draw(unitcircle);<br /> draw(scale(2) * unitcircle);<br /> for(int d = 90; d &lt; 360 + 90; d += 72){<br /> draw(2 * dir(d) -- dir(d));<br /> }<br /> <br /> real s = 4;<br /> dot(1 * dir( 90), linewidth(s));<br /> dot(1 * dir(162), linewidth(s));<br /> dot(1 * dir(234), linewidth(s));<br /> dot(1 * dir(306), linewidth(s));<br /> dot(1 * dir(378), linewidth(s));<br /> dot(2 * dir(378), linewidth(s));<br /> dot(2 * dir(306), linewidth(s));<br /> dot(2 * dir(234), linewidth(s));<br /> dot(2 * dir(162), linewidth(s));<br /> dot(2 * dir( 90), linewidth(s));<br /> <br /> defaultpen(fontsize(10pt));<br /> real r = 0.05;<br /> label(&quot;$A$&quot;, (1-r) * dir( 90), -dir( 90));<br /> label(&quot;$B$&quot;, (1-r) * dir(162), -dir(162));<br /> label(&quot;$C$&quot;, (1-r) * dir(234), -dir(234));<br /> label(&quot;$D$&quot;, (1-r) * dir(306), -dir(306));<br /> label(&quot;$E$&quot;, (1-r) * dir(378), -dir(378));<br /> label(&quot;$F$&quot;, (2+r) * dir(378), dir(378));<br /> label(&quot;$G$&quot;, (2+r) * dir(306), dir(306));<br /> label(&quot;$H$&quot;, (2+r) * dir(234), dir(234));<br /> label(&quot;$I$&quot;, (2+r) * dir(162), dir(162));<br /> label(&quot;$J$&quot;, (2+r) * dir( 90), dir( 90));<br /> &lt;/asy&gt;&lt;/center&gt;<br /> ==Solution==<br /> We divide this up into casework. The &quot;directions&quot; the bug can go are &lt;math&gt;\text{Clockwise}&lt;/math&gt;, &lt;math&gt;\text{Counter-Clockwise}&lt;/math&gt;, and &lt;math&gt;\text{Switching}&lt;/math&gt;. Let an &lt;math&gt;I&lt;/math&gt; signal going clockwise (because it has to be in the ''inner'' circle), an &lt;math&gt;O&lt;/math&gt; signal going counter-clockwise, and an &lt;math&gt;S&lt;/math&gt; switching between inner and outer circles. An example string of length fifteen that gets the bug back to &lt;math&gt;A&lt;/math&gt; would be &lt;math&gt;ISSIIISOOSISSII&lt;/math&gt;.<br /> For the bug to end up back at &lt;math&gt;A&lt;/math&gt;, the difference between the number of &lt;math&gt;I&lt;/math&gt;'s and &lt;math&gt;O&lt;/math&gt;'s must be a multiple of &lt;math&gt;5&lt;/math&gt;.<br /> ;Case 1 -- There are 15 more &lt;math&gt;I&lt;/math&gt;'s than &lt;math&gt;O&lt;/math&gt;'s.<br /> :There is clearly &lt;math&gt;1&lt;/math&gt; way for this to happen.<br /> <br /> ;Case 2 -- There are &lt;math&gt;5&lt;/math&gt; more &lt;math&gt;I&lt;/math&gt;'s than &lt;math&gt;O&lt;/math&gt;'s.<br /> :We split this case up into several sub-cases based on the number of &lt;math&gt;S&lt;/math&gt;'s.<br /> :;Sub-case 1 -- There are &lt;math&gt;10&lt;/math&gt; &lt;math&gt;S&lt;/math&gt;'s and &lt;math&gt;5&lt;/math&gt; &lt;math&gt;I&lt;/math&gt;'s.<br /> ::Notice that the number of ways to order the &lt;math&gt;I&lt;/math&gt;'s and &lt;math&gt;O&lt;/math&gt;'s are independent assortments because the &lt;math&gt;I&lt;/math&gt;'s must be in the &quot;even&quot; spaces between &lt;math&gt;S&lt;/math&gt;'s (i.e. before the 1st &lt;math&gt;S&lt;/math&gt;, between the 2nd and 3rd &lt;math&gt;S&lt;/math&gt;'s, etc.), while the &lt;math&gt;O&lt;/math&gt;'s must be in the &quot;odd&quot; spaces.<br /> <br /> ::There are &lt;math&gt;6&lt;/math&gt; places to put the &lt;math&gt;I&lt;/math&gt;'s (after the 0th, 2nd, 4th, 6th, 8th, and 10th &lt;math&gt;S&lt;/math&gt;'s), and &lt;math&gt;4&lt;/math&gt; places to put the (0) &lt;math&gt;O&lt;/math&gt;'s. We use stars and bars to get an answer of &lt;math&gt;\binom{10}{5}\binom{4}{0}&lt;/math&gt;<br /> :;Sub-case 2 -- There are &lt;math&gt;8&lt;/math&gt; &lt;math&gt;S&lt;/math&gt;'s, &lt;math&gt;6&lt;/math&gt; &lt;math&gt;I&lt;/math&gt;'s, and &lt;math&gt;1&lt;/math&gt; &lt;math&gt;O&lt;/math&gt;.<br /> ::Similarly and by using stars and bars, we get an amount of &lt;math&gt;\binom{10}{4}\binom{4}{1}&lt;/math&gt;<br /> :All the other sub-cases are similar, with a total of &lt;math&gt;\binom{10}{5}\binom{4}{0}+\binom{10}{4}\binom{4}{1}+\cdots+\binom{10}{1}\binom{4}{4}=\binom{14}{5}=2002&lt;/math&gt; by [https://artofproblemsolving.com/wiki/index.php?title=Combinatorial_identity#Vandermonde.27s_Identity Vandermonde's Identity].<br /> <br /> ;Case 3 -- There are &lt;math&gt;5&lt;/math&gt; more &lt;math&gt;O&lt;/math&gt;'s than &lt;math&gt;I&lt;/math&gt;'s.<br /> :This case is similar to the other case.<br /> :Here is an example of a sub-case for this case.<br /> :;Sub-case: There are &lt;math&gt;10&lt;/math&gt; &lt;math&gt;S&lt;/math&gt;'s and &lt;math&gt;5&lt;/math&gt; &lt;math&gt;O&lt;/math&gt;'s.<br /> ::There are &lt;math&gt;\binom{9}{4}\binom{5}{0}&lt;/math&gt; ways to do this.<br /> :We can see now that the pattern is going to be &lt;math&gt;\binom{9}{4}\binom{5}{0}+\binom{9}{3}\binom{5}{1}+\cdots+\binom{9}{0}\binom{5}{4}=\binom{14}{4}=1001&lt;/math&gt;.<br /> <br /> <br /> So, the total number of ways is &lt;math&gt;1+2002+1001=3\boxed{004}&lt;/math&gt;<br /> <br /> == Solution 2 (Official MAA)==<br /> Note that the set of sequences of moves the bug makes is in bijective correspondence with the set of strings of &lt;math&gt;X&lt;/math&gt;s and &lt;math&gt;Y&lt;/math&gt;s of length &lt;math&gt;15&lt;/math&gt;, where &lt;math&gt;X&lt;/math&gt; denotes a move which is either counterclockwise or inward along a spoke and &lt;math&gt;Y&lt;/math&gt; denotes a move which is either clockwise or outward along a spoke. (The proof of this basically boils down to the fact that which one depends on whether the bug is on the inner wheel or the outer wheel.) Now the condition that the bug ends at A implies that the difference between the number of &lt;math&gt;X&lt;/math&gt;s and the number of &lt;math&gt;Y&lt;/math&gt;s is a multiple of &lt;math&gt;5&lt;/math&gt;, and so we must have either &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;9&lt;/math&gt;, or &lt;math&gt;14&lt;/math&gt; &lt;math&gt;X&lt;/math&gt;s within the first fourteen moves with the last move being an &lt;math&gt;X&lt;/math&gt;. This implies the answer is &lt;cmath&gt;\binom{14}4+\binom{14}9+\binom{14}{14} = 3004\equiv \boxed{004}\pmod{1000}.&lt;/cmath&gt;<br /> <br /> ==Solution 3 (Similar to Solution 2 but with slightly modified wording so it is easier to understand)==<br /> Let an O signal a move that ends in the outer circle and I signal a move that ends in the Inner circle. Now notice that for a string of 15 moves to end at A, the difference between O's and I's in the string must be a multiple of 5.<br /> <br /> 15 I: Trivially 1 case.<br /> <br /> 5 O+10 I: Since the string has to end in an I for the bug to land on A, there are a total of &lt;math&gt;\binom{14}{5}=2002&lt;/math&gt; ways to put 5 O's in the first 14 moves.<br /> <br /> 10 O+5 I: Similarly there are &lt;math&gt;\binom{14}{4}=1001&lt;/math&gt; ways to put &lt;math&gt;5-1=4&lt;/math&gt; I's in the first 14 moves.<br /> <br /> 15 O: Impossible since the string has to end with an I.<br /> <br /> This brings us an answer of &lt;math&gt;1+2002+1001=3\boxed{004}&lt;/math&gt;<br /> <br /> -solution by mathleticguyyy-<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Crazyeyemoody907