https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Blehlivesonearth&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-21T03:36:58Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_12&diff=149625 2021 AIME I Problems/Problem 12 2021-03-16T22:56:09Z <p>Blehlivesonearth: /* Solution */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;A_1A_2A_3...A_{12}&lt;/math&gt; be a dodecagon (12-gon). Three frogs initially sit at &lt;math&gt;A_4,A_8,&lt;/math&gt; and &lt;math&gt;A_{12}&lt;/math&gt;. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is &lt;math&gt;\frac mn&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> The expected number of steps depends on the distance between the frogs, not on the order in which these distances appear. Let &lt;math&gt;E(a,b,c)&lt;/math&gt; where &lt;math&gt;a+b+c=9&lt;/math&gt; denote the expected number of steps that it takes for two frogs to meet if traversing in clockwise or counterclockwise order, the frogs are &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; vertices apart. Then<br /> <br /> &lt;math&gt;E(3,3,3)=1+\frac{1}{4} E(3,3,3)+\frac{3}{4} E(1,3,5)&lt;/math&gt;, giving &lt;math&gt;E(3,3,3)=\frac{4}{3}+E(1,3,5)&lt;/math&gt;; (1)<br /> <br /> &lt;math&gt;E(1,3,5)=1+\frac{1}{8}E(1,1,7)+\frac{1}{2}E(1,3,5)+\frac{1}{8}E(3,3,3)&lt;/math&gt;, giving &lt;math&gt;E(1,3,5)=2+\frac{1}{4}E(1,1,7)+\frac{1}{4}E(3,3,3)&lt;/math&gt;; (2)<br /> <br /> &lt;math&gt;E(1,1,7)=1+\frac{1}{4}E(1,1,7)+\frac{1}{4}E(1,3,5)&lt;/math&gt;, giving &lt;math&gt;E(1,1,7)=\frac{4}{3}+\frac{1}{3}E(1,3,5)&lt;/math&gt;; (3)<br /> <br /> Plug in (1) and (3) into (2), we see that &lt;math&gt;E(1,3,5)=4&lt;/math&gt;. &lt;math&gt;E(3,3,3)=\frac{4}{3}+4=\frac{16}{3}&lt;/math&gt;. Each step is one minute. The answer is &lt;math&gt;16+3=\boxed{19}&lt;/math&gt;.<br /> <br /> -Ross Gao<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_7&diff=143714 2018 AIME II Problems/Problem 7 2021-01-29T04:26:24Z <p>Blehlivesonearth: /* Solution 3 */</p> <hr /> <div>==Problem 7==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has side lengths &lt;math&gt;AB = 9&lt;/math&gt;, &lt;math&gt;BC =&lt;/math&gt; &lt;math&gt;5\sqrt{3}&lt;/math&gt;, and &lt;math&gt;AC = 12&lt;/math&gt;. Points &lt;math&gt;A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B&lt;/math&gt; are on segment &lt;math&gt;\overline{AB}&lt;/math&gt; with &lt;math&gt;P_{k}&lt;/math&gt; between &lt;math&gt;P_{k-1}&lt;/math&gt; and &lt;math&gt;P_{k+1}&lt;/math&gt; for &lt;math&gt;k = 1, 2, ..., 2449&lt;/math&gt;, and points &lt;math&gt;A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C&lt;/math&gt; are on segment &lt;math&gt;\overline{AC}&lt;/math&gt; with &lt;math&gt;Q_{k}&lt;/math&gt; between &lt;math&gt;Q_{k-1}&lt;/math&gt; and &lt;math&gt;Q_{k+1}&lt;/math&gt; for &lt;math&gt;k = 1, 2, ..., 2449&lt;/math&gt;. Furthermore, each segment &lt;math&gt;\overline{P_{k}Q_{k}}&lt;/math&gt;, &lt;math&gt;k = 1, 2, ..., 2449&lt;/math&gt;, is parallel to &lt;math&gt;\overline{BC}&lt;/math&gt;. The segments cut the triangle into &lt;math&gt;2450&lt;/math&gt; regions, consisting of &lt;math&gt;2449&lt;/math&gt; trapezoids and &lt;math&gt;1&lt;/math&gt; triangle. Each of the &lt;math&gt;2450&lt;/math&gt; regions has the same area. Find the number of segments &lt;math&gt;\overline{P_{k}Q_{k}}&lt;/math&gt;, &lt;math&gt;k = 1, 2, ..., 2450&lt;/math&gt;, that have rational length.<br /> <br /> == Solution 1 ==<br /> For each &lt;math&gt;k&lt;/math&gt; between &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;2450&lt;/math&gt;, the area of the trapezoid with &lt;math&gt;\overline{P_kQ_k}&lt;/math&gt; as its bottom base is the difference between the areas of two triangles, both similar to &lt;math&gt;\triangle{ABC}&lt;/math&gt;. Let &lt;math&gt;d_k&lt;/math&gt; be the length of segment &lt;math&gt;\overline{P_kQ_k}&lt;/math&gt;. The area of the trapezoid with bases &lt;math&gt;\overline{P_{k-1}Q_{k-1}}&lt;/math&gt; and &lt;math&gt;P_kQ_k&lt;/math&gt; is &lt;math&gt;\left(\frac{d_k}{5\sqrt{3}}\right)^2 - \left(\frac{d_{k-1}}{5\sqrt{3}}\right)^2 = \frac{d_k^2-d_{k-1}^2}{75}&lt;/math&gt; times the area of &lt;math&gt;\triangle{ABC}&lt;/math&gt;. (This logic also applies to the topmost triangle if we notice that &lt;math&gt;d_0 = 0&lt;/math&gt;.) However, we also know that the area of each shape is &lt;math&gt;\frac{1}{2450}&lt;/math&gt; times the area of &lt;math&gt;\triangle{ABC}&lt;/math&gt;. We then have &lt;math&gt;\frac{d_k^2-d_{k-1}^2}{75} = \frac{1}{2450}&lt;/math&gt;. Simplifying, &lt;math&gt;d_k^2-d_{k-1}^2 = \frac{3}{98}&lt;/math&gt;. However, we know that &lt;math&gt;d_0^2 = 0&lt;/math&gt;, so &lt;math&gt;d_1^2 = \frac{3}{98}&lt;/math&gt;, and in general, &lt;math&gt;d_k^2 = \frac{3k}{98}&lt;/math&gt; and &lt;math&gt;d_k = \frac{\sqrt{\frac{3k}{2}}}{7}&lt;/math&gt;. The smallest &lt;math&gt;k&lt;/math&gt; that gives a rational &lt;math&gt;d_k&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt;, so &lt;math&gt;d_k&lt;/math&gt; is rational if and only if &lt;math&gt;k = 6n^2&lt;/math&gt; for some integer &lt;math&gt;n&lt;/math&gt;.The largest &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;6n^2&lt;/math&gt; is less than &lt;math&gt;2450&lt;/math&gt; is &lt;math&gt;20&lt;/math&gt;, so &lt;math&gt;k&lt;/math&gt; has &lt;math&gt;\boxed{020}&lt;/math&gt; possible values.<br /> <br /> Solution by zeroman<br /> <br /> ==Solution 2==<br /> We have that there are &lt;math&gt;2449&lt;/math&gt; trapezoids and &lt;math&gt;1&lt;/math&gt; triangle of equal area, with that one triangle being &lt;math&gt;AP_1Q_1&lt;/math&gt;. Notice, if we &quot;stack&quot; the trapezoids on top of &lt;math&gt;\bigtriangleup AP_1Q_1&lt;/math&gt; the way they already are, we'd create a similar triangle, all of which are similar to &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;, and since the trapezoids and &lt;math&gt;\bigtriangleup AP_1Q_1&lt;/math&gt; have equal area, each of these similar triangles, &lt;math&gt;AP_kQ_k&lt;/math&gt; have area &lt;math&gt;\frac{k}{2450}\left[ ABC\right]&lt;/math&gt;, and so &lt;math&gt;\frac{\left[ AP_kQ_k\right]}{\left[ABC\right]}=\frac{k}{2450}&lt;/math&gt;. We want the ratio of the side lengths &lt;math&gt;P_kQ_k:BC&lt;/math&gt;. Since area is a 2-dimensional unit of measurement, and side lengths are 1-dimensional, the ratio is simply the square root of the areas, or &lt;math&gt;\frac{P_kQ_k}{BC}=\sqrt{\frac{k}{2450}} \implies P_kQ_k=BC\cdot \sqrt{\frac{k}{2450}}=5\sqrt{3}\cdot\sqrt{\frac{k}{2450}}=\frac{1}{7}\cdot \sqrt{\frac{3k}{2}}=\frac{3}{7}\sqrt{\frac{k}{6}} \implies k=6n^2&lt;2450 \implies 0&lt;n\leq 20&lt;/math&gt;, so there are &lt;math&gt;\boxed{020}&lt;/math&gt; solutions. <br /> <br /> Solution by ktong<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;T_1&lt;/math&gt; stand for &lt;math&gt;AP_1Q_1&lt;/math&gt;, and &lt;math&gt;T_k = AP_kQ_k&lt;/math&gt;. All triangles &lt;math&gt;T&lt;/math&gt; are similar by AA. Let the area of &lt;math&gt;T_1&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt;. The next trapezoid will also have an area of &lt;math&gt;x&lt;/math&gt;, as given. Therefore, &lt;math&gt;T_k&lt;/math&gt; has an area of &lt;math&gt;\sqrt{k}x&lt;/math&gt;. The ratio of the areas is equal to the square of the scale factor for any plane figure and its image. Therefore, &lt;math&gt;AP_k=AP_1\cdot \sqrt{k}&lt;/math&gt;, and the same if &lt;math&gt;Q&lt;/math&gt; is substituted for &lt;math&gt;P&lt;/math&gt; throughout. We want the side &lt;math&gt;P_kQ_k&lt;/math&gt; to be rational. Setting up proportions: &lt;cmath&gt;5\sqrt{3} : \sqrt{2450}=35\sqrt{2}&lt;/cmath&gt; &lt;cmath&gt;\sqrt{6} : 14&lt;/cmath&gt; which shows that &lt;math&gt;x=\frac{\sqrt{6}}{14}&lt;/math&gt;. In order for &lt;math&gt;\sqrt{k}x&lt;/math&gt; to be rational, &lt;math&gt;\sqrt{k}&lt;/math&gt; must be some rational multiple of &lt;math&gt;\sqrt{6}&lt;/math&gt;. This is achieved at &lt;math&gt;k=\sqrt{6}, 2\sqrt{6}, ... , 20\sqrt{6}&lt;/math&gt;. We end there as &lt;math&gt;21\sqrt{6}=\sqrt{2646}&lt;/math&gt;. There are 20 numbers from 1 to 20, so there are &lt;math&gt;\boxed{020}&lt;/math&gt; solutions.<br /> <br /> Solution by a1b2<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=II|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_6&diff=143694 2017 AIME II Problems/Problem 6 2021-01-29T02:11:29Z <p>Blehlivesonearth: /* Solution 4 (Bounding) */</p> <hr /> <div>==Problem==<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;\sqrt{n^2+85n+2017}&lt;/math&gt; is an integer.<br /> <br /> ==Solution 1==<br /> Manipulating the given expression, &lt;math&gt;\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}&lt;/math&gt;. The expression under the radical must be an square number for the entire expression to be an integer, so &lt;math&gt;(2n+85)^2+843=s^2&lt;/math&gt;. Rearranging, &lt;math&gt;s^2-(2n+85)^2=843&lt;/math&gt;. By difference of squares, &lt;math&gt;(s-(2n+85))(s+(2n+85))=1\times843=3\times281&lt;/math&gt;. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, &lt;math&gt;2n+85&lt;/math&gt; is found to be &lt;math&gt;421&lt;/math&gt; and &lt;math&gt;139&lt;/math&gt;. The two values of &lt;math&gt;n&lt;/math&gt; that satisfy one of the equations are &lt;math&gt;168&lt;/math&gt; and &lt;math&gt;27&lt;/math&gt;. Summing these together gives us the answer ; &lt;math&gt;168+27=\boxed{195}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Clearly, the result when &lt;math&gt;n&lt;/math&gt; is plugged into the given expression is larger than &lt;math&gt;n&lt;/math&gt; itself. Let &lt;math&gt;x&lt;/math&gt; be the positive difference between that result and &lt;math&gt;n&lt;/math&gt;, so that &lt;math&gt;\sqrt{n^2+85n+2017}=n+x&lt;/math&gt;. Squaring both sides and canceling the &lt;math&gt;n^2&lt;/math&gt; terms gives &lt;math&gt;85n+2017=2xn+x^2&lt;/math&gt;. Combining like terms, &lt;math&gt;(85-2x)n=x^2-2017&lt;/math&gt;, so<br /> <br /> &lt;cmath&gt;n=\frac{x^2-2017}{85-2x}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;n&lt;/math&gt; is positive, there are two cases, which are simple (luckily). Remembering that &lt;math&gt;x&lt;/math&gt; is a positive integer, then &lt;math&gt;x^2-2017&lt;/math&gt; and &lt;math&gt;85-2x&lt;/math&gt; are either both positive or both negative. The smallest value for which &lt;math&gt;x^2&gt;2017&lt;/math&gt; is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that &lt;math&gt;x&lt;45&lt;/math&gt; (from the numerator) and &lt;math&gt;85-2x&lt;0&lt;/math&gt;, which means &lt;math&gt;x&gt;42&lt;/math&gt;. This only gives two solutions, &lt;math&gt;x=43, 44&lt;/math&gt;. Plugging these into the expression for &lt;math&gt;n&lt;/math&gt;, we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is &lt;math&gt;168+27=\boxed{195}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Abuse the discriminant)==<br /> <br /> Let the integer given by the square root be represented by &lt;math&gt;x&lt;/math&gt;. Then &lt;math&gt;0 = n^2 + 85n + 2017 - x^2&lt;/math&gt;. For this to have rational solutions for &lt;math&gt;n&lt;/math&gt; (checking whether they are integers is done later), the discriminant of this quadratic must be a perfect square. (This can be easily shown using the quadratic formula.)<br /> <br /> Thus, &lt;math&gt;b^2 - 4ac = 7225 + 4x^2 - 8068 = y^2&lt;/math&gt; for some integer &lt;math&gt;y&lt;/math&gt;. Then &lt;math&gt;4x^2 - 843 = y^2&lt;/math&gt;. Rearranging this equation yields that &lt;math&gt;843 = (2x+y)(2x-y)&lt;/math&gt;. Noticing that there are 2 factor pairs of &lt;math&gt;843&lt;/math&gt;, namely, &lt;math&gt;1*843&lt;/math&gt; and &lt;math&gt;3*281&lt;/math&gt;, there are 2 systems to solve for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; that create rational &lt;math&gt;n&lt;/math&gt;. These yield solutions &lt;math&gt;(x,y)&lt;/math&gt; of &lt;math&gt;(211, 421)&lt;/math&gt; and &lt;math&gt;(71, 139)&lt;/math&gt;.<br /> <br /> The solution to the initial quadratic in &lt;math&gt;n&lt;/math&gt; must then be &lt;math&gt;\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}&lt;/math&gt;. Noticing that for each value of &lt;math&gt;x&lt;/math&gt; that has rational solutions for &lt;math&gt;n&lt;/math&gt;, the corresponding value of the square root of the discriminant is &lt;math&gt;y&lt;/math&gt;, the formula can be rewritten as &lt;math&gt;n = \frac{-85 \pm y}{2}&lt;/math&gt;. One solution is &lt;math&gt;\frac{421 - 85}{2} = 168&lt;/math&gt; and the other solution is &lt;math&gt;\frac{139 - 85}{2} = 27&lt;/math&gt;. Thus the answer is &lt;math&gt;168 + 27 = \boxed{195}&lt;/math&gt; as both rational solutions are integers.<br /> <br /> ==Solution 4 (Squeezing/Sandwich method)==<br /> <br /> Notice that &lt;math&gt;(n+42)^2= n^2+84n+1764&lt;/math&gt;. Also note that &lt;math&gt;(n+45)^2= n^2+90n+2025&lt;/math&gt;. Thus, &lt;cmath&gt;(n+42)^2&lt; n^2+85n+2017&lt;(n+45)^2&lt;/cmath&gt; where &lt;math&gt;n^2+85n+2017&lt;/math&gt; is a perfect square. Hence,&lt;cmath&gt;n^2+85n+2017= (n+43)^2&lt;/cmath&gt; or &lt;cmath&gt;n^2+85n+2017= (n+44)^2.&lt;/cmath&gt; Solving the two equations yields the two solutions &lt;math&gt;n= 168, 27&lt;/math&gt;. Therefore, our answer is &lt;math&gt;\boxed{195}&lt;/math&gt;.<br /> <br /> ==Solution 5 (Using factors)==<br /> <br /> Let the expression be equal to &lt;math&gt;a&lt;/math&gt;. This expression can be factored into &lt;math&gt;\sqrt{(n+44)^2-3n+81}&lt;/math&gt;. Then square both sides, and the expression becomes &lt;math&gt;(n+44)^2-3n+81=a^2&lt;/math&gt;. We have a difference of two squares. Rearranging, we have &lt;math&gt;(n+44+a)(n+44-a)=3(n-27)&lt;/math&gt;. By inspection, the only possible values for &lt;math&gt;(n+44-a)&lt;/math&gt; are 0 and 1. When &lt;math&gt;(n+44-a)=0&lt;/math&gt;, we must have &lt;math&gt;n-27=0&lt;/math&gt;. Therefore, &lt;math&gt;27&lt;/math&gt; is a solution. When we have &lt;math&gt;(n+44-a)=1&lt;/math&gt;, so &lt;math&gt;n=a-43&lt;/math&gt;. Plugging this back to &lt;math&gt;(n+44+a)=3(n-27)&lt;/math&gt; (since &lt;math&gt;(n+44-a)=1&lt;/math&gt;), we find that &lt;math&gt;a=211 \implies n=168&lt;/math&gt;. Thus, the answer is &lt;math&gt;27+168= \boxed{195}&lt;/math&gt;.<br /> <br /> '''-RootThreeOverTwo'''<br /> <br /> ==Solution 6==<br /> <br /> Ignore the square root for now. This expression can be factored into &lt;math&gt;(n+44)^2-3n+81&lt;/math&gt;. Just by inspection, when &lt;math&gt;n=27&lt;/math&gt;, the expression becomes &lt;math&gt;71^2&lt;/math&gt;, so &lt;math&gt;27&lt;/math&gt; is a solution. Proceed as Solution 5 to find the other solution(s).<br /> <br /> ==Solution 7 (alternative factoring)==<br /> More intuitive, but a little bit slower considering the decimals.<br /> <br /> <br /> Label the entire given expression as k^2.<br /> <br /> Instinctively we can do a crude completion of the square, resulting in k^2 = (n+42.5)^2+210.75<br /> Rearrange the equation to get a difference of squares.<br /> <br /> k^2-(n+42.5)^2 = 210.75<br /> <br /> (k+n+42.5)(k-n-42.5) = 210.75<br /> <br /> Factor 21,075 to get 3^1,5^2, and 281^1<br /> <br /> Now the two factors given are either divided by 10 each or one being divided by 100. Let's start with the former case.<br /> <br /> If you try 281*3/10 and 5*5/10, you quickly realize that n becomes negative. Naturally, you will realize you want the number's difference to be larger. Try 281*5/10 and 3*5/10. This gives an answer of 27 for x. The next largest possibility also works, giving an n of 168. As you rise, some numbers don't work because it results in an n that is not an integer, as in the example of 281*5*5/10 and 3/10. <br /> <br /> Now if you continue on with the next case, where one factor is divided by 100, very swiftly will you realize most don't work simply because the difference is too small, or it doesn't give an integer. It helps a lot when you realize that the decimal does not end in a 5, the answer will not be an integer. After a few short tests, we get &lt;math&gt;168+27=\boxed{195}&lt;/math&gt;.<br /> <br /> <br /> <br /> -jackshi2006<br /> <br /> ==Video Solution==<br /> <br /> https://youtu.be/Z23Yz05eblY<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II&diff=143692 2020 AIME II 2021-01-29T01:55:39Z <p>Blehlivesonearth: </p> <hr /> <div>'''2020 [[AIME|AIME II]]''' problems and solutions. The test was offered on June 6, 2020 for students who have took 2020 AIME I and students who were planning to take the cancelled 2020 AIME II on March 26, 2020. The first link will contain the full set of test problems. The rest will contain each individual problem and its solution.<br /> <br /> * [[2020 AIME II Problems|Entire Test]]<br /> * [[2020 AIME II Answer Key|Answer Key]]<br /> ** [[2020 AIME II Problems/Problem 1|Problem 1]] <br /> ** [[2020 AIME II Problems/Problem 2|Problem 2]]<br /> ** [[2020 AIME II Problems/Problem 3|Problem 3]]<br /> ** [[2020 AIME II Problems/Problem 4|Problem 4]]<br /> ** [[2020 AIME II Problems/Problem 5|Problem 5]]<br /> ** [[2020 AIME II Problems/Problem 6|Problem 6]]<br /> ** [[2020 AIME II Problems/Problem 7|Problem 7]]<br /> ** [[2020 AIME II Problems/Problem 8|Problem 8]]<br /> ** [[2020 AIME II Problems/Problem 9|Problem 9]]<br /> ** [[2020 AIME II Problems/Problem 10|Problem 10]]<br /> ** [[2020 AIME II Problems/Problem 11|Problem 11]]<br /> ** [[2020 AIME II Problems/Problem 12|Problem 12]]<br /> ** [[2020 AIME II Problems/Problem 13|Problem 13]]<br /> ** [[2020 AIME II Problems/Problem 14|Problem 14]]<br /> ** [[2020 AIME II Problems/Problem 15|Problem 15]]<br /> <br /> ==See also==<br /> {{AIME box|year=2020|n=II|before=[[2020 AIME I]]|after=[[2021 AIME I]]}}<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II&diff=143691 2020 AIME II 2021-01-29T01:54:06Z <p>Blehlivesonearth: </p> <hr /> <div>'''2020 [[AIME|AIME II]]''' problems and solutions. The test was offered on June 6, 2020 for students who have took 2020 AIME I and students who were planning to take the cancelled 2020 AIME II on March 26, 2020. The first link will contain the full set of test problems. The rest will contain each individual problem and its solution.<br /> <br /> * [[2020 AIME II Problems|Entire Test]]<br /> * [[2020 AIME II Answer Key|Answer Key]]<br /> ** [[2020 AIME II Problems/Problem 1|Problem 1]] <br /> ** [[2020 AIME II Problems/Problem 2|Problem 2]]<br /> ** [[2020 AIME II Problems/Problem 3|Problem 3]]<br /> ** [[2020 AIME II Problems/Problem 4|Problem 4]]<br /> ** [[2020 AIME II Problems/Problem 5|Problem 5]]<br /> ** [[2020 AIME II Problems/Problem 6|Problem 6]]<br /> ** [[2020 AIME II Problems/Problem 7|Problem 7]]<br /> ** [[2020 AIME II Problems/Problem 8|Problem 8]]<br /> ** [[2020 AIME II Problems/Problem 9|Problem 9]]<br /> ** [[2020 AIME II Problems/Problem 10|Problem 10]]<br /> ** [[2020 AIME II Problems/Problem 11|Problem 11]]<br /> ** [[2020 AIME II Problems/Problem 12|Problem 12]]<br /> ** [[2020 AIME II Problems/Problem 13|Problem 13]]<br /> ** [[2020 AIME II Problems/Problem 14|Problem 14]]<br /> ** [[2020 AIME II Problems/Problem 15|Problem 15]]<br /> <br /> {{AIME box|year=2020|n=II|before=[[2020 AIME I]]|after=[[2021 AIME I]]}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_23&diff=142714 2014 AMC 12B Problems/Problem 23 2021-01-19T04:58:48Z <p>Blehlivesonearth: /* Sidenote */</p> <hr /> <div>==Problem==<br /> The number &lt;math&gt;2017&lt;/math&gt; is prime. Let &lt;math&gt;S = \sum \limits_{k=0}^{62} \dbinom{2014}{k}&lt;/math&gt;. What is the remainder when &lt;math&gt;S&lt;/math&gt; is divided by &lt;math&gt;2017?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }32\qquad<br /> \textbf{(B) }684\qquad<br /> \textbf{(C) }1024\qquad<br /> \textbf{(D) }1576\qquad<br /> \textbf{(E) }2016\qquad&lt;/math&gt;<br /> <br /> ==Solution==<br /> Note that &lt;math&gt;2014\equiv -3 \mod2017&lt;/math&gt;. We have for &lt;math&gt;k\ge1&lt;/math&gt;<br /> &lt;cmath&gt;\dbinom{2014}{k}\equiv \frac{(-3)(-4)(-5)....(-2-k)}{k!}\mod 2017&lt;/cmath&gt; <br /> &lt;cmath&gt;\equiv (-1)^k\dbinom{k+2}{k} \mod 2017&lt;/cmath&gt;<br /> &lt;cmath&gt;\equiv (-1)^k\dbinom{k+2}{2} \mod 2017&lt;/cmath&gt;<br /> Therefore <br /> &lt;cmath&gt;\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017&lt;/cmath&gt;<br /> This is simply an alternating series of triangular numbers that goes like this: &lt;math&gt;1-3+6-10+15-21....&lt;/math&gt;<br /> After finding the first few sums of the series, it becomes apparent that <br /> &lt;cmath&gt;\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv -\left(\frac{n+1}{2} \right) \left(\frac{n+1}{2}+1 \right) \mod 2017 \textnormal{ if n is odd}&lt;/cmath&gt;<br /> and <br /> &lt;cmath&gt;\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv \left(\frac{n}{2}+1 \right)^2 \mod 2017 \textnormal{ if n is even}&lt;/cmath&gt;<br /> Obviously, &lt;math&gt;62&lt;/math&gt; falls in the second category, so our desired value is <br /> &lt;cmath&gt;\left(\frac{62}{2}+1 \right)^2 = 32^2 = \boxed{\textbf{(C)}\ 1024}&lt;/cmath&gt;<br /> <br /> ===Sidenote===<br /> Another way to finish, using the fact that &lt;math&gt;\dbinom{k+2}{2} = 1 + 2 + \dots + (k+1)&lt;/math&gt; (Hockey-Stick Identity):<br /> &lt;cmath&gt;\begin{align*}<br /> \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2}<br /> &amp;\equiv \sum \limits_{k=1}^{63}(-1)^k (1 + 2 + \dots + k) \\<br /> &amp;\equiv 1 - (1+2) + (1+2+3) - (1+2+3+4) + \dots + (1 + \dots + 63) \\<br /> &amp;\equiv 1 + 3 + 5 + \dots + 63 \\<br /> &amp;\equiv \boxed{1024} \mod 2017<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2014|ab=B|num-b=22|num-a=24}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_23&diff=142591 2015 AMC 12A Problems/Problem 23 2021-01-18T06:17:39Z <p>Blehlivesonearth: </p> <hr /> <div>{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #23]] and [[2015 AMC 10A Problems|2015 AMC 10A #25]]}}<br /> ==Problem==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be a square of side length 1. Two points are chosen independently at random on the sides of &lt;math&gt;S&lt;/math&gt;. The probability that the straight-line distance between the points is at least &lt;math&gt;\frac12&lt;/math&gt; is &lt;math&gt;\frac{a-b\pi}{c}&lt;/math&gt;, where &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are positive integers and &lt;math&gt;\text{gcd}(a,b,c) = 1&lt;/math&gt;. What is &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 59 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 61 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 63&lt;/math&gt;<br /> <br /> ==Solution==<br /> Divide the boundary of the square into halves, thereby forming 8 segments. Without loss of generality, let the first point &lt;math&gt;A&lt;/math&gt; be in the bottom-left segment. Then, it is easy to see that any point in the 5 segments not bordering the bottom-left segment will be distance at least &lt;math&gt;\dfrac{1}{2}&lt;/math&gt; apart from &lt;math&gt;A&lt;/math&gt;. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least 0.5 apart from &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;\dfrac{0 + 1}{2} = \dfrac{1}{2}&lt;/math&gt; because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)<br /> <br /> If the second point &lt;math&gt;B&lt;/math&gt; is on the left-bottom segment, then if &lt;math&gt;A&lt;/math&gt; is distance &lt;math&gt;x&lt;/math&gt; away from the left-bottom vertex, then &lt;math&gt;B&lt;/math&gt; must be at least &lt;math&gt;\dfrac{1}{2} - \sqrt{0.25 - x^2}&lt;/math&gt; away from that same vertex. Thus, using an averaging argument we find that the probability in this case is<br /> &lt;cmath&gt;\frac{1}{\frac{1}{2}^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4\left(\frac{1}{4} - \frac{\pi}{16}\right) = 1 - \frac{\pi}{4}.&lt;/cmath&gt;<br /> <br /> (Alternatively, one can equate the problem to finding all valid &lt;math&gt;(x, y)&lt;/math&gt; with &lt;math&gt;0 &lt; x, y &lt; \dfrac{1}{2}&lt;/math&gt; such that &lt;math&gt;x^2 + y^2 \ge \dfrac{1}{4}&lt;/math&gt;, i.e. (x, y) is outside the unit circle with radius 0.5.)<br /> <br /> Thus, averaging the probabilities gives<br /> &lt;cmath&gt;P = \frac{1}{8} \left(5 + \frac{1}{2} + 1 - \frac{\pi}{4}\right) = \frac{1}{32} (26 - \pi).&lt;/cmath&gt;<br /> <br /> Our answer is &lt;math&gt;\textbf{(A)}&lt;/math&gt;.<br /> <br /> ==Case Solution==<br /> Fix one of the points on a SIDE. There are three cases: the other point is on the same side, a peripheral side, or the opposite side, with probability &lt;math&gt;0.25, 0.5, 0.25&lt;/math&gt;, respectively.<br /> <br /> Opposite side: Probability is obviously &lt;math&gt;1&lt;/math&gt;, no matter what. <br /> <br /> Same side: Pretend the points are on a line with coordinates &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;. If &lt;math&gt;|a-b| \le \frac{1}{2}&lt;/math&gt;, drawing a graph will give probability &lt;math&gt;\frac{1}{4}&lt;/math&gt;.<br /> <br /> Peripheral side: superimpose a coordinate system over the points; call them &lt;math&gt;(x, 0)&lt;/math&gt; and &lt;math&gt;(0, y)&lt;/math&gt;. WLOG set &lt;math&gt;x, y &gt;= 0&lt;/math&gt; and &lt;math&gt;x, y &lt;= 1&lt;/math&gt;. We need &lt;math&gt;x^2+y^2&gt;0.25&lt;/math&gt;, and drawing the coordinate system with bounds &lt;math&gt;(0, 0), (1, 0), (0, 1), (1, 1)&lt;/math&gt; gives probability &lt;math&gt;1-\frac{\pi}{16}&lt;/math&gt; that the distance between the points is &lt;math&gt;&gt;0.5&lt;/math&gt;.<br /> <br /> Adding these up and finding the fraction gives us &lt;math&gt;\frac{1}{32} (26 - \pi)&lt;/math&gt; for an answer of &lt;math&gt; \textbf{(A)}\ 59 &lt;/math&gt;.<br /> <br /> ==Solution 3 (Average Function Value/Quick Faux Integration)==<br /> WLOG, let the first point be on the bottom side of the square. The points where the second point could exist are outside a circle of radius 0.5 centered on the first point. The parts of the square that lie in this circle are the distance from the point to the closest side of the square &lt;math&gt;n&lt;/math&gt;, the distance from the point to the outside of the circle (the radius &lt;math&gt;0.5&lt;/math&gt;), and any portion of the nearest side that lies within the circle as represented by the Pythagorean Theorem &lt;math&gt;\sqrt{\frac{1}{4}-n^2}&lt;/math&gt;. Thus, the total length the second point can exist in can be represented by &lt;math&gt;f(n)=4-(0.5+n+\sqrt{\frac{1}{4}-n^2})&lt;/math&gt;. Distributing, &lt;math&gt;f(n)=3.5-n-\sqrt{\frac{1}{4}-n^2}&lt;/math&gt;.<br /> <br /> Then, we can find the average of this function through calculus (wow more calc?). This formula is as follows,<br /> &lt;cmath&gt;\frac{1}{a_{f} -a_{i}} \int_{a_{i}}^{a_{f}} f(x)\;dx&lt;/cmath&gt;<br /> <br /> For this case, the limits of integration are &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;0.5&lt;/math&gt; (&lt;math&gt;0\leq n\leq 0.5&lt;/math&gt;). Then, we have,<br /> <br /> &lt;cmath&gt;2\int_{0}^{\frac{1}{2}} 3.5-n-\sqrt{\frac{1}{4}-n^2}\;dn&lt;/cmath&gt;<br /> &lt;cmath&gt;2(\int_{0}^{\frac{1}{2}} 3.5\;dn -\int_{0}^{\frac{1}{2}}n\;dn -\int_{0}^{\frac{1}{2}}\sqrt{\frac{1}{4}-n^2}\;dn) &lt;/cmath&gt;<br /> <br /> <br /> <br /> Even if you don't know how to integrate, as long as you know the idea of integration, you can figure these out. Graphing the first two, you can see that the first is a rectangle of length &lt;math&gt;0.5&lt;/math&gt; and width &lt;math&gt;3.5&lt;/math&gt;. The second is an isosceles right triangle of leg length &lt;math&gt;0.5&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\int_{0}^{\frac{1}{2}} 3.5\;dn=\frac{7}{4}, \int_{0}^{\frac{1}{2}}n\;dn=\frac{1}{8}&lt;/cmath&gt;<br /> <br /> Recognize that the third integral is a semicircle of radius &lt;math&gt;0.5&lt;/math&gt; and centered at the origin. This is where &lt;math&gt;\pi&lt;/math&gt; comes in. From &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;0.5&lt;/math&gt;, the integral is simply a quarter circle. &lt;math&gt;\frac{\frac{1}{2}^2\pi}{4}=\frac{\pi}{16}&lt;/math&gt;.<br /> &lt;cmath&gt;\int_{0}^{\frac{1}{2}}\sqrt{\frac{1}{4}-n^2}\;dn)=\frac{\pi}{16}&lt;/cmath&gt;<br /> <br /> If you want me to actually integrate these, look below. Do note that this is for those that have a limited knowledge of integration or those that have little time but are being very clever.<br /> <br /> Overall, where second point could lie to satisfy the problem is a length of &lt;math&gt;2(\frac{7}{4}-\frac{\pi}{16}-\frac{1}{8})=\frac{26-\pi}{8}&lt;/math&gt;. By contrast, the total length where it could lie is the perimeter of the square &lt;math&gt;4&lt;/math&gt;. So the possible points that the second point could be make up &lt;math&gt;\frac{\frac{26-\pi}{8}}{4}=\frac{26-\pi}{32}&lt;/math&gt; of the square's perimeter. Obviously, &lt;math&gt;\gcd(32, 26, 1)=1&lt;/math&gt;. &lt;math&gt;32+26+1=59\implies\boxed{A}&lt;/math&gt;<br /> <br /> ''Sorry for the long explanation!''<br /> '''Actual Integration as Promised'''<br /> &lt;cmath&gt;2(\int_{0}^{\frac{1}{2}} 3.5\;dn -\int_{0}^{\frac{1}{2}}n\;dn -\int_{0}^{\frac{1}{2}}\sqrt{\frac{1}{4}-n^2}\;dn)&lt;/cmath&gt;<br /> <br /> First two integrals easily done by power rule<br /> &lt;math&gt;\int_{0}^{\frac{1}{2}} 3.5\;dn=3.5\int_{0}^{\frac{1}{2}} 1\;dn=3.5n\Big|_0^{\frac{1}{2}}=7/4&lt;/math&gt;<br /> &lt;math&gt;\int_{0}^{\frac{1}{2}}n\;dn= \frac{n^2}{2}\Big|_0^{\frac{1}{2}}=1/8&lt;/math&gt;<br /> <br /> <br /> Last integral by trig substitution (long)<br /> <br /> &lt;math&gt;\int_{0}^{\frac{1}{2}}\sqrt{\frac{1}{4}-n^2}\;dn=\frac{1}{2}\int_{0}^{\frac{1}{2}}\sqrt{1-(2n)^2}&lt;/math&gt;<br /> <br /> If &lt;math&gt;\sin(u)=2n&lt;/math&gt;, then &lt;math&gt;dn=\frac{\cos(u)}{2}\;du&lt;/math&gt; (differentiate both sides). <br /> <br /> Then, &lt;math&gt;\frac{1}{4}\int_{n=0}^{n=\frac{1}{4}}\cos(u)\sqrt{1-\sin(u)^2}\;du=\frac{1}{4}\int_{n=0}^{n=\frac{1}{2}}\cos(u)\sqrt{\cos^2(u)}\;du=\frac{1}{2}\int_{n=0}^{n=\frac{1}{2}}\cos^2(u)\;du&lt;/math&gt;<br /> <br /> This is a known integral that can be derived from further trig identities (specifically double angle). For the sake of brevity, <br /> <br /> &lt;math&gt;\frac{1}{4}\int_{n=0}^{n=\frac{1}{2}}\cos^2(u)\;du= \frac{1}{4}(\frac{2u + \sin(2u)}{4})\Big|_{x=0}^{x=\frac{1}{2}}=\frac{2u+\sin(2u)}{16}\Big|_{x=0}^{x=\frac{1}{2}}&lt;/math&gt;.<br /> <br /> Convert the limits: &lt;math&gt;\sin(u)=2\cdot0\implies u_1=0&lt;/math&gt; and &lt;math&gt;\sin(u)=2\cdot\frac{1}{2} \implies u_2=\frac{\pi}{2}&lt;/math&gt;<br /> <br /> Finally, we have &lt;math&gt;\frac{2u+\sin(2u)}{16}\Big|_{0}^{\frac{\pi}{2}}=\frac{\pi+\sin(\pi)}{16}-0=\frac{\pi}{16}&lt;/math&gt;<br /> <br /> <br /> ~ Solution By BJHHar<br /> <br /> ==Solution 4 (Extension, CALCULUS)==<br /> Set the problem up similarly to in solution 1, where we split the square into 8 segments. Notice that each segment is the same, so WLOG use any one of them. For the purposes of this solution, I will assume the segment we use starts at (0, 0) and ends at (0.5, 0). The square I use will have vertices of &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;(1,0)&lt;/math&gt;, &lt;math&gt;(0,1)&lt;/math&gt;, and &lt;math&gt;(1,1)&lt;/math&gt;. <br /> <br /> A way we can figure out when the distance is at least &lt;math&gt;0.5&lt;/math&gt; is if we figure out when it isn't &lt;math&gt;0.5&lt;/math&gt;. Let's pick a point on our segment and denote it with &lt;math&gt;(x, 0)&lt;/math&gt;. Then, there are three ways a second point can be within the boundaries of this first point. Either, it is to the left of it (if possible), to the right of it, or it is on the segment that forms a right angle with it. <br /> <br /> Obviously, there is &lt;math&gt;x&lt;/math&gt; distance if the point is to the left of the point &lt;math&gt;x&lt;/math&gt;. Since we denoted this segment to be from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;0.5&lt;/math&gt;, then there will always be &lt;math&gt;0.5&lt;/math&gt; distance to the right of the point &lt;math&gt;x&lt;/math&gt; (as that's the maximum that we are trying to figure out).<br /> <br /> The difficult part is finding the total length on the segment that is perpendicular to our segment. However, since the square has a right angle, we can first find that the segment should have a length of &lt;math&gt;\sqrt{0.25-x^2}&lt;/math&gt; (by the Pythagorean theorem, with hypotenuse 0.5 and one leg being x).<br /> <br /> Now that we have our three distances, all we need to do is find the average value of them. We can best do this with &quot;Integral/Interval&quot;, and so we take the integral from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;0.5&lt;/math&gt; of &lt;math&gt;x+0.5+\sqrt{0.25-x^2}&lt;/math&gt; and then divide it by &lt;math&gt;0.5&lt;/math&gt; (the interval). To integrate &lt;math&gt;\sqrt{0.25-x^2}&lt;/math&gt; by hand, we want to pull out the &lt;math&gt;1/4&lt;/math&gt;, and then apply u-sub and our integration rules to find the answer. <br /> <br /> We get &lt;math&gt;0.75 - \frac{\pi}{16} \cdot 2&lt;/math&gt;. Simplify this into &lt;math&gt;\frac{6-\pi}{8}&lt;/math&gt;. <br /> <br /> Now we are on our last stage. Proceed to make this equivalent to what the question is asking, as we have found the probability that the second point is within a distance of &lt;math&gt;0.5&lt;/math&gt;, whereas the question asks for at least a distance of &lt;math&gt;0.5&lt;/math&gt; (so more than). We can do this simply by doing &lt;math&gt;4- \frac{6-\pi}{8}&lt;/math&gt; (as &lt;math&gt;4&lt;/math&gt; is the total amount of length). This equates to &lt;math&gt;\frac{26-\pi}{8}&lt;/math&gt; and then we divide by &lt;math&gt;4&lt;/math&gt; as that is the total amount of length (remember this is probability). <br /> <br /> Thus, we get &lt;math&gt;\frac{26-\pi}{32}&lt;/math&gt; for our probability, and so the answer is &lt;math&gt;26 + 1 + 32 = \textbf{(A)}\ 59&lt;/math&gt;.<br /> <br /> Some notes: I tried to explain everything but it's quite difficult to explain - there is a way of non-calculus (like there always is) that I think was mentioned above, something with circles (since the thing under the square root is just &lt;math&gt;x^2+y^2 = \frac{1}{4}&lt;/math&gt;, so finding the average value of that isn't difficult). <br /> <br /> IronicNinja~ Edited by AngelaLZ~<br /> <br /> ==Solution 5 (Area)==<br /> Choose a certain side for one of the points to be on. Let the distance from the point to the vertex on its left be &lt;math&gt;x&lt;/math&gt;<br /> <br /> We split this into two cases: <br /> <br /> Case 1: &lt;math&gt;0\leq x\leq \frac12&lt;/math&gt;:<br /> <br /> The total length of the segments for which the other point can be on such that the straight-line distance between the points is less than &lt;math&gt;\frac12&lt;/math&gt; is &lt;cmath&gt;\sqrt{\frac14-x^2}+x+\frac12.&lt;/cmath&gt; We can graph this in the Cartesian plane and find the area of the region below the curve and above the line &lt;math&gt;y=0&lt;/math&gt;.<br /> <br /> Case 2: &lt;math&gt;\frac12&lt; x\leq 1&lt;/math&gt;:<br /> <br /> This is basically Case 1 but flipped over the line &lt;math&gt;x=\frac12&lt;/math&gt;.<br /> <br /> So our total probability is 1 minus the area of the graph over the total area (4, perimeter of square). Notice that the desired area of the region below the curve we found earlier is the sum of a quarter circle with radius &lt;math&gt;\frac12&lt;/math&gt; and centered at &lt;math&gt;(0,0)&lt;/math&gt; and a trapezoid with height &lt;math&gt;\frac12&lt;/math&gt; and bases of length &lt;math&gt;\frac12&lt;/math&gt; and &lt;math&gt;\frac32&lt;/math&gt;. Adding this all up then multiplying by 2, we have &lt;cmath&gt;\frac{\pi}{8}+\frac34&lt;/cmath&gt; and then the probability of the desired result would be &lt;cmath&gt;1-\frac{\pi+6}{32}=\frac{26-\pi}{32}&lt;/cmath&gt; and our answer is &lt;math&gt;26+1+32=\boxed{59}&lt;/math&gt;. ~caroline2023<br /> <br /> === Video Solution by Richard Rusczyk ===<br /> <br /> https://artofproblemsolving.com/videos/amc/2015amc12a/399<br /> <br /> ~ dolphin7<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2015|ab=A|num-b=22|num-a=24}}<br /> {{AMC10 box|year=2015|ab=A|num-b=24|after=Last Problem}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_23&diff=142585 2015 AMC 12B Problems/Problem 23 2021-01-18T04:17:10Z <p>Blehlivesonearth: </p> <hr /> <div>{{duplicate|[[2015 AMC 12B Problems|2015 AMC 12B #23]] and [[2015 AMC 10B Problems|2015 AMC 10B #25]]}}<br /> ==Problem==<br /> <br /> A rectangular box measures &lt;math&gt;a \times b \times c&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are integers and &lt;math&gt;1\leq a \leq b \leq c&lt;/math&gt;. The volume and the surface area of the box are numerically equal. How many ordered triples &lt;math&gt;(a,b,c)&lt;/math&gt; are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> We need &lt;cmath&gt;abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).&lt;/cmath&gt;<br /> Since &lt;math&gt;ab, ac \le bc&lt;/math&gt;, we get &lt;math&gt;abc \le 6bc&lt;/math&gt;. Thus &lt;math&gt;a\le 6&lt;/math&gt;. From the second equation we see that &lt;math&gt;a &gt; 2&lt;/math&gt;. Thus &lt;math&gt;a\in \{3, 4, 5, 6\}&lt;/math&gt;. <br /> <br /> *If &lt;math&gt;a=3&lt;/math&gt; we need &lt;math&gt;bc = 6(b+c) \Rightarrow (b-6)(c-6)=36&lt;/math&gt;. We get '''five''' roots &lt;math&gt;\{(3, 7, 42), (3, 8, 24), (3,9,18), (3, 10, 15), (3,12,12)\}.&lt;/math&gt;<br /> *If &lt;math&gt;a=4&lt;/math&gt; we need &lt;math&gt;bc = 4(b+c) \Rightarrow (b-4)(c-4)=16&lt;/math&gt;. We get '''three''' roots &lt;math&gt;\{(4,5,20), (4,6,12), (4,8,8)\}&lt;/math&gt;. <br /> *If &lt;math&gt;a=5&lt;/math&gt; we need &lt;math&gt;3bc = 10(b+c)&lt;/math&gt;, which is the same as &lt;math&gt;9bc=30(b+c)\Rightarrow (3b-10)(3c-10)=100&lt;/math&gt;. We get only '''one''' root (corresponding to &lt;math&gt;100=5\cdot 20&lt;/math&gt;) &lt;math&gt;(5,5,10)&lt;/math&gt;. <br /> *If &lt;math&gt;a=6&lt;/math&gt; we need &lt;math&gt;4bc = 12(b+c)&lt;/math&gt;. Then &lt;math&gt;(b-3)(c-3)=9&lt;/math&gt;. We get ''' one''' root &lt;math&gt;(6,6,6)&lt;/math&gt;. <br /> Thus, there are &lt;math&gt;5+3+1+1 = \boxed{\textbf{(B)}\; 10}&lt;/math&gt; solutions.<br /> <br /> ==Solution 2==<br /> <br /> The surface area is &lt;math&gt;2(ab+bc+ca)&lt;/math&gt;, and the volume is &lt;math&gt;abc&lt;/math&gt;, so equating the two yields <br /> <br /> &lt;cmath&gt;2(ab+bc+ca)=abc.&lt;/cmath&gt;<br /> <br /> Divide both sides by &lt;math&gt;2abc&lt;/math&gt; to obtain &lt;cmath&gt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.&lt;/cmath&gt;<br /> <br /> First consider the bound of the variable &lt;math&gt;a&lt;/math&gt;. Since &lt;math&gt;\frac{1}{a}&lt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},&lt;/math&gt; we have &lt;math&gt;a&gt;2&lt;/math&gt;, or &lt;math&gt;a\geqslant3&lt;/math&gt;.<br /> <br /> Also note that &lt;math&gt;c \geq b \geq a &gt; 0&lt;/math&gt;, hence &lt;math&gt;\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}&lt;/math&gt;.<br /> Thus, &lt;math&gt;\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}&lt;/math&gt;, so &lt;math&gt;a \leq 6&lt;/math&gt;.<br /> <br /> So we have &lt;math&gt;a=3, 4, 5&lt;/math&gt; or &lt;math&gt;6&lt;/math&gt;.<br /> <br /> Before the casework, let's consider the possible range for &lt;math&gt;b&lt;/math&gt; if &lt;math&gt;\frac{1}{b}+\frac{1}{c}=k&gt;0&lt;/math&gt;. From &lt;math&gt;\frac{1}{b}&lt;k&lt;/math&gt;, we have &lt;math&gt;b&gt;\frac{1}{k}&lt;/math&gt;. From &lt;math&gt;\frac{2}{b} \geq \frac{1}{b}+\frac{1}{c}=k&lt;/math&gt;, we have &lt;math&gt;b \leq \frac{2}{k}&lt;/math&gt;. Thus &lt;math&gt;\frac{1}{k}&lt;b \leq \frac{2}{k}&lt;/math&gt;.<br /> <br /> When &lt;math&gt;a=3&lt;/math&gt;, we get &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{6}&lt;/math&gt;, so &lt;math&gt;b=7, 8, 9, 10, 11, 12&lt;/math&gt;. We find the solutions &lt;math&gt;(a, b, c)=(3, 7, 42)&lt;/math&gt;, &lt;math&gt;(3, 8, 24)&lt;/math&gt;, &lt;math&gt;(3, 9, 18)&lt;/math&gt;, &lt;math&gt;(3, 10, 15)&lt;/math&gt;, &lt;math&gt;(3, 12, 12)&lt;/math&gt;, for a total of &lt;math&gt;5&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=4&lt;/math&gt;, we get &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{4}&lt;/math&gt;, so &lt;math&gt;b=5, 6, 7, 8&lt;/math&gt;. We find the solutions &lt;math&gt;(a, b, c)=(4, 5, 20)&lt;/math&gt;, &lt;math&gt;(4, 6, 12)&lt;/math&gt;, &lt;math&gt;(4, 8, 8)&lt;/math&gt;, for a total of &lt;math&gt;3&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=5&lt;/math&gt;, we get &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{3}{10}&lt;/math&gt;, so &lt;math&gt;b=5, 6&lt;/math&gt;. The only solution in this case is &lt;math&gt;(a, b, c)=(5, 5, 10)&lt;/math&gt;.<br /> <br /> When &lt;math&gt;a=6&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; is forced to be &lt;math&gt;6&lt;/math&gt;, and thus &lt;math&gt;(a, b, c)=(6, 6, 6)&lt;/math&gt;.<br /> <br /> Thus, there are &lt;math&gt;5+3+1+1 = \boxed{\textbf{(B)}\; 10}&lt;/math&gt; solutions.<br /> <br /> ==Note==<br /> This is also AMC 10B Problem 25, but the pages are separate. Someone should fix this.<br /> EDIT: fixed it, but someone help with the link<br /> <br /> EDIT #2: fixed all<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|after=Last Question|num-b=24}}<br /> {{AMC12 box|year=2015|ab=B|num-a=24|num-b=22}}<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_23&diff=142490 2015 AMC 12B Problems/Problem 23 2021-01-17T23:14:56Z <p>Blehlivesonearth: </p> <hr /> <div>==Problem==<br /> {{duplicate|[[2015 AMC 12B Problems|2015 AMC 12B #23]] and [[2015 AMC 10B Problems|2015 AMC 10B #25]]}}<br /> A rectangular box measures &lt;math&gt;a \times b \times c&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are integers and &lt;math&gt;1\leq a \leq b \leq c&lt;/math&gt;. The volume and the surface area of the box are numerically equal. How many ordered triples &lt;math&gt;(a,b,c)&lt;/math&gt; are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> We need &lt;cmath&gt;abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).&lt;/cmath&gt;<br /> Since &lt;math&gt;ab, ac \le bc&lt;/math&gt;, we get &lt;math&gt;abc \le 6bc&lt;/math&gt;. Thus &lt;math&gt;a\le 6&lt;/math&gt;. From the second equation we see that &lt;math&gt;a &gt; 2&lt;/math&gt;. Thus &lt;math&gt;a\in \{3, 4, 5, 6\}&lt;/math&gt;. <br /> <br /> *If &lt;math&gt;a=3&lt;/math&gt; we need &lt;math&gt;bc = 6(b+c) \Rightarrow (b-6)(c-6)=36&lt;/math&gt;. We get '''five''' roots &lt;math&gt;\{(3, 7, 42), (3, 8, 24), (3,9,18), (3, 10, 15), (3,12,12)\}.&lt;/math&gt;<br /> *If &lt;math&gt;a=4&lt;/math&gt; we need &lt;math&gt;bc = 4(b+c) \Rightarrow (b-4)(c-4)=16&lt;/math&gt;. We get '''three''' roots &lt;math&gt;\{(4,5,20), (4,6,12), (4,8,8)\}&lt;/math&gt;. <br /> *If &lt;math&gt;a=5&lt;/math&gt; we need &lt;math&gt;3bc = 10(b+c)&lt;/math&gt;, which is the same as &lt;math&gt;9bc=30(b+c)\Rightarrow (3b-10)(3c-10)=100&lt;/math&gt;. We get only '''one''' root (corresponding to &lt;math&gt;100=5\cdot 20&lt;/math&gt;) &lt;math&gt;(5,5,10)&lt;/math&gt;. <br /> *If &lt;math&gt;a=6&lt;/math&gt; we need &lt;math&gt;4bc = 12(b+c)&lt;/math&gt;. Then &lt;math&gt;(b-3)(c-3)=9&lt;/math&gt;. We get ''' one''' root &lt;math&gt;(6,6,6)&lt;/math&gt;. <br /> Thus, there are &lt;math&gt;5+3+1+1 = \boxed{\textbf{(B)}\; 10}&lt;/math&gt; solutions.<br /> <br /> ==Solution 2==<br /> <br /> The surface area is &lt;math&gt;2(ab+bc+ca)&lt;/math&gt;, and the volume is &lt;math&gt;abc&lt;/math&gt;, so equating the two yields <br /> <br /> &lt;cmath&gt;2(ab+bc+ca)=abc.&lt;/cmath&gt;<br /> <br /> Divide both sides by &lt;math&gt;2abc&lt;/math&gt; to obtain &lt;cmath&gt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.&lt;/cmath&gt;<br /> <br /> First consider the bound of the variable &lt;math&gt;a&lt;/math&gt;. Since &lt;math&gt;\frac{1}{a}&lt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},&lt;/math&gt; we have &lt;math&gt;a&gt;2&lt;/math&gt;, or &lt;math&gt;a\geqslant3&lt;/math&gt;.<br /> <br /> Also note that &lt;math&gt;c \geq b \geq a &gt; 0&lt;/math&gt;, hence &lt;math&gt;\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}&lt;/math&gt;.<br /> Thus, &lt;math&gt;\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}&lt;/math&gt;, so &lt;math&gt;a \leq 6&lt;/math&gt;.<br /> <br /> So we have &lt;math&gt;a=3, 4, 5&lt;/math&gt; or &lt;math&gt;6&lt;/math&gt;.<br /> <br /> Before the casework, let's consider the possible range for &lt;math&gt;b&lt;/math&gt; if &lt;math&gt;\frac{1}{b}+\frac{1}{c}=k&gt;0&lt;/math&gt;. From &lt;math&gt;\frac{1}{b}&lt;k&lt;/math&gt;, we have &lt;math&gt;b&gt;\frac{1}{k}&lt;/math&gt;. From &lt;math&gt;\frac{2}{b} \geq \frac{1}{b}+\frac{1}{c}=k&lt;/math&gt;, we have &lt;math&gt;b \leq \frac{2}{k}&lt;/math&gt;. Thus &lt;math&gt;\frac{1}{k}&lt;b \leq \frac{2}{k}&lt;/math&gt;.<br /> <br /> When &lt;math&gt;a=3&lt;/math&gt;, we get &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{6}&lt;/math&gt;, so &lt;math&gt;b=7, 8, 9, 10, 11, 12&lt;/math&gt;. We find the solutions &lt;math&gt;(a, b, c)=(3, 7, 42)&lt;/math&gt;, &lt;math&gt;(3, 8, 24)&lt;/math&gt;, &lt;math&gt;(3, 9, 18)&lt;/math&gt;, &lt;math&gt;(3, 10, 15)&lt;/math&gt;, &lt;math&gt;(3, 12, 12)&lt;/math&gt;, for a total of &lt;math&gt;5&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=4&lt;/math&gt;, we get &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{4}&lt;/math&gt;, so &lt;math&gt;b=5, 6, 7, 8&lt;/math&gt;. We find the solutions &lt;math&gt;(a, b, c)=(4, 5, 20)&lt;/math&gt;, &lt;math&gt;(4, 6, 12)&lt;/math&gt;, &lt;math&gt;(4, 8, 8)&lt;/math&gt;, for a total of &lt;math&gt;3&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=5&lt;/math&gt;, we get &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{3}{10}&lt;/math&gt;, so &lt;math&gt;b=5, 6&lt;/math&gt;. The only solution in this case is &lt;math&gt;(a, b, c)=(5, 5, 10)&lt;/math&gt;.<br /> <br /> When &lt;math&gt;a=6&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; is forced to be &lt;math&gt;6&lt;/math&gt;, and thus &lt;math&gt;(a, b, c)=(6, 6, 6)&lt;/math&gt;.<br /> <br /> Thus, there are &lt;math&gt;5+3+1+1 = \boxed{\textbf{(B)}\; 10}&lt;/math&gt; solutions.<br /> <br /> ==Note==<br /> This is also AMC 10B Problem 25, but the pages are separate. Someone should fix this.<br /> EDIT: fixed it, but someone help with the link<br /> EDIT #2: fixed all<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|after=Last Question|num-b=24}}<br /> {{AMC12 box|year=2015|ab=B|num-a=24|num-b=22}}<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_23&diff=142424 2017 AMC 12B Problems/Problem 23 2021-01-17T06:30:13Z <p>Blehlivesonearth: /* Solution 2 */</p> <hr /> <div>==Problem 23==<br /> The graph of &lt;math&gt;y=f(x)&lt;/math&gt;, where &lt;math&gt;f(x)&lt;/math&gt; is a polynomial of degree &lt;math&gt;3&lt;/math&gt;, contains points &lt;math&gt;A(2,4)&lt;/math&gt;, &lt;math&gt;B(3,9)&lt;/math&gt;, and &lt;math&gt;C(4,16)&lt;/math&gt;. Lines &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; intersect the graph again at points &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt;, respectively, and the sum of the &lt;math&gt;x&lt;/math&gt;-coordinates of &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; is 24. What is &lt;math&gt;f(0)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\quad {-2} \qquad \qquad \textbf{(B)}\quad 0 \qquad\qquad \textbf{(C)}\quad 2 \qquad\qquad \textbf{(D)}\quad \dfrac{24}5 \qquad\qquad\textbf{(E)}\quad 8&lt;/math&gt;<br /> <br /> ==Solution==<br /> First, we can define &lt;math&gt;f(x) = a(x-2)(x-3)(x-4) +x^2&lt;/math&gt;, which contains points &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;. Now we find that lines &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; are defined by the equations &lt;math&gt;y = 5x - 6&lt;/math&gt;, &lt;math&gt;y= 6x-8&lt;/math&gt;, and &lt;math&gt;y=7x-12&lt;/math&gt; respectively. Since we want to find the &lt;math&gt;x&lt;/math&gt;-coordinates of the intersections of these lines and &lt;math&gt;f(x)&lt;/math&gt;, we set each of them to &lt;math&gt;f(x)&lt;/math&gt;, and synthetically divide by the solutions we already know exist (eg. if we were looking at line &lt;math&gt;AB&lt;/math&gt;, we would synthetically divide by the solutions &lt;math&gt;x=2&lt;/math&gt; and &lt;math&gt;x=3&lt;/math&gt;, because we already know &lt;math&gt;AB&lt;/math&gt; intersects the graph at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, which have &lt;math&gt;x&lt;/math&gt;-coordinates of &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;). After completing this process on all three lines, we get that the &lt;math&gt;x&lt;/math&gt;-coordinates of &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; are &lt;math&gt;\frac{4a-1}{a}&lt;/math&gt;, &lt;math&gt;\frac{3a-1}{a}&lt;/math&gt;, and &lt;math&gt;\frac{2a-1}{a}&lt;/math&gt; respectively. Adding these together, we get &lt;math&gt;\frac{9a-3}{a} = 24&lt;/math&gt; which gives us &lt;math&gt;a = -\frac{1}{5}&lt;/math&gt;. Substituting this back into the original equation, we get &lt;math&gt;f(x) = -\frac{1}{5}(x-2)(x-3)(x-4) + x^2&lt;/math&gt;, and &lt;math&gt;f(0) = -\frac{1}{5}(-2)(-3)(-4) + 0 = \boxed{\textbf{(D)}\frac{24}{5}}&lt;/math&gt;<br /> <br /> Solution by vedadehhc<br /> <br /> ==Solution 2==<br /> &lt;math&gt;\boxed{\textbf{No need to find the equations for the lines, really.}}&lt;/math&gt; First of all, &lt;math&gt;f(x) = a(x-2)(x-3)(x-4) +x^2&lt;/math&gt;. Let's say the line &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;y=bx+c&lt;/math&gt;, and &lt;math&gt;x_1&lt;/math&gt; is the &lt;math&gt;x&lt;/math&gt; coordinate of the third intersection, then &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;x_1&lt;/math&gt; are the three roots of &lt;math&gt;f(x) - bx-c&lt;/math&gt;. Apparently the value of &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; have no effect on the sum of the 3 roots, because the coefficient of the &lt;math&gt;x^2&lt;/math&gt; term is always &lt;math&gt;-9a+1&lt;/math&gt;. So we have, <br /> &lt;cmath&gt; \frac{9a-1}{a} = 2+3 + x_1=3+4+x_2 = 2+4+x_3&lt;/cmath&gt;<br /> Add them up we have <br /> &lt;cmath&gt; 3\frac{9a-1}{a} = 18 + x_1+x_2+x_3 = 18 +24&lt;/cmath&gt;<br /> Solve it, we get &lt;math&gt;a = -\frac{1}{5}&lt;/math&gt;.<br /> &lt;math&gt;\boxed{\textbf{(D)}\frac{24}{5}}&lt;/math&gt;.<br /> <br /> - Mathdummy<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_26&diff=142384 1981 AHSME Problems/Problem 26 2021-01-17T04:38:48Z <p>Blehlivesonearth: </p> <hr /> <div>== Problem ==<br /> <br /> Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is &lt;math&gt;\frac{1}{6}&lt;/math&gt;, independent of the outcome of any other toss.) <br /> &lt;math&gt;\textbf{(A) } \frac{1}{3} \textbf{(B) } \frac{2}{9} \textbf{(C) } \frac{5}{18} \textbf{(D) } \frac{25}{91} \textbf{(E) } \frac{36}{91}&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> The probability that Carol wins during the first cycle through is &lt;math&gt;\frac{5}{6}*\frac{5}{6}*\frac{1}{6}&lt;/math&gt;, and the probability that Carol wins on the second cycle through is &lt;math&gt;\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}&lt;/math&gt;. It is clear that this is an infinite geometric sequence, and we must find the sum of it in order to find the answer to this question. Thus we set up the equation: &lt;math&gt;\frac{\frac{25}{216}}{1-\frac{125}{216}}&lt;/math&gt;, or &lt;math&gt;\frac{\frac{25}{216}}{\frac{91}{216}}&lt;/math&gt;, which simplifies into &lt;math&gt;\boxed{\textbf{(D) } \frac{25}{91}}&lt;/math&gt;<br /> <br /> == See also ==<br /> <br /> * [[AMC 12 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> {{AHSME box|year=1981|before=[[1980 AHSME]]|after=[[1982 AHSME]]}} <br /> <br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_26&diff=142383 1981 AHSME Problems/Problem 26 2021-01-17T04:37:30Z <p>Blehlivesonearth: </p> <hr /> <div>== Problem ==<br /> <br /> Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is &lt;math&gt;\frac{1}{6}&lt;/math&gt;, independent of the outcome of any other toss.) <br /> &lt;math&gt;\textbf{(A) } \frac{1}{3} \textbf{(B) } \frac{2}{9} \textbf{(C) } \frac{5}{18} \textbf{(D) } \frac{25}{91} \textbf{(E) } \frac{36}{91}&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> The probability that Carol wins during the first cycle through is &lt;math&gt;\frac{5}{6}*\frac{5}{6}*\frac{1}{6}&lt;/math&gt;, and the probability that Carol wins on the second cycle through is &lt;math&gt;\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}&lt;/math&gt;. It is clear that this is an infinite geometric sequence, and we must find the sum of it in order to find the answer to this question. Thus we set up the equation: &lt;math&gt;\frac{\frac{25}{216}}{1-\frac{125}{216}}&lt;/math&gt;, or &lt;math&gt;\frac{\frac{25}{216}}{\frac{91}{216}}&lt;/math&gt;, which simplifies into &lt;math&gt;\boxed{\textbf{(D) } \frac{25}{91}}&lt;/math&gt;</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_26&diff=142381 1981 AHSME Problems/Problem 26 2021-01-17T04:32:46Z <p>Blehlivesonearth: /* Problem */</p> <hr /> <div>== Problem ==<br /> <br /> Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is &lt;math&gt;\frac{1}{6}&lt;/math&gt;, independent of the outcome of any other toss.) <br /> &lt;math&gt;\\<br /> \textbf{(A) } \frac{25}{91}&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> The probability that Carol wins during the first cycle through is &lt;math&gt;\frac{5}{6}*\frac{5}{6}*\frac{1}{6}&lt;/math&gt;, and the probability that Carol wins on the second cycle through is &lt;math&gt;\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}&lt;/math&gt;. It is clear that this is an infinite geometric sequence, and we must find the sum of it in order to find the answer to this question. Thus we set up the equation: &lt;math&gt;\frac{\frac{25}{216}}{1-\frac{125}{216}}&lt;/math&gt;, or &lt;math&gt;\frac{\frac{25}{216}}{\frac{91}{216}}&lt;/math&gt;, which simplifies into &lt;math&gt;\boxed{\textbf{(A) } \frac{25}{91}}&lt;/math&gt;</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A&diff=142228 2022 AMC 12A 2021-01-16T19:28:03Z <p>Blehlivesonearth: </p> <hr /> <div>{{AMC12 Problems|year=2022|ab=A}}<br /> ==Problem 1==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=B|before=[[2021 AMC 12B Problems]]|after=[[2022 AMC 12B Problems]]}}<br /> <br /> [[Category:AMC 12 Problems]]<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A&diff=142227 2022 AMC 12A 2021-01-16T19:27:36Z <p>Blehlivesonearth: </p> <hr /> <div>{{AMC12 Problems|year=2022|ab=B}}<br /> ==Problem 1==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=B|before=[[2021 AMC 12B Problems]]|after=[[2022 AMC 12B Problems]]}}<br /> <br /> [[Category:AMC 12 Problems]]<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12B&diff=142226 2022 AMC 12B 2021-01-16T19:25:29Z <p>Blehlivesonearth: </p> <hr /> <div>Be patient, the 2022 AMC 12B will come after February 2022.</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=AMC_12_Problems_and_Solutions&diff=142224 AMC 12 Problems and Solutions 2021-01-16T17:59:53Z <p>Blehlivesonearth: </p> <hr /> <div>[[AMC 12]] problems and solutions. <br /> <br /> {| class=&quot;wikitable&quot; style=&quot;text-align:center&quot;<br /> |-<br /> ! Year || Test A || Test B<br /> |-<br /> | 2022 || [[2022 AMC 12A | AMC 12A]] || [[2022 AMC 12B | AMC 12B]]<br /> |-<br /> | 2021 || [[2021 AMC 12A | AMC 12A]] || [[2021_AMC_12B | AMC 12B]]<br /> |-<br /> | 2020 || [[2020_AMC_12A | AMC 12A]] || [[2020 AMC 12B | AMC 12B]]<br /> |-<br /> | 2019 || [[2019 AMC 12A | AMC 12A]] || [[2019 AMC 12B | AMC 12B]]<br /> |-<br /> | 2018 || [[2018 AMC 12A | AMC 12A]] || [[2018 AMC 12B | AMC 12B]]<br /> |-<br /> | 2017 || [[2017 AMC 12A | AMC 12A]] || [[2017 AMC 12B | AMC 12B]]<br /> |-<br /> | 2016 || [[2016 AMC 12A | AMC 12A]] || [[2016 AMC 12B | AMC 12B]]<br /> |-<br /> | 2015 || [[2015 AMC 12A | AMC 12A]] || [[2015 AMC 12B | AMC 12B]]<br /> |-<br /> | 2014 || [[2014 AMC 12A | AMC 12A]] || [[2014 AMC 12B | AMC 12B]]<br /> |-<br /> | 2013 || [[2013 AMC 12A | AMC 12A]] || [[2013 AMC 12B | AMC 12B]]<br /> |-<br /> | 2012 || [[2012 AMC 12A | AMC 12A]] || [[2012 AMC 12B | AMC 12B]]<br /> |-<br /> | 2011 || [[2011 AMC 12A | AMC 12A]] || [[2011 AMC 12B | AMC 12B]]<br /> |-<br /> | 2010 || [[2010 AMC 12A | AMC 12A]] || [[2010 AMC 12B | AMC 12B]]<br /> |-<br /> | 2009 || [[2009 AMC 12A | AMC 12A]] || [[2009 AMC 12B | AMC 12B]]<br /> |-<br /> | 2008 || [[2008 AMC 12A | AMC 12A]] || [[2008 AMC 12B | AMC 12B]]<br /> |-<br /> | 2007 || [[2007 AMC 12A | AMC 12A]] || [[2007 AMC 12B | AMC 12B]]<br /> |-<br /> | 2006 || [[2006 AMC 12A | AMC 12A]] || [[2006 AMC 12B | AMC 12B]]<br /> |-<br /> | 2005 || [[2005 AMC 12A | AMC 12A]] || [[2005 AMC 12B | AMC 12B]]<br /> |-<br /> | 2004 || [[2004 AMC 12A | AMC 12A]] || [[2004 AMC 12B | AMC 12B]]<br /> |-<br /> | 2003 || [[2003 AMC 12A | AMC 12A]] || [[2003 AMC 12B | AMC 12B]]<br /> |-<br /> | 2002 || [[2002 AMC 12A | AMC 12A]] || [[2002 AMC 12B | AMC 12B]]<br /> |-<br /> | 2001<br /> | colspan=&quot;2&quot;| [[2001 AMC 12 | AMC 12]]<br /> |-<br /> | 2000<br /> | colspan=&quot;2&quot;| [[2000 AMC 12 | AMC 12]]<br /> |}<br /> <br /> == Resources ==<br /> * [[American Mathematics Competitions]]<br /> * [[AMC Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Math Contest Problems]]</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=AMC_12_Problems_and_Solutions&diff=142223 AMC 12 Problems and Solutions 2021-01-16T17:58:57Z <p>Blehlivesonearth: </p> <hr /> <div>[[AMC 12]] problems and solutions. <br /> <br /> {| class=&quot;wikitable&quot; style=&quot;text-align:center&quot;<br /> |-<br /> ! Year || Test A || Test B<br /> |-<br /> | 2021 || [[2022 AMC 12A | AMC 12A]]<br /> |-<br /> | 2021 || [[2021 AMC 12A | AMC 12A]] || [[2021_AMC_12B | AMC 12B]]<br /> |-<br /> | 2020 || [[2020_AMC_12A | AMC 12A]] || [[2020 AMC 12B | AMC 12B]]<br /> |-<br /> | 2019 || [[2019 AMC 12A | AMC 12A]] || [[2019 AMC 12B | AMC 12B]]<br /> |-<br /> | 2018 || [[2018 AMC 12A | AMC 12A]] || [[2018 AMC 12B | AMC 12B]]<br /> |-<br /> | 2017 || [[2017 AMC 12A | AMC 12A]] || [[2017 AMC 12B | AMC 12B]]<br /> |-<br /> | 2016 || [[2016 AMC 12A | AMC 12A]] || [[2016 AMC 12B | AMC 12B]]<br /> |-<br /> | 2015 || [[2015 AMC 12A | AMC 12A]] || [[2015 AMC 12B | AMC 12B]]<br /> |-<br /> | 2014 || [[2014 AMC 12A | AMC 12A]] || [[2014 AMC 12B | AMC 12B]]<br /> |-<br /> | 2013 || [[2013 AMC 12A | AMC 12A]] || [[2013 AMC 12B | AMC 12B]]<br /> |-<br /> | 2012 || [[2012 AMC 12A | AMC 12A]] || [[2012 AMC 12B | AMC 12B]]<br /> |-<br /> | 2011 || [[2011 AMC 12A | AMC 12A]] || [[2011 AMC 12B | AMC 12B]]<br /> |-<br /> | 2010 || [[2010 AMC 12A | AMC 12A]] || [[2010 AMC 12B | AMC 12B]]<br /> |-<br /> | 2009 || [[2009 AMC 12A | AMC 12A]] || [[2009 AMC 12B | AMC 12B]]<br /> |-<br /> | 2008 || [[2008 AMC 12A | AMC 12A]] || [[2008 AMC 12B | AMC 12B]]<br /> |-<br /> | 2007 || [[2007 AMC 12A | AMC 12A]] || [[2007 AMC 12B | AMC 12B]]<br /> |-<br /> | 2006 || [[2006 AMC 12A | AMC 12A]] || [[2006 AMC 12B | AMC 12B]]<br /> |-<br /> | 2005 || [[2005 AMC 12A | AMC 12A]] || [[2005 AMC 12B | AMC 12B]]<br /> |-<br /> | 2004 || [[2004 AMC 12A | AMC 12A]] || [[2004 AMC 12B | AMC 12B]]<br /> |-<br /> | 2003 || [[2003 AMC 12A | AMC 12A]] || [[2003 AMC 12B | AMC 12B]]<br /> |-<br /> | 2002 || [[2002 AMC 12A | AMC 12A]] || [[2002 AMC 12B | AMC 12B]]<br /> |-<br /> | 2001<br /> | colspan=&quot;2&quot;| [[2001 AMC 12 | AMC 12]]<br /> |-<br /> | 2000<br /> | colspan=&quot;2&quot;| [[2000 AMC 12 | AMC 12]]<br /> |}<br /> <br /> == Resources ==<br /> * [[American Mathematics Competitions]]<br /> * [[AMC Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Math Contest Problems]]</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems&diff=142220 2022 AMC 12A Problems 2021-01-16T16:56:30Z <p>Blehlivesonearth: Created page with &quot;{{AMC12 Problems|year=2022|ab=B}} ==Problem 1== These problems will not be posted until the 2022 AMC 12A is released. Solution ==Problem...&quot;</p> <hr /> <div>{{AMC12 Problems|year=2022|ab=B}}<br /> ==Problem 1==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> These problems will not be posted until the 2022 AMC 12A is released.<br /> <br /> [[2022 AMC 12A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=B|before=[[2021 AMC 12B Problems]]|after=[[2022 AMC 12B Problems]]}}<br /> <br /> [[Category:AMC 12 Problems]]<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=AIME_Problems_and_Solutions&diff=141514 AIME Problems and Solutions 2021-01-04T23:42:41Z <p>Blehlivesonearth: </p> <hr /> <div>This is a list of all [[AIME]] exams in the AoPSWiki. Some of them contain complete questions and solutions, others complete questions, and others are lacking both questions and solutions. Many of these problems and solutions are also available in the [http://www.artofproblemsolving.com/Forum/resources.php?c=182 AoPS Resources] section. If you find problems that are in the Resources section which are not in the AoPSWiki, please consider adding them. Also, if you notice that a problem in the Wiki differs from the original wording, feel free to correct it. Finally, additions to and improvements on the solutions in the AoPSWiki are always welcome.<br /> <br /> {| class=&quot;wikitable&quot; style=&quot;text-align:center&quot;<br /> |-<br /> ! Year || Test I || Test II<br /> |-<br /> | 2021 || [[2021 AIME I | AIME I]] || [[2021 AIME II | AIME II]]<br /> |-<br /> | 2020 || [[2020 AIME I | AIME I]] || [[2020 AIME II | AOIME]]<br /> |-<br /> | 2019 || [[2019 AIME I | AIME I]] || [[2019 AIME II | AIME II]]<br /> |-<br /> | 2018 || [[2018 AIME I | AIME I]] || [[2018 AIME II | AIME II]]<br /> |-<br /> | 2017 || [[2017 AIME I | AIME I]] || [[2017 AIME II | AIME II]]<br /> |-<br /> | 2016 || [[2016 AIME I | AIME I]] || [[2016 AIME II | AIME II]]<br /> |-<br /> | 2015 || [[2015 AIME I | AIME I]] || [[2015 AIME II | AIME II]]<br /> |-<br /> | 2014 || [[2014 AIME I | AIME I]] || [[2014 AIME II | AIME II]]<br /> |-<br /> | 2013 || [[2013 AIME I | AIME I]] || [[2013 AIME II | AIME II]]<br /> |-<br /> | 2012 || [[2012 AIME I | AIME I]] || [[2012 AIME II | AIME II]]<br /> |-<br /> | 2011 || [[2011 AIME I | AIME I]] || [[2011 AIME II | AIME II]]<br /> |-<br /> | 2010 || [[2010 AIME I | AIME I]] || [[2010 AIME II | AIME II]]<br /> |-<br /> | 2009 || [[2009 AIME I | AIME I]] || [[2009 AIME II | AIME II]]<br /> |-<br /> | 2008 || [[2008 AIME I | AIME I]] || [[2008 AIME II | AIME II]]<br /> |-<br /> | 2007 || [[2007 AIME I | AIME I]] || [[2007 AIME II | AIME II]]<br /> |-<br /> | 2006 || [[2006 AIME I | AIME I]] || [[2006 AIME II | AIME II]]<br /> |-<br /> | 2005 || [[2005 AIME I | AIME I]] || [[2005 AIME II | AIME II]]<br /> |-<br /> | 2004 || [[2004 AIME I | AIME I]] || [[2004 AIME II | AIME II]]<br /> |-<br /> | 2003 || [[2003 AIME I | AIME I]] || [[2003 AIME II | AIME II]]<br /> |-<br /> | 2002 || [[2002 AIME I | AIME I]] || [[2002 AIME II | AIME II]]<br /> |-<br /> | 2001 || [[2001 AIME I | AIME I]] || [[2001 AIME II | AIME II]]<br /> |-<br /> | 2000 || [[2000 AIME I | AIME I]] || [[2000 AIME II | AIME II]]<br /> |-<br /> | 1999<br /> | colspan=&quot;2&quot;| [[1999 AIME | AIME]]<br /> |-<br /> | 1998<br /> | colspan=&quot;2&quot;| [[1998 AIME | AIME]]<br /> |-<br /> | 1997<br /> | colspan=&quot;2&quot;| [[1997 AIME | AIME]] <br /> |-<br /> | 1996<br /> | colspan=&quot;2&quot;| [[1996 AIME | AIME]]<br /> |-<br /> | 1995<br /> | colspan=&quot;2&quot;| [[1995 AIME | AIME]]<br /> |-<br /> | 1994<br /> | colspan=&quot;2&quot;| [[1994 AIME | AIME]]<br /> |-<br /> | 1993<br /> | colspan=&quot;2&quot;| [[1993 AIME | AIME]]<br /> |-<br /> | 1992<br /> | colspan=&quot;2&quot;| [[1992 AIME | AIME]]<br /> |-<br /> | 1991<br /> | colspan=&quot;2&quot;| [[1991 AIME | AIME]]<br /> |-<br /> | 1990<br /> | colspan=&quot;2&quot;| [[1990 AIME | AIME]]<br /> |-<br /> | 1989<br /> | colspan=&quot;2&quot;| [[1989 AIME | AIME]]<br /> |-<br /> | 1988 <br /> | colspan=&quot;2&quot;| [[1988 AIME | AIME]]<br /> |-<br /> | 1987<br /> | colspan=&quot;2&quot;| [[1987 AIME | AIME]]<br /> |-<br /> | 1986<br /> | colspan=&quot;2&quot;| [[1986 AIME | AIME]]<br /> |-<br /> | 1985<br /> | colspan=&quot;2&quot;| [[1985 AIME | AIME]] <br /> |-<br /> | 1984 <br /> | colspan=&quot;2&quot;| [[1984 AIME | AIME]] <br /> |-<br /> | 1983<br /> | colspan=&quot;2&quot;| [[1983 AIME | AIME]]<br /> |}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_8&diff=141286 2017 AIME I Problems/Problem 8 2021-01-02T02:26:58Z <p>Blehlivesonearth: /* Solution 2 (Trig Bash) */</p> <hr /> <div>==Problem 8==<br /> Two real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are chosen independently and uniformly at random from the interval &lt;math&gt;(0, 75)&lt;/math&gt;. Let &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; be two points on the plane with &lt;math&gt;OP = 200&lt;/math&gt;. Let &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; be on the same side of line &lt;math&gt;OP&lt;/math&gt; such that the degree measures of &lt;math&gt;\angle POQ&lt;/math&gt; and &lt;math&gt;\angle POR&lt;/math&gt; are &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; respectively, and &lt;math&gt;\angle OQP&lt;/math&gt; and &lt;math&gt;\angle ORP&lt;/math&gt; are both right angles. The probability that &lt;math&gt;QR \leq 100&lt;/math&gt; is equal to &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Noting that &lt;math&gt;\angle OQP&lt;/math&gt; and &lt;math&gt;\angle ORP&lt;/math&gt; are right angles, we realize that we can draw a semicircle with diameter &lt;math&gt;\overline{OP}&lt;/math&gt; and points &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; on the semicircle. Since the radius of the semicircle is &lt;math&gt;100&lt;/math&gt;, if &lt;math&gt;\overline{QR} \leq 100&lt;/math&gt;, then &lt;math&gt;\overarc{QR}&lt;/math&gt; must be less than or equal to &lt;math&gt;60^{\circ}&lt;/math&gt;.<br /> <br /> This simplifies the problem greatly. Since the degree measure of an angle on a circle is simply half the degree measure of its subtended arc, the problem is simply asking:<br /> <br /> Given &lt;math&gt;a, b&lt;/math&gt; such that &lt;math&gt;0&lt;a, b&lt;75&lt;/math&gt;, what is the probability that &lt;math&gt;|a-b| \leq 30&lt;/math&gt;? <br /> Through simple geometric probability, we get that &lt;math&gt;P = \frac{16}{25}&lt;/math&gt;.<br /> <br /> The answer is &lt;math&gt;16+25=\boxed{041}&lt;/math&gt;<br /> <br /> ~IYN~<br /> <br /> ==Solution 2 (Trig Bash)==<br /> Put &lt;math&gt;\triangle POQ&lt;/math&gt; and &lt;math&gt;\triangle POR&lt;/math&gt; with &lt;math&gt;O&lt;/math&gt; on the origin and the triangles on the &lt;math&gt;1^{st}&lt;/math&gt; quadrant.<br /> The coordinates of &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; is &lt;math&gt;(200 \cos^{2}a,200 \cos a\sin a )&lt;/math&gt;, &lt;math&gt;(200\cos^{2}b,200\cos(b)\sin b)&lt;/math&gt;. So &lt;math&gt;PQ^{2}&lt;/math&gt; = &lt;math&gt;(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2}&lt;/math&gt;, which we want to be less then &lt;math&gt;100^{2}&lt;/math&gt;.<br /> So &lt;math&gt;(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2} &lt;= 100^{2} &lt;/math&gt;<br /> &lt;cmath&gt;(\cos^{2} a - \cos^{2} b)^{2} +(\cos a \sin a - \cos b \sin b)^{2} \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;\cos^{4} a + \cos^{4} b - 2\cos^{2} a \cos^{2} b +\cos^{2}a \sin^{2} a + \cos^{2} b \sin^{2} b - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;\cos^{2} a(\cos^{2} a + \sin^{2} a)+\cos^{2} b(\cos^{2} b+\sin^{2} b) - 2\cos^{2} a \cos^{2} b- 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;\cos^{2} a(1-\cos^{2} b)+\cos^{2} b(1-\cos^{2} a) - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;(\cos a\sin b)^{2} +(\cos b\sin a)^{2} - 2 (\cos a \sin b)(\cos b \sin a)\le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;(\cos a\sin b-\cos b\sin a)^{2}\le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt; \sin^{2} (b-a) \le \frac{1}{4} &lt;/cmath&gt;<br /> So we want &lt;math&gt; -\frac{1}{2} \le \sin (b-a) \le \frac{1}{2} &lt;/math&gt;, which is equivalent to &lt;math&gt; -30 \le b-a \le 30&lt;/math&gt; or &lt;math&gt; 150 \le b-a \le 210&lt;/math&gt;. The second inequality is impossible so we only consider what the first inequality does to our &lt;math&gt;75&lt;/math&gt; by &lt;math&gt;75&lt;/math&gt; box in the &lt;math&gt;ab&lt;/math&gt; plane. This cuts off two isosceles right triangles from opposite corners with side lengths &lt;math&gt;45&lt;/math&gt; from the &lt;math&gt;75&lt;/math&gt; by &lt;math&gt;75&lt;/math&gt; box. Hence the probability is &lt;math&gt;1-\frac{45^2}{75^2} = 1- \frac{9}{25}=\frac{16}{25}&lt;/math&gt; and the answer is &lt;math&gt;16+25 = \boxed{41}&lt;/math&gt;<br /> <br /> Solution by Leesisi<br /> <br /> ==Solution 3 (Quicker Trig)==<br /> &lt;asy&gt;<br /> pair O, P, Q, R;<br /> draw(circle(O, 10));<br /> O = (10, 0);<br /> P = (-10, 0);<br /> Q = (10*cos(pi/3), 10*sin(pi/3));<br /> R = (10*cos(5*pi/6), 10*sin(5*pi/6));<br /> dot(Q);<br /> dot(O);<br /> dot(P);<br /> dot(R);<br /> draw(P--O--Q--P--R--O);<br /> draw(Q--R, red);<br /> label(&quot;$O$&quot;, O, 2*E);<br /> label(&quot;$P$&quot;, P, 2*W);<br /> label(&quot;$Q$&quot;, Q, NE);<br /> label(&quot;$R$&quot;, R, NW);<br /> label(&quot;$200$&quot;, (0,0), 2*S);<br /> label(&quot;$x$&quot;, (Q+R)/2, N);<br /> draw(rightanglemark(O, Q, P, 38));<br /> draw(rightanglemark(O, R, P, 38));<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;QR=x.&lt;/math&gt; Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: &lt;math&gt;OQ = 200 \cos a, PQ = 200 \sin a, PR = 200 \sin b, OR = 200 \cos b.&lt;/math&gt; Now observe that quadrilateral &lt;math&gt;OQRP&lt;/math&gt; is a [[cyclic quadrilateral]]. Thus, we are able to apply [[Ptolemy's Theorem]] to it:<br /> &lt;cmath&gt;200 x + (200 \cos a) (200 \sin b) = (200 \sin a) (200 \cos b),&lt;/cmath&gt;<br /> &lt;cmath&gt;x + 200 (\cos a \sin b) = 200 (\sin a \cos b),&lt;/cmath&gt;<br /> &lt;cmath&gt;x = 200(\sin a \cos b - \sin b \cos a),&lt;/cmath&gt;<br /> &lt;cmath&gt;x = 200 \sin(a-b).&lt;/cmath&gt;<br /> We want &lt;math&gt;|x| \le 100&lt;/math&gt; (the absolute value comes from the fact that &lt;math&gt;a&lt;/math&gt; is not necessarily greater than &lt;math&gt;b,&lt;/math&gt; so we cannot assume that &lt;math&gt;Q&lt;/math&gt; is to the right of &lt;math&gt;R&lt;/math&gt; as in the diagram), so we substitute:<br /> &lt;cmath&gt;|200 \sin(a-b)| \le 100,&lt;/cmath&gt;<br /> &lt;cmath&gt;|\sin(a-b)| \le \frac{1}{2},&lt;/cmath&gt;<br /> &lt;cmath&gt;|a-b| \le 30 ^\circ,&lt;/cmath&gt;<br /> &lt;cmath&gt;-30 \le a-b \le 30.&lt;/cmath&gt;<br /> By simple geometric probability (see Solution 2 for complete explanation), &lt;math&gt;\frac{m}{n} = 1 - \frac{2025}{5625} = 1 - \frac{9}{25} = \frac{16}{25},&lt;/math&gt; so &lt;math&gt;m+n = \boxed{041}.&lt;/math&gt;<br /> <br /> ~burunduchok<br /> <br /> ==Solution 4==<br /> <br /> Scale the circle down from radius 100 (diameter 200) to radius 6 (diameter 12). Then we want the probability that &lt;math&gt;PQ \le 6&lt;/math&gt;. Now note that all possible &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; lie on a &lt;math&gt;5\pi&lt;/math&gt; interval on the circumference of the circle. But for &lt;math&gt;PQ&lt;6&lt;/math&gt;, &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; must be less than &lt;math&gt;2\pi&lt;/math&gt; apart on the circumference of the circle. Simple geometric probability gives us &lt;math&gt;\frac{16}{25}&lt;/math&gt;, so the answer is &lt;math&gt;41&lt;/math&gt;. (Professor-Mom)<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_6&diff=141146 2012 AIME II Problems/Problem 6 2020-12-31T06:34:40Z <p>Blehlivesonearth: /* Solution 2 */</p> <hr /> <div>== Problem 6 ==<br /> Let &lt;math&gt;z=a+bi&lt;/math&gt; be the complex number with &lt;math&gt;\vert z \vert = 5&lt;/math&gt; and &lt;math&gt;b &gt; 0&lt;/math&gt; such that the distance between &lt;math&gt;(1+2i)z^3&lt;/math&gt; and &lt;math&gt;z^5&lt;/math&gt; is maximized, and let &lt;math&gt;z^4 = c+di&lt;/math&gt;. Find &lt;math&gt;c+d&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Let's consider the maximization constraint first: we want to maximize the value of &lt;math&gt;|z^5 - (1+2i)z^3|&lt;/math&gt;<br /> Simplifying, we have<br /> <br /> &lt;math&gt;|z^3| * |z^2 - (1+2i)|&lt;/math&gt;<br /> <br /> &lt;math&gt;=|z|^3 * |z^2 - (1+2i)|&lt;/math&gt;<br /> <br /> &lt;math&gt;=125|z^2 - (1+2i)|&lt;/math&gt;<br /> <br /> Thus we only need to maximize the value of &lt;math&gt;|z^2 - (1+2i)|&lt;/math&gt;.<br /> <br /> To maximize this value, we must have that &lt;math&gt;z^2&lt;/math&gt; is in the opposite direction of &lt;math&gt;1+2i&lt;/math&gt;. The unit vector in the complex plane in the desired direction is &lt;math&gt;\frac{-1}{\sqrt{5}} + \frac{-2}{\sqrt{5}} i&lt;/math&gt;. Furthermore, we know that the magnitude of &lt;math&gt;z^2&lt;/math&gt; is &lt;math&gt;25&lt;/math&gt;, because the magnitude of &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;5&lt;/math&gt;. From this information, we can find that &lt;math&gt;z^2 = \sqrt{5} (-5 - 10i)&lt;/math&gt;<br /> <br /> Squaring, we get &lt;math&gt;z^4 = 5 (25 - 100 + 100i) = -375 + 500i&lt;/math&gt;. Finally, &lt;math&gt;c+d = -375 + 500 = \boxed{125}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> WLOG, let &lt;math&gt;z_{1}=5(\cos{\theta_{1}}+i\sin{\theta_{1}})&lt;/math&gt; and<br /> <br /> &lt;math&gt;z_{2}=1+2i=\sqrt{5}(\cos{\theta_{2}+i\sin{\theta_{2}}})&lt;/math&gt;<br /> <br /> This means that<br /> <br /> &lt;math&gt;z_{1}^3=125(\cos{3\theta_{1}}+i\sin{3\theta_{1}})&lt;/math&gt;<br /> <br /> &lt;math&gt;z_{1}^4=625(\cos{4\theta_{1}}+i\sin{4\theta_{1}})&lt;/math&gt;<br /> <br /> Hence, this means that<br /> <br /> &lt;math&gt;z_{2}z_{1}^3=125\sqrt{5}(\cos({\theta_{2}+3\theta_{1}})+i\sin({\theta_{2}+3\theta_{1}}))&lt;/math&gt;<br /> <br /> And<br /> <br /> &lt;math&gt;z_{1}^5=3125(\cos{5\theta_{1}}+i\sin{5\theta_{1}})&lt;/math&gt;<br /> <br /> Now, common sense tells us that the distance between these two complex numbers is maxed when they both are points satisfying the equation of the line &lt;math&gt;yi=mx&lt;/math&gt;, or when they are each a &lt;math&gt;180^{\circ}&lt;/math&gt; rotation away from each other.<br /> <br /> Hence, we must have that &lt;math&gt;5\theta_{1}=3\theta_{1}+\theta_{2}+180^{\circ}\implies\theta_{1}=\frac{\theta_{2}+180^{\circ}}{2}&lt;/math&gt;<br /> <br /> Now, plug this back into &lt;math&gt;z_{1}^4&lt;/math&gt;(if you want to know why, reread what we want in the problem!)<br /> <br /> So now, we have that<br /> &lt;math&gt;z_{1}^4=625(\cos{2\theta_{2}}+i\sin{2\theta_{2}})&lt;/math&gt;<br /> <br /> Notice that &lt;math&gt;\cos\theta_{2}=\frac{1}{\sqrt{5}}&lt;/math&gt; and &lt;math&gt;\sin\theta_{2}=\frac{2}{\sqrt{5}}&lt;/math&gt;<br /> <br /> Then, we have that &lt;math&gt;\cos{2\theta_{2}}=\cos^2{\theta_{2}}-\sin^2{\theta_{2}}=-\frac{3}{5}&lt;/math&gt; and &lt;math&gt;\sin{2\theta_{2}}=2\sin{\theta_{2}}\cos{\theta_{2}}=\frac{4}{5}&lt;/math&gt;<br /> <br /> Finally, plugging back in, we find that &lt;math&gt;z_{1}^4=625(-\frac{3}{5}+\frac{4i}{5})=-375+500i&lt;/math&gt;<br /> <br /> &lt;math&gt;-375+500=\boxed{125}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_11&diff=141110 2015 AIME II Problems/Problem 11 2020-12-31T02:31:50Z <p>Blehlivesonearth: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> The circumcircle of acute &lt;math&gt;\triangle ABC&lt;/math&gt; has center &lt;math&gt;O&lt;/math&gt;. The line passing through point &lt;math&gt;O&lt;/math&gt; perpendicular to &lt;math&gt;\overline{OB}&lt;/math&gt; intersects lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, respectively. Also &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=4&lt;/math&gt;, &lt;math&gt;BQ=4.5&lt;/math&gt;, and &lt;math&gt;BP=\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Diagram==<br /> &lt;asy&gt;<br /> unitsize(30);<br /> draw(Circle((0,0),3));<br /> pair A,B,C,O, Q, P, M, N;<br /> A=(2.5, -sqrt(11/4));<br /> B=(-2.5, -sqrt(11/4));<br /> C=(-1.96, 2.28);<br /> Q=(-1.89, 2.81);<br /> P=(1.13, -1.68);<br /> O=origin;<br /> M=foot(O,C,B);<br /> N=foot(O,A,B);<br /> draw(A--B--C--cycle);<br /> label(&quot;$A$&quot;,A,SE);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,NW);<br /> label(&quot;$Q$&quot;,Q,NW);<br /> dot(O);<br /> label(&quot;$O$&quot;,O,NE);<br /> label(&quot;$M$&quot;,M,W);<br /> label(&quot;$N$&quot;,N,S);<br /> label(&quot;$P$&quot;,P,S);<br /> draw(B--O);<br /> draw(C--Q);<br /> draw(Q--O);<br /> draw(O--C);<br /> draw(O--A);<br /> draw(O--P);<br /> draw(O--M, dashed);<br /> draw(O--N, dashed);<br /> draw(rightanglemark((-2.5, -sqrt(11/4)),(0,0),(-1.89, 2.81),5));<br /> draw(rightanglemark(O,N,B,5));<br /> draw(rightanglemark(B,O,P,5));<br /> draw(rightanglemark(O,M,C,5));<br /> &lt;/asy&gt;<br /> ===Solution 1===<br /> Call &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; the feet of the altitudes from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt;, respectively. Let &lt;math&gt;OB = r&lt;/math&gt; . Notice that &lt;math&gt;\triangle{OMB} \sim \triangle{QOB}&lt;/math&gt; because both are right triangles, and &lt;math&gt;\angle{OBQ} \cong \angle{OBM}&lt;/math&gt;. By &lt;math&gt;\frac{MB}{BO}=\frac{BO}{BQ}&lt;/math&gt;, &lt;math&gt;MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}&lt;/math&gt;. However, since &lt;math&gt;O&lt;/math&gt; is the circumcenter of triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;OM&lt;/math&gt; is a perpendicular bisector by the definition of a circumcenter. Hence, &lt;math&gt;\frac{r^2}{4.5} = 2 \implies r = 3&lt;/math&gt;. Since we know &lt;math&gt;BN=\frac{5}{2}&lt;/math&gt; and &lt;math&gt;\triangle BOP \sim \triangle BNO&lt;/math&gt;, we have &lt;math&gt;\frac{BP}{3} = \frac{3}{\frac{5}{2}}&lt;/math&gt;. Thus, &lt;math&gt;BP = \frac{18}{5}&lt;/math&gt;. &lt;math&gt;m + n=\boxed{023}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Notice that &lt;math&gt;\angle{CBO}=90-A&lt;/math&gt;, so &lt;math&gt;\angle{BQO}=A&lt;/math&gt;. From this we get that &lt;math&gt;\triangle{BPQ}\sim \triangle{BCA}&lt;/math&gt;. So &lt;math&gt;\dfrac{BP}{BC}=\dfrac{BQ}{BA}&lt;/math&gt;, plugging in the given values we get &lt;math&gt;\dfrac{BP}{4}=\dfrac{4.5}{5}&lt;/math&gt;, so &lt;math&gt;BP=\dfrac{18}{5}&lt;/math&gt;, and &lt;math&gt;m+n=\boxed{023}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> Let &lt;math&gt;r=BO&lt;/math&gt;. Drawing perpendiculars, &lt;math&gt;BM=MC=2&lt;/math&gt; and &lt;math&gt;BN=NA=2.5&lt;/math&gt;. From there, &lt;math&gt;OM=\sqrt{r^2-4}&lt;/math&gt;. Thus, &lt;math&gt;OQ=\frac{\sqrt{4r^2+9}}{2}&lt;/math&gt;. Using &lt;math&gt;\triangle{BOQ}&lt;/math&gt;, we get &lt;math&gt;r=3&lt;/math&gt;. Now let's find &lt;math&gt;NP&lt;/math&gt;. After some calculations with &lt;math&gt;\triangle{BON}&lt;/math&gt; ~ &lt;math&gt;\triangle{OPN}&lt;/math&gt;, &lt;math&gt;{NP=11/10}&lt;/math&gt;. Therefore, &lt;math&gt;BP=\frac{5}{2}+\frac{11}{10}=18/5&lt;/math&gt;. &lt;math&gt;18+5=\boxed{023}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> Let &lt;math&gt;\angle{BQO}=\alpha&lt;/math&gt;. Extend &lt;math&gt;OB&lt;/math&gt; to touch the circumcircle at a point &lt;math&gt;K&lt;/math&gt;. Then, note that &lt;math&gt;\angle{KAC}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha&lt;/math&gt;. But since &lt;math&gt;BK&lt;/math&gt; is a diameter, &lt;math&gt;\angle{KAB}=90^\circ&lt;/math&gt;, implying &lt;math&gt;\angle{CAB}=\alpha&lt;/math&gt;. It follows that &lt;math&gt;APCQ&lt;/math&gt; is a cyclic quadrilateral.<br /> <br /> Let &lt;math&gt;BP=x&lt;/math&gt;. By Power of a Point, &lt;cmath&gt;5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.&lt;/cmath&gt;The answer is &lt;math&gt;18+5=\boxed{023}&lt;/math&gt;.<br /> <br /> ===Solution 5===<br /> Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.<br /> <br /> Denote the circumradius of &lt;math&gt;ABC&lt;/math&gt; to be &lt;math&gt;R&lt;/math&gt;, the circumcircle of &lt;math&gt;ABC&lt;/math&gt; to be &lt;math&gt;O&lt;/math&gt;, and the shortest distance from &lt;math&gt;Q&lt;/math&gt; to circle &lt;math&gt;O&lt;/math&gt; to be &lt;math&gt;x&lt;/math&gt;. <br /> <br /> Using Power of a Point on &lt;math&gt;Q&lt;/math&gt; relative to circle &lt;math&gt;O&lt;/math&gt;, we get that &lt;math&gt;x(x+2r) = 0.5 \cdot 4.5 = \frac{9}{4}&lt;/math&gt;. Using Pythagorean Theorem on triangle &lt;math&gt;QOB&lt;/math&gt; to get &lt;math&gt;(x + r)^2 + r^2 = \frac{81}{4}&lt;/math&gt;. Subtracting the first equation from the second, we get that &lt;math&gt;2r^2 = 18&lt;/math&gt; and therefore &lt;math&gt;r = 3&lt;/math&gt;. Now, set &lt;math&gt;\cos{ABC} = y&lt;/math&gt;. Using law of cosines on &lt;math&gt;ABC&lt;/math&gt; to find &lt;math&gt;AC&lt;/math&gt; in terms of &lt;math&gt;y&lt;/math&gt; and plugging that into the extended law of sines, we get &lt;math&gt;\frac{\sqrt{4^2 + 5^2 - 2 \cdot 4 \cdot 5 x}}{\sqrt{1 - x^2}} = 2R = 6&lt;/math&gt;. Squaring both sides and cross multiplying, we get &lt;math&gt;36x^2 - 40x + 5 = 0&lt;/math&gt;. Now, we get &lt;math&gt;x = \frac{10 \pm \sqrt{55}}{18}&lt;/math&gt; using quadratic formula. If you drew a decent diagram, &lt;math&gt;B&lt;/math&gt; is acute and therefore &lt;math&gt;x = \frac{10 + \sqrt{55}}{18}&lt;/math&gt;(You can also try plugging in both in the end and seeing which gives a rational solution). Note that &lt;math&gt;BP = 3\frac{1}{\sin{OPB}} = \frac{3}{\cos{\angle ABC - \angle QBO}}.&lt;/math&gt; Using the cosine addition formula and then plugging in what we know about &lt;math&gt;QBO&lt;/math&gt;, we get that &lt;math&gt;BP = \frac{162}{2\cos{B} + \sqrt{5}\sin{B}}&lt;/math&gt;. Now, the hard part is to find what &lt;math&gt;\sin{B}&lt;/math&gt; is. We therefore want &lt;math&gt;\frac{\sqrt{324 - (10 + \sqrt{55})^2}}{18} = \frac{\sqrt{169 - 20\sqrt{55}}}{18}&lt;/math&gt;. For the numerator, by inspection &lt;math&gt;(a + b\sqrt{55})^2&lt;/math&gt; will not work for integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. The other case is if there is &lt;math&gt;(a\sqrt{5} + b\sqrt{11})^2&lt;/math&gt;. By inspection, &lt;math&gt;5\sqrt{5} - 2\sqrt{11}&lt;/math&gt; works. Therefore, plugging all this in yields the answer, &lt;math&gt;\frac{18}{5} \rightarrow \boxed{23}&lt;/math&gt;. Solution by hyxue<br /> <br /> ===Solution 6===<br /> &lt;asy&gt; <br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(15cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -4.7673964645097335, xmax = 9.475267639476614, ymin = -1.6884766592324019, ymax = 6.385449160754665; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw(circle((0.7129306199257198,2.4781596958650733), 3.000319171815248), linewidth(2) + wrwrwr); <br /> draw((0.7129306199257198,2.4781596958650733)--(3.178984115621537,0.7692140299269852), linewidth(2) + wrwrwr); <br /> draw((xmin, 1.4430262733614363*xmin + 1.4493820802284032)--(xmax, 1.4430262733614363*xmax + 1.4493820802284032), linewidth(2) + wrwrwr); /* line */<br /> draw((xmin, -0.020161290322580634*xmin + 0.8333064516129032)--(xmax, -0.020161290322580634*xmax + 0.8333064516129032), linewidth(2) + wrwrwr); /* line */<br /> draw((xmin, -8.047527437688247*xmin + 26.352175924366414)--(xmax, -8.047527437688247*xmax + 26.352175924366414), linewidth(2) + wrwrwr); /* line */<br /> draw((xmin, -2.5113572383524088*xmin + 8.752778799300463)--(xmax, -2.5113572383524088*xmax + 8.752778799300463), linewidth(2) + wrwrwr); /* line */<br /> draw((xmin, 0.12426176956126818*xmin + 2.389569675458691)--(xmax, 0.12426176956126818*xmax + 2.389569675458691), linewidth(2) + wrwrwr); /* line */<br /> draw(circle((1.9173376033752174,4.895608471162773), 0.7842529827808445), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((-1.82,0.87),dotstyle); <br /> label(&quot;$A$&quot;, (-1.7801363959463627,0.965838014692327), NE * labelscalefactor); <br /> dot((3.178984115621537,0.7692140299269852),dotstyle); <br /> label(&quot;$B$&quot;, (3.2140445236332655,0.8641046996638531), NE * labelscalefactor); <br /> dot((2.6857306099246263,4.738685150758791),dotstyle); <br /> label(&quot;$C$&quot;, (2.7238749148597092,4.831703985774336), NE * labelscalefactor); <br /> dot((0.7129306199257198,2.4781596958650733),linewidth(4pt) + dotstyle); <br /> label(&quot;$O$&quot;, (0.7539479965810783,2.556577122410283), NE * labelscalefactor); <br /> dot((-0.42105034508654754,0.8417953698606159),linewidth(4pt) + dotstyle); <br /> label(&quot;$P$&quot;, (-0.38361543510094825,0.9195955987702934), NE * labelscalefactor); <br /> dot((2.6239558409689123,5.235819298886746),linewidth(4pt) + dotstyle); <br /> label(&quot;$Q$&quot;, (2.6591355325688624,5.312625111363486), NE * labelscalefactor); <br /> dot((1.3292769824200672,5.414489427724579),linewidth(4pt) + dotstyle); <br /> label(&quot;$A'$&quot;, (1.3643478867519216,5.488346291867214), NE * labelscalefactor); <br /> dot((1.8469115849379867,4.11452402186953),linewidth(4pt) + dotstyle); <br /> label(&quot;$P'$&quot;, (1.8822629450786978,4.184310162865866), NE * labelscalefactor); <br /> dot((2.5624172335003985,5.731052930966743),linewidth(4pt) + dotstyle); <br /> label(&quot;$D$&quot;, (2.603644633462422,5.802794720137042), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> &lt;/asy&gt;<br /> Reflect &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;P&lt;/math&gt; across &lt;math&gt;OB&lt;/math&gt; to points &lt;math&gt;A'&lt;/math&gt; and &lt;math&gt;P'&lt;/math&gt;, respectively with &lt;math&gt;A'&lt;/math&gt; on the circle and &lt;math&gt;P, O, P'&lt;/math&gt; collinear. Now, &lt;math&gt;\angle A'CQ = 180^{\circ} - \angle A'CB = \angle A'AB = \angle P'PB&lt;/math&gt; by parallel lines. From here, &lt;math&gt;\angle P'PB = \angle PP'B = \angle A'P'Q&lt;/math&gt; as &lt;math&gt;P, P', Q&lt;/math&gt; collinear. From here, &lt;math&gt;A'P'QC&lt;/math&gt; is cyclic, and by power of a point we obtain &lt;math&gt;\frac{18}{5} \implies \boxed{023}&lt;/math&gt;.<br /> ~awang11's sol<br /> <br /> ==See also==<br /> {{AIME box|year=2015|n=II|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_15&diff=140882 2012 AIME II Problems/Problem 15 2020-12-29T03:07:50Z <p>Blehlivesonearth: /* Solution 4 */</p> <hr /> <div>== Problem 15 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=7&lt;/math&gt;, and &lt;math&gt;AC=3&lt;/math&gt;. The bisector of angle &lt;math&gt;A&lt;/math&gt; meets side &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and circle &lt;math&gt;\omega&lt;/math&gt; at a second point &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;\gamma&lt;/math&gt; be the circle with diameter &lt;math&gt;\overline{DE}&lt;/math&gt;. Circles &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; meet at &lt;math&gt;E&lt;/math&gt; and a second point &lt;math&gt;F&lt;/math&gt;. Then &lt;math&gt;AF^2 = \frac mn&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> <br /> ==Quick Solution using Olympiad Terms ==<br /> <br /> Take a force-overlaid inversion about &lt;math&gt;A&lt;/math&gt; and note &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; map to each other. As &lt;math&gt;DE&lt;/math&gt; was originally the diameter of &lt;math&gt;\gamma&lt;/math&gt;, &lt;math&gt;DE&lt;/math&gt; is still the diameter of &lt;math&gt;\gamma&lt;/math&gt;. Thus &lt;math&gt;\gamma&lt;/math&gt; is preserved. Note that the midpoint &lt;math&gt;M&lt;/math&gt; of &lt;math&gt;BC&lt;/math&gt; lies on &lt;math&gt;\gamma&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;\omega&lt;/math&gt; are swapped. Thus points &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt; map to each other, and are isogonal. It follows that &lt;math&gt;AF&lt;/math&gt; is a symmedian of &lt;math&gt;\triangle{ABC}&lt;/math&gt;, or that &lt;math&gt;ABFC&lt;/math&gt; is harmonic. Then &lt;math&gt;(AB)(FC)=(BF)(CA)&lt;/math&gt;, and thus we can let &lt;math&gt;BF=5x, CF=3x&lt;/math&gt; for some &lt;math&gt;x&lt;/math&gt;. By the LoC, it is easy to see &lt;math&gt;\angle{BAC}=120^\circ&lt;/math&gt; so &lt;math&gt;(5x)^2+(3x)^2-2\cos{60^\circ}(5x)(3x)=49&lt;/math&gt;. Solving gives &lt;math&gt;x^2=\frac{49}{19}&lt;/math&gt;, from which by Ptolemy's we see &lt;math&gt;AF=\frac{30}{\sqrt{19}}&lt;/math&gt;. We conclude the answer is &lt;math&gt;900+19=\boxed{919}&lt;/math&gt;, as desired.<br /> <br /> '''- Emathmaster'''<br /> <br /> Quick Side Note: You might be wondering what the motivation for this solution is. Most of the people who've done EGMO Chapter 8 should recognize this as problem 8.32 (2009 Russian Olympiad) with the computational finish afterwards.<br /> Now if you haven't done this, but still know what inversion is, here's the motivation. We'd see that it's kinda hard to angle chase, and if we could, it would still be a bit hard to apply (you could use trig, but it won't be so clean most likely). If you give up after realizing that angle chasing won't work, you'd likely go in a similar approach to Solution 1 (below) or maybe be a bit more insightful and go with the elementary solution above.<br /> <br /> Finally, we notice there's circles! Classic setup for inversion! Since we're involving an angle-bisector, the first thing that comes to mind is a force overlaid inversion described in Lemma 8.16 of EGMO (where we invert with radius &lt;math&gt;\sqrt{AB \cdot AC}&lt;/math&gt; and center &lt;math&gt;A&lt;/math&gt;, then reflect over the &lt;math&gt;A&lt;/math&gt;-angle bisector, which fixes &lt;math&gt;B, C&lt;/math&gt;). We try applying this to the problem, and it's fruitful - we end up with this solution.<br /> -MSC<br /> <br /> == Solution 1==<br /> Use the angle bisector theorem to find &lt;math&gt;CD=\frac{21}{8}&lt;/math&gt;, &lt;math&gt;BD=\frac{35}{8}&lt;/math&gt;, and use Stewart's Theorem to find &lt;math&gt;AD=\frac{15}{8}&lt;/math&gt;. Use Power of the Point to find &lt;math&gt;DE=\frac{49}{8}&lt;/math&gt;, and so &lt;math&gt;AE=8&lt;/math&gt;. Use law of cosines to find &lt;math&gt;\angle CAD = \frac{\pi} {3}&lt;/math&gt;, hence &lt;math&gt;\angle BAD = \frac{\pi}{3}&lt;/math&gt; as well, and &lt;math&gt;\triangle BCE&lt;/math&gt; is equilateral, so &lt;math&gt;BC=CE=BE=7&lt;/math&gt;.<br /> <br /> I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot \cos \angle AFE.&lt;/math&gt; (1)<br /> <br /> &lt;math&gt;AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot \cos \angle AEF.&lt;/math&gt; Adding these two and simplifying we get:<br /> <br /> &lt;math&gt;EF = AF \cdot \cos \angle AFE + AE \cdot \cos \angle AEF&lt;/math&gt; (2). Ah, but &lt;math&gt;\angle AFE = \angle ACE&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; lies on &lt;math&gt;\omega&lt;/math&gt;), and we can find &lt;math&gt;cos \angle ACE&lt;/math&gt; using the law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot \cos \angle ACE&lt;/math&gt;, and plugging in &lt;math&gt;AE = 8, AC = 3, BE = BC = 7,&lt;/math&gt; we get &lt;math&gt;\cos \angle ACE = -1/7 = \cos \angle AFE&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;\angle AEF = \angle DEF&lt;/math&gt;, and &lt;math&gt;\angle DFE = \pi/2&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; is on the circle &lt;math&gt;\gamma&lt;/math&gt; with diameter &lt;math&gt;DE&lt;/math&gt;), so &lt;math&gt;\cos \angle AEF = EF/DE = 8 \cdot EF/49&lt;/math&gt;. <br /> <br /> Plugging in all our values into equation (2), we get:<br /> <br /> &lt;math&gt;EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}&lt;/math&gt;, or &lt;math&gt;EF = \frac{7}{15} \cdot AF&lt;/math&gt;.<br /> <br /> Finally, we plug this into equation (1), yielding:<br /> <br /> &lt;math&gt;8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot \frac{-1}{7}&lt;/math&gt;. Thus,<br /> <br /> &lt;math&gt;64 = \frac{AF^2}{225} \cdot (225+49+30),&lt;/math&gt; or &lt;math&gt;AF^2 = \frac{900}{19}.&lt;/math&gt; The answer is &lt;math&gt;\boxed{919}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> <br /> Let &lt;math&gt;a = BC&lt;/math&gt;, &lt;math&gt;b = CA&lt;/math&gt;, &lt;math&gt;c = AB&lt;/math&gt; for convenience. We claim that &lt;math&gt;AF&lt;/math&gt; is a symmedian. Indeed, let &lt;math&gt;M&lt;/math&gt; be the midpoint of segment &lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;\angle EAB=\angle EAC&lt;/math&gt;, it follows that &lt;math&gt;EB = EC&lt;/math&gt; and consequently &lt;math&gt;EM\perp BC&lt;/math&gt;. Therefore, &lt;math&gt;M\in \gamma&lt;/math&gt;. Now let &lt;math&gt;G = FD\cap \omega&lt;/math&gt;. Since &lt;math&gt;EG&lt;/math&gt; is a diameter, &lt;math&gt;G&lt;/math&gt; lies on the perpendicular bisector of &lt;math&gt;BC&lt;/math&gt;; hence &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt; are collinear. From &lt;math&gt;\angle DAG = \angle DMG = 90&lt;/math&gt;, it immediately follows that quadrilateral &lt;math&gt;ADMG&lt;/math&gt; is cyclic. Therefore, &lt;math&gt;\angle MAD = \angle MGD=\angle EAF&lt;/math&gt;, implying that &lt;math&gt;AF&lt;/math&gt; is a symmedian, as claimed.<br /> <br /> The rest is standard; here's a quick way to finish. From above, quadrilateral &lt;math&gt;ABFC&lt;/math&gt; is harmonic, so &lt;math&gt;\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}&lt;/math&gt;. In conjunction with &lt;math&gt;\triangle ABF\sim\triangle AMC&lt;/math&gt;, it follows that &lt;math&gt;AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}&lt;/math&gt;. (Notice that this holds for all triangles &lt;math&gt;ABC&lt;/math&gt;.) To finish, substitute &lt;math&gt;a = 7&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=5&lt;/math&gt; to obtain &lt;math&gt;AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}&lt;/math&gt; as before.<br /> <br /> '''-Solution by thecmd999'''<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> size(6cm);<br /> pair E,X,B,C,A,D,M,F,R,I;<br /> real z=sqrt(3)*14/3;<br /> real y=2*sqrt(3)/21;<br /> real x=224*sqrt(3)/57;<br /> E=(z,0);<br /> X=(0,0);<br /> D=(sqrt(3)*7/6,-7/8);<br /> M=(sqrt(3)*7/6,0);<br /> B=z/2*dir(60);<br /> C=z/2*dir(300);<br /> A=(y,-8/7);<br /> F=(x,-sqrt(3)*x/4);<br /> R=circumcenter(A,B,C);<br /> I=circumcenter(M,E,F);<br /> draw(E--X);<br /> draw(A--E);<br /> draw(A--B);<br /> draw(A--C);<br /> draw(B--C);<br /> draw(A--F);<br /> draw(X--F);<br /> draw(E--F);<br /> draw(circumcircle(A,B,C));<br /> draw(circumcircle(M,F,E));<br /> dot(D);<br /> dot(F);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(E);<br /> dot(X);<br /> dot(R);<br /> dot(I);<br /> label(&quot;$A$&quot;,A,dir(220));<br /> label(&quot;$B$&quot;,B,dir(110));<br /> label(&quot;$C$&quot;,C,dir(250));<br /> label(&quot;$D$&quot;,D,dir(60));<br /> label(&quot;$E$&quot;,E,dir(0));<br /> label(&quot;$F$&quot;,F,dir(315));<br /> label(&quot;$X$&quot;,X,dir(180));<br /> &lt;/asy&gt;<br /> First of all, use the [[Angle Bisector Theorem]] to find that &lt;math&gt;BD=35/8&lt;/math&gt; and &lt;math&gt;CD=21/8&lt;/math&gt;, and use [[Stewart's Theorem]] to find that &lt;math&gt;AD=15/8&lt;/math&gt;. Then use [[Power of a Point Theorem|Power of a Point]] to find that &lt;math&gt;DE=49/8&lt;/math&gt;. Then use the [[Circumradius|circumradius of a triangle]] formula to find that the length of the circumradius of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{7\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;DE&lt;/math&gt; is the diameter of circle &lt;math&gt;\gamma&lt;/math&gt;, &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;. Extending &lt;math&gt;DF&lt;/math&gt; to intersect circle &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt;, we find that &lt;math&gt;XE&lt;/math&gt; is the diameter of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; (since &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;). Therefore, &lt;math&gt;XE=\frac{14\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;EF=x&lt;/math&gt;, &lt;math&gt;XD=a&lt;/math&gt;, and &lt;math&gt;DF=b&lt;/math&gt;. Then, by the [[Pythagorean Theorem]],<br /> <br /> &lt;cmath&gt;x^2+b^2=\left(\frac{49}{8}\right)^2=\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> and<br /> <br /> &lt;cmath&gt;x^2+(a+b)^2=\left(\frac{14\sqrt{3}}{3}\right)^2=\frac{196}{3}.&lt;/cmath&gt;<br /> <br /> Subtracting the first equation from the second, the &lt;math&gt;x^2&lt;/math&gt; term cancels out and we obtain:<br /> <br /> &lt;cmath&gt;(a+b)^2-b^2=\frac{196}{3}-\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2+2ab = \frac{5341}{192}.&lt;/cmath&gt;<br /> <br /> By Power of a Point, &lt;math&gt;ab=BD \cdot DC=735/64=2205/192&lt;/math&gt;, so<br /> <br /> &lt;cmath&gt;a^2+2 \cdot \frac{2205}{192}=\frac{5341}{192}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2=\frac{931}{192}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a=XD&lt;/math&gt;, &lt;math&gt;XD=\frac{7\sqrt{19}}{8\sqrt{3}}&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\angle EXF&lt;/math&gt; and &lt;math&gt;\angle EAF&lt;/math&gt; intercept the same arc in circle &lt;math&gt;\omega&lt;/math&gt; and the same goes for &lt;math&gt;\angle XFA&lt;/math&gt; and &lt;math&gt;\angle XEA&lt;/math&gt;, &lt;math&gt;\angle EXF\cong\angle EAF&lt;/math&gt; and &lt;math&gt;\angle XFA\cong\angle XEA&lt;/math&gt;. Therefore, &lt;math&gt;\triangle XDE\sim\triangle ADF&lt;/math&gt; by [[Similarity (geometry)|AA Similarity]]. Since side lengths in similar triangles are proportional,<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{\frac{14\sqrt{3}}{3}}{\frac{7\sqrt{19}}{8\sqrt{3}}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{16}{\sqrt{19}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF \cdot \sqrt{19} = 30&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF = \frac{30}{\sqrt{19}}.&lt;/cmath&gt;<br /> <br /> However, the problem asks for &lt;math&gt;AF^2&lt;/math&gt;, so &lt;math&gt;AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> ==Solution 4==<br /> It can be verified with law of cosines that &lt;math&gt;\angle BAC=120^\circ.&lt;/math&gt; Also, &lt;math&gt;E&lt;/math&gt; is the midpoint of major arc &lt;math&gt;BC&lt;/math&gt; so &lt;math&gt;BE=CE,&lt;/math&gt; and &lt;math&gt;\angle BEC=60^\circ.&lt;/math&gt; Thus &lt;math&gt;CBE&lt;/math&gt; is equilateral. Notice now that &lt;math&gt;\angle BFC=\angle BFE= 60.&lt;/math&gt; But &lt;math&gt;\angle DFE=90&lt;/math&gt; so &lt;math&gt;FD&lt;/math&gt; bisects &lt;math&gt;\angle BFC.&lt;/math&gt; Thus, &lt;math&gt;\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.&lt;/math&gt; <br /> <br /> Let &lt;math&gt;BF=5a, CF=3a.&lt;/math&gt; By law of cosines on &lt;math&gt;BFC&lt;/math&gt; we find &lt;math&gt;a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.&lt;/math&gt; But by ptolemy on &lt;math&gt;BFCA&lt;/math&gt;, &lt;math&gt;15a+15a=7*AF,&lt;/math&gt; so &lt;math&gt;AF= \frac{30}{\sqrt{19}},&lt;/math&gt; so &lt;math&gt;AF^2=\frac{900}{19}&lt;/math&gt; and the answer is &lt;math&gt;900+19=\boxed{919}&lt;/math&gt;<br /> <br /> ~abacadaea<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=14|after=Last Problem}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_14&diff=138500 2021 AMC 10A Problems/Problem 14 2020-11-26T00:05:34Z <p>Blehlivesonearth: Created page with &quot;The problem will be released in 115 days.&quot;</p> <hr /> <div>The problem will be released in 115 days.</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_13&diff=138499 2021 AMC 10A Problems/Problem 13 2020-11-26T00:05:25Z <p>Blehlivesonearth: Created page with &quot;The problem will be released in 115 days.&quot;</p> <hr /> <div>The problem will be released in 115 days.</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_11&diff=138498 2021 AMC 10A Problems/Problem 11 2020-11-26T00:05:04Z <p>Blehlivesonearth: Created page with &quot;The problem will be released in 115 days.&quot;</p> <hr /> <div>The problem will be released in 115 days.</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_10&diff=138497 2021 AMC 10A Problems/Problem 10 2020-11-26T00:04:49Z <p>Blehlivesonearth: Created page with &quot;The problem will be released in 115 days.&quot;</p> <hr /> <div>The problem will be released in 115 days.</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_9&diff=138496 2021 AMC 10A Problems/Problem 9 2020-11-26T00:04:36Z <p>Blehlivesonearth: Created page with &quot;The problem will be released in 115 days.&quot;</p> <hr /> <div>The problem will be released in 115 days.</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_8&diff=138495 2021 AMC 10A Problems/Problem 8 2020-11-26T00:04:22Z <p>Blehlivesonearth: Created page with &quot;The problem will be released in 115 days.&quot;</p> <hr /> <div>The problem will be released in 115 days.</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_15&diff=135848 2006 AIME I Problems/Problem 15 2020-10-26T04:33:59Z <p>Blehlivesonearth: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Given that a sequence satisfies &lt;math&gt; x_0=0 &lt;/math&gt; and &lt;math&gt; |x_k|=|x_{k-1}+3| &lt;/math&gt; for all integers &lt;math&gt; k\ge 1, &lt;/math&gt; find the minimum possible value of &lt;math&gt; |x_1+x_2+\cdots+x_{2006}|. &lt;/math&gt;<br /> <br /> __TOC__<br /> == Solutions ==<br /> === Solution 1 ===<br /> Suppose &lt;math&gt;b_{i} = \frac {x_{i}}3&lt;/math&gt;.<br /> We have<br /> &lt;cmath&gt;<br /> \sum_{i = 1}^{2006}b_{i}^{2} = \sum_{i = 0}^{2005}(b_{i} + 1)^{2} = \sum_{i = 0}^{2005}(b_{i}^{2} + 2b_{i} + 1)<br /> &lt;/cmath&gt;<br /> So<br /> &lt;cmath&gt;<br /> \sum_{i = 0}^{2005}b_{i} = \frac {b_{2006}^{2} - 2006}2<br /> &lt;/cmath&gt;<br /> Now<br /> &lt;cmath&gt;<br /> \sum_{i = 1}^{2006}b_{i} = \frac {b_{2006}^{2} + 2b_{2006} - 2006}2<br /> &lt;/cmath&gt;<br /> Therefore<br /> &lt;cmath&gt;<br /> \left|\sum_{i = 1}^{2006}b_{i}\right| = \left|\frac {(b_{2006} + 1)^{2} - 2007}2\right|\geq \frac {2025 - 2007}{2} = 9<br /> &lt;/cmath&gt;<br /> So<br /> &lt;cmath&gt;<br /> \left|\sum_{i = 1}^{2006}x_{i}\right|\geq \boxed{027}<br /> &lt;/cmath&gt;<br /> <br /> === Solution 2 ===<br /> First, we state that iff &lt;math&gt;x_{i - 1}\ge0&lt;/math&gt;, &lt;math&gt;|x_i| = |x_{i - 1}| + 3&lt;/math&gt; and iff &lt;math&gt;x_{i - 1} &lt; 0&lt;/math&gt;, &lt;math&gt;|x_i| = |x_{i - 1}| - 3&lt;/math&gt;. Now suppose &lt;math&gt;x_i = x_j&lt;/math&gt; for some &lt;math&gt;0\le i &lt; j\le2006&lt;/math&gt;. Now, this means that &lt;math&gt;|x_i| = |x_j|&lt;/math&gt;, and so the number of positive numbers in the set &lt;math&gt;\{x_i,x_{i + 1},\ldots,x_{j - 1}\}&lt;/math&gt; equals the number of negative numbers. Now pair the numbers in this list up in the following way: Whenever a positive and a negative number are adjacent in this progression, pair them up and remove them from this list. We claim that every pair will sum to -3.<br /> <br /> If the positive number comes first, then the negative number will have a magnitude three greater, so this is true. If the negative number comes first, then the positive number will have magnitude three smaller, and this will also be true. Now let us examine what happens when we remove those two from the sequence. WLOG, let the numbers be &lt;math&gt;x_k&lt;/math&gt; and &lt;math&gt;x_{k + 1}&lt;/math&gt;. Since one is positive and the other is negative, &lt;math&gt;|x_{k + 2}| = |x_{k + 1}|\pm3 = |x_k|\pm3\mp3 = |x_k| = |x_{k - 1} + 3|&lt;/math&gt;. So the new sequence works under the same criteria as the old one. In this way, we can pair all of the numbers up in this subsequence so the sums of the pairs are -3. Thus, the average of these numbers will be -3/2 for all subsequences that start and end with the same number (not including one of those).<br /> <br /> Now, take all of the repeating subsequences out of the original sequence. The only thing that will be left will be a sequence &lt;math&gt;\{0,3,6,9,\cdots,3k\}&lt;/math&gt; for some even &lt;math&gt;k&lt;/math&gt;. Since we started with 2006 terms, we removed &lt;math&gt;2006 - k&lt;/math&gt; (an even number) with an average of -3/2. Thus, the sum of both this remaining sequence and the removed stuff is &lt;math&gt;(2006 - k)( - 3/2) + \sum_{i = 1}^k3k = (3/2)(k - 2006 + k(k + 1)) = 3/2(k^2 + 2k - 2006)&lt;/math&gt;. This must be minimized, so we find the roots: &lt;math&gt;k^2 + 2k = 2006\implies (k + 1)^2 = 2007&lt;/math&gt; and &lt;math&gt;44^2 = 1936 &lt; 2007 &lt; 2025 = 45^2&lt;/math&gt;. Plugging in &lt;math&gt;k = 44&lt;/math&gt; yields &lt;math&gt;(3/2)(2025 - 2007) = 27&lt;/math&gt; (and &lt;math&gt;k = 42&lt;/math&gt; yields &lt;math&gt;- 237&lt;/math&gt;, a worse result). Thus, &lt;math&gt;\fbox{027}&lt;/math&gt; is the closest to zero this sum can get.<br /> <br /> === Solution 3 ===<br /> We know &lt;math&gt;|x_k| = |x_{k - 1} + 3|&lt;/math&gt;. We get rid of the absolute value by squaring both sides: &lt;math&gt;{x_k}^2 = {x_{k - 1}}^2 + 6{x_{k - 1}} + 9\Rightarrow {x_k}^2 - {x_{k - 1}}^2 = 6{x_{k - 1}} + 9&lt;/math&gt;. So we set this up:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*} {x_1}^2 - {x_0}^2 &amp; = 6{x_0} + 9 \\<br /> {x_2}^2 - {x_1}^2 &amp; = 6{x_1} + 9 \\<br /> &amp; \vdots \\<br /> {x_{2007}}^2 - {x_{2006}}^2 &amp; = 6{x_{2006}} + 9<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> There are &lt;math&gt;2007&lt;/math&gt; equations. Sum them. We get:<br /> &lt;math&gt;{x_{2007}}^2 = 6\left(x_1 + x_2 + \cdots + x_{2006}\right) + 9\cdot{2007}&lt;/math&gt;<br /> <br /> So &lt;math&gt;|x_1 + x_2 + \cdots + x_{2006}| = \frac {\left|{x_{2007}}^2 - 9\cdot{2007}\right|}{6}&lt;/math&gt;<br /> <br /> We know &lt;math&gt;3\ |\ x_{2007}&lt;/math&gt; and we want to minimize &lt;math&gt;\left|{x_{2007}}^2 - 9\cdot{2007}\right|&lt;/math&gt;, so &lt;math&gt;x_{2007}&lt;/math&gt; must be &lt;math&gt;3\cdot{45}&lt;/math&gt; for it to be minimal (&lt;math&gt;45^2 = 2025&lt;/math&gt; which is closest to &lt;math&gt;2007&lt;/math&gt;). <br /> <br /> This means that &lt;math&gt;|x_1 + x_2 + \cdots + x_{2006}| = \left|\frac {9(2025 - 2007)}{6}\right| = \boxed{027}&lt;/math&gt;<br /> <br /> === Solution 4 ===<br /> <br /> Playing around with a couple numbers, we see that we can generate the sequence &lt;math&gt;0, 3, -6, 3, -6, \cdots&lt;/math&gt;, and we can also generate the sequence &lt;math&gt;3, 6, 9, 12, \cdots&lt;/math&gt; after each &lt;math&gt;-6&lt;/math&gt; value. Thus, we will apply this to try and find some bounds. We can test if the first &lt;math&gt;1000&lt;/math&gt; pairs of numbers each sum up to &lt;math&gt;-3&lt;/math&gt;, and the rest form an arithmetic sequence, if the first &lt;math&gt;990&lt;/math&gt; pairs sum up to &lt;math&gt;-3&lt;/math&gt;, and so on. When we get to &lt;math&gt;980&lt;/math&gt;, we find that &lt;math&gt;980(-3) + 3 + 6 + \cdots + 3\cdot 46 = 303&lt;/math&gt;. If we shift the number of pairs up by &lt;math&gt;1&lt;/math&gt;, we get &lt;math&gt;981(-3) + 3 + 6 + \cdots + 3\cdot 44 = \boxed{027}&lt;/math&gt;. - Spacesam<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_15&diff=135847 2006 AIME I Problems/Problem 15 2020-10-26T04:33:44Z <p>Blehlivesonearth: /* Solutions */</p> <hr /> <div>== Problem ==<br /> Given that a sequence satisfies &lt;math&gt; x_0=0 &lt;/math&gt; and &lt;math&gt; |x_k|=|x_{k-1}+3| &lt;/math&gt; for all integers &lt;math&gt; k\ge 1, &lt;/math&gt; find the minimum possible value of &lt;math&gt; |x_1+x_2+\cdots+x_{2006}|. &lt;/math&gt;<br /> <br /> __TOC__<br /> == Solutions ==<br /> === Solution 1 ===<br /> Suppose &lt;math&gt;b_{i} = \frac {x_{i}}3&lt;/math&gt;.<br /> We have<br /> &lt;cmath&gt;<br /> \sum_{i = 1}^{2006}b_{i}^{2} = \sum_{i = 0}^{2005}(b_{i} + 1)^{2} = \sum_{i = 0}^{2005}(b_{i}^{2} + 2b_{i} + 1)<br /> &lt;/cmath&gt;<br /> So<br /> &lt;cmath&gt;<br /> \sum_{i = 0}^{2005}b_{i} = \frac {b_{2006}^{2} - 2006}2<br /> &lt;/cmath&gt;<br /> Now<br /> &lt;cmath&gt;<br /> \sum_{i = 1}^{2006}b_{i} = \frac {b_{2006}^{2} + 2b_{2006} - 2006}2<br /> &lt;/cmath&gt;<br /> Therefore<br /> &lt;cmath&gt;<br /> \left|\sum_{i = 1}^{2006}b_{i}\right| = \left|\frac {(b_{2006} + 1)^{2} - 2007}2\right|\geq \frac {2025 - 2007}{2} = 9<br /> &lt;/cmath&gt;<br /> So<br /> &lt;cmath&gt;<br /> \left|\sum_{i = 1}^{2006}x_{i}\right|\geq \boxed{027}<br /> &lt;/cmath&gt;<br /> <br /> === Solution 2 ===<br /> First, we state that iff &lt;math&gt;x_{i - 1}\ge0&lt;/math&gt;, &lt;math&gt;|x_i| = |x_{i - 1}| + 3&lt;/math&gt; and iff &lt;math&gt;x_{i - 1} &lt; 0&lt;/math&gt;, &lt;math&gt;|x_i| = |x_{i - 1}| - 3&lt;/math&gt;. Now suppose &lt;math&gt;x_i = x_j&lt;/math&gt; for some &lt;math&gt;0\le i &lt; j\le2006&lt;/math&gt;. Now, this means that &lt;math&gt;|x_i| = |x_j|&lt;/math&gt;, and so the number of positive numbers in the set &lt;math&gt;\{x_i,x_{i + 1},\ldots,x_{j - 1}\}&lt;/math&gt; equals the number of negative numbers. Now pair the numbers in this list up in the following way: Whenever a positive and a negative number are adjacent in this progression, pair them up and remove them from this list. We claim that every pair will sum to -3.<br /> <br /> If the positive number comes first, then the negative number will have a magnitude three greater, so this is true. If the negative number comes first, then the positive number will have magnitude three smaller, and this will also be true. Now let us examine what happens when we remove those two from the sequence. WLOG, let the numbers be &lt;math&gt;x_k&lt;/math&gt; and &lt;math&gt;x_{k + 1}&lt;/math&gt;. Since one is positive and the other is negative, &lt;math&gt;|x_{k + 2}| = |x_{k + 1}|\pm3 = |x_k|\pm3\mp3 = |x_k| = |x_{k - 1} + 3|&lt;/math&gt;. So the new sequence works under the same criteria as the old one. In this way, we can pair all of the numbers up in this subsequence so the sums of the pairs are -3. Thus, the average of these numbers will be -3/2 for all subsequences that start and end with the same number (not including one of those).<br /> <br /> Now, take all of the repeating subsequences out of the original sequence. The only thing that will be left will be a sequence &lt;math&gt;\{0,3,6,9,\cdots,3k\}&lt;/math&gt; for some even &lt;math&gt;k&lt;/math&gt;. Since we started with 2006 terms, we removed &lt;math&gt;2006 - k&lt;/math&gt; (an even number) with an average of -3/2. Thus, the sum of both this remaining sequence and the removed stuff is &lt;math&gt;(2006 - k)( - 3/2) + \sum_{i = 1}^k3k = (3/2)(k - 2006 + k(k + 1)) = 3/2(k^2 + 2k - 2006)&lt;/math&gt;. This must be minimized, so we find the roots: &lt;math&gt;k^2 + 2k = 2006\implies (k + 1)^2 = 2007&lt;/math&gt; and &lt;math&gt;44^2 = 1936 &lt; 2007 &lt; 2025 = 45^2&lt;/math&gt;. Plugging in &lt;math&gt;k = 44&lt;/math&gt; yields &lt;math&gt;(3/2)(2025 - 2007) = 27&lt;/math&gt; (and &lt;math&gt;k = 42&lt;/math&gt; yields &lt;math&gt;- 237&lt;/math&gt;, a worse result). Thus, &lt;math&gt;\fbox{027}&lt;/math&gt; is the closest to zero this sum can get.<br /> <br /> === Solution 3 ===<br /> We know &lt;math&gt;|x_k| = |x_{k - 1} + 3|&lt;/math&gt;. We get rid of the absolute value by squaring both sides: &lt;math&gt;{x_k}^2 = {x_{k - 1}}^2 + 6{x_{k - 1}} + 9\Rightarrow {x_k}^2 - {x_{k - 1}}^2 = 6{x_{k - 1}} + 9&lt;/math&gt;. So we set this up:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*} {x_1}^2 - {x_0}^2 &amp; = 6{x_0} + 9 \\<br /> {x_2}^2 - {x_1}^2 &amp; = 6{x_1} + 9 \\<br /> &amp; \vdots \\<br /> {x_{2007}}^2 - {x_{2006}}^2 &amp; = 6{x_{2006}} + 9<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> There are &lt;math&gt;2007&lt;/math&gt; equations. Sum them. We get:<br /> &lt;math&gt;{x_{2007}}^2 = 6\left(x_1 + x_2 + \cdots + x_{2006}\right) + 9\cdot{2007}&lt;/math&gt;<br /> <br /> So &lt;math&gt;|x_1 + x_2 + \cdots + x_{2006}| = \frac {\left|{x_{2007}}^2 - 9\cdot{2007}\right|}{6}&lt;/math&gt;<br /> <br /> We know &lt;math&gt;3\ |\ x_{2007}&lt;/math&gt; and we want to minimize &lt;math&gt;\left|{x_{2007}}^2 - 9\cdot{2007}\right|&lt;/math&gt;, so &lt;math&gt;x_{2007}&lt;/math&gt; must be &lt;math&gt;3\cdot{45}&lt;/math&gt; for it to be minimal (&lt;math&gt;45^2 = 2025&lt;/math&gt; which is closest to &lt;math&gt;2007&lt;/math&gt;). <br /> <br /> This means that &lt;math&gt;|x_1 + x_2 + \cdots + x_{2006}| = \left|\frac {9(2025 - 2007)}{6}\right| = \boxed{027}&lt;/math&gt;<br /> <br /> === Solution 4 ===<br /> <br /> Playing around with a couple numbers, we see that we can generate the sequence &lt;math&gt;0, 3, -6, 3, -6, \cdots&lt;/math&gt;, and we can also generate the sequence &lt;math&gt;3, 6, 9, 12, \cdots&lt;/math&gt; after each &lt;math&gt;-6&lt;/math&gt; value. Thus, we will apply this to try and find some bounds. We can test if the first &lt;math&gt;1000&lt;/math&gt; pairs of numbers each sum up to &lt;math&gt;-3&lt;/math&gt;, and the rest form an arithmetic sequence, if the first &lt;math&gt;990&lt;/math&gt; pairs sum up to &lt;math&gt;-3&lt;/math&gt;, and so on. When we get to &lt;math&gt;980&lt;/math&gt;, we find that &lt;math&gt;980(-3) + 3 + 6 + \cdots + 3\cdot 46 = 303&lt;/math&gt;. If we shift the number of pairs up by &lt;math&gt;1&lt;/math&gt;, we get &lt;math&gt;981(-3) + 3 + 6 + \cdots + 3\cdot 44 = \boxed{027}&lt;/math&gt;. - Spacesam<br /> <br /> == Solution 4 ==<br /> Playing around with a couple numbers, we see that we can generate the sequence &lt;math&gt;0, 3, -6, 3, -6, \cdots&lt;/math&gt;, and we can also generate the sequence &lt;math&gt;3, 6, 9, 12, \cdots&lt;/math&gt; after each &lt;math&gt;-6&lt;/math&gt; value. Thus, we will apply this to try and find some bounds. We can test if the first &lt;math&gt;1000&lt;/math&gt; pairs of numbers each sum up to &lt;math&gt;-3&lt;/math&gt;, and the rest form an arithmetic sequence, if the first &lt;math&gt;990&lt;/math&gt; pairs sum up to &lt;math&gt;-3&lt;/math&gt;, and so on. When we get to &lt;math&gt;980&lt;/math&gt;, we find that &lt;math&gt;980(-3) + 3 + 6 + \cdots + 3\cdot 46 = 303&lt;/math&gt;. If we shift the number of pairs up by &lt;math&gt;1&lt;/math&gt;, we get &lt;math&gt;981(-3) + 3 + 6 + \cdots + 3\cdot 44 = \boxed{027}&lt;/math&gt;. - Spacesam<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_15&diff=135846 2006 AIME I Problems/Problem 15 2020-10-26T04:32:48Z <p>Blehlivesonearth: /* Solution */</p> <hr /> <div>== Problem ==<br /> Given that a sequence satisfies &lt;math&gt; x_0=0 &lt;/math&gt; and &lt;math&gt; |x_k|=|x_{k-1}+3| &lt;/math&gt; for all integers &lt;math&gt; k\ge 1, &lt;/math&gt; find the minimum possible value of &lt;math&gt; |x_1+x_2+\cdots+x_{2006}|. &lt;/math&gt;<br /> <br /> __TOC__<br /> == Solutions ==<br /> === Solution 1 ===<br /> Suppose &lt;math&gt;b_{i} = \frac {x_{i}}3&lt;/math&gt;.<br /> We have<br /> &lt;cmath&gt;<br /> \sum_{i = 1}^{2006}b_{i}^{2} = \sum_{i = 0}^{2005}(b_{i} + 1)^{2} = \sum_{i = 0}^{2005}(b_{i}^{2} + 2b_{i} + 1)<br /> &lt;/cmath&gt;<br /> So<br /> &lt;cmath&gt;<br /> \sum_{i = 0}^{2005}b_{i} = \frac {b_{2006}^{2} - 2006}2<br /> &lt;/cmath&gt;<br /> Now<br /> &lt;cmath&gt;<br /> \sum_{i = 1}^{2006}b_{i} = \frac {b_{2006}^{2} + 2b_{2006} - 2006}2<br /> &lt;/cmath&gt;<br /> Therefore<br /> &lt;cmath&gt;<br /> \left|\sum_{i = 1}^{2006}b_{i}\right| = \left|\frac {(b_{2006} + 1)^{2} - 2007}2\right|\geq \frac {2025 - 2007}{2} = 9<br /> &lt;/cmath&gt;<br /> So<br /> &lt;cmath&gt;<br /> \left|\sum_{i = 1}^{2006}x_{i}\right|\geq \boxed{027}<br /> &lt;/cmath&gt;<br /> <br /> === Solution 2 ===<br /> First, we state that iff &lt;math&gt;x_{i - 1}\ge0&lt;/math&gt;, &lt;math&gt;|x_i| = |x_{i - 1}| + 3&lt;/math&gt; and iff &lt;math&gt;x_{i - 1} &lt; 0&lt;/math&gt;, &lt;math&gt;|x_i| = |x_{i - 1}| - 3&lt;/math&gt;. Now suppose &lt;math&gt;x_i = x_j&lt;/math&gt; for some &lt;math&gt;0\le i &lt; j\le2006&lt;/math&gt;. Now, this means that &lt;math&gt;|x_i| = |x_j|&lt;/math&gt;, and so the number of positive numbers in the set &lt;math&gt;\{x_i,x_{i + 1},\ldots,x_{j - 1}\}&lt;/math&gt; equals the number of negative numbers. Now pair the numbers in this list up in the following way: Whenever a positive and a negative number are adjacent in this progression, pair them up and remove them from this list. We claim that every pair will sum to -3.<br /> <br /> If the positive number comes first, then the negative number will have a magnitude three greater, so this is true. If the negative number comes first, then the positive number will have magnitude three smaller, and this will also be true. Now let us examine what happens when we remove those two from the sequence. WLOG, let the numbers be &lt;math&gt;x_k&lt;/math&gt; and &lt;math&gt;x_{k + 1}&lt;/math&gt;. Since one is positive and the other is negative, &lt;math&gt;|x_{k + 2}| = |x_{k + 1}|\pm3 = |x_k|\pm3\mp3 = |x_k| = |x_{k - 1} + 3|&lt;/math&gt;. So the new sequence works under the same criteria as the old one. In this way, we can pair all of the numbers up in this subsequence so the sums of the pairs are -3. Thus, the average of these numbers will be -3/2 for all subsequences that start and end with the same number (not including one of those).<br /> <br /> Now, take all of the repeating subsequences out of the original sequence. The only thing that will be left will be a sequence &lt;math&gt;\{0,3,6,9,\cdots,3k\}&lt;/math&gt; for some even &lt;math&gt;k&lt;/math&gt;. Since we started with 2006 terms, we removed &lt;math&gt;2006 - k&lt;/math&gt; (an even number) with an average of -3/2. Thus, the sum of both this remaining sequence and the removed stuff is &lt;math&gt;(2006 - k)( - 3/2) + \sum_{i = 1}^k3k = (3/2)(k - 2006 + k(k + 1)) = 3/2(k^2 + 2k - 2006)&lt;/math&gt;. This must be minimized, so we find the roots: &lt;math&gt;k^2 + 2k = 2006\implies (k + 1)^2 = 2007&lt;/math&gt; and &lt;math&gt;44^2 = 1936 &lt; 2007 &lt; 2025 = 45^2&lt;/math&gt;. Plugging in &lt;math&gt;k = 44&lt;/math&gt; yields &lt;math&gt;(3/2)(2025 - 2007) = 27&lt;/math&gt; (and &lt;math&gt;k = 42&lt;/math&gt; yields &lt;math&gt;- 237&lt;/math&gt;, a worse result). Thus, &lt;math&gt;\fbox{027}&lt;/math&gt; is the closest to zero this sum can get.<br /> <br /> === Solution 3 ===<br /> We know &lt;math&gt;|x_k| = |x_{k - 1} + 3|&lt;/math&gt;. We get rid of the absolute value by squaring both sides: &lt;math&gt;{x_k}^2 = {x_{k - 1}}^2 + 6{x_{k - 1}} + 9\Rightarrow {x_k}^2 - {x_{k - 1}}^2 = 6{x_{k - 1}} + 9&lt;/math&gt;. So we set this up:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*} {x_1}^2 - {x_0}^2 &amp; = 6{x_0} + 9 \\<br /> {x_2}^2 - {x_1}^2 &amp; = 6{x_1} + 9 \\<br /> &amp; \vdots \\<br /> {x_{2007}}^2 - {x_{2006}}^2 &amp; = 6{x_{2006}} + 9<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> There are &lt;math&gt;2007&lt;/math&gt; equations. Sum them. We get:<br /> &lt;math&gt;{x_{2007}}^2 = 6\left(x_1 + x_2 + \cdots + x_{2006}\right) + 9\cdot{2007}&lt;/math&gt;<br /> <br /> So &lt;math&gt;|x_1 + x_2 + \cdots + x_{2006}| = \frac {\left|{x_{2007}}^2 - 9\cdot{2007}\right|}{6}&lt;/math&gt;<br /> <br /> We know &lt;math&gt;3\ |\ x_{2007}&lt;/math&gt; and we want to minimize &lt;math&gt;\left|{x_{2007}}^2 - 9\cdot{2007}\right|&lt;/math&gt;, so &lt;math&gt;x_{2007}&lt;/math&gt; must be &lt;math&gt;3\cdot{45}&lt;/math&gt; for it to be minimal (&lt;math&gt;45^2 = 2025&lt;/math&gt; which is closest to &lt;math&gt;2007&lt;/math&gt;). <br /> <br /> This means that &lt;math&gt;|x_1 + x_2 + \cdots + x_{2006}| = \left|\frac {9(2025 - 2007)}{6}\right| = \boxed{027}&lt;/math&gt;<br /> <br /> == Solution 4 ==<br /> Playing around with a couple numbers, we see that we can generate the sequence &lt;math&gt;0, 3, -6, 3, -6, \cdots&lt;/math&gt;, and we can also generate the sequence &lt;math&gt;3, 6, 9, 12, \cdots&lt;/math&gt; after each &lt;math&gt;-6&lt;/math&gt; value. Thus, we will apply this to try and find some bounds. We can test if the first &lt;math&gt;1000&lt;/math&gt; pairs of numbers each sum up to &lt;math&gt;-3&lt;/math&gt;, and the rest form an arithmetic sequence, if the first &lt;math&gt;990&lt;/math&gt; pairs sum up to &lt;math&gt;-3&lt;/math&gt;, and so on. When we get to &lt;math&gt;980&lt;/math&gt;, we find that &lt;math&gt;980(-3) + 3 + 6 + \cdots + 3\cdot 46 = 303&lt;/math&gt;. If we shift the number of pairs up by &lt;math&gt;1&lt;/math&gt;, we get &lt;math&gt;981(-3) + 3 + 6 + \cdots + 3\cdot 44 = \boxed{027}&lt;/math&gt;. - Spacesam<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_11&diff=135845 2006 AIME II Problems/Problem 11 2020-10-26T04:11:54Z <p>Blehlivesonearth: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> A [[sequence]] is defined as follows &lt;math&gt; a_1=a_2=a_3=1, &lt;/math&gt; and, for all positive integers &lt;math&gt; n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. &lt;/math&gt; Given that &lt;math&gt; a_{28}=6090307, a_{29}=11201821, &lt;/math&gt; and &lt;math&gt; a_{30}=20603361, &lt;/math&gt; find the [[remainder]] when &lt;math&gt;\sum^{28}_{k=1} a_k &lt;/math&gt; is divided by 1000.<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Define the sum as &lt;math&gt;s&lt;/math&gt;. Since &lt;math&gt;a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} &lt;/math&gt;, the sum will be:<br /> &lt;center&gt;&lt;math&gt;s = a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\<br /> s = a_{28} + \left(\sum^{30}_{k=4} a_{k} - \sum^{29}_{k=3} a_{k}\right) - \left(\sum^{28}_{k=2} a_{k}\right)\\<br /> s = a_{28} + (a_{30} - a_{3}) - \left(\sum^{28}_{k=2} a_{k}\right) = a_{28} + a_{30} - a_{3} - (s - a_{1})\\<br /> s = -s + a_{28} + a_{30}<br /> &lt;/math&gt;&lt;/center&gt;<br /> <br /> Thus &lt;math&gt;s = \frac{a_{28} + a_{30}}{2}&lt;/math&gt;, and &lt;math&gt;a_{28},\,a_{30}&lt;/math&gt; are both given; the last four digits of their sum is &lt;math&gt;3668&lt;/math&gt;, and half of that is &lt;math&gt;1834&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{834}&lt;/math&gt;.<br /> <br /> Solution by an anonymous user.<br /> <br /> === Solution 2 (bash) ===<br /> <br /> Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:<br /> <br /> &lt;math&gt;a_{1}\equiv 1 \pmod {1000}<br /> \newline<br /> a_{2}\equiv 1 \pmod {1000}<br /> \newline<br /> a_{3}\equiv 1 \pmod {1000}<br /> \newline<br /> a_{4}\equiv 3 \pmod {1000}<br /> \newline<br /> a_{5}\equiv 5 \pmod {1000}<br /> \newline<br /> \cdots <br /> \newline<br /> a_{25} \equiv 793 \pmod {1000}<br /> \newline<br /> a_{26} \equiv 281 \pmod {1000}<br /> \newline<br /> a_{27} \equiv 233 \pmod {1000}<br /> \newline<br /> a_{28} \equiv 307 \pmod {1000}&lt;/math&gt;<br /> <br /> Adding all the residues shows the sum is congruent to &lt;math&gt;\boxed{834}&lt;/math&gt; mod 1000.<br /> <br /> ~ I-_-I<br /> <br /> === Solution 3 (some guessing involved)/&quot;Engineer's Induction&quot; ===<br /> All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given &lt;math&gt;a_{28}, a_{29}, &lt;/math&gt; and &lt;math&gt;a_{30}&lt;/math&gt;, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some &lt;math&gt;p, q, r&lt;/math&gt; such that &lt;math&gt;\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}&lt;/math&gt;. From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that &lt;math&gt;(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})&lt;/math&gt;, at least for the first few terms. From this, we have that &lt;math&gt;\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}}(\mod 1000)&lt;/math&gt;.<br /> <br /> Solution by zeroman; clarified by srisainandan6<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_11&diff=135844 2006 AIME II Problems/Problem 11 2020-10-26T04:11:27Z <p>Blehlivesonearth: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> A [[sequence]] is defined as follows &lt;math&gt; a_1=a_2=a_3=1, &lt;/math&gt; and, for all positive integers &lt;math&gt; n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. &lt;/math&gt; Given that &lt;math&gt; a_{28}=6090307, a_{29}=11201821, &lt;/math&gt; and &lt;math&gt; a_{30}=20603361, &lt;/math&gt; find the [[remainder]] when &lt;math&gt;\sum^{28}_{k=1} a_k &lt;/math&gt; is divided by 1000.<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Define the sum as &lt;math&gt;s&lt;/math&gt;. Since &lt;math&gt;a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} &lt;/math&gt;, the sum will be:<br /> &lt;center&gt;&lt;math&gt;s = a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\<br /> s = a_{28} + \left(\sum^{30}_{k=4} a_{k} - \sum^{29}_{k=3} a_{k}\right) - \left(\sum^{28}_{k=2} a_{k}\right)\\<br /> s = a_{28} + (a_{30} - a_{3}) - \left(\sum^{28}_{k=2} a_{k}\right) = a_{28} + a_{30} - a_{3} - (s - a_{1})\\<br /> s = -s + a_{28} + a_{30}<br /> &lt;/math&gt;&lt;/center&gt;<br /> <br /> Thus &lt;math&gt;s = \frac{a_{28} + a_{30}}{2}&lt;/math&gt;, and &lt;math&gt;a_{28},\,a_{30}&lt;/math&gt; are both given; the last four digits of their sum is &lt;math&gt;3668&lt;/math&gt;, and half of that is &lt;math&gt;1834&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{834}&lt;/math&gt;.<br /> <br /> by an anonymous user<br /> <br /> === Solution 2 (bash) ===<br /> <br /> Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:<br /> <br /> &lt;math&gt;a_{1}\equiv 1 \pmod {1000}<br /> \newline<br /> a_{2}\equiv 1 \pmod {1000}<br /> \newline<br /> a_{3}\equiv 1 \pmod {1000}<br /> \newline<br /> a_{4}\equiv 3 \pmod {1000}<br /> \newline<br /> a_{5}\equiv 5 \pmod {1000}<br /> \newline<br /> \cdots <br /> \newline<br /> a_{25} \equiv 793 \pmod {1000}<br /> \newline<br /> a_{26} \equiv 281 \pmod {1000}<br /> \newline<br /> a_{27} \equiv 233 \pmod {1000}<br /> \newline<br /> a_{28} \equiv 307 \pmod {1000}&lt;/math&gt;<br /> <br /> Adding all the residues shows the sum is congruent to &lt;math&gt;\boxed{834}&lt;/math&gt; mod 1000.<br /> <br /> ~ I-_-I<br /> <br /> === Solution 3 (some guessing involved)/&quot;Engineer's Induction&quot; ===<br /> All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given &lt;math&gt;a_{28}, a_{29}, &lt;/math&gt; and &lt;math&gt;a_{30}&lt;/math&gt;, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some &lt;math&gt;p, q, r&lt;/math&gt; such that &lt;math&gt;\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}&lt;/math&gt;. From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that &lt;math&gt;(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})&lt;/math&gt;, at least for the first few terms. From this, we have that &lt;math&gt;\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}}(\mod 1000)&lt;/math&gt;.<br /> <br /> Solution by zeroman; clarified by srisainandan6<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_5&diff=135789 1986 AIME Problems/Problem 5 2020-10-25T03:43:45Z <p>Blehlivesonearth: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> What is that largest [[positive integer]] &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;n^3+100&lt;/math&gt; is [[divisible]] by &lt;math&gt;n+10&lt;/math&gt;?<br /> <br /> == Solution 1 ==<br /> If &lt;math&gt;n+10 \mid n^3+100&lt;/math&gt;, &lt;math&gt;\gcd(n^3+100,n+10)=n+10&lt;/math&gt;. Using the [[Euclidean algorithm]], we have &lt;math&gt;\gcd(n^3+100,n+10)= \gcd(-10n^2+100,n+10)&lt;/math&gt; &lt;math&gt;= \gcd(100n+100,n+10)&lt;/math&gt; &lt;math&gt;= \gcd(-900,n+10)&lt;/math&gt;, so &lt;math&gt;n+10&lt;/math&gt; must divide &lt;math&gt;900&lt;/math&gt;. The greatest [[integer]] &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;n+10&lt;/math&gt; divides &lt;math&gt;900&lt;/math&gt; is &lt;math&gt;\boxed{890}&lt;/math&gt;; we can double-check manually and we find that indeed &lt;math&gt;900\mid 890^3+100&lt;/math&gt;.<br /> <br /> == Solution 2 (Simple) ==<br /> Let &lt;math&gt;n+10=k&lt;/math&gt;, then &lt;math&gt;n=k-10&lt;/math&gt;. Then &lt;math&gt;n^3+100 = k^3-30k^2+300k-900&lt;/math&gt; Therefore, &lt;math&gt;900&lt;/math&gt; must be divisible by &lt;math&gt;k&lt;/math&gt;, which is largest when &lt;math&gt;k=900&lt;/math&gt; and &lt;math&gt;n=\boxed{890}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> In a similar manner, we can apply synthetic division. We are looking for &lt;math&gt;\frac{n^3 + 100}{n + 10} = n^2 - 10n + 100 - \frac{900}{n + 10}&lt;/math&gt;. Again, &lt;math&gt;n + 10&lt;/math&gt; must be a factor of &lt;math&gt;900 \Longrightarrow n = \boxed{890}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> The key to this problem is to realize that &lt;math&gt;n+10 \mid n^3 +1000&lt;/math&gt; for all &lt;math&gt;n&lt;/math&gt;. Since we are asked to find the maximum possible &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;n+10 \mid n^3 +100&lt;/math&gt;, we have: &lt;math&gt;n+10 \mid ((n^3 +1000) - (n^3 +100) \longrightarrow n+10 \mid 900&lt;/math&gt;. This is because of the property that states that if &lt;math&gt;a \mid b&lt;/math&gt; and &lt;math&gt;a \mid c&lt;/math&gt;, then &lt;math&gt;a \mid b \pm c&lt;/math&gt;. Since, the largest factor of 900 is itself we have: &lt;math&gt;n+10=900 \Longrightarrow \boxed{n = 890}&lt;/math&gt;<br /> <br /> ~qwertysri987<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_13&diff=135766 1985 AIME Problems/Problem 13 2020-10-24T22:19:16Z <p>Blehlivesonearth: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> The numbers in the [[sequence]] &lt;math&gt;101&lt;/math&gt;, &lt;math&gt;104&lt;/math&gt;, &lt;math&gt;109&lt;/math&gt;, &lt;math&gt;116&lt;/math&gt;,&lt;math&gt;\ldots&lt;/math&gt; are of the form &lt;math&gt;a_n=100+n^2&lt;/math&gt;, where &lt;math&gt;n=1,2,3,\ldots&lt;/math&gt; For each &lt;math&gt;n&lt;/math&gt;, let &lt;math&gt;d_n&lt;/math&gt; be the greatest common divisor of &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;a_{n+1}&lt;/math&gt;. Find the maximum value of &lt;math&gt;d_n&lt;/math&gt; as &lt;math&gt;n&lt;/math&gt; ranges through the [[positive integer]]s.<br /> <br /> == Solution 1==<br /> If &lt;math&gt;(x,y)&lt;/math&gt; denotes the [[greatest common divisor]] of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, then we have &lt;math&gt;d_n=(a_n,a_{n+1})=(100+n^2,100+n^2+2n+1)&lt;/math&gt;. Now assuming that &lt;math&gt;d_n&lt;/math&gt; [[divisor | divides]] &lt;math&gt;100+n^2&lt;/math&gt;, it must divide &lt;math&gt;2n+1&lt;/math&gt; if it is going to divide the entire [[expression]] &lt;math&gt;100+n^2+2n+1&lt;/math&gt;.<br /> <br /> Thus the [[equation]] turns into &lt;math&gt;d_n=(100+n^2,2n+1)&lt;/math&gt;. Now note that since &lt;math&gt;2n+1&lt;/math&gt; is [[odd integer | odd]] for [[integer | integral]] &lt;math&gt;n&lt;/math&gt;, we can multiply the left integer, &lt;math&gt;100+n^2&lt;/math&gt;, by a power of two without affecting the greatest common divisor. Since the &lt;math&gt;n^2&lt;/math&gt; term is quite restrictive, let's multiply by &lt;math&gt;4&lt;/math&gt; so that we can get a &lt;math&gt;(2n+1)^2&lt;/math&gt; in there.<br /> <br /> So &lt;math&gt;d_n=(4n^2+400,2n+1)=((2n+1)^2-4n+399,2n+1)=(-4n+399,2n+1)&lt;/math&gt;. It simplified the way we wanted it to!<br /> Now using similar techniques we can write &lt;math&gt;d_n=(-2(2n+1)+401,2n+1)=(401,2n+1)&lt;/math&gt;. Thus &lt;math&gt;d_n&lt;/math&gt; must divide &lt;math&gt;\boxed{401}&lt;/math&gt; for every single &lt;math&gt;n&lt;/math&gt;. This means the largest possible value for &lt;math&gt;d_n&lt;/math&gt; is &lt;math&gt;401&lt;/math&gt;, and we see that it can be achieved when &lt;math&gt;n = 200&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> We know that &lt;math&gt;a_n = 100+n^2&lt;/math&gt; and &lt;math&gt;a_{n+1} = 100+(n+1)^2 = 100+ n^2+2n+1&lt;/math&gt;. Since we want to find the GCD of &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;a_{n+1}&lt;/math&gt;, we can use the [[Euclidean algorithm]]:<br /> <br /> &lt;math&gt;a_{n+1}-a_n = 2n+1&lt;/math&gt;<br /> <br /> <br /> Now, the question is to find the GCD of &lt;math&gt;2n+1&lt;/math&gt; and &lt;math&gt;100+n^2&lt;/math&gt;. We subtract &lt;math&gt;2n+1&lt;/math&gt; 100 times from &lt;math&gt;100+n^2&lt;/math&gt;. This leaves us with &lt;math&gt;n^2-200n&lt;/math&gt;. We want this to equal 0, so solving for &lt;math&gt;n&lt;/math&gt; gives us &lt;math&gt;n=200&lt;/math&gt;. The last remainder is 0, thus &lt;math&gt;200*2+1 = \boxed{401}&lt;/math&gt; is our GCD.<br /> == Solution 3==<br /> If Solution 2 is not entirely obvious, our answer is the max possible range of &lt;math&gt;\frac{x(x-200)}{2x+1}&lt;/math&gt;. Using the Euclidean Algorithm on &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;2x+1&lt;/math&gt; yields that they are relatively prime. Thus, the only way the GCD will not be 1 is if the&lt;math&gt; x-200&lt;/math&gt; term share factors with the &lt;math&gt;2x+1&lt;/math&gt;. Using the Euclidean Algorithm, &lt;math&gt;\gcd(x-200,2x+1)=\gcd(x-200,2x+1-2(x-200))=\gcd(x-200,401)&lt;/math&gt;. Thus, the max GCD is 401.<br /> <br /> ==Solution 4==<br /> We can just plug in Euclidean algorithm, to go from &lt;math&gt;\gcd(n^2 + 100, n^2 + 2n + 101)&lt;/math&gt; to &lt;math&gt;\gcd(n^2 + 100, 2n + 1)&lt;/math&gt; to &lt;math&gt;\gcd(n^2 + 100 - 100(2n + 1), 2n + 1)&lt;/math&gt; to get &lt;math&gt;\gcd(n^2 - 200n, 2n + 1)&lt;/math&gt;. Now we know that no matter what, &lt;math&gt;n&lt;/math&gt; is relatively prime to &lt;math&gt;2n + 1&lt;/math&gt;. Therefore the equation can be simplified to: &lt;math&gt;\gcd(n - 200, 2n + 1)&lt;/math&gt;. Subtracting &lt;math&gt;2n - 400&lt;/math&gt; from &lt;math&gt;2n + 1&lt;/math&gt; results in &lt;math&gt;\gcd(n - 200,401)&lt;/math&gt;. The greatest possible value of this is &lt;math&gt;\boxed{401}&lt;/math&gt;, an happens when &lt;math&gt;n \equiv 200 \pmod{401}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1985|num-b=12|num-a=14}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=Expression&diff=134983 Expression 2020-10-13T00:17:14Z <p>Blehlivesonearth: </p> <hr /> <div>{{stub}}<br /> <br /> In [[mathematics]], an '''expression''' is any meaningful combination of symbols. What this means exactly varies depending on the mathematical context. For instance, arithmetic expressions typically consist of [[number]]s, [[variable]]s, and [[operator]]s, arranged in a sensible way. Thus, &lt;math&gt;3 - \frac x4&lt;/math&gt; is an arithmetic expression, while &lt;math&gt;7 \times + 43&lt;/math&gt; is not. There are no equal signs in expressions.<br /> <br /> Again depending on context, one is often interested in finding equivalences between expressions of various sorts. In standard arithmetic, for instance, the two expressions &lt;math&gt;(3x + 4) - x + 2&lt;/math&gt; is equivalent to the expression &lt;math&gt;2x + 6&lt;/math&gt;. This is represented by the use of an equal sign, &lt;math&gt;3x + 4 -x +2 = 2x - 6&lt;/math&gt;. In other branches of mathematics, other symbols are sometimes used, especially the symbol &lt;math&gt;\equiv&lt;/math&gt;.<br /> <br /> Note that in arithmetic, an equality like the one above is ''not'' an expression. In [[mathematical logic]], however, arithmetic equations often ''are'' expressions. For instance, &lt;math&gt;3x + 2 = 11&lt;/math&gt; is a valid expression in [[Peano arithmetic]] (with a proper interpretation of symbols), and is logically equivalent to the expression &lt;math&gt;x = 3&lt;/math&gt;. We might write this equivalence of expressions as &lt;math&gt;\left(3x +2 = 11\right) \Longleftrightarrow \left(x = 3\right)&lt;/math&gt;. Just like the equality of two arithmetic expressions is not an arithmetic expression, this equivalence of logical expressions is not a logical expression.<br /> <br /> == See Also ==<br /> * [[Algebra]]<br /> * [[Order of operations]]</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=1978_IMO_Problems/Problem_2&diff=133739 1978 IMO Problems/Problem 2 2020-09-17T03:11:49Z <p>Blehlivesonearth: Created page with &quot;yeet&quot;</p> <hr /> <div>yeet</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=1978_IMO_Problems/Problem_3&diff=133738 1978 IMO Problems/Problem 3 2020-09-17T03:11:39Z <p>Blehlivesonearth: Created page with &quot;yeet&quot;</p> <hr /> <div>yeet</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=1978_IMO_Problems/Problem_4&diff=133737 1978 IMO Problems/Problem 4 2020-09-17T03:11:32Z <p>Blehlivesonearth: Created page with &quot;yeet&quot;</p> <hr /> <div>yeet</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=1978_IMO_Problems/Problem_5&diff=133736 1978 IMO Problems/Problem 5 2020-09-17T03:11:12Z <p>Blehlivesonearth: Created page with &quot;yeet&quot;</p> <hr /> <div>yeet</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_17&diff=132162 2019 AMC 12B Problems/Problem 17 2020-08-19T20:26:44Z <p>Blehlivesonearth: /* Solution 4 (Quick and Easy) */</p> <hr /> <div>==Problem==<br /> <br /> How many nonzero complex numbers &lt;math&gt;z&lt;/math&gt; have the property that &lt;math&gt;0, z,&lt;/math&gt; and &lt;math&gt;z^3,&lt;/math&gt; when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\text{infinitely many}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Convert &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;z^3&lt;/math&gt; into modulus-argument (polar) form, giving &lt;math&gt;z=r\text{cis}(\theta)&lt;/math&gt; for some &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;\theta&lt;/math&gt;. Thus, by De Moivre's Theorem, &lt;math&gt;z^3=r^3\text{cis}(3\theta)&lt;/math&gt;. Since the distance from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;r&lt;/math&gt;, and the triangle is equilateral, the distance from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;z^3&lt;/math&gt; must also be &lt;math&gt;r&lt;/math&gt;, so &lt;math&gt;r^3=r&lt;/math&gt;, giving &lt;math&gt;r=1&lt;/math&gt;. (We know &lt;math&gt;r \neq 0&lt;/math&gt; since the problem statement specifies that &lt;math&gt;z&lt;/math&gt; must be nonzero.)<br /> <br /> Now, to get from &lt;math&gt;z&lt;/math&gt; to &lt;math&gt;z^3&lt;/math&gt;, which should be a rotation of &lt;math&gt;120^{\circ}&lt;/math&gt; if the triangle is equilateral, we multiply by &lt;math&gt;z^2 = r^2\text{cis}(2\theta)&lt;/math&gt;, again using De Moivre's Theorem. Thus we require &lt;math&gt;2\theta=\pm\frac{\pi}{3} + 2\pi k&lt;/math&gt; (where &lt;math&gt;k&lt;/math&gt; can be any integer). If &lt;math&gt;0 &lt; \theta &lt; \frac{\pi}{2}&lt;/math&gt;, we must have &lt;math&gt;\theta=\frac{\pi}{6}&lt;/math&gt;, while if &lt;math&gt;\frac{\pi}{2} \leq \theta &lt; \pi&lt;/math&gt;, we must have &lt;math&gt;\theta = \frac{5\pi}{6}&lt;/math&gt;. Hence there are &lt;math&gt;2&lt;/math&gt; values that work for &lt;math&gt;0 &lt; \theta &lt; \pi&lt;/math&gt;. By symmetry, the interval &lt;math&gt;\pi \leq \theta &lt; 2\pi&lt;/math&gt; will also give &lt;math&gt;2&lt;/math&gt; solutions. The answer is thus &lt;math&gt;2 + 2 = \boxed{\textbf{(D) }4}&lt;/math&gt;.<br /> <br /> ''Note'': Here's a graph showing how &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;z^3&lt;/math&gt; move as &lt;math&gt;\theta&lt;/math&gt; increases: https://www.desmos.com/calculator/xtnpzoqkgs.<br /> <br /> ==Solution 2 (Quick Look)==<br /> As before, &lt;math&gt;r=1&lt;/math&gt;. Represent &lt;math&gt;z&lt;/math&gt; in polar form. By De Moivre's Theorem, &lt;math&gt;z^3=\text{cis}(3\theta)&lt;/math&gt;. To form an equilateral triangle, their difference in angle must be &lt;math&gt;\frac{\pi}{3}&lt;/math&gt;, so <br /> &lt;cmath&gt;\frac{\text{cis}(3\theta)}{\text{cis}(\theta)}=\text{cis}(2\theta)=\text{cis}(\pm\frac{\pi}{3})&lt;/cmath&gt; <br /> From the polar form of &lt;math&gt;z&lt;/math&gt;, we know that &lt;math&gt;0\geq\theta\leq2\pi&lt;/math&gt;, so &lt;math&gt;\text{cis}(2\theta)&lt;/math&gt; cycles in a circle twice. By contrast, &lt;math&gt;\pm\frac{\pi}{3}&lt;/math&gt; represent &lt;math&gt;2&lt;/math&gt; fixed, distinct points. Thus, &lt;math&gt;\text{cis}(2\theta)&lt;/math&gt; intersects these points twice each&lt;math&gt;\implies\boxed{\textbf{(D) }4}&lt;/math&gt;<br /> <br /> ''Visual: https://www.desmos.com/calculator/rnpxzns0jn''<br /> <br /> <br /> To be more rigorous, you can find the &lt;math&gt;4&lt;/math&gt; solutions. &lt;math&gt;\text{cis}(2\theta)&lt;/math&gt; cycles twice, so &lt;math&gt;2\theta=\pm\frac{\pi}{3}+2n\pi&lt;/math&gt;, where &lt;math&gt;n=0,1&lt;/math&gt;. Then, &lt;math&gt;\theta=\frac{\pi}{6}&lt;/math&gt;, &lt;math&gt;\frac{7\pi}{6}&lt;/math&gt;, &lt;math&gt;\frac{5\pi}{6}&lt;/math&gt;, &lt;math&gt;\frac{11\pi}{6}&lt;/math&gt;. Substitute those values into &lt;math&gt;z&lt;/math&gt; and check that they are valid. &lt;math&gt;\implies\boxed{\textbf{(D) }4}&lt;/math&gt;<br /> <br /> (Solution by BJHHar)<br /> <br /> ==Solution 3==<br /> <br /> For the triangle to be equilateral, the vector from &lt;math&gt;z&lt;/math&gt; to &lt;math&gt;z^3&lt;/math&gt;, i.e &lt;math&gt;z^3 - z&lt;/math&gt;, must be a &lt;math&gt;60^{\circ}&lt;/math&gt; rotation of the vector from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;z&lt;/math&gt;, i.e. just &lt;math&gt;z&lt;/math&gt;. Thus we must have<br /> <br /> &lt;cmath&gt;\frac{(z^3-z)}{(z-0)}=\text{cis}{(\pi/3)} \text{ or } \text{cis}(5\pi/3)&lt;/cmath&gt;<br /> <br /> Simplifying gives <br /> &lt;cmath&gt;z^2-1= \text{cis}(\pi/3) \text{ or } z^2-1= \text{cis}(5\pi/3)&lt;/cmath&gt;<br /> so<br /> &lt;cmath&gt;z^2=1+\text{cis}(\pi/3) \text{ or } z^2=1+\text{cis}(5\pi/3)&lt;/cmath&gt;<br /> <br /> Since any nonzero complex number will have two square roots, each equation gives two solutions. Thus, as before, the total number of possible values of &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(D) }4}&lt;/math&gt;.<br /> <br /> ==Solution 4 (Quick and Easy) ==<br /> <br /> Since the complex numbers &lt;math&gt;0,z,&lt;/math&gt; and &lt;math&gt;z^3&lt;/math&gt; form an equilateral triangle in the complex plane, we note that either &lt;math&gt;z^3&lt;/math&gt; is a 60 degrees counterclockwise rotation about the origin from &lt;math&gt;z&lt;/math&gt; or &lt;math&gt;z&lt;/math&gt; is a 60 degrees counterclockwise rotation about the origin from &lt;math&gt;z^3&lt;/math&gt;.<br /> <br /> Therefore, we note that either &lt;math&gt;z^3 = z \text{cis} 60^\circ{}&lt;/math&gt; or &lt;math&gt;z \text{cis}(-60^\circ{}) = z^3&lt;/math&gt;<br /> <br /> The first equation in &lt;math&gt;z&lt;/math&gt; (meaning &lt;math&gt;z^3 = z \text{cis} 60^\circ{}&lt;/math&gt;) gives us: &lt;math&gt;z^2 = cis 60^\circ{}&lt;/math&gt;, which gives 2 solutions in &lt;math&gt;z&lt;/math&gt;.<br /> <br /> The second equation in &lt;math&gt;z&lt;/math&gt; (which is &lt;math&gt;z \text{cis} (-60^\circ{}) = z^3&lt;/math&gt;) gives us &lt;math&gt;z^2 = (\text{cis} -60^\circ{})&lt;/math&gt;, which must give another 2 solutions in &lt;math&gt;z&lt;/math&gt;.<br /> <br /> Therefore, there are &lt;math&gt;\boxed{(D) 4}&lt;/math&gt; solutions for &lt;math&gt;z&lt;/math&gt;. (Professor-Mom) (Minor edit: mathbw225<br /> <br /> Note: The motivation for this method came from an older AIME problem, namely https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8.<br /> <br /> ==Video Solution==<br /> <br /> For those who prefer a video: https://www.youtube.com/watch?v=uBL80yd1ihc<br /> ==See Also==<br /> {{AMC12 box|year=2019|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_25&diff=132160 2019 AMC 12B Problems/Problem 25 2020-08-19T19:36:12Z <p>Blehlivesonearth: /* Solution 1 (vectors) */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral with &lt;math&gt;BC=2&lt;/math&gt; and &lt;math&gt;CD=6.&lt;/math&gt; Suppose that the centroids of &lt;math&gt;\triangle ABC,\triangle BCD,&lt;/math&gt; and &lt;math&gt;\triangle ACD&lt;/math&gt; form the vertices of an equilateral triangle. What is the maximum possible value of the area of &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 27 \qquad\textbf{(B) } 16\sqrt3 \qquad\textbf{(C) } 12+10\sqrt3 \qquad\textbf{(D) } 9+12\sqrt3 \qquad\textbf{(E) } 30&lt;/math&gt;<br /> <br /> ==Solution 1 (vectors)==<br /> Place an origin at &lt;math&gt;A&lt;/math&gt;, and assign position vectors of &lt;math&gt;B = \vec{p}&lt;/math&gt; and &lt;math&gt;D = \vec{q}&lt;/math&gt;. Since &lt;math&gt;AB&lt;/math&gt; is not parallel to &lt;math&gt;AD&lt;/math&gt;, vectors &lt;math&gt;\vec{p}&lt;/math&gt; and &lt;math&gt;\vec{q}&lt;/math&gt; are linearly independent, so we can write &lt;math&gt;C = m\vec{p} + n\vec{q}&lt;/math&gt; for some constants &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. Now, recall that the centroid of a triangle &lt;math&gt;\triangle XYZ&lt;/math&gt; has position vector &lt;math&gt;\frac{1}{3}\left(\vec{x}+\vec{y}+\vec{z}\right)&lt;/math&gt;. <br /> <br /> Thus the centroid of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;g_1 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}n\vec{q}&lt;/math&gt;; the centroid of &lt;math&gt;\triangle BCD&lt;/math&gt; is &lt;math&gt;g_2 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}(n+1)\vec{q}&lt;/math&gt;; and the centroid of &lt;math&gt;\triangle ACD&lt;/math&gt; is &lt;math&gt;g_3 = \frac{1}{3}m\vec{p} + \frac{1}{3}(n+1)\vec{q}&lt;/math&gt;. <br /> <br /> Hence &lt;math&gt;\overrightarrow{G_{1}G_{2}} = \frac{1}{3}\vec{q}&lt;/math&gt;, &lt;math&gt;\overrightarrow{G_{2}G_{3}} = -\frac{1}{3}\vec{p}&lt;/math&gt;, and &lt;math&gt;\overrightarrow{G_{3}G_{1}} = \frac{1}{3}\vec{p} - \frac{1}{3}\vec{q}&lt;/math&gt;. For &lt;math&gt;\triangle G_{1}G_{2}G_{3}&lt;/math&gt; to be equilateral, we need &lt;math&gt;\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{2}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{q}\right| \Rightarrow AB = AD&lt;/math&gt;. Further, &lt;math&gt;\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{1}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{p} - \vec{q}\right| = BD&lt;/math&gt;. Hence we have &lt;math&gt;AB = AD = BD&lt;/math&gt;, so &lt;math&gt;\triangle ABD&lt;/math&gt; is equilateral.<br /> <br /> Now let the side length of &lt;math&gt;\triangle ABD&lt;/math&gt; be &lt;math&gt;k&lt;/math&gt;, and let &lt;math&gt;\angle BCD = \theta&lt;/math&gt;. By the Law of Cosines in &lt;math&gt;\triangle BCD&lt;/math&gt;, we have &lt;math&gt;k^2 = 2^2 + 6^2 - 2 \cdot 2 \cdot 6 \cdot \cos{\theta} = 40 - 24\cos{\theta}&lt;/math&gt;. Since &lt;math&gt;\triangle ABD&lt;/math&gt; is equilateral, its area is &lt;math&gt;\frac{\sqrt{3}}{4}k^2 = 10\sqrt{3} - 6\sqrt{3}\cos{\theta}&lt;/math&gt;, while the area of &lt;math&gt;\triangle BCD&lt;/math&gt; is &lt;math&gt;\frac{1}{2} \cdot 2 \cdot 6 \cdot \sin{\theta} = 6 \sin{\theta}&lt;/math&gt;. Thus the total area of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;10\sqrt{3} + 6\left(\sin{\theta} - \sqrt{3}\cos{\theta}\right) = 10\sqrt{3} + 12\left(\frac{1}{2} \sin{\theta} - \frac{\sqrt{3}}{2}\cos{\theta}\right) = 10\sqrt{3}+12\sin{\left(\theta-60^{\circ}\right)}&lt;/math&gt;, where in the last step we used the subtraction formula for &lt;math&gt;\sin&lt;/math&gt;. Alternatively, we can use calculus to find the local maximum. Observe that &lt;math&gt;\sin{\left(\theta-60^{\circ}\right)}&lt;/math&gt; has maximum value &lt;math&gt;1&lt;/math&gt; when e.g. &lt;math&gt;\theta = 150^{\circ}&lt;/math&gt;, which is a valid configuration, so the maximum area is &lt;math&gt;10\sqrt{3} + 12(1) = \boxed{\textbf{(C) } 12+10\sqrt3}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Let &lt;math&gt;G_1&lt;/math&gt;, &lt;math&gt;G_2&lt;/math&gt;, &lt;math&gt;G_3&lt;/math&gt; be the centroids of &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;BCD&lt;/math&gt;, and &lt;math&gt;CDA&lt;/math&gt; respectively, and let &lt;math&gt;M&lt;/math&gt; be the midpoint of &lt;math&gt;BC&lt;/math&gt;. &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;G_1&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; are collinear due to well-known properties of the centroid. Likewise, &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;G_2&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; are collinear as well. Because (as is also well-known) &lt;math&gt;3MG_1 = AM&lt;/math&gt; and &lt;math&gt;3MG_2 = DM&lt;/math&gt;, we have &lt;math&gt;\triangle MG_1G_2\sim\triangle MAD&lt;/math&gt;. This implies that &lt;math&gt;AD&lt;/math&gt; is parallel to &lt;math&gt;G_1G_2&lt;/math&gt;, and in terms of lengths, &lt;math&gt;AD = 3G_1G_2&lt;/math&gt;. (SAS Similarity)<br /> <br /> We can apply the same argument to the pair of triangles &lt;math&gt;\triangle BCD&lt;/math&gt; and &lt;math&gt;\triangle ACD&lt;/math&gt;, concluding that &lt;math&gt;AB&lt;/math&gt; is parallel to &lt;math&gt;G_2G_3&lt;/math&gt; and &lt;math&gt;AB = 3G_2G_3&lt;/math&gt;. Because &lt;math&gt;3G_1G_2 = 3G_2G_3&lt;/math&gt; (due to the triangle being equilateral), &lt;math&gt;AB = AD&lt;/math&gt;, and the pair of parallel lines preserve the &lt;math&gt;60^{\circ}&lt;/math&gt; angle, meaning &lt;math&gt;\angle BAD = 60^\circ&lt;/math&gt;. Therefore &lt;math&gt;\triangle BAD&lt;/math&gt; is equilateral.<br /> <br /> At this point, we can finish as in Solution 1, or, to avoid using trigonometry, we can continue as follows:<br /> <br /> Let &lt;math&gt;BD = 2x&lt;/math&gt;, where &lt;math&gt;2 &lt; x &lt; 4&lt;/math&gt; due to the Triangle Inequality in &lt;math&gt;\triangle BCD&lt;/math&gt;. By breaking the quadrilateral into &lt;math&gt;\triangle ABD&lt;/math&gt; and &lt;math&gt;\triangle BCD&lt;/math&gt;, we can create an expression for the area of &lt;math&gt;ABCD&lt;/math&gt;. We use the formula for the area of an equilateral triangle given its side length to find the area of &lt;math&gt;\triangle ABD&lt;/math&gt; and Heron's formula to find the area of &lt;math&gt;\triangle BCD&lt;/math&gt;.<br /> <br /> After simplifying,<br /> <br /> &lt;cmath&gt;[ABCD] = x^2\sqrt 3 + \sqrt{36 - (x^2-10)^2}&lt;/cmath&gt;<br /> <br /> Substituting &lt;math&gt;k = x^2 - 10&lt;/math&gt;, the expression becomes<br /> <br /> &lt;cmath&gt;[ABCD] = k\sqrt{3} + \sqrt{36 - k^2} + 10\sqrt{3}&lt;/cmath&gt;<br /> <br /> We can ignore the &lt;math&gt;10\sqrt{3}&lt;/math&gt; for now and focus on &lt;math&gt;k\sqrt{3} + \sqrt{36 - k^2}&lt;/math&gt;.<br /> <br /> By the Cauchy-Schwarz inequality,<br /> <br /> &lt;cmath&gt;\left(k\sqrt 3 + \sqrt{36-k^2}\right)^2 \leq \left(\left(\sqrt{3}\right)^2+1^2\right)\left(\left(k\right)^2 + \left(\sqrt{36-k^2}\right)^2\right)&lt;/cmath&gt;<br /> <br /> The RHS simplifies to &lt;math&gt;12^2&lt;/math&gt;, meaning the maximum value of &lt;math&gt;k\sqrt{3} + \sqrt{36 - k^2}&lt;/math&gt; is &lt;math&gt;12&lt;/math&gt;.<br /> <br /> Thus the maximum possible area of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C) }12 + 10\sqrt{3}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2019|ab=B|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_17&diff=132067 2019 AMC 12A Problems/Problem 17 2020-08-18T20:53:50Z <p>Blehlivesonearth: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;s_k&lt;/math&gt; denote the sum of the &lt;math&gt;\textit{k}&lt;/math&gt;th powers of the roots of the polynomial &lt;math&gt;x^3-5x^2+8x-13&lt;/math&gt;. In particular, &lt;math&gt;s_0=3&lt;/math&gt;, &lt;math&gt;s_1=5&lt;/math&gt;, and &lt;math&gt;s_2=9&lt;/math&gt;. Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; be real numbers such that &lt;math&gt;s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}&lt;/math&gt; for &lt;math&gt;k = 2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;....&lt;/math&gt; What is &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Applying Newton's Sums (see [https://artofproblemsolving.com/wiki/index.php/Newton%27s_Sums this link]), we have&lt;cmath&gt;s_{k+1}+(-5)s_k+(8)s_{k-1}+(-13)s_{k-2}=0,&lt;/cmath&gt;so&lt;cmath&gt;s_{k+1}=5s_k-8s_{k-1}+13s_{k-2},&lt;/cmath&gt;we get the answer as &lt;math&gt;5+(-8)+13=10&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Let &lt;math&gt;p, q&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; be the roots of the polynomial. Then,<br /> <br /> &lt;math&gt;p^3 - 5p^2 + 8p - 13 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;q^3 - 5q^2 + 8q - 13 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;r^3 - 5r^2 + 8r - 13 = 0&lt;/math&gt;<br /> <br /> Adding these three equations, we get<br /> <br /> &lt;math&gt;(p^3 + q^3 + r^3) - 5(p^2 + q^2 + r^2) + 8(p + q + r) - 39 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;s_3 - 5s_2 + 8s_1 = 39&lt;/math&gt;<br /> <br /> &lt;math&gt;39&lt;/math&gt; can be written as &lt;math&gt;13s_0&lt;/math&gt;, giving<br /> <br /> &lt;math&gt;s_3 = 5s_2 - 8s_1 + 13s_0&lt;/math&gt;<br /> <br /> We are given that &lt;math&gt;s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}&lt;/math&gt; is satisfied for &lt;math&gt;k = 2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;....&lt;/math&gt;, meaning it must be satisfied when &lt;math&gt;k = 2&lt;/math&gt;, giving us &lt;math&gt;s_3 = a \, s_2 + b \, s_1 + c \, s_0&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;a = 5, b = -8&lt;/math&gt;, and &lt;math&gt;c = 13&lt;/math&gt; by matching coefficients.<br /> <br /> &lt;math&gt;5 - 8 + 13 = \boxed{\textbf{(D) } 10}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;p, q&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; be the roots of the polynomial. By Viète's Formulae, we have<br /> <br /> &lt;math&gt;p+q+r = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;pq+qr+rp = 8&lt;/math&gt;<br /> <br /> &lt;math&gt;pqr=13&lt;/math&gt;.<br /> <br /> We know &lt;math&gt;s_k = p^k + q^k + r^k&lt;/math&gt;. Consider &lt;math&gt;(p+q+r)(s_k) =5s_k&lt;/math&gt;.<br /> <br /> &lt;math&gt;5s_k = [p^{k+1} + q^{k+1} + r^{k+1}] + p^k q + p^k r + pq^k + q^k r + pr^k + qr^k&lt;/math&gt;<br /> <br /> Using &lt;math&gt;pqr = 13&lt;/math&gt; and &lt;math&gt;s_{k-2} = p^{k-2} + q^{k-2} + r^{k-2}&lt;/math&gt;, we see<br /> &lt;math&gt;13s_{k-2} = p^{k-1}qr + pq^{k-1}r + pqr^{k-1}&lt;/math&gt;.<br /> <br /> We have &lt;cmath&gt;\begin{split} 5s_k + 13s_{k-2} &amp;= s_{k+1} + (p^k q + p^k r + p^{k-1}qr) + (pq^k + pq^{k-1}r + q^k r) + (pqr^{k-1} + pr^k + qr^k) \\<br /> &amp;= s_{k+1} + p^{k-1} (pq + pr + qr) + q^{k-1} (pq + pr + qr) + r^{k-1} (pq + pr + qr) \\<br /> &amp;= s_{k+1} + (p^{k-1} + q^{k-1} + r^{k-1})(pq + pr + qr) \\<br /> &amp;= 5s_k + 13s_{k-2} = s_{k+1} + 8s_{k-1}\end{split}&lt;/cmath&gt;<br /> <br /> Rearrange to get<br /> &lt;math&gt;s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2}&lt;/math&gt;<br /> <br /> So, &lt;math&gt;a+ b + c = 5 -8 + 13 = \boxed{\textbf{(D) } 10}&lt;/math&gt;.<br /> <br /> -gregwwl<br /> <br /> ==Video Solution==<br /> For those who want a video solution: https://www.youtube.com/watch?v=tAS_DbKmtzI<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2019|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_25&diff=132062 2019 AMC 10A Problems/Problem 25 2020-08-18T19:17:03Z <p>Blehlivesonearth: /* See Also */</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #25]] and [[2019 AMC 12A Problems|2019 AMC 12A #24]]}}<br /> <br /> ==Problem==<br /> <br /> For how many integers &lt;math&gt;n&lt;/math&gt; between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;50&lt;/math&gt;, inclusive, is &lt;cmath&gt;\frac{(n^2-1)!}{(n!)^n}&lt;/cmath&gt; an integer? (Recall that &lt;math&gt;0! = 1&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> The main insight is that <br /> <br /> &lt;cmath&gt;\frac{(n^2)!}{(n!)^{n+1}}&lt;/cmath&gt; <br /> <br /> is always an integer. This is true because it is precisely the number of ways to split up &lt;math&gt;n^2&lt;/math&gt; objects into &lt;math&gt;n&lt;/math&gt; unordered groups of size &lt;math&gt;n&lt;/math&gt;. Thus,<br /> <br /> &lt;cmath&gt;\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}&lt;/cmath&gt;<br /> <br /> is an integer if &lt;math&gt;n^2 \mid n!&lt;/math&gt;, or in other words, if &lt;math&gt;n \mid (n-1)!&lt;/math&gt;. This condition is false precisely when &lt;math&gt;n=4&lt;/math&gt; or &lt;math&gt;n&lt;/math&gt; is prime, by [[Wilson's Theorem]]. There are &lt;math&gt;15&lt;/math&gt; primes between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;50&lt;/math&gt;, inclusive, so there are &lt;math&gt;15 + 1 = 16&lt;/math&gt; terms for which<br /> <br /> &lt;cmath&gt;\frac{(n^2-1)!}{(n!)^{n}}&lt;/cmath&gt;<br /> <br /> is potentially not an integer. It can be easily verified that the above expression is not an integer for &lt;math&gt;n=4&lt;/math&gt; as there are more factors of &lt;math&gt;2&lt;/math&gt; in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime &lt;math&gt;n=p&lt;/math&gt;, as there are more factors of p in the denominator than the numerator. Thus all &lt;math&gt;16&lt;/math&gt; values of n make the expression not an integer and the answer is &lt;math&gt;50-16=\boxed{\mathbf{(D)}\ 34}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> We can use the P-Adic Valuation (more info could be found here: [[Mathematicial notation]]) of n to solve this problem (recall the P-Adic Valuation of 'n' is denoted by &lt;math&gt;v_p (n)&lt;/math&gt; and is defined as the greatest power of some prime 'p' that divides n. For example, &lt;math&gt;v_2 (6)=1&lt;/math&gt; or &lt;math&gt;v_7 (245)=2&lt;/math&gt; .) Using Legendre's formula, we know that :<br /> <br /> &lt;cmath&gt; v_p (n!)= \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor &lt;/cmath&gt;<br /> <br /> Seeing factorials involved in the problem, this prompts us to use Legendre's formula where n is a power of a prime.<br /> <br /> We also know that , &lt;math&gt;v_p (m^n) = n \cdot v_p (m)&lt;/math&gt; .<br /> Knowing that &lt;math&gt;a\mid b&lt;/math&gt; if &lt;math&gt;v_p (a) \le v_p (b)&lt;/math&gt; , we have that :<br /> <br /> &lt;cmath&gt; n \cdot v_p (n!) \le v_p ((n^2 -1 )!) &lt;/cmath&gt; and we must find all n for which this is true.<br /> <br /> If we plug in &lt;math&gt;n=p&lt;/math&gt;, by Legendre's we get two equations:<br /> <br /> &lt;cmath&gt; v_p ((n^2 -1)!) = \sum_{i=1}^\infty \lfloor \frac {n^2 -1}{p^i} \rfloor = (p-1)+0+...+0 = p-1 &lt;/cmath&gt;<br /> <br /> And we also get :<br /> <br /> &lt;cmath&gt; v_p ((n!)^n) = n \cdot v_p (n!)= n \cdot \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor = p \cdot ( 1+0+...0) = p &lt;/cmath&gt;<br /> <br /> But we are asked to prove that &lt;math&gt; n \cdot v_p (n!) \le v_p ((n^2 -1 )!) \Longrightarrow p \le p-1 &lt;/math&gt; which is false for all 'n' where n is prime.<br /> <br /> Now we try the same for &lt;math&gt;n=p^2&lt;/math&gt; , where p is a prime. By Legendre we arrive at:<br /> <br /> &lt;cmath&gt;v_p ((p^4 -1)!) = p^3 + p^2 + p -3&lt;/cmath&gt; and &lt;cmath&gt;p^2 \cdot v_p (p^2 !) = p^3 + p^2 &lt;/cmath&gt;<br /> <br /> Then we get:<br /> <br /> &lt;cmath&gt; p^2 \cdot v_p (p!) \le v_p ((n^4 -1)!) \Longrightarrow p^3 + p^2 \le p^3 + p^2 + p -3 &lt;/cmath&gt; Which is true for all primes except for 2, so &lt;math&gt;2^2 = 4&lt;/math&gt; doesn't work. It can easily be verified that for all &lt;math&gt;n=p^i&lt;/math&gt; where &lt;math&gt;i&lt;/math&gt; is an integer greater than 2, satisfies the inequality :&lt;cmath&gt; n \cdot v_p (n!) \le v_p ((n^2 -1 )!).&lt;/cmath&gt;<br /> <br /> Therefore, there are 16 values that don't work and &lt;math&gt; 50-16 = \boxed{\mathbf{(D)}\ 34}&lt;/math&gt; values that work.<br /> <br /> ~qwertysri987<br /> <br /> ==See Also==<br /> Video Solution by Richard Rusczyk:<br /> https://www.youtube.com/watch?v=9klaWnZojq0<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=24|after=Last Problem}}<br /> {{AMC12 box|year=2019|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_18&diff=132046 2019 AMC 10A Problems/Problem 18 2020-08-18T12:03:27Z <p>Blehlivesonearth: /* Solution 5 */</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #18]] and [[2019 AMC 12A Problems|2019 AMC 12A #11]]}}<br /> <br /> ==Problem==<br /> <br /> For some positive integer &lt;math&gt;k&lt;/math&gt;, the repeating base-&lt;math&gt;k&lt;/math&gt; representation of the (base-ten) fraction &lt;math&gt;\frac{7}{51}&lt;/math&gt; is &lt;math&gt;0.\overline{23}_k = 0.232323..._k&lt;/math&gt;. What is &lt;math&gt;k&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 /> ==Solution 1==<br /> <br /> We can expand the fraction &lt;math&gt;0.\overline{23}_k&lt;/math&gt; as follows: &lt;math&gt;0.\overline{23}_k = 2\cdot k^{-1} + 3 \cdot k^{-2} + 2 \cdot k^{-3} + 3 \cdot k^{-4} + ...&lt;/math&gt; Notice that this is equivalent to <br /> &lt;cmath&gt;2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + ... )&lt;/cmath&gt;<br /> <br /> By summing the geometric series and simplifying, we have &lt;math&gt;\frac{2k+3}{k^2-1} = \frac{7}{51}&lt;/math&gt;. Solving this quadratic equation (or simply testing the answer choices) yields the answer &lt;math&gt;k = \boxed{\textbf{(D) }16}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Let &lt;math&gt;a = 0.2323\dots_k&lt;/math&gt;. Therefore, &lt;math&gt;k^2a=23.2323\dots_k&lt;/math&gt;.<br /> <br /> From this, we see that &lt;math&gt;k^2a-a=23_k&lt;/math&gt;, so &lt;math&gt;a = \frac{23_k}{k^2-1} = \frac{2k+3}{k^2-1} = \frac{7}{51}&lt;/math&gt;.<br /> <br /> Now, similar to in Solution 1, we can either test if &lt;math&gt;2k+3&lt;/math&gt; is a multiple of 7 with the answer choices, or actually solve the quadratic, so that the answer is &lt;math&gt;\boxed{\textbf{(D) }16}&lt;/math&gt;.<br /> <br /> ==Solution 3 (bash)==<br /> We can simply plug in all the answer choices as values of &lt;math&gt;k&lt;/math&gt;, and see which one works. After legendary, amazingly, historically great calculations, this eventually gives us &lt;math&gt;\boxed{\textbf{(D) }16}&lt;/math&gt; as the answer.<br /> <br /> -ellpet<br /> <br /> ==Solution 4==<br /> Just as in Solution 1, we arrive at the equation &lt;math&gt;\frac{2k+3}{k^2-1}=\frac{7}{51}&lt;/math&gt;.<br /> <br /> We can now rewrite this as &lt;math&gt;\frac{2k+3}{(k-1)(k+1)}=\frac{7}{51}=\frac{7}{3\cdot 17}&lt;/math&gt;. Notice that &lt;math&gt;2k+3=2(k+1)+1=2(k-1)+5&lt;/math&gt;. As &lt;math&gt;17&lt;/math&gt; is a prime, we therefore must have that one of &lt;math&gt;k-1&lt;/math&gt; and &lt;math&gt;k+1&lt;/math&gt; is divisible by &lt;math&gt;17&lt;/math&gt;. Now, checking each of the answer choices, this gives &lt;math&gt;\boxed{\textbf{(D) }16}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Assuming you are familiar with the rules for basic repeating decimals, &lt;math&gt;0.232323... = \frac{23}{99}&lt;/math&gt;. Now we want our base, &lt;math&gt;k&lt;/math&gt;, to conform to &lt;math&gt;23\equiv7\pmod k&lt;/math&gt; and &lt;math&gt;99\equiv51\pmod k&lt;/math&gt;, the reason being that we wish to convert the number from base &lt;math&gt;10&lt;/math&gt; to base &lt;math&gt;k&lt;/math&gt;. Given the first equation, we know that &lt;math&gt;k&lt;/math&gt; must equal 9, 16, 23, or generally, &lt;math&gt;7n+2&lt;/math&gt;. The only number in this set that is one of the multiple choices is &lt;math&gt;16&lt;/math&gt;. When we test this on the second equation, &lt;math&gt;99\equiv51\pmod k&lt;/math&gt;, it comes to be true. Therefore, our answer is &lt;math&gt;\boxed{\textbf{(D) }16}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> For those who want a video solution: https://www.youtube.com/watch?v=DFfRJolhwN0<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=17|num-a=19}}<br /> {{AMC12 box|year=2019|ab=A|num-b=10|num-a=12}}<br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=1988_USAMO_Problems/Problem_4&diff=131995 1988 USAMO Problems/Problem 4 2020-08-17T23:32:25Z <p>Blehlivesonearth: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> &lt;math&gt;\Delta ABC&lt;/math&gt; is a triangle with incenter &lt;math&gt;I&lt;/math&gt;. Show that the circumcenters of &lt;math&gt;\Delta IAB&lt;/math&gt;, &lt;math&gt;\Delta IBC&lt;/math&gt;, and &lt;math&gt;\Delta ICA&lt;/math&gt; lie on a circle whose center is the circumcenter of &lt;math&gt;\Delta ABC&lt;/math&gt;.<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> Let the circumcenters of &lt;math&gt;\Delta IAB&lt;/math&gt;, &lt;math&gt;\Delta IBC&lt;/math&gt;, and &lt;math&gt;\Delta ICA&lt;/math&gt; be &lt;math&gt;O_c&lt;/math&gt;, &lt;math&gt;O_a&lt;/math&gt;, and &lt;math&gt;O_b&lt;/math&gt;, respectively. It then suffices to show that &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;O_a&lt;/math&gt;, &lt;math&gt;O_b&lt;/math&gt;, and &lt;math&gt;O_c&lt;/math&gt; are concyclic.<br /> <br /> We shall prove that quadrilateral &lt;math&gt;ABO_aC&lt;/math&gt; is cyclic first. Let &lt;math&gt;\angle BAC=\alpha&lt;/math&gt;, &lt;math&gt;\angle CBA=\beta&lt;/math&gt;, and &lt;math&gt;\angle ACB=\gamma&lt;/math&gt;. Then &lt;math&gt;\angle ICB=\gamma/2&lt;/math&gt; and &lt;math&gt;\angle IBC=\beta/2&lt;/math&gt;. Therefore minor arc &lt;math&gt;\overarc{BIC}&lt;/math&gt; in the circumcircle of &lt;math&gt;IBC&lt;/math&gt; has a degree measure of &lt;math&gt;\beta+\gamma&lt;/math&gt;. This shows that &lt;math&gt;\angle CO_aB=\beta+\gamma&lt;/math&gt;, implying that &lt;math&gt;\angle BAC+\angle BO_aC=\alpha+\beta+\gamma=180^{\circ}&lt;/math&gt;. Therefore quadrilateral &lt;math&gt;ABO_aC&lt;/math&gt; is cyclic.<br /> <br /> This shows that point &lt;math&gt;O_a&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta ABC&lt;/math&gt;. Analagous proofs show that &lt;math&gt;O_b&lt;/math&gt; and &lt;math&gt;O_c&lt;/math&gt; are also on the circumcircle of &lt;math&gt;ABC&lt;/math&gt;, which completes the proof. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> Let &lt;math&gt;M&lt;/math&gt; denote the midpoint of arc &lt;math&gt;AC&lt;/math&gt;. It is well known that &lt;math&gt;M&lt;/math&gt; is equidistant from &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;I&lt;/math&gt; (to check, prove &lt;math&gt;\angle IAM = \angle AIM = \frac{\angle BAC + \angle ABC}{2}&lt;/math&gt;), so that &lt;math&gt;M&lt;/math&gt; is the circumcenter of &lt;math&gt;AIC&lt;/math&gt;. Similar results hold for &lt;math&gt;BIC&lt;/math&gt; and &lt;math&gt;CIA&lt;/math&gt;, and hence &lt;math&gt;O_c&lt;/math&gt;, &lt;math&gt;O_a&lt;/math&gt;, and &lt;math&gt;O_b&lt;/math&gt; all lie on the circumcircle of &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> Extend &lt;math&gt;CI&lt;/math&gt; to point &lt;math&gt;L&lt;/math&gt; on &lt;math&gt;(ABC)&lt;/math&gt;. By The Incenter-Excenter Lemma, B, I, A are all concyclic. Thus, L is the circumcenter of triangle &lt;math&gt;IAB&lt;/math&gt;. In other words, &lt;math&gt;L=O_c&lt;/math&gt;, so &lt;math&gt;O_c&lt;/math&gt; is on &lt;math&gt;(ABC)&lt;/math&gt;. Similarly, we can show that &lt;math&gt;O_a&lt;/math&gt; and &lt;math&gt;O_b&lt;/math&gt; are on &lt;math&gt;(ABC)&lt;/math&gt;, and thus, &lt;math&gt;A,B,C,O_a,O_b,O_c&lt;/math&gt; are all concyclic. It follows that the circumcenters are equal.<br /> <br /> ==See Also==<br /> {{USAMO box|year=1988|num-b=3|num-a=5}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Olympiad Geometry Problems]]</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_17&diff=131966 2020 AMC 10B Problems/Problem 17 2020-08-17T05:06:39Z <p>Blehlivesonearth: /* Solution */</p> <hr /> <div>{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #17]] and [[2020 AMC 12B Problems|2020 AMC 12B #15]]}}<br /> <br /> ==Problem==<br /> <br /> There are &lt;math&gt;10&lt;/math&gt; people standing equally spaced around a circle. Each person knows exactly &lt;math&gt;3&lt;/math&gt; of the other &lt;math&gt;9&lt;/math&gt; people: the &lt;math&gt;2&lt;/math&gt; people standing next to her or him, as well as the person directly across the circle. How many ways are there for the &lt;math&gt;10&lt;/math&gt; people to split up into &lt;math&gt;5&lt;/math&gt; pairs so that the members of each pair know each other?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15&lt;/math&gt;<br /> <br /> ==Solution==<br /> Consider the 10 people to be standing in a circle, where two people opposite each other form a diameter of the circle.<br /> <br /> Let us use casework on the number of diameters.<br /> <br /> Case 1: &lt;math&gt;0&lt;/math&gt; diameters<br /> <br /> There are &lt;math&gt;2&lt;/math&gt; ways: either &lt;math&gt;1&lt;/math&gt; pairs with &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt; pairs with &lt;math&gt;4&lt;/math&gt;, and so on or &lt;math&gt;10&lt;/math&gt; pairs with &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt; pairs with &lt;math&gt;3&lt;/math&gt;, etc.<br /> <br /> Case 2: &lt;math&gt;1&lt;/math&gt; diameter<br /> <br /> There are &lt;math&gt;5&lt;/math&gt; possible diameters to draw (everyone else pairs with the person next to them).<br /> <br /> Note that there cannot be &lt;math&gt;2&lt;/math&gt; diameters since there would be one person on either side that will not have a pair adjacent to them. The only scenario forced is when the two people on either side would be paired up across a diameter. Thus, a contradiction will arise.<br /> <br /> Case 3: &lt;math&gt;3&lt;/math&gt; diameters<br /> <br /> There are &lt;math&gt;5&lt;/math&gt; possible diameters to draw. <br /> <br /> Note that there cannot be a case with &lt;math&gt;4&lt;/math&gt; diameters because then there would have to be &lt;math&gt;5&lt;/math&gt; diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises.<br /> <br /> Case 4: &lt;math&gt;5&lt;/math&gt; diameters<br /> <br /> There is only &lt;math&gt;1&lt;/math&gt; way to do this.<br /> <br /> Thus, in total there are &lt;math&gt;2+5+5+1=\boxed{13}&lt;/math&gt; possible ways.<br /> <br /> ==Video Solution==<br /> https://youtu.be/3BvJeZU3T-M?t=419 (for AMC 10)<br /> https://youtu.be/0xgTR3UEqbQ (for AMC 12)<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=16|num-a=18}}<br /> {{AMC12 box|year=2020|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Blehlivesonearth https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_23&diff=131965 2020 AMC 12B Problems/Problem 23 2020-08-17T03:33:17Z <p>Blehlivesonearth: /* Solution */</p> <hr /> <div><br /> <br /> ==Problem 23==<br /> <br /> How many integers &lt;math&gt;n \geq 2&lt;/math&gt; are there such that whenever &lt;math&gt;z_1, z_2, ..., z_n&lt;/math&gt; are complex numbers such that<br /> <br /> &lt;cmath&gt;|z_1| = |z_2| = ... = |z_n| = 1 \text{ and } z_1 + z_2 + ... + z_n = 0,&lt;/cmath&gt;<br /> then the numbers &lt;math&gt;z_1, z_2, ..., z_n&lt;/math&gt; are equally spaced on the unit circle in the complex plane?<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 /> [[2020 AMC 12B Problems/Problem 23|Solution]]<br /> <br /> ==Solution==<br /> <br /> For &lt;math&gt;n=2&lt;/math&gt;, we see that if &lt;math&gt;z_{1}+z_{2}=0&lt;/math&gt;, then &lt;math&gt;z_{1}=-z_{2}&lt;/math&gt;, so they are evenly spaced along the unit circle.<br /> <br /> For &lt;math&gt;n=3&lt;/math&gt;, WLOG, we can set &lt;math&gt;z_{1}=1&lt;/math&gt;. Notice that now &lt;math&gt;\Re(z_{2}+z_{3})=-1&lt;/math&gt; and &lt;math&gt;\Im\{z_{2}\}=-\Im\{z_{3}\}&lt;/math&gt;. This forces &lt;math&gt;z_{2}&lt;/math&gt; and &lt;math&gt;z_{3}&lt;/math&gt; to be equal to &lt;math&gt;e^{i\frac{2\pi}{3}}&lt;/math&gt; and &lt;math&gt;e^{-i\frac{2\pi}{3}}&lt;/math&gt;, meaning that all three are equally spaced along the unit circle.<br /> <br /> We can now show that we can construct complex numbers when &lt;math&gt;n\geq 4&lt;/math&gt; that do not satisfy the conditions in the problem.<br /> <br /> Suppose that the condition in the problem holds for some &lt;math&gt;n=k&lt;/math&gt;. We can now add two points &lt;math&gt;z_{k+1}&lt;/math&gt; and &lt;math&gt;z_{k+2}&lt;/math&gt; anywhere on the unit circle such that &lt;math&gt;z_{k+1}=-z_{k+2}&lt;/math&gt;, which will break the condition. Now that we have shown that &lt;math&gt;n=2&lt;/math&gt; and &lt;math&gt;n=3&lt;/math&gt; works, by this construction, any &lt;math&gt;n\geq 4&lt;/math&gt; does not work, making the answer &lt;math&gt;\boxed{\textbf{(B)} 2}&lt;/math&gt;.<br /> <br /> -Solution by Qqqwerw<br /> <br /> ==Solution 1.5==<br /> (for people not used to notation)<br /> <br /> Note: I personally don't think the question was very well worded; it should explicitly state that if the first two conditions/equations are satisfied, then the final condition MUST be satisfied as well. The question can be misinterpreted as &quot;how many different sizes of a list of complex numbers exist such that at least one satisfies the conditions&quot; (in which case the answer would be infinitely many).<br /> <br /> The problem asks for the number of list lengths that satisfy the three conditions given; i.e. how many different sizes can the list of complex numbers be such that the elements in the list always satisfy the conditions. This means that for a list of length 3, for example, all possible lists have to satisfy the conditions. If even one variation of a list of length 3 does not satisfy the conditions, then 3 is not an acceptable value of n.<br /> <br /> A complex number can be written in the form &lt;math&gt;a + bi&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; represents the real part and &lt;math&gt;b&lt;/math&gt; represents the coefficient of the imaginary part. If a complex number is plotted on the complex plane, then &lt;math&gt;a&lt;/math&gt; becomes the x-coordinate, and &lt;math&gt;b&lt;/math&gt; becomes the y-coordinate.<br /> <br /> The first condition states that the absolute value of each complex number is equal to 1; i.e. the point (a, b) is a distance of one from the origin. If all of the complex numbers are a distance of 1 from the origin, they must all be on the unit circle. This condition simply allows us to be able to check condition 3, think of it as a sort of &quot;given&quot;.<br /> <br /> The second condition states that the sum of all the complex numbers is equal to 0. Therefore, the sum of all the x-coordinates and the y-coordinates will yield 0. However, we know that real numbers and complex numbers cannot be simplified together (assuming that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are real numbers). This means that the real parts of the complex numbers, i.e. the x-coordinates, must sum to 0 by themselves. The same is true for the imaginary parts (the y-coordinates).<br /> <br /> The third condition states that the complex numbers must be evenly spaced along the unit circle. This condition can be used to determine acceptable values of x.<br /> <br /> Start with &lt;math&gt;n = 2&lt;/math&gt;. If we have 2 complex numbers, in order to satisfy the second condition, the x coordinates must be opposites. The same is true for the y-coordinates. After drawing (or imagining) we find that the two points are directly across from one another on the unit circle. Upon further thought, we find that whenever we have an even number of points, if they are paired up in this way, they will always satisfy the first and second conditions. However, it is possible to arrange these pairs of points (whenever we have at least 2 pairs) such as the points are not evenly spaced out. Therefore, any even value of &lt;math&gt;n&lt;/math&gt; greater than 2 will not work (since we proved that not all variations satisfy the last condition).<br /> <br /> Now we consider &lt;math&gt;n = 3&lt;/math&gt;. It is much easier to consider 2 numbers instead of 3, so we will assume that &lt;math&gt;z_{1}=1&lt;/math&gt;, or &lt;math&gt;(1, 0)&lt;/math&gt; (we can do this because a circle has rotational symmetry). Since the y-coordinate of the first number is 0, the y-coordinates of the other 2 numbers must be opposites. Since the x coordinate of the first number is 1, the x-coordinates of the other 2 numbers must sum to -1. Using this knowledge, we find that the coordinates of the other 2 numbers are<br /> <br /> &lt;math&gt;-\frac{1}{2}&lt;/math&gt; &lt;math&gt;+&lt;/math&gt; &lt;math&gt;\sqrt{3}&lt;/math&gt; and &lt;math&gt;-\frac{1}{2}&lt;/math&gt; &lt;math&gt;-&lt;/math&gt; &lt;math&gt;\sqrt{3}&lt;/math&gt;<br /> <br /> After calculating angle measurements, we see that the 3 points are, indeed, equally spaced apart. Therefore 3 is an acceptable value of &lt;math&gt;n&lt;/math&gt; (all variations of &lt;math&gt;n = 3&lt;/math&gt; are simply rotations of this specific case, and therefore satisfy all of the conditions).<br /> <br /> Lastly we consider the odd numbers. If we start with the same arrangement of points as above, we see that adding pairs of opposite points will continue to satisfy conditions 1 and 2, but NOT ALWAYS condition 3. Therefore, any odd number greater than 3 is not<br /> an acceptable value of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> So we have &lt;math&gt;n = 2&lt;/math&gt; and &lt;math&gt;n = 3&lt;/math&gt; for a total of 2 possibilities &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;\boxed{B}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> On The Spot STEM: https://www.youtube.com/watch?v=JOgSOni5HhM<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Blehlivesonearth