https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Sigmapie&feedformat=atom AoPS Wiki - User contributions [en] 2023-02-06T16:48:09Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_15&diff=187046 2011 AIME I Problems/Problem 15 2023-01-19T00:23:42Z <p>Sigmapie: </p> <hr /> <div>== Problem ==<br /> For some integer &lt;math&gt;m&lt;/math&gt;, the polynomial &lt;math&gt;x^3 - 2011x + m&lt;/math&gt; has the three integer roots &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;. Find &lt;math&gt;|a| + |b| + |c|&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> From Vieta's formulas, we know that &lt;math&gt;a+b+c = 0&lt;/math&gt;, and &lt;math&gt;ab+bc+ac = -2011&lt;/math&gt;. Thus &lt;math&gt;a = -(b+c)&lt;/math&gt;. All three of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are non-zero: say, if &lt;math&gt;a=0&lt;/math&gt;, then &lt;math&gt;b=-c=\pm\sqrt{2011}&lt;/math&gt; (which is not an integer). &lt;math&gt;\textsc{wlog}&lt;/math&gt;, let &lt;math&gt;|a| \ge |b| \ge |c|&lt;/math&gt;. If &lt;math&gt;a &gt; 0&lt;/math&gt;, then &lt;math&gt;b,c &lt; 0&lt;/math&gt; and if &lt;math&gt;a &lt; 0&lt;/math&gt;, then &lt;math&gt;b,c &gt; 0&lt;/math&gt;. We have &lt;cmath&gt;-2011=ab+bc+ac = a(b+c)+bc = -a^2+bc&lt;/cmath&gt;<br /> Thus &lt;math&gt;a^2 = 2011 + bc&lt;/math&gt;. We know that &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; have the same sign. So &lt;math&gt;|a| \ge 45 = \lceil \sqrt{2011} \rceil&lt;/math&gt;. <br /> <br /> Also, if we fix &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b+c&lt;/math&gt; is fixed, so &lt;math&gt;bc&lt;/math&gt; is maximized when &lt;math&gt;b = c&lt;/math&gt; . Hence, &lt;cmath&gt;2011 = a^2 - bc &gt; \tfrac{3}{4}a^2 \qquad \Longrightarrow \qquad a ^2 &lt; \tfrac{4}{3}\cdot 2011 = 2681+\tfrac{1}{3}&lt;/cmath&gt;<br /> So &lt;math&gt;|a| \le 51&lt;/math&gt;. Thus we have bounded &lt;math&gt;a&lt;/math&gt; as &lt;math&gt;45\le |a| \le 51&lt;/math&gt;, i.e. &lt;math&gt;45\le |b+c| \le 51&lt;/math&gt; since &lt;math&gt;a=-(b+c)&lt;/math&gt;. Let's analyze &lt;math&gt;bc=(b+c)^2-2011&lt;/math&gt;. Here is a table: <br /> <br /> &lt;table border = 1 cellspacing = 0 cellpadding = 5 style = &quot;text-align:center;&quot;&gt;<br /> <br /> &lt;tr&gt;&lt;th&gt;&lt;math&gt;|a|&lt;/math&gt;&lt;/th&gt;&lt;th&gt;&lt;math&gt;bc=2011-a^2&lt;/math&gt;&lt;/th&gt;&lt;/tr&gt;<br /> <br /> &lt;tr&gt;&lt;td&gt;&lt;math&gt;45&lt;/math&gt;&lt;/td&gt;&lt;td&gt;&lt;math&gt;14&lt;/math&gt;&lt;/td&gt;&lt;/tr&gt;<br /> &lt;tr&gt;&lt;td&gt;&lt;math&gt;46&lt;/math&gt;&lt;/td&gt;&lt;td&gt;&lt;math&gt;105&lt;/math&gt;&lt;/td&gt;&lt;/tr&gt;<br /> &lt;tr&gt;&lt;td&gt;&lt;math&gt;47&lt;/math&gt;&lt;/td&gt;&lt;td&gt;&lt;math&gt;198&lt;/math&gt;&lt;/td&gt;&lt;/tr&gt;<br /> &lt;tr&gt;&lt;td&gt;&lt;math&gt;48&lt;/math&gt;&lt;/td&gt;&lt;td&gt;&lt;math&gt;293&lt;/math&gt;&lt;/td&gt;&lt;/tr&gt;<br /> &lt;tr&gt;&lt;td&gt;&lt;math&gt;49&lt;/math&gt;&lt;/td&gt;&lt;td&gt;&lt;math&gt;390&lt;/math&gt;&lt;/td&gt;&lt;/tr&gt;<br /> &lt;/table&gt;<br /> &lt;br /&gt;<br /> <br /> We can tell we don't need to bother with &lt;math&gt;45&lt;/math&gt;, <br /> <br /> &lt;math&gt;105 = (3)(5)(7)&lt;/math&gt;, So &lt;math&gt;46&lt;/math&gt; won't work. &lt;math&gt;198/47 &gt; 4&lt;/math&gt;, <br /> <br /> &lt;math&gt;198&lt;/math&gt; is not divisible by &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;198/6 = 33&lt;/math&gt;, which is too small to get &lt;math&gt;47&lt;/math&gt;.<br /> <br /> &lt;math&gt;293/48 &gt; 6&lt;/math&gt;, &lt;math&gt;293&lt;/math&gt; is not divisible by &lt;math&gt;7&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;, we can clearly tell that &lt;math&gt;10&lt;/math&gt; is too much.<br /> <br /> <br /> Hence, &lt;math&gt;|a| = 49&lt;/math&gt;, &lt;math&gt;a^2 -2011 = 390&lt;/math&gt;. &lt;math&gt;b = 39&lt;/math&gt;, &lt;math&gt;c = 10&lt;/math&gt;.<br /> <br /> Answer: &lt;math&gt;\boxed{098}&lt;/math&gt;<br /> <br /> == Solution 2==<br /> <br /> <br /> Starting off like the previous solution, we know that &lt;math&gt;a + b + c = 0&lt;/math&gt;, and &lt;math&gt;ab + bc + ac = -2011&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;c = -b-a&lt;/math&gt;.<br /> <br /> Substituting, &lt;math&gt;ab + b(-b-a) + a(-b-a) = ab-b^2-ab-ab-a^2 = -2011&lt;/math&gt;.<br /> <br /> Factoring the perfect square, we get: &lt;math&gt;ab-(b+a)^2=-2011&lt;/math&gt; or &lt;math&gt;(b+a)^2-ab=2011&lt;/math&gt;.<br /> <br /> Therefore, a sum (&lt;math&gt;a+b&lt;/math&gt;) squared minus a product (&lt;math&gt;ab&lt;/math&gt;) gives &lt;math&gt;2011&lt;/math&gt;..<br /> <br /> &lt;br/&gt;<br /> <br /> We can guess and check different &lt;math&gt;a+b&lt;/math&gt;’s starting with &lt;math&gt;45&lt;/math&gt; since &lt;math&gt;44^2 &lt; 2011&lt;/math&gt;.<br /> <br /> &lt;math&gt;45^2 = 2025&lt;/math&gt; therefore &lt;math&gt;ab = 2025-2011 = 14&lt;/math&gt;. <br /> <br /> Since no factors of &lt;math&gt;14&lt;/math&gt; can sum to &lt;math&gt;45&lt;/math&gt; (&lt;math&gt;1+14&lt;/math&gt; being the largest sum), a + b cannot equal &lt;math&gt;45&lt;/math&gt;.<br /> <br /> &lt;math&gt;46^2 = 2116&lt;/math&gt; making &lt;math&gt;ab = 105 = 3 * 5 * 7&lt;/math&gt;.<br /> <br /> &lt;math&gt;5 * 7 + 3 &lt; 46&lt;/math&gt; and &lt;math&gt;3 * 5 * 7 &gt; 46&lt;/math&gt; so &lt;math&gt;46&lt;/math&gt; cannot work either.<br /> <br /> &lt;br/&gt;<br /> <br /> We can continue to do this until we reach &lt;math&gt;49&lt;/math&gt;.<br /> <br /> &lt;math&gt;49^2 = 2401&lt;/math&gt; making &lt;math&gt;ab = 390 = 2 * 3 * 5* 13&lt;/math&gt;.<br /> <br /> &lt;math&gt;3 * 13 + 2* 5 = 49&lt;/math&gt;, so one root is &lt;math&gt;10&lt;/math&gt; and another is &lt;math&gt;39&lt;/math&gt;. The roots sum to zero, so the last root must be &lt;math&gt;-49&lt;/math&gt;.<br /> <br /> &lt;br/&gt;<br /> <br /> &lt;math&gt;|-49|+10+39 = \boxed{098}&lt;/math&gt;.<br /> <br /> <br /> == Solution 3 ==<br /> Let us first note the obvious that is derived from Vieta's formulas: &lt;math&gt;a+b+c=0, ab+bc+ac=-2011&lt;/math&gt;. Now, due to the first equation, let us say that &lt;math&gt;a+b=-c&lt;/math&gt;, meaning that &lt;math&gt;a,b&gt;0&lt;/math&gt; and &lt;math&gt;c&lt;0&lt;/math&gt;. Now, since both &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are greater than 0, their absolute values are both equal to &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, respectively. Since &lt;math&gt;c&lt;/math&gt; is less than 0, it equals &lt;math&gt;-a-b&lt;/math&gt;. Therefore, &lt;math&gt;|c|=|-a-b|=a+b&lt;/math&gt;, meaning &lt;math&gt;|a|+|b|+|c|=2(a+b)&lt;/math&gt;. We now apply Newton's sums to get that &lt;math&gt;a^2+b^2+ab=2011&lt;/math&gt;,or &lt;math&gt;(a+b)^2-ab=2011&lt;/math&gt;. Solving, we find that &lt;math&gt;49^2-390&lt;/math&gt; satisfies this, meaning &lt;math&gt;a+b=49&lt;/math&gt;, so &lt;math&gt;2(a+b)=\boxed{098}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We have<br /> <br /> &lt;math&gt;(x-a)\cdot (x-b)\cdot (x-c)=x^3-(a+b+c)x+(ab+ac+bc)x-abc&lt;/math&gt; <br /> <br /> As a result, we have<br /> <br /> &lt;math&gt;a+b+c=0&lt;/math&gt;<br /> <br /> &lt;math&gt;ab+bc+ac=-2011&lt;/math&gt;<br /> <br /> &lt;math&gt;abc=-m&lt;/math&gt;<br /> <br /> So, &lt;math&gt;a=-b-c&lt;/math&gt;<br /> <br /> As a result, &lt;math&gt;ab+bc+ac=(-b-c)b+(-b-c)c+bc=-b^2-c^2-bc=-2011&lt;/math&gt;<br /> <br /> Solve &lt;math&gt;b=\frac {-c+\sqrt{c^2-4(c^2-2011)}}{2}&lt;/math&gt; and <br /> &lt;math&gt;\Delta =8044-3c^2=k^2&lt;/math&gt;, where &lt;math&gt;k&lt;/math&gt; is an integer<br /> <br /> Cause &lt;math&gt;89&lt;\sqrt{8044}&lt;90&lt;/math&gt;<br /> <br /> So, after we tried for &lt;math&gt;2&lt;/math&gt; times, we get &lt;math&gt;k=88&lt;/math&gt; and &lt;math&gt;c=10&lt;/math&gt;<br /> <br /> then &lt;math&gt;b=39&lt;/math&gt;, &lt;math&gt;a=-b-c=-49&lt;/math&gt;<br /> <br /> As a result, &lt;math&gt;|a|+|b|+|c|=10+39+49=\boxed{098}&lt;/math&gt;<br /> <br /> ==Solution 5 (mod to help bash)==<br /> First, derive the equations &lt;math&gt;a=-b-c&lt;/math&gt; and &lt;math&gt;ab+bc+ca=-2011\implies b^2+bc+c^2=2011&lt;/math&gt;. Since the product is negative, &lt;math&gt;a&lt;/math&gt; is negative, and &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; positive. Now, a simple mod 3 testing of all cases shows that &lt;math&gt;b\equiv \{1,2\} \pmod{3}&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; has the repective value. We can choose &lt;math&gt;b&lt;/math&gt; not congruent to 0, make sure you see why. Now, we bash on values of &lt;math&gt;b&lt;/math&gt;, testing the quadratic function to see if &lt;math&gt;c&lt;/math&gt; is positive. You can also use a delta argument like solution 4, but this is simpler. We get that for &lt;math&gt;b=10&lt;/math&gt;, &lt;math&gt;c=39, -49&lt;/math&gt;. Choosing &lt;math&gt;c&lt;/math&gt; positive we get &lt;math&gt;a=-49&lt;/math&gt;, so &lt;math&gt;|a|+|b|+|c|=10+29+39=\boxed{098}&lt;/math&gt;<br /> ~firebolt360<br /> <br /> ==Solution 6==<br /> Note that &lt;math&gt;-c=b+a&lt;/math&gt;, so &lt;math&gt;c^2=a^2+2ab+b^2&lt;/math&gt;, or &lt;math&gt;-c^2+ab=-a^2-ab-b^2&lt;/math&gt;. Also, &lt;math&gt;ab+bc+ca=-2011&lt;/math&gt;, so &lt;math&gt;(a+b)c+ab=-c^2+ab=-2011&lt;/math&gt;. Substituting &lt;math&gt;-c^2+ab=-a^2-ab-b^2&lt;/math&gt;, we can obtain &lt;math&gt;a^2+ab+b^2=2011&lt;/math&gt;, or &lt;math&gt;\frac{a^3-b^3}{a-b}=2011&lt;/math&gt;. If it is not known that &lt;math&gt;2011&lt;/math&gt; is prime, it may be proved in &lt;math&gt;5&lt;/math&gt; minutes or so by checking all primes up to &lt;math&gt;43&lt;/math&gt;. If &lt;math&gt;2011&lt;/math&gt; divided either of &lt;math&gt;a, b&lt;/math&gt;, then in order for &lt;math&gt;a^3-b^3&lt;/math&gt; to contain an extra copy of &lt;math&gt;2011&lt;/math&gt;, both &lt;math&gt;a, b&lt;/math&gt; would need to be divisible by &lt;math&gt;2021&lt;/math&gt;. But then &lt;math&gt;c&lt;/math&gt; would also be divisible by &lt;math&gt;2011&lt;/math&gt;, and the sum &lt;math&gt;ab+bc+ca&lt;/math&gt; would clearly be divisible by &lt;math&gt;2011^2&lt;/math&gt;. <br /> <br /> By LTE, &lt;math&gt;v_{2011}(a^3-b^3)=v_{2011}(a-b)&lt;/math&gt; if &lt;math&gt;a-b&lt;/math&gt; is divisible by &lt;math&gt;2011&lt;/math&gt; and neither &lt;math&gt;a,b&lt;/math&gt; are divisible by &lt;math&gt;2011&lt;/math&gt;. Thus, the only possibility remaining is if &lt;math&gt;a-b&lt;/math&gt; did not divide &lt;math&gt;2011&lt;/math&gt;. Let &lt;math&gt;a=k+b&lt;/math&gt;. Then, we have &lt;math&gt;(b+k)^3-b^3=2011k&lt;/math&gt;. Rearranging gives &lt;math&gt;3b(b+k)=2011-k^2&lt;/math&gt;. As in the above solutions, we may eliminate certain values of &lt;math&gt;k&lt;/math&gt; by using mods. Then, we may test values until we obtain &lt;math&gt;k=29&lt;/math&gt;, and &lt;math&gt;a=10&lt;/math&gt;. Thus, &lt;math&gt;b=39&lt;/math&gt;, &lt;math&gt;c=-49&lt;/math&gt;, and our answer is &lt;math&gt;49+39+10=098&lt;/math&gt;.<br /> ==Video Solution==<br /> <br /> https://www.youtube.com/watch?v=QNbfAu5rdJI&amp;t=26s ~ MathEx<br /> <br /> == See also ==<br /> <br /> {{AIME box|year=2011|num-b=14|after=Last Problem|n=I}}<br /> <br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_15&diff=187038 2011 AIME I Problems/Problem 15 2023-01-18T19:17:30Z <p>Sigmapie: </p> <hr /> <div>== Problem ==<br /> For some integer &lt;math&gt;m&lt;/math&gt;, the polynomial &lt;math&gt;x^3 - 2011x + m&lt;/math&gt; has the three integer roots &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;. Find &lt;math&gt;|a| + |b| + |c|&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> From Vieta's formulas, we know that &lt;math&gt;a+b+c = 0&lt;/math&gt;, and &lt;math&gt;ab+bc+ac = -2011&lt;/math&gt;. Thus &lt;math&gt;a = -(b+c)&lt;/math&gt;. All three of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are non-zero: say, if &lt;math&gt;a=0&lt;/math&gt;, then &lt;math&gt;b=-c=\pm\sqrt{2011}&lt;/math&gt; (which is not an integer). &lt;math&gt;\textsc{wlog}&lt;/math&gt;, let &lt;math&gt;|a| \ge |b| \ge |c|&lt;/math&gt;. If &lt;math&gt;a &gt; 0&lt;/math&gt;, then &lt;math&gt;b,c &lt; 0&lt;/math&gt; and if &lt;math&gt;a &lt; 0&lt;/math&gt;, then &lt;math&gt;b,c &gt; 0&lt;/math&gt;. We have &lt;cmath&gt;-2011=ab+bc+ac = a(b+c)+bc = -a^2+bc&lt;/cmath&gt;<br /> Thus &lt;math&gt;a^2 = 2011 + bc&lt;/math&gt;. We know that &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; have the same sign. So &lt;math&gt;|a| \ge 45 = \lceil \sqrt{2011} \rceil&lt;/math&gt;. <br /> <br /> Also, if we fix &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b+c&lt;/math&gt; is fixed, so &lt;math&gt;bc&lt;/math&gt; is maximized when &lt;math&gt;b = c&lt;/math&gt; . Hence, &lt;cmath&gt;2011 = a^2 - bc &gt; \tfrac{3}{4}a^2 \qquad \Longrightarrow \qquad a ^2 &lt; \tfrac{4}{3}\cdot 2011 = 2681+\tfrac{1}{3}&lt;/cmath&gt;<br /> So &lt;math&gt;|a| \le 51&lt;/math&gt;. Thus we have bounded &lt;math&gt;a&lt;/math&gt; as &lt;math&gt;45\le |a| \le 51&lt;/math&gt;, i.e. &lt;math&gt;45\le |b+c| \le 51&lt;/math&gt; since &lt;math&gt;a=-(b+c)&lt;/math&gt;. Let's analyze &lt;math&gt;bc=(b+c)^2-2011&lt;/math&gt;. Here is a table: <br /> <br /> &lt;table border = 1 cellspacing = 0 cellpadding = 5 style = &quot;text-align:center;&quot;&gt;<br /> <br /> &lt;tr&gt;&lt;th&gt;&lt;math&gt;|a|&lt;/math&gt;&lt;/th&gt;&lt;th&gt;&lt;math&gt;bc=2011-a^2&lt;/math&gt;&lt;/th&gt;&lt;/tr&gt;<br /> <br /> &lt;tr&gt;&lt;td&gt;&lt;math&gt;45&lt;/math&gt;&lt;/td&gt;&lt;td&gt;&lt;math&gt;14&lt;/math&gt;&lt;/td&gt;&lt;/tr&gt;<br /> &lt;tr&gt;&lt;td&gt;&lt;math&gt;46&lt;/math&gt;&lt;/td&gt;&lt;td&gt;&lt;math&gt;105&lt;/math&gt;&lt;/td&gt;&lt;/tr&gt;<br /> &lt;tr&gt;&lt;td&gt;&lt;math&gt;47&lt;/math&gt;&lt;/td&gt;&lt;td&gt;&lt;math&gt;198&lt;/math&gt;&lt;/td&gt;&lt;/tr&gt;<br /> &lt;tr&gt;&lt;td&gt;&lt;math&gt;48&lt;/math&gt;&lt;/td&gt;&lt;td&gt;&lt;math&gt;293&lt;/math&gt;&lt;/td&gt;&lt;/tr&gt;<br /> &lt;tr&gt;&lt;td&gt;&lt;math&gt;49&lt;/math&gt;&lt;/td&gt;&lt;td&gt;&lt;math&gt;390&lt;/math&gt;&lt;/td&gt;&lt;/tr&gt;<br /> &lt;/table&gt;<br /> &lt;br /&gt;<br /> <br /> We can tell we don't need to bother with &lt;math&gt;45&lt;/math&gt;, <br /> <br /> &lt;math&gt;105 = (3)(5)(7)&lt;/math&gt;, So &lt;math&gt;46&lt;/math&gt; won't work. &lt;math&gt;198/47 &gt; 4&lt;/math&gt;, <br /> <br /> &lt;math&gt;198&lt;/math&gt; is not divisible by &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;198/6 = 33&lt;/math&gt;, which is too small to get &lt;math&gt;47&lt;/math&gt;.<br /> <br /> &lt;math&gt;293/48 &gt; 6&lt;/math&gt;, &lt;math&gt;293&lt;/math&gt; is not divisible by &lt;math&gt;7&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;, we can clearly tell that &lt;math&gt;10&lt;/math&gt; is too much.<br /> <br /> <br /> Hence, &lt;math&gt;|a| = 49&lt;/math&gt;, &lt;math&gt;a^2 -2011 = 390&lt;/math&gt;. &lt;math&gt;b = 39&lt;/math&gt;, &lt;math&gt;c = 10&lt;/math&gt;.<br /> <br /> Answer: &lt;math&gt;\boxed{098}&lt;/math&gt;<br /> <br /> == Solution 2==<br /> <br /> <br /> Starting off like the previous solution, we know that &lt;math&gt;a + b + c = 0&lt;/math&gt;, and &lt;math&gt;ab + bc + ac = -2011&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;c = -b-a&lt;/math&gt;.<br /> <br /> Substituting, &lt;math&gt;ab + b(-b-a) + a(-b-a) = ab-b^2-ab-ab-a^2 = -2011&lt;/math&gt;.<br /> <br /> Factoring the perfect square, we get: &lt;math&gt;ab-(b+a)^2=-2011&lt;/math&gt; or &lt;math&gt;(b+a)^2-ab=2011&lt;/math&gt;.<br /> <br /> Therefore, a sum (&lt;math&gt;a+b&lt;/math&gt;) squared minus a product (&lt;math&gt;ab&lt;/math&gt;) gives &lt;math&gt;2011&lt;/math&gt;..<br /> <br /> &lt;br/&gt;<br /> <br /> We can guess and check different &lt;math&gt;a+b&lt;/math&gt;’s starting with &lt;math&gt;45&lt;/math&gt; since &lt;math&gt;44^2 &lt; 2011&lt;/math&gt;.<br /> <br /> &lt;math&gt;45^2 = 2025&lt;/math&gt; therefore &lt;math&gt;ab = 2025-2011 = 14&lt;/math&gt;. <br /> <br /> Since no factors of &lt;math&gt;14&lt;/math&gt; can sum to &lt;math&gt;45&lt;/math&gt; (&lt;math&gt;1+14&lt;/math&gt; being the largest sum), a + b cannot equal &lt;math&gt;45&lt;/math&gt;.<br /> <br /> &lt;math&gt;46^2 = 2116&lt;/math&gt; making &lt;math&gt;ab = 105 = 3 * 5 * 7&lt;/math&gt;.<br /> <br /> &lt;math&gt;5 * 7 + 3 &lt; 46&lt;/math&gt; and &lt;math&gt;3 * 5 * 7 &gt; 46&lt;/math&gt; so &lt;math&gt;46&lt;/math&gt; cannot work either.<br /> <br /> &lt;br/&gt;<br /> <br /> We can continue to do this until we reach &lt;math&gt;49&lt;/math&gt;.<br /> <br /> &lt;math&gt;49^2 = 2401&lt;/math&gt; making &lt;math&gt;ab = 390 = 2 * 3 * 5* 13&lt;/math&gt;.<br /> <br /> &lt;math&gt;3 * 13 + 2* 5 = 49&lt;/math&gt;, so one root is &lt;math&gt;10&lt;/math&gt; and another is &lt;math&gt;39&lt;/math&gt;. The roots sum to zero, so the last root must be &lt;math&gt;-49&lt;/math&gt;.<br /> <br /> &lt;br/&gt;<br /> <br /> &lt;math&gt;|-49|+10+39 = \boxed{098}&lt;/math&gt;.<br /> <br /> <br /> == Solution 3 ==<br /> Let us first note the obvious that is derived from Vieta's formulas: &lt;math&gt;a+b+c=0, ab+bc+ac=-2011&lt;/math&gt;. Now, due to the first equation, let us say that &lt;math&gt;a+b=-c&lt;/math&gt;, meaning that &lt;math&gt;a,b&gt;0&lt;/math&gt; and &lt;math&gt;c&lt;0&lt;/math&gt;. Now, since both &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are greater than 0, their absolute values are both equal to &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, respectively. Since &lt;math&gt;c&lt;/math&gt; is less than 0, it equals &lt;math&gt;-a-b&lt;/math&gt;. Therefore, &lt;math&gt;|c|=|-a-b|=a+b&lt;/math&gt;, meaning &lt;math&gt;|a|+|b|+|c|=2(a+b)&lt;/math&gt;. We now apply Newton's sums to get that &lt;math&gt;a^2+b^2+ab=2011&lt;/math&gt;,or &lt;math&gt;(a+b)^2-ab=2011&lt;/math&gt;. Solving, we find that &lt;math&gt;49^2-390&lt;/math&gt; satisfies this, meaning &lt;math&gt;a+b=49&lt;/math&gt;, so &lt;math&gt;2(a+b)=\boxed{098}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We have<br /> <br /> &lt;math&gt;(x-a)\cdot (x-b)\cdot (x-c)=x^3-(a+b+c)x+(ab+ac+bc)x-abc&lt;/math&gt; <br /> <br /> As a result, we have<br /> <br /> &lt;math&gt;a+b+c=0&lt;/math&gt;<br /> <br /> &lt;math&gt;ab+bc+ac=-2011&lt;/math&gt;<br /> <br /> &lt;math&gt;abc=-m&lt;/math&gt;<br /> <br /> So, &lt;math&gt;a=-b-c&lt;/math&gt;<br /> <br /> As a result, &lt;math&gt;ab+bc+ac=(-b-c)b+(-b-c)c+bc=-b^2-c^2-bc=-2011&lt;/math&gt;<br /> <br /> Solve &lt;math&gt;b=\frac {-c+\sqrt{c^2-4(c^2-2011)}}{2}&lt;/math&gt; and <br /> &lt;math&gt;\Delta =8044-3c^2=k^2&lt;/math&gt;, where &lt;math&gt;k&lt;/math&gt; is an integer<br /> <br /> Cause &lt;math&gt;89&lt;\sqrt{8044}&lt;90&lt;/math&gt;<br /> <br /> So, after we tried for &lt;math&gt;2&lt;/math&gt; times, we get &lt;math&gt;k=88&lt;/math&gt; and &lt;math&gt;c=10&lt;/math&gt;<br /> <br /> then &lt;math&gt;b=39&lt;/math&gt;, &lt;math&gt;a=-b-c=-49&lt;/math&gt;<br /> <br /> As a result, &lt;math&gt;|a|+|b|+|c|=10+39+49=\boxed{098}&lt;/math&gt;<br /> <br /> ==Solution 5 (mod to help bash)==<br /> First, derive the equations &lt;math&gt;a=-b-c&lt;/math&gt; and &lt;math&gt;ab+bc+ca=-2011\implies b^2+bc+c^2=2011&lt;/math&gt;. Since the product is negative, &lt;math&gt;a&lt;/math&gt; is negative, and &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; positive. Now, a simple mod 3 testing of all cases shows that &lt;math&gt;b\equiv \{1,2\} \pmod{3}&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; has the repective value. We can choose &lt;math&gt;b&lt;/math&gt; not congruent to 0, make sure you see why. Now, we bash on values of &lt;math&gt;b&lt;/math&gt;, testing the quadratic function to see if &lt;math&gt;c&lt;/math&gt; is positive. You can also use a delta argument like solution 4, but this is simpler. We get that for &lt;math&gt;b=10&lt;/math&gt;, &lt;math&gt;c=39, -49&lt;/math&gt;. Choosing &lt;math&gt;c&lt;/math&gt; positive we get &lt;math&gt;a=-49&lt;/math&gt;, so &lt;math&gt;|a|+|b|+|c|=10+29+39=\boxed{098}&lt;/math&gt;<br /> ~firebolt360<br /> <br /> ==Solution 6==<br /> Note that &lt;math&gt;-c=b+a&lt;/math&gt;, so &lt;math&gt;c^2=a^2+2ab+b^2&lt;/math&gt;, or &lt;math&gt;-c^2+ab=-a^2-ab-b^2&lt;/math&gt;. Also, &lt;math&gt;ab+bc+ca=-2011&lt;/math&gt;, so &lt;math&gt;(a+b)c+ab=-c^2+ab=-2011&lt;/math&gt;. Substituting &lt;math&gt;-c^2+ab=-a^2-ab-b^2&lt;/math&gt;, we can obtain &lt;math&gt;a^2+ab+b^2=2011&lt;/math&gt;, or &lt;math&gt;\frac{a^3-b^3}{a-b}=2011&lt;/math&gt;. If it is not known that &lt;math&gt;2011&lt;/math&gt; is prime, it may be proved in &lt;math&gt;5&lt;/math&gt; minutes or so by checking all primes up to &lt;math&gt;43&lt;/math&gt;. If &lt;math&gt;2011&lt;/math&gt; divided either of &lt;math&gt;a, b&lt;/math&gt;, then in order for &lt;math&gt;a^3-b^3&lt;/math&gt; to contain an extra copy of &lt;math&gt;2011&lt;/math&gt;, both &lt;math&gt;a, b&lt;/math&gt; would need to be divisible by &lt;math&gt;2021&lt;/math&gt;. But then &lt;math&gt;c&lt;/math&gt; would also be divisible by &lt;math&gt;2011&lt;/math&gt;, and the sum &lt;math&gt;ab+bc+ca&lt;/math&gt; would clearly be divisible by &lt;math&gt;2011^2&lt;/math&gt;. <br /> <br /> By LTE, &lt;math&gt;v_2011(a^3-b^3)=v_2011(a-b)&lt;/math&gt; if &lt;math&gt;a-b&lt;/math&gt; is divisible by &lt;math&gt;2011&lt;/math&gt; and neither &lt;math&gt;a,b&lt;/math&gt; are divisible by &lt;math&gt;2011&lt;/math&gt;. Thus, the only possibility remaining is if &lt;math&gt;a-b&lt;/math&gt; did not divide &lt;math&gt;2011&lt;/math&gt;. Let &lt;math&gt;a=k+b&lt;/math&gt;. Then, we have &lt;math&gt;(b+k)^3-b^3=2011k&lt;/math&gt;. Rearranging gives &lt;math&gt;3b(b+k)=2011-k^2&lt;/math&gt;. As in the above solutions, we may eliminate certain values of &lt;math&gt;k&lt;/math&gt; by using mods. Then, we may test values until we obtain &lt;math&gt;k=29&lt;/math&gt;, and &lt;math&gt;a=10&lt;/math&gt;. Thus, &lt;math&gt;b=39&lt;/math&gt;, &lt;math&gt;c=-49&lt;/math&gt;, and our answer is &lt;math&gt;49+39+10=098&lt;/math&gt;.<br /> ==Video Solution==<br /> <br /> https://www.youtube.com/watch?v=QNbfAu5rdJI&amp;t=26s ~ MathEx<br /> <br /> == See also ==<br /> <br /> {{AIME box|year=2011|num-b=14|after=Last Problem|n=I}}<br /> <br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_15&diff=184051 2022 AIME II Problems/Problem 15 2022-12-15T23:39:21Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> <br /> Two externally tangent circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; have centers &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt;, respectively. A third circle &lt;math&gt;\Omega&lt;/math&gt; passing through &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; intersects &lt;math&gt;\omega_1&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, as shown. Suppose that &lt;math&gt;AB = 2&lt;/math&gt;, &lt;math&gt;O_1O_2 = 15&lt;/math&gt;, &lt;math&gt;CD = 16&lt;/math&gt;, and &lt;math&gt;ABO_1CDO_2&lt;/math&gt; is a convex hexagon. Find the area of this hexagon.<br /> &lt;asy&gt;<br /> import geometry;<br /> size(10cm);<br /> point O1=(0,0),O2=(15,0),B=9*dir(30);<br /> circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B);<br /> point A=intersectionpoints(o,w2),D=intersectionpoints(o,w2),C=intersectionpoints(o,w1);<br /> filldraw(A--B--O1--C--D--O2--cycle,0.2*red+white,black);<br /> draw(w1);<br /> draw(w2);<br /> draw(O1--O2,dashed);<br /> draw(o);<br /> dot(O1);<br /> dot(O2);<br /> dot(A);<br /> dot(D);<br /> dot(C);<br /> dot(B);<br /> label(&quot;$\omega_1$&quot;,8*dir(110),SW);<br /> label(&quot;$\omega_2$&quot;,5*dir(70)+(15,0),SE);<br /> label(&quot;$O_1$&quot;,O1,W);<br /> label(&quot;$O_2$&quot;,O2,E);<br /> label(&quot;$B$&quot;,B,N+1/2*E);<br /> label(&quot;$A$&quot;,A,N+1/2*W);<br /> label(&quot;$C$&quot;,C,S+1/4*W);<br /> label(&quot;$D$&quot;,D,S+1/4*E);<br /> label(&quot;$15$&quot;,midpoint(O1--O2),N);<br /> label(&quot;$16$&quot;,midpoint(C--D),N);<br /> label(&quot;$2$&quot;,midpoint(A--B),S);<br /> label(&quot;$\Omega$&quot;,o.C+(o.r-1)*dir(270));<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> First observe that &lt;math&gt;AO_2 = O_2D&lt;/math&gt; and &lt;math&gt;BO_1 = O_1C&lt;/math&gt;. Let points &lt;math&gt;A'&lt;/math&gt; and &lt;math&gt;B'&lt;/math&gt; be the reflections of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, respectively, about the perpendicular bisector of &lt;math&gt;\overline{O_1O_2}&lt;/math&gt;. Then quadrilaterals &lt;math&gt;ABO_1O_2&lt;/math&gt; and &lt;math&gt;A'B'O_2O_1&lt;/math&gt; are congruent, so hexagons &lt;math&gt;ABO_1CDO_2&lt;/math&gt; and &lt;math&gt;B'A'O_1CDO_2&lt;/math&gt; have the same area. Furthermore, triangles &lt;math&gt;DO_2B'&lt;/math&gt; and &lt;math&gt;A'O_1C&lt;/math&gt; are congruent, so &lt;math&gt;B'D = A'C&lt;/math&gt; and quadrilateral &lt;math&gt;B'A'CD&lt;/math&gt; is an isosceles trapezoid.<br /> &lt;asy&gt;<br /> import olympiad;<br /> size(180);<br /> defaultpen(linewidth(0.7));<br /> pair Bp = dir(105), Ap = dir(75), O1 = dir(25), C = dir(320), D = dir(220), O2 = dir(175);<br /> draw(unitcircle^^Bp--Ap--O1--C--D--O2--cycle);<br /> label(&quot;$B'$&quot;,Bp,dir(origin--Bp));<br /> label(&quot;$A'$&quot;,Ap,dir(origin--Ap));<br /> label(&quot;$O_1$&quot;,O1,dir(origin--O1));<br /> label(&quot;$C$&quot;,C,dir(origin--C));<br /> label(&quot;$D$&quot;,D,dir(origin--D));<br /> label(&quot;$O_2$&quot;,O2,dir(origin--O2));<br /> draw(O2--O1,linetype(&quot;4 4&quot;));<br /> draw(Bp--D^^Ap--C,linetype(&quot;2 2&quot;));<br /> &lt;/asy&gt;<br /> Next, remark that &lt;math&gt;A'O_1 = DO_2&lt;/math&gt;, so quadrilateral &lt;math&gt;A'O_1DO_2&lt;/math&gt; is also an isosceles trapezoid; in turn, &lt;math&gt;A'D = O_1O_2 = 15&lt;/math&gt;, and similarly &lt;math&gt;B'C = 15&lt;/math&gt;. Thus, Ptolmey's theorem on &lt;math&gt;B'A'CD&lt;/math&gt; yields &lt;math&gt;B'D\cdot A'C + 2\cdot 16 = 15^2&lt;/math&gt;, whence &lt;math&gt;B'D = A'C = \sqrt{193}&lt;/math&gt;. Let &lt;math&gt;\alpha = \angle B'A'D&lt;/math&gt;. The Law of Cosines on triangle &lt;math&gt;B'A'D&lt;/math&gt; yields<br /> &lt;cmath&gt;\cos\alpha = \frac{15^2 + 2^2 - (\sqrt{193})^2}{2\cdot 2\cdot 15} = \frac{36}{60} = \frac 35,&lt;/cmath&gt; and hence &lt;math&gt;\sin\alpha = \tfrac 45&lt;/math&gt;. Thus the distance between bases &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; is &lt;math&gt;12&lt;/math&gt; (in fact, &lt;math&gt;\triangle B'A'D&lt;/math&gt; is a &lt;math&gt;9-12-15&lt;/math&gt; triangle with a &lt;math&gt;7-12-\sqrt{193}&lt;/math&gt; triangle removed), which implies the area of &lt;math&gt;B'A'CD&lt;/math&gt; is &lt;math&gt;\tfrac12\cdot 12\cdot(2+16) = 108&lt;/math&gt;.<br /> <br /> Now let &lt;math&gt;O_1C = O_2B' = r_1&lt;/math&gt; and &lt;math&gt;O_2D = O_1A' = r_2&lt;/math&gt;; the tangency of circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; implies &lt;math&gt;r_1 + r_2 = 15&lt;/math&gt;. Furthermore, angles &lt;math&gt;B'O_2D&lt;/math&gt; and &lt;math&gt;B'A'D&lt;/math&gt; are opposite angles in cyclic quadrilateral &lt;math&gt;A'B'O_2D&lt;/math&gt;, which implies the measure of angle &lt;math&gt;B'O_2D&lt;/math&gt; is &lt;math&gt;180^\circ - \alpha&lt;/math&gt;. Therefore, the Law of Cosines applied to triangle &lt;math&gt;\triangle B'O_2D&lt;/math&gt; yields<br /> &lt;cmath&gt;\begin{align*}<br /> 193 &amp;= r_1^2 + r_2^2 - 2r_1r_2(-\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \tfrac45r_1r_2\\<br /> &amp;= (r_1+r_2)^2 - \tfrac45 r_1r_2 = 225 - \tfrac45r_1r_2.<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;r_1r_2 = 40&lt;/math&gt;, and so the area of triangle &lt;math&gt;B'O_2D&lt;/math&gt; is &lt;math&gt;\tfrac12r_1r_2\sin\alpha = 16&lt;/math&gt;.<br /> <br /> Thus, the area of hexagon &lt;math&gt;ABO_{1}CDO_{2}&lt;/math&gt; is &lt;math&gt;108 + 2\cdot 16 = \boxed{140}&lt;/math&gt;.<br /> <br /> ~djmathman<br /> <br /> ==Solution 2==<br /> <br /> Denote by &lt;math&gt;O&lt;/math&gt; the center of &lt;math&gt;\Omega&lt;/math&gt;.<br /> Denote by &lt;math&gt;r&lt;/math&gt; the radius of &lt;math&gt;\Omega&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;O_1&lt;/math&gt;, &lt;math&gt;O_2&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt; are all on circle &lt;math&gt;\Omega&lt;/math&gt;.<br /> <br /> Denote &lt;math&gt;\angle O_1 O O_2 = 2 \theta&lt;/math&gt;.<br /> Denote &lt;math&gt;\angle O_1 O B = \alpha&lt;/math&gt;.<br /> Denote &lt;math&gt;\angle O_2 O A = \beta&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are on circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\Omega&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; is a perpendicular bisector of &lt;math&gt;O_1 O&lt;/math&gt;. Hence, &lt;math&gt;\angle O_1 O C = \alpha&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; are on circles &lt;math&gt;\omega_2&lt;/math&gt; and &lt;math&gt;\Omega&lt;/math&gt;, &lt;math&gt;AD&lt;/math&gt; is a perpendicular bisector of &lt;math&gt;O_2 O&lt;/math&gt;. Hence, &lt;math&gt;\angle O_2 O D = \beta&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\triangle O O_1 O_2&lt;/math&gt;, <br /> &lt;cmath&gt;<br /> $<br /> O_1 O_2 = 2 r \sin \theta . <br />$<br /> &lt;/cmath&gt;<br /> <br /> Hence, <br /> &lt;cmath&gt;<br /> $<br /> 2 r \sin \theta = 15 . <br />$<br /> &lt;/cmath&gt;<br /> <br /> In &lt;math&gt;\triangle O AB&lt;/math&gt;, <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> AB &amp; = 2 r \sin \frac{2 \theta - \alpha - \beta}{2} \\<br /> &amp; = 2 r \sin \theta \cos \frac{\alpha + \beta}{2}<br /> - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\<br /> &amp; = 15 \cos \frac{\alpha + \beta}{2}<br /> - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} .<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Hence,<br /> &lt;cmath&gt;<br /> $<br /> 15 \cos \frac{\alpha + \beta}{2}<br /> - 2 r \cos \theta \sin \frac{\alpha + \beta}{2}<br /> = 2 . \hspace{1cm} (1)<br />$<br /> &lt;/cmath&gt;<br /> <br /> <br /> In &lt;math&gt;\triangle O CD&lt;/math&gt;, <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> CD &amp; = 2 r \sin \frac{360^\circ - 2 \theta - \alpha - \beta}{2} \\<br /> &amp; = 2 r \sin \left( \theta + \frac{\alpha + \beta}{2} \right) \\<br /> &amp; = 2 r \sin \theta \cos \frac{\alpha + \beta}{2}<br /> + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\<br /> &amp; = 15 \cos \frac{\alpha + \beta}{2}<br /> + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} .<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Hence,<br /> &lt;cmath&gt;<br /> $<br /> 15 \cos \frac{\alpha + \beta}{2}<br /> + 2 r \cos \theta \sin \frac{\alpha + \beta}{2}<br /> = 16 . \hspace{1cm} (2)<br />$<br /> &lt;/cmath&gt;<br /> <br /> Taking &lt;math&gt;\frac{(1) + (2)}{30}&lt;/math&gt;, we get &lt;math&gt;\cos \frac{\alpha + \beta}{2} = \frac{3}{5}&lt;/math&gt;.<br /> Thus, &lt;math&gt;\sin \frac{\alpha + \beta}{2} = \frac{4}{5}&lt;/math&gt;.<br /> <br /> Taking these into (1), we get &lt;math&gt;2 r \cos \theta = \frac{35}{4}&lt;/math&gt;.<br /> Hence,<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> 2 r &amp; = \sqrt{ \left( 2 r \sin \theta \right)^2 + \left( 2 r \cos \theta \right)^2} \\<br /> &amp; = \frac{5}{4} \sqrt{193} . <br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Hence, &lt;math&gt;\cos \theta = \frac{7}{\sqrt{193}}&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\triangle O O_1 B&lt;/math&gt;, <br /> &lt;cmath&gt;<br /> $<br /> O_1 B = 2 r \sin \frac{\alpha}{2} . <br />$<br /> &lt;/cmath&gt;<br /> <br /> In &lt;math&gt;\triangle O O_2 A&lt;/math&gt;, by applying the law of sines, we get<br /> &lt;cmath&gt;<br /> $<br /> O_2 A = 2 r \sin \frac{\beta}{2} . <br />$<br /> &lt;/cmath&gt;<br /> <br /> Because circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; are externally tangent, &lt;math&gt;B&lt;/math&gt; is on circle &lt;math&gt;\omega_1&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt; is on circle &lt;math&gt;\omega_2&lt;/math&gt;,<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> O_1 O_2 &amp; = O_1 B + O_2 A \\<br /> &amp; = 2 r \sin \frac{\alpha}{2} + 2 r \sin \frac{\beta}{2} \\<br /> &amp; = 2 r \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right) . <br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;\sin \frac{\alpha}{2} + \sin \frac{\beta}{2} = \frac{12}{\sqrt{193}}&lt;/math&gt;.<br /> <br /> Now, we compute &lt;math&gt;\sin \alpha&lt;/math&gt; and &lt;math&gt;\sin \beta&lt;/math&gt;.<br /> <br /> Recall &lt;math&gt;\cos \frac{\alpha + \beta}{2} = \frac{3}{5}&lt;/math&gt; and &lt;math&gt;\sin \frac{\alpha + \beta}{2} = \frac{4}{5}&lt;/math&gt;.<br /> Thus, &lt;math&gt;e^{i \frac{\alpha}{2}} e^{i \frac{\beta}{2}} = e^{i \frac{\alpha + \beta}{2}} = \frac{3}{5} + i \frac{4}{5}&lt;/math&gt;.<br /> <br /> We also have<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \sin \frac{\alpha}{2} + \sin \frac{\beta}{2}<br /> &amp; = \frac{1}{2i} \left( e^{i \frac{\alpha}{2}} - e^{-i \frac{\alpha}{2}} <br /> + e^{i \frac{\beta}{2}} - e^{-i \frac{\beta}{2}} \right) \\<br /> &amp; = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} } \right)<br /> \left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right)<br /> . <br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Thus,<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \sin \alpha + \sin \beta <br /> &amp; = \frac{1}{2i} \left( e^{i \alpha} - e^{-i \alpha}<br /> + e^{i \beta} - e^{-i \beta} \right) \\<br /> &amp; = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right)<br /> \left( e^{i \alpha} + e^{i \beta} \right) \\<br /> &amp; = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right)<br /> \left(<br /> \left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right)^2<br /> - 2 e^{i \frac{\alpha + \beta}{2}}<br /> \right) \\<br /> &amp; = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right)<br /> \left(<br /> \left( \frac{2 i \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)}{1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} }} \right)^2<br /> - 2 e^{i \frac{\alpha + \beta}{2}}<br /> \right) \\<br /> &amp; = - \frac{1}{i} \left( e^{i \frac{\alpha + \beta}{2}} - e^{-i \frac{\alpha + \beta}{2}} \right)<br /> \left(<br /> \frac{2 \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)^2}<br /> {e^{i \frac{\alpha + \beta}{2}} + e^{-i \frac{\alpha + \beta}{2}} - 2 }<br /> + 1<br /> \right) \\<br /> &amp; = \frac{167 \cdot 8}{193 \cdot 5 } .<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Therefore,<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> {\rm Area} \ ABO_1CDO_2<br /> &amp; = {\rm Area} \ \triangle O_3 AB + {\rm Area} \ \triangle O_3 BO_1<br /> + {\rm Area} \ \triangle O_3 O_1 C \\<br /> &amp; \quad + {\rm Area} \ \triangle O_3 C D<br /> + {\rm Area} \ \triangle O_3 D O_2 + {\rm Area} \ \triangle O_3 O_2 A \\<br /> &amp; = \frac{1}{2} r^2 \left(<br /> \sin \left( 2 \theta - \alpha - \beta \right) + \sin \alpha + \sin \alpha<br /> + \sin \left( 360^\circ - 2 \theta - \alpha - \beta \right)<br /> + \sin \beta + \sin \beta \right) \\<br /> &amp; = \frac{1}{2} r^2 \left(<br /> \sin \left( 2 \theta - \alpha - \beta \right)<br /> - \sin \left( 2 \theta + \alpha + \beta \right)<br /> + 2 \sin \alpha + 2 \sin \beta <br /> \right) \\<br /> &amp; = r^2 \left(<br /> - \cos 2 \theta \sin \left( \alpha + \beta \right)<br /> + \sin \alpha + \sin \beta<br /> \right) \\<br /> &amp; = r^2 \left( \left( 1 - 2 \cos^2 \theta \right) 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha + \beta}{2} <br /> + \sin \alpha + \sin \beta \right) \\<br /> &amp; = \boxed{\textbf{(140) }} .<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> ~Steven Chen (www.professorchenedu.com)<br /> <br /> ==Solution 3== <br /> [[File:AIME 2022 15.png|400px|right]]<br /> Let points &lt;math&gt;A'&lt;/math&gt; and &lt;math&gt;B'&lt;/math&gt; be the reflections of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B,&lt;/math&gt; respectively, about the perpendicular bisector of &lt;math&gt;O_1 O_2.&lt;/math&gt; <br /> &lt;cmath&gt;B'O_2 = BO_1 = O_1 P = O_1 C,&lt;/cmath&gt;<br /> &lt;cmath&gt;A'O_1 = AO_2 = O_2 P = O_2 D.&lt;/cmath&gt;<br /> We establish the equality of the arcs and conclude that the corresponding chords are equal<br /> &lt;cmath&gt;\overset{\Large\frown} {CO_1} + \overset{\Large\frown} {A'O_1} +\overset{\Large\frown} {A'B'} = \overset{\Large\frown} {B'O_2} +\overset{\Large\frown} {A'O_1} +\overset{\Large\frown} {A'B'} =\overset{\Large\frown} {B'O_2} +\overset{\Large\frown} {DO_2} +\overset{\Large\frown} {A'B'}&lt;/cmath&gt;<br /> &lt;cmath&gt; \implies A'D = B'C = O_1 O_2 = 15.&lt;/cmath&gt;<br /> Similarly &lt;math&gt;A'C = B'D \implies \triangle A'CO_1 = \triangle B'DO_2.&lt;/math&gt;<br /> <br /> Ptolmey's theorem on &lt;math&gt;A'CDB'&lt;/math&gt; yields &lt;cmath&gt;B'D \cdot A'C + A'B' \cdot CD = A'D \cdot B'C \implies&lt;/cmath&gt;<br /> &lt;cmath&gt; B'D^2 + 2 \cdot 16 = 15^2 \implies B'D = A'C = \sqrt{193}.&lt;/cmath&gt;<br /> The area of the trapezoid &lt;math&gt;A'CDB'&lt;/math&gt; is equal to the area of an isosceles triangle with sides <br /> &lt;math&gt;A'D = B'C = 15&lt;/math&gt; and &lt;math&gt;A'B' + CD = 18.&lt;/math&gt;<br /> <br /> The height of this triangle is &lt;math&gt;\sqrt{15^2-9^2} = 12.&lt;/math&gt; The area of &lt;math&gt;A'CDB'&lt;/math&gt; is &lt;math&gt;108.&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\sin \angle B'CD = \frac{12}{15} = \frac{4}{5},&lt;/cmath&gt;<br /> &lt;cmath&gt;\angle B'CD + \angle B'O_2 D = 180^o \implies \sin \angle B'O_2 D = \frac{4}{5}.&lt;/cmath&gt;<br /> <br /> Denote &lt;math&gt;\angle B'O_2 D = 2\alpha.&lt;/math&gt;<br /> &lt;math&gt;\angle B'O_2 D &gt; \frac{\pi}{2},&lt;/math&gt; hence &lt;math&gt;\cos \angle B'O_2 D = \cos 2\alpha = -\frac{3}{5}.&lt;/math&gt; <br /> &lt;cmath&gt;\tan \alpha =\frac { \sin 2 \alpha}{1+\cos 2 \alpha} = \frac {4/5}{1 - 3/5}=2.&lt;/cmath&gt;<br /> <br /> Semiperimeter of &lt;math&gt;\triangle B'O_2 D &lt;/math&gt; is &lt;math&gt;s = \frac {15 + \sqrt{193}}{2}.&lt;/math&gt;<br /> <br /> The distance from the vertex &lt;math&gt;O_2&lt;/math&gt; to the tangent points of the inscribed circle of the triangle &lt;math&gt;B'O_2 D&lt;/math&gt; is equal &lt;math&gt;s – B'D = \frac{15 – \sqrt{193}}{2}.&lt;/math&gt;<br /> <br /> The radius of the inscribed circle is &lt;math&gt;r = (s – B'D) \tan \alpha.&lt;/math&gt;<br /> <br /> The area of triangle &lt;math&gt;B'O_2 D&lt;/math&gt; is &lt;math&gt;[B'O_2 D] = sr = s (s – B'D) \tan \alpha = \frac {15^2 – 193}{2} = 16.&lt;/math&gt;<br /> <br /> The hexagon &lt;math&gt;ABO_1 CDO_2&lt;/math&gt; has the same area as hexagon &lt;math&gt;B'A'O_1 CDO_2.&lt;/math&gt; <br /> <br /> The area of hexagon &lt;math&gt;B'A'O_1 CDO_2&lt;/math&gt; is equal to the sum of the area of the trapezoid &lt;math&gt;A'CDB'&lt;/math&gt; and the areas of two equal triangles &lt;math&gt;B'O_2 D&lt;/math&gt; and &lt;math&gt;A'O_1 C,&lt;/math&gt; so the area of the hexagon &lt;math&gt;ABO_1 CDO_2&lt;/math&gt; is &lt;cmath&gt;108 + 16 + 16 = \boxed{140}.&lt;/cmath&gt;<br /> <br /> '''vladimir.shelomovskii@gmail.com, vvsss'''<br /> <br /> ==See Also==<br /> {{AIME box|year=2022|n=II|num-b=14|after=Last Problem}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_15&diff=184050 2022 AIME II Problems/Problem 15 2022-12-15T23:37:55Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> <br /> Two externally tangent circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; have centers &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt;, respectively. A third circle &lt;math&gt;\Omega&lt;/math&gt; passing through &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; intersects &lt;math&gt;\omega_1&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, as shown. Suppose that &lt;math&gt;AB = 2&lt;/math&gt;, &lt;math&gt;O_1O_2 = 15&lt;/math&gt;, &lt;math&gt;CD = 16&lt;/math&gt;, and &lt;math&gt;ABO_1CDO_2&lt;/math&gt; is a convex hexagon. Find the area of this hexagon.<br /> &lt;asy&gt;<br /> import geometry;<br /> size(10cm);<br /> point O1=(0,0),O2=(15,0),B=9*dir(30);<br /> circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B);<br /> point A=intersectionpoints(o,w2),D=intersectionpoints(o,w2),C=intersectionpoints(o,w1);<br /> filldraw(A--B--O1--C--D--O2--cycle,0.2*red+white,black);<br /> draw(w1);<br /> draw(w2);<br /> draw(O1--O2,dashed);<br /> draw(o);<br /> dot(O1);<br /> dot(O2);<br /> dot(A);<br /> dot(D);<br /> dot(C);<br /> dot(B);<br /> label(&quot;$\omega_1$&quot;,8*dir(110),SW);<br /> label(&quot;$\omega_2$&quot;,5*dir(70)+(15,0),SE);<br /> label(&quot;$O_1$&quot;,O1,W);<br /> label(&quot;$O_2$&quot;,O2,E);<br /> label(&quot;$B$&quot;,B,N+1/2*E);<br /> label(&quot;$A$&quot;,A,N+1/2*W);<br /> label(&quot;$C$&quot;,C,S+1/4*W);<br /> label(&quot;$D$&quot;,D,S+1/4*E);<br /> label(&quot;$15$&quot;,midpoint(O1--O2),N);<br /> label(&quot;$16$&quot;,midpoint(C--D),N);<br /> label(&quot;$2$&quot;,midpoint(A--B),S);<br /> label(&quot;$\Omega$&quot;,o.C+(o.r-1)*dir(270));<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> First observe that &lt;math&gt;AO_2 = O_2D&lt;/math&gt; and &lt;math&gt;BO_1 = O_1C&lt;/math&gt;. Let points &lt;math&gt;A'&lt;/math&gt; and &lt;math&gt;B'&lt;/math&gt; be the reflections of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, respectively, about the perpendicular bisector of &lt;math&gt;\overline{O_1O_2}&lt;/math&gt;. Then quadrilaterals &lt;math&gt;ABO_1O_2&lt;/math&gt; and &lt;math&gt;A'B'O_2O_1&lt;/math&gt; are congruent, so hexagons &lt;math&gt;ABO_1CDO_2&lt;/math&gt; and &lt;math&gt;B'A'O_1CDO_2&lt;/math&gt; have the same area. Furthermore, triangles &lt;math&gt;DO_2B'&lt;/math&gt; and &lt;math&gt;A'O_1C&lt;/math&gt; are congruent, so &lt;math&gt;B'D = A'C&lt;/math&gt; and quadrilateral &lt;math&gt;B'A'CD&lt;/math&gt; is an isosceles trapezoid.<br /> &lt;asy&gt;<br /> import olympiad;<br /> size(180);<br /> defaultpen(linewidth(0.7));<br /> pair Bp = dir(105), Ap = dir(75), O1 = dir(25), C = dir(320), D = dir(220), O2 = dir(175);<br /> draw(unitcircle^^Bp--Ap--O1--C--D--O2--cycle);<br /> label(&quot;$B'$&quot;,Bp,dir(origin--Bp));<br /> label(&quot;$A'$&quot;,Ap,dir(origin--Ap));<br /> label(&quot;$O_1$&quot;,O1,dir(origin--O1));<br /> label(&quot;$C$&quot;,C,dir(origin--C));<br /> label(&quot;$D$&quot;,D,dir(origin--D));<br /> label(&quot;$O_2$&quot;,O2,dir(origin--O2));<br /> draw(O2--O1,linetype(&quot;4 4&quot;));<br /> draw(Bp--D^^Ap--C,linetype(&quot;2 2&quot;));<br /> &lt;/asy&gt;<br /> Next, remark that &lt;math&gt;A'O_1 = DO_2&lt;/math&gt;, so quadrilateral &lt;math&gt;A'O_1DO_2&lt;/math&gt; is also an isosceles trapezoid; in turn, &lt;math&gt;A'D = O_1O_2 = 15&lt;/math&gt;, and similarly &lt;math&gt;B'C = 15&lt;/math&gt;. Thus, Ptolmey's theorem on &lt;math&gt;B'A'CD&lt;/math&gt; yields &lt;math&gt;B'D\cdot A'C + 2\cdot 16 = 15^2&lt;/math&gt;, whence &lt;math&gt;B'D = A'C = \sqrt{193}&lt;/math&gt;. Let &lt;math&gt;\alpha = \angle B'A'D&lt;/math&gt;. The Law of Cosines on triangle &lt;math&gt;B'A'D&lt;/math&gt; yields<br /> &lt;cmath&gt;\cos\alpha = \frac{15^2 + 2^2 - (\sqrt{193})^2}{2\cdot 2\cdot 15} = \frac{36}{60} = \frac 35,&lt;/cmath&gt; and hence &lt;math&gt;\sin\alpha = \tfrac 45&lt;/math&gt;. Thus the distance between bases &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; is &lt;math&gt;12&lt;/math&gt; (in fact, &lt;math&gt;\triangle B'A'D&lt;/math&gt; is a &lt;math&gt;9-12-15&lt;/math&gt; triangle with a &lt;math&gt;7-12-\sqrt{193}&lt;/math&gt; triangle removed), which implies the area of &lt;math&gt;B'A'CD&lt;/math&gt; is &lt;math&gt;\tfrac12\cdot 12\cdot(2+16) = 108&lt;/math&gt;.<br /> <br /> Now let &lt;math&gt;O_1C = O_2B' = r_1&lt;/math&gt; and &lt;math&gt;O_2D = O_1A' = r_2&lt;/math&gt;; the tangency of circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; implies &lt;math&gt;r_1 + r_2 = 15&lt;/math&gt;. Furthermore, angles &lt;math&gt;B'O_2D&lt;/math&gt; and &lt;math&gt;B'A'D&lt;/math&gt; are opposite angles in cyclic quadrilateral &lt;math&gt;A'B'O_2D&lt;/math&gt;, which implies the measure of angle &lt;math&gt;B'O_2D&lt;/math&gt; is &lt;math&gt;180^\circ - \alpha&lt;/math&gt;. Therefore, the Law of Cosines applied to triangle &lt;math&gt;\triangle B'O_2D&lt;/math&gt; yields<br /> &lt;cmath&gt;\begin{align*}<br /> 193 &amp;= r_1^2 + r_2^2 - 2r_1r_2(-\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \tfrac45r_1r_2\\<br /> &amp;= (r_1+r_2)^2 - \tfrac45 r_1r_2 = 225 - \tfrac45r_1r_2.<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;r_1r_2 = 40&lt;/math&gt;, and so the area of triangle &lt;math&gt;B'O_2D&lt;/math&gt; is &lt;math&gt;\tfrac12r_1r_2\sin\alpha = 16&lt;/math&gt;.<br /> <br /> Thus, the area of hexagon &lt;math&gt;ABO_{1}CDO_{2}&lt;/math&gt; is &lt;math&gt;108 + 2\cdot 16 = \boxed{140}&lt;/math&gt;.<br /> <br /> ~djmathman<br /> <br /> ==Solution 2==<br /> <br /> Denote by &lt;math&gt;O&lt;/math&gt; the center of &lt;math&gt;\Omega&lt;/math&gt;.<br /> Denote by &lt;math&gt;r&lt;/math&gt; the radius of &lt;math&gt;\Omega&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;O_1&lt;/math&gt;, &lt;math&gt;O_2&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt; are all on circle &lt;math&gt;\Omega&lt;/math&gt;.<br /> <br /> Denote &lt;math&gt;\angle O_1 O O_2 = 2 \theta&lt;/math&gt;.<br /> Denote &lt;math&gt;\angle O_1 O B = \alpha&lt;/math&gt;.<br /> Denote &lt;math&gt;\angle O_2 O A = \beta&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are on circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\Omega&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; is a perpendicular bisector of &lt;math&gt;O_1 O&lt;/math&gt;. Hence, &lt;math&gt;\angle O_1 O C = \alpha&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; are on circles &lt;math&gt;\omega_2&lt;/math&gt; and &lt;math&gt;\Omega&lt;/math&gt;, &lt;math&gt;AD&lt;/math&gt; is a perpendicular bisector of &lt;math&gt;O_2 O&lt;/math&gt;. Hence, &lt;math&gt;\angle O_2 O D = \beta&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\triangle O O_1 O_2&lt;/math&gt;, <br /> &lt;cmath&gt;<br /> $<br /> O_1 O_2 = 2 r \sin \theta . <br />$<br /> &lt;/cmath&gt;<br /> <br /> Hence, <br /> &lt;cmath&gt;<br /> $<br /> 2 r \sin \theta = 15 . <br />$<br /> &lt;/cmath&gt;<br /> <br /> In &lt;math&gt;\triangle O AB&lt;/math&gt;, <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> AB &amp; = 2 r \sin \frac{2 \theta - \alpha - \beta}{2} \\<br /> &amp; = 2 r \sin \theta \cos \frac{\alpha + \beta}{2}<br /> - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\<br /> &amp; = 15 \cos \frac{\alpha + \beta}{2}<br /> - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} .<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Hence,<br /> &lt;cmath&gt;<br /> $<br /> 15 \cos \frac{\alpha + \beta}{2}<br /> - 2 r \cos \theta \sin \frac{\alpha + \beta}{2}<br /> = 2 . \hspace{1cm} (1)<br />$<br /> &lt;/cmath&gt;<br /> <br /> <br /> In &lt;math&gt;\triangle O CD&lt;/math&gt;, <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> CD &amp; = 2 r \sin \frac{360^\circ - 2 \theta - \alpha - \beta}{2} \\<br /> &amp; = 2 r \sin \left( \theta + \frac{\alpha + \beta}{2} \right) \\<br /> &amp; = 2 r \sin \theta \cos \frac{\alpha + \beta}{2}<br /> + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\<br /> &amp; = 15 \cos \frac{\alpha + \beta}{2}<br /> + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} .<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Hence,<br /> &lt;cmath&gt;<br /> $<br /> 15 \cos \frac{\alpha + \beta}{2}<br /> + 2 r \cos \theta \sin \frac{\alpha + \beta}{2}<br /> = 16 . \hspace{1cm} (2)<br />$<br /> &lt;/cmath&gt;<br /> <br /> Taking &lt;math&gt;\frac{(1) + (2)}{30}&lt;/math&gt;, we get &lt;math&gt;\cos \frac{\alpha + \beta}{2} = \frac{3}{5}&lt;/math&gt;.<br /> Thus, &lt;math&gt;\sin \frac{\alpha + \beta}{2} = \frac{4}{5}&lt;/math&gt;.<br /> <br /> Taking these into (1), we get &lt;math&gt;2 r \cos \theta = \frac{35}{4}&lt;/math&gt;.<br /> Hence,<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> 2 r &amp; = \sqrt{ \left( 2 r \sin \theta \right)^2 + \left( 2 r \cos \theta \right)^2} \\<br /> &amp; = \frac{5}{4} \sqrt{193} . <br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Hence, &lt;math&gt;\cos \theta = \frac{7}{\sqrt{193}}&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\triangle O O_1 B&lt;/math&gt;, <br /> &lt;cmath&gt;<br /> $<br /> O_1 B = 2 r \sin \frac{\alpha}{2} . <br />$<br /> &lt;/cmath&gt;<br /> <br /> In &lt;math&gt;\triangle O O_2 A&lt;/math&gt;, by applying the law of sines, we get<br /> &lt;cmath&gt;<br /> $<br /> O_2 A = 2 r \sin \frac{\beta}{2} . <br />$<br /> &lt;/cmath&gt;<br /> <br /> Because circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; are externally tangent, &lt;math&gt;B&lt;/math&gt; is on circle &lt;math&gt;\omega_1&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt; is on circle &lt;math&gt;\omega_2&lt;/math&gt;,<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> O_1 O_2 &amp; = O_1 B + O_2 A \\<br /> &amp; = 2 r \sin \frac{\alpha}{2} + 2 r \sin \frac{\beta}{2} \\<br /> &amp; = 2 r \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right) . <br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;\sin \frac{\alpha}{2} + \sin \frac{\beta}{2} = \frac{12}{\sqrt{193}}&lt;/math&gt;.<br /> <br /> Now, we compute &lt;math&gt;\sin \alpha&lt;/math&gt; and &lt;math&gt;\sin \beta&lt;/math&gt;.<br /> <br /> Recall &lt;math&gt;\cos \frac{\alpha + \beta}{2} = \frac{3}{5}&lt;/math&gt; and &lt;math&gt;\sin \frac{\alpha + \beta}{2} = \frac{4}{5}&lt;/math&gt;.<br /> Thus, &lt;math&gt;e^{i \frac{\alpha}{2}} e^{i \frac{\beta}{2}} = e^{i \frac{\alpha + \beta}{2}} = \frac{3}{5} + i \frac{4}{5}&lt;/math&gt;.<br /> <br /> We also have<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \sin \frac{\alpha}{2} + \sin \frac{\beta}{2}<br /> &amp; = \frac{1}{2i} \left( e^{i \frac{\alpha}{2}} - e^{-i \frac{\alpha}{2}} <br /> + e^{i \frac{\beta}{2}} - e^{-i \frac{\beta}{2}} \right) \\<br /> &amp; = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} } \right)<br /> \left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right)<br /> . <br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Thus,<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \sin \alpha + \sin \beta <br /> &amp; = \frac{1}{2i} \left( e^{i \alpha} - e^{-i \alpha}<br /> + e^{i \beta} - e^{-i \beta} \right) \\<br /> &amp; = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right)<br /> \left( e^{i \alpha} + e^{i \beta} \right) \\<br /> &amp; = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right)<br /> \left(<br /> \left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right)^2<br /> - 2 e^{i \frac{\alpha + \beta}{2}}<br /> \right) \\<br /> &amp; = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right)<br /> \left(<br /> \left( \frac{2 i \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)}{1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} }} \right)^2<br /> - 2 e^{i \frac{\alpha + \beta}{2}}<br /> \right) \\<br /> &amp; = - \frac{1}{i} \left( e^{i \frac{\alpha + \beta}{2}} - e^{-i \frac{\alpha + \beta}{2}} \right)<br /> \left(<br /> \frac{2 \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)^2}<br /> {e^{i \frac{\alpha + \beta}{2}} + e^{-i \frac{\alpha + \beta}{2}} - 2 }<br /> + 1<br /> \right) \\<br /> &amp; = \frac{167 \cdot 8}{193 \cdot 5 } .<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Therefore,<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> {\rm Area} \ ABO_1CDO_2<br /> &amp; = {\rm Area} \ \triangle O_3 AB + {\rm Area} \ \triangle O_3 BO_1<br /> + {\rm Area} \ \triangle O_3 O_1 C \\<br /> &amp; \quad + {\rm Area} \ \triangle O_3 C D<br /> + {\rm Area} \ \triangle O_3 D O_2 + {\rm Area} \ \triangle O_3 O_2 A \\<br /> &amp; = \frac{1}{2} r^2 \left(<br /> \sin \left( 2 \theta - \alpha - \beta \right) + \sin \alpha + \sin \alpha<br /> + \sin \left( 360^\circ - 2 \theta - \alpha - \beta \right)<br /> + \sin \beta + \sin \beta \right) \\<br /> &amp; = \frac{1}{2} r^2 \left(<br /> \sin \left( 2 \theta - \alpha - \beta \right)<br /> - \sin \left( 2 \theta + \alpha + \beta \right)<br /> + 2 \sin \alpha + 2 \sin \beta <br /> \right) \\<br /> &amp; = r^2 \left(<br /> - \cos 2 \theta \sin \left( \alpha + \beta \right)<br /> + \sin \alpha + \sin \beta<br /> \right) \\<br /> &amp; = r^2 \left( \left( 1 - 2 \cos^2 \theta \right) 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha + \beta}{2} <br /> + \sin \alpha + \sin \beta \right) \\<br /> &amp; = \boxed{\textbf{(140) }} .<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> ~Steven Chen (www.professorchenedu.com)<br /> <br /> ==Solution 3== <br /> [[File:AIME 2022 15.png|400px|right]]<br /> Let points &lt;math&gt;A'&lt;/math&gt; and &lt;math&gt;B'&lt;/math&gt; be the reflections of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B,&lt;/math&gt; respectively, about the perpendicular bisector of &lt;math&gt;O_1 O_2.&lt;/math&gt; <br /> &lt;cmath&gt;B'O_2 = BO_1 = O_1 P = O_1 C,&lt;/cmath&gt;<br /> &lt;cmath&gt;A'O_1 = AO_2 = O_2 P = O_2 D.&lt;/cmath&gt;<br /> We establish the equality of the arcs and conclude that the corresponding chords are equal<br /> &lt;cmath&gt;\overset{\Large\frown} {CO_1} + \overset{\Large\frown} {A'O_1} +\overset{\Large\frown} {A'B'} = \overset{\Large\frown} {B'O_2} +\overset{\Large\frown} {A'O_1} +\overset{\Large\frown} {A'B'} =\overset{\Large\frown} {B'O_2} +\overset{\Large\frown} {DO_2} +\overset{\Large\frown} {A'B'}&lt;/cmath&gt;<br /> &lt;cmath&gt; \implies A'D = B'C = O_1 O_2 = 15.&lt;/cmath&gt;<br /> Similarly &lt;math&gt;A'C = B'D \implies \triangle A'CO_1 = \triangle B'DO_2.&lt;/math&gt;<br /> <br /> Ptolmey's theorem on &lt;math&gt;A'CDB'&lt;/math&gt; yields &lt;cmath&gt;B'D \cdot A'C + A'B' \cdot CD = A'D \cdot B'C \implies&lt;/cmath&gt;<br /> &lt;cmath&gt; B'D^2 + 2 \cdot 16 = 15^2 \implies B'D = A'C = \sqrt{193}.&lt;/cmath&gt;<br /> The area of the trapezoid &lt;math&gt;A'CDB'&lt;/math&gt; is equal to the area of an isosceles triangle with sides <br /> &lt;math&gt;A'D = B'C = 15&lt;/math&gt; and &lt;math&gt;A'B' + CD = 18.&lt;/math&gt;<br /> <br /> The height of this triangle is &lt;math&gt;\sqrt{15^2-9^2} = 12.&lt;/math&gt; The area of &lt;math&gt;A'CDB'&lt;/math&gt; is &lt;math&gt;108.&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\sin \angle B'CD = \frac{12}{15} = \frac{4}{5},&lt;/cmath&gt;<br /> &lt;cmath&gt;\angle B'CD + \angle B'O_2 D = 180^o \implies \sin \angle B'O_2 D = \frac{4}{5}.&lt;/cmath&gt;<br /> <br /> Denote &lt;math&gt;\angle B'O_2 D = 2\alpha.&lt;/math&gt;<br /> &lt;math&gt;\angle B'O_2 D &gt; \frac{\pi}{2},&lt;/math&gt; hence &lt;math&gt;\cos \angle B'O_2 D = \cos 2\alpha = -\frac{3}{5}.&lt;/math&gt; <br /> &lt;cmath&gt;\tan \alpha =\frac { \sin 2 \alpha}{1+\cos 2 \alpha} = \frac {4/5}{1 - 3/5}=2.&lt;/cmath&gt;<br /> <br /> Semiperimeter of &lt;math&gt;\triangle B'O_2 D &lt;/math&gt; is &lt;math&gt;s = \frac {15 + \sqrt{193}}{2}.&lt;/math&gt;<br /> <br /> The distance from the vertex &lt;math&gt;O_2&lt;/math&gt; to the tangent points of the inscribed circle of the triangle &lt;math&gt;B'O_2 D&lt;/math&gt; is equal &lt;math&gt;s – B'D = \frac{15 – \sqrt{193}}{2}.&lt;/math&gt;<br /> <br /> The radius of the inscribed circle is &lt;math&gt;r = (s – B'D) \tan \alpha.&lt;/math&gt;<br /> <br /> The area of triangle &lt;math&gt;B'O_2 D&lt;/math&gt; is &lt;math&gt;[B'O_2 D] = sr = s (s – B'D) \tan \alpha = \frac {15^2 – 193}{2} = 16.&lt;/math&gt;<br /> <br /> The hexagon &lt;math&gt;ABO_1 CDO_2&lt;/math&gt; has the same area as hexagon &lt;math&gt;B'A'O_1 CDO_2.&lt;/math&gt; <br /> <br /> The area of hexagon &lt;math&gt;B'A'O_1 CDO_2&lt;/math&gt; is equal to the sum of the area of the trapezoid &lt;math&gt;A'CDB'&lt;/math&gt; and the areas of two equal triangles &lt;math&gt;B'O_2 D&lt;/math&gt; and &lt;math&gt;A'O_1 C,&lt;/math&gt; so the area of the hexagon &lt;math&gt;ABO_1 CDO_2&lt;/math&gt; is &lt;cmath&gt;108 + 16 + 16 = \boxed{140}.&lt;/cmath&gt;<br /> <br /> '''vladimir.shelomovskii@gmail.com, vvsss'''<br /> <br /> ==Solution 4==<br /> <br /> ==See Also==<br /> {{AIME box|year=2022|n=II|num-b=14|after=Last Problem}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_11&diff=182754 2005 AIME II Problems/Problem 11 2022-11-21T22:52:18Z <p>Sigmapie: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;m &lt;/math&gt; be a positive integer, and let &lt;math&gt; a_0, a_1,\ldots,a_m &lt;/math&gt; be a sequence of reals such that &lt;math&gt;a_0 = 37, a_1 = 72, a_m = 0, &lt;/math&gt; and &lt;math&gt; a_{k+1} = a_{k-1} - \frac 3{a_k} &lt;/math&gt; for &lt;math&gt; k = 1,2,\ldots, m-1. &lt;/math&gt; Find &lt;math&gt;m. &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> For &lt;math&gt;0 &lt; k &lt; m&lt;/math&gt;, we have<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;a_{k}a_{k+1} = a_{k-1}a_{k} - 3 &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> Thus the product &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; is a [[monovariant]]: it decreases by 3 each time &lt;math&gt;k&lt;/math&gt; increases by 1. For &lt;math&gt;k = 0&lt;/math&gt; we have &lt;math&gt;a_{k}a_{k+1} = 37\cdot 72&lt;/math&gt;, so when &lt;math&gt;k = \frac{37 \cdot 72}{3} = 888&lt;/math&gt;, &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; will be zero for the first time, which implies that &lt;math&gt;m = \boxed{889}&lt;/math&gt;, our answer.<br /> <br /> <br /> Note: In order for &lt;math&gt;a_{m} = 0&lt;/math&gt; we need &lt;math&gt;a_{m-2}a_{m-1}=3&lt;/math&gt; simply by the recursion definition.<br /> <br /> ==Solution 2==<br /> <br /> Plugging in &lt;math&gt;k = m-1&lt;/math&gt; to the given relation, we get &lt;math&gt;0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}&lt;/math&gt;. Inspecting the value of &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; for small values of &lt;math&gt;k&lt;/math&gt;, we see that &lt;math&gt;a_{k}a_{k+1} = 37\cdot 72 - 3k&lt;/math&gt;. Setting the RHS of this equation equal to &lt;math&gt;3&lt;/math&gt;, we find that &lt;math&gt;m&lt;/math&gt; must be &lt;math&gt; \boxed{889}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> ==Induction Proof==<br /> As above, we experiment with some values of &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt;, conjecturing that &lt;math&gt;a_{m-p}a_{m-p-1}&lt;/math&gt; = &lt;math&gt;3p&lt;/math&gt; ,where &lt;math&gt;m&lt;/math&gt; is a positive integer and so is &lt;math&gt;p&lt;/math&gt;, and we prove this formally using induction. The base case is for &lt;math&gt;p = 1&lt;/math&gt;, &lt;math&gt;a_{m} = a_{m-2} - 3/a_{m-1}&lt;/math&gt; Since &lt;math&gt;a_{m} = 0 &lt;/math&gt;, &lt;math&gt;a_{m-1}a_{m-2} = 3&lt;/math&gt;; from the recursion given in the problem &lt;math&gt;a_{m-p+1} = a_{m-p-1} - 3/a_{m-p}&lt;/math&gt;, so &lt;math&gt;a_{m-p+1} = 3p/a_{m-p} - 3/a_{m-p} = 3(p-1)/a_{m-p}&lt;/math&gt;, so &lt;math&gt;a_{m-p}a_{m-p+1} = a_{m-(p-1)}a_{m-(p-1)-1} = 3(p-1)&lt;/math&gt;, hence proving our formula by induction.<br /> ~USAMO2023<br /> <br /> ==Solution 3 (Telescoping)==<br /> Note that &lt;math&gt; a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3 &lt;/math&gt;. Then, we can generate a sum of series of equations such that &lt;math&gt;\sum_{k=1}^{m-1} a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3(m-1)&lt;/math&gt;. Then, note that all but the first and last terms on the LHS cancel out, leaving us with &lt;math&gt;a_m\cdot a_{m-1} - a_1\cdot a_0 = -3(m-1)&lt;/math&gt;. Plugging in &lt;math&gt;a_m=0&lt;/math&gt;, &lt;math&gt;a_0=37&lt;/math&gt;, &lt;math&gt;a_1=72&lt;/math&gt;, we have &lt;math&gt;-37\cdot 72 = -3(m-1)&lt;/math&gt;. Solving for &lt;math&gt;m&lt;/math&gt; gives &lt;math&gt;m=\boxed{889}&lt;/math&gt;. <br /> ~sigma<br /> <br /> <br /> ==Video solution==<br /> <br /> https://www.youtube.com/watch?v=JfxNr7lv7iQ<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_11&diff=182753 2005 AIME II Problems/Problem 11 2022-11-21T22:51:29Z <p>Sigmapie: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;m &lt;/math&gt; be a positive integer, and let &lt;math&gt; a_0, a_1,\ldots,a_m &lt;/math&gt; be a sequence of reals such that &lt;math&gt;a_0 = 37, a_1 = 72, a_m = 0, &lt;/math&gt; and &lt;math&gt; a_{k+1} = a_{k-1} - \frac 3{a_k} &lt;/math&gt; for &lt;math&gt; k = 1,2,\ldots, m-1. &lt;/math&gt; Find &lt;math&gt;m. &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> For &lt;math&gt;0 &lt; k &lt; m&lt;/math&gt;, we have<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;a_{k}a_{k+1} = a_{k-1}a_{k} - 3 &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> Thus the product &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; is a [[monovariant]]: it decreases by 3 each time &lt;math&gt;k&lt;/math&gt; increases by 1. For &lt;math&gt;k = 0&lt;/math&gt; we have &lt;math&gt;a_{k}a_{k+1} = 37\cdot 72&lt;/math&gt;, so when &lt;math&gt;k = \frac{37 \cdot 72}{3} = 888&lt;/math&gt;, &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; will be zero for the first time, which implies that &lt;math&gt;m = \boxed{889}&lt;/math&gt;, our answer.<br /> <br /> <br /> Note: In order for &lt;math&gt;a_{m} = 0&lt;/math&gt; we need &lt;math&gt;a_{m-2}a_{m-1}=3&lt;/math&gt; simply by the recursion definition.<br /> <br /> ==Solution 2==<br /> <br /> Plugging in &lt;math&gt;k = m-1&lt;/math&gt; to the given relation, we get &lt;math&gt;0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}&lt;/math&gt;. Inspecting the value of &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; for small values of &lt;math&gt;k&lt;/math&gt;, we see that &lt;math&gt;a_{k}a_{k+1} = 37\cdot 72 - 3k&lt;/math&gt;. Setting the RHS of this equation equal to &lt;math&gt;3&lt;/math&gt;, we find that &lt;math&gt;m&lt;/math&gt; must be &lt;math&gt; \boxed{889}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> ==Induction Proof==<br /> As above, we experiment with some values of &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt;, conjecturing that &lt;math&gt;a_{m-p}a_{m-p-1}&lt;/math&gt; = &lt;math&gt;3p&lt;/math&gt; ,where &lt;math&gt;m&lt;/math&gt; is a positive integer and so is &lt;math&gt;p&lt;/math&gt;, and we prove this formally using induction. The base case is for &lt;math&gt;p = 1&lt;/math&gt;, &lt;math&gt;a_{m} = a_{m-2} - 3/a_{m-1}&lt;/math&gt; Since &lt;math&gt;a_{m} = 0 &lt;/math&gt;, &lt;math&gt;a_{m-1}a_{m-2} = 3&lt;/math&gt;; from the recursion given in the problem &lt;math&gt;a_{m-p+1} = a_{m-p-1} - 3/a_{m-p}&lt;/math&gt;, so &lt;math&gt;a_{m-p+1} = 3p/a_{m-p} - 3/a_{m-p} = 3(p-1)/a_{m-p}&lt;/math&gt;, so &lt;math&gt;a_{m-p}a_{m-p+1} = a_{m-(p-1)}a_{m-(p-1)-1} = 3(p-1)&lt;/math&gt;, hence proving our formula by induction.<br /> ~USAMO2023<br /> <br /> ==Solution 3 (Telescoping)==<br /> Note that &lt;math&gt; a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3 &lt;/math&gt;. Then, we can generate a series of equations such that &lt;math&gt;\sum_{k=1}^{m-1} a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3(m-1)&lt;/math&gt;. Then, note that all but the first and last terms on the LHS cancel out, leaving us with &lt;math&gt;a_m\cdot a_{m-1} - a_1\cdot a_0 = -3(m-1)&lt;/math&gt;. Plugging in &lt;math&gt;a_m=0&lt;/math&gt;, &lt;math&gt;a_0=37&lt;/math&gt;, &lt;math&gt;a_1=72&lt;/math&gt;, we have &lt;math&gt;-37\cdot 72 = -3(m-1)&lt;/math&gt;. Solving for &lt;math&gt;m&lt;/math&gt; gives &lt;math&gt;m=\boxed{889}&lt;/math&gt;. <br /> ~sigma<br /> <br /> <br /> ==Video solution==<br /> <br /> https://www.youtube.com/watch?v=JfxNr7lv7iQ<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_11&diff=182752 2005 AIME II Problems/Problem 11 2022-11-21T22:51:12Z <p>Sigmapie: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;m &lt;/math&gt; be a positive integer, and let &lt;math&gt; a_0, a_1,\ldots,a_m &lt;/math&gt; be a sequence of reals such that &lt;math&gt;a_0 = 37, a_1 = 72, a_m = 0, &lt;/math&gt; and &lt;math&gt; a_{k+1} = a_{k-1} - \frac 3{a_k} &lt;/math&gt; for &lt;math&gt; k = 1,2,\ldots, m-1. &lt;/math&gt; Find &lt;math&gt;m. &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> For &lt;math&gt;0 &lt; k &lt; m&lt;/math&gt;, we have<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;a_{k}a_{k+1} = a_{k-1}a_{k} - 3 &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> Thus the product &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; is a [[monovariant]]: it decreases by 3 each time &lt;math&gt;k&lt;/math&gt; increases by 1. For &lt;math&gt;k = 0&lt;/math&gt; we have &lt;math&gt;a_{k}a_{k+1} = 37\cdot 72&lt;/math&gt;, so when &lt;math&gt;k = \frac{37 \cdot 72}{3} = 888&lt;/math&gt;, &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; will be zero for the first time, which implies that &lt;math&gt;m = \boxed{889}&lt;/math&gt;, our answer.<br /> <br /> <br /> Note: In order for &lt;math&gt;a_{m} = 0&lt;/math&gt; we need &lt;math&gt;a_{m-2}a_{m-1}=3&lt;/math&gt; simply by the recursion definition.<br /> <br /> ==Solution 2==<br /> <br /> Plugging in &lt;math&gt;k = m-1&lt;/math&gt; to the given relation, we get &lt;math&gt;0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}&lt;/math&gt;. Inspecting the value of &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; for small values of &lt;math&gt;k&lt;/math&gt;, we see that &lt;math&gt;a_{k}a_{k+1} = 37\cdot 72 - 3k&lt;/math&gt;. Setting the RHS of this equation equal to &lt;math&gt;3&lt;/math&gt;, we find that &lt;math&gt;m&lt;/math&gt; must be &lt;math&gt; \boxed{889}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> ==Induction Proof==<br /> As above, we experiment with some values of &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt;, conjecturing that &lt;math&gt;a_{m-p}a_{m-p-1}&lt;/math&gt; = &lt;math&gt;3p&lt;/math&gt; ,where &lt;math&gt;m&lt;/math&gt; is a positive integer and so is &lt;math&gt;p&lt;/math&gt;, and we prove this formally using induction. The base case is for &lt;math&gt;p = 1&lt;/math&gt;, &lt;math&gt;a_{m} = a_{m-2} - 3/a_{m-1}&lt;/math&gt; Since &lt;math&gt;a_{m} = 0 &lt;/math&gt;, &lt;math&gt;a_{m-1}a_{m-2} = 3&lt;/math&gt;; from the recursion given in the problem &lt;math&gt;a_{m-p+1} = a_{m-p-1} - 3/a_{m-p}&lt;/math&gt;, so &lt;math&gt;a_{m-p+1} = 3p/a_{m-p} - 3/a_{m-p} = 3(p-1)/a_{m-p}&lt;/math&gt;, so &lt;math&gt;a_{m-p}a_{m-p+1} = a_{m-(p-1)}a_{m-(p-1)-1} = 3(p-1)&lt;/math&gt;, hence proving our formula by induction.<br /> ~USAMO2023<br /> <br /> ==Solution 3 (Telescoping)==<br /> Note that &lt;math&gt; a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3 &lt;/math&gt;. Then, we can generate a series of equations such that &lt;math&gt;\sum_{k=1}^{m-1} a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3(m-1)&lt;/math&gt;. Then, note that all but the first and last terms on the LHS cancel out, leaving us with &lt;math&gt;a_m\cdot a_{m-1} - a_1\cdot a_0 = -3(m-1)&lt;/math&gt;. Plugging in &lt;math&gt;a_m=0&lt;/math&gt;, &lt;math&gt;a_0=37&lt;/math&gt;, &lt;math&gt;a_1=72&lt;/math&gt;, we have -37\cdot 72 = -3(m-1)&lt;math&gt;. Solving for &lt;/math&gt;m&lt;math&gt; gives &lt;/math&gt;m=\boxed{889}. <br /> ~sigma<br /> <br /> <br /> ==Video solution==<br /> <br /> https://www.youtube.com/watch?v=JfxNr7lv7iQ<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_17&diff=181587 2022 AMC 12A Problems/Problem 17 2022-11-16T04:13:21Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> <br /> Supppose &lt;math&gt;a&lt;/math&gt; is a real number such that the equation &lt;cmath&gt;a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}&lt;/cmath&gt;<br /> has more than one solution in the interval &lt;math&gt;(0, \pi)&lt;/math&gt;. The set of all such &lt;math&gt;a&lt;/math&gt; that can be written<br /> in the form &lt;cmath&gt;(p,q) \cup (q,r),&lt;/cmath&gt;<br /> where &lt;math&gt;p, q,&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are real numbers with &lt;math&gt;p &lt; q&lt; r&lt;/math&gt;. What is &lt;math&gt;p+q+r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -4 \qquad \textbf{(B) } -1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> We are given that &lt;math&gt;a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}&lt;/math&gt;<br /> <br /> Using the sine double angle formula combine with the fact that &lt;math&gt;\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}-1)&lt;/math&gt;, which can be derived using sine angle addition with &lt;math&gt;\sin{(2x + x)}&lt;/math&gt;, we have &lt;cmath&gt;a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)&lt;/cmath&gt;<br /> Since &lt;math&gt;\sin{x} \ne 0&lt;/math&gt; as it is on the open interval &lt;math&gt;(0, \pi)&lt;/math&gt;, we can divide out &lt;math&gt;\sin{x}&lt;/math&gt; from both sides, leaving us with &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt;<br /> Now, distributing &lt;math&gt;a&lt;/math&gt; and rearranging, we achieve the equation &lt;cmath&gt;4\cos^2{x} - 2a\cos{x} - (1+a) = 0&lt;/cmath&gt; which is a quadratic in &lt;math&gt;\cos{x}&lt;/math&gt;.<br /> <br /> Applying the quadratic formula to solve for &lt;math&gt;\cos{x}&lt;/math&gt;, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}&lt;/cmath&gt; and expanding the terms under the radical, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}&lt;/cmath&gt;<br /> Factoring, since &lt;math&gt;4a^2+16a+16 = (2a+4)^2&lt;/math&gt;, we can simplify our expression even further to &lt;cmath&gt;\cos{x} =\frac{a\pm(a+2)}{4}&lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - DavidHovey<br /> <br /> ==Solution 2==<br /> <br /> We can optimize from the step from &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt; in solution 1 by writing<br /> <br /> &lt;cmath&gt;a = \frac{4\cos^2{x}-1}{1+2\cos{x}} = 2\cos x - 1&lt;/cmath&gt;<br /> <br /> and then get<br /> &lt;cmath&gt;<br /> \cos x = \frac{a+1}{2}.<br /> &lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - Dan<br /> <br /> ==Solution 3==<br /> Use the sum to product formula to obtain &lt;math&gt;2a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{3x}&lt;/math&gt;. Use the double angle formula on the RHS to obtain &lt;math&gt;a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{\frac{3x}{2}}\cos{\frac{3x}{2}}&lt;/math&gt;. From here, it is obvious that &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt; is always a solution, and thus we divide by &lt;math&gt;\sin{\frac{3x}{2}}&lt;/math&gt; to get &lt;cmath&gt;a\cdot\cos{\frac{x}{2}}=\cos{\frac{3x}{2}}&lt;/cmath&gt; We wish to find all &lt;math&gt;a&lt;/math&gt; such that there is at least one more solution to this equation distinct from &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;. Letting &lt;math&gt;y=\cos{\frac{x}{2}}&lt;/math&gt;, and noting that &lt;math&gt;\cos{\frac{3x}{2}}=4y^3-3y&lt;/math&gt;, we can rearrange our equation to &lt;math&gt;4y^3=y(3+a)&lt;/math&gt; The smallest value &lt;math&gt;x&lt;/math&gt; where &lt;math&gt;y=0&lt;/math&gt; is &lt;math&gt;\pi&lt;/math&gt;, which is not in our domain so we divide by &lt;math&gt;y&lt;/math&gt; to obtain &lt;math&gt;4y^2=a+3&lt;/math&gt;. By the trivial inequality, &lt;math&gt;a+3\ge{0}&lt;/math&gt;. Furthermore, &lt;math&gt;y\neq{0}&lt;/math&gt;, so &lt;math&gt;a+3&gt;0&lt;/math&gt;. Also, if &lt;math&gt;a=-2&lt;/math&gt;, then the solution to this equation would be shared with &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;, so there would only be one distinct solution. Finally, because &lt;math&gt;y\le{1}&lt;/math&gt; due to the restrictions of a sine wave, and that &lt;math&gt;y\neq{1}&lt;/math&gt; due to the restrictions on &lt;math&gt;x&lt;/math&gt;, we have &lt;math&gt;-3&lt;a&lt;1&lt;/math&gt; with &lt;math&gt;a\neq{-2}&lt;/math&gt;. Thus, &lt;math&gt;p=-3,q=-2, r=1&lt;/math&gt;, so our final answer is &lt;math&gt;-3+(-2)+1=\boxed{\textbf{(A) -4}}&lt;/math&gt;. <br /> <br /> ~sigma<br /> <br /> ==Video Solution 1 (Quick and Simple)==<br /> https://youtu.be/Tl5hBEkHzbA<br /> <br /> ~Education, the Study of Everything<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2022|ab=A|num-b=16|num-a=18}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_17&diff=181586 2022 AMC 12A Problems/Problem 17 2022-11-16T04:10:48Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> <br /> Supppose &lt;math&gt;a&lt;/math&gt; is a real number such that the equation &lt;cmath&gt;a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}&lt;/cmath&gt;<br /> has more than one solution in the interval &lt;math&gt;(0, \pi)&lt;/math&gt;. The set of all such &lt;math&gt;a&lt;/math&gt; that can be written<br /> in the form &lt;cmath&gt;(p,q) \cup (q,r),&lt;/cmath&gt;<br /> where &lt;math&gt;p, q,&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are real numbers with &lt;math&gt;p &lt; q&lt; r&lt;/math&gt;. What is &lt;math&gt;p+q+r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -4 \qquad \textbf{(B) } -1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> We are given that &lt;math&gt;a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}&lt;/math&gt;<br /> <br /> Using the sine double angle formula combine with the fact that &lt;math&gt;\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}-1)&lt;/math&gt;, which can be derived using sine angle addition with &lt;math&gt;\sin{(2x + x)}&lt;/math&gt;, we have &lt;cmath&gt;a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)&lt;/cmath&gt;<br /> Since &lt;math&gt;\sin{x} \ne 0&lt;/math&gt; as it is on the open interval &lt;math&gt;(0, \pi)&lt;/math&gt;, we can divide out &lt;math&gt;\sin{x}&lt;/math&gt; from both sides, leaving us with &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt;<br /> Now, distributing &lt;math&gt;a&lt;/math&gt; and rearranging, we achieve the equation &lt;cmath&gt;4\cos^2{x} - 2a\cos{x} - (1+a) = 0&lt;/cmath&gt; which is a quadratic in &lt;math&gt;\cos{x}&lt;/math&gt;.<br /> <br /> Applying the quadratic formula to solve for &lt;math&gt;\cos{x}&lt;/math&gt;, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}&lt;/cmath&gt; and expanding the terms under the radical, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}&lt;/cmath&gt;<br /> Factoring, since &lt;math&gt;4a^2+16a+16 = (2a+4)^2&lt;/math&gt;, we can simplify our expression even further to &lt;cmath&gt;\cos{x} =\frac{a\pm(a+2)}{4}&lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - DavidHovey<br /> <br /> ==Solution 2==<br /> <br /> We can optimize from the step from &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt; in solution 1 by writing<br /> <br /> &lt;cmath&gt;a = \frac{4\cos^2{x}-1}{1+2\cos{x}} = 2\cos x - 1&lt;/cmath&gt;<br /> <br /> and then get<br /> &lt;cmath&gt;<br /> \cos x = \frac{a+1}{2}.<br /> &lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - Dan<br /> <br /> ==Solution 3==<br /> Use the sum to product formula to obtain &lt;math&gt;2a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{3x}&lt;/math&gt;. Use the double angle formula on the RHS to obtain &lt;math&gt;a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{\frac{3x}{2}}\cos{\frac{3x}{2}}&lt;/math&gt;. From here, it is obvious that &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt; is always a solution, and thus we divide by &lt;math&gt;\sin{\frac{3x}{2}}&lt;/math&gt; to get &lt;cmath&gt;a\cdot\cos{\frac{x}{2}}=2\cos{\frac{3x}{2}}&lt;/cmath&gt; We wish to find all &lt;math&gt;a&lt;/math&gt; such that there is at least one more solution to this equation distinct from &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;. Letting &lt;math&gt;y=\cos{\frac{x}{2}}&lt;/math&gt;, and noting that &lt;math&gt;\cos{\frac{3x}{2}}=4y^3-3y&lt;/math&gt;, we can rearrange our equation to &lt;math&gt;4y^3=y(3+a)&lt;/math&gt; The smallest value &lt;math&gt;x&lt;/math&gt; where &lt;math&gt;y=0&lt;/math&gt; is &lt;math&gt;\pi&lt;/math&gt;, which is not in our domain so we divide by &lt;math&gt;y&lt;/math&gt; to obtain &lt;math&gt;4y^2=a+3&lt;/math&gt;. By the trivial inequality, &lt;math&gt;a+3\ge{0}&lt;/math&gt;. Furthermore, &lt;math&gt;y\neq{0}&lt;/math&gt;, so &lt;math&gt;a+3&gt;0&lt;/math&gt;. Also, if &lt;math&gt;a=-2&lt;/math&gt;, then the solution to this equation would be shared with &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;, so there would only be one distinct solution. Finally, because &lt;math&gt;y\le{1}&lt;/math&gt; due to the restrictions of a sine wave, and that &lt;math&gt;y\neq{1}&lt;/math&gt; due to the restrictions on &lt;math&gt;x&lt;/math&gt;, we have &lt;math&gt;-3&lt;a&lt;1&lt;/math&gt; with &lt;math&gt;a\neq{-2}&lt;/math&gt;. Thus, &lt;math&gt;p=-3,q=-2, r=1&lt;/math&gt;, so our final answer is &lt;math&gt;-3+(-2)+1=\boxed{\textbf{(A) -4}}&lt;/math&gt;. <br /> <br /> ~sigma<br /> <br /> ==Video Solution 1 (Quick and Simple)==<br /> https://youtu.be/Tl5hBEkHzbA<br /> <br /> ~Education, the Study of Everything<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2022|ab=A|num-b=16|num-a=18}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_17&diff=181585 2022 AMC 12A Problems/Problem 17 2022-11-16T04:08:29Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> <br /> Supppose &lt;math&gt;a&lt;/math&gt; is a real number such that the equation &lt;cmath&gt;a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}&lt;/cmath&gt;<br /> has more than one solution in the interval &lt;math&gt;(0, \pi)&lt;/math&gt;. The set of all such &lt;math&gt;a&lt;/math&gt; that can be written<br /> in the form &lt;cmath&gt;(p,q) \cup (q,r),&lt;/cmath&gt;<br /> where &lt;math&gt;p, q,&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are real numbers with &lt;math&gt;p &lt; q&lt; r&lt;/math&gt;. What is &lt;math&gt;p+q+r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -4 \qquad \textbf{(B) } -1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> We are given that &lt;math&gt;a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}&lt;/math&gt;<br /> <br /> Using the sine double angle formula combine with the fact that &lt;math&gt;\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}-1)&lt;/math&gt;, which can be derived using sine angle addition with &lt;math&gt;\sin{(2x + x)}&lt;/math&gt;, we have &lt;cmath&gt;a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)&lt;/cmath&gt;<br /> Since &lt;math&gt;\sin{x} \ne 0&lt;/math&gt; as it is on the open interval &lt;math&gt;(0, \pi)&lt;/math&gt;, we can divide out &lt;math&gt;\sin{x}&lt;/math&gt; from both sides, leaving us with &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt;<br /> Now, distributing &lt;math&gt;a&lt;/math&gt; and rearranging, we achieve the equation &lt;cmath&gt;4\cos^2{x} - 2a\cos{x} - (1+a) = 0&lt;/cmath&gt; which is a quadratic in &lt;math&gt;\cos{x}&lt;/math&gt;.<br /> <br /> Applying the quadratic formula to solve for &lt;math&gt;\cos{x}&lt;/math&gt;, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}&lt;/cmath&gt; and expanding the terms under the radical, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}&lt;/cmath&gt;<br /> Factoring, since &lt;math&gt;4a^2+16a+16 = (2a+4)^2&lt;/math&gt;, we can simplify our expression even further to &lt;cmath&gt;\cos{x} =\frac{a\pm(a+2)}{4}&lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - DavidHovey<br /> <br /> ==Solution 2==<br /> <br /> We can optimize from the step from &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt; in solution 1 by writing<br /> <br /> &lt;cmath&gt;a = \frac{4\cos^2{x}-1}{1+2\cos{x}} = 2\cos x - 1&lt;/cmath&gt;<br /> <br /> and then get<br /> &lt;cmath&gt;<br /> \cos x = \frac{a+1}{2}.<br /> &lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - Dan<br /> <br /> ==Solution 3==<br /> Use the sum to product formula to obtain &lt;math&gt;2a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{3x}&lt;/math&gt;. Use the double angle formula on the RHS to obtain &lt;math&gt;a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{\frac{3x}{2}}\cos{\frac{3x}{2}}&lt;/math&gt;. From here, it is obvious that &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt; is always a solution, and thus we divide by &lt;math&gt;\sin{\frac{3x}{2}}&lt;/math&gt; to get &lt;cmath&gt;a\cdot\cos{\frac{x}{2}}=2\cos{\frac{3x}{2}}&lt;/cmath&gt; We wish to find all &lt;math&gt;a&lt;/math&gt; such that there is at least one more solution to this equation distinct from &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;. Letting &lt;math&gt;y=\cos{\frac{x}{2}}&lt;/math&gt;, and noting that &lt;math&gt;\cos{\frac{3x}{2}}=4y^3-3y&lt;/math&gt;, we can rearrange our equation to &lt;math&gt;4y^3=y(3+a)&lt;/math&gt; The smallest value &lt;math&gt;x&lt;/math&gt; where &lt;math&gt;y=0&lt;/math&gt; is &lt;math&gt;\pi&lt;/math&gt;, which is not in our domain so we divide by &lt;math&gt;y&lt;/math&gt; to obtain &lt;math&gt;4y^2=a+3&lt;/math&gt;. By the trivial inequality, &lt;math&gt;a+3\ge{0}&lt;/math&gt;. Furthermore, &lt;math&gt;y\neq{0}&lt;/math&gt;, so &lt;math&gt;a+3&gt;0&lt;/math&gt;. Also, if &lt;math&gt;a=-2&lt;/math&gt;, then the solution to this equation would be shared with &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;, so there would only be one distinct solution. Finally, because &lt;math&gt;y&lt;1&lt;/math&gt; due to the restrictions of a sine wave, and that &lt;math&gt;y\neq{1}&lt;/math&gt; due to the restrictions on &lt;math&gt;x&lt;/math&gt;, we have &lt;math&gt;-3&lt;a&lt;1&lt;/math&gt; with &lt;math&gt;a\neq{-2}&lt;/math&gt;. Thus, &lt;math&gt;p=-3,q=-2, r=1&lt;/math&gt;, so our final answer is &lt;math&gt;-3+(-2)+1=\boxed{\textbf{(A) -4}}&lt;/math&gt;. <br /> <br /> ~sigma<br /> <br /> ==Video Solution 1 (Quick and Simple)==<br /> https://youtu.be/Tl5hBEkHzbA<br /> <br /> ~Education, the Study of Everything<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2022|ab=A|num-b=16|num-a=18}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_17&diff=181584 2022 AMC 12A Problems/Problem 17 2022-11-16T04:07:45Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> <br /> Supppose &lt;math&gt;a&lt;/math&gt; is a real number such that the equation &lt;cmath&gt;a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}&lt;/cmath&gt;<br /> has more than one solution in the interval &lt;math&gt;(0, \pi)&lt;/math&gt;. The set of all such &lt;math&gt;a&lt;/math&gt; that can be written<br /> in the form &lt;cmath&gt;(p,q) \cup (q,r),&lt;/cmath&gt;<br /> where &lt;math&gt;p, q,&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are real numbers with &lt;math&gt;p &lt; q&lt; r&lt;/math&gt;. What is &lt;math&gt;p+q+r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -4 \qquad \textbf{(B) } -1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> We are given that &lt;math&gt;a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}&lt;/math&gt;<br /> <br /> Using the sine double angle formula combine with the fact that &lt;math&gt;\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}-1)&lt;/math&gt;, which can be derived using sine angle addition with &lt;math&gt;\sin{(2x + x)}&lt;/math&gt;, we have &lt;cmath&gt;a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)&lt;/cmath&gt;<br /> Since &lt;math&gt;\sin{x} \ne 0&lt;/math&gt; as it is on the open interval &lt;math&gt;(0, \pi)&lt;/math&gt;, we can divide out &lt;math&gt;\sin{x}&lt;/math&gt; from both sides, leaving us with &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt;<br /> Now, distributing &lt;math&gt;a&lt;/math&gt; and rearranging, we achieve the equation &lt;cmath&gt;4\cos^2{x} - 2a\cos{x} - (1+a) = 0&lt;/cmath&gt; which is a quadratic in &lt;math&gt;\cos{x}&lt;/math&gt;.<br /> <br /> Applying the quadratic formula to solve for &lt;math&gt;\cos{x}&lt;/math&gt;, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}&lt;/cmath&gt; and expanding the terms under the radical, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}&lt;/cmath&gt;<br /> Factoring, since &lt;math&gt;4a^2+16a+16 = (2a+4)^2&lt;/math&gt;, we can simplify our expression even further to &lt;cmath&gt;\cos{x} =\frac{a\pm(a+2)}{4}&lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - DavidHovey<br /> <br /> ==Solution 2==<br /> <br /> We can optimize from the step from &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt; in solution 1 by writing<br /> <br /> &lt;cmath&gt;a = \frac{4\cos^2{x}-1}{1+2\cos{x}} = 2\cos x - 1&lt;/cmath&gt;<br /> <br /> and then get<br /> &lt;cmath&gt;<br /> \cos x = \frac{a+1}{2}.<br /> &lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - Dan<br /> <br /> ==Solution 3==<br /> Use the sum to product formula to obtain &lt;math&gt;2a\cdot\sin{\frac{3\pi*x}{2}}\cos{\frac{1\pi*x}{2}}=\sin{3\pi*x}&lt;/math&gt;. Use the double angle formula on the RHS to obtain &lt;math&gt;a\cdot\sin{\frac{3\pi*x}{2}}\cos{\frac{1\pi*x}{2}}=\sin{\frac{3\pi*x}{2}}\cos{\frac{3\pi*x}{2}}&lt;/math&gt;. From here, it is obvious that &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt; is always a solution, and thus we divide by &lt;math&gt;\sin{\frac{3\pi*x}{2}}&lt;/math&gt; to get &lt;cmath&gt;a\cdot\cos{\frac{1\pi*x}{2}}=2\cos{\frac{3\pi*x}{2}}&lt;/cmath&gt; We wish to find all &lt;math&gt;a&lt;/math&gt; such that there is at least one more solution to this equation distinct from &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;. Letting &lt;math&gt;y=\cos{\frac{1\pi*x}{2}}&lt;/math&gt;, and noting that &lt;math&gt;\cos{\frac{3\pi*x}{2}}=4y^3-3y&lt;/math&gt;, we can rearrange our equation to &lt;math&gt;4y^3=y(3+a)&lt;/math&gt; The smallest value &lt;math&gt;x&lt;/math&gt; where &lt;math&gt;y=0&lt;/math&gt; is &lt;math&gt;\pi&lt;/math&gt;, which is not in our domain so we divide by &lt;math&gt;y&lt;/math&gt; to obtain &lt;math&gt;4y^2=a+3&lt;/math&gt;. By the trivial inequality, &lt;math&gt;a+3\ge{0}&lt;/math&gt;. Furthermore, &lt;math&gt;y\neq{0}&lt;/math&gt;, so &lt;math&gt;a+3&gt;0&lt;/math&gt;. Also, if &lt;math&gt;a=-2&lt;/math&gt;, then the solution to this equation would be shared with &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;, so there would only be one distinct solution. Finally, because &lt;math&gt;y&lt;1&lt;/math&gt; due to the restrictions of a sine wave, and that &lt;math&gt;y\neq{1}&lt;/math&gt; due to the restrictions on &lt;math&gt;x&lt;/math&gt;, we have &lt;math&gt;-3&lt;a&lt;1&lt;/math&gt; with &lt;math&gt;a\neq{-2}&lt;/math&gt;. Thus, &lt;math&gt;p=-3,q=-2, r=1&lt;/math&gt;, so our final answer is &lt;math&gt;-3+(-2)+1=\boxed{\textbf{(A) -4}}&lt;/math&gt;. <br /> <br /> ~sigma<br /> <br /> ==Video Solution 1 (Quick and Simple)==<br /> https://youtu.be/Tl5hBEkHzbA<br /> <br /> ~Education, the Study of Everything<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2022|ab=A|num-b=16|num-a=18}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_17&diff=181583 2022 AMC 12A Problems/Problem 17 2022-11-16T04:07:22Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> <br /> Supppose &lt;math&gt;a&lt;/math&gt; is a real number such that the equation &lt;cmath&gt;a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}&lt;/cmath&gt;<br /> has more than one solution in the interval &lt;math&gt;(0, \pi)&lt;/math&gt;. The set of all such &lt;math&gt;a&lt;/math&gt; that can be written<br /> in the form &lt;cmath&gt;(p,q) \cup (q,r),&lt;/cmath&gt;<br /> where &lt;math&gt;p, q,&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are real numbers with &lt;math&gt;p &lt; q&lt; r&lt;/math&gt;. What is &lt;math&gt;p+q+r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -4 \qquad \textbf{(B) } -1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> We are given that &lt;math&gt;a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}&lt;/math&gt;<br /> <br /> Using the sine double angle formula combine with the fact that &lt;math&gt;\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}-1)&lt;/math&gt;, which can be derived using sine angle addition with &lt;math&gt;\sin{(2x + x)}&lt;/math&gt;, we have &lt;cmath&gt;a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)&lt;/cmath&gt;<br /> Since &lt;math&gt;\sin{x} \ne 0&lt;/math&gt; as it is on the open interval &lt;math&gt;(0, \pi)&lt;/math&gt;, we can divide out &lt;math&gt;\sin{x}&lt;/math&gt; from both sides, leaving us with &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt;<br /> Now, distributing &lt;math&gt;a&lt;/math&gt; and rearranging, we achieve the equation &lt;cmath&gt;4\cos^2{x} - 2a\cos{x} - (1+a) = 0&lt;/cmath&gt; which is a quadratic in &lt;math&gt;\cos{x}&lt;/math&gt;.<br /> <br /> Applying the quadratic formula to solve for &lt;math&gt;\cos{x}&lt;/math&gt;, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}&lt;/cmath&gt; and expanding the terms under the radical, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}&lt;/cmath&gt;<br /> Factoring, since &lt;math&gt;4a^2+16a+16 = (2a+4)^2&lt;/math&gt;, we can simplify our expression even further to &lt;cmath&gt;\cos{x} =\frac{a\pm(a+2)}{4}&lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - DavidHovey<br /> <br /> ==Solution 2==<br /> <br /> We can optimize from the step from &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt; in solution 1 by writing<br /> <br /> &lt;cmath&gt;a = \frac{4\cos^2{x}-1}{1+2\cos{x}} = 2\cos x - 1&lt;/cmath&gt;<br /> <br /> and then get<br /> &lt;cmath&gt;<br /> \cos x = \frac{a+1}{2}.<br /> &lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - Dan<br /> <br /> ==Solution 3==<br /> Use the sum to product formula to obtain &lt;math&gt;2a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{3x}&lt;/math&gt;. Use the double angle formula on the RHS to obtain &lt;math&gt;a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{\frac{3x}{2}}\cos{\frac{3x}{2}}&lt;/math&gt;. From here, it is obvious that &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt; is always a solution, and thus we divide by &lt;math&gt;\sin{\frac{3x}{2}}&lt;/math&gt; to get &lt;cmath&gt;a\cdot\cos{\frac{x}{2}}=2\cos{\frac{3x}{2}}&lt;/cmath&gt; We wish to find all &lt;math&gt;a&lt;/math&gt; such that there is at least one more solution to this equation distinct from &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;. Letting &lt;math&gt;y=\cos{\frac{x}{2}}&lt;/math&gt;, and noting that &lt;math&gt;\cos{\frac{3x}{2}}=4y^3-3y&lt;/math&gt;, we can rearrange our equation to &lt;math&gt;4y^3=y(3+a)&lt;/math&gt; The smallest value &lt;math&gt;x&lt;/math&gt; where &lt;math&gt;y=0&lt;/math&gt; is &lt;math&gt;\pi&lt;/math&gt;, which is not in our domain so we divide by &lt;math&gt;y&lt;/math&gt; to obtain &lt;math&gt;4y^2=a+3&lt;/math&gt;. By the trivial inequality, &lt;math&gt;a+3\ge{0}&lt;/math&gt;. Furthermore, &lt;math&gt;y\neq{0}&lt;/math&gt;, so &lt;math&gt;a+3&gt;0&lt;/math&gt;. Also, if &lt;math&gt;a=-2&lt;/math&gt;, then the solution to this equation would be shared with &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;, so there would only be one distinct solution. Finally, because &lt;math&gt;y&lt;1&lt;/math&gt; due to the restrictions of a sine wave, and that &lt;math&gt;y\neq{1}&lt;/math&gt; due to the restrictions on &lt;math&gt;x&lt;/math&gt;, we have &lt;math&gt;-3&lt;a&lt;1&lt;/math&gt; with &lt;math&gt;a\neq{-2}&lt;/math&gt;. Thus, &lt;math&gt;p=-3,q=-2, r=1&lt;/math&gt;, so our final answer is &lt;math&gt;-3+(-2)+1=\boxed{\textbf{(A) -4}}&lt;/math&gt;. <br /> <br /> ~sigma<br /> <br /> ==Video Solution 1 (Quick and Simple)==<br /> https://youtu.be/Tl5hBEkHzbA<br /> <br /> ~Education, the Study of Everything<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2022|ab=A|num-b=16|num-a=18}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_17&diff=181582 2022 AMC 12A Problems/Problem 17 2022-11-16T04:05:08Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> <br /> Supppose &lt;math&gt;a&lt;/math&gt; is a real number such that the equation &lt;cmath&gt;a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}&lt;/cmath&gt;<br /> has more than one solution in the interval &lt;math&gt;(0, \pi)&lt;/math&gt;. The set of all such &lt;math&gt;a&lt;/math&gt; that can be written<br /> in the form &lt;cmath&gt;(p,q) \cup (q,r),&lt;/cmath&gt;<br /> where &lt;math&gt;p, q,&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are real numbers with &lt;math&gt;p &lt; q&lt; r&lt;/math&gt;. What is &lt;math&gt;p+q+r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -4 \qquad \textbf{(B) } -1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> We are given that &lt;math&gt;a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}&lt;/math&gt;<br /> <br /> Using the sine double angle formula combine with the fact that &lt;math&gt;\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}-1)&lt;/math&gt;, which can be derived using sine angle addition with &lt;math&gt;\sin{(2x + x)}&lt;/math&gt;, we have &lt;cmath&gt;a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)&lt;/cmath&gt;<br /> Since &lt;math&gt;\sin{x} \ne 0&lt;/math&gt; as it is on the open interval &lt;math&gt;(0, \pi)&lt;/math&gt;, we can divide out &lt;math&gt;\sin{x}&lt;/math&gt; from both sides, leaving us with &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt;<br /> Now, distributing &lt;math&gt;a&lt;/math&gt; and rearranging, we achieve the equation &lt;cmath&gt;4\cos^2{x} - 2a\cos{x} - (1+a) = 0&lt;/cmath&gt; which is a quadratic in &lt;math&gt;\cos{x}&lt;/math&gt;.<br /> <br /> Applying the quadratic formula to solve for &lt;math&gt;\cos{x}&lt;/math&gt;, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}&lt;/cmath&gt; and expanding the terms under the radical, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}&lt;/cmath&gt;<br /> Factoring, since &lt;math&gt;4a^2+16a+16 = (2a+4)^2&lt;/math&gt;, we can simplify our expression even further to &lt;cmath&gt;\cos{x} =\frac{a\pm(a+2)}{4}&lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - DavidHovey<br /> <br /> ==Solution 2==<br /> <br /> We can optimize from the step from &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt; in solution 1 by writing<br /> <br /> &lt;cmath&gt;a = \frac{4\cos^2{x}-1}{1+2\cos{x}} = 2\cos x - 1&lt;/cmath&gt;<br /> <br /> and then get<br /> &lt;cmath&gt;<br /> \cos x = \frac{a+1}{2}.<br /> &lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - Dan<br /> <br /> ==Solution 3==<br /> Use the sum to product formula to obtain &lt;math&gt;2a\cdot\sin{\frac{3\pi*x}{2}}\cos{\frac{1\pi*x}{2}}=\sin{3\pi*x}&lt;/math&gt;. Use the double angle formula on the RHS to obtain &lt;math&gt;a\cdot\sin{\frac{3\pi*x}{2}}\cos{\frac{1\pi*x}{2}}=\sin{\frac{3\pi*x}{2}}\cos{\frac{3\pi*x}{2}}&lt;/math&gt;. From here, it is obvious that &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt; is always a solution, and thus we divide by &lt;math&gt;\sin{\frac{3\pi*x}{2}}&lt;/math&gt; to get &lt;cmath&gt;a\cdot\cos{\frac{1\pi*x}{2}}=2\cos{\frac{3\pi*x}{2}}&lt;/cmath&gt; We wish to find all &lt;math&gt;a&lt;/math&gt; such that there is at least one more solution to this equation distinct from &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;. Letting &lt;math&gt;y=\cos{\frac{1\pi*x}{2}}&lt;/math&gt;, and noting that &lt;math&gt;\cos{\frac{3\pi*x}{2}}=4y^3-3y&lt;/math&gt;, we can rearrange our equation to &lt;math&gt;4y^3=y(3+a)&lt;/math&gt; The smallest value &lt;math&gt;x&lt;/math&gt; where &lt;math&gt;y=0&lt;/math&gt; is &lt;math&gt;\pi&lt;/math&gt;, which is not in our domain so we divide by &lt;math&gt;y&lt;/math&gt; to obtain &lt;math&gt;4y^2=a+3&lt;/math&gt;. By the trivial inequality, &lt;math&gt;a+3\ge{0}&lt;/math&gt;. Furthermore, &lt;math&gt;y\neq{0}&lt;/math&gt;, so &lt;math&gt;a+3&gt;0&lt;/math&gt;. Also, if &lt;math&gt;a=-2&lt;/math&gt;, then the solution to this equation would be shared with &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;, so there would only be one distinct solution. Finally, because &lt;math&gt;y&lt;1&lt;/math&gt; due to the restrictions of a sine wave, and that &lt;math&gt;y\neq{1}&lt;/math&gt; due to the restrictions on &lt;math&gt;x&lt;/math&gt;, we have &lt;math&gt;-3&lt;a&lt;1&lt;/math&gt; with &lt;math&gt;a\neq{-2}&lt;/math&gt;. Thus, &lt;math&gt;p=-3,q=-2, r=1&lt;/math&gt;, so our final answer is &lt;math&gt;-3+(-2)+1=\boxed{\textbf{(A) -4}}&lt;/math&gt;. <br /> <br /> ~sigma<br /> <br /> ==Video Solution 1 (Quick and Simple)==<br /> https://youtu.be/Tl5hBEkHzbA<br /> <br /> ~Education, the Study of Everything<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2022|ab=A|num-b=16|num-a=18}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_17&diff=181581 2022 AMC 12A Problems/Problem 17 2022-11-16T03:57:37Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> <br /> Supppose &lt;math&gt;a&lt;/math&gt; is a real number such that the equation &lt;cmath&gt;a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}&lt;/cmath&gt;<br /> has more than one solution in the interval &lt;math&gt;(0, \pi)&lt;/math&gt;. The set of all such &lt;math&gt;a&lt;/math&gt; that can be written<br /> in the form &lt;cmath&gt;(p,q) \cup (q,r),&lt;/cmath&gt;<br /> where &lt;math&gt;p, q,&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are real numbers with &lt;math&gt;p &lt; q&lt; r&lt;/math&gt;. What is &lt;math&gt;p+q+r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -4 \qquad \textbf{(B) } -1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> We are given that &lt;math&gt;a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}&lt;/math&gt;<br /> <br /> Using the sine double angle formula combine with the fact that &lt;math&gt;\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}-1)&lt;/math&gt;, which can be derived using sine angle addition with &lt;math&gt;\sin{(2x + x)}&lt;/math&gt;, we have &lt;cmath&gt;a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)&lt;/cmath&gt;<br /> Since &lt;math&gt;\sin{x} \ne 0&lt;/math&gt; as it is on the open interval &lt;math&gt;(0, \pi)&lt;/math&gt;, we can divide out &lt;math&gt;\sin{x}&lt;/math&gt; from both sides, leaving us with &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt;<br /> Now, distributing &lt;math&gt;a&lt;/math&gt; and rearranging, we achieve the equation &lt;cmath&gt;4\cos^2{x} - 2a\cos{x} - (1+a) = 0&lt;/cmath&gt; which is a quadratic in &lt;math&gt;\cos{x}&lt;/math&gt;.<br /> <br /> Applying the quadratic formula to solve for &lt;math&gt;\cos{x}&lt;/math&gt;, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}&lt;/cmath&gt; and expanding the terms under the radical, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}&lt;/cmath&gt;<br /> Factoring, since &lt;math&gt;4a^2+16a+16 = (2a+4)^2&lt;/math&gt;, we can simplify our expression even further to &lt;cmath&gt;\cos{x} =\frac{a\pm(a+2)}{4}&lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - DavidHovey<br /> <br /> ==Solution 2==<br /> <br /> We can optimize from the step from &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt; in solution 1 by writing<br /> <br /> &lt;cmath&gt;a = \frac{4\cos^2{x}-1}{1+2\cos{x}} = 2\cos x - 1&lt;/cmath&gt;<br /> <br /> and then get<br /> &lt;cmath&gt;<br /> \cos x = \frac{a+1}{2}.<br /> &lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - Dan<br /> <br /> ==Solution 3==<br /> Use the sum to product formula to obtain &lt;math&gt;a\cdot\sin{\frac{3\pi*x}{2}}\cos{\frac{1\pi*x}{2}}=\sin{3\pi*x}&lt;/math&gt;. Use the double angle formula on the RHS to obtain &lt;math&gt;a\cdot\sin{\frac{3\pi*x}{2}}\cos{\frac{1\pi*x}{2}}=\sin{\frac{3\pi*x}{2}}\cos{\frac{3\pi*x}{2}}&lt;/math&gt;. From here, it is obvious that &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt; is always a solution, and thus we divide by &lt;math&gt;\sin{\frac{3\pi*x}{2}}&lt;/math&gt; to get &lt;cmath&gt;a\cdot\cos{\frac{1\pi*x}{2}}=\cos{\frac{3\pi*x}{2}}&lt;/cmath&gt; We wish to find all &lt;math&gt;a&lt;/math&gt; such that there is at least one more solution to this equation distinct from &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;. Letting &lt;math&gt;y=\cos{\frac{1\pi*x}{2}}&lt;/math&gt;, and noting that &lt;math&gt;\cos{\frac{3\pi*x}{2}}=4y^3-3y&lt;/math&gt;, we can rearrange our equation to &lt;math&gt;4y^3=y(3+a)&lt;/math&gt; The smallest value &lt;math&gt;x&lt;/math&gt; where &lt;math&gt;y=0&lt;/math&gt; is &lt;math&gt;\pi&lt;/math&gt;, which is not in our domain so we divide by &lt;math&gt;y&lt;/math&gt; to obtain &lt;math&gt;4y^2=a+3&lt;/math&gt;. By the trivial inequality, &lt;math&gt;a+3\ge{0}&lt;/math&gt;. Furthermore, &lt;math&gt;y\neq{0}&lt;/math&gt;, so &lt;math&gt;a+3&gt;0&lt;/math&gt;. Also, if &lt;math&gt;a=-2&lt;/math&gt;, then the solution to this equation would be shared with &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;, so there would only be one distinct solution. Finally, because &lt;math&gt;y&lt;1&lt;/math&gt; due to the restrictions of a sine wave, and that &lt;math&gt;y\neq{1}&lt;/math&gt; due to the restrictions on &lt;math&gt;x&lt;/math&gt;, we have &lt;math&gt;-3&lt;a&lt;1&lt;/math&gt; with &lt;math&gt;a\neq{-2}&lt;/math&gt;. Thus, &lt;math&gt;p=-3,q=-2, r=1&lt;/math&gt;, so our final answer is &lt;math&gt;-3+(-2)+1=\boxed{\textbf{(A) -4}}&lt;/math&gt;. <br /> <br /> ~sigma<br /> <br /> ==Video Solution 1 (Quick and Simple)==<br /> https://youtu.be/Tl5hBEkHzbA<br /> <br /> ~Education, the Study of Everything<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2022|ab=A|num-b=16|num-a=18}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_17&diff=181580 2022 AMC 12A Problems/Problem 17 2022-11-16T03:57:21Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> <br /> Supppose &lt;math&gt;a&lt;/math&gt; is a real number such that the equation &lt;cmath&gt;a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}&lt;/cmath&gt;<br /> has more than one solution in the interval &lt;math&gt;(0, \pi)&lt;/math&gt;. The set of all such &lt;math&gt;a&lt;/math&gt; that can be written<br /> in the form &lt;cmath&gt;(p,q) \cup (q,r),&lt;/cmath&gt;<br /> where &lt;math&gt;p, q,&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are real numbers with &lt;math&gt;p &lt; q&lt; r&lt;/math&gt;. What is &lt;math&gt;p+q+r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -4 \qquad \textbf{(B) } -1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> We are given that &lt;math&gt;a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}&lt;/math&gt;<br /> <br /> Using the sine double angle formula combine with the fact that &lt;math&gt;\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}-1)&lt;/math&gt;, which can be derived using sine angle addition with &lt;math&gt;\sin{(2x + x)}&lt;/math&gt;, we have &lt;cmath&gt;a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)&lt;/cmath&gt;<br /> Since &lt;math&gt;\sin{x} \ne 0&lt;/math&gt; as it is on the open interval &lt;math&gt;(0, \pi)&lt;/math&gt;, we can divide out &lt;math&gt;\sin{x}&lt;/math&gt; from both sides, leaving us with &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt;<br /> Now, distributing &lt;math&gt;a&lt;/math&gt; and rearranging, we achieve the equation &lt;cmath&gt;4\cos^2{x} - 2a\cos{x} - (1+a) = 0&lt;/cmath&gt; which is a quadratic in &lt;math&gt;\cos{x}&lt;/math&gt;.<br /> <br /> Applying the quadratic formula to solve for &lt;math&gt;\cos{x}&lt;/math&gt;, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}&lt;/cmath&gt; and expanding the terms under the radical, we get &lt;cmath&gt;\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}&lt;/cmath&gt;<br /> Factoring, since &lt;math&gt;4a^2+16a+16 = (2a+4)^2&lt;/math&gt;, we can simplify our expression even further to &lt;cmath&gt;\cos{x} =\frac{a\pm(a+2)}{4}&lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - DavidHovey<br /> <br /> ==Solution 2==<br /> <br /> We can optimize from the step from &lt;cmath&gt;a\cdot(1+2\cos{x})=4\cos^2{x}-1&lt;/cmath&gt; in solution 1 by writing<br /> <br /> &lt;cmath&gt;a = \frac{4\cos^2{x}-1}{1+2\cos{x}} = 2\cos x - 1&lt;/cmath&gt;<br /> <br /> and then get<br /> &lt;cmath&gt;<br /> \cos x = \frac{a+1}{2}.<br /> &lt;/cmath&gt;<br /> <br /> Now, solving for our two solutions, &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\cos{x} = \frac{a+1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cos{x} = -\frac{1}{2}&lt;/math&gt; yields a solution that is valid for all &lt;math&gt;a&lt;/math&gt;, that being &lt;math&gt;x = \frac{2\pi}{3}&lt;/math&gt;, we must now solve for the case where &lt;math&gt;\frac{a+1}{2}&lt;/math&gt; yields a valid value.<br /> <br /> As &lt;math&gt;x\in (0, \pi)&lt;/math&gt;, &lt;math&gt;\cos{x}\in (-1, 1)&lt;/math&gt;, and therefore &lt;math&gt;\frac{a+1}{2}\in (-1, 1)&lt;/math&gt;, and &lt;math&gt;a\in(-3,1)&lt;/math&gt;.<br /> <br /> There is one more case we must consider inside this interval though, the case where &lt;math&gt;\frac{a+1}{2} = -\frac{1}{2}&lt;/math&gt;, as this would lead to a double root for &lt;math&gt;\cos{x}&lt;/math&gt;, yielding only one valid solution for &lt;math&gt;x&lt;/math&gt;. Solving for this case, &lt;math&gt; a \ne -2&lt;/math&gt;.<br /> <br /> Therefore, combining this fact with our solution interval, &lt;math&gt;a\in(-3, -2) \cup (-2, 1)&lt;/math&gt;, so the answer is &lt;math&gt;-3-2+1 = \boxed{\textbf{(A) -4}}&lt;/math&gt;<br /> <br /> - Dan<br /> <br /> ==Solution 3==<br /> Use the sum to product formula to obtain &lt;math&gt;a\cdot\sin{\frac{3\pi*x}{2}}\cos{\frac{1\pi*x}{2}}=\sin{3\pi*x}&lt;/math&gt;. Use the double angle formula on the RHS to obtain &lt;math&gt;a\cdot\sin{\frac{3\pi*x}{2}}\cos{\frac{1\pi*x}{2}}=\sin{\frac{3\pi*x}{2}}\cos{\frac{3\pi*x}{2}}&lt;/math&gt;. From here, it is obvious that &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt; is always a solution, and thus we divide by &lt;math&gt;\sin{\frac{3\pi*x}{2}}&lt;/math&gt; to get &lt;cmath&gt;a\cdot\cos{\frac{1\pi*x}{2}}=\cos{\frac{3\pi*x}{2}}&lt;/cmath&gt; We wish to find all &lt;math&gt;a&lt;/math&gt; such that there is at least one more solution to this equation distinct from &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;. Letting &lt;math&gt;y=\cos{\frac{1\pi*x}{2}}&lt;/math&gt;, and noting that &lt;math&gt;\cos{\frac{3\pi*x}{2}}=4y^3-3y&lt;/math&gt;, we can rearrange our equation to &lt;math&gt;4y^3=y(3+a)&lt;/math&gt; The smallest value &lt;math&gt;x&lt;/math&gt; where &lt;math&gt;y=0&lt;/math&gt; is &lt;math&gt;\pi&lt;/math&gt;, which is not in our domain so we divide by &lt;math&gt;y&lt;/math&gt; to obtain &lt;math&gt;4y^2=a+3&lt;/math&gt;. By the trivial inequality, &lt;math&gt;a+3\ge{0}&lt;/math&gt;. Furthermore, &lt;math&gt;y\neq{0}&lt;/math&gt;, so &lt;math&gt;a+3&gt;0&lt;/math&gt;. Also, if &lt;math&gt;a=-2&lt;/math&gt;, then the solution to this equation would be shared with &lt;math&gt;x=\frac{2\pi}{3}&lt;/math&gt;, so there would only be one distinct solution. Finally, because &lt;math&gt;y&lt;1&lt;/math&gt; due to the restrictions of a sine wave, and that &lt;math&gt;y\neq{1}&lt;/math&gt; due to the restrictions on &lt;math&gt;x&lt;/math&gt;, we have &lt;math&gt;-3&lt;a&lt;1&lt;/math&gt; with &lt;math&gt;a\neq{-2}&lt;/math&gt;. Thus, &lt;math&gt;p=-3,q=-2, r=1&lt;/math&gt;, so our final answer is &lt;math&gt;-3+(-2)+1=\boxed{\textbf{(A) -4}}&lt;/math&gt;. <br /> <br /> ==Video Solution 1 (Quick and Simple)==<br /> https://youtu.be/Tl5hBEkHzbA<br /> <br /> ~Education, the Study of Everything<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2022|ab=A|num-b=16|num-a=18}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_22&diff=180995 2022 AMC 12A Problems/Problem 22 2022-11-12T16:44:33Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;c&lt;/math&gt; be a real number, and let &lt;math&gt;z_1&lt;/math&gt; and &lt;math&gt;z_2&lt;/math&gt; be the two complex numbers satisfying the equation<br /> &lt;math&gt;z^2 - cz + 10 = 0&lt;/math&gt;. Points &lt;math&gt;z_1&lt;/math&gt;, &lt;math&gt;z_2&lt;/math&gt;, &lt;math&gt;\frac{1}{z_1}&lt;/math&gt;, and &lt;math&gt;\frac{1}{z_2}&lt;/math&gt; are the vertices of (convex) quadrilateral &lt;math&gt;Q&lt;/math&gt; in the complex plane. When the area of &lt;math&gt;Q&lt;/math&gt; obtains its maximum possible value, &lt;math&gt;c&lt;/math&gt; is closest to which of the following?<br /> <br /> ==Solution==<br /> <br /> Because &lt;math&gt;c&lt;/math&gt; is real, &lt;math&gt;z_2 = \bar z_1&lt;/math&gt;.<br /> We have<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> 10 &amp; = z_1 z_2 \\<br /> &amp; = z_1 \bar z_1 \\<br /> &amp; = |z_1|^2 ,<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> where the first equality follows from Vieta's formula.<br /> <br /> Thus, &lt;math&gt;|z_1| = \sqrt{10}&lt;/math&gt;.<br /> <br /> We have<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> c &amp; = z_1 + z_2 \\<br /> &amp; = z_1 + \bar z_1 \\<br /> &amp; = 2 {\rm Re} \ z_1 ,<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> where the first equality follows from Vieta's formula.<br /> <br /> Thus, &lt;math&gt;{\rm Re} \ z_1 = \frac{c}{2}&lt;/math&gt;.<br /> <br /> We have<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \frac{1}{z_1} &amp; = \frac{1}{10} \frac{10}{z_1} \\<br /> &amp; = \frac{1}{10} \frac{z_1 z_2}{z_1} \\<br /> &amp; = \frac{z_2}{10} \\<br /> &amp; = \frac{\bar z_1}{10} .<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> where the second equality follows from Vieta's formula.<br /> <br /> We have<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \frac{1}{z_2} &amp; = \frac{1}{10} \frac{10}{z_2} \\<br /> &amp; = \frac{1}{10} \frac{z_1 z_2}{z_2} \\<br /> &amp; = \frac{z_1}{10} .<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> where the second equality follows from Vieta's formula.<br /> <br /> Therefore,<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> {\rm Area} \ Q<br /> &amp; = \frac{1}{2} \left| {\rm Re} \ z_1 \right|<br /> \cdot 2 \left| {\rm Im} \ z_1 \right|<br /> \cdot \left( 1 - \frac{1}{10^2} \right) \\<br /> &amp; = \frac{1}{2} |c| \sqrt{10 - \frac{c^2}{4}} \left( 1 - \frac{1}{10^2} \right) \\<br /> &amp; = \frac{1 - \frac{1}{10^2}}{4}<br /> \sqrt{c^2 \left( 40 - c^2 \right)} \\<br /> &amp; \leq \frac{1 - \frac{1}{10^2}}{4} \cdot \frac{c^2 + \left( 40 - c^2 \right)}{2} \\<br /> &amp; = \frac{1 - \frac{1}{10^2}}{4} \cdot 20 ,<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> where the inequality follows from the AM-GM inequality, and it is augmented to an equality if and only if &lt;math&gt;c^2 = 40 - c^2&lt;/math&gt;.<br /> Thus, &lt;math&gt;|c| = 2 \sqrt{5} \approx \boxed{\textbf{(A) 4.5}}&lt;/math&gt;.<br /> <br /> ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br /> <br /> ==Solution 2==<br /> Because &lt;math&gt;z^2 - cz + 10 = 0&lt;/math&gt;, notice that &lt;math&gt;|z_1||z_2|=|10|=10&lt;/math&gt;. Furthermore, note that because &lt;math&gt;c&lt;/math&gt; is real, &lt;math&gt;z_2=\bar z_1&lt;/math&gt;. Thus, &lt;math&gt;\frac{1}{z_1}=\frac{\bar z_1}{z_1\cdot{\bar z_1}}=\frac{z_2}{|z_1|^2}=\frac{z_2}{100}&lt;/math&gt;. Similarly, &lt;math&gt;\frac{1}{z_2}=\frac{z_1}{100}&lt;/math&gt;. On the complex coordinate plane, let &lt;math&gt;z_1=A_2&lt;/math&gt;, &lt;math&gt;z_2=B_2&lt;/math&gt;,&lt;math&gt;\frac{1}{z_2}=A_1&lt;/math&gt;, &lt;math&gt;\frac{1}{z_1}=B_1&lt;/math&gt;. Notice how &lt;math&gt;OA_1B_1&lt;/math&gt; is similar to &lt;math&gt;OA_2B_2&lt;/math&gt;. Thus, the area of &lt;math&gt;A_1B_1B_2B_1&lt;/math&gt; is &lt;math&gt;(k)(OA_2B_2)&lt;/math&gt; for some constant &lt;math&gt;k&lt;/math&gt;, and &lt;math&gt;OA_2B_2 = &lt;/math&gt;<br /> (In progress)<br /> <br /> ==Video Solution==<br /> <br /> https://youtu.be/pcB2sg7Ag58<br /> <br /> ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2022|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems&diff=180976 2022 AMC 12A Problems 2022-11-12T16:17:18Z <p>Sigmapie: /* Problem 23 */</p> <hr /> <div>{{AMC12 Problems|year=2022|ab=A}}<br /> ==Problem 1==<br /> <br /> What is the value of &lt;cmath&gt;3+\frac{1}{3+\frac{1}{3+\frac13}}?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)}\ \frac{31}{10}\qquad\textbf{(B)}\ \frac{49}{15}\qquad\textbf{(C)}\ \frac{33}{10}\qquad\textbf{(D)}\ \frac{109}{33}\qquad\textbf{(E)}\ \frac{15}{4}&lt;/math&gt;<br /> <br /> [[2022 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> The sum of three numbers is &lt;math&gt;96.&lt;/math&gt; The first number is &lt;math&gt;6&lt;/math&gt; times the third number, and the third number is &lt;math&gt;40&lt;/math&gt; less than the second number. What is the absolute value of the difference between the first and second numbers?<br /> <br /> &lt;math&gt;\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5&lt;/math&gt;<br /> <br /> [[2022 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> These problems will be posted once the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> The least common multiple of a positive divisor &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;18&lt;/math&gt; is &lt;math&gt;180&lt;/math&gt;, and the greatest common divisor of &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;45&lt;/math&gt; is &lt;math&gt;15&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12&lt;/math&gt;<br /> <br /> [[2022 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> The &lt;math&gt;\textit{taxicab distance}&lt;/math&gt; between points &lt;math&gt;(x_1, y_1)&lt;/math&gt; and &lt;math&gt;(x_2, y_2)&lt;/math&gt; in the coordinate plane is given by &lt;math&gt;|x_1 - x_2| + |y_1 - y_2|&lt;/math&gt;. For how many points &lt;math&gt;P&lt;/math&gt; with integer coordinates is the taxicab distance between &lt;math&gt;P&lt;/math&gt; and the origin less than or equal to &lt;math&gt;20&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \qquad\textbf{(C)} \, 841 \qquad\textbf{(D)} \, 921 \qquad\textbf{(E)} \, 924 &lt;/math&gt;<br /> <br /> <br /> [[2022 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> A data set consists of &lt;math&gt;6&lt;/math&gt; not distinct) positive integers: &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt;. The<br /> average (arithmetic mean) of the &lt;math&gt;6&lt;/math&gt; numbers equals a value in the data set. What is<br /> the sum of all positive values of &lt;math&gt;X&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 10 \qquad \textbf{(B) } 26 \qquad \textbf{(C) } 32 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 40&lt;/math&gt;<br /> <br /> [[2022 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> A rectangle is partitioned into &lt;math&gt;5&lt;/math&gt; regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible?<br /> <br /> &lt;asy&gt; size(5.5cm); draw((0,0)--(0,2)--(2,2)--(2,0)--cycle); draw((2,0)--(8,0)--(8,2)--(2,2)--cycle); draw((8,0)--(12,0)--(12,2)--(8,2)--cycle); draw((0,2)--(6,2)--(6,4)--(0,4)--cycle); draw((6,2)--(12,2)--(12,4)--(6,4)--cycle); &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }120\qquad\textbf{(B) }270\qquad\textbf{(C) }360\qquad\textbf{(D) }540\qquad\textbf{(E) }720&lt;/math&gt;<br /> <br /> [[2022 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> The infinite product<br /> &lt;cmath&gt;\sqrt{10} \cdot \sqrt{\sqrt{10}} \cdot \sqrt{\sqrt{\sqrt{10}}} \ldots&lt;/cmath&gt;<br /> evaluates to a real number. What is that number?<br /> <br /> &lt;math&gt;\textbf{(A) }\sqrt{10}\qquad\textbf{(B) }\sqrt{100}\qquad\textbf{(C) }\sqrt{1000}\qquad\textbf{(D) }10\qquad\textbf{(E) }10\sqrt{10}&lt;/math&gt;<br /> <br /> [[2022 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> On Halloween &lt;math&gt;31&lt;/math&gt; children walked into the principal's office asking for candy. They<br /> can be classified into three types: Some always lie; some always tell the truth; and<br /> some alternately lie and tell the truth. The alternaters arbitrarily choose their first<br /> response, either a lie or the truth, but each subsequent statement has the opposite<br /> truth value from its predecessor. The principal asked everyone the same three<br /> questions in this order.<br /> <br /> &quot;Are you a truth-teller?&quot; The principal gave a piece of candy to each of the &lt;math&gt;22&lt;/math&gt;<br /> children who answered yes.<br /> <br /> &quot;Are you an alternater?&quot; The principal gave a piece of candy to each of the &lt;math&gt;15&lt;/math&gt;<br /> children who answered yes.<br /> <br /> &quot;Are you a liar?&quot; The principal gave a piece of candy to each of the &lt;math&gt;9&lt;/math&gt; children who<br /> answered yes.<br /> <br /> How many pieces of candy in all did the principal give to the children who always<br /> tell the truth?<br /> <br /> &lt;math&gt;\textbf{(A) } 7 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 31&lt;/math&gt;<br /> <br /> [[2022 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> How many ways are there to split the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;14&lt;/math&gt; into &lt;math&gt;7&lt;/math&gt; pairs such that in each pair, the greater number is at least &lt;math&gt;2&lt;/math&gt; times the lesser number?<br /> <br /> &lt;math&gt;\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144&lt;/math&gt;<br /> <br /> [[2022 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> What is the product of all real numbers &lt;math&gt;x&lt;/math&gt; such that the distance on the number line between &lt;math&gt;\log_6x&lt;/math&gt; and &lt;math&gt;\log_69&lt;/math&gt; is twice the distance on the number line between &lt;math&gt;\log_610&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 10 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 25 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 81&lt;/math&gt;<br /> <br /> <br /> [[2022 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Let &lt;math&gt;M&lt;/math&gt; be the midpoint of &lt;math&gt;AB&lt;/math&gt; in regular tetrahedron &lt;math&gt;ABCD&lt;/math&gt;. What is &lt;math&gt;\cos(\angle CMD)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac14 \qquad \textbf{(B) } \frac13 \qquad \textbf{(C) } \frac25 \qquad \textbf{(D) } \frac12 \qquad \textbf{(E) } \frac{\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> [[2022 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Let &lt;math&gt;\mathcal{R}&lt;/math&gt; be the region in the complex plane consisting of all complex numbers &lt;math&gt;z&lt;/math&gt; that can be written as the sum of complex numbers &lt;math&gt;z_1&lt;/math&gt; and &lt;math&gt;z_2&lt;/math&gt;, where &lt;math&gt;z_1&lt;/math&gt; lies on the segment with endpoints &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;4i&lt;/math&gt;, and &lt;math&gt;z_2&lt;/math&gt; has magnitude at most &lt;math&gt;1&lt;/math&gt;. What integer is closest to the area of &lt;math&gt;\mathcal{R}&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17&lt;/math&gt;<br /> <br /> <br /> [[2022 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> These problems will be posted once the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> These problems will be posted once the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> These problems will be posted once the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> These problems will be posted once the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> These problems will be posted once the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> These problems will be posted once the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> These problems will be posted once the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> These problems will be posted once the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> These problems will be posted once the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Let &lt;math&gt;h_n&lt;/math&gt; and &lt;math&gt;k_n&lt;/math&gt; be the unique relatively prime positive integers such that &lt;cmath&gt;\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...\frac{1}{n}=\frac{h_n}{k_n}.&lt;/cmath&gt; Let &lt;math&gt;L_n&lt;/math&gt; denote the least common multiple of the numbers &lt;math&gt;1, 2, 3, ..., n&lt;/math&gt;. For how many integers with &lt;math&gt;1\le{n}\le{22}&lt;/math&gt; is &lt;math&gt;k_n&lt;L_n&lt;/math&gt;?<br /> <br /> <br /> <br /> <br /> <br /> [[2022 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> How many strings of length &lt;math&gt;5&lt;/math&gt; formed from the digits &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt; are there such that for each &lt;math&gt;j \in \{1,2,3,4\}&lt;/math&gt;, at least &lt;math&gt;j&lt;/math&gt; of the digits are less than &lt;math&gt;j&lt;/math&gt;? (For example, &lt;math&gt;02214&lt;/math&gt; satisfies this condition<br /> because it contains at least &lt;math&gt;1&lt;/math&gt; digit less than &lt;math&gt;1&lt;/math&gt;, at least &lt;math&gt;2&lt;/math&gt; digits less than &lt;math&gt;2&lt;/math&gt;, at least &lt;math&gt;3&lt;/math&gt; digits less<br /> than &lt;math&gt;3&lt;/math&gt;, and at least &lt;math&gt;4&lt;/math&gt; digits less than &lt;math&gt;4&lt;/math&gt;. The string &lt;math&gt;23404&lt;/math&gt; does not satisfy the condition because it<br /> does not contain at least &lt;math&gt;2&lt;/math&gt; digits less than &lt;math&gt;2&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\textbf{(A)} \, 500 \qquad\textbf{(B)} \, 625 \qquad\textbf{(C)} \, 1089 \qquad\textbf{(D)} \, 1199 \qquad\textbf{(E)} \, 1296 &lt;/math&gt;<br /> <br /> <br /> [[2022 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> A circle with integer radius &lt;math&gt;r&lt;/math&gt; is centered at &lt;math&gt;(r, r)&lt;/math&gt;. Distinct line segments of length &lt;math&gt;c_i&lt;/math&gt; connect points &lt;math&gt;(0, a_i)&lt;/math&gt; to &lt;math&gt;(b_i, 0)&lt;/math&gt; for &lt;math&gt;1 \le i \le 14&lt;/math&gt; and are tangent to the circle, where &lt;math&gt;a_i&lt;/math&gt;, &lt;math&gt;b_i&lt;/math&gt;, and &lt;math&gt;c_i&lt;/math&gt; are all positive integers and &lt;math&gt;c_1 \le c_2 \le \cdots \le c_{14}&lt;/math&gt;. What is the ratio &lt;math&gt;\frac{c_{14}}{c_1}&lt;/math&gt; for the least possible value of &lt;math&gt;r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{21}{5} \qquad\textbf{(B)} ~\frac{85}{13} \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~\frac{39}{5} \qquad\textbf{(E)} ~17 &lt;/math&gt;<br /> <br /> [[2022 AMC 12A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC12 box|year=2022|ab=A|before=[[2021 AMC 12B Problems]]|after=[[2022 AMC 12B Problems]]}}<br /> <br /> [[Category:AMC 12 Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_12&diff=180956 2022 AMC 10A Problems/Problem 12 2022-11-12T14:54:43Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> <br /> On Halloween &lt;math&gt;31&lt;/math&gt; children walked into the principal's office asking for candy. They<br /> can be classified into three types: Some always lie; some always tell the truth; and<br /> some alternately lie and tell the truth. The alternaters arbitrarily choose their first<br /> response, either a lie or the truth, but each subsequent statement has the opposite<br /> truth value from its predecessor. The principal asked everyone the same three<br /> questions in this order.<br /> <br /> &quot;Are you a truth-teller?&quot; The principal gave a piece of candy to each of the &lt;math&gt;22&lt;/math&gt;<br /> children who answered yes.<br /> <br /> &quot;Are you an alternater?&quot; The principal gave a piece of candy to each of the &lt;math&gt;15&lt;/math&gt;<br /> children who answered yes.<br /> <br /> &quot;Are you a liar?&quot; The principal gave a piece of candy to each of the &lt;math&gt;9&lt;/math&gt; children who<br /> answered yes.<br /> <br /> How many pieces of candy in all did the principal give to the children who always<br /> tell the truth?<br /> <br /> &lt;math&gt;\textbf{(A) } 7 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 31&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Consider when the principal asks &quot;Are you a liar?&quot;: The truth tellers truthfully say no, and the liars lie and say no. This leaves only alternaters who lie on this question to answer yes. Thus, all 9 children that answered yes are alternaters that falsely answer question 1 and 3, and truthfully answer question 2. The rest of the alternaters, however many there are, have the opposite behavior.<br /> <br /> Consider the second question, &quot;Are you an alternater?&quot;: The truth tellers again answer no, the liars falsely answer yes, and alternaters that truthfully answer also say yes. From the previous part, we know that 9 alternaters truthfully answer here. Because only liars and 9 alternaters answer yes, we can deduce that there are &lt;math&gt;15-9=6&lt;/math&gt; liars.<br /> <br /> Consider the first question, &quot;Are you a truth teller?&quot;: Truth tellers say yes, liars also say yes, and alternaters that lie on this question also say yes. From the first part, we know that 9 alternaters lie here. From the previous part, we know that there are 6 liars. Because only the number of truth tellers is unknown here, we can deduce that there are &lt;math&gt;22-9-6=7&lt;/math&gt; truth tellers.<br /> <br /> The final question is how many pieces of candy did the principal give to truth tellers. Because truth tellers only answer yes on the first question, we know that all 7 of them said yes once, resulting in &lt;math&gt;\boxed{\textbf{(A) } 7}&lt;/math&gt; pieces of candy.<br /> <br /> - phuang1024<br /> <br /> ==Solution 2==<br /> On the first question, truth tellers would say yes. Liars would say yes because they are not truth tellers and thus will say the opposite of no. Some alternators may lie on this question too, meaning they say yes. <br /> <br /> On the second question, Liars would say yes because they are not alternators and thus will say the opposite of no. Alternators only say yes to this question if and only if they said yes to the previous question. Thus, the difference between the amount of people that said yes first and that said yes second is the amount of truth tellers. &lt;math&gt;22-15=7&lt;/math&gt;. Because it is obvious that truth tellers only say yes on the first question, our answer is &lt;math&gt;\boxed{7}&lt;/math&gt; ~sigma<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2022|ab=A|num-b=11|num-a=13}}<br /> {{AMC12 box|year=2022|ab=A|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems&diff=180955 2022 AMC 10A Problems 2022-11-12T14:47:48Z <p>Sigmapie: /* Problem 18 */</p> <hr /> <div>{{AMC10 Problems|year=2022|ab=A}}<br /> <br /> ==Problem 1==<br /> <br /> What is the value of &lt;cmath&gt;3+\frac{1}{3+\frac{1}{3+\frac13}}?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)}\ \frac{31}{10}\qquad\textbf{(B)}\ \frac{49}{15}\qquad\textbf{(C)}\ \frac{33}{10}\qquad\textbf{(D)}\ \frac{109}{33}\qquad\textbf{(E)}\ \frac{15}{4}&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Mike cycled &lt;math&gt;15&lt;/math&gt; laps in &lt;math&gt;57&lt;/math&gt; minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first &lt;math&gt;27&lt;/math&gt; minutes?<br /> <br /> &lt;math&gt;\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> The sum of three numbers is &lt;math&gt;96.&lt;/math&gt; The first number is &lt;math&gt;6&lt;/math&gt; times the third number, and the third number is &lt;math&gt;40&lt;/math&gt; less than the second number. What is the absolute value of the difference between the first and second numbers?<br /> <br /> &lt;math&gt;\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> In some countries, automobile fuel efficiency is measured in liters per &lt;math&gt;100&lt;/math&gt; kilometers while other countries use miles per gallon. Suppose that 1 kilometer equals &lt;math&gt;m&lt;/math&gt; miles, and &lt;math&gt;1&lt;/math&gt; gallon equals &lt;math&gt;l&lt;/math&gt; liters. Which of the following gives the fuel efficiency in liters per &lt;math&gt;100&lt;/math&gt; kilometers for a car that gets &lt;math&gt;x&lt;/math&gt; miles per gallon?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{x}{100lm} \qquad \textbf{(B) } \frac{xlm}{100} \qquad \textbf{(C) } \frac{lm}{100x} \qquad \textbf{(D) } \frac{100}{xlm} \qquad \textbf{(E) } \frac{100lm}{x}&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Square &lt;math&gt;ABCD&lt;/math&gt; has side length &lt;math&gt;1&lt;/math&gt;. Points &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;Q&lt;/math&gt;, &lt;math&gt;R&lt;/math&gt;, and &lt;math&gt;S&lt;/math&gt; each lie on a side of &lt;math&gt;ABCD&lt;/math&gt; such that &lt;math&gt;APQCRS&lt;/math&gt; is an equilateral convex hexagon with side length &lt;math&gt;s&lt;/math&gt;. What is &lt;math&gt;s&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{\sqrt{2}}{3} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } 2 - \sqrt{2} \qquad \textbf{(D) } 1 - \frac{\sqrt{2}}{4} \qquad \textbf{(E) } \frac{2}{3}&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Which expression is equal to &lt;cmath&gt;\left|a-2-\sqrt{(a-1)^2}\right|&lt;/cmath&gt; for &lt;math&gt;a&lt;0?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 3-2a \qquad \textbf{(B) } 1-a \qquad \textbf{(C) } 1 \qquad \textbf{(D) } a+1 \qquad \textbf{(E) } 3&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> The least common multiple of a positive divisor &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;18&lt;/math&gt; is &lt;math&gt;180&lt;/math&gt;, and the greatest common divisor of &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;45&lt;/math&gt; is &lt;math&gt;15&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> A data set consists of &lt;math&gt;6&lt;/math&gt; not distinct) positive integers: &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt;. The<br /> average (arithmetic mean) of the &lt;math&gt;6&lt;/math&gt; numbers equals a value in the data set. What is<br /> the sum of all positive values of &lt;math&gt;X&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 10 \qquad \textbf{(B) } 26 \qquad \textbf{(C) } 32 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 40&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> A rectangle is partitioned into &lt;math&gt;5&lt;/math&gt; regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible?<br /> <br /> &lt;asy&gt; size(5.5cm); draw((0,0)--(0,2)--(2,2)--(2,0)--cycle); draw((2,0)--(8,0)--(8,2)--(2,2)--cycle); draw((8,0)--(12,0)--(12,2)--(8,2)--cycle); draw((0,2)--(6,2)--(6,4)--(0,4)--cycle); draw((6,2)--(12,2)--(12,4)--(6,4)--cycle); &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }120\qquad\textbf{(B) }270\qquad\textbf{(C) }360\qquad\textbf{(D) }540\qquad\textbf{(E) }720&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> XXX<br /> <br /> &lt;math&gt;\textbf{(A) } X \qquad \textbf{(B) } X \qquad \textbf{(C) } X \qquad \textbf{(D) } X \qquad \textbf{(E) } X&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> Ted mistakenly wrote &lt;math&gt;2^m\cdot\sqrt{\frac{1}{4096}}&lt;/math&gt; as &lt;math&gt;2\cdot\sqrt[m]{\frac{1}{4096}}.&lt;/math&gt; What is the sum of all real numbers &lt;math&gt;m&lt;/math&gt; for which these two expressions have the same value?<br /> <br /> &lt;math&gt;\textbf{(A) } 5 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> On Halloween &lt;math&gt;31&lt;/math&gt; children walked into the principal's office asking for candy. They<br /> can be classified into three types: Some always lie; some always tell the truth; and<br /> some alternately lie and tell the truth. The alternaters arbitrarily choose their first<br /> response, either a lie or the truth, but each subsequent statement has the opposite<br /> truth value from its predecessor. The principal asked everyone the same three<br /> questions in this order.<br /> <br /> &quot;Are you a truth-teller?&quot; The principal gave a piece of candy to each of the &lt;math&gt;22&lt;/math&gt;<br /> children who answered yes.<br /> <br /> &quot;Are you an alternater?&quot; The principal gave a piece of candy to each of the &lt;math&gt;15&lt;/math&gt;<br /> children who answered yes.<br /> <br /> &quot;Are you a liar?&quot; The principal gave a piece of candy to each of the &lt;math&gt;9&lt;/math&gt; children who<br /> answered yes.<br /> <br /> How many pieces of candy in all did the principal give to the children who always<br /> tell the truth?<br /> <br /> &lt;math&gt;\textbf{(A) } 7 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 31&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be a scalene triangle. Point &lt;math&gt;P&lt;/math&gt; lies on &lt;math&gt;\overline{BC}&lt;/math&gt; so that &lt;math&gt;\overline{AP}&lt;/math&gt; bisects &lt;math&gt;\angle BAC.&lt;/math&gt; The line through &lt;math&gt;B&lt;/math&gt; perpendicular to &lt;math&gt;\overline{AP}&lt;/math&gt; intersects the line through &lt;math&gt;A&lt;/math&gt; parallel to &lt;math&gt;\overline{BC}&lt;/math&gt; at point &lt;math&gt;D.&lt;/math&gt; Suppose &lt;math&gt;BP=2&lt;/math&gt; and &lt;math&gt;PC=3.&lt;/math&gt; What is &lt;math&gt;AD?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> How many ways are there to split the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;14&lt;/math&gt; into &lt;math&gt;7&lt;/math&gt; pairs such that in each pair, the greater number is at least &lt;math&gt;2&lt;/math&gt; times the lesser number?<br /> <br /> &lt;math&gt;\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; with side lengths &lt;math&gt;AB=7, BC=24, CD=20, DA=15&lt;/math&gt; is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form &lt;math&gt;\frac{a\pi-b}{c},&lt;/math&gt; where &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are positive integers such that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; have no common prime factor. What is &lt;math&gt;a+b+c?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 260 \qquad \textbf{(B) } 855 \qquad \textbf{(C) } 1235 \qquad \textbf{(D) } 1565 \qquad \textbf{(E) } 1997&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> The roots of the polynomial &lt;math&gt;10x^3 - 39x^2 + 29x - 6&lt;/math&gt; are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by 2 units. What is the volume of the new box?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}{5} \qquad \textbf{(C) } \frac{81}{5} \qquad \textbf{(D) } 30 \qquad \textbf{(E) } 48&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> How many three-digit positive integers &lt;math&gt;\underline{a} \ \underline{b} \ \underline{c}&lt;/math&gt; are there whose nonzero digits &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; satisfy<br /> &lt;cmath&gt;0.\overline{\underline{a}~\underline{b}~\underline{c}} = \frac{1}{3} (0.\overline{a} + 0.\overline{b} + 0.\overline{c})?&lt;/cmath&gt;<br /> (The bar indicates repetition, thus &lt;math&gt;0.\overline{\underline{a}~\underline{b}~\underline{c}}&lt;/math&gt; in the infinite repeating decimal &lt;math&gt;0.\underline{a}~\underline{b}~\underline{c}~\underline{a}~\underline{b}~\underline{c}~\cdots&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A) } 9 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 11 \qquad \textbf{(D) } 13 \qquad \textbf{(E) } 14&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Let &lt;math&gt;T_k&lt;/math&gt; be the transformation of the coordinate plane that first rotates the plane &lt;math&gt;k&lt;/math&gt; degrees counterclockwise around the origin and then reflects the plane across the &lt;math&gt;y&lt;/math&gt;-axis. What is the least positive integer &lt;math&gt;n&lt;/math&gt; such that performing the sequence of transformations &lt;math&gt;T_1, T_2, T_3,...,T_n&lt;/math&gt; returns the point &lt;math&gt;(1, 0)&lt;/math&gt; back to itself?<br /> <br /> &lt;math&gt;\textbf{(A) } 359 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 719 \qquad \textbf{(D) } 720 \qquad \textbf{(E) } 721 &lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> XXX<br /> <br /> &lt;math&gt;\textbf{(A) } X \qquad \textbf{(B) } X \qquad \textbf{(C) } X \qquad \textbf{(D) } X \qquad \textbf{(E) } X&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> <br /> A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are &lt;math&gt;57&lt;/math&gt;, &lt;math&gt;60&lt;/math&gt;, and &lt;math&gt;91&lt;/math&gt;. What is the fourth term of this sequence? <br /> <br /> &lt;math&gt;\textbf{(A) } X \qquad \textbf{(B) } X \qquad \textbf{(C) } X \qquad \textbf{(D) } X \qquad \textbf{(E) } 206&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> A bowl is formed by attaching four regular hexagons of side &lt;math&gt;1&lt;/math&gt; to a square of side &lt;math&gt;1&lt;/math&gt;. The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl? <br /> <br /> &lt;math&gt;\textbf{(A) } X \qquad \textbf{(B) } X \qquad \textbf{(C) } X \qquad \textbf{(D) } X \qquad \textbf{(E) } X&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> Suppose that 13 cards numbered &lt;math&gt;1, 2, 3, \cdots, 13&lt;/math&gt; are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards 1, 2, 3 are picked up on the first pass, 4 and 5 on the second pass, 6 on the third pass, 7, 8, 9, 10 on the fourth pass, and 11, 12, 13 on the fifth pass. For how many of the &lt;math&gt;13!&lt;/math&gt; possible orderings of the cards will the &lt;math&gt;13&lt;/math&gt; cards be picked up in exactly two passes?<br /> <br /> XXX [Image]<br /> <br /> &lt;math&gt;\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Isosceles trapezoid &lt;math&gt;ABCD&lt;/math&gt; has parallel sides &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{BC},&lt;/math&gt; with &lt;math&gt;BC &lt; AD&lt;/math&gt; and &lt;math&gt;AB = CD.&lt;/math&gt; There is a point &lt;math&gt;P&lt;/math&gt; in the plane such that &lt;math&gt;PA=1, PB=2, PC=3,&lt;/math&gt; and &lt;math&gt;PD=4.&lt;/math&gt; What is &lt;math&gt;\tfrac{BC}{AD}?&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{4}\qquad\textbf{(B) }\frac{1}{3}\qquad\textbf{(C) }\frac{1}{2}\qquad\textbf{(D) }\frac{2}{3}\qquad\textbf{(E) }\frac{3}{4}&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> How many strings of length 5 formed from the digits 0, 1, 2, 3, 4 are there such that for each &lt;math&gt;j \in \{1,2,3,4\}&lt;/math&gt;, at least &lt;math&gt;j&lt;/math&gt; of the digits are less than &lt;math&gt;j&lt;/math&gt;? (For example, &lt;math&gt;02214&lt;/math&gt; satisfies this condition because it contains at least 1 digit less than 1, at least 2 digits less than 2, at least 3 digits less than 3, and at least 4 digits less than 4. The string &lt;math&gt;23404&lt;/math&gt; does not satisfy the condition because it does not contain at least 2 digits less than 2.)<br /> <br /> <br /> &lt;math&gt;\textbf{(A) } 500 \qquad \textbf{(B) } 625 \qquad \textbf{(C) } 1089 \qquad \textbf{(D) } 1199 \qquad \textbf{(E) } 1296&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> XXX<br /> <br /> &lt;math&gt;\textbf{(A) } X \qquad \textbf{(B) } X \qquad \textbf{(C) } X \qquad \textbf{(D) } X \qquad \textbf{(E) } X&lt;/math&gt;<br /> <br /> [[2022 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2020|ab=A|before=[[2021 Fall AMC 10B Problems]]|after=[[2022 AMC 10B Problems]]}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_24&diff=180493 2013 AMC 12A Problems/Problem 24 2022-11-08T19:54:14Z <p>Sigmapie: </p> <hr /> <div>== Problem==<br /> <br /> Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \frac{553}{715} \qquad \textbf{(B)} \ \frac{443}{572} \qquad \textbf{(C)} \ \frac{111}{143} \qquad \textbf{(D)} \ \frac{81}{104} \qquad \textbf{(E)} \ \frac{223}{286}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Suppose &lt;math&gt;p&lt;/math&gt; is the answer. We calculate &lt;math&gt;1-p&lt;/math&gt;.<br /> <br /> Assume that the circumradius of the 12-gon is &lt;math&gt;1&lt;/math&gt;, and the 6 different lengths are &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_6&lt;/math&gt;, in increasing order. Then<br /> <br /> &lt;math&gt;a_k = 2\sin ( \frac{k\pi}{12} )&lt;/math&gt;. <br /> <br /> So &lt;math&gt;a_1=(\sqrt{6}-\sqrt{2})/2 \approx 0.5&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_2=1&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_3=\sqrt{2}\approx 1.4&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_4=\sqrt{3}\approx 1.7&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_5=(\sqrt{6}+\sqrt{2})/2 = a_1 + a_3 &lt;/math&gt;,<br /> <br /> &lt;math&gt;a_6 = 2&lt;/math&gt;.<br /> <br /> <br /> <br /> Now, Consider the following inequalities:<br /> <br /> &lt;math&gt;a_3&gt;2a_1 &gt; a_2&lt;/math&gt;<br /> <br /> &lt;math&gt;a_4&gt; a_1 + a_2&gt;a_3&lt;/math&gt; <br /> <br /> &lt;math&gt;a_4&lt;a_1 + a_3=a_5&lt;/math&gt;<br /> <br /> &lt;math&gt;a_1 + a_4 &gt; a_6&lt;/math&gt;<br /> <br /> &lt;math&gt;2a_2 = 2 = a_6&lt;/math&gt;. Thus any two segments with at least one them longer than &lt;math&gt;a_2&lt;/math&gt; have a sum greater than &lt;math&gt;a_6&lt;/math&gt;.<br /> <br /> Therefore, all triples (in increasing order) that can't be the side lengths of a triangle are the following. Note that x-y-z means &lt;math&gt;(a_x, a_y, a_z)&lt;/math&gt;:<br /> <br /> 1-1-3, 1-1-4, 1-1-5, 1-1-6,<br /> 1-2-4, 1-2-5, 1-2-6,<br /> 1-3-5, 1-3-6,<br /> 2-2-6<br /> <br /> Note that there are &lt;math&gt;12&lt;/math&gt; segments of each length of &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_5&lt;/math&gt;, respectively, and &lt;math&gt;6&lt;/math&gt; segments of length &lt;math&gt;a_6&lt;/math&gt;. There are &lt;math&gt;66&lt;/math&gt; segments in total.<br /> <br /> In the above list there are &lt;math&gt;3&lt;/math&gt; triples of the type a-a-b without ''6'', &lt;math&gt;2&lt;/math&gt; triples of a-a-6 where a is not ''6'', &lt;math&gt;3&lt;/math&gt; triples of a-b-c without ''6'', and &lt;math&gt;2&lt;/math&gt; triples of a-b-6 where a, b are not ''6''. So,<br /> <br /> &lt;cmath&gt;1-p = \frac{1}{66\cdot 65\cdot 64} ( 3\cdot 3 \cdot 12\cdot 11\cdot 12 + 2\cdot 3 \cdot 12\cdot 11\cdot 6 + 3\cdot 6\cdot 12^3 + 2\cdot 6 \cdot 12^2 \cdot 6)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt; = \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286} &lt;/cmath&gt;<br /> <br /> So &lt;math&gt;p = 223/286&lt;/math&gt;.<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2013amc12a/364<br /> <br /> ~dolphin7<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_24&diff=180492 2013 AMC 12A Problems/Problem 24 2022-11-08T19:53:36Z <p>Sigmapie: </p> <hr /> <div>== Problem==<br /> <br /> Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \frac{553}{715} \qquad \textbf{(B)} \ \frac{443}{572} \qquad \textbf{(C)} \ \frac{111}{143} \qquad \textbf{(D)} \ \frac{81}{104} \qquad \textbf{(E)} \ \frac{223}{286}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Suppose &lt;math&gt;p&lt;/math&gt; is the answer. We calculate &lt;math&gt;1-p&lt;/math&gt;.<br /> <br /> Assume that the circumradius of the 12-gon is &lt;math&gt;1&lt;/math&gt;, and the 6 different lengths are &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_6&lt;/math&gt;, in increasing order. Then<br /> <br /> &lt;math&gt;a_k = 2\sin ( \frac{k\pi}{12} )&lt;/math&gt;. <br /> <br /> So &lt;math&gt;a_1=(\sqrt{6}-\sqrt{2})/2 \approx 0.5&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_2=1&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_3=\sqrt{2}\approx 1.4&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_4=\sqrt{3}\approx 1.7&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_5=(\sqrt{6}+\sqrt{2})/2 = a_1 + a_3 &lt;/math&gt;,<br /> <br /> &lt;math&gt;a_6 = 2&lt;/math&gt;.<br /> <br /> <br /> <br /> Now, Consider the following inequalities:<br /> <br /> &lt;math&gt;a_3&gt;2a_1 &gt; a_2&lt;/math&gt;<br /> <br /> &lt;math&gt;a_4&gt; a_1 + a_2&gt;a_3&lt;/math&gt; <br /> <br /> &lt;math&gt;a_4&lt;a_1 + a_3=a_5&lt;/math&gt;<br /> <br /> &lt;math&gt;a_1 + a_4 &gt; a_6&lt;/math&gt;<br /> <br /> &lt;math&gt;2a_2 = 2 = a_6&lt;/math&gt;. Thus any two segments with at least one them longer than &lt;math&gt;a_2&lt;/math&gt; have a sum greater than &lt;math&gt;a_6&lt;/math&gt;.<br /> <br /> Therefore, all triples (in increasing order) that can't be the side lengths of a triangle are the following. Note that x-y-z means &lt;math&gt;(a_x, a_y, a_z)&lt;/math&gt;:<br /> <br /> 1-1-3, 1-1-4, 1-1-5, 1-1-6,<br /> 1-2-4, 1-2-5, 1-2-6,<br /> 1-3-5, 1-3-6,<br /> 2-2-6<br /> <br /> Note that there are &lt;math&gt;12&lt;/math&gt; segments of each length of &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_5&lt;/math&gt;, respectively, and &lt;math&gt;6&lt;/math&gt; segments of length &lt;math&gt;a_6&lt;/math&gt;. There are &lt;math&gt;66&lt;/math&gt; segments in total.<br /> <br /> In the above list there are &lt;math&gt;3&lt;/math&gt; triples of the type a-a-b without ''6'', &lt;math&gt;2&lt;/math&gt; triples of a-a-6 where a is not ''6'', &lt;math&gt;3&lt;/math&gt; triples of a-b-c without ''6'', and &lt;math&gt;2&lt;/math&gt; triples of a-b-6 where a, b are not ''6''. So,<br /> <br /> &lt;cmath&gt;1-p = \frac{1}{66\cdot 65\cdot 64} ( 3\cdot 3 \cdot 12\cdot 11\cdot 12 + 2\cdot 3 \cdot 12\cdot 11\cdot 6 + 3\cdot 6\cdot 12^3 + 2\cdot 6 \cdot 12^2 \cdot 6)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt; = \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286} &lt;/cmath&gt;<br /> <br /> So &lt;math&gt;p = 223/286&lt;/math&gt;.<br /> <br /> ==Solution 1 ALT==<br /> Just like Solution 1, we find &lt;math&gt;1-p&lt;/math&gt;. Let &lt;math&gt;a_n&lt;/math&gt; be the length of the diagonal that connects vertices that are &lt;math&gt;n&lt;/math&gt; sides apart. By the law of Cosines, we have &lt;cmath&gt;a_n^2=1+2a_{n-1}^2&lt;/cmath&gt; WLOG let &lt;math&gt;a_1 = 1&lt;/math&gt;, then it is a well known fact that &lt;math&gt;a_n=\sqrt{2^n-1}&lt;/math&gt; holds as an explicit formula. Thus we have &lt;math&gt;a_1=1, a_2=\sqrt{3}, a_3=\sqrt{7}, a_4=\sqrt{15}, a_5=\sqrt{31}, a_6=\sqrt{63}&lt;/math&gt;<br /> <br /> Now continue as in Solution 1. <br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2013amc12a/364<br /> <br /> ~dolphin7<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_24&diff=180491 2013 AMC 12A Problems/Problem 24 2022-11-08T19:53:11Z <p>Sigmapie: Undo revision 180490 by Sigmapie (talk)</p> <hr /> <div>== Problem==<br /> <br /> Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \frac{553}{715} \qquad \textbf{(B)} \ \frac{443}{572} \qquad \textbf{(C)} \ \frac{111}{143} \qquad \textbf{(D)} \ \frac{81}{104} \qquad \textbf{(E)} \ \frac{223}{286}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Suppose &lt;math&gt;p&lt;/math&gt; is the answer. We calculate &lt;math&gt;1-p&lt;/math&gt;.<br /> <br /> Assume that the circumradius of the 12-gon is &lt;math&gt;1&lt;/math&gt;, and the 6 different lengths are &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_6&lt;/math&gt;, in increasing order. Then<br /> <br /> &lt;math&gt;a_k = 2\sin ( \frac{k\pi}{12} )&lt;/math&gt;. <br /> <br /> So &lt;math&gt;a_1=(\sqrt{6}-\sqrt{2})/2 \approx 0.5&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_2=1&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_3=\sqrt{2}\approx 1.4&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_4=\sqrt{3}\approx 1.7&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_5=(\sqrt{6}+\sqrt{2})/2 = a_1 + a_3 &lt;/math&gt;,<br /> <br /> &lt;math&gt;a_6 = 2&lt;/math&gt;.<br /> <br /> <br /> <br /> Now, Consider the following inequalities:<br /> <br /> &lt;math&gt;a_3&gt;2a_1 &gt; a_2&lt;/math&gt;<br /> <br /> &lt;math&gt;a_4&gt; a_1 + a_2&gt;a_3&lt;/math&gt; <br /> <br /> &lt;math&gt;a_4&lt;a_1 + a_3=a_5&lt;/math&gt;<br /> <br /> &lt;math&gt;a_1 + a_4 &gt; a_6&lt;/math&gt;<br /> <br /> &lt;math&gt;2a_2 = 2 = a_6&lt;/math&gt;. Thus any two segments with at least one them longer than &lt;math&gt;a_2&lt;/math&gt; have a sum greater than &lt;math&gt;a_6&lt;/math&gt;.<br /> <br /> Therefore, all triples (in increasing order) that can't be the side lengths of a triangle are the following. Note that x-y-z means &lt;math&gt;(a_x, a_y, a_z)&lt;/math&gt;:<br /> <br /> 1-1-3, 1-1-4, 1-1-5, 1-1-6,<br /> 1-2-4, 1-2-5, 1-2-6,<br /> 1-3-5, 1-3-6,<br /> 2-2-6<br /> <br /> Note that there are &lt;math&gt;12&lt;/math&gt; segments of each length of &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_5&lt;/math&gt;, respectively, and &lt;math&gt;6&lt;/math&gt; segments of length &lt;math&gt;a_6&lt;/math&gt;. There are &lt;math&gt;66&lt;/math&gt; segments in total.<br /> <br /> In the above list there are &lt;math&gt;3&lt;/math&gt; triples of the type a-a-b without ''6'', &lt;math&gt;2&lt;/math&gt; triples of a-a-6 where a is not ''6'', &lt;math&gt;3&lt;/math&gt; triples of a-b-c without ''6'', and &lt;math&gt;2&lt;/math&gt; triples of a-b-6 where a, b are not ''6''. So,<br /> <br /> &lt;cmath&gt;1-p = \frac{1}{66\cdot 65\cdot 64} ( 3\cdot 3 \cdot 12\cdot 11\cdot 12 + 2\cdot 3 \cdot 12\cdot 11\cdot 6 + 3\cdot 6\cdot 12^3 + 2\cdot 6 \cdot 12^2 \cdot 6)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt; = \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286} &lt;/cmath&gt;<br /> <br /> So &lt;math&gt;p = 223/286&lt;/math&gt;.<br /> <br /> ==Solution 1 ALT==<br /> Just like Solution 1, we find &lt;math&gt;1-p&lt;/math&gt;. Let &lt;math&gt;a_n&lt;/math&gt; be the length of the diagonal that connects vertices that are &lt;math&gt;n&lt;/math&gt; sides apart. By the law of Cosines, we have &lt;cmath&gt;a_n^2=1+2a_n^2&lt;/cmath&gt; WLOG let &lt;math&gt;a_1 = 1&lt;/math&gt;, then it is a well known fact that &lt;math&gt;a_n=\sqrt{2^n-1}&lt;/math&gt; holds as an explicit formula. Thus we have &lt;math&gt;a_1=1, a_2=\sqrt{3}, a_3=\sqrt{7}, a_4=\sqrt{15}, a_5=\sqrt{31}, a_6=\sqrt{63}&lt;/math&gt;<br /> <br /> Now continue as in Solution 1. <br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2013amc12a/364<br /> <br /> ~dolphin7<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_24&diff=180490 2013 AMC 12A Problems/Problem 24 2022-11-08T19:52:37Z <p>Sigmapie: </p> <hr /> <div>== Problem==<br /> <br /> Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \frac{553}{715} \qquad \textbf{(B)} \ \frac{443}{572} \qquad \textbf{(C)} \ \frac{111}{143} \qquad \textbf{(D)} \ \frac{81}{104} \qquad \textbf{(E)} \ \frac{223}{286}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Suppose &lt;math&gt;p&lt;/math&gt; is the answer. We calculate &lt;math&gt;1-p&lt;/math&gt;.<br /> <br /> Assume that the circumradius of the 12-gon is &lt;math&gt;1&lt;/math&gt;, and the 6 different lengths are &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_6&lt;/math&gt;, in increasing order. Then<br /> <br /> &lt;math&gt;a_k = 2\sin ( \frac{k\pi}{12} )&lt;/math&gt;. <br /> <br /> So &lt;math&gt;a_1=(\sqrt{6}-\sqrt{2})/2 \approx 0.5&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_2=1&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_3=\sqrt{2}\approx 1.4&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_4=\sqrt{3}\approx 1.7&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_5=(\sqrt{6}+\sqrt{2})/2 = a_1 + a_3 &lt;/math&gt;,<br /> <br /> &lt;math&gt;a_6 = 2&lt;/math&gt;.<br /> <br /> <br /> <br /> Now, Consider the following inequalities:<br /> <br /> &lt;math&gt;a_3&gt;2a_1 &gt; a_2&lt;/math&gt;<br /> <br /> &lt;math&gt;a_4&gt; a_1 + a_2&gt;a_3&lt;/math&gt; <br /> <br /> &lt;math&gt;a_4&lt;a_1 + a_3=a_5&lt;/math&gt;<br /> <br /> &lt;math&gt;a_1 + a_4 &gt; a_6&lt;/math&gt;<br /> <br /> &lt;math&gt;2a_2 = 2 = a_6&lt;/math&gt;. Thus any two segments with at least one them longer than &lt;math&gt;a_2&lt;/math&gt; have a sum greater than &lt;math&gt;a_6&lt;/math&gt;.<br /> <br /> Therefore, all triples (in increasing order) that can't be the side lengths of a triangle are the following. Note that x-y-z means &lt;math&gt;(a_x, a_y, a_z)&lt;/math&gt;:<br /> <br /> 1-1-3, 1-1-4, 1-1-5, 1-1-6,<br /> 1-2-4, 1-2-5, 1-2-6,<br /> 1-3-5, 1-3-6,<br /> 2-2-6<br /> <br /> Note that there are &lt;math&gt;12&lt;/math&gt; segments of each length of &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_5&lt;/math&gt;, respectively, and &lt;math&gt;6&lt;/math&gt; segments of length &lt;math&gt;a_6&lt;/math&gt;. There are &lt;math&gt;66&lt;/math&gt; segments in total.<br /> <br /> In the above list there are &lt;math&gt;3&lt;/math&gt; triples of the type a-a-b without ''6'', &lt;math&gt;2&lt;/math&gt; triples of a-a-6 where a is not ''6'', &lt;math&gt;3&lt;/math&gt; triples of a-b-c without ''6'', and &lt;math&gt;2&lt;/math&gt; triples of a-b-6 where a, b are not ''6''. So,<br /> <br /> &lt;cmath&gt;1-p = \frac{1}{66\cdot 65\cdot 64} ( 3\cdot 3 \cdot 12\cdot 11\cdot 12 + 2\cdot 3 \cdot 12\cdot 11\cdot 6 + 3\cdot 6\cdot 12^3 + 2\cdot 6 \cdot 12^2 \cdot 6)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt; = \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286} &lt;/cmath&gt;<br /> <br /> So &lt;math&gt;p = 223/286&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Just like Solution 1, we find &lt;math&gt;1-p&lt;/math&gt;. Let &lt;math&gt;a_n&lt;/math&gt; be the length of the diagonal that connects vertices that are &lt;math&gt;n&lt;/math&gt; sides apart. By the law of Cosines, we have &lt;cmath&gt;a_n^2=1+2a_n^2&lt;/cmath&gt;<br /> Now continue as in Solution 1. <br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2013amc12a/364<br /> <br /> ~dolphin7<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_24&diff=180489 2013 AMC 12A Problems/Problem 24 2022-11-08T19:52:28Z <p>Sigmapie: </p> <hr /> <div>== Problem==<br /> <br /> Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \frac{553}{715} \qquad \textbf{(B)} \ \frac{443}{572} \qquad \textbf{(C)} \ \frac{111}{143} \qquad \textbf{(D)} \ \frac{81}{104} \qquad \textbf{(E)} \ \frac{223}{286}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Suppose &lt;math&gt;p&lt;/math&gt; is the answer. We calculate &lt;math&gt;1-p&lt;/math&gt;.<br /> <br /> Assume that the circumradius of the 12-gon is &lt;math&gt;1&lt;/math&gt;, and the 6 different lengths are &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_6&lt;/math&gt;, in increasing order. Then<br /> <br /> &lt;math&gt;a_k = 2\sin ( \frac{k\pi}{12} )&lt;/math&gt;. <br /> <br /> So &lt;math&gt;a_1=(\sqrt{6}-\sqrt{2})/2 \approx 0.5&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_2=1&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_3=\sqrt{2}\approx 1.4&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_4=\sqrt{3}\approx 1.7&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_5=(\sqrt{6}+\sqrt{2})/2 = a_1 + a_3 &lt;/math&gt;,<br /> <br /> &lt;math&gt;a_6 = 2&lt;/math&gt;.<br /> <br /> <br /> <br /> Now, Consider the following inequalities:<br /> <br /> &lt;math&gt;a_3&gt;2a_1 &gt; a_2&lt;/math&gt;<br /> <br /> &lt;math&gt;a_4&gt; a_1 + a_2&gt;a_3&lt;/math&gt; <br /> <br /> &lt;math&gt;a_4&lt;a_1 + a_3=a_5&lt;/math&gt;<br /> <br /> &lt;math&gt;a_1 + a_4 &gt; a_6&lt;/math&gt;<br /> <br /> &lt;math&gt;2a_2 = 2 = a_6&lt;/math&gt;. Thus any two segments with at least one them longer than &lt;math&gt;a_2&lt;/math&gt; have a sum greater than &lt;math&gt;a_6&lt;/math&gt;.<br /> <br /> Therefore, all triples (in increasing order) that can't be the side lengths of a triangle are the following. Note that x-y-z means &lt;math&gt;(a_x, a_y, a_z)&lt;/math&gt;:<br /> <br /> 1-1-3, 1-1-4, 1-1-5, 1-1-6,<br /> 1-2-4, 1-2-5, 1-2-6,<br /> 1-3-5, 1-3-6,<br /> 2-2-6<br /> <br /> Note that there are &lt;math&gt;12&lt;/math&gt; segments of each length of &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_5&lt;/math&gt;, respectively, and &lt;math&gt;6&lt;/math&gt; segments of length &lt;math&gt;a_6&lt;/math&gt;. There are &lt;math&gt;66&lt;/math&gt; segments in total.<br /> <br /> In the above list there are &lt;math&gt;3&lt;/math&gt; triples of the type a-a-b without ''6'', &lt;math&gt;2&lt;/math&gt; triples of a-a-6 where a is not ''6'', &lt;math&gt;3&lt;/math&gt; triples of a-b-c without ''6'', and &lt;math&gt;2&lt;/math&gt; triples of a-b-6 where a, b are not ''6''. So,<br /> <br /> &lt;cmath&gt;1-p = \frac{1}{66\cdot 65\cdot 64} ( 3\cdot 3 \cdot 12\cdot 11\cdot 12 + 2\cdot 3 \cdot 12\cdot 11\cdot 6 + 3\cdot 6\cdot 12^3 + 2\cdot 6 \cdot 12^2 \cdot 6)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt; = \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286} &lt;/cmath&gt;<br /> <br /> So &lt;math&gt;p = 223/286&lt;/math&gt;.<br /> <br /> ==Solution 1 ALT==<br /> Just like Solution 1, we find &lt;math&gt;1-p&lt;/math&gt;. Let &lt;math&gt;a_n&lt;/math&gt; be the length of the diagonal that connects vertices that are &lt;math&gt;n&lt;/math&gt; sides apart. By the law of Cosines, we have &lt;cmath&gt;a_n^2=1+2a_n^2&lt;/cmath&gt; WLOG let &lt;math&gt;a_1 = 1&lt;/math&gt;, then it is a well known fact that &lt;math&gt;a_n=\sqrt{2^n-1}&lt;/math&gt; holds as an explicit formula. Thus we have &lt;math&gt;a_1=1, a_2=\sqrt{3}, a_3=\sqrt{7}, a_4=\sqrt{15}, a_5=\sqrt{31}, a_6=\sqrt{63}&lt;/math&gt;<br /> <br /> Now continue as in Solution 1. <br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2013amc12a/364<br /> <br /> ~dolphin7<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_24&diff=180488 2013 AMC 12A Problems/Problem 24 2022-11-08T19:50:03Z <p>Sigmapie: </p> <hr /> <div>== Problem==<br /> <br /> Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \frac{553}{715} \qquad \textbf{(B)} \ \frac{443}{572} \qquad \textbf{(C)} \ \frac{111}{143} \qquad \textbf{(D)} \ \frac{81}{104} \qquad \textbf{(E)} \ \frac{223}{286}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Suppose &lt;math&gt;p&lt;/math&gt; is the answer. We calculate &lt;math&gt;1-p&lt;/math&gt;.<br /> <br /> Assume that the circumradius of the 12-gon is &lt;math&gt;1&lt;/math&gt;, and the 6 different lengths are &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_6&lt;/math&gt;, in increasing order. Then<br /> <br /> &lt;math&gt;a_k = 2\sin ( \frac{k\pi}{12} )&lt;/math&gt;. <br /> <br /> So &lt;math&gt;a_1=(\sqrt{6}-\sqrt{2})/2 \approx 0.5&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_2=1&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_3=\sqrt{2}\approx 1.4&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_4=\sqrt{3}\approx 1.7&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_5=(\sqrt{6}+\sqrt{2})/2 = a_1 + a_3 &lt;/math&gt;,<br /> <br /> &lt;math&gt;a_6 = 2&lt;/math&gt;.<br /> <br /> <br /> <br /> Now, Consider the following inequalities:<br /> <br /> &lt;math&gt;a_3&gt;2a_1 &gt; a_2&lt;/math&gt;<br /> <br /> &lt;math&gt;a_4&gt; a_1 + a_2&gt;a_3&lt;/math&gt; <br /> <br /> &lt;math&gt;a_4&lt;a_1 + a_3=a_5&lt;/math&gt;<br /> <br /> &lt;math&gt;a_1 + a_4 &gt; a_6&lt;/math&gt;<br /> <br /> &lt;math&gt;2a_2 = 2 = a_6&lt;/math&gt;. Thus any two segments with at least one them longer than &lt;math&gt;a_2&lt;/math&gt; have a sum greater than &lt;math&gt;a_6&lt;/math&gt;.<br /> <br /> Therefore, all triples (in increasing order) that can't be the side lengths of a triangle are the following. Note that x-y-z means &lt;math&gt;(a_x, a_y, a_z)&lt;/math&gt;:<br /> <br /> 1-1-3, 1-1-4, 1-1-5, 1-1-6,<br /> 1-2-4, 1-2-5, 1-2-6,<br /> 1-3-5, 1-3-6,<br /> 2-2-6<br /> <br /> Note that there are &lt;math&gt;12&lt;/math&gt; segments of each length of &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_5&lt;/math&gt;, respectively, and &lt;math&gt;6&lt;/math&gt; segments of length &lt;math&gt;a_6&lt;/math&gt;. There are &lt;math&gt;66&lt;/math&gt; segments in total.<br /> <br /> In the above list there are &lt;math&gt;3&lt;/math&gt; triples of the type a-a-b without ''6'', &lt;math&gt;2&lt;/math&gt; triples of a-a-6 where a is not ''6'', &lt;math&gt;3&lt;/math&gt; triples of a-b-c without ''6'', and &lt;math&gt;2&lt;/math&gt; triples of a-b-6 where a, b are not ''6''. So,<br /> <br /> &lt;cmath&gt;1-p = \frac{1}{66\cdot 65\cdot 64} ( 3\cdot 3 \cdot 12\cdot 11\cdot 12 + 2\cdot 3 \cdot 12\cdot 11\cdot 6 + 3\cdot 6\cdot 12^3 + 2\cdot 6 \cdot 12^2 \cdot 6)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt; = \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286} &lt;/cmath&gt;<br /> <br /> So &lt;math&gt;p = 223/286&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Just like Solution 1, we find &lt;math&gt;1-p&lt;/math&gt;. Let &lt;math&gt;a_n&lt;/math&gt; be the length of the diagonal that connects vertices that are &lt;math&gt;n&lt;/math&gt; sides apart. By the law of Cosines, we have &lt;cmath&gt;a_n^2=1+2a_n^2&lt;/cmath&gt;<br /> Now continue as in Solution 1. <br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2013amc12a/364<br /> <br /> ~dolphin7<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_24&diff=180487 2013 AMC 12A Problems/Problem 24 2022-11-08T19:49:35Z <p>Sigmapie: </p> <hr /> <div>== Problem==<br /> <br /> Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \frac{553}{715} \qquad \textbf{(B)} \ \frac{443}{572} \qquad \textbf{(C)} \ \frac{111}{143} \qquad \textbf{(D)} \ \frac{81}{104} \qquad \textbf{(E)} \ \frac{223}{286}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Suppose &lt;math&gt;p&lt;/math&gt; is the answer. We calculate &lt;math&gt;1-p&lt;/math&gt;.<br /> <br /> Assume that the circumradius of the 12-gon is &lt;math&gt;1&lt;/math&gt;, and the 6 different lengths are &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_6&lt;/math&gt;, in increasing order. Then<br /> <br /> &lt;math&gt;a_k = 2\sin ( \frac{k\pi}{12} )&lt;/math&gt;. <br /> <br /> So &lt;math&gt;a_1=(\sqrt{6}-\sqrt{2})/2 \approx 0.5&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_2=1&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_3=\sqrt{2}\approx 1.4&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_4=\sqrt{3}\approx 1.7&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_5=(\sqrt{6}+\sqrt{2})/2 = a_1 + a_3 &lt;/math&gt;,<br /> <br /> &lt;math&gt;a_6 = 2&lt;/math&gt;.<br /> <br /> <br /> <br /> Now, Consider the following inequalities:<br /> <br /> &lt;math&gt;a_3&gt;2a_1 &gt; a_2&lt;/math&gt;<br /> <br /> &lt;math&gt;a_4&gt; a_1 + a_2&gt;a_3&lt;/math&gt; <br /> <br /> &lt;math&gt;a_4&lt;a_1 + a_3=a_5&lt;/math&gt;<br /> <br /> &lt;math&gt;a_1 + a_4 &gt; a_6&lt;/math&gt;<br /> <br /> &lt;math&gt;2a_2 = 2 = a_6&lt;/math&gt;. Thus any two segments with at least one them longer than &lt;math&gt;a_2&lt;/math&gt; have a sum greater than &lt;math&gt;a_6&lt;/math&gt;.<br /> <br /> Therefore, all triples (in increasing order) that can't be the side lengths of a triangle are the following. Note that x-y-z means &lt;math&gt;(a_x, a_y, a_z)&lt;/math&gt;:<br /> <br /> 1-1-3, 1-1-4, 1-1-5, 1-1-6,<br /> 1-2-4, 1-2-5, 1-2-6,<br /> 1-3-5, 1-3-6,<br /> 2-2-6<br /> <br /> Note that there are &lt;math&gt;12&lt;/math&gt; segments of each length of &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_5&lt;/math&gt;, respectively, and &lt;math&gt;6&lt;/math&gt; segments of length &lt;math&gt;a_6&lt;/math&gt;. There are &lt;math&gt;66&lt;/math&gt; segments in total.<br /> <br /> In the above list there are &lt;math&gt;3&lt;/math&gt; triples of the type a-a-b without ''6'', &lt;math&gt;2&lt;/math&gt; triples of a-a-6 where a is not ''6'', &lt;math&gt;3&lt;/math&gt; triples of a-b-c without ''6'', and &lt;math&gt;2&lt;/math&gt; triples of a-b-6 where a, b are not ''6''. So,<br /> <br /> &lt;cmath&gt;1-p = \frac{1}{66\cdot 65\cdot 64} ( 3\cdot 3 \cdot 12\cdot 11\cdot 12 + 2\cdot 3 \cdot 12\cdot 11\cdot 6 + 3\cdot 6\cdot 12^3 + 2\cdot 6 \cdot 12^2 \cdot 6)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt; = \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286} &lt;/cmath&gt;<br /> <br /> So &lt;math&gt;p = 223/286&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Just like Solution 1, we find &lt;math&gt;1-p&lt;/math&gt;. Let &lt;math&gt;a_n&lt;/math&gt; be the length of the diagonal that connects vertices that are &lt;math&gt;n&lt;/math&gt; sides apart. By the law of Cosines, we have &lt;cmath&gt;a_n^2=1+2a_n^2&lt;/cmath&gt; WLOG &lt;math&gt;a_1=1&lt;/math&gt;, then we can use the well known fact that &lt;math&gt;a_n=\sqrt{2^n-1}&lt;/math&gt; holds, thus &lt;math&gt;a_2=\sqrt{3}, a_2=\sqrt{7}, a_2=\sqrt{15}, a_2=\sqrt{31}, a_6=\sqrt{63}&lt;/math&gt;. Let the sides of the triangle be &lt;math&gt;p\le{r}\le{s}&lt;/math&gt;. It is obvious that if &lt;math&gt;x+y\ge{z}&lt;/math&gt;, then &lt;math&gt;a_x+a_y&gt;a_z&lt;/math&gt; too. Thus, we need only check &lt;math&gt;p=1, p=2&lt;/math&gt;.<br /> If &lt;math&gt;p=1&lt;/math&gt;: The ordered pairs &lt;math&gt;(r,s) = (1,3), (1,4), (1,5), (1,6), (2, 4), (2, 5), (2, 6), (3, 5), (3, 6)&lt;/math&gt; do not form a triangle for a total of &lt;math&gt;9&lt;/math&gt; cases. <br /> <br /> Now continue as in Solution 1. <br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2013amc12a/364<br /> <br /> ~dolphin7<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=Talk:2019_AMC_10C_Problems/Problem_22&diff=180242 Talk:2019 AMC 10C Problems/Problem 22 2022-11-06T00:17:43Z <p>Sigmapie: Created page with &quot;The original solution was flawed, so I edited it.&quot;</p> <hr /> <div>The original solution was flawed, so I edited it.</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10C_Problems/Problem_22&diff=180240 2019 AMC 10C Problems/Problem 22 2022-11-06T00:16:35Z <p>Sigmapie: </p> <hr /> <div>===Problem===<br /> <br /> &lt;math&gt;100&lt;/math&gt; chicks are sitting in a circle. The first chick in the circle says the number &lt;math&gt;1&lt;/math&gt;, then the chick &lt;math&gt;2&lt;/math&gt; seats away from the first chick says the number &lt;math&gt;2&lt;/math&gt;, then the chick &lt;math&gt;3&lt;/math&gt; seats away from the chicken that said the number &lt;math&gt;2&lt;/math&gt; says the number &lt;math&gt;3&lt;/math&gt;, and so on. The process will always go clockwise. Some of the chicks in the circle will say more than one number while others might not even say a number at all. The process stops when the &lt;math&gt;1001&lt;/math&gt;th number is said. How many numbers would the chick that said &lt;math&gt;1001&lt;/math&gt; have said by that point (including &lt;math&gt;1001&lt;/math&gt;)?<br /> <br /> <br /> &lt;math&gt;\mathrm{(A) \ } 20\qquad \mathrm{(B) \ } 21\qquad \mathrm{(C) \ } 30\qquad \mathrm{(D) \ } 31\qquad \mathrm{(E) \ } 41&lt;/math&gt;<br /> <br /> ===Solution===<br /> <br /> The &lt;math&gt;n^{th}&lt;/math&gt; number is said by chick &lt;math&gt;\frac{n(n+1)}{2} \pmod{100}&lt;/math&gt;. Note that when &lt;math&gt;n=1001&lt;/math&gt;, the first chick said the number. Thus, we are looking for &lt;math&gt;\frac{n(n+1)}{2}\equiv{2} \pmod{100}&lt;/math&gt;, or &lt;math&gt;\frac{(n+2)(n-1)}{2}\equiv{0} \pmod{100}&lt;/math&gt;. This is equivalent to stating that &lt;math&gt;(n+2)(n-1)\equiv{0} \pmod{8}&lt;/math&gt; and &lt;math&gt;(n+2)(n-1)\equiv{0} \pmod{25}&lt;/math&gt;. It is clear that there are &lt;math&gt;2*2=4&lt;/math&gt; numbers that the first chick says every &lt;math&gt;8*25=200&lt;/math&gt; numbers said by all chicks, so our answer is &lt;math&gt;4\cdot{1000/200}+1=\boxed{21}&lt;/math&gt;</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10C_Problems/Problem_22&diff=180239 2019 AMC 10C Problems/Problem 22 2022-11-06T00:15:55Z <p>Sigmapie: </p> <hr /> <div>===Problem===<br /> <br /> &lt;math&gt;100&lt;/math&gt; chicks are sitting in a circle. The first chick in the circle says the number &lt;math&gt;1&lt;/math&gt;, then the chick &lt;math&gt;2&lt;/math&gt; seats away from the first chick says the number &lt;math&gt;2&lt;/math&gt;, then the chick &lt;math&gt;3&lt;/math&gt; seats away from the chicken that said the number &lt;math&gt;2&lt;/math&gt; says the number &lt;math&gt;3&lt;/math&gt;, and so on. The process will always go clockwise. Some of the chicks in the circle will say more than one number while others might not even say a number at all. The process stops when the &lt;math&gt;1001&lt;/math&gt;th number is said. How many numbers would the chick that said &lt;math&gt;1001&lt;/math&gt; have said by that point (including &lt;math&gt;1001&lt;/math&gt;)?<br /> <br /> <br /> &lt;math&gt;\mathrm{(A) \ } 20\qquad \mathrm{(B) \ } 21\qquad \mathrm{(C) \ } 30\qquad \mathrm{(D) \ } 31\qquad \mathrm{(E) \ } 41&lt;/math&gt;<br /> <br /> ===Solution===<br /> <br /> The &lt;math&gt;n^{th}&lt;/math&gt; number is said by chick &lt;math&gt;\frac{n(n+1)}{2} \pmod{100}&lt;/math&gt;. Note that when &lt;math&gt;n=1001&lt;/math&gt;, the first chick said the number. Thus, we are looking for &lt;math&gt;\frac{n(n+1)}{2}\equiv{2} \pmod{100}&lt;/math&gt;, or &lt;math&gt;\frac{(n+2)(n-1)}{2}\equiv{0} \pmod{100}&lt;/math&gt;. This is equivalent to stating that &lt;math&gt;(n+2)(n-1)\equiv{0} \pmod{8}&lt;/math&gt; and &lt;math&gt;(n+2)(n-1)\equiv{0} \pmod{25}&lt;/math&gt;. It is clear that there are &lt;math&gt;4&lt;/math&gt; numbers that the first chick says every &lt;math&gt;200&lt;/math&gt; numbers said by all chicks, so our answer is &lt;math&gt;4\cdot{1000/200}+1=\boxed{21}&lt;/math&gt;</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10C_Problems/Problem_22&diff=180238 2019 AMC 10C Problems/Problem 22 2022-11-06T00:15:11Z <p>Sigmapie: </p> <hr /> <div>===Problem===<br /> <br /> &lt;math&gt;100&lt;/math&gt; chicks are sitting in a circle. The first chick in the circle says the number &lt;math&gt;1&lt;/math&gt;, then the chick &lt;math&gt;2&lt;/math&gt; seats away from the first chick says the number &lt;math&gt;2&lt;/math&gt;, then the chick &lt;math&gt;3&lt;/math&gt; seats away from the chicken that said the number &lt;math&gt;2&lt;/math&gt; says the number &lt;math&gt;3&lt;/math&gt;, and so on. The process will always go clockwise. Some of the chicks in the circle will say more than one number while others might not even say a number at all. The process stops when the &lt;math&gt;1001&lt;/math&gt;th number is said. How many numbers would the chick that said &lt;math&gt;1001&lt;/math&gt; have said by that point (including &lt;math&gt;1001&lt;/math&gt;)?<br /> <br /> <br /> &lt;math&gt;\mathrm{(A) \ } 20\qquad \mathrm{(B) \ } 21\qquad \mathrm{(C) \ } 30\qquad \mathrm{(D) \ } 31\qquad \mathrm{(E) \ } 41&lt;/math&gt;<br /> <br /> ===Solution===<br /> <br /> The &lt;math&gt;n^{th}&lt;/math&gt; number is said by the chick &lt;math&gt;\frac{n(n+1)}{2}-1\pmod{100}&lt;/math&gt; away from the first chick. This expression is &lt;math&gt;0&lt;/math&gt; when &lt;math&gt;n=1001&lt;/math&gt;, so we are looking when &lt;math&gt;n(n+1)\equiv 2\pmod{4}&lt;/math&gt; and &lt;math&gt;n(n+1)\equiv 2\pmod{25}&lt;/math&gt;. The first congruence is satisfied when &lt;math&gt;n\equiv 1,2\pmod{4}&lt;/math&gt; and the second congruence is satisfied when &lt;math&gt;n\equiv 1\pmod{25}&lt;/math&gt;, so the answer is &lt;math&gt;21&lt;/math&gt;.<br /> <br /> Edit: This solution is flawed, although it does lead to the correct answer. <br /> <br /> ===Solution 2 (Correct Version)===<br /> The &lt;math&gt;n^{th}&lt;/math&gt; number is said by chick &lt;math&gt;\frac{n(n+1)}{2} \pmod{100}&lt;/math&gt;. Note that when &lt;math&gt;n=1001&lt;/math&gt;, the first chick said the number. Thus, we are looking for &lt;math&gt;\frac{n(n+1)}{2}\equiv{2} \pmod{100}&lt;/math&gt;, or &lt;math&gt;\frac{(n+2)(n-1)}{2}\equiv{0} \pmod{100}&lt;/math&gt;. This is equivalent to stating that &lt;math&gt;(n+2)(n-1)\equiv{0} \pmod{8}&lt;/math&gt; and &lt;math&gt;(n+2)(n-1)\equiv{0} \pmod{25}&lt;/math&gt;. It is clear that there are &lt;math&gt;4&lt;/math&gt; numbers that the first chick says every &lt;math&gt;200&lt;/math&gt; numbers said by all chicks, so our answer is &lt;math&gt;4\cdot{1000/200}+1=\boxed{21}&lt;/math&gt;</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_16&diff=179746 2021 AMC 12B Problems/Problem 16 2022-10-29T23:37:50Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;g(x)&lt;/math&gt; be a polynomial with leading coefficient &lt;math&gt;1,&lt;/math&gt; whose three roots are the reciprocals of the three roots of &lt;math&gt;f(x)=x^3+ax^2+bx+c,&lt;/math&gt; where &lt;math&gt;1&lt;a&lt;b&lt;c.&lt;/math&gt; What is &lt;math&gt;g(1)&lt;/math&gt; in terms of &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\frac{a+b+c}{c^2} \qquad \textbf{(E) }\frac{1+a+b+c}{a+b+c}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that &lt;math&gt;f(1/x)&lt;/math&gt; has the same roots as &lt;math&gt;g(x)&lt;/math&gt;, if it is multiplied by some monomial so that the leading term is &lt;math&gt;x^3&lt;/math&gt; they will be equal. We have<br /> &lt;cmath&gt;f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c&lt;/cmath&gt;<br /> so we can see that<br /> &lt;cmath&gt;g(x) = \frac{x^3}{c}f(1/x)&lt;/cmath&gt;<br /> Therefore<br /> &lt;cmath&gt;g(1) = \frac{1}{c}f(1) = \boxed{\textbf{(A) }\frac{1+a+b+c}c}&lt;/cmath&gt;<br /> <br /> ==Solution 2 (Vieta's bash)==<br /> Let the three roots of &lt;math&gt;f(x)&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;. (Here e does NOT mean 2.7182818...)<br /> We know that &lt;math&gt;a=-(d+e+f)&lt;/math&gt;, &lt;math&gt;b=de+ef+df&lt;/math&gt;, and &lt;math&gt;c=-def&lt;/math&gt;, and that &lt;math&gt;g(1)=1-\frac{1}{d}-\frac{1}{e}-\frac{1}{f}+\frac{1}{de}+\frac{1}{ef}+\frac{1}{df}-\frac{1}{def}&lt;/math&gt; (Vieta's). This is equal to &lt;math&gt;\frac{def-de-df-ef+d+e+f-1}{def}&lt;/math&gt;, which equals &lt;math&gt;\boxed{(\textbf{A}) \frac{1+a+b+c}{c}}&lt;/math&gt;. -dstanz5<br /> <br /> ==Solution 3 (Fakesolve) ==<br /> <br /> Because the problem doesn't specify what the coefficients of the polynomial actually are, we can just plug in any arbitrary polynomial that satisfies the constraints. Let's take &lt;math&gt;f(x) = (x+5)^3 = x^3+15x^2+75x+125&lt;/math&gt;. Then &lt;math&gt;f(x)&lt;/math&gt; has a triple root of &lt;math&gt;x = -5&lt;/math&gt;. Then &lt;math&gt;g(x)&lt;/math&gt; has a triple root of &lt;math&gt;-\frac{1}{5}&lt;/math&gt;, and it's monic, so &lt;math&gt;g(x) = \left(x+\frac{1}{5}\right)^3 = \frac{125x^3+75x^2+15x+1}{125}&lt;/math&gt;. We can see that this is &lt;math&gt;\frac{1+a+b+c}{c}&lt;/math&gt;, which is answer choice &lt;math&gt;\boxed{(A)}&lt;/math&gt;.<br /> <br /> -Darren Yao<br /> <br /> ==Solution 4==<br /> <br /> If we let &lt;math&gt;p, q, &lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; be the roots of &lt;math&gt;f(x)&lt;/math&gt;, &lt;math&gt;f(x) = (x-p)(x-q)(x-r)&lt;/math&gt; and &lt;math&gt;g(x) = (x-\frac{1}{p})(x-\frac{1}{q})(x-\frac{1}{r})&lt;/math&gt;. The requested value, &lt;math&gt;g(1)&lt;/math&gt;, is then<br /> &lt;cmath&gt;(1-\frac{1}{p})(1-\frac{1}{q})(1-\frac{1}{r}) = \frac{(p-1)(q-1)(r-1)}{pqr}&lt;/cmath&gt;<br /> The numerator is &lt;math&gt;-f(1)&lt;/math&gt; (using the product form of &lt;math&gt;f(x)&lt;/math&gt; ) and the denominator is &lt;math&gt;-c&lt;/math&gt;, so the answer is<br /> &lt;cmath&gt;\frac{f(1)}{c} = \boxed{(\textbf{A}) \frac{1+a+b+c}{c}}&lt;/cmath&gt;<br /> <br /> - gting<br /> <br /> ==Solution 5 (good at guessing)==<br /> g(1) = sum of coefficients. if its (x-r)(x-s)(x-t), then it becomes (x-1/r)(x-1/s)(x-1/t)<br /> so -rst becomes -1/rst so c becomes 1/c. Also, there is x^3 so the answer must include 1. the only answer having both of these is A.<br /> <br /> ~smellyman<br /> <br /> ==Solution 6==<br /> It is well known that reversing the order of the coefficients of a polynomial turns each root into its corresponding reciprocal. Thus, a polynomial with the desired roots may be written as &lt;math&gt;cx^3+bx^2+a+1&lt;/math&gt;. As the problem statement asks for a monic polynomial, our answer is &lt;cmath&gt;\boxed{(\textbf{A}) \frac{1+a+b+c}{c}}&lt;/cmath&gt;<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtube.com/watch?v=vCEJzhDRUoU<br /> <br /> == Video Solution by OmegaLearn (Vieta's Formula) ==<br /> https://youtu.be/afrGHNo_JcY<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=p4iCAZRUESs<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=15|num-a=17}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_16&diff=179745 2021 AMC 12B Problems/Problem 16 2022-10-29T23:36:47Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;g(x)&lt;/math&gt; be a polynomial with leading coefficient &lt;math&gt;1,&lt;/math&gt; whose three roots are the reciprocals of the three roots of &lt;math&gt;f(x)=x^3+ax^2+bx+c,&lt;/math&gt; where &lt;math&gt;1&lt;a&lt;b&lt;c.&lt;/math&gt; What is &lt;math&gt;g(1)&lt;/math&gt; in terms of &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\frac{a+b+c}{c^2} \qquad \textbf{(E) }\frac{1+a+b+c}{a+b+c}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that &lt;math&gt;f(1/x)&lt;/math&gt; has the same roots as &lt;math&gt;g(x)&lt;/math&gt;, if it is multiplied by some monomial so that the leading term is &lt;math&gt;x^3&lt;/math&gt; they will be equal. We have<br /> &lt;cmath&gt;f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c&lt;/cmath&gt;<br /> so we can see that<br /> &lt;cmath&gt;g(x) = \frac{x^3}{c}f(1/x)&lt;/cmath&gt;<br /> Therefore<br /> &lt;cmath&gt;g(1) = \frac{1}{c}f(1) = \boxed{\textbf{(A) }\frac{1+a+b+c}c}&lt;/cmath&gt;<br /> <br /> ==Solution 2 (Vieta's bash)==<br /> Let the three roots of &lt;math&gt;f(x)&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;. (Here e does NOT mean 2.7182818...)<br /> We know that &lt;math&gt;a=-(d+e+f)&lt;/math&gt;, &lt;math&gt;b=de+ef+df&lt;/math&gt;, and &lt;math&gt;c=-def&lt;/math&gt;, and that &lt;math&gt;g(1)=1-\frac{1}{d}-\frac{1}{e}-\frac{1}{f}+\frac{1}{de}+\frac{1}{ef}+\frac{1}{df}-\frac{1}{def}&lt;/math&gt; (Vieta's). This is equal to &lt;math&gt;\frac{def-de-df-ef+d+e+f-1}{def}&lt;/math&gt;, which equals &lt;math&gt;\boxed{(\textbf{A}) \frac{1+a+b+c}{c}}&lt;/math&gt;. -dstanz5<br /> <br /> ==Solution 3 (Fakesolve) ==<br /> <br /> Because the problem doesn't specify what the coefficients of the polynomial actually are, we can just plug in any arbitrary polynomial that satisfies the constraints. Let's take &lt;math&gt;f(x) = (x+5)^3 = x^3+15x^2+75x+125&lt;/math&gt;. Then &lt;math&gt;f(x)&lt;/math&gt; has a triple root of &lt;math&gt;x = -5&lt;/math&gt;. Then &lt;math&gt;g(x)&lt;/math&gt; has a triple root of &lt;math&gt;-\frac{1}{5}&lt;/math&gt;, and it's monic, so &lt;math&gt;g(x) = \left(x+\frac{1}{5}\right)^3 = \frac{125x^3+75x^2+15x+1}{125}&lt;/math&gt;. We can see that this is &lt;math&gt;\frac{1+a+b+c}{c}&lt;/math&gt;, which is answer choice &lt;math&gt;\boxed{(A)}&lt;/math&gt;.<br /> <br /> -Darren Yao<br /> <br /> ==Solution 4==<br /> <br /> If we let &lt;math&gt;p, q, &lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; be the roots of &lt;math&gt;f(x)&lt;/math&gt;, &lt;math&gt;f(x) = (x-p)(x-q)(x-r)&lt;/math&gt; and &lt;math&gt;g(x) = (x-\frac{1}{p})(x-\frac{1}{q})(x-\frac{1}{r})&lt;/math&gt;. The requested value, &lt;math&gt;g(1)&lt;/math&gt;, is then<br /> &lt;cmath&gt;(1-\frac{1}{p})(1-\frac{1}{q})(1-\frac{1}{r}) = \frac{(p-1)(q-1)(r-1)}{pqr}&lt;/cmath&gt;<br /> The numerator is &lt;math&gt;-f(1)&lt;/math&gt; (using the product form of &lt;math&gt;f(x)&lt;/math&gt; ) and the denominator is &lt;math&gt;-c&lt;/math&gt;, so the answer is<br /> &lt;cmath&gt;\frac{f(1)}{c} = \boxed{(\textbf{A}) \frac{1+a+b+c}{c}}&lt;/cmath&gt;<br /> <br /> - gting<br /> <br /> ==Solution 5 (good at guessing)==<br /> g(1) = sum of coefficients. if its (x-r)(x-s)(x-t), then it becomes (x-1/r)(x-1/s)(x-1/t)<br /> so -rst becomes -1/rst so c becomes 1/c. Also, there is x^3 so the answer must include 1. the only answer having both of these is A.<br /> <br /> ~smellyman<br /> <br /> ==Solution 6==<br /> It is well known that reversing the order of the coefficients of a polynomial turns each root into its corresponding reciprocal. Thus, a polynomial with the desired roots may be written as &lt;math&gt;cx^3+bx^2+a+1&lt;/math&gt;. As the problem statement asks for a monic polynomial, our answer is &lt;cmath&gt;\frac{f(1)}{c} = \boxed{(\textbf{A}) \frac{1+a+b+c}{c}}&lt;/cmath&gt;<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtube.com/watch?v=vCEJzhDRUoU<br /> <br /> == Video Solution by OmegaLearn (Vieta's Formula) ==<br /> https://youtu.be/afrGHNo_JcY<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=p4iCAZRUESs<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=15|num-a=17}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_16&diff=179744 2021 AMC 12B Problems/Problem 16 2022-10-29T23:35:40Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;g(x)&lt;/math&gt; be a polynomial with leading coefficient &lt;math&gt;1,&lt;/math&gt; whose three roots are the reciprocals of the three roots of &lt;math&gt;f(x)=x^3+ax^2+bx+c,&lt;/math&gt; where &lt;math&gt;1&lt;a&lt;b&lt;c.&lt;/math&gt; What is &lt;math&gt;g(1)&lt;/math&gt; in terms of &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\frac{a+b+c}{c^2} \qquad \textbf{(E) }\frac{1+a+b+c}{a+b+c}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that &lt;math&gt;f(1/x)&lt;/math&gt; has the same roots as &lt;math&gt;g(x)&lt;/math&gt;, if it is multiplied by some monomial so that the leading term is &lt;math&gt;x^3&lt;/math&gt; they will be equal. We have<br /> &lt;cmath&gt;f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c&lt;/cmath&gt;<br /> so we can see that<br /> &lt;cmath&gt;g(x) = \frac{x^3}{c}f(1/x)&lt;/cmath&gt;<br /> Therefore<br /> &lt;cmath&gt;g(1) = \frac{1}{c}f(1) = \boxed{\textbf{(A) }\frac{1+a+b+c}c}&lt;/cmath&gt;<br /> <br /> ==Solution 2 (Vieta's bash)==<br /> Let the three roots of &lt;math&gt;f(x)&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;. (Here e does NOT mean 2.7182818...)<br /> We know that &lt;math&gt;a=-(d+e+f)&lt;/math&gt;, &lt;math&gt;b=de+ef+df&lt;/math&gt;, and &lt;math&gt;c=-def&lt;/math&gt;, and that &lt;math&gt;g(1)=1-\frac{1}{d}-\frac{1}{e}-\frac{1}{f}+\frac{1}{de}+\frac{1}{ef}+\frac{1}{df}-\frac{1}{def}&lt;/math&gt; (Vieta's). This is equal to &lt;math&gt;\frac{def-de-df-ef+d+e+f-1}{def}&lt;/math&gt;, which equals &lt;math&gt;\boxed{(\textbf{A}) \frac{1+a+b+c}{c}}&lt;/math&gt;. -dstanz5<br /> <br /> ==Solution 3 (Fakesolve) ==<br /> <br /> Because the problem doesn't specify what the coefficients of the polynomial actually are, we can just plug in any arbitrary polynomial that satisfies the constraints. Let's take &lt;math&gt;f(x) = (x+5)^3 = x^3+15x^2+75x+125&lt;/math&gt;. Then &lt;math&gt;f(x)&lt;/math&gt; has a triple root of &lt;math&gt;x = -5&lt;/math&gt;. Then &lt;math&gt;g(x)&lt;/math&gt; has a triple root of &lt;math&gt;-\frac{1}{5}&lt;/math&gt;, and it's monic, so &lt;math&gt;g(x) = \left(x+\frac{1}{5}\right)^3 = \frac{125x^3+75x^2+15x+1}{125}&lt;/math&gt;. We can see that this is &lt;math&gt;\frac{1+a+b+c}{c}&lt;/math&gt;, which is answer choice &lt;math&gt;\boxed{(A)}&lt;/math&gt;.<br /> <br /> -Darren Yao<br /> <br /> ==Solution 4==<br /> <br /> If we let &lt;math&gt;p, q, &lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; be the roots of &lt;math&gt;f(x)&lt;/math&gt;, &lt;math&gt;f(x) = (x-p)(x-q)(x-r)&lt;/math&gt; and &lt;math&gt;g(x) = (x-\frac{1}{p})(x-\frac{1}{q})(x-\frac{1}{r})&lt;/math&gt;. The requested value, &lt;math&gt;g(1)&lt;/math&gt;, is then<br /> &lt;cmath&gt;(1-\frac{1}{p})(1-\frac{1}{q})(1-\frac{1}{r}) = \frac{(p-1)(q-1)(r-1)}{pqr}&lt;/cmath&gt;<br /> The numerator is &lt;math&gt;-f(1)&lt;/math&gt; (using the product form of &lt;math&gt;f(x)&lt;/math&gt; ) and the denominator is &lt;math&gt;-c&lt;/math&gt;, so the answer is<br /> &lt;cmath&gt;\frac{f(1)}{c} = \boxed{(\textbf{A}) \frac{1+a+b+c}{c}}&lt;/cmath&gt;<br /> <br /> - gting<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtube.com/watch?v=vCEJzhDRUoU<br /> <br /> == Video Solution by OmegaLearn (Vieta's Formula) ==<br /> https://youtu.be/afrGHNo_JcY<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Solution 5 (good at guessing)==<br /> g(1) = sum of coefficients. if its (x-r)(x-s)(x-t), then it becomes (x-1/r)(x-1/s)(x-1/t)<br /> so -rst becomes -1/rst so c becomes 1/c. Also, there is x^3 so the answer must include 1. the only answer having both of these is A.<br /> <br /> ==Solution 6==<br /> It is well known that reversing the order of the coefficients of a polynomial turns each root into its corresponding reciprocal. Thus, a polynomial with the desired roots may be written as &lt;math&gt;cx^3+bx^2+a+1&lt;/math&gt;. As the problem statement asks for a monic polynomial, our answer is &lt;cmath&gt;\frac{f(1)}{c} = \boxed{(\textbf{A}) \frac{1+a+b+c}{c}}&lt;/cmath&gt;<br /> <br /> ~smellyman<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=p4iCAZRUESs<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=15|num-a=17}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_13&diff=179719 2020 AIME II Problems/Problem 13 2022-10-29T20:09:10Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> Convex pentagon &lt;math&gt;ABCDE&lt;/math&gt; has side lengths &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=CD=DE=6&lt;/math&gt;, and &lt;math&gt;EA=7&lt;/math&gt;. Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of &lt;math&gt;ABCDE&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Assume the incircle touches &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;CD&lt;/math&gt;, &lt;math&gt;DE&lt;/math&gt;, &lt;math&gt;EA&lt;/math&gt; at &lt;math&gt;P,Q,R,S,T&lt;/math&gt; respectively. Then let &lt;math&gt;PB=x=BQ=RD=SD&lt;/math&gt;, &lt;math&gt;ET=y=ES=CR=CQ&lt;/math&gt;, &lt;math&gt;AP=AT=z&lt;/math&gt;. So we have &lt;math&gt;x+y=6&lt;/math&gt;, &lt;math&gt;x+z=5&lt;/math&gt; and &lt;math&gt;y+z&lt;/math&gt;=7, solve it we have &lt;math&gt;x=2&lt;/math&gt;, &lt;math&gt;z=3&lt;/math&gt;, &lt;math&gt;y=4&lt;/math&gt;. Let the center of the incircle be &lt;math&gt;I&lt;/math&gt;, by SAS we can proof triangle &lt;math&gt;BIQ&lt;/math&gt; is congruent to triangle &lt;math&gt;DIS&lt;/math&gt;, and triangle &lt;math&gt;CIR&lt;/math&gt; is congruent to triangle &lt;math&gt;SIE&lt;/math&gt;. Then we have &lt;math&gt;\angle AED=\angle BCD&lt;/math&gt;, &lt;math&gt;\angle ABC=\angle CDE&lt;/math&gt;. Extend &lt;math&gt;CD&lt;/math&gt;, cross ray &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;M&lt;/math&gt;, ray &lt;math&gt;AE&lt;/math&gt; at &lt;math&gt;N&lt;/math&gt;, then by AAS we have triangle &lt;math&gt;END&lt;/math&gt; is congruent to triangle &lt;math&gt;BMC&lt;/math&gt;. Thus &lt;math&gt;\angle M=\angle N&lt;/math&gt;. Let &lt;math&gt;EN=MC=a&lt;/math&gt;, then &lt;math&gt;BM=DN=a+2&lt;/math&gt;. So by law of cosine in triangle &lt;math&gt;END&lt;/math&gt; and triangle &lt;math&gt;ANM&lt;/math&gt; we can obtain &lt;cmath&gt;\frac{2a+8}{2(a+7)}=\cos N=\frac{a^2+(a+2)^2-36}{2a(a+2)}&lt;/cmath&gt;, solved it gives us &lt;math&gt;a=8&lt;/math&gt;, which yield triangle &lt;math&gt;ANM&lt;/math&gt; to be a triangle with side length 15, 15, 24, draw a height from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;NM&lt;/math&gt; divides it into two triangles with side lengths 9, 12, 15, so the area of triangle &lt;math&gt;ANM&lt;/math&gt; is 108. Triangle &lt;math&gt;END&lt;/math&gt; is a triangle with side lengths 6, 8, 10, so the area of two of them is 48, so the area of pentagon is &lt;math&gt;108-48=\boxed{60}&lt;/math&gt;.<br /> <br /> -Fanyuchen20020715<br /> <br /> ==Solution 2 (Complex Bash)==<br /> Suppose that the circle intersects &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, &lt;math&gt;\overline{DE}&lt;/math&gt;, and &lt;math&gt;\overline{EA}&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;Q&lt;/math&gt;, &lt;math&gt;R&lt;/math&gt;, &lt;math&gt;S&lt;/math&gt;, and &lt;math&gt;T&lt;/math&gt; respectively. Then &lt;math&gt;AT = AP = a&lt;/math&gt;, &lt;math&gt;BP = BQ = b&lt;/math&gt;, &lt;math&gt;CQ = CR = c&lt;/math&gt;, &lt;math&gt;DR = DS = d&lt;/math&gt;, and &lt;math&gt;ES = ET = e&lt;/math&gt;. So &lt;math&gt;a + b = 5&lt;/math&gt;, &lt;math&gt;b + c = 6&lt;/math&gt;, &lt;math&gt;c + d = 6&lt;/math&gt;, &lt;math&gt;d + e = 6&lt;/math&gt;, and &lt;math&gt;e + a = 7&lt;/math&gt;. Then &lt;math&gt;2a + 2b + 2c + 2d + 2e = 30&lt;/math&gt;, so &lt;math&gt;a + b + c + d + e= 15&lt;/math&gt;. Then we can solve for each individually. &lt;math&gt;a = 3&lt;/math&gt;, &lt;math&gt;b = 2&lt;/math&gt;, &lt;math&gt;c = 4&lt;/math&gt;, &lt;math&gt;d = 2&lt;/math&gt;, and &lt;math&gt;e = 4&lt;/math&gt;. To find the radius, we notice that &lt;math&gt;4 \arctan(\frac{2}{r}) + 4 \arctan(\frac{4}{r}) + 2 \arctan (\frac{3}{r}) = 360 ^ \circ&lt;/math&gt;, or &lt;math&gt;2 \arctan(\frac{2}{r}) + 2 \arctan(\frac{4}{r}) + \arctan (\frac{3}{r}) = 180 ^ \circ&lt;/math&gt;. Each of these angles in this could be represented by complex numbers. When two complex numbers are multiplied, their angles add up to create the angle of the resulting complex number. Thus, &lt;math&gt;(r + 2i)^2 \cdot (r + 4i)^2 \cdot (r + 3i)&lt;/math&gt; is real. Expanding, we get:<br /> &lt;math&gt;(r^2 + 4ir - 4)(r^2 + 8ir -16)(r + 3i)&lt;/math&gt;, then <br /> &lt;math&gt;(r^4 + 12ir^3 - 52r^2 - 96ir + 64)(r + 3i)&lt;/math&gt;.<br /> On the last expanding, we only multiply the reals with the imaginaries and vice versa, because we only care that the imaginary component equals 0.<br /> &lt;math&gt;15ir^4 - 252ir^2 + 192i = 0&lt;/math&gt;.<br /> &lt;math&gt;5r^4 - 84r^2 + 64 = 0&lt;/math&gt;<br /> &lt;math&gt;(5r^2 - 4)(r^2 - 16) = 0&lt;/math&gt;.<br /> &lt;math&gt;r&lt;/math&gt; must equal 4, as r cannot be negative or be approximately equal to 1. <br /> Thus, the area of &lt;math&gt;ABCDE&lt;/math&gt; is &lt;math&gt;4 \cdot (a + b + c + d + e) = 4 \cdot 15 = \boxed{60}&lt;/math&gt;<br /> <br /> -nihao4112<br /> <br /> ==Solution 3 (Guess)==<br /> This pentagon is very close to a regular pentagon with side lengths &lt;math&gt;6&lt;/math&gt;. The area of a regular pentagon with side lengths &lt;math&gt;s&lt;/math&gt; is &lt;math&gt;\frac{5s^2}{4\sqrt{5-2\sqrt{5}}}&lt;/math&gt;. &lt;math&gt;5-2\sqrt{5}&lt;/math&gt; is slightly greater than &lt;math&gt;\frac{1}{2}&lt;/math&gt; given that &lt;math&gt;2\sqrt{5}&lt;/math&gt; is slightly less than &lt;math&gt;\frac{9}{2}&lt;/math&gt;. &lt;math&gt;4\sqrt{5-2\sqrt{5}}&lt;/math&gt; is then slightly greater than &lt;math&gt;2\sqrt{2}&lt;/math&gt;. We will approximate that to be &lt;math&gt;2.9&lt;/math&gt;. The area is now roughly &lt;math&gt;\frac{180}{2.9}&lt;/math&gt;, but because the actual pentagon is not regular, but has the same perimeter of the regular one that we are comparing to we can say that this is an overestimate on the area and turn the &lt;math&gt;2.9&lt;/math&gt; into &lt;math&gt;3&lt;/math&gt; thus turning the area into &lt;math&gt;\frac{180}{3}&lt;/math&gt; which is &lt;math&gt;60&lt;/math&gt; and since &lt;math&gt;60&lt;/math&gt; is a multiple of the semiperimeter &lt;math&gt;15&lt;/math&gt;, we can safely say that the answer is most likely &lt;math&gt;\boxed{60}&lt;/math&gt;.<br /> <br /> ~Lopkiloinm<br /> <br /> ==Solution 4 (Guess)==<br /> Because the AIME answers have to be a whole number it would meant the radius of the circle have to be a whole number, thus by drawing the diagram and experimenting, we can safely say the radius is 4 and the answer is 60<br /> <br /> (Edit: While the guess would be technically correct, the assumption that the radius would have to be a whole number for the ans to be a whole number is wrong)<br /> <br /> By EtherealMidnight<br /> <br /> ==Solution 5 (Official MAA 1)==<br /> Let &lt;math&gt;\omega&lt;/math&gt; be the inscribed circle, &lt;math&gt;I&lt;/math&gt; be its center, and &lt;math&gt;r&lt;/math&gt; be its radius. The area of &lt;math&gt;ABCDE&lt;/math&gt; is equal to its semiperimeter, &lt;math&gt;15,&lt;/math&gt; times &lt;math&gt;r&lt;/math&gt;, so the problem is reduced to finding &lt;math&gt;r&lt;/math&gt;. Let &lt;math&gt;a&lt;/math&gt; be the length of the tangent segment from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;\omega&lt;/math&gt;, and analogously define &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt;. Then &lt;math&gt;a+b=5&lt;/math&gt;, &lt;math&gt;b+c= c+d=d+e=6&lt;/math&gt;, and &lt;math&gt;e+a=7&lt;/math&gt;, with a total of &lt;math&gt;a+b+c+d+e=15&lt;/math&gt;. Hence &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=d=2&lt;/math&gt;, and &lt;math&gt;c=e=4&lt;/math&gt;. It follows that &lt;math&gt;\angle B= \angle D&lt;/math&gt; and &lt;math&gt;\angle C= \angle E&lt;/math&gt;. Let &lt;math&gt;Q&lt;/math&gt; be the point where &lt;math&gt;\omega&lt;/math&gt; is tangent to &lt;math&gt;\overline{CD}&lt;/math&gt;. Then &lt;math&gt;\angle IAE = \angle IAB =\frac{1}{2}\angle A&lt;/math&gt;. The sum of the internal angles in polygons &lt;math&gt;ABCQI&lt;/math&gt; and &lt;math&gt;AIQDE&lt;/math&gt; are equal, so &lt;math&gt;\angle IAE + \angle AIQ + \angle IQD + \angle D + \angle E = \angle IAB + \angle B + \angle C + \angle CQI + \angle QIA&lt;/math&gt;, which implies that &lt;math&gt;\angle AIQ&lt;/math&gt; must be &lt;math&gt;180^\circ&lt;/math&gt;. Therefore points &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt;, and &lt;math&gt;Q&lt;/math&gt; are collinear.<br /> &lt;asy&gt;<br /> defaultpen(fontsize(8pt));<br /> unitsize(0.025cm);<br /> <br /> pair[] vertices = {(0,0), (5,0), (8.6,4.8), (3.8,8.4), (-1.96, 6.72)};<br /> string[] labels = {&quot;A$&quot;, &quot;$B$&quot;, &quot;$C$&quot;, &quot;$D$&quot;, &quot;$E$&quot;};<br /> pair[] dirs = {SW, SE,E, N, NW};<br /> <br /> <br /> string[] smallLabels = {&quot;$a$&quot;, &quot;$b$&quot;, &quot;$c$&quot;, &quot;$d$&quot;, &quot;$e$&quot;};<br /> <br /> pair I = (3,4); <br /> real rad = 4;<br /> pair Q = foot(I, vertices, vertices);<br /> <br /> pair[] interpoints = {}; <br /> for(int i =0; i&lt;vertices.length; ++i){<br /> interpoints.push(foot(I, vertices[i], vertices[(i+1)%vertices.length]));<br /> } <br /> <br /> <br /> for(int i = 0; i&lt; vertices.length; ++i){<br /> draw(vertices[i]--vertices[(i+1)%vertices.length]);<br /> dot(labels[i],vertices[i],dirs[i]);<br /> draw(I--vertices[i]);<br /> }<br /> draw(Circle(I, rad));<br /> dot(&quot;$I$&quot;, I, dir(200)); <br /> draw(I--Q); <br /> dot(&quot;$Q$&quot;, Q, NE);<br /> <br /> for(int i = 0; i &lt; vertices.length; ++i){ <br /> label(smallLabels[i], vertices[i] --interpoints[i]); <br /> //dot(interpoints[i], blue);<br /> label(smallLabels[i], interpoints[(i-1)%vertices.length] -- vertices[i]);<br /> }<br /> <br /> &lt;/asy&gt;<br /> Because &lt;math&gt;\overline{AQ} \perp \overline{CD}&lt;/math&gt;, it follows that&lt;cmath&gt;AC^2-AD^2=CQ^2-DQ^2=c^2-d^2=12.&lt;/cmath&gt;Another expression for &lt;math&gt;AC^2-AD^2&lt;/math&gt; can be found as follows. Note that &lt;math&gt;\tan \left(\frac{\angle B}{2}\right) = \frac{r}{2}&lt;/math&gt; and &lt;math&gt;\tan \left(\frac{\angle E}{2}\right) = \frac{r}{4}&lt;/math&gt;, so<br /> &lt;cmath&gt;\cos (\angle B) =\frac{1-\tan^2 \left(\frac{\angle B}{2}\right)}{1+\tan^2 \left(\frac{\angle B}{2}\right)} = \frac{4-r^2}{4+r^2}&lt;/cmath&gt;and<br /> &lt;cmath&gt;\cos (\angle E) = \frac{1-\tan^2 \left(\frac{\angle E}{2}\right)}{1+\tan^2 \left(\frac{\angle E}{2}\right)}= \frac{16-r^2}{16+r^2}.&lt;/cmath&gt;Applying the Law of Cosines to &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle AED&lt;/math&gt; gives<br /> &lt;cmath&gt;AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos (\angle B) = 5^2+6^2-2 \cdot 5 \cdot 6 \cdot \frac{4-r^2}{4+r^2}&lt;/cmath&gt;<br /> and<br /> &lt;cmath&gt;AD^2=AE^2+DE^2-2 \cdot AE \cdot DE \cdot \cos(\angle E) = 7^2+6^2-2 \cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}.&lt;/cmath&gt;<br /> Hence<br /> <br /> &lt;cmath&gt;12=AC^2- AD^2= 5^2-2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2} -7^2+2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2},&lt;/cmath&gt;<br /> yielding<br /> &lt;cmath&gt;2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}- 2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2}= 36;&lt;/cmath&gt;<br /> equivalently<br /> &lt;cmath&gt;7(16-r^2)(4+r^2)-5(4-r^2)(16+r^2) = 3(16+r^2)(4+r^2).&lt;/cmath&gt;<br /> Substituting &lt;math&gt;x=r^2&lt;/math&gt; gives the quadratic equation &lt;math&gt;5x^2-84x+64=0&lt;/math&gt;, with solutions &lt;math&gt;\frac{42 - 38}{5}=\frac45&lt;/math&gt;, and &lt;math&gt;\frac{42 + 38}{5}= 16&lt;/math&gt;. The solution &lt;math&gt;r^2=\frac45&lt;/math&gt; corresponds to a five-pointed star, which is not convex. Indeed, if &lt;math&gt;r&lt;3&lt;/math&gt;, then &lt;math&gt; \tan \left(\frac{\angle A}{2}\right)&lt;/math&gt;, &lt;math&gt;\tan \left(\frac{\angle C}{2}\right)&lt;/math&gt;, and &lt;math&gt;\tan \left(\frac{\angle E}{2}\right)&lt;/math&gt; are less than &lt;math&gt;1,&lt;/math&gt; implying that &lt;math&gt;\angle A&lt;/math&gt;, &lt;math&gt;\angle C&lt;/math&gt;, and &lt;math&gt;\angle E&lt;/math&gt; are acute, which cannot happen in a convex pentagon. Thus &lt;math&gt;r^2=16&lt;/math&gt; and &lt;math&gt;r=4&lt;/math&gt;. The requested area is &lt;math&gt;15\cdot4 = \boxed{60}&lt;/math&gt;.<br /> <br /> ==Solution 6 (Official MAA 2)==<br /> Define &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, &lt;math&gt;e&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; as in Solution 5. Then, as in Solution 5, &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=d=2&lt;/math&gt;, &lt;math&gt;c=e=4&lt;/math&gt;, &lt;math&gt;\angle B= \angle D&lt;/math&gt;, and &lt;math&gt;\angle C= \angle E&lt;/math&gt;. Let &lt;math&gt;\alpha =\frac{\angle A}{2}&lt;/math&gt;, &lt;math&gt;\beta = \frac{\angle B}{2}&lt;/math&gt;, and &lt;math&gt;\gamma=\frac{\angle C}{2}&lt;/math&gt;. It follows that &lt;math&gt;540^{\circ} = 2\alpha + 4 \beta + 4 \gamma&lt;/math&gt;, so &lt;math&gt;270^{\circ} = \alpha + 2\beta + 2 \gamma&lt;/math&gt;. Thus<br /> &lt;cmath&gt;\tan(2\beta + 2 \gamma) = \frac{1}{\tan \alpha},&lt;/cmath&gt;<br /> &lt;math&gt;\tan(\beta) = \frac{r}{2}&lt;/math&gt;, &lt;math&gt;\tan(\gamma) = \frac{r}{4}&lt;/math&gt;, and &lt;math&gt;\tan(\alpha) = \frac {r}{3}&lt;/math&gt;. By the Tangent Addition Formula,<br /> &lt;cmath&gt;\tan(\beta +\gamma) = \frac{6r}{8-r^2}&lt;/cmath&gt;<br /> and<br /> &lt;cmath&gt;\tan(2\beta + 2\gamma) = \frac{\frac{12r}{8-r^2}}{1-\frac{36r^2}{(8-r^2)^2}} = \frac{12r(8-r^2)}{(8-r^2)^2-36r^2}.&lt;/cmath&gt;<br /> Therefore<br /> &lt;cmath&gt;\frac{12r(8-r^2)}{(8-r^2)^2-36r^2} = \frac{3}{r},&lt;/cmath&gt;<br /> which simplifies to &lt;math&gt;5r^4 - 84r^2 + 64 = 0&lt;/math&gt;. Then the solution proceeds as in Solution 5.<br /> <br /> ==Solution 7 (Official MAA 3)==<br /> Define &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, &lt;math&gt;e&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; as in Solution 5. Note that<br /> &lt;cmath&gt;\arctan\left(\frac{a}{r}\right) + \arctan\left(\frac{b}{r}\right) + \arctan\left(\frac{c}{r}\right) + \arctan\left(\frac{d}{r}\right) + \arctan\left(\frac{e}{r}\right) = 180^{\circ}.&lt;/cmath&gt;<br /> Hence<br /> &lt;cmath&gt;\operatorname{Arg}(r + 3i) + 2\cdot \operatorname{Arg}(r + 2i) + 2\cdot \operatorname{Arg}(r + 4i) = 180^{\circ}.&lt;/cmath&gt;<br /> Therefore<br /> &lt;cmath&gt;\operatorname{Im} \big( (r + 3i)(r+2i)^2(r+4i)^2 \big) = 0.&lt;/cmath&gt;<br /> Simplifying this equation gives the same quadratic equation in &lt;math&gt;r^2&lt;/math&gt; as in Solution 5.<br /> <br /> ==Solution 8 (The same circle)==<br /> [[File:2020 AIME II 13.png|500px|right]]<br /> Notation shown on diagram. As in solution 5, we get &lt;math&gt;\overline{AQ} \perp \overline{CD}, AG = 3, GB = 2, CQ = 4&lt;/math&gt; and so on.<br /> <br /> Let &lt;math&gt; \overline{AB}&lt;/math&gt; cross &lt;math&gt; \overline{CD}&lt;/math&gt; at &lt;math&gt;F, \overline{AE}&lt;/math&gt; cross &lt;math&gt; \overline{CD}&lt;/math&gt; at &lt;math&gt;F', CF = x.&lt;/math&gt;<br /> &lt;math&gt;FQ = FG \implies FB = CF + CQ – BG = x + 2.&lt;/math&gt;<br /> &lt;math&gt;\angle BAQ = \angle EAQ \implies DF' = x + 2, EF' = x.&lt;/math&gt;<br /> <br /> Triangle &lt;math&gt;\triangle AFF'&lt;/math&gt; has semiperimeter &lt;math&gt;s = 2x + 11.&lt;/math&gt;<br /> <br /> The radius of &lt;i&gt;&lt;b&gt;incircle&lt;/b&gt;&lt;/i&gt; &lt;math&gt;\omega&lt;/math&gt; is <br /> &lt;math&gt;r =\sqrt{\frac{s-FF’}{s}}(s-AF) = \sqrt{\frac{3}{2x +11}}(x+4). &lt;/math&gt;<br /> <br /> Triangle &lt;math&gt;\triangle BCF&lt;/math&gt; has semiperimeter &lt;math&gt;s = x + 4.&lt;/math&gt;<br /> <br /> The radius of &lt;i&gt;&lt;b&gt;excircle &lt;/b&gt;&lt;/i&gt; &lt;math&gt;\omega&lt;/math&gt; is <br /> &lt;math&gt;r = \sqrt{\frac{s(s-BF)(s-CF)}{s-BC}} = \sqrt{ \frac{(x+4)\cdot 2 \cdot 4}{x - 2}}.&lt;/math&gt; <br /> <br /> It is the same radius, therefore<br /> &lt;cmath&gt; \sqrt{\frac{3}{2x +11}}(x+4) = \sqrt{\frac{8(x+4)}{x – 2}} \implies \frac {3(x+4)}{2x+11} = \frac {8}{x-2} \implies (x-8)(3x + 14) = 0 \implies x = 8, r = 4.&lt;/cmath&gt;<br /> <br /> Then the solution proceeds as in Solution 5.<br /> <br /> '''vladimir.shelomovskii@gmail.com, vvsss'''<br /> <br /> ==Video Solution 1==<br /> https://youtu.be/bz5N-jI2e0U?t=327<br /> <br /> ==Video Solution 2==<br /> https://youtu.be/_fwkGTdMd8U<br /> <br /> {{AIME box|year=2020|n=II|num-b=12|num-a=14}}<br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_16&diff=178944 2018 AMC 10B Problems/Problem 16 2022-10-09T03:57:00Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a_1,a_2,\dots,a_{2018}&lt;/math&gt; be a strictly increasing sequence of positive integers such that &lt;cmath&gt;a_1+a_2+\cdots+a_{2018}=2018^{2018}.&lt;/cmath&gt;<br /> What is the remainder when &lt;math&gt;a_1^3+a_2^3+\cdots+a_{2018}^3&lt;/math&gt; is divided by &lt;math&gt;6&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> One could simply list out all the residues to the third power &lt;math&gt;\mod 6&lt;/math&gt;. (Edit: Euler's totient theorem is not a valid approach to showing that they are all congruent &lt;math&gt;\mod 6&lt;/math&gt;. This is due to the fact that &lt;math&gt;a_k&lt;/math&gt; need not be relatively prime to &lt;math&gt;6&lt;/math&gt;. Edit 2: It can be if one splits up mod 6 into mod 2 and 3)<br /> <br /> Therefore the answer is congruent to &lt;math&gt;2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ (E)4}&lt;/math&gt;<br /> <br /> Note from Williamgolly: We can WLOG assume &lt;math&gt;a_1,a_2... a_{2017} \equiv 0 \pmod 6&lt;/math&gt; and have &lt;math&gt;a_{2018} \equiv 2 \pmod 6&lt;/math&gt; to make life easier.<br /> <br /> ==Solution 2==<br /> <br /> Note that &lt;math&gt;\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;<br /> a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2({2018}^{2018}-a_1)+3a_2^2({2018}^{2018}-a_2)+\cdots+3a_{2018}^2({2018}^{2018}-a_{2018})<br /> \equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6<br /> &lt;/math&gt;<br /> Therefore, &lt;math&gt;-2(a_1^3+a_2^3+\cdots+a_{2018}^3)\equiv \left(2018^{2018}\right)^3\equiv\left( 2^{2018}\right)^3\equiv 4^3\equiv 4\pmod{6}&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;a_1^3+a_2^3+\cdots+a_{2018}^3\equiv 1\pmod 3&lt;/math&gt;. However, since cubing preserves parity, and the sum of the individual terms is even, the some of the cubes is also even, and our answer is &lt;math&gt;\boxed{\text{(E) }4}&lt;/math&gt;<br /> <br /> ==Solution 3 (Partial Proof)==<br /> First, we can assume that the problem will have a consistent answer for all possible values of &lt;math&gt;a_1&lt;/math&gt;. For the purpose of this solution, we will assume that &lt;math&gt;a_1 = 1&lt;/math&gt;.<br /> <br /> We first note that &lt;math&gt;1^3+2^3+...+n^3 = (1+2+...+n)^2&lt;/math&gt;. So what we are trying to find is what &lt;math&gt;\left(2018^{2018}\right)^2=\left(2018^{4036}\right)&lt;/math&gt; mod &lt;math&gt;6&lt;/math&gt;. We start by noting that &lt;math&gt;2018&lt;/math&gt; is congruent to &lt;math&gt;2 \pmod{6}&lt;/math&gt;. So we are trying to find &lt;math&gt;\left(2^{4036}\right) \pmod{6}&lt;/math&gt;. Instead of trying to do this with some number theory skills, we could just look for a pattern. We start with small powers of &lt;math&gt;2&lt;/math&gt; and see that &lt;math&gt;2^1&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt; mod &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;2^2&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt; mod &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;2^3&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt; mod &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;2^4&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt; mod &lt;math&gt;6&lt;/math&gt;, and so on... So we see that since &lt;math&gt;\left(2^{4036}\right)&lt;/math&gt; has an even power, it must be congruent to &lt;math&gt;4 \pmod{6}&lt;/math&gt;, thus giving our answer &lt;math&gt;\boxed{\text{(E) }4}&lt;/math&gt;. You can prove this pattern using mods. But I thought this was easier.<br /> <br /> -TheMagician<br /> <br /> ==Solution 4 (Lazy solution)==<br /> First, we can assume that the problem will have a consistent answer for all possible values of &lt;math&gt;a_1&lt;/math&gt;. For the purpose of this solution, assume &lt;math&gt;a_1, a_2, ... a_{2017}&lt;/math&gt; are multiples of 6 and find &lt;math&gt;2018^{2018} \pmod{6}&lt;/math&gt; (which happens to be &lt;math&gt;4&lt;/math&gt;). Then &lt;math&gt;{a_1}^3 + ... + {a_{2018}}^3&lt;/math&gt; is congruent to &lt;math&gt;64 \pmod{6}&lt;/math&gt; or just &lt;math&gt;\boxed{\textbf{(E)} 4}&lt;/math&gt;. <br /> <br /> -Patrick4President<br /> <br /> ~minor edit made by CatachuKetchup~<br /> <br /> ==Solution 5 (Even Lazier Solution)==<br /> Due to the large amounts of variables in the problem, and the fact that the test is only 75 minutes, you can assume that the answer is probably just &lt;math&gt;2018^{2018} \pmod{6} &lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(E)} 4}&lt;/math&gt;.<br /> <br /> ~ Zeeshan12 [Be warned that this technique is not recommended for all problems and you should use it as a last resort]<br /> <br /> ==Algebraic Insight into Given Property==<br /> Mods is a good way to prove &lt;math&gt;a^3 \equiv a \pmod6&lt;/math&gt;: residues are simply &lt;math&gt;3, \pm 2, \pm 1, 0&lt;/math&gt;. Only &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; are necessary to check.<br /> Another way is to observe that &lt;math&gt;a^3-a&lt;/math&gt; factors into &lt;math&gt;(a-1)a(a+1)&lt;/math&gt;. Any &lt;math&gt;k&lt;/math&gt; consecutive numbers must be a multiple of &lt;math&gt;k&lt;/math&gt;, so &lt;math&gt;a^3-a&lt;/math&gt; is both divisible by &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;. This provides an algebraic method for proving &lt;math&gt;a^3 \equiv a \pmod6&lt;/math&gt; for all &lt;math&gt;a&lt;/math&gt;.<br /> <br /> ==Video Solution 1==<br /> With Modular Arithmetic Intro<br /> https://www.youtube.com/watch?v=wbv3TArroSs<br /> <br /> ~IceMatrix<br /> <br /> ==Video Solution 2==<br /> https://www.youtube.com/watch?v=SRjZ6B5DR74<br /> <br /> ~bunny1<br /> <br /> == Video Solution 3==<br /> https://youtu.be/4_x1sgcQCp4?t=112<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_13&diff=178943 2018 AMC 10B Problems/Problem 13 2022-10-09T03:51:54Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> How many of the first &lt;math&gt;2018&lt;/math&gt; numbers in the sequence &lt;math&gt;101, 1001, 10001, 100001, \dots&lt;/math&gt; are divisible by &lt;math&gt;101&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }253 \qquad<br /> \textbf{(B) }504 \qquad<br /> \textbf{(C) }505 \qquad<br /> \textbf{(D) }506 \qquad<br /> \textbf{(E) }1009 \qquad<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> The number &lt;math&gt;10^n+1&lt;/math&gt; is divisible by 101 if and only if &lt;math&gt;10^n\equiv -1\pmod{101}&lt;/math&gt;. We note that &lt;math&gt;(10,10^2,10^3,10^4)\equiv (10,-1,-10,1)\pmod{101}&lt;/math&gt;, so the powers of 10 are 4-periodic mod 101.<br /> <br /> It follows that &lt;math&gt;10^n\equiv -1\pmod{101}&lt;/math&gt; if and only if &lt;math&gt;n\equiv 2\pmod 4&lt;/math&gt;.<br /> <br /> In the given list, &lt;math&gt;10^2+1,10^3+1,10^4+1,\dots,10^{2019}+1&lt;/math&gt;, the desired exponents are &lt;math&gt;2,6,10,\dots,2018&lt;/math&gt;, and there are &lt;math&gt;\dfrac{2020}{4}=\boxed{\textbf{(C) } 505}&lt;/math&gt; numbers in that list.<br /> &lt;br&gt;<br /> <br /> ==Solution 2==<br /> Note that &lt;math&gt;10^{2k}+1&lt;/math&gt; for some odd &lt;math&gt;k&lt;/math&gt; will suffice &lt;math&gt;\mod {101}&lt;/math&gt;. Each &lt;math&gt;2k \in \{2,6,10,\dots,2018\}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) } 505}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> If we divide each number by &lt;math&gt;101&lt;/math&gt;, we see a pattern occuring in every 4 numbers. &lt;math&gt;101, 1000001, 10000000001, \dots&lt;/math&gt;. We divide &lt;math&gt;2018&lt;/math&gt; by &lt;math&gt;4&lt;/math&gt; to get &lt;math&gt;504&lt;/math&gt; with &lt;math&gt;2&lt;/math&gt; left over. Looking at our pattern of four numbers from above, the first number is divisible by &lt;math&gt;101&lt;/math&gt;. This means that the first of the &lt;math&gt;2&lt;/math&gt; left over will be divisible by &lt;math&gt;101&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{\textbf{(C) } 505}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Note that &lt;math&gt;909&lt;/math&gt; is divisible by &lt;math&gt;101&lt;/math&gt;, and thus &lt;math&gt;9999&lt;/math&gt; is too. We know that &lt;math&gt;101&lt;/math&gt; is divisible and &lt;math&gt;1001&lt;/math&gt; isn't so let us start from &lt;math&gt;10001&lt;/math&gt;. We subtract &lt;math&gt;9999&lt;/math&gt; to get 2. Likewise from &lt;math&gt;100001&lt;/math&gt; we subtract, but we instead subtract &lt;math&gt;9999&lt;/math&gt; times &lt;math&gt;10&lt;/math&gt; or &lt;math&gt;99990&lt;/math&gt; to get &lt;math&gt;11&lt;/math&gt;. We do it again and multiply the 9's by &lt;math&gt;10&lt;/math&gt; to get &lt;math&gt;101&lt;/math&gt;. Following the same knowledge, we can use mod &lt;math&gt;101&lt;/math&gt; to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is &lt;math&gt;{0,-9,-99 ( 2),11, 0, ...}&lt;/math&gt;. Thus it repeats every four. Consider the sequence after the 1st term and we have 2017 numbers. Divide &lt;math&gt;2017&lt;/math&gt; by four to get &lt;math&gt;504&lt;/math&gt; remainder &lt;math&gt;1&lt;/math&gt;. Thus the answer is &lt;math&gt;504&lt;/math&gt; plus the 1st term or &lt;math&gt;\boxed{\textbf{(C) } 505}&lt;/math&gt;.<br /> ==Solution 5==<br /> Note that &lt;math&gt;101=x^2+1&lt;/math&gt; and &lt;math&gt;100...0001=x^n+1&lt;/math&gt;, where &lt;math&gt;x=10&lt;/math&gt;. We have that &lt;math&gt;\frac{x^n+1}{x^2+1}&lt;/math&gt; must have a remainder of &lt;math&gt;0&lt;/math&gt;. By the remainder theorem, the roots of &lt;math&gt;x^2+1&lt;/math&gt; must also be roots of &lt;math&gt;x^n+1&lt;/math&gt;. Plugging in &lt;math&gt;i,-i&lt;/math&gt; to &lt;math&gt;x^n+1&lt;/math&gt; yields that &lt;math&gt;n\equiv2\mod{4}&lt;/math&gt;. Because the sequence starts with &lt;math&gt;10^2+1&lt;/math&gt;, the answer is &lt;math&gt;\lceil 2018/4 \rceil=\boxed{\textbf{(C) } 505}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/zfChnbMGLVQ?t=5122<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_13&diff=178941 2018 AMC 10B Problems/Problem 13 2022-10-09T03:50:35Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> How many of the first &lt;math&gt;2018&lt;/math&gt; numbers in the sequence &lt;math&gt;101, 1001, 10001, 100001, \dots&lt;/math&gt; are divisible by &lt;math&gt;101&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }253 \qquad<br /> \textbf{(B) }504 \qquad<br /> \textbf{(C) }505 \qquad<br /> \textbf{(D) }506 \qquad<br /> \textbf{(E) }1009 \qquad<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> The number &lt;math&gt;10^n+1&lt;/math&gt; is divisible by 101 if and only if &lt;math&gt;10^n\equiv -1\pmod{101}&lt;/math&gt;. We note that &lt;math&gt;(10,10^2,10^3,10^4)\equiv (10,-1,-10,1)\pmod{101}&lt;/math&gt;, so the powers of 10 are 4-periodic mod 101.<br /> <br /> It follows that &lt;math&gt;10^n\equiv -1\pmod{101}&lt;/math&gt; if and only if &lt;math&gt;n\equiv 2\pmod 4&lt;/math&gt;.<br /> <br /> In the given list, &lt;math&gt;10^2+1,10^3+1,10^4+1,\dots,10^{2019}+1&lt;/math&gt;, the desired exponents are &lt;math&gt;2,6,10,\dots,2018&lt;/math&gt;, and there are &lt;math&gt;\dfrac{2020}{4}=\boxed{\textbf{(C) } 505}&lt;/math&gt; numbers in that list.<br /> &lt;br&gt;<br /> <br /> ==Solution 2==<br /> Note that &lt;math&gt;10^{2k}+1&lt;/math&gt; for some odd &lt;math&gt;k&lt;/math&gt; will suffice &lt;math&gt;\mod {101}&lt;/math&gt;. Each &lt;math&gt;2k \in \{2,6,10,\dots,2018\}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) } 505}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> If we divide each number by &lt;math&gt;101&lt;/math&gt;, we see a pattern occuring in every 4 numbers. &lt;math&gt;101, 1000001, 10000000001, \dots&lt;/math&gt;. We divide &lt;math&gt;2018&lt;/math&gt; by &lt;math&gt;4&lt;/math&gt; to get &lt;math&gt;504&lt;/math&gt; with &lt;math&gt;2&lt;/math&gt; left over. Looking at our pattern of four numbers from above, the first number is divisible by &lt;math&gt;101&lt;/math&gt;. This means that the first of the &lt;math&gt;2&lt;/math&gt; left over will be divisible by &lt;math&gt;101&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{\textbf{(C) } 505}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Note that &lt;math&gt;909&lt;/math&gt; is divisible by &lt;math&gt;101&lt;/math&gt;, and thus &lt;math&gt;9999&lt;/math&gt; is too. We know that &lt;math&gt;101&lt;/math&gt; is divisible and &lt;math&gt;1001&lt;/math&gt; isn't so let us start from &lt;math&gt;10001&lt;/math&gt;. We subtract &lt;math&gt;9999&lt;/math&gt; to get 2. Likewise from &lt;math&gt;100001&lt;/math&gt; we subtract, but we instead subtract &lt;math&gt;9999&lt;/math&gt; times &lt;math&gt;10&lt;/math&gt; or &lt;math&gt;99990&lt;/math&gt; to get &lt;math&gt;11&lt;/math&gt;. We do it again and multiply the 9's by &lt;math&gt;10&lt;/math&gt; to get &lt;math&gt;101&lt;/math&gt;. Following the same knowledge, we can use mod &lt;math&gt;101&lt;/math&gt; to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is &lt;math&gt;{0,-9,-99 ( 2),11, 0, ...}&lt;/math&gt;. Thus it repeats every four. Consider the sequence after the 1st term and we have 2017 numbers. Divide &lt;math&gt;2017&lt;/math&gt; by four to get &lt;math&gt;504&lt;/math&gt; remainder &lt;math&gt;1&lt;/math&gt;. Thus the answer is &lt;math&gt;504&lt;/math&gt; plus the 1st term or &lt;math&gt;\boxed{\textbf{(C) } 505}&lt;/math&gt;.<br /> ==Solution 5==<br /> Note that &lt;math&gt;101=x^2+1&lt;/math&gt; and &lt;math&gt;100...0001=x^n+1&lt;/math&gt;, where &lt;math&gt;x=10&lt;/math&gt;. We have that &lt;math&gt;\frac{x^n+1}{x^2+1}&lt;/math&gt; must have a remainder of &lt;math&gt;0&lt;/math&gt;. By the remainder theorem, the roots of &lt;math&gt;x^2+1&lt;/math&gt; must also be roots of &lt;math&gt;x^n+1&lt;/math&gt;. Plugging in &lt;math&gt;i,-i&lt;/math&gt; to &lt;math&gt;x^n+1&lt;/math&gt; yields that &lt;math&gt;n\equiv2\mod{4}&lt;/math&gt;. Because the sequence starts with &lt;math&gt;10^2+1&lt;/math&gt;, the answer is &lt;math&gt;\ceil{2018/4}=\boxed{505}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/zfChnbMGLVQ?t=5122<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_14&diff=178430 2015 AMC 10A Problems/Problem 14 2022-09-25T02:24:16Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> <br /> The diagram below shows the circular face of a clock with radius &lt;math&gt;20&lt;/math&gt; cm and a circular disk with radius &lt;math&gt;10&lt;/math&gt; cm externally tangent to the clock face at &lt;math&gt;12&lt;/math&gt; o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?<br /> <br /> &lt;asy&gt;size(170);defaultpen(linewidth(0.9)+fontsize(13pt));draw(unitcircle^^circle((0,1.5),0.5));<br /> path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle;<br /> for(int i=1;i&lt;=12;i=i+1){draw(0.9*dir(90-30*i)--dir(90-30*i));label(&quot;$&quot;+(string) i+&quot;$&quot;,0.78*dir(90-30*i));}<br /> dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A) }\text{2 o' clock} \qquad\textbf{(B) }\text{3 o' clock} \qquad\textbf{(C) }\text{4 o' clock} \qquad\textbf{(D) }\text{6 o' clock} \qquad\textbf{(E) }\text{8 o' clock}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is &lt;math&gt;\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;. Similarly, the arrow would be pointing downward at 6:00. It would already have completed three 180 degree turns. Therefore, two 180 degree turns would be completed at 4:00.<br /> <br /> ==Solution 2==<br /> The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. &lt;math&gt;\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so, therefore, the disk must travel its circumference before the arrow goes up. Its circumference is &lt;math&gt;20\pi&lt;/math&gt;, so that is &lt;math&gt;20\pi&lt;/math&gt; traveled on a &lt;math&gt;60\pi&lt;/math&gt; arrow path. This is a ratio of 1/3, so the angle it carves is 120 degrees, which leads us to the correct answer of 4 o' clock. &lt;math&gt;\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> Suppose that the small disk also had a clock face on it, and that both disks were toothed wheels, free to rotate around their centers. The part of the picture where they engage would look like this:<br /> &lt;asy&gt;<br /> fill(arc((0,0),2,30,150)--cycle,lightgrey);<br /> draw(arc((0,0),2,30,150));<br /> draw(1.8*dir(90)--2*dir(90));<br /> draw(1.8*dir(60)--2*dir(60));<br /> label(&quot;12&quot;,1.56*dir(90));<br /> label(&quot;1&quot;,1.56*dir(60));<br /> <br /> draw(arc((0,3),1,-15,-165));<br /> draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3));<br /> draw(0.9*dir(-60)+(0,3)--dir(-60)+(0,3));<br /> draw(0.9*dir(-30)+(0,3)--dir(-30)+(0,3));<br /> label(&quot;6&quot;,.74*dir(-90)+(0,3));<br /> label(&quot;5&quot;,.74*dir(-60)+(0,3));<br /> label(&quot;4&quot;,.74*dir(-30)+(0,3));<br /> &lt;/asy&gt;<br /> The small cog has half the radius, and therefore half the circumference. If the large cog turns &lt;math&gt;30^\circ&lt;/math&gt; anticlockwise (i.e. 1 hour), the small cog turns &lt;math&gt;60^\circ&lt;/math&gt; clockwise (i.e. 2 hours).<br /> &lt;asy&gt;<br /> fill(arc((0,0),2,30,150)--cycle,lightgrey);<br /> draw(arc((0,0),2,30,150),EndArcArrow);label(&quot;$30^\circ$&quot;,2*dir(150),W);<br /> draw(1.8*dir(120)--2*dir(120));<br /> draw(1.8*dir(90)--2*dir(90));<br /> label(rotate(30)*&quot;12&quot;,1.56*dir(120));<br /> label(rotate(30)*&quot;1&quot;,1.56*dir(90));<br /> <br /> draw(arc((0,3),1,-15,-165),EndArcArrow);label(&quot;$60^\circ$&quot;,dir(-165)+(0,3),W);<br /> draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3));<br /> draw(0.9*dir(-120)+(0,3)--dir(-120)+(0,3));<br /> draw(0.9*dir(-150)+(0,3)--dir(-150)+(0,3));<br /> label(rotate(-60)*&quot;6&quot;,.74*dir(-150)+(0,3));<br /> label(rotate(-60)*&quot;5&quot;,.74*dir(-120)+(0,3));<br /> label(rotate(-60)*&quot;4&quot;,.74*dir(-90)+(0,3));<br /> &lt;/asy&gt;<br /> However, in the original problem the large cog does not rotate; it stays where it is. Therefore we must turn the whole diagram above &lt;math&gt;30^\circ&lt;/math&gt; clockwise to see what happens when the small cog rolls around it.<br /> &lt;asy&gt;<br /> fill(arc((0,0),2,30,150)--cycle,lightgrey);<br /> draw(arc((0,0),2,30,150));<br /> draw(1.8*dir(90)--2*dir(90));<br /> draw(1.8*dir(60)--2*dir(60));<br /> label(&quot;12&quot;,1.56*dir(90));<br /> label(&quot;1&quot;,1.56*dir(60));<br /> <br /> pair c=(1.5,sqrt(27)/2);<br /> draw(arc(c,1,0,-200),EndArcArrow);label(&quot;$90^\circ$&quot;,dir(-180)+c,W);<br /> draw(0.9*dir(-120)+c--dir(-120)+c);<br /> draw(0.9*dir(-150)+c--dir(-150)+c);<br /> draw(0.9*dir(-180)+c--dir(-180)+c);<br /> label(rotate(-90)*&quot;6&quot;,.74*dir(-180)+c);<br /> label(rotate(-90)*&quot;5&quot;,.74*dir(-150)+c);<br /> label(rotate(-90)*&quot;4&quot;,.74*dir(-120)+c);<br /> &lt;/asy&gt;<br /> It turns out that, when the point of tangency moves &lt;math&gt;30^\circ&lt;/math&gt; clockwise (one hour), from our point of view the small disk rotates &lt;math&gt;90^\circ&lt;/math&gt; clockwise (three hours) around its center. Thus, for the small disk to perform a complete rotation of &lt;math&gt;360^\circ&lt;/math&gt; (twelve hours) around its center from our point of view, the point of tangency must move round four hours. So the answer is &lt;math&gt;\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;<br /> <br /> <br /> ==Solution 5==<br /> We unfold the big clock and draw the small clock. We know that the small clocks arrow will be facing up again at 6 o clock on the unfolded big clock. We now draw corresponding arrows at each hour and fold it back again.<br /> <br /> Williamgolly<br /> *Edited by ab2024<br /> <br /> ==Solution 6==<br /> <br /> We can approach this problem with angle measures. As the circumference of the disk is &lt;math&gt;10\pi,&lt;/math&gt; and the clock &lt;math&gt;20 \pi,&lt;/math&gt; we have that 30 degrees, or the angular measure between hours, of the disk is only 15 degrees of the clock. This yields that every two hour ticks that the clock rotates, on the third one, the ticks will meet. However, the disk must rotate 360 degrees in order to come back to its original position, so the angular measure that the disk has covered relative to the clock is simply &lt;math&gt;12 \cdot 15 \cdot \frac{2}{3},&lt;/math&gt; or &lt;math&gt;120^\circ&lt;/math&gt; from the 12 starting point, so &lt;math&gt;\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;.<br /> <br /> ~ab2024<br /> <br /> ==Solution 7==<br /> If the big clock were a flat plane, then the smaller clock could travel &lt;math&gt;\dfrac{40\pi}{20\pi}=2&lt;/math&gt; full revolutions.<br /> <br /> But we also need to account for rotation. If we mark a red dot on the bottom of the small clock/bottom of the arrow, and then drag it around the clock, the direction of the arrow would still change. After traveling around the big clock, the small clock would travel &lt;math&gt;1&lt;/math&gt; full rotation.<br /> <br /> Considering these two movements, the small clock travels 3 full rotations around the big clock so the arrow is next pointing upwards at &lt;math&gt;\dfrac{12 \text{ o'clock}}{3}=\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;.<br /> <br /> ~BakedPotato66<br /> ==Solution 8==<br /> The center of rotation is in the center of the smaller circle, but extends to the center of the larger circle. That means the circumference of the circle in relation to the arrow is &lt;math&gt;60 \pi&lt;/math&gt;. The other circle is &lt;math&gt;20 \pi&lt;/math&gt; and so that is &lt;math&gt;\frac{1}{3}&lt;/math&gt;. So &lt;math&gt;\frac{12}{3} = 4&lt;/math&gt; which is &lt;math&gt;\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;.<br /> <br /> ==Solution 9 (Astronomy)==<br /> Let the center of the disk be Planet X with orbit eccentricity &lt;math&gt;0&lt;/math&gt; and the center of the clock be the Sun. Note that the question would then be asking for the solar day of Planet X, rather than the sidereal day. Because planet X is rotating around its axis in the same direction as it is revolving around the Sun, the solar day can be calculated as &lt;math&gt;d_{solar}=\frac{1}{\frac{1}{y}+\frac{1}{d_{siderial}}}&lt;/math&gt;, where &lt;math&gt;y&lt;/math&gt; is the length of the year of Planet X and &lt;math&gt;d_{siderial}&lt;/math&gt; is one sidereal day. It is easy to see that &lt;math&gt;y=12&lt;/math&gt; and &lt;math&gt;d_{siderial}=6&lt;/math&gt;, therefore the answer is &lt;math&gt;\frac{1}{\frac{1}{12}+\frac{1}{6}}=\frac{1}{\frac{1}{4}}=\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;<br /> <br /> ~sigmapie<br /> <br /> <br /> ==Video Solution 1==<br /> https://youtu.be/hPEgOYA4Xq8<br /> <br /> ~savannahsolver<br /> <br /> ==Video Solution 2==<br /> Video Solution: https://www.youtube.com/watch?v=kCsQbTi3iLU<br /> <br /> ==See Also==<br /> 2014 AIME I #10: https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_10<br /> <br /> Coin Rotation Paradox: https://en.wikipedia.org/wiki/Coin_rotation_paradox<br /> <br /> {{AMC10 box|year=2015|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Geometry Problems]]</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_14&diff=178429 2015 AMC 10A Problems/Problem 14 2022-09-25T02:23:11Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> <br /> The diagram below shows the circular face of a clock with radius &lt;math&gt;20&lt;/math&gt; cm and a circular disk with radius &lt;math&gt;10&lt;/math&gt; cm externally tangent to the clock face at &lt;math&gt;12&lt;/math&gt; o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?<br /> <br /> &lt;asy&gt;size(170);defaultpen(linewidth(0.9)+fontsize(13pt));draw(unitcircle^^circle((0,1.5),0.5));<br /> path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle;<br /> for(int i=1;i&lt;=12;i=i+1){draw(0.9*dir(90-30*i)--dir(90-30*i));label(&quot;$&quot;+(string) i+&quot;$&quot;,0.78*dir(90-30*i));}<br /> dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A) }\text{2 o' clock} \qquad\textbf{(B) }\text{3 o' clock} \qquad\textbf{(C) }\text{4 o' clock} \qquad\textbf{(D) }\text{6 o' clock} \qquad\textbf{(E) }\text{8 o' clock}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is &lt;math&gt;\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;. Similarly, the arrow would be pointing downward at 6:00. It would already have completed three 180 degree turns. Therefore, two 180 degree turns would be completed at 4:00.<br /> <br /> ==Solution 2==<br /> The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. &lt;math&gt;\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so, therefore, the disk must travel its circumference before the arrow goes up. Its circumference is &lt;math&gt;20\pi&lt;/math&gt;, so that is &lt;math&gt;20\pi&lt;/math&gt; traveled on a &lt;math&gt;60\pi&lt;/math&gt; arrow path. This is a ratio of 1/3, so the angle it carves is 120 degrees, which leads us to the correct answer of 4 o' clock. &lt;math&gt;\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> Suppose that the small disk also had a clock face on it, and that both disks were toothed wheels, free to rotate around their centers. The part of the picture where they engage would look like this:<br /> &lt;asy&gt;<br /> fill(arc((0,0),2,30,150)--cycle,lightgrey);<br /> draw(arc((0,0),2,30,150));<br /> draw(1.8*dir(90)--2*dir(90));<br /> draw(1.8*dir(60)--2*dir(60));<br /> label(&quot;12&quot;,1.56*dir(90));<br /> label(&quot;1&quot;,1.56*dir(60));<br /> <br /> draw(arc((0,3),1,-15,-165));<br /> draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3));<br /> draw(0.9*dir(-60)+(0,3)--dir(-60)+(0,3));<br /> draw(0.9*dir(-30)+(0,3)--dir(-30)+(0,3));<br /> label(&quot;6&quot;,.74*dir(-90)+(0,3));<br /> label(&quot;5&quot;,.74*dir(-60)+(0,3));<br /> label(&quot;4&quot;,.74*dir(-30)+(0,3));<br /> &lt;/asy&gt;<br /> The small cog has half the radius, and therefore half the circumference. If the large cog turns &lt;math&gt;30^\circ&lt;/math&gt; anticlockwise (i.e. 1 hour), the small cog turns &lt;math&gt;60^\circ&lt;/math&gt; clockwise (i.e. 2 hours).<br /> &lt;asy&gt;<br /> fill(arc((0,0),2,30,150)--cycle,lightgrey);<br /> draw(arc((0,0),2,30,150),EndArcArrow);label(&quot;$30^\circ$&quot;,2*dir(150),W);<br /> draw(1.8*dir(120)--2*dir(120));<br /> draw(1.8*dir(90)--2*dir(90));<br /> label(rotate(30)*&quot;12&quot;,1.56*dir(120));<br /> label(rotate(30)*&quot;1&quot;,1.56*dir(90));<br /> <br /> draw(arc((0,3),1,-15,-165),EndArcArrow);label(&quot;$60^\circ$&quot;,dir(-165)+(0,3),W);<br /> draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3));<br /> draw(0.9*dir(-120)+(0,3)--dir(-120)+(0,3));<br /> draw(0.9*dir(-150)+(0,3)--dir(-150)+(0,3));<br /> label(rotate(-60)*&quot;6&quot;,.74*dir(-150)+(0,3));<br /> label(rotate(-60)*&quot;5&quot;,.74*dir(-120)+(0,3));<br /> label(rotate(-60)*&quot;4&quot;,.74*dir(-90)+(0,3));<br /> &lt;/asy&gt;<br /> However, in the original problem the large cog does not rotate; it stays where it is. Therefore we must turn the whole diagram above &lt;math&gt;30^\circ&lt;/math&gt; clockwise to see what happens when the small cog rolls around it.<br /> &lt;asy&gt;<br /> fill(arc((0,0),2,30,150)--cycle,lightgrey);<br /> draw(arc((0,0),2,30,150));<br /> draw(1.8*dir(90)--2*dir(90));<br /> draw(1.8*dir(60)--2*dir(60));<br /> label(&quot;12&quot;,1.56*dir(90));<br /> label(&quot;1&quot;,1.56*dir(60));<br /> <br /> pair c=(1.5,sqrt(27)/2);<br /> draw(arc(c,1,0,-200),EndArcArrow);label(&quot;$90^\circ$&quot;,dir(-180)+c,W);<br /> draw(0.9*dir(-120)+c--dir(-120)+c);<br /> draw(0.9*dir(-150)+c--dir(-150)+c);<br /> draw(0.9*dir(-180)+c--dir(-180)+c);<br /> label(rotate(-90)*&quot;6&quot;,.74*dir(-180)+c);<br /> label(rotate(-90)*&quot;5&quot;,.74*dir(-150)+c);<br /> label(rotate(-90)*&quot;4&quot;,.74*dir(-120)+c);<br /> &lt;/asy&gt;<br /> It turns out that, when the point of tangency moves &lt;math&gt;30^\circ&lt;/math&gt; clockwise (one hour), from our point of view the small disk rotates &lt;math&gt;90^\circ&lt;/math&gt; clockwise (three hours) around its center. Thus, for the small disk to perform a complete rotation of &lt;math&gt;360^\circ&lt;/math&gt; (twelve hours) around its center from our point of view, the point of tangency must move round four hours. So the answer is &lt;math&gt;\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;<br /> <br /> <br /> ==Solution 5==<br /> We unfold the big clock and draw the small clock. We know that the small clocks arrow will be facing up again at 6 o clock on the unfolded big clock. We now draw corresponding arrows at each hour and fold it back again.<br /> <br /> Williamgolly<br /> *Edited by ab2024<br /> <br /> ==Solution 6==<br /> <br /> We can approach this problem with angle measures. As the circumference of the disk is &lt;math&gt;10\pi,&lt;/math&gt; and the clock &lt;math&gt;20 \pi,&lt;/math&gt; we have that 30 degrees, or the angular measure between hours, of the disk is only 15 degrees of the clock. This yields that every two hour ticks that the clock rotates, on the third one, the ticks will meet. However, the disk must rotate 360 degrees in order to come back to its original position, so the angular measure that the disk has covered relative to the clock is simply &lt;math&gt;12 \cdot 15 \cdot \frac{2}{3},&lt;/math&gt; or &lt;math&gt;120^\circ&lt;/math&gt; from the 12 starting point, so &lt;math&gt;\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;.<br /> <br /> ~ab2024<br /> <br /> ==Solution 7==<br /> If the big clock were a flat plane, then the smaller clock could travel &lt;math&gt;\dfrac{40\pi}{20\pi}=2&lt;/math&gt; full revolutions.<br /> <br /> But we also need to account for rotation. If we mark a red dot on the bottom of the small clock/bottom of the arrow, and then drag it around the clock, the direction of the arrow would still change. After traveling around the big clock, the small clock would travel &lt;math&gt;1&lt;/math&gt; full rotation.<br /> <br /> Considering these two movements, the small clock travels 3 full rotations around the big clock so the arrow is next pointing upwards at &lt;math&gt;\dfrac{12 \text{ o'clock}}{3}=\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;.<br /> <br /> ~BakedPotato66<br /> ==Solution 8==<br /> The center of rotation is in the center of the smaller circle, but extends to the center of the larger circle. That means the circumference of the circle in relation to the arrow is &lt;math&gt;60 \pi&lt;/math&gt;. The other circle is &lt;math&gt;20 \pi&lt;/math&gt; and so that is &lt;math&gt;\frac{1}{3}&lt;/math&gt;. So &lt;math&gt;\frac{12}{3} = 4&lt;/math&gt; which is &lt;math&gt;\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;.<br /> <br /> ==Solution 9 (Astronomy)==<br /> Let the center of the disk be Planet X with orbit eccentricity &lt;math&gt;0&lt;/math&gt; and the center of the clock be the Sun. Note that the question would then be asking for the solar day of Planet X, rather than the sidereal day. Because planet X is rotating around its axis in the same direction as it is revolving around the Sun, the solar day can be calculated as &lt;math&gt;d_{solar}=\frac{1}{\frac{1}{y}+\frac{1}{d_{siderial}}}&lt;/math&gt;, where &lt;math&gt;y&lt;/math&gt; is the length of the year of Planet X and &lt;math&gt;d_{siderial}&lt;/math&gt; is one sidereal day. It is easy to see that &lt;math&gt;y=12&lt;/math&gt; and &lt;math&gt;d_{siderial}=6&lt;/math&gt;, therefore the answer is &lt;math&gt;\frac{1}{\frac{1}{12}+\frac{1}{6}}=\frac{1}{\frac{1}{4}}=\boxed{\textbf{(C) }4 \ \text{o' clock}}&lt;/math&gt;$<br /> <br /> <br /> ==Video Solution 1==<br /> https://youtu.be/hPEgOYA4Xq8<br /> <br /> ~savannahsolver<br /> <br /> ==Video Solution 2==<br /> Video Solution: https://www.youtube.com/watch?v=kCsQbTi3iLU<br /> <br /> ==See Also==<br /> 2014 AIME I #10: https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_10<br /> <br /> Coin Rotation Paradox: https://en.wikipedia.org/wiki/Coin_rotation_paradox<br /> <br /> {{AMC10 box|year=2015|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Geometry Problems]]</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_23&diff=178267 2011 AMC 10A Problems/Problem 23 2022-09-16T20:01:39Z <p>Sigmapie: </p> <hr /> <div>== Problem ==<br /> Seven students count from 1 to 1000 as follows:<br /> <br /> Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br /> <br /> Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br /> <br /> Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br /> <br /> Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br /> <br /> Finally, George says the only number that no one else says.<br /> <br /> What number does George say?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> First look at the numbers Alice says. &lt;math&gt;1, 3, 4, 6, 7, 9 \cdots&lt;/math&gt; skipping every number that is congruent to &lt;math&gt;2 \pmod 3&lt;/math&gt;. Thus, Barbara says those numbers EXCEPT every second - being &lt;math&gt;2 + 3^1 \equiv 5 \pmod{3^2=9}&lt;/math&gt;. So Barbara skips every number congruent to &lt;math&gt;5 \pmod 9&lt;/math&gt;. We continue and see: <br /> <br /> Alice skips &lt;math&gt;2 \pmod 3&lt;/math&gt;, Barbara skips &lt;math&gt;5 \pmod 9&lt;/math&gt;, Candice skips &lt;math&gt;14 \pmod {27}&lt;/math&gt;, Debbie skips &lt;math&gt;41 \pmod {81}&lt;/math&gt;, Eliza skips &lt;math&gt;122 \pmod {243}&lt;/math&gt;, and Fatima skips &lt;math&gt;365 \pmod {729}&lt;/math&gt;.<br /> <br /> Since the only number congruent to &lt;math&gt;365 \pmod {729}&lt;/math&gt; and less than &lt;math&gt;1,000&lt;/math&gt; is &lt;math&gt;365&lt;/math&gt;, the correct answer is &lt;math&gt; \boxed{365\ \mathbf{(C)}} &lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> After Alice says all her numbers, the numbers not mentioned yet are &lt;cmath&gt;\text{Alice: } 2,5,8,11,14,17,\cdots,998.&lt;/cmath&gt;After Barbara says all her numbers, the numbers that haven't been said yet are &lt;cmath&gt;\text{Barbara: } 5,14,23,32,41,50,\cdots,995.&lt;/cmath&gt;After Candice, the list is &lt;cmath&gt;\text{Candice: } 14,41,68,\cdots,986.&lt;/cmath&gt;Notice how each list is an arithmetic sequence where the common ratio is thrice the common ratio of the previous list and the first term is the second term of the previous list. Now that the pattern is clear, we construct the rest of the lists: &lt;cmath&gt;\text{Debbie: } 41,122,203,\cdots,959&lt;/cmath&gt;&lt;cmath&gt;\text{Eliza: } 122,365,608,878&lt;/cmath&gt;&lt;cmath&gt;\text{Fatima: } 365&lt;/cmath&gt;<br /> <br /> Thus, George says &lt;math&gt;\boxed{\textbf{(C) } 365}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> Notice that Alice has skipped the numbers &lt;math&gt;3n-1&lt;/math&gt; for &lt;math&gt;n=1,2,3,...,333&lt;/math&gt;.<br /> Namely, &lt;cmath&gt;3\cdot1-1,3\cdot2-1,3\cdot3-1,...,3\cdot333-1&lt;/cmath&gt;<br /> Thus the numbers that Barbara skips are &lt;cmath&gt;3\cdot2-1,3\cdot5-1,3\cdot8-1,...&lt;/cmath&gt;<br /> or in a more general expression, &lt;math&gt;3(3n-1)-1&lt;/math&gt; for &lt;math&gt;n=1,2,3,...&lt;/math&gt;.<br /> Namely, &lt;cmath&gt;3\cdot(3\cdot1-1)-1,3\cdot(3\cdot2-1)-1,3\cdot(3\cdot3-1)-1,...&lt;/cmath&gt;<br /> Repeating the pattern until George, we have the first number he says,<br /> &lt;cmath&gt;3(3(3(3(3(3\cdot1-1)-1)-1)-1)-1)-1= \boxed{\textbf{(C) } 365}&lt;/cmath&gt;<br /> In addition, note that the second number George says exceeds &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> == Solution 4 (Answer Choices) ==<br /> Similar to Solution 1 we find that the only number that George can say must leave a remainder of &lt;math&gt;2&lt;/math&gt; when divided by &lt;math&gt;3&lt;/math&gt;, and that it must also leave a remainder of &lt;math&gt;5&lt;/math&gt; when divided by &lt;math&gt;9&lt;/math&gt;. Since we as human beings are usually lazy, and that MAA provides answer choices, we check all the possible numbers and find that our answer is &lt;math&gt;\boxed{\textbf{(C) } 365}&lt;/math&gt;.<br /> <br /> == Solution 5==<br /> Every integer from 1 to 1000 can be written in the form &lt;math&gt;t_n+1&lt;/math&gt; in base 10, where &lt;math&gt;t_n&lt;/math&gt; is a trinary integer with no more than 7 significant digits. The important insight is that person &lt;math&gt;1\le{k}\le6&lt;/math&gt; will not say &lt;math&gt;n&lt;/math&gt; if and only if the &lt;math&gt;k&lt;/math&gt;th digit from the right of &lt;math&gt;t_n&lt;/math&gt; is 1. Therefore, the last 6 digits of &lt;math&gt;t_n&lt;/math&gt; must be 1, as the first 6 people never said the number. The only options for &lt;math&gt;t_n&lt;/math&gt; are thus &lt;math&gt;2111111&lt;/math&gt;, &lt;math&gt;1111111&lt;/math&gt;, and &lt;math&gt;0111111&lt;/math&gt;. But, since George only said one number, the first two must have been too big for it to be &lt;math&gt;\le1000&lt;/math&gt;. Our answer is therefore &lt;math&gt;111111_3+1=\frac{3^6-1}{3-1}+1=364+1=\boxed{\textbf{(C) } 365}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_23&diff=178266 2011 AMC 10A Problems/Problem 23 2022-09-16T20:00:35Z <p>Sigmapie: </p> <hr /> <div>== Problem ==<br /> Seven students count from 1 to 1000 as follows:<br /> <br /> Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br /> <br /> Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br /> <br /> Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br /> <br /> Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br /> <br /> Finally, George says the only number that no one else says.<br /> <br /> What number does George say?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> First look at the numbers Alice says. &lt;math&gt;1, 3, 4, 6, 7, 9 \cdots&lt;/math&gt; skipping every number that is congruent to &lt;math&gt;2 \pmod 3&lt;/math&gt;. Thus, Barbara says those numbers EXCEPT every second - being &lt;math&gt;2 + 3^1 \equiv 5 \pmod{3^2=9}&lt;/math&gt;. So Barbara skips every number congruent to &lt;math&gt;5 \pmod 9&lt;/math&gt;. We continue and see: <br /> <br /> Alice skips &lt;math&gt;2 \pmod 3&lt;/math&gt;, Barbara skips &lt;math&gt;5 \pmod 9&lt;/math&gt;, Candice skips &lt;math&gt;14 \pmod {27}&lt;/math&gt;, Debbie skips &lt;math&gt;41 \pmod {81}&lt;/math&gt;, Eliza skips &lt;math&gt;122 \pmod {243}&lt;/math&gt;, and Fatima skips &lt;math&gt;365 \pmod {729}&lt;/math&gt;.<br /> <br /> Since the only number congruent to &lt;math&gt;365 \pmod {729}&lt;/math&gt; and less than &lt;math&gt;1,000&lt;/math&gt; is &lt;math&gt;365&lt;/math&gt;, the correct answer is &lt;math&gt; \boxed{365\ \mathbf{(C)}} &lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> After Alice says all her numbers, the numbers not mentioned yet are &lt;cmath&gt;\text{Alice: } 2,5,8,11,14,17,\cdots,998.&lt;/cmath&gt;After Barbara says all her numbers, the numbers that haven't been said yet are &lt;cmath&gt;\text{Barbara: } 5,14,23,32,41,50,\cdots,995.&lt;/cmath&gt;After Candice, the list is &lt;cmath&gt;\text{Candice: } 14,41,68,\cdots,986.&lt;/cmath&gt;Notice how each list is an arithmetic sequence where the common ratio is thrice the common ratio of the previous list and the first term is the second term of the previous list. Now that the pattern is clear, we construct the rest of the lists: &lt;cmath&gt;\text{Debbie: } 41,122,203,\cdots,959&lt;/cmath&gt;&lt;cmath&gt;\text{Eliza: } 122,365,608,878&lt;/cmath&gt;&lt;cmath&gt;\text{Fatima: } 365&lt;/cmath&gt;<br /> <br /> Thus, George says &lt;math&gt;\boxed{\textbf{(C) } 365}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> Notice that Alice has skipped the numbers &lt;math&gt;3n-1&lt;/math&gt; for &lt;math&gt;n=1,2,3,...,333&lt;/math&gt;.<br /> Namely, &lt;cmath&gt;3\cdot1-1,3\cdot2-1,3\cdot3-1,...,3\cdot333-1&lt;/cmath&gt;<br /> Thus the numbers that Barbara skips are &lt;cmath&gt;3\cdot2-1,3\cdot5-1,3\cdot8-1,...&lt;/cmath&gt;<br /> or in a more general expression, &lt;math&gt;3(3n-1)-1&lt;/math&gt; for &lt;math&gt;n=1,2,3,...&lt;/math&gt;.<br /> Namely, &lt;cmath&gt;3\cdot(3\cdot1-1)-1,3\cdot(3\cdot2-1)-1,3\cdot(3\cdot3-1)-1,...&lt;/cmath&gt;<br /> Repeating the pattern until George, we have the first number he says,<br /> &lt;cmath&gt;3(3(3(3(3(3\cdot1-1)-1)-1)-1)-1)-1= \boxed{\textbf{(C) } 365}&lt;/cmath&gt;<br /> In addition, note that the second number George says exceeds &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> == Solution 4 (Answer Choices) ==<br /> Similar to Solution 1 we find that the only number that George can say must leave a remainder of &lt;math&gt;2&lt;/math&gt; when divided by &lt;math&gt;3&lt;/math&gt;, and that it must also leave a remainder of &lt;math&gt;5&lt;/math&gt; when divided by &lt;math&gt;9&lt;/math&gt;. Since we as human beings are usually lazy, and that MAA provides answer choices, we check all the possible numbers and find that our answer is &lt;math&gt;\boxed{\textbf{(C) } 365}&lt;/math&gt;.<br /> <br /> == Solution 5==<br /> Every integer from 1 to 1000 can be written in the form &lt;math&gt;t_n+1&lt;/math&gt; in base 10, where &lt;math&gt;t_n&lt;/math&gt; is a trinary integer with no more than 7 significant digits. The important insight is that person &lt;math&gt;1\le{k}\le6&lt;/math&gt; will not say &lt;math&gt;n&lt;/math&gt; if the &lt;math&gt;k&lt;/math&gt;th digit from the right of &lt;math&gt;t_n&lt;/math&gt; is 1. Therefore, the last 6 digits of &lt;math&gt;t_n&lt;/math&gt; must be 1, as the first 6 people never said the number. The only options for &lt;math&gt;t_n&lt;/math&gt; are thus &lt;math&gt;2111111&lt;/math&gt;, &lt;math&gt;1111111&lt;/math&gt;, and &lt;math&gt;0111111&lt;/math&gt;. But, since George only said one number, the first two must have been too big for it to be &lt;math&gt;\le1000&lt;/math&gt;. Our answer is therefore &lt;math&gt;111111_3+1=\frac{3^6-1}{3-1}+1=364+1=\boxed{\textbf{(C) } 365}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_23&diff=178265 2011 AMC 10A Problems/Problem 23 2022-09-16T19:59:47Z <p>Sigmapie: </p> <hr /> <div>== Problem ==<br /> Seven students count from 1 to 1000 as follows:<br /> <br /> Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br /> <br /> Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br /> <br /> Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br /> <br /> Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br /> <br /> Finally, George says the only number that no one else says.<br /> <br /> What number does George say?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> First look at the numbers Alice says. &lt;math&gt;1, 3, 4, 6, 7, 9 \cdots&lt;/math&gt; skipping every number that is congruent to &lt;math&gt;2 \pmod 3&lt;/math&gt;. Thus, Barbara says those numbers EXCEPT every second - being &lt;math&gt;2 + 3^1 \equiv 5 \pmod{3^2=9}&lt;/math&gt;. So Barbara skips every number congruent to &lt;math&gt;5 \pmod 9&lt;/math&gt;. We continue and see: <br /> <br /> Alice skips &lt;math&gt;2 \pmod 3&lt;/math&gt;, Barbara skips &lt;math&gt;5 \pmod 9&lt;/math&gt;, Candice skips &lt;math&gt;14 \pmod {27}&lt;/math&gt;, Debbie skips &lt;math&gt;41 \pmod {81}&lt;/math&gt;, Eliza skips &lt;math&gt;122 \pmod {243}&lt;/math&gt;, and Fatima skips &lt;math&gt;365 \pmod {729}&lt;/math&gt;.<br /> <br /> Since the only number congruent to &lt;math&gt;365 \pmod {729}&lt;/math&gt; and less than &lt;math&gt;1,000&lt;/math&gt; is &lt;math&gt;365&lt;/math&gt;, the correct answer is &lt;math&gt; \boxed{365\ \mathbf{(C)}} &lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> After Alice says all her numbers, the numbers not mentioned yet are &lt;cmath&gt;\text{Alice: } 2,5,8,11,14,17,\cdots,998.&lt;/cmath&gt;After Barbara says all her numbers, the numbers that haven't been said yet are &lt;cmath&gt;\text{Barbara: } 5,14,23,32,41,50,\cdots,995.&lt;/cmath&gt;After Candice, the list is &lt;cmath&gt;\text{Candice: } 14,41,68,\cdots,986.&lt;/cmath&gt;Notice how each list is an arithmetic sequence where the common ratio is thrice the common ratio of the previous list and the first term is the second term of the previous list. Now that the pattern is clear, we construct the rest of the lists: &lt;cmath&gt;\text{Debbie: } 41,122,203,\cdots,959&lt;/cmath&gt;&lt;cmath&gt;\text{Eliza: } 122,365,608,878&lt;/cmath&gt;&lt;cmath&gt;\text{Fatima: } 365&lt;/cmath&gt;<br /> <br /> Thus, George says &lt;math&gt;\boxed{\textbf{(C) } 365}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> Notice that Alice has skipped the numbers &lt;math&gt;3n-1&lt;/math&gt; for &lt;math&gt;n=1,2,3,...,333&lt;/math&gt;.<br /> Namely, &lt;cmath&gt;3\cdot1-1,3\cdot2-1,3\cdot3-1,...,3\cdot333-1&lt;/cmath&gt;<br /> Thus the numbers that Barbara skips are &lt;cmath&gt;3\cdot2-1,3\cdot5-1,3\cdot8-1,...&lt;/cmath&gt;<br /> or in a more general expression, &lt;math&gt;3(3n-1)-1&lt;/math&gt; for &lt;math&gt;n=1,2,3,...&lt;/math&gt;.<br /> Namely, &lt;cmath&gt;3\cdot(3\cdot1-1)-1,3\cdot(3\cdot2-1)-1,3\cdot(3\cdot3-1)-1,...&lt;/cmath&gt;<br /> Repeating the pattern until George, we have the first number he says,<br /> &lt;cmath&gt;3(3(3(3(3(3\cdot1-1)-1)-1)-1)-1)-1= \boxed{\textbf{(C) } 365}&lt;/cmath&gt;<br /> In addition, note that the second number George says exceeds &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> == Solution 4 (Answer Choices) ==<br /> Similar to Solution 1 we find that the only number that George can say must leave a remainder of &lt;math&gt;2&lt;/math&gt; when divided by &lt;math&gt;3&lt;/math&gt;, and that it must also leave a remainder of &lt;math&gt;5&lt;/math&gt; when divided by &lt;math&gt;9&lt;/math&gt;. Since we as human beings are usually lazy, and that MAA provides answer choices, we check all the possible numbers and find that our answer is &lt;math&gt;\boxed{\textbf{(C) } 365}&lt;/math&gt;.<br /> <br /> == Solution 5==<br /> Every integer from 1 to 1000 can be written in the form &lt;math&gt;t_n+1&lt;/math&gt; in base 10, where &lt;math&gt;t_n&lt;/math&gt; is a trinary integer with no more than 7 significant digits. The important insight is that person &lt;math&gt;1\ge{k}\ge6&lt;/math&gt; will not say &lt;math&gt;n&lt;/math&gt; if the &lt;math&gt;k&lt;/math&gt;th digit from the right of &lt;math&gt;t_n&lt;/math&gt; is 1. Therefore, the last 6 digits of &lt;math&gt;t_n&lt;/math&gt; must be 1, as the first 6 people never said the number. The only options for &lt;math&gt;t_n&lt;/math&gt; are thus &lt;math&gt;2111111&lt;/math&gt;, &lt;math&gt;1111111&lt;/math&gt;, and &lt;math&gt;0111111&lt;/math&gt;. But, since George only said one number, the first two must have been too big for it to be &lt;math&gt;\le1000&lt;/math&gt;. Our answer is therefore &lt;math&gt;111111_3+1=\frac{3^6-1}{3-1}+1=364+1=\boxed{\textbf{(C) } 365}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_23&diff=178264 2011 AMC 10A Problems/Problem 23 2022-09-16T19:56:28Z <p>Sigmapie: </p> <hr /> <div>== Problem ==<br /> Seven students count from 1 to 1000 as follows:<br /> <br /> Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br /> <br /> Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br /> <br /> Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br /> <br /> Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br /> <br /> Finally, George says the only number that no one else says.<br /> <br /> What number does George say?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> First look at the numbers Alice says. &lt;math&gt;1, 3, 4, 6, 7, 9 \cdots&lt;/math&gt; skipping every number that is congruent to &lt;math&gt;2 \pmod 3&lt;/math&gt;. Thus, Barbara says those numbers EXCEPT every second - being &lt;math&gt;2 + 3^1 \equiv 5 \pmod{3^2=9}&lt;/math&gt;. So Barbara skips every number congruent to &lt;math&gt;5 \pmod 9&lt;/math&gt;. We continue and see: <br /> <br /> Alice skips &lt;math&gt;2 \pmod 3&lt;/math&gt;, Barbara skips &lt;math&gt;5 \pmod 9&lt;/math&gt;, Candice skips &lt;math&gt;14 \pmod {27}&lt;/math&gt;, Debbie skips &lt;math&gt;41 \pmod {81}&lt;/math&gt;, Eliza skips &lt;math&gt;122 \pmod {243}&lt;/math&gt;, and Fatima skips &lt;math&gt;365 \pmod {729}&lt;/math&gt;.<br /> <br /> Since the only number congruent to &lt;math&gt;365 \pmod {729}&lt;/math&gt; and less than &lt;math&gt;1,000&lt;/math&gt; is &lt;math&gt;365&lt;/math&gt;, the correct answer is &lt;math&gt; \boxed{365\ \mathbf{(C)}} &lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> After Alice says all her numbers, the numbers not mentioned yet are &lt;cmath&gt;\text{Alice: } 2,5,8,11,14,17,\cdots,998.&lt;/cmath&gt;After Barbara says all her numbers, the numbers that haven't been said yet are &lt;cmath&gt;\text{Barbara: } 5,14,23,32,41,50,\cdots,995.&lt;/cmath&gt;After Candice, the list is &lt;cmath&gt;\text{Candice: } 14,41,68,\cdots,986.&lt;/cmath&gt;Notice how each list is an arithmetic sequence where the common ratio is thrice the common ratio of the previous list and the first term is the second term of the previous list. Now that the pattern is clear, we construct the rest of the lists: &lt;cmath&gt;\text{Debbie: } 41,122,203,\cdots,959&lt;/cmath&gt;&lt;cmath&gt;\text{Eliza: } 122,365,608,878&lt;/cmath&gt;&lt;cmath&gt;\text{Fatima: } 365&lt;/cmath&gt;<br /> <br /> Thus, George says &lt;math&gt;\boxed{\textbf{(C) } 365}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> Notice that Alice has skipped the numbers &lt;math&gt;3n-1&lt;/math&gt; for &lt;math&gt;n=1,2,3,...,333&lt;/math&gt;.<br /> Namely, &lt;cmath&gt;3\cdot1-1,3\cdot2-1,3\cdot3-1,...,3\cdot333-1&lt;/cmath&gt;<br /> Thus the numbers that Barbara skips are &lt;cmath&gt;3\cdot2-1,3\cdot5-1,3\cdot8-1,...&lt;/cmath&gt;<br /> or in a more general expression, &lt;math&gt;3(3n-1)-1&lt;/math&gt; for &lt;math&gt;n=1,2,3,...&lt;/math&gt;.<br /> Namely, &lt;cmath&gt;3\cdot(3\cdot1-1)-1,3\cdot(3\cdot2-1)-1,3\cdot(3\cdot3-1)-1,...&lt;/cmath&gt;<br /> Repeating the pattern until George, we have the first number he says,<br /> &lt;cmath&gt;3(3(3(3(3(3\cdot1-1)-1)-1)-1)-1)-1= \boxed{\textbf{(C) } 365}&lt;/cmath&gt;<br /> In addition, note that the second number George says exceeds &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> == Solution 4 (Answer Choices) ==<br /> Similar to Solution 1 we find that the only number that George can say must leave a remainder of &lt;math&gt;2&lt;/math&gt; when divided by &lt;math&gt;3&lt;/math&gt;, and that it must also leave a remainder of &lt;math&gt;5&lt;/math&gt; when divided by &lt;math&gt;9&lt;/math&gt;. Since we as human beings are usually lazy, and that MAA provides answer choices, we check all the possible numbers and find that our answer is &lt;math&gt;\boxed{\textbf{(C) } 365}&lt;/math&gt;.<br /> <br /> == Solution 5==<br /> Every integer from 1 to 1000 can be written in the form &lt;math&gt;t_n+1&lt;/math&gt; in base 10, where &lt;math&gt;t_n&lt;/math&gt; is a trinary integer with no more than 7 significant digits. The important insight is that person &lt;math&gt;k&lt;/math&gt; will not say &lt;math&gt;n&lt;/math&gt; if the &lt;math&gt;k&lt;/math&gt;th digit from the right of &lt;math&gt;t_n&lt;/math&gt; is 1. Therefore, the last 6 digits of &lt;math&gt;t_n&lt;/math&gt; must be 1, as the first 6 people never said the number. The only options for &lt;math&gt;t_n&lt;/math&gt; are thus &lt;math&gt;2111111&lt;/math&gt;, &lt;math&gt;1111111&lt;/math&gt;, and &lt;math&gt;0111111&lt;/math&gt;. But, since George only said one number, the first two must have been too big for it to be &lt;math&gt;\le1000&lt;/math&gt;. Our answer is therefore &lt;math&gt;111111_3+1=\frac{3^6-1}{3-1}+1=364+1=\boxed{\textbf{(C) } 365}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_18&diff=178211 2009 AMC 12A Problems/Problem 18 2022-09-14T23:16:59Z <p>Sigmapie: </p> <hr /> <div>{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #18]] and [[2009 AMC 10A Problems|2009 AMC 10A #25]]}}<br /> <br /> == Problem ==<br /> For &lt;math&gt;k &gt; 0&lt;/math&gt;, let &lt;math&gt;I_k = 10\ldots 064&lt;/math&gt;, where there are &lt;math&gt;k&lt;/math&gt; zeros between the &lt;math&gt;1&lt;/math&gt; and the &lt;math&gt;6&lt;/math&gt;. Let &lt;math&gt;N(k)&lt;/math&gt; be the number of factors of &lt;math&gt;2&lt;/math&gt; in the prime factorization of &lt;math&gt;I_k&lt;/math&gt;. What is the maximum value of &lt;math&gt;N(k)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> The number &lt;math&gt;I_k&lt;/math&gt; can be written as &lt;math&gt;10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6&lt;/math&gt;.<br /> <br /> For &lt;math&gt;k\in\{1,2,3\}&lt;/math&gt; we have &lt;math&gt;I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)&lt;/math&gt;. The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have &lt;math&gt;N(k)=k+2\leq 5&lt;/math&gt;.<br /> <br /> For &lt;math&gt;k&gt;4&lt;/math&gt; we have &lt;math&gt;I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)&lt;/math&gt;. For &lt;math&gt;k&gt;4&lt;/math&gt; the value in the parentheses is odd, hence &lt;math&gt;N(k)=6&lt;/math&gt;.<br /> <br /> This leaves the case &lt;math&gt;k=4&lt;/math&gt;. We have &lt;math&gt;I_4 = 2^6 \left( 5^6 + 1 \right)&lt;/math&gt;. The value &lt;math&gt;5^6 + 1&lt;/math&gt; is obviously even. And as &lt;math&gt;5\equiv 1 \pmod 4&lt;/math&gt;, we have &lt;math&gt;5^6 \equiv 1 \pmod 4&lt;/math&gt;, and therefore &lt;math&gt;5^6 + 1 \equiv 2 \pmod 4&lt;/math&gt;. Hence the largest power of &lt;math&gt;2&lt;/math&gt; that divides &lt;math&gt;5^6+1&lt;/math&gt; is &lt;math&gt;2^1&lt;/math&gt;, and this gives us the desired maximum of the function &lt;math&gt;N&lt;/math&gt;: &lt;math&gt;N(4) = \boxed{7}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Notice that &lt;math&gt;2&lt;/math&gt; is a prime factor of an integer &lt;math&gt;n&lt;/math&gt; if and only if &lt;math&gt;n&lt;/math&gt; is even. Therefore, given any sufficiently high positive integral value of &lt;math&gt;k&lt;/math&gt;, dividing &lt;math&gt;I_k&lt;/math&gt; by &lt;math&gt;2^6&lt;/math&gt; yields a terminal digit of zero, and dividing by 2 again leaves us with &lt;math&gt;2^7 \cdot a = I_k&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; is an odd integer.<br /> Observe then that &lt;math&gt;\boxed{\textbf{(B)}7}&lt;/math&gt; must be the maximum value for &lt;math&gt;N(k)&lt;/math&gt; because whatever value we choose for &lt;math&gt;k&lt;/math&gt;, &lt;math&gt;N(k)&lt;/math&gt; must be less than or equal to &lt;math&gt;7&lt;/math&gt;.<br /> <br /> &quot;Isn't this solution incomplete because we need to show that &lt;math&gt;N(k) = 7&lt;/math&gt; can be reached?&quot;<br /> <br /> An example of 7 being reached is 1000064. 1000064 divided by &lt;math&gt;2^7=128&lt;/math&gt; is 7813.<br /> <br /> In fact, 1000064 is the ONLY &lt;math&gt;N(k)&lt;/math&gt; that satisfies &lt;math&gt;7&lt;/math&gt;. All others are 6 or lower, because if there are more zeroes, to be divisible by 128 (&lt;math&gt;2^7&lt;/math&gt;), the last 7 digits must be divisible by 128, but 64 isn't. Meanwhile, if there are less zeroes, we can test by division that they do not work.<br /> <br /> == Solution 3 ==<br /> <br /> As in the first solution, the number &lt;math&gt;I_k&lt;/math&gt; can be written as &lt;math&gt;10^{k+2} + 64 = 5^{k+2} 2^{k+2} + 2^6&lt;/math&gt;. Factor &lt;math&gt;2^6&lt;/math&gt; out of the expression to get &lt;math&gt;I_k = 2^6(1+2^{k-4}5^{k+2})&lt;/math&gt;.<br /> <br /> You can now easily see that &lt;math&gt;I_k&lt;/math&gt; is divisible by at least 6 factors of two, from &lt;math&gt;2^6&lt;/math&gt;. Any other factors of two will come from the expression &lt;math&gt;(1+2^{k-4}5^{k+2})&lt;/math&gt;. <br /> <br /> Make the substitution: &lt;math&gt;x=k-4&lt;/math&gt;.<br /> You now have &lt;math&gt;(1+2^{x}5^{x+6}) = (1+10^x 5^6)&lt;/math&gt;<br /> <br /> &lt;math&gt;5^6=15625&lt;/math&gt;, so the expression becomes &lt;math&gt;(1+15625\cdot10^x)&lt;/math&gt; This is valid when &lt;math&gt;x&gt;-4&lt;/math&gt;. <br /> <br /> Obviously, if &lt;math&gt;x&lt;/math&gt; is negative, the expression will be fractional and not contain any extra factors of two.<br /> <br /> If &lt;math&gt;x&gt;0&lt;/math&gt;, the value &lt;math&gt;15625\cdot10^x&lt;/math&gt; will end in a zero. When 1 is added to the expression, the expression will end in 1. Since numbers divisible by 2 end in 0,2,4,6, or 8, the expression is not divisible by 2 and will not contribute to the total.<br /> <br /> If &lt;math&gt;x=0&lt;/math&gt;, the expression evaluates to &lt;math&gt;15626&lt;/math&gt;. Dividing out twos, you find that this value is only divisible by one factor of 2.<br /> <br /> The six factors of two from earlier add to this factor of two to give &lt;math&gt;\textbf{(B)}\ 7\qquad&lt;/math&gt;<br /> <br /> Solution written by coolak<br /> <br /> <br /> ==Solution 4==<br /> Similar to the other solutions, notice that &lt;math&gt;I_k&lt;/math&gt; can be written as &lt;math&gt;10^{k+2}+64 \Rightarrow 2^{k+2}5^{k+2}+2^6&lt;/math&gt;. Factoring out &lt;math&gt;2^6&lt;/math&gt; we see that<br /> <br /> &lt;math&gt;I_k = 2^6(2^{k-4}5^{k+2}+1)&lt;/math&gt;<br /> <br /> Notice that for &lt;math&gt;k &lt; 4&lt;/math&gt;, &lt;math&gt;2^{k-4}&lt;/math&gt; will not be an integer, and will &quot;steal&quot; some &lt;math&gt;2&lt;/math&gt;'s from the &lt;math&gt;2^6&lt;/math&gt;. We don't want this to happen, since we want to maximize the exponent of &lt;math&gt;2&lt;/math&gt;. We start by considering &lt;math&gt;k = 4&lt;/math&gt;. Then<br /> <br /> &lt;math&gt;I_k = 2^6(*5^6+1) \Rightarrow 2^6(5^6+1)&lt;/math&gt;<br /> <br /> &lt;math&gt;5^6&lt;/math&gt; is an odd number; more specifically, it ends in &lt;math&gt;25&lt;/math&gt; (all powers of &lt;math&gt;5&lt;/math&gt; after &lt;math&gt;5^1&lt;/math&gt; end in &lt;math&gt;25&lt;/math&gt;). Therefore the value in the parentheses will be some large number that ends in &lt;math&gt;26&lt;/math&gt;. Considering the rules of divisibility, we find that &lt;math&gt;5^6+1&lt;/math&gt; is even, so it is divisible by &lt;math&gt;2&lt;/math&gt;. Now our exponent of &lt;math&gt;2&lt;/math&gt; is at &lt;math&gt;7&lt;/math&gt;. But the divisibility rule for &lt;math&gt;4&lt;/math&gt; is the last &lt;math&gt;2&lt;/math&gt; digits of the number must be divisible by &lt;math&gt;4&lt;/math&gt;. We know the last digits: &lt;math&gt;26&lt;/math&gt;, which is not divisible by &lt;math&gt;4&lt;/math&gt;. Therefore &lt;math&gt;5^6 + 1&lt;/math&gt; is divisible by &lt;math&gt;2&lt;/math&gt;, but not &lt;math&gt;4&lt;/math&gt;. Testing more values of &lt;math&gt;k&lt;/math&gt;, we find that for &lt;math&gt;k \ge 5&lt;/math&gt;, the last digit becomes &lt;math&gt;1&lt;/math&gt;, which means it is not even divisible by &lt;math&gt;2&lt;/math&gt;. Therefore the highest possible exponent of &lt;math&gt;2&lt;/math&gt; that we can reach is &lt;math&gt;7 \Rightarrow \boxed{\text{B}}&lt;/math&gt;.<br /> <br /> ==Solution 5 (LTE)==<br /> Let &lt;math&gt;m=k+2&lt;/math&gt;. &lt;math&gt;v_2(10^m+2^6)=6&lt;/math&gt; if &lt;math&gt;m&gt;6&lt;/math&gt; and &lt;math&gt;v_2(10^m+2^6)=m&lt;/math&gt; if &lt;math&gt;m&lt;6&lt;/math&gt;. <br /> However, if &lt;math&gt;m=6&lt;/math&gt;, then &lt;math&gt;v_2(10^6+2^6)=v_2(2^6(5^6+1))=6+v_2(5^6+1)&lt;/math&gt;. By LTE, &lt;math&gt;v_2(5^6-1)=v_2(5-1)+v_2(5+1)+v_2(6)-1=2+1+1-1=3&lt;/math&gt;. Since &lt;math&gt;v_2(5^6-1)=3&lt;/math&gt;, &lt;math&gt;v_2(5^6+1)&lt;/math&gt; must equal &lt;math&gt;1&lt;/math&gt;. So, the answer is &lt;math&gt;6+1=7 \Rightarrow \boxed{\text{B}}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=A|num-b=17|num-a=19}}<br /> {{AMC10 box|year=2009|ab=A|num-b=24|after=Last Question}}<br /> <br /> [[Category: Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_18&diff=178210 2009 AMC 12A Problems/Problem 18 2022-09-14T23:16:41Z <p>Sigmapie: </p> <hr /> <div>{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #18]] and [[2009 AMC 10A Problems|2009 AMC 10A #25]]}}<br /> <br /> == Problem ==<br /> For &lt;math&gt;k &gt; 0&lt;/math&gt;, let &lt;math&gt;I_k = 10\ldots 064&lt;/math&gt;, where there are &lt;math&gt;k&lt;/math&gt; zeros between the &lt;math&gt;1&lt;/math&gt; and the &lt;math&gt;6&lt;/math&gt;. Let &lt;math&gt;N(k)&lt;/math&gt; be the number of factors of &lt;math&gt;2&lt;/math&gt; in the prime factorization of &lt;math&gt;I_k&lt;/math&gt;. What is the maximum value of &lt;math&gt;N(k)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> The number &lt;math&gt;I_k&lt;/math&gt; can be written as &lt;math&gt;10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6&lt;/math&gt;.<br /> <br /> For &lt;math&gt;k\in\{1,2,3\}&lt;/math&gt; we have &lt;math&gt;I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)&lt;/math&gt;. The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have &lt;math&gt;N(k)=k+2\leq 5&lt;/math&gt;.<br /> <br /> For &lt;math&gt;k&gt;4&lt;/math&gt; we have &lt;math&gt;I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)&lt;/math&gt;. For &lt;math&gt;k&gt;4&lt;/math&gt; the value in the parentheses is odd, hence &lt;math&gt;N(k)=6&lt;/math&gt;.<br /> <br /> This leaves the case &lt;math&gt;k=4&lt;/math&gt;. We have &lt;math&gt;I_4 = 2^6 \left( 5^6 + 1 \right)&lt;/math&gt;. The value &lt;math&gt;5^6 + 1&lt;/math&gt; is obviously even. And as &lt;math&gt;5\equiv 1 \pmod 4&lt;/math&gt;, we have &lt;math&gt;5^6 \equiv 1 \pmod 4&lt;/math&gt;, and therefore &lt;math&gt;5^6 + 1 \equiv 2 \pmod 4&lt;/math&gt;. Hence the largest power of &lt;math&gt;2&lt;/math&gt; that divides &lt;math&gt;5^6+1&lt;/math&gt; is &lt;math&gt;2^1&lt;/math&gt;, and this gives us the desired maximum of the function &lt;math&gt;N&lt;/math&gt;: &lt;math&gt;N(4) = \boxed{7}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Notice that &lt;math&gt;2&lt;/math&gt; is a prime factor of an integer &lt;math&gt;n&lt;/math&gt; if and only if &lt;math&gt;n&lt;/math&gt; is even. Therefore, given any sufficiently high positive integral value of &lt;math&gt;k&lt;/math&gt;, dividing &lt;math&gt;I_k&lt;/math&gt; by &lt;math&gt;2^6&lt;/math&gt; yields a terminal digit of zero, and dividing by 2 again leaves us with &lt;math&gt;2^7 \cdot a = I_k&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; is an odd integer.<br /> Observe then that &lt;math&gt;\boxed{\textbf{(B)}7}&lt;/math&gt; must be the maximum value for &lt;math&gt;N(k)&lt;/math&gt; because whatever value we choose for &lt;math&gt;k&lt;/math&gt;, &lt;math&gt;N(k)&lt;/math&gt; must be less than or equal to &lt;math&gt;7&lt;/math&gt;.<br /> <br /> &quot;Isn't this solution incomplete because we need to show that &lt;math&gt;N(k) = 7&lt;/math&gt; can be reached?&quot;<br /> <br /> An example of 7 being reached is 1000064. 1000064 divided by &lt;math&gt;2^7=128&lt;/math&gt; is 7813.<br /> <br /> In fact, 1000064 is the ONLY &lt;math&gt;N(k)&lt;/math&gt; that satisfies &lt;math&gt;7&lt;/math&gt;. All others are 6 or lower, because if there are more zeroes, to be divisible by 128 (&lt;math&gt;2^7&lt;/math&gt;), the last 7 digits must be divisible by 128, but 64 isn't. Meanwhile, if there are less zeroes, we can test by division that they do not work.<br /> <br /> == Solution 3 ==<br /> <br /> As in the first solution, the number &lt;math&gt;I_k&lt;/math&gt; can be written as &lt;math&gt;10^{k+2} + 64 = 5^{k+2} 2^{k+2} + 2^6&lt;/math&gt;. Factor &lt;math&gt;2^6&lt;/math&gt; out of the expression to get &lt;math&gt;I_k = 2^6(1+2^{k-4}5^{k+2})&lt;/math&gt;.<br /> <br /> You can now easily see that &lt;math&gt;I_k&lt;/math&gt; is divisible by at least 6 factors of two, from &lt;math&gt;2^6&lt;/math&gt;. Any other factors of two will come from the expression &lt;math&gt;(1+2^{k-4}5^{k+2})&lt;/math&gt;. <br /> <br /> Make the substitution: &lt;math&gt;x=k-4&lt;/math&gt;.<br /> You now have &lt;math&gt;(1+2^{x}5^{x+6}) = (1+10^x 5^6)&lt;/math&gt;<br /> <br /> &lt;math&gt;5^6=15625&lt;/math&gt;, so the expression becomes &lt;math&gt;(1+15625\cdot10^x)&lt;/math&gt; This is valid when &lt;math&gt;x&gt;-4&lt;/math&gt;. <br /> <br /> Obviously, if &lt;math&gt;x&lt;/math&gt; is negative, the expression will be fractional and not contain any extra factors of two.<br /> <br /> If &lt;math&gt;x&gt;0&lt;/math&gt;, the value &lt;math&gt;15625\cdot10^x&lt;/math&gt; will end in a zero. When 1 is added to the expression, the expression will end in 1. Since numbers divisible by 2 end in 0,2,4,6, or 8, the expression is not divisible by 2 and will not contribute to the total.<br /> <br /> If &lt;math&gt;x=0&lt;/math&gt;, the expression evaluates to &lt;math&gt;15626&lt;/math&gt;. Dividing out twos, you find that this value is only divisible by one factor of 2.<br /> <br /> The six factors of two from earlier add to this factor of two to give &lt;math&gt;\textbf{(B)}\ 7\qquad&lt;/math&gt;<br /> <br /> Solution written by coolak<br /> <br /> <br /> ==Solution 4==<br /> Similar to the other solutions, notice that &lt;math&gt;I_k&lt;/math&gt; can be written as &lt;math&gt;10^{k+2}+64 \Rightarrow 2^{k+2}5^{k+2}+2^6&lt;/math&gt;. Factoring out &lt;math&gt;2^6&lt;/math&gt; we see that<br /> <br /> &lt;math&gt;I_k = 2^6(2^{k-4}5^{k+2}+1)&lt;/math&gt;<br /> <br /> Notice that for &lt;math&gt;k &lt; 4&lt;/math&gt;, &lt;math&gt;2^{k-4}&lt;/math&gt; will not be an integer, and will &quot;steal&quot; some &lt;math&gt;2&lt;/math&gt;'s from the &lt;math&gt;2^6&lt;/math&gt;. We don't want this to happen, since we want to maximize the exponent of &lt;math&gt;2&lt;/math&gt;. We start by considering &lt;math&gt;k = 4&lt;/math&gt;. Then<br /> <br /> &lt;math&gt;I_k = 2^6(*5^6+1) \Rightarrow 2^6(5^6+1)&lt;/math&gt;<br /> <br /> &lt;math&gt;5^6&lt;/math&gt; is an odd number; more specifically, it ends in &lt;math&gt;25&lt;/math&gt; (all powers of &lt;math&gt;5&lt;/math&gt; after &lt;math&gt;5^1&lt;/math&gt; end in &lt;math&gt;25&lt;/math&gt;). Therefore the value in the parentheses will be some large number that ends in &lt;math&gt;26&lt;/math&gt;. Considering the rules of divisibility, we find that &lt;math&gt;5^6+1&lt;/math&gt; is even, so it is divisible by &lt;math&gt;2&lt;/math&gt;. Now our exponent of &lt;math&gt;2&lt;/math&gt; is at &lt;math&gt;7&lt;/math&gt;. But the divisibility rule for &lt;math&gt;4&lt;/math&gt; is the last &lt;math&gt;2&lt;/math&gt; digits of the number must be divisible by &lt;math&gt;4&lt;/math&gt;. We know the last digits: &lt;math&gt;26&lt;/math&gt;, which is not divisible by &lt;math&gt;4&lt;/math&gt;. Therefore &lt;math&gt;5^6 + 1&lt;/math&gt; is divisible by &lt;math&gt;2&lt;/math&gt;, but not &lt;math&gt;4&lt;/math&gt;. Testing more values of &lt;math&gt;k&lt;/math&gt;, we find that for &lt;math&gt;k \ge 5&lt;/math&gt;, the last digit becomes &lt;math&gt;1&lt;/math&gt;, which means it is not even divisible by &lt;math&gt;2&lt;/math&gt;. Therefore the highest possible exponent of &lt;math&gt;2&lt;/math&gt; that we can reach is &lt;math&gt;7 \Rightarrow \boxed{\text{B}}&lt;/math&gt;.<br /> <br /> ==Solution 5 (LTE)==<br /> Let &lt;math&gt;m=k+2&lt;/math&gt;. v_2(10^m+2^6)=6&lt;math&gt; if &lt;/math&gt;m&gt;6&lt;math&gt; and &lt;/math&gt;v_2(10^m+2^6)=m&lt;math&gt; if &lt;/math&gt;m&lt;6&lt;math&gt;. <br /> However, if &lt;/math&gt;m=6&lt;math&gt;, then &lt;/math&gt;v_2(10^6+2^6)=v_2(2^6(5^6+1))=6+v_2(5^6+1)&lt;math&gt;. By LTE, &lt;/math&gt;v_2(5^6-1)=v_2(5-1)+v_2(5+1)+v_2(6)-1=2+1+1-1=3&lt;math&gt;. Since &lt;/math&gt;v_2(5^6-1)=3&lt;math&gt;, &lt;/math&gt;v_2(5^6+1)&lt;math&gt; must equal &lt;/math&gt;1&lt;math&gt;. So, the answer is &lt;/math&gt;6+1=7 \Rightarrow \boxed{\text{B}}\$<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=A|num-b=17|num-a=19}}<br /> {{AMC10 box|year=2009|ab=A|num-b=24|after=Last Question}}<br /> <br /> [[Category: Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_18&diff=178209 2009 AMC 12A Problems/Problem 18 2022-09-14T23:14:42Z <p>Sigmapie: </p> <hr /> <div>{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #18]] and [[2009 AMC 10A Problems|2009 AMC 10A #25]]}}<br /> <br /> == Problem ==<br /> For &lt;math&gt;k &gt; 0&lt;/math&gt;, let &lt;math&gt;I_k = 10\ldots 064&lt;/math&gt;, where there are &lt;math&gt;k&lt;/math&gt; zeros between the &lt;math&gt;1&lt;/math&gt; and the &lt;math&gt;6&lt;/math&gt;. Let &lt;math&gt;N(k)&lt;/math&gt; be the number of factors of &lt;math&gt;2&lt;/math&gt; in the prime factorization of &lt;math&gt;I_k&lt;/math&gt;. What is the maximum value of &lt;math&gt;N(k)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> The number &lt;math&gt;I_k&lt;/math&gt; can be written as &lt;math&gt;10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6&lt;/math&gt;.<br /> <br /> For &lt;math&gt;k\in\{1,2,3\}&lt;/math&gt; we have &lt;math&gt;I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)&lt;/math&gt;. The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have &lt;math&gt;N(k)=k+2\leq 5&lt;/math&gt;.<br /> <br /> For &lt;math&gt;k&gt;4&lt;/math&gt; we have &lt;math&gt;I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)&lt;/math&gt;. For &lt;math&gt;k&gt;4&lt;/math&gt; the value in the parentheses is odd, hence &lt;math&gt;N(k)=6&lt;/math&gt;.<br /> <br /> This leaves the case &lt;math&gt;k=4&lt;/math&gt;. We have &lt;math&gt;I_4 = 2^6 \left( 5^6 + 1 \right)&lt;/math&gt;. The value &lt;math&gt;5^6 + 1&lt;/math&gt; is obviously even. And as &lt;math&gt;5\equiv 1 \pmod 4&lt;/math&gt;, we have &lt;math&gt;5^6 \equiv 1 \pmod 4&lt;/math&gt;, and therefore &lt;math&gt;5^6 + 1 \equiv 2 \pmod 4&lt;/math&gt;. Hence the largest power of &lt;math&gt;2&lt;/math&gt; that divides &lt;math&gt;5^6+1&lt;/math&gt; is &lt;math&gt;2^1&lt;/math&gt;, and this gives us the desired maximum of the function &lt;math&gt;N&lt;/math&gt;: &lt;math&gt;N(4) = \boxed{7}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Notice that &lt;math&gt;2&lt;/math&gt; is a prime factor of an integer &lt;math&gt;n&lt;/math&gt; if and only if &lt;math&gt;n&lt;/math&gt; is even. Therefore, given any sufficiently high positive integral value of &lt;math&gt;k&lt;/math&gt;, dividing &lt;math&gt;I_k&lt;/math&gt; by &lt;math&gt;2^6&lt;/math&gt; yields a terminal digit of zero, and dividing by 2 again leaves us with &lt;math&gt;2^7 \cdot a = I_k&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; is an odd integer.<br /> Observe then that &lt;math&gt;\boxed{\textbf{(B)}7}&lt;/math&gt; must be the maximum value for &lt;math&gt;N(k)&lt;/math&gt; because whatever value we choose for &lt;math&gt;k&lt;/math&gt;, &lt;math&gt;N(k)&lt;/math&gt; must be less than or equal to &lt;math&gt;7&lt;/math&gt;.<br /> <br /> &quot;Isn't this solution incomplete because we need to show that &lt;math&gt;N(k) = 7&lt;/math&gt; can be reached?&quot;<br /> <br /> An example of 7 being reached is 1000064. 1000064 divided by &lt;math&gt;2^7=128&lt;/math&gt; is 7813.<br /> <br /> In fact, 1000064 is the ONLY &lt;math&gt;N(k)&lt;/math&gt; that satisfies &lt;math&gt;7&lt;/math&gt;. All others are 6 or lower, because if there are more zeroes, to be divisible by 128 (&lt;math&gt;2^7&lt;/math&gt;), the last 7 digits must be divisible by 128, but 64 isn't. Meanwhile, if there are less zeroes, we can test by division that they do not work.<br /> <br /> == Solution 3 ==<br /> <br /> As in the first solution, the number &lt;math&gt;I_k&lt;/math&gt; can be written as &lt;math&gt;10^{k+2} + 64 = 5^{k+2} 2^{k+2} + 2^6&lt;/math&gt;. Factor &lt;math&gt;2^6&lt;/math&gt; out of the expression to get &lt;math&gt;I_k = 2^6(1+2^{k-4}5^{k+2})&lt;/math&gt;.<br /> <br /> You can now easily see that &lt;math&gt;I_k&lt;/math&gt; is divisible by at least 6 factors of two, from &lt;math&gt;2^6&lt;/math&gt;. Any other factors of two will come from the expression &lt;math&gt;(1+2^{k-4}5^{k+2})&lt;/math&gt;. <br /> <br /> Make the substitution: &lt;math&gt;x=k-4&lt;/math&gt;.<br /> You now have &lt;math&gt;(1+2^{x}5^{x+6}) = (1+10^x 5^6)&lt;/math&gt;<br /> <br /> &lt;math&gt;5^6=15625&lt;/math&gt;, so the expression becomes &lt;math&gt;(1+15625\cdot10^x)&lt;/math&gt; This is valid when &lt;math&gt;x&gt;-4&lt;/math&gt;. <br /> <br /> Obviously, if &lt;math&gt;x&lt;/math&gt; is negative, the expression will be fractional and not contain any extra factors of two.<br /> <br /> If &lt;math&gt;x&gt;0&lt;/math&gt;, the value &lt;math&gt;15625\cdot10^x&lt;/math&gt; will end in a zero. When 1 is added to the expression, the expression will end in 1. Since numbers divisible by 2 end in 0,2,4,6, or 8, the expression is not divisible by 2 and will not contribute to the total.<br /> <br /> If &lt;math&gt;x=0&lt;/math&gt;, the expression evaluates to &lt;math&gt;15626&lt;/math&gt;. Dividing out twos, you find that this value is only divisible by one factor of 2.<br /> <br /> The six factors of two from earlier add to this factor of two to give &lt;math&gt;\textbf{(B)}\ 7\qquad&lt;/math&gt;<br /> <br /> Solution written by coolak<br /> <br /> <br /> ==Solution 4==<br /> Similar to the other solutions, notice that &lt;math&gt;I_k&lt;/math&gt; can be written as &lt;math&gt;10^{k+2}+64 \Rightarrow 2^{k+2}5^{k+2}+2^6&lt;/math&gt;. Factoring out &lt;math&gt;2^6&lt;/math&gt; we see that<br /> <br /> &lt;math&gt;I_k = 2^6(2^{k-4}5^{k+2}+1)&lt;/math&gt;<br /> <br /> Notice that for &lt;math&gt;k &lt; 4&lt;/math&gt;, &lt;math&gt;2^{k-4}&lt;/math&gt; will not be an integer, and will &quot;steal&quot; some &lt;math&gt;2&lt;/math&gt;'s from the &lt;math&gt;2^6&lt;/math&gt;. We don't want this to happen, since we want to maximize the exponent of &lt;math&gt;2&lt;/math&gt;. We start by considering &lt;math&gt;k = 4&lt;/math&gt;. Then<br /> <br /> &lt;math&gt;I_k = 2^6(*5^6+1) \Rightarrow 2^6(5^6+1)&lt;/math&gt;<br /> <br /> &lt;math&gt;5^6&lt;/math&gt; is an odd number; more specifically, it ends in &lt;math&gt;25&lt;/math&gt; (all powers of &lt;math&gt;5&lt;/math&gt; after &lt;math&gt;5^1&lt;/math&gt; end in &lt;math&gt;25&lt;/math&gt;). Therefore the value in the parentheses will be some large number that ends in &lt;math&gt;26&lt;/math&gt;. Considering the rules of divisibility, we find that &lt;math&gt;5^6+1&lt;/math&gt; is even, so it is divisible by &lt;math&gt;2&lt;/math&gt;. Now our exponent of &lt;math&gt;2&lt;/math&gt; is at &lt;math&gt;7&lt;/math&gt;. But the divisibility rule for &lt;math&gt;4&lt;/math&gt; is the last &lt;math&gt;2&lt;/math&gt; digits of the number must be divisible by &lt;math&gt;4&lt;/math&gt;. We know the last digits: &lt;math&gt;26&lt;/math&gt;, which is not divisible by &lt;math&gt;4&lt;/math&gt;. Therefore &lt;math&gt;5^6 + 1&lt;/math&gt; is divisible by &lt;math&gt;2&lt;/math&gt;, but not &lt;math&gt;4&lt;/math&gt;. Testing more values of &lt;math&gt;k&lt;/math&gt;, we find that for &lt;math&gt;k \ge 5&lt;/math&gt;, the last digit becomes &lt;math&gt;1&lt;/math&gt;, which means it is not even divisible by &lt;math&gt;2&lt;/math&gt;. Therefore the highest possible exponent of &lt;math&gt;2&lt;/math&gt; that we can reach is &lt;math&gt;7 \Rightarrow \boxed{\text{B}}&lt;/math&gt;.<br /> <br /> ==Solution 5 (LTE)==<br /> &lt;math&gt;v_2(10^k+2^6)=6&lt;/math&gt; if &lt;math&gt;k&gt;6&lt;/math&gt; and &lt;math&gt;v_2(10^k+2^6)=k&lt;/math&gt; if &lt;math&gt;k&lt;6&lt;/math&gt;. <br /> However, if &lt;math&gt;k=6&lt;/math&gt;, then &lt;math&gt;v_2(10^6+2^6)=v_2(2^6(5^6+1))=6+v_2(5^6+1)&lt;/math&gt;. By LTE, &lt;math&gt;v_2(5^6-1)=v_2(5-1)+v_2(5+1)+v_2(6)-1=2+1+1-1=3&lt;/math&gt;. Since &lt;math&gt;v_2(5^6-1)=3&lt;/math&gt;, &lt;math&gt;v_2(5^6+1)&lt;/math&gt; must equal &lt;math&gt;1&lt;/math&gt;. So, the answer is &lt;math&gt;6+1=7 \Rightarrow \boxed{\text{B}}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=A|num-b=17|num-a=19}}<br /> {{AMC10 box|year=2009|ab=A|num-b=24|after=Last Question}}<br /> <br /> [[Category: Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_18&diff=178208 2009 AMC 12A Problems/Problem 18 2022-09-14T23:13:21Z <p>Sigmapie: </p> <hr /> <div>{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #18]] and [[2009 AMC 10A Problems|2009 AMC 10A #25]]}}<br /> <br /> == Problem ==<br /> For &lt;math&gt;k &gt; 0&lt;/math&gt;, let &lt;math&gt;I_k = 10\ldots 064&lt;/math&gt;, where there are &lt;math&gt;k&lt;/math&gt; zeros between the &lt;math&gt;1&lt;/math&gt; and the &lt;math&gt;6&lt;/math&gt;. Let &lt;math&gt;N(k)&lt;/math&gt; be the number of factors of &lt;math&gt;2&lt;/math&gt; in the prime factorization of &lt;math&gt;I_k&lt;/math&gt;. What is the maximum value of &lt;math&gt;N(k)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> The number &lt;math&gt;I_k&lt;/math&gt; can be written as &lt;math&gt;10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6&lt;/math&gt;.<br /> <br /> For &lt;math&gt;k\in\{1,2,3\}&lt;/math&gt; we have &lt;math&gt;I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)&lt;/math&gt;. The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have &lt;math&gt;N(k)=k+2\leq 5&lt;/math&gt;.<br /> <br /> For &lt;math&gt;k&gt;4&lt;/math&gt; we have &lt;math&gt;I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)&lt;/math&gt;. For &lt;math&gt;k&gt;4&lt;/math&gt; the value in the parentheses is odd, hence &lt;math&gt;N(k)=6&lt;/math&gt;.<br /> <br /> This leaves the case &lt;math&gt;k=4&lt;/math&gt;. We have &lt;math&gt;I_4 = 2^6 \left( 5^6 + 1 \right)&lt;/math&gt;. The value &lt;math&gt;5^6 + 1&lt;/math&gt; is obviously even. And as &lt;math&gt;5\equiv 1 \pmod 4&lt;/math&gt;, we have &lt;math&gt;5^6 \equiv 1 \pmod 4&lt;/math&gt;, and therefore &lt;math&gt;5^6 + 1 \equiv 2 \pmod 4&lt;/math&gt;. Hence the largest power of &lt;math&gt;2&lt;/math&gt; that divides &lt;math&gt;5^6+1&lt;/math&gt; is &lt;math&gt;2^1&lt;/math&gt;, and this gives us the desired maximum of the function &lt;math&gt;N&lt;/math&gt;: &lt;math&gt;N(4) = \boxed{7}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Notice that &lt;math&gt;2&lt;/math&gt; is a prime factor of an integer &lt;math&gt;n&lt;/math&gt; if and only if &lt;math&gt;n&lt;/math&gt; is even. Therefore, given any sufficiently high positive integral value of &lt;math&gt;k&lt;/math&gt;, dividing &lt;math&gt;I_k&lt;/math&gt; by &lt;math&gt;2^6&lt;/math&gt; yields a terminal digit of zero, and dividing by 2 again leaves us with &lt;math&gt;2^7 \cdot a = I_k&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; is an odd integer.<br /> Observe then that &lt;math&gt;\boxed{\textbf{(B)}7}&lt;/math&gt; must be the maximum value for &lt;math&gt;N(k)&lt;/math&gt; because whatever value we choose for &lt;math&gt;k&lt;/math&gt;, &lt;math&gt;N(k)&lt;/math&gt; must be less than or equal to &lt;math&gt;7&lt;/math&gt;.<br /> <br /> &quot;Isn't this solution incomplete because we need to show that &lt;math&gt;N(k) = 7&lt;/math&gt; can be reached?&quot;<br /> <br /> An example of 7 being reached is 1000064. 1000064 divided by &lt;math&gt;2^7=128&lt;/math&gt; is 7813.<br /> <br /> In fact, 1000064 is the ONLY &lt;math&gt;N(k)&lt;/math&gt; that satisfies &lt;math&gt;7&lt;/math&gt;. All others are 6 or lower, because if there are more zeroes, to be divisible by 128 (&lt;math&gt;2^7&lt;/math&gt;), the last 7 digits must be divisible by 128, but 64 isn't. Meanwhile, if there are less zeroes, we can test by division that they do not work.<br /> <br /> == Solution 3 ==<br /> <br /> As in the first solution, the number &lt;math&gt;I_k&lt;/math&gt; can be written as &lt;math&gt;10^{k+2} + 64 = 5^{k+2} 2^{k+2} + 2^6&lt;/math&gt;. Factor &lt;math&gt;2^6&lt;/math&gt; out of the expression to get &lt;math&gt;I_k = 2^6(1+2^{k-4}5^{k+2})&lt;/math&gt;.<br /> <br /> You can now easily see that &lt;math&gt;I_k&lt;/math&gt; is divisible by at least 6 factors of two, from &lt;math&gt;2^6&lt;/math&gt;. Any other factors of two will come from the expression &lt;math&gt;(1+2^{k-4}5^{k+2})&lt;/math&gt;. <br /> <br /> Make the substitution: &lt;math&gt;x=k-4&lt;/math&gt;.<br /> You now have &lt;math&gt;(1+2^{x}5^{x+6}) = (1+10^x 5^6)&lt;/math&gt;<br /> <br /> &lt;math&gt;5^6=15625&lt;/math&gt;, so the expression becomes &lt;math&gt;(1+15625\cdot10^x)&lt;/math&gt; This is valid when &lt;math&gt;x&gt;-4&lt;/math&gt;. <br /> <br /> Obviously, if &lt;math&gt;x&lt;/math&gt; is negative, the expression will be fractional and not contain any extra factors of two.<br /> <br /> If &lt;math&gt;x&gt;0&lt;/math&gt;, the value &lt;math&gt;15625\cdot10^x&lt;/math&gt; will end in a zero. When 1 is added to the expression, the expression will end in 1. Since numbers divisible by 2 end in 0,2,4,6, or 8, the expression is not divisible by 2 and will not contribute to the total.<br /> <br /> If &lt;math&gt;x=0&lt;/math&gt;, the expression evaluates to &lt;math&gt;15626&lt;/math&gt;. Dividing out twos, you find that this value is only divisible by one factor of 2.<br /> <br /> The six factors of two from earlier add to this factor of two to give &lt;math&gt;\textbf{(B)}\ 7\qquad&lt;/math&gt;<br /> <br /> Solution written by coolak<br /> <br /> <br /> ==Solution 4==<br /> Similar to the other solutions, notice that &lt;math&gt;I_k&lt;/math&gt; can be written as &lt;math&gt;10^{k+2}+64 \Rightarrow 2^{k+2}5^{k+2}+2^6&lt;/math&gt;. Factoring out &lt;math&gt;2^6&lt;/math&gt; we see that<br /> <br /> &lt;math&gt;I_k = 2^6(2^{k-4}5^{k+2}+1)&lt;/math&gt;<br /> <br /> Notice that for &lt;math&gt;k &lt; 4&lt;/math&gt;, &lt;math&gt;2^{k-4}&lt;/math&gt; will not be an integer, and will &quot;steal&quot; some &lt;math&gt;2&lt;/math&gt;'s from the &lt;math&gt;2^6&lt;/math&gt;. We don't want this to happen, since we want to maximize the exponent of &lt;math&gt;2&lt;/math&gt;. We start by considering &lt;math&gt;k = 4&lt;/math&gt;. Then<br /> <br /> &lt;math&gt;I_k = 2^6(*5^6+1) \Rightarrow 2^6(5^6+1)&lt;/math&gt;<br /> <br /> &lt;math&gt;5^6&lt;/math&gt; is an odd number; more specifically, it ends in &lt;math&gt;25&lt;/math&gt; (all powers of &lt;math&gt;5&lt;/math&gt; after &lt;math&gt;5^1&lt;/math&gt; end in &lt;math&gt;25&lt;/math&gt;). Therefore the value in the parentheses will be some large number that ends in &lt;math&gt;26&lt;/math&gt;. Considering the rules of divisibility, we find that &lt;math&gt;5^6+1&lt;/math&gt; is even, so it is divisible by &lt;math&gt;2&lt;/math&gt;. Now our exponent of &lt;math&gt;2&lt;/math&gt; is at &lt;math&gt;7&lt;/math&gt;. But the divisibility rule for &lt;math&gt;4&lt;/math&gt; is the last &lt;math&gt;2&lt;/math&gt; digits of the number must be divisible by &lt;math&gt;4&lt;/math&gt;. We know the last digits: &lt;math&gt;26&lt;/math&gt;, which is not divisible by &lt;math&gt;4&lt;/math&gt;. Therefore &lt;math&gt;5^6 + 1&lt;/math&gt; is divisible by &lt;math&gt;2&lt;/math&gt;, but not &lt;math&gt;4&lt;/math&gt;. Testing more values of &lt;math&gt;k&lt;/math&gt;, we find that for &lt;math&gt;k \ge 5&lt;/math&gt;, the last digit becomes &lt;math&gt;1&lt;/math&gt;, which means it is not even divisible by &lt;math&gt;2&lt;/math&gt;. Therefore the highest possible exponent of &lt;math&gt;2&lt;/math&gt; that we can reach is &lt;math&gt;7 \Rightarrow \boxed{\text{B}}&lt;/math&gt;.<br /> <br /> ==Solution 5 (LTE)==<br /> &lt;math&gt;v_2(10^k+2^6)=6&lt;/math&gt; if &lt;math&gt;k&gt;6&lt;/math&gt; and &lt;math&gt;v_2(10^k+2^6)=k&lt;/math&gt; if &lt;math&gt;k&lt;6&lt;/math&gt;. <br /> However, if &lt;math&gt;k=6&lt;/math&gt;, then &lt;math&gt;v_2(10^6+2^6)=v_2(2^6(5^6+1))=6+v_2(5^6+1)&lt;/math&gt;. By LTE, &lt;math&gt;v_2(5^6-1)=v_2(5-1)+v_2(5+1)+v_2(6)-1=2+1+1-1=3&lt;/math&gt;. Since &lt;math&gt;v_2(5^6-1)=3&lt;/math&gt;, &lt;math&gt;v_2(5^6+1)&lt;/math&gt; must equal &lt;math&gt;1&lt;/math&gt;. So, the answer is &lt;math&gt;6+1=\boxed{7}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=A|num-b=17|num-a=19}}<br /> {{AMC10 box|year=2009|ab=A|num-b=24|after=Last Question}}<br /> <br /> [[Category: Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_18&diff=178207 2009 AMC 12A Problems/Problem 18 2022-09-14T23:12:05Z <p>Sigmapie: </p> <hr /> <div>{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #18]] and [[2009 AMC 10A Problems|2009 AMC 10A #25]]}}<br /> <br /> == Problem ==<br /> For &lt;math&gt;k &gt; 0&lt;/math&gt;, let &lt;math&gt;I_k = 10\ldots 064&lt;/math&gt;, where there are &lt;math&gt;k&lt;/math&gt; zeros between the &lt;math&gt;1&lt;/math&gt; and the &lt;math&gt;6&lt;/math&gt;. Let &lt;math&gt;N(k)&lt;/math&gt; be the number of factors of &lt;math&gt;2&lt;/math&gt; in the prime factorization of &lt;math&gt;I_k&lt;/math&gt;. What is the maximum value of &lt;math&gt;N(k)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> The number &lt;math&gt;I_k&lt;/math&gt; can be written as &lt;math&gt;10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6&lt;/math&gt;.<br /> <br /> For &lt;math&gt;k\in\{1,2,3\}&lt;/math&gt; we have &lt;math&gt;I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)&lt;/math&gt;. The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have &lt;math&gt;N(k)=k+2\leq 5&lt;/math&gt;.<br /> <br /> For &lt;math&gt;k&gt;4&lt;/math&gt; we have &lt;math&gt;I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)&lt;/math&gt;. For &lt;math&gt;k&gt;4&lt;/math&gt; the value in the parentheses is odd, hence &lt;math&gt;N(k)=6&lt;/math&gt;.<br /> <br /> This leaves the case &lt;math&gt;k=4&lt;/math&gt;. We have &lt;math&gt;I_4 = 2^6 \left( 5^6 + 1 \right)&lt;/math&gt;. The value &lt;math&gt;5^6 + 1&lt;/math&gt; is obviously even. And as &lt;math&gt;5\equiv 1 \pmod 4&lt;/math&gt;, we have &lt;math&gt;5^6 \equiv 1 \pmod 4&lt;/math&gt;, and therefore &lt;math&gt;5^6 + 1 \equiv 2 \pmod 4&lt;/math&gt;. Hence the largest power of &lt;math&gt;2&lt;/math&gt; that divides &lt;math&gt;5^6+1&lt;/math&gt; is &lt;math&gt;2^1&lt;/math&gt;, and this gives us the desired maximum of the function &lt;math&gt;N&lt;/math&gt;: &lt;math&gt;N(4) = \boxed{7}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Notice that &lt;math&gt;2&lt;/math&gt; is a prime factor of an integer &lt;math&gt;n&lt;/math&gt; if and only if &lt;math&gt;n&lt;/math&gt; is even. Therefore, given any sufficiently high positive integral value of &lt;math&gt;k&lt;/math&gt;, dividing &lt;math&gt;I_k&lt;/math&gt; by &lt;math&gt;2^6&lt;/math&gt; yields a terminal digit of zero, and dividing by 2 again leaves us with &lt;math&gt;2^7 \cdot a = I_k&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; is an odd integer.<br /> Observe then that &lt;math&gt;\boxed{\textbf{(B)}7}&lt;/math&gt; must be the maximum value for &lt;math&gt;N(k)&lt;/math&gt; because whatever value we choose for &lt;math&gt;k&lt;/math&gt;, &lt;math&gt;N(k)&lt;/math&gt; must be less than or equal to &lt;math&gt;7&lt;/math&gt;.<br /> <br /> &quot;Isn't this solution incomplete because we need to show that &lt;math&gt;N(k) = 7&lt;/math&gt; can be reached?&quot;<br /> <br /> An example of 7 being reached is 1000064. 1000064 divided by &lt;math&gt;2^7=128&lt;/math&gt; is 7813.<br /> <br /> In fact, 1000064 is the ONLY &lt;math&gt;N(k)&lt;/math&gt; that satisfies &lt;math&gt;7&lt;/math&gt;. All others are 6 or lower, because if there are more zeroes, to be divisible by 128 (&lt;math&gt;2^7&lt;/math&gt;), the last 7 digits must be divisible by 128, but 64 isn't. Meanwhile, if there are less zeroes, we can test by division that they do not work.<br /> <br /> == Solution 3 ==<br /> <br /> As in the first solution, the number &lt;math&gt;I_k&lt;/math&gt; can be written as &lt;math&gt;10^{k+2} + 64 = 5^{k+2} 2^{k+2} + 2^6&lt;/math&gt;. Factor &lt;math&gt;2^6&lt;/math&gt; out of the expression to get &lt;math&gt;I_k = 2^6(1+2^{k-4}5^{k+2})&lt;/math&gt;.<br /> <br /> You can now easily see that &lt;math&gt;I_k&lt;/math&gt; is divisible by at least 6 factors of two, from &lt;math&gt;2^6&lt;/math&gt;. Any other factors of two will come from the expression &lt;math&gt;(1+2^{k-4}5^{k+2})&lt;/math&gt;. <br /> <br /> Make the substitution: &lt;math&gt;x=k-4&lt;/math&gt;.<br /> You now have &lt;math&gt;(1+2^{x}5^{x+6}) = (1+10^x 5^6)&lt;/math&gt;<br /> <br /> &lt;math&gt;5^6=15625&lt;/math&gt;, so the expression becomes &lt;math&gt;(1+15625\cdot10^x)&lt;/math&gt; This is valid when &lt;math&gt;x&gt;-4&lt;/math&gt;. <br /> <br /> Obviously, if &lt;math&gt;x&lt;/math&gt; is negative, the expression will be fractional and not contain any extra factors of two.<br /> <br /> If &lt;math&gt;x&gt;0&lt;/math&gt;, the value &lt;math&gt;15625\cdot10^x&lt;/math&gt; will end in a zero. When 1 is added to the expression, the expression will end in 1. Since numbers divisible by 2 end in 0,2,4,6, or 8, the expression is not divisible by 2 and will not contribute to the total.<br /> <br /> If &lt;math&gt;x=0&lt;/math&gt;, the expression evaluates to &lt;math&gt;15626&lt;/math&gt;. Dividing out twos, you find that this value is only divisible by one factor of 2.<br /> <br /> The six factors of two from earlier add to this factor of two to give &lt;math&gt;\textbf{(B)}\ 7\qquad&lt;/math&gt;<br /> <br /> Solution written by coolak<br /> <br /> <br /> ==Solution 4==<br /> Similar to the other solutions, notice that &lt;math&gt;I_k&lt;/math&gt; can be written as &lt;math&gt;10^{k+2}+64 \Rightarrow 2^{k+2}5^{k+2}+2^6&lt;/math&gt;. Factoring out &lt;math&gt;2^6&lt;/math&gt; we see that<br /> <br /> &lt;math&gt;I_k = 2^6(2^{k-4}5^{k+2}+1)&lt;/math&gt;<br /> <br /> Notice that for &lt;math&gt;k &lt; 4&lt;/math&gt;, &lt;math&gt;2^{k-4}&lt;/math&gt; will not be an integer, and will &quot;steal&quot; some &lt;math&gt;2&lt;/math&gt;'s from the &lt;math&gt;2^6&lt;/math&gt;. We don't want this to happen, since we want to maximize the exponent of &lt;math&gt;2&lt;/math&gt;. We start by considering &lt;math&gt;k = 4&lt;/math&gt;. Then<br /> <br /> &lt;math&gt;I_k = 2^6(*5^6+1) \Rightarrow 2^6(5^6+1)&lt;/math&gt;<br /> <br /> &lt;math&gt;5^6&lt;/math&gt; is an odd number; more specifically, it ends in &lt;math&gt;25&lt;/math&gt; (all powers of &lt;math&gt;5&lt;/math&gt; after &lt;math&gt;5^1&lt;/math&gt; end in &lt;math&gt;25&lt;/math&gt;). Therefore the value in the parentheses will be some large number that ends in &lt;math&gt;26&lt;/math&gt;. Considering the rules of divisibility, we find that &lt;math&gt;5^6+1&lt;/math&gt; is even, so it is divisible by &lt;math&gt;2&lt;/math&gt;. Now our exponent of &lt;math&gt;2&lt;/math&gt; is at &lt;math&gt;7&lt;/math&gt;. But the divisibility rule for &lt;math&gt;4&lt;/math&gt; is the last &lt;math&gt;2&lt;/math&gt; digits of the number must be divisible by &lt;math&gt;4&lt;/math&gt;. We know the last digits: &lt;math&gt;26&lt;/math&gt;, which is not divisible by &lt;math&gt;4&lt;/math&gt;. Therefore &lt;math&gt;5^6 + 1&lt;/math&gt; is divisible by &lt;math&gt;2&lt;/math&gt;, but not &lt;math&gt;4&lt;/math&gt;. Testing more values of &lt;math&gt;k&lt;/math&gt;, we find that for &lt;math&gt;k \ge 5&lt;/math&gt;, the last digit becomes &lt;math&gt;1&lt;/math&gt;, which means it is not even divisible by &lt;math&gt;2&lt;/math&gt;. Therefore the highest possible exponent of &lt;math&gt;2&lt;/math&gt; that we can reach is &lt;math&gt;7 \Rightarrow \boxed{\text{B}}&lt;/math&gt;.<br /> <br /> ==Solution 5 (LTE)==<br /> &lt;math&gt;v_2(10^k+2^6)=6&lt;/math&gt; if &lt;math&gt;k&gt;6&lt;/math&gt; and &lt;math&gt;v_2(10^k+2^6)=k&lt;/math&gt; if &lt;math&gt;k&lt;6&lt;/math&gt;. <br /> However, if &lt;math&gt;k=6&lt;/math&gt;, then &lt;math&gt;v_2(10^6+2^6)=6+v_2(5^6+1)&lt;/math&gt;. By LTE, &lt;math&gt;v_2(5^6-1)=v_2(5-1)+v_2(5+1)+v_2(6)-1=2+1+1-1=3&lt;/math&gt;. Since &lt;math&gt;v_2(5^6-1)=3&lt;/math&gt;, &lt;math&gt;v_2(5^6+1)&lt;/math&gt; must equal &lt;math&gt;1&lt;/math&gt;. So, the answer is &lt;math&gt;6+1=\boxed{7}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=A|num-b=17|num-a=19}}<br /> {{AMC10 box|year=2009|ab=A|num-b=24|after=Last Question}}<br /> <br /> [[Category: Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_13&diff=175914 2006 AIME II Problems/Problem 13 2022-07-14T14:40:14Z <p>Sigmapie: </p> <hr /> <div>== Problem ==<br /> How many integers &lt;math&gt; N &lt;/math&gt; less than &lt;math&gt;1000&lt;/math&gt; can be written as the sum of &lt;math&gt; j &lt;/math&gt; consecutive positive odd integers from exactly 5 values of &lt;math&gt; j\ge 1 &lt;/math&gt;?<br /> <br /> == Solution ==<br /> Let the first odd integer be &lt;math&gt;2n+1&lt;/math&gt;, &lt;math&gt;n\geq 0&lt;/math&gt;. Then the final odd integer is &lt;math&gt;2n+1 + 2(j-1) = 2(n+j) - 1&lt;/math&gt;. The odd integers form an [[arithmetic sequence]] with sum &lt;math&gt;N = j\left(\frac{(2n+1) + (2(n+j)-1)}{2}\right) = j(2n+j)&lt;/math&gt;. Thus, &lt;math&gt;j&lt;/math&gt; is a factor of &lt;math&gt;N&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;n\geq 0&lt;/math&gt;, it follows that &lt;math&gt;2n+j \geq j&lt;/math&gt; and &lt;math&gt;j\leq \sqrt{N}&lt;/math&gt;.<br /> <br /> Since there are exactly &lt;math&gt;5&lt;/math&gt; values of &lt;math&gt;j&lt;/math&gt; that satisfy the equation, there must be either &lt;math&gt;9&lt;/math&gt; or &lt;math&gt;10&lt;/math&gt; factors of &lt;math&gt;N&lt;/math&gt;. This means &lt;math&gt;N=p_1^2p_2^2&lt;/math&gt; or &lt;math&gt;N=p_1p_2^4&lt;/math&gt;. Unfortunately, we cannot simply observe prime factorizations of &lt;math&gt;N&lt;/math&gt; because the factor &lt;math&gt;(2n+j)&lt;/math&gt; does not cover all integers for any given value of &lt;math&gt;j&lt;/math&gt;.<br /> <br /> Instead we do some casework:<br /> <br /> *If &lt;math&gt;N&lt;/math&gt; is odd, then &lt;math&gt;j&lt;/math&gt; must also be odd. For every odd value of &lt;math&gt;j&lt;/math&gt;, &lt;math&gt;2n+j&lt;/math&gt; is also odd, making this case valid for all odd &lt;math&gt;j&lt;/math&gt;. Looking at the forms above and the bound of &lt;math&gt;1000&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt; must be<br /> &lt;cmath&gt;(3^2\cdot5^2),\ (3^2\cdot7^2),\ (3^4\cdot5),\ (3^4\cdot7),\ (3^4\cdot 11)&lt;/cmath&gt;<br /> :Those give &lt;math&gt;5&lt;/math&gt; possibilities for odd &lt;math&gt;N&lt;/math&gt;.<br /> *If &lt;math&gt;N&lt;/math&gt; is even, then &lt;math&gt;j&lt;/math&gt; must also be even. Substituting &lt;math&gt;j=2k&lt;/math&gt;, we get<br /> &lt;cmath&gt;N = 4k(n+k) \Longrightarrow \frac{N}{4} = k(n+k)&lt;/cmath&gt;<br /> <br /> :Now we can just look at all the [[prime factorization]]s since &lt;math&gt;(n+k)&lt;/math&gt; cover the integers for any &lt;math&gt;k&lt;/math&gt;. Note that our upper bound is now &lt;math&gt;250&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\frac{N}{4} = (2^2\cdot3^2),(2^2\cdot5^2),(2^2\cdot7^2), (3^2\cdot5^2), (2^4\cdot3), (2^4\cdot5), (2^4\cdot7), (2^4\cdot11), (2^4\cdot13), (3^4\cdot2)&lt;/cmath&gt;<br /> <br /> :Those give &lt;math&gt;10&lt;/math&gt; possibilities for even &lt;math&gt;N&lt;/math&gt;.<br /> <br /> The total number of integers &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;5 + 10 = \boxed{15}&lt;/math&gt;.<br /> == Solution 2==<br /> Let the largest odd number below the sequence be the &lt;math&gt;q&lt;/math&gt;th positive odd number, and the largest odd number in the sequence be the &lt;math&gt;p&lt;/math&gt;th positive odd number. Therefore, the sum is &lt;math&gt;p^2-q^2=(p+q)(p-q)&lt;/math&gt; by sum of consecutive odd numbers. Note that &lt;math&gt;p+q&lt;/math&gt; and &lt;math&gt;p-q&lt;/math&gt; have the same parity, and &lt;math&gt;q&lt;/math&gt; can equal &lt;math&gt;0&lt;/math&gt;. We then perform casework based on the parity of &lt;math&gt;p-q&lt;/math&gt;. <br /> <br /> If &lt;math&gt;p-q&lt;/math&gt; is odd, then &lt;math&gt;p^2-q^2&lt;/math&gt; must always be odd. Therefore, to have 5 pairs of odd factors, we must have either &lt;math&gt;9&lt;/math&gt; (in which case the number is a perfect square) or &lt;math&gt;10&lt;/math&gt; factors. Considering the upper bound, the only way this can happen is &lt;math&gt;p_1^4\cdot{p_2}&lt;/math&gt; or &lt;math&gt;p_1^2\cdot{p_2^2}&lt;/math&gt;. N must then be one of<br /> &lt;cmath&gt;(3^2\cdot5^2),\ (3^2\cdot7^2),\ (3^4\cdot5),\ (3^4\cdot7),\ (3^4\cdot 11)&lt;/cmath&gt;<br /> So, there are &lt;math&gt;5&lt;/math&gt; solutions when &lt;math&gt;(p+q)(p-q)&lt;/math&gt; is odd. <br /> <br /> If &lt;math&gt;p-q&lt;/math&gt; is even, then &lt;math&gt;(p+q)(p-q)&lt;/math&gt; must have at least two factors of &lt;math&gt;2&lt;/math&gt;, so we can rewrite the expression as &lt;math&gt;4(k)(k-q)&lt;/math&gt; where &lt;math&gt;k=\frac{p+q}{2}&lt;/math&gt;. We can disregard the &lt;math&gt;4&lt;/math&gt; by dividing by &lt;math&gt;4&lt;/math&gt; and restricting our upper bound to &lt;math&gt;250&lt;/math&gt;. Since &lt;math&gt;k&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; don't have to be the same parity, we can include all cases less than 250 that have 9 or 10 factors. We then have <br /> &lt;cmath&gt;(2^2\cdot3^2),(2^2\cdot5^2),(2^2\cdot7^2), (3^2\cdot5^2), (2^4\cdot3), (2^4\cdot5), (2^4\cdot7), (2^4\cdot11), (2^4\cdot13), (3^4\cdot2)&lt;/cmath&gt;<br /> as the possibilities. <br /> <br /> Therefore, there are &lt;math&gt;10+5=\boxed{015}&lt;/math&gt; possibilities for &lt;math&gt;p^2-q^2=N&lt;/math&gt; ~sigma<br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Sigmapie https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_20&diff=175913 2007 AMC 8 Problems/Problem 20 2022-07-14T14:08:17Z <p>Sigmapie: </p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;Before the district play, the Unicorns had won &lt;math&gt;45&lt;/math&gt;% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> At the beginning of the problem, the Unicorns had played &lt;math&gt;y&lt;/math&gt; games and they had won &lt;math&gt;x&lt;/math&gt; of these games. From the information given in the problem, we can say that &lt;math&gt;\frac{x}{y}=0.45.&lt;/math&gt; Next, the Unicorns win 6 more games and lose 2 more, for a total of &lt;math&gt;6+2=8&lt;/math&gt; games played during district play. We are told that they end the season having won half of their games, or &lt;math&gt;0.5 &lt;/math&gt; of their games. We can write another equation: &lt;math&gt;\frac{x+6}{y+8}=0.5.&lt;/math&gt; This gives us a system of equations:<br /> &lt;math&gt;\frac{x}{y}=0.45&lt;/math&gt; and &lt;math&gt;\frac{x+6}{y+8}=0.5.&lt;/math&gt;<br /> We first multiply both sides of the first equation by &lt;math&gt;y&lt;/math&gt; to get &lt;math&gt;x=0.45y.&lt;/math&gt; Then, we multiply both sides of the second equation by &lt;math&gt;(y+8)&lt;/math&gt; to get &lt;math&gt;x+6=0.5(y+8).&lt;/math&gt; Applying the Distributive Property gives yields &lt;math&gt;x+6=0.5y+4.&lt;/math&gt; Now we substitute &lt;math&gt;0.45y&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; to get &lt;math&gt;0.45y+6=0.5y+4.&lt;/math&gt; Solving gives us &lt;math&gt;y=40.&lt;/math&gt; Since the problem asks for the total number of games, we add on the last 8 games to get the solution &lt;math&gt;\boxed{\textbf{(A)}\ 48}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Simplifying 45% to &lt;math&gt;\frac{9}{20}&lt;/math&gt;, we see that the numbers of games before district play are a multiple of 20. After that the Aces played 8 more games to the total number of games is in the form of 20x+8 where x is any positive integer. The only answer choice is &lt;math&gt;\boxed{48}&lt;/math&gt;, which is 20(2)+8.<br /> <br /> -harsha12345<br /> <br /> ==Solution 3==<br /> First we simplify &lt;math&gt;45&lt;/math&gt;% to &lt;math&gt;\frac{9}{20}&lt;/math&gt;. Ratio of won to total is &lt;math&gt;\frac{9}{20}&lt;/math&gt;, but ratio of total number won to total number played is &lt;math&gt;\frac{9x}{20x}&lt;/math&gt; for some &lt;math&gt;x&lt;/math&gt;. After they won 6 more games and lost 2 more games the number of games they won is &lt;math&gt;9x+6&lt;/math&gt;, and the total number of games is &lt;math&gt;20x+8&lt;/math&gt;. Turning it into a fraction we get &lt;math&gt;\frac{9x+6}{20x+8}=\frac{1}{2}&lt;/math&gt;, so solving for &lt;math&gt;x&lt;/math&gt; we get &lt;math&gt;x=2.&lt;/math&gt; Plugging in 2 for &lt;math&gt;x&lt;/math&gt; we get &lt;math&gt;20(2)+8=\boxed{48}&lt;/math&gt;.<br /> <br /> -harsha12345<br /> <br /> ==Solution 4==<br /> Because 45% can be simplified to &lt;math&gt;9/20&lt;/math&gt;, and we know that we cannot play a fraction amount of games, we know that the amount of games before district play is divisible by 20. After district play, there was &lt;math&gt;8&lt;/math&gt; games, so in total there must be &lt;math&gt;20x+8&lt;/math&gt;. The only answer in this format is &lt;math&gt;\boxed{\mathrm{(A)}48}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Let &lt;math&gt;n&lt;/math&gt; be the number of pre-district games. Therefore, we can write the percentage of total games won as a weighted average, namely &lt;math&gt;.45(n)+.75(8)=(n+8)(.5)&lt;/math&gt;. Solving this equation for &lt;math&gt;n&lt;/math&gt; gives &lt;math&gt;40&lt;/math&gt;, but since the problem asked for all games, the answer is &lt;math&gt;n+8=40+8=\boxed{\mathrm{(A)}48}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2007|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Sigmapie