https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ivanthekingiii&feedformat=atom AoPS Wiki - User contributions [en] 2020-10-21T18:33:42Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_20&diff=90100 2016 AMC 10B Problems/Problem 20 2018-01-31T03:26:55Z <p>Ivanthekingiii: /* Solution 2: Geometric */</p> <hr /> <div>==Problem==<br /> <br /> A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius &lt;math&gt;2&lt;/math&gt; centered at &lt;math&gt;A(2,2)&lt;/math&gt; to the circle of radius &lt;math&gt;3&lt;/math&gt; centered at &lt;math&gt;A’(5,6)&lt;/math&gt;. What distance does the origin &lt;math&gt;O(0,0)&lt;/math&gt;, move under this transformation?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> ==Solution 1: Algebraic==<br /> The center of dilation must lie on the line &lt;math&gt;A A'&lt;/math&gt;, which can be expressed &lt;math&gt;y = \dfrac{4x}{3} - \dfrac{2}{3}&lt;/math&gt;. Also, the ratio of dilation must be equal to &lt;math&gt;\dfrac{3}{2}&lt;/math&gt;, which is the ratio of the radii of the circles. Thus, we are looking for a point &lt;math&gt;(x,y)&lt;/math&gt; such that &lt;math&gt;\dfrac{3}{2} \left( 2 - x \right) = 5 - x&lt;/math&gt; (for the &lt;math&gt;x&lt;/math&gt;-coordinates), and &lt;math&gt;\dfrac{3}{2} \left( 2 - y \right) = 6 - y&lt;/math&gt;. Solving these, we get &lt;math&gt;x = -4&lt;/math&gt; and &lt;math&gt;y = - 6&lt;/math&gt;. This means that any point &lt;math&gt;(a,b)&lt;/math&gt; on the plane will dilate to the point &lt;math&gt;\left( \dfrac{3}{2} (a + 4) - 4, \dfrac{3}{2} (b + 6) - 6 \right)&lt;/math&gt;, which means that the point &lt;math&gt;(0,0)&lt;/math&gt; dilates to &lt;math&gt;\left( 6 - 4, 9 - 6 \right) = (2,3)&lt;/math&gt;. Thus, the origin moves &lt;math&gt;\sqrt{2^2 + 3^2} = \boxed{\sqrt{13}}&lt;/math&gt; units.<br /> <br /> ==Solution 2: Geometric==<br /> &lt;asy&gt; /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */<br /> /* by adihaya */<br /> import graph; size(13cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -7., xmax = 9., ymin = -7., ymax = 9.6; /* image dimensions */<br /> pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pen qqzzff = rgb(0.,0.6,1.); pen ffwwqq = rgb(1.,0.4,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); <br /> pair O = (0.,0.), A = (2.,2.), B = (2.,0.), C = (3.209655293155585,3.5927128026548525), D = (2.,4.), F = (-3.999634206191805,-5.999512274922407), G = (-3.999634206191812,-5.9995122749224175); <br /> /* by adihaya */<br /> draw((2.482722020656878,0.)--(2.4827220206568783,0.4827220206568779)--(2.,0.48272202065687797)--B--cycle, qqwuqq); <br /> draw((5.482722020656878,0.)--(5.482722020656878,0.48272202065687797)--(5.,0.48272202065687797)--(5.,0.)--cycle, qqwuqq); <br /> Label laxis; laxis.p = fontsize(10); <br /> xaxis(xmin, xmax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); <br /> yaxis(ymin, ymax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ <br /> /* draw figures */<br /> draw(shift(A) * scale(2., 2.)*unitcircle); <br /> draw(shift((5.,6.)) * scale(3.000060969351735, 3.000060969351735)*unitcircle); <br /> draw((xmin, 1.3333333333333333*xmin-0.6666666666666666)--(xmax, 1.3333333333333333*xmax-0.6666666666666666)); /* line */<br /> draw((2.,ymin)--(2.,ymax)); /* line */<br /> draw((5.,ymin)--(5.,ymax)); /* line */<br /> draw((xmin, 1.6666869897839114*xmin + 0.6666260204321771)--(xmax, 1.6666869897839114*xmax + 0.6666260204321771)); /* line */<br /> draw(O--F, qqzzff); <br /> draw(F--A, ffwwqq); <br /> /* dots and labels */<br /> dot(O,blue); <br /> label(&quot;$O$&quot;, (0.08696973475182286,0.23426871275979863), NE * labelscalefactor,blue); <br /> dot(A,blue); <br /> label(&quot;$A$&quot;, (2.089474351594523,2.23677332960249), NE * labelscalefactor,blue); <br /> dot((5.,6.),blue); <br /> label(&quot;$A'$&quot;, (5.093231276858573,6.2190268290055695), NE * labelscalefactor,blue); <br /> dot(B,xdxdff); <br /> label(&quot;$B$&quot;, (2.089474351594523,0.23426871275979863), NE * labelscalefactor,xdxdff); <br /> label(&quot;$c$&quot;, (0.9971991060439592,3.2607813723061394), NE * labelscalefactor); <br /> dot(C,xdxdff); <br /> label(&quot;$C$&quot;, (3.2955282685566036,3.829674729363722), NE * labelscalefactor,xdxdff); <br /> label(&quot;$d$&quot;, (3.477574142815031,8.107752774436745), NE * labelscalefactor); <br /> label(&quot;$a$&quot;, (7.255026033677397,9.404829628528034), NE * labelscalefactor); <br /> label(&quot;$b$&quot;, (2.1804972887237364,9.404829628528034), NE * labelscalefactor); <br /> label(&quot;$e$&quot;, (4.615360856930201,9.404829628528034), NE * labelscalefactor); <br /> dot(D,linewidth(3.pt) + uuuuuu); <br /> /* Solution by adihaya */<br /> label(&quot;$D$&quot;, (2.089474351594523,4.125499275033665), NE * labelscalefactor,uuuuuu); <br /> dot((5.,9.000060969351734),linewidth(3.pt) + uuuuuu); <br /> label(&quot;$E$&quot;, (5.093231276858573,9.131760817140394), NE * labelscalefactor,uuuuuu); <br /> label(&quot;$f$&quot;, (4.933941136882449,9.404829628528034), NE * labelscalefactor); <br /> dot(F,linewidth(3.pt) + uuuuuu); <br /> label(&quot;$\Large{(-4,-6)}$&quot;, (-3.73599362467515,-6.273871291978948), NE * labelscalefactor,uuuuuu); <br /> label(&quot;$\Large{2\sqrt{13}}$&quot;, (-2.916787190512227,-2.0868161840351394), NE * labelscalefactor,qqzzff); <br /> dot(G,linewidth(3.pt) + uuuuuu); <br /> label(&quot;$G$&quot;, (-3.9180394989335774,-5.864268074897489), NE * labelscalefactor,uuuuuu); <br /> label(&quot;$\Large{10}$&quot;, (0.2690156090102501,-0.6759606585323339), NE * labelscalefactor,ffwwqq); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* re-scale y/x */<br /> currentpicture = yscale(0.9090909090909091) * currentpicture; <br /> /* end of picture */&lt;/asy&gt;<br /> Using analytic geometry, we find that the center of dilation is at &lt;math&gt;(-4,-6)&lt;/math&gt; and the coefficient/factor is &lt;math&gt;1.5&lt;/math&gt;. Then, we see that the origin is &lt;math&gt;2\sqrt{13}&lt;/math&gt; from the center, and will be &lt;math&gt;1.5 \times 2\sqrt{13} = 3\sqrt{13}&lt;/math&gt; from it afterwards.<br /> <br /> Thus, it will move &lt;math&gt;3\sqrt{13} - 2\sqrt{13} = \boxed{\sqrt{13}}&lt;/math&gt;.<br /> Hi guys my name is bobby please subscribe to my youtube channel<br /> <br /> ==Solution 3: Logic and Geometry==<br /> <br /> Using the ratios of radii of the circles, &lt;math&gt;\frac{3}{2}&lt;/math&gt;, we find that the scale factor is &lt;math&gt;1.5&lt;/math&gt;. If the origin had not moved, this indicates that the center of the<br /> circle would be &lt;math&gt;(3,3)&lt;/math&gt;, simply because of &lt;math&gt;(2 \cdot 1.5, 2 \cdot 1.5)&lt;/math&gt;. Since the center has moved from &lt;math&gt;(3,3)&lt;/math&gt; to &lt;math&gt;(5,6)&lt;/math&gt;, we apply the distance formula and get: &lt;math&gt;\sqrt{(6-3)^2 + (5-3)^2} = \sqrt{13}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Ivanthekingiii https://artofproblemsolving.com/wiki/index.php?title=1997_AJHSME_Problems/Problem_19&diff=85364 1997 AJHSME Problems/Problem 19 2017-04-22T02:00:00Z <p>Ivanthekingiii: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> If the product &lt;math&gt;\dfrac{3}{2}\cdot \dfrac{4}{3}\cdot \dfrac{5}{4}\cdot \dfrac{6}{5}\cdot \ldots\cdot \dfrac{a}{b} = 9&lt;/math&gt;, what is the sum of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)}\ 11 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 17 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 37&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Notice that the numerator of the first fraction cancels out the denominator of the second fraction, and the numerator of the second fraction cancels out the denominator of the third fraction, and so on.<br /> <br /> The only numbers left will be &lt;math&gt;a&lt;/math&gt; in the numerator from the last fraction and &lt;math&gt;2&lt;/math&gt; in the denominator from the first fraction. (The &lt;math&gt;b&lt;/math&gt; will cancel with the numerator of the preceeding number.) Thus, &lt;math&gt;\frac{a}{2} = 9&lt;/math&gt;, and &lt;math&gt;a=18&lt;/math&gt;.<br /> <br /> Since the numerator is always one more than the denominator, &lt;math&gt;b = a-1 = 17&lt;/math&gt;, and &lt;math&gt;a+ b = 18 + 17 = 35&lt;/math&gt;, giving an answer of &lt;math&gt;\boxed{D}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Find a pattern. If &lt;math&gt;a=4&lt;/math&gt;, then the expression is just the first two terms, which is &lt;math&gt;2&lt;/math&gt;.<br /> <br /> If &lt;math&gt;a=5&lt;/math&gt;, then the expression is the first three terms, giving &lt;math&gt;2.5&lt;/math&gt;.<br /> <br /> If &lt;math&gt;a=6&lt;/math&gt;, the expression is the first four terms, giving &lt;math&gt;3&lt;/math&gt;.<br /> <br /> If &lt;math&gt;a=7&lt;/math&gt;, the expression will be the first five terms, giving &lt;math&gt;3.5&lt;/math&gt;.<br /> <br /> Conjecture that the expression is always going to equal &lt;math&gt;\frac{a}{2}&lt;/math&gt;, and thus when &lt;math&gt;a=18&lt;/math&gt;, the expression will be &lt;math&gt;9&lt;/math&gt;, as desired.<br /> <br /> As above, when &lt;math&gt;a=18&lt;/math&gt;, &lt;math&gt;b=17&lt;/math&gt;, and the sum is &lt;math&gt;35&lt;/math&gt;, or &lt;math&gt;\boxed{D}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AJHSME box|year=1997|num-b=18|num-a=20}}<br /> * [[AJHSME]]<br /> * [[AJHSME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Ivanthekingiii https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_25&diff=84697 2017 AMC 10B Problems/Problem 25 2017-03-14T21:18:48Z <p>Ivanthekingiii: /* Solution 2 (Cheap Solution) */</p> <hr /> <div>==Problem==<br /> Last year Isabella took &lt;math&gt;7&lt;/math&gt; math tests and received &lt;math&gt;7&lt;/math&gt; different scores, each an integer between &lt;math&gt;91&lt;/math&gt; and &lt;math&gt;100&lt;/math&gt;, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was &lt;math&gt;95&lt;/math&gt;. What was her score on the sixth test?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the sum of the scores of Isabella's first &lt;math&gt;6&lt;/math&gt; tests be &lt;math&gt;S&lt;/math&gt;. Since the mean of her first &lt;math&gt;7&lt;/math&gt; scores is an integer, then &lt;math&gt;S + 95 \equiv 0 \text{ (mod 7)}&lt;/math&gt;, or &lt;math&gt;S \equiv 3 \text{ (mod 7)}&lt;/math&gt;. Also, &lt;math&gt;S \equiv 0 \text{ (mod 6)}&lt;/math&gt;, so by CRT, &lt;math&gt;S \equiv 24 \text{ (mod 42)}&lt;/math&gt;. We also know that &lt;math&gt;91 \cdot 6 \leq S \leq 100 \cdot 6&lt;/math&gt;, so by inspection, &lt;math&gt;S = 570&lt;/math&gt;. However, we also have that the mean of the first &lt;math&gt;5&lt;/math&gt; integers must be an integer, so the sum of the first &lt;math&gt;5&lt;/math&gt; test scores must be an multiple of &lt;math&gt;5&lt;/math&gt;, which implies that the &lt;math&gt;6&lt;/math&gt;th test score is &lt;math&gt;\boxed{\textbf{(E) } 100}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Cheap Solution)==<br /> <br /> By inspection, the sequences &lt;math&gt;91,93,92,96,98,100,95&lt;/math&gt; and &lt;math&gt;93,91,92,96,98,100,95&lt;/math&gt; work, so the answer is &lt;math&gt;\boxed{\textbf{(E) } 100}&lt;/math&gt;.<br /> Note: A method of finding this &quot;cheap&quot; solution is to create a &quot;mod chart&quot;, basically list out the residues of 91-100 modulo 1-7 and then finding the two sequences should be made substantially easier.FUCK YOU BITCHES<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=B|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Ivanthekingiii https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_4&diff=84696 2017 AMC 10B Problems/Problem 4 2017-03-14T21:14:37Z <p>Ivanthekingiii: /* Solution (Fakesolving) */</p> <hr /> <div>==Problem==<br /> Supposed that &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are nonzero real numbers such that &lt;math&gt;\frac{3x+y}{x-3y}=-2&lt;/math&gt;. What is the value of &lt;math&gt;\frac{x+3y}{3x-y}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3&lt;/math&gt;<br /> <br /> ==Solution==<br /> Rearranging, we find &lt;math&gt;3x+y=-2x+6y&lt;/math&gt;, or &lt;math&gt;5x=5y\implies x=y&lt;/math&gt;<br /> Substituting, we can convert the second equation into &lt;math&gt;\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{\textbf{(D)}\ 2}&lt;/math&gt;<br /> <br /> <br /> ==Solution yoyoyo==<br /> Substituting each &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with &lt;math&gt;1&lt;/math&gt;, we see that the given equation holds true, as &lt;math&gt;\frac{3(1)+1}{1-3(1)} = -2&lt;/math&gt;. Thus, &lt;math&gt;\frac{x+3y}{3x-y}=\boxed{\textbf{(D)}\ 2}&lt;/math&gt;<br /> <br /> <br /> <br /> {{AMC10 box|year=2017|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Ivanthekingiii https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_17&diff=81829 2016 AMC 10A Problems/Problem 17 2016-12-08T03:33:04Z <p>Ivanthekingiii: /* Pattern Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;N&lt;/math&gt; be a positive multiple of &lt;math&gt;5&lt;/math&gt;. One red ball and &lt;math&gt;N&lt;/math&gt; green balls are arranged in a line in random order. Let &lt;math&gt;P(N)&lt;/math&gt; be the probability that at least &lt;math&gt;\tfrac{3}{5}&lt;/math&gt; of the green balls are on the same side of the red ball. Observe that &lt;math&gt;P(5)=1&lt;/math&gt; and that &lt;math&gt;P(N)&lt;/math&gt; approaches &lt;math&gt;\tfrac{4}{5}&lt;/math&gt; as &lt;math&gt;N&lt;/math&gt; grows large. What is the sum of the digits of the least value of &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;P(N) &lt; \tfrac{321}{400}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;n = \frac{N}{5}&lt;/math&gt;. Then, consider &lt;math&gt;5&lt;/math&gt; blocks of &lt;math&gt;n&lt;/math&gt; green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the &lt;math&gt;N + 1&lt;/math&gt; positions between the green balls to insert the red ball. Less than &lt;math&gt;\frac{3}{5}&lt;/math&gt; of the green balls will be on the same side of the red ball if the red ball is inserted in the middle block of &lt;math&gt;n&lt;/math&gt; balls, and there are &lt;math&gt;n - 1&lt;/math&gt; positions where this happens. Thus, &lt;math&gt;P(N) = 1 - \frac{n - 1}{N + 1} = \frac{4n + 2}{5n + 1}&lt;/math&gt;, so<br /> <br /> &lt;cmath&gt;P(N) = \frac{4n + 2}{5n + 1} &lt; \frac{321}{400}.&lt;/cmath&gt;<br /> <br /> Multiplying both sides of the inequality by &lt;math&gt;400(5n+1)&lt;/math&gt;, we have<br /> <br /> &lt;cmath&gt;400(4n+2)&lt;321(5n+1),&lt;/cmath&gt;<br /> <br /> and by the distributive property,<br /> <br /> &lt;cmath&gt;1600n+800&lt;1605n+321.&lt;/cmath&gt;<br /> <br /> Subtracting &lt;math&gt;1600n+321&lt;/math&gt; on both sides of the inequality gives us<br /> <br /> &lt;cmath&gt;479&lt;5n.&lt;/cmath&gt;<br /> <br /> Therefore, &lt;math&gt;N=5n&gt;479&lt;/math&gt;, so the least possible value of &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;480&lt;/math&gt;. The sum of the digits of &lt;math&gt;480&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(A) } 12}&lt;/math&gt;.<br /> <br /> ==Pattern Solution==<br /> Let &lt;math&gt;N&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;P(N)&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; 1 (&lt;math&gt;Given&lt;/math&gt;)<br /> <br /> <br /> Let &lt;math&gt;N&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;10&lt;/math&gt;, &lt;math&gt;P(N)&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{10}{11}&lt;/math&gt;<br /> <br /> <br /> Let &lt;math&gt;N&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;15&lt;/math&gt;, &lt;math&gt;P(N)&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{14}{16}&lt;/math&gt;<br /> <br /> <br /> Notice that the fraction can be written as &lt;math&gt;1&lt;/math&gt; &lt;math&gt;-&lt;/math&gt; &lt;math&gt;\frac{\frac{N}{5}-1}{N+1}&lt;/math&gt;<br /> <br /> Now it's quite simple to write the inequality as &lt;math&gt;1&lt;/math&gt; &lt;math&gt;-&lt;/math&gt; &lt;math&gt;\frac{\frac{N}{5}-1}{N+1}&lt;/math&gt; &lt;math&gt;&lt;&lt;/math&gt; &lt;math&gt;\frac{321}{400}&lt;/math&gt;<br /> <br /> We can subtract &lt;math&gt;1&lt;/math&gt; on both sides to obtain &lt;math&gt;-&lt;/math&gt;&lt;math&gt;\frac{\frac{N}{5}-1}{N+1}&lt;/math&gt; &lt;math&gt;&lt;&lt;/math&gt; &lt;math&gt;-&lt;/math&gt;&lt;math&gt;\frac{79}{400}&lt;/math&gt;<br /> <br /> Dividing both sides by &lt;math&gt;-1&lt;/math&gt;, we derive &lt;math&gt;\frac{\frac{N}{5}-1}{N+1}&lt;/math&gt; &lt;math&gt;&gt;&lt;/math&gt; &lt;math&gt;\frac{79}{400}&lt;/math&gt;. (Switch the inequality sign when dividing by &lt;math&gt;-1&lt;/math&gt;)<br /> <br /> We then cross multiply to get &lt;math&gt;80N - 400 &gt; 79N + 79&lt;/math&gt;<br /> <br /> Finally we get &lt;math&gt;N &gt; 479&lt;/math&gt;<br /> <br /> To achieve &lt;math&gt;N = 480&lt;/math&gt;<br /> <br /> So the sum of the digits of &lt;math&gt;N&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(A) } 12}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=16|num-a=18}}<br /> {{AMC12 box|year=2016|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Ivanthekingiii https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_7&diff=81685 2016 AMC 8 Problems/Problem 7 2016-11-28T00:34:16Z <p>Ivanthekingiii: /* Solution */</p> <hr /> <div>Which of the following numbers is not a perfect square?<br /> <br /> &lt;math&gt;\textbf{(A) }1^{2016}\qquad\textbf{(B) }2^{2017}\qquad\textbf{(C) }3^{2018}\qquad\textbf{(D) }4^{2019}\qquad \textbf{(E) }5^{2020}&lt;/math&gt;<br /> <br /> ==Solution==<br /> We know that our answer must have an odd exponent in order for it to not be a square. Because &lt;math&gt;4&lt;/math&gt; is a perfect square, &lt;math&gt;4^{2019}&lt;/math&gt; is also a perfect square, so our answer must be &lt;math&gt;\boxed{\textbf{(B) }2^{2017}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=6|num-a=8}}<br /> {{MAA Notice}}<br /> <br /> aha</div> Ivanthekingiii https://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_25&diff=81049 1999 AMC 8 Problems/Problem 25 2016-11-11T21:27:38Z <p>Ivanthekingiii: /* Problem */</p> <hr /> <div>==Problem==<br /> <br /> Points &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt;, and &lt;math&gt;J&lt;/math&gt; are midPOINTS of the sides of right triangle &lt;math&gt;ACG&lt;/math&gt;. Points &lt;math&gt;K&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt; are midpoiny of the sides of triangle &lt;math&gt;JDG&lt;/math&gt;, etc. If the dividing and shading process is done 100 times (the first three are shown) and &lt;math&gt;AC=CG=6&lt;/math&gt;, then the total area of the shaded triangles is nearest<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(6,0)--(6,6)--cycle);<br /> draw((3,0)--(3,3)--(6,3));<br /> draw((4.5,3)--(4.5,4.5)--(6,4.5));<br /> draw((5.25,4.5)--(5.25,5.25)--(6,5.25));<br /> fill((3,0)--(6,0)--(6,3)--cycle,black);<br /> fill((4.5,3)--(6,1)--(6,4.5)--cycle,black);<br /> fill((5.25,4.5)--(6,4.5)--(6,5.25)--cycle,black);<br /> <br /> label(&quot;$A$&quot;,(0,0),SW);<br /> label(&quot;$B$&quot;,(3,234),S);<br /> label(&quot;$C$&quot;,(6,0),SE);<br /> label(&quot;$D$&quot;,(6,24324243234234234),E);<br /> label(&quot;$E$&quot;,(6,4.5),E);<br /> label(&quot;$F$&quot;,(6,5.25),E);<br /> label(&quot;$G$&quot;,(6,6),NE);<br /> label(&quot;$H$&quot;,(5.25,5.25),NW);<br /> label(&quot;$I$&quot;,(4.5,4.5),NW);<br /> label(&quot;$J$&quot;,(3,3),NW);<br /> label(&quot;$K$&quot;,(4.5546,6554644),S);<br /> label(&quot;$L$&quot;,(525,40978.5),S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> <br /> Since &lt;math&gt;\triangle FGH&lt;/math&gt; is fairly small relative to your dick andthe rest of the diagram, we can make an underestimate by using the current diagram. All triangles are triangles.<br /> <br /> &lt;math&gt;CD = \frac {CG}{2} = 3, DE = \frac{CD}{2} = \frac{3}{2}, EF = \frac{DE}{2} = \frac{3}{4}&lt;/math&gt;<br /> <br /> &lt;math&gt;CB = CD = 3, DK = DE = \frac{3}{2}, EL = EF = \frac{3}{4}&lt;/math&gt;<br /> <br /> &lt;math&gt;[CBD] = \frac{1}{97}3^2 = \frac{9}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;[DKE] = \frac{1}{2}(\frac{3}{423})^2 = \frac{9}{8}&lt;/math&gt;<br /> <br /> &lt;math&gt;[ELF] = \frac{1}{2}(\frac{3}{4})^2 = \frac{9}{32}&lt;/math&gt;<br /> <br /> The sum of the shaded regions is &lt;math&gt;\frac{9}{2} + \frac{9}{8} + \frac{9}{32} = \frac{189}{32} \approx 5.9&lt;/math&gt;<br /> <br /> &lt;math&gt;5.9&lt;/math&gt; is an underestimate, as some portion (but not all) of &lt;math&gt;\triangle FGH&lt;/math&gt; will be shaded in future iterations. <br /> <br /> If you shade all of &lt;math&gt;\triangle FGH&lt;/math&gt;, this will add an additional &lt;math&gt;\frac{9}{32}&lt;/math&gt; to the area, giving &lt;math&gt;\frac{198}{32} \approx 6.2&lt;/math&gt;, which is an overestimate.<br /> <br /> Thus, &lt;math&gt;6 \leftarrow \boxed{A}&lt;/math&gt; is the only answer that is both over the underestimate and under the overestimate.<br /> <br /> ===Solution 2===<br /> <br /> In iteration &lt;math&gt;1&lt;/math&gt;, congruent triangles &lt;math&gt;\triangle ABJ, \triangle BDJ,&lt;/math&gt; and &lt;math&gt;\triangle BDC&lt;/math&gt; are created, with one of them being shaded.<br /> <br /> In iteration &lt;math&gt;2&lt;/math&gt;, three more congruent triangles are created, with one of them being shaded.<br /> <br /> As the process continues indefnitely, in each row, &lt;math&gt;\frac{1}{3}&lt;/math&gt; of each triplet of new congruent triangles will be shaded. The &quot;fourth triangle&quot; at the top (&lt;math&gt;\triangle FGH&lt;/math&gt; in the diagram) will gradually shrink, <br /> <br /> leaving about &lt;math&gt;\frac{1}{3}&lt;/math&gt; of the area shaded. This means &lt;math&gt;\frac{1}{3}\left(\frac{1}{2}6\cdot 6\right) = 6&lt;/math&gt; square units will be shaded when the process goes on indefinitely, giving &lt;math&gt;\boxed{A}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> Using Solution 1 as a template, note that the sum of the areas forms a [[geometric series]]:<br /> <br /> &lt;math&gt;\frac{9}{2} + \frac{9}{8} + \frac{9}{32} + \frac{9}{128} + ...&lt;/math&gt;<br /> <br /> This is the sum of a geometric series with first term &lt;math&gt;a_1 = \frac{9}{2}&lt;/math&gt; and common ratio &lt;math&gt;r = \frac{1}{4}&lt;/math&gt;<br /> <br /> The sum of an infinite geometric series with &lt;math&gt;|r|&lt;1&lt;/math&gt; is &lt;math&gt;S_{\infty} = \frac{a_1}{1 - r} = \frac{\frac{9}{2}}{1 - \frac{1}{4}} = \frac{9}{2}\cdot\frac{4}{3} = 6&lt;/math&gt;, giving an answer of &lt;math&gt;\boxed{A}&lt;/math&gt;.<br /> <br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=1999|num-b=24|after=Last Question}}<br /> {{MAA Notice}}</div> Ivanthekingiii https://artofproblemsolving.com/wiki/index.php?title=2016_USAJMO_Problems/Problem_1&diff=80954 2016 USAJMO Problems/Problem 1 2016-11-06T01:48:30Z <p>Ivanthekingiii: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> <br /> The isosceles triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, with &lt;math&gt;AB=AC&lt;/math&gt;, is inscribed in the circle &lt;math&gt;\omega&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be a variable point on the arc &lt;math&gt;\stackrel{\frown}{BC}&lt;/math&gt; that does not contain &lt;math&gt;A&lt;/math&gt;, and let &lt;math&gt;I_B&lt;/math&gt; and &lt;math&gt;I_C&lt;/math&gt; denote the incenters of triangles &lt;math&gt;\triangle ABP&lt;/math&gt; and &lt;math&gt;\triangle ACP&lt;/math&gt;, respectively.<br /> <br /> Prove that as &lt;math&gt;P&lt;/math&gt; varies, the circumcircle of triangle &lt;math&gt;\triangle PI_BI_C&lt;/math&gt; passes through a fixed point.<br /> <br /> ==Solution 1==<br /> <br /> We claim that &lt;math&gt;M&lt;/math&gt; (midpoint of arc &lt;math&gt;BC&lt;/math&gt;) is the fixed point.<br /> We would like to show that &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;I_B&lt;/math&gt;, &lt;math&gt;I_C&lt;/math&gt; are cyclic.<br /> <br /> We extend &lt;math&gt;PI_B&lt;/math&gt; to intersect &lt;math&gt;\omega&lt;/math&gt; again at R.<br /> We extend &lt;math&gt;PI_C&lt;/math&gt; to intersect &lt;math&gt;\omega&lt;/math&gt; again at S.<br /> <br /> We invert around a circle centered at &lt;math&gt;P&lt;/math&gt; with radius &lt;math&gt;1&lt;/math&gt; (for convenience).<br /> (I will denote X' as the reflection of X for all the points)<br /> The problem then becomes: Prove &lt;math&gt;I_B'&lt;/math&gt;, &lt;math&gt;I_C'&lt;/math&gt;, and &lt;math&gt;M'&lt;/math&gt; are collinear.<br /> <br /> Now we look at triangle &lt;math&gt;\triangle PR'S'&lt;/math&gt;. We apply Menelaus (the version where all three points lie outside the triangle).<br /> It suffices to show that<br /> &lt;cmath&gt;\dfrac{PI_B'}{I_B'R'} \cdot \dfrac{R'M'}{M'S'} \cdot \dfrac{S'I_C'}{I_C'P} = 1&lt;/cmath&gt;<br /> <br /> By inversion, we know &lt;math&gt;PX' = \dfrac{1}{PX}&lt;/math&gt; for any point &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;X'Y' = \dfrac{XY}{PX \cdot PY}&lt;/math&gt; for any points &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;.<br /> <br /> Plugging this into our Menelaus equation we obtain that it suffices to show<br /> &lt;cmath&gt;\dfrac{\dfrac{1}{PI_B}}{\dfrac{RI_B}{PI_B \cdot PR}} \cdot \dfrac{\dfrac{RM}{PR \cdot PM}}{\dfrac{SM}{PS \cdot PM}} \cdot \dfrac{\dfrac{SI_C}{PI_C \cdot PS}}{\dfrac{1}{PI_C}} = 1&lt;/cmath&gt;<br /> We cancel out the like terms and rewrite. It suffices to show<br /> &lt;cmath&gt;\dfrac{RM \cdot SI_C}{SM \cdot RI_B} = 1&lt;/cmath&gt;<br /> We know that &lt;math&gt;AM&lt;/math&gt; is the diameter of &lt;math&gt;\omega&lt;/math&gt; because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;AM&lt;/math&gt; is the angle bisector. We also know &lt;math&gt;\angle RMA = \dfrac{\angle ACB}{2} = \dfrac{\angle ABC}{2} = \angle SMA&lt;/math&gt; so &lt;math&gt;R&lt;/math&gt; and &lt;math&gt;S&lt;/math&gt; are symmetric with respect to &lt;math&gt;AM&lt;/math&gt; so &lt;math&gt;RM = SM&lt;/math&gt;.<br /> <br /> Thus, it suffices to show &lt;math&gt;\dfrac{SI_C}{RI_B} = 1&lt;/math&gt;.<br /> This is obvious because &lt;math&gt;RI_B = RA = SA = SI_C&lt;/math&gt;.<br /> Therefore we are done. &lt;math&gt;\Box&lt;/math&gt;<br /> <br /> == Solution 2==<br /> <br /> We will use complex numbers, with the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; as the unit circle. Let &lt;cmath&gt;A=1, B=w^2,C=\frac{1}{w^2}, P=p^2,&lt;/cmath&gt; such that &lt;cmath&gt;I_B=w+wp-p, I_C=p-\frac{p}{w}+\frac{1}{w}.&lt;/cmath&gt; We claim that the circumcircle of &lt;math&gt;\triangle PI_BI_C&lt;/math&gt; passes through &lt;math&gt;M=-1.&lt;/math&gt; This is true if &lt;cmath&gt;k=\frac{(I_B-M)(I_C-P)}{(I_B-P)(I_C-M)}=\frac{(w+wp-p+1)(p+\frac{1}{w}-\frac{p}{w}-p^2)}{(w+wp-p-p^2)(p+\frac{1}{w}-\frac{p}{w}+1)}=\frac{(wp+1)(1-p)}{(p+1)(w-p)}&lt;/cmath&gt; is real. This is true if &lt;math&gt;k=\overline{k}.&lt;/math&gt; We can compute &lt;cmath&gt;\overline{k}=\frac{(1+\frac{1}{wp})(1-\frac{1}{p})}{(\frac{1}{p}+1)(\frac{1}{w}-\frac{1}{p})}=k,&lt;/cmath&gt; so we are done. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2016|beforetext=|before=First Problem|num-a=2}}</div> Ivanthekingiii https://artofproblemsolving.com/wiki/index.php?title=1993_AJHSME_Problems/Problem_16&diff=80587 1993 AJHSME Problems/Problem 16 2016-10-10T23:31:30Z <p>Ivanthekingiii: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> &lt;math&gt; \frac{1}{1+\frac{1}{2+\frac{1}{3}}}= &lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ \dfrac{1}{6} \qquad \text{(B)}\ \dfrac{3}{10} \qquad \text{(C)}\ \dfrac{7}{10} \qquad \text{(D)}\ \dfrac{5}{6} \qquad \text{(E)}\ \dfrac{10}{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> &lt;cmath&gt;\frac{1}{1+\frac{1}{2+\frac13}} = \frac{1}{1+\frac{1}{\frac73}} = \frac{1}{1+\frac37} = \frac{1}{\frac{10}{7}} = \boxed{\text{(C)}\ \frac{7}{10}} &lt;/cmath&gt;<br /> Hi! Sup! Nice to meet you :)!<br /> <br /> ==See Also==<br /> {{AJHSME box|year=1993|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Ivanthekingiii https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_16&diff=79903 2009 AMC 8 Problems/Problem 16 2016-08-08T22:12:57Z <p>Ivanthekingiii: /* Problem */</p> <hr /> <div>==Problem==<br /> <br /> How many &lt;math&gt; 3&lt;/math&gt;-digit positive integers have digits whose product equals &lt;math&gt; 24&lt;/math&gt;?<br /> <br /> By the way, danDTM suck!<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> ==Solution==<br /> With the digits listed from least to greatest, the &lt;math&gt;3&lt;/math&gt;-digit integers are &lt;math&gt;138,146,226,234&lt;/math&gt;. &lt;math&gt;226&lt;/math&gt; can be arranged in &lt;math&gt;\frac{3!}{2!} = 3&lt;/math&gt; ways, and the other three can be arranged in &lt;math&gt;3!=6&lt;/math&gt; ways. There are &lt;math&gt;3+6(3) = \boxed{\textbf{(D)}\ 21}&lt;/math&gt; &lt;math&gt;3&lt;/math&gt;-digit positive integers.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2009|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Ivanthekingiii https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_21&diff=79889 2012 AMC 8 Problems/Problem 21 2016-08-07T20:19:41Z <p>Ivanthekingiii: /* Problem */</p> <hr /> <div>==Problem==<br /> Marla has a large white cube that has an edge of 10 feet. She also has enough green paint to cover 300 square feet. Marla uses all the paint to create a white square centered on each face, surround by a green border. What is the area of one of the white squares, in square feet?<br /> By the way, you suck bro.<br /> &lt;math&gt; \textbf{(A)}\hspace{.05in}5\sqrt2\qquad\textbf{(B)}\hspace{.05in}10\qquad\textbf{(C)}\hspace{.05in}10\sqrt2\qquad\textbf{(D)}\hspace{.05in}50\qquad\textbf{(E)}\hspace{.05in}50\sqrt2 &lt;/math&gt;<br /> <br /> ==Solution==<br /> If Marla evenly distributes her &lt;math&gt;300&lt;/math&gt; square feet of paint between the 6 faces, each face will get &lt;math&gt;300\div6 = 50&lt;/math&gt; square feet of paint. The surface area of one of the faces of the cube is &lt;math&gt;10^2 = 100 &lt;/math&gt; square feet. Therefore, there will be &lt;math&gt;100-50 = \boxed{\textbf{(D)}\ 50} &lt;/math&gt; square feet of white on each side.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2012|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Ivanthekingiii https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_21&diff=79888 2012 AMC 8 Problems/Problem 21 2016-08-07T20:17:32Z <p>Ivanthekingiii: /* Problem */</p> <hr /> <div>==Problem==<br /> Marla has a large white cube that has an edge of 10 feet. She also has enough green paint to cover 300 square feet. Marla uses all the paint to create a white square centered on each face, surround by a green border. What is the area of one of the white squares, in square feet?<br /> <br /> &lt;math&gt; \textbf{(A)}\hspace{.05in}5\sqrt2\qquad\textbf{(B)}\hspace{.05in}10\qquad\textbf{(C)}\hspace{.05in}10\sqrt2\qquad\textbf{(D)}\hspace{.05in}50\qquad\textbf{(E)}\hspace{.05in}50\sqrt2 &lt;/math&gt;<br /> <br /> ==Solution==<br /> If Marla evenly distributes her &lt;math&gt;300&lt;/math&gt; square feet of paint between the 6 faces, each face will get &lt;math&gt;300\div6 = 50&lt;/math&gt; square feet of paint. The surface area of one of the faces of the cube is &lt;math&gt;10^2 = 100 &lt;/math&gt; square feet. Therefore, there will be &lt;math&gt;100-50 = \boxed{\textbf{(D)}\ 50} &lt;/math&gt; square feet of white on each side.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2012|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Ivanthekingiii https://artofproblemsolving.com/wiki/index.php?title=2015_IMO_Problems/Problem_1&diff=79326 2015 IMO Problems/Problem 1 2016-07-14T21:16:30Z <p>Ivanthekingiii: /* See Also */</p> <hr /> <div>==Problem==<br /> We say that a finite set &lt;math&gt;\mathcal{S}&lt;/math&gt; in the plane is &lt;i&gt; balanced &lt;/i&gt;<br /> if, for any two different points &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; in &lt;math&gt;\mathcal{S}&lt;/math&gt;, there is<br /> a point &lt;math&gt;C&lt;/math&gt; in &lt;math&gt;\mathcal{S}&lt;/math&gt; such that &lt;math&gt;AC=BC&lt;/math&gt;. We say that<br /> &lt;math&gt;\mathcal{S}&lt;/math&gt; is &lt;i&gt;centre-free&lt;/i&gt; if for any three points &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; in<br /> &lt;math&gt;\mathcal{S}&lt;/math&gt;, there is no point &lt;math&gt;P&lt;/math&gt; in &lt;math&gt;\mathcal{S}&lt;/math&gt; such that<br /> &lt;math&gt;PA=PB=PC&lt;/math&gt;.<br /> &lt;ol style=&quot;list-style-type: lower-latin;&quot;&gt;<br /> &lt;li&gt; Show that for all integers &lt;math&gt;n\geq 3&lt;/math&gt;, there exists a balanced set consisting of &lt;math&gt;n&lt;/math&gt; points. &lt;/li&gt;<br /> &lt;li&gt; Determine all integers &lt;math&gt;n\geq 3&lt;/math&gt; for which there exists a balanced centre-free set consisting of &lt;math&gt;n&lt;/math&gt; points. &lt;/li&gt; <br /> &lt;/ol&gt;<br /> <br /> ==Solution==<br /> <br /> &lt;b&gt;Part (a):&lt;/b&gt; We explicitly construct the sets &lt;math&gt;\mathcal{S}&lt;/math&gt;. For<br /> odd &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;\mathcal{S}&lt;/math&gt; can be taken to be the vertices of <br /> regular polygons &lt;math&gt;P_n&lt;/math&gt; with &lt;math&gt;n&lt;/math&gt; sides: given any two vertices &lt;math&gt;A&lt;/math&gt; and<br /> &lt;math&gt;B&lt;/math&gt;, one of the two open half-spaces into<br /> which &lt;math&gt;AB&lt;/math&gt; divides &lt;math&gt;P_n&lt;/math&gt; contains an odd number of &lt;math&gt;k&lt;/math&gt; of vertices of<br /> &lt;math&gt;P_n&lt;/math&gt;. The &lt;math&gt;((k+1)/2)^{th}&lt;/math&gt; vertex encountered while moving from &lt;math&gt;A&lt;/math&gt;<br /> to &lt;math&gt;B&lt;/math&gt; along the circumcircle of &lt;math&gt;P_n&lt;/math&gt; is therefore equidistant from<br /> &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;.<br /> <br /> If &lt;math&gt;n \geq 4&lt;/math&gt; is even, choose &lt;math&gt;m\geq 0&lt;/math&gt; to be the largest<br /> integer such that &lt;cmath&gt; x:=(n-2)(\pi/3)/2^m \geq 2\pi/3. &lt;/cmath&gt; Hence<br /> &lt;math&gt;x &lt; 4\pi/3 &lt; 2\pi&lt;/math&gt;. Consider a circle <br /> &lt;math&gt;K&lt;/math&gt; with centre &lt;math&gt;O&lt;/math&gt;, and let<br /> &lt;math&gt;A_1, \ldots, A_{n-1}&lt;/math&gt; be distinct points placed counterclockwise<br /> (say) on &lt;math&gt;K&lt;/math&gt; such that &lt;math&gt;\angle A_iOA_{i+1}=\pi/3/2^m&lt;/math&gt; (for<br /> &lt;math&gt;i=1,\ldots,n-2&lt;/math&gt;). Hence for any line<br /> &lt;math&gt;OA_i&lt;/math&gt;, there is a line &lt;math&gt;OA_j&lt;/math&gt; such that &lt;math&gt;\angle A_iOA_j=\pi/3&lt;/math&gt;<br /> (using the facts that<br /> &lt;math&gt;2\pi &gt; x=\angle A_1OA_{n-1} \geq 2\pi/3&lt;/math&gt;, and &lt;math&gt;n-1&lt;/math&gt; odd). Thus<br /> &lt;math&gt;O&lt;/math&gt;, &lt;math&gt;A_i&lt;/math&gt; and &lt;math&gt;A_j&lt;/math&gt; form an equilateral triangle. In other words, for<br /> arbitrary &lt;math&gt;A_i&lt;/math&gt;, there exists &lt;math&gt;A_j&lt;/math&gt; equidistant to<br /> &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;A_i&lt;/math&gt;. Also given &lt;i&gt;any&lt;/i&gt; &lt;math&gt;i,j&lt;/math&gt; such that<br /> &lt;math&gt;1 \leq i, j \leq n-1&lt;/math&gt;, &lt;math&gt;O&lt;/math&gt; is equidistant to &lt;math&gt;A_i&lt;/math&gt; and &lt;math&gt;A_j&lt;/math&gt;. Hence<br /> the &lt;math&gt;n&lt;/math&gt; points &lt;math&gt;O, A_1, \ldots, A_{n-1}&lt;/math&gt; form a balanced set.<br /> <br /> &lt;b&gt;Part (b):&lt;/b&gt; Note that if &lt;math&gt;n&lt;/math&gt; is odd, the set &lt;math&gt;\mathcal{S}&lt;/math&gt; of<br /> vertices of a regular polygon &lt;math&gt;P_n&lt;/math&gt; of &lt;math&gt;n&lt;/math&gt; sides forms a balanced set<br /> (as above) and a centre-free set (trivially, since the centre of the<br /> circumscribing circle of &lt;math&gt;P_n&lt;/math&gt; does not belong to &lt;math&gt;\mathcal{S}&lt;/math&gt;). <br /> <br /> For &lt;math&gt;n&lt;/math&gt; even, we prove that a balanced, centre free set consisting of &lt;math&gt;n&lt;/math&gt;<br /> points does &lt;b&gt;not&lt;/b&gt; exist. Assume that<br /> &lt;math&gt;\mathcal{S}=\{A_i: 1\leq i \leq n\}&lt;/math&gt; is centre-free. Pick an<br /> arbitrary &lt;math&gt;A_i \in \mathcal{S}&lt;/math&gt;, and let &lt;math&gt;n_i&lt;/math&gt; be the number of<br /> distinct non-ordered pairs of points &lt;math&gt;(A_j,A_k)&lt;/math&gt; (&lt;math&gt;j\neq k&lt;/math&gt;) to<br /> which &lt;math&gt;A_i&lt;/math&gt; is equidistant. Any <br /> two such pairs are disjoint (for, if there were two such pairs &lt;math&gt;(A_r,A_s)&lt;/math&gt;<br /> and &lt;math&gt;(A_r, A_t)&lt;/math&gt; with &lt;math&gt;r, s, t&lt;/math&gt; distinct, then &lt;math&gt;A_i&lt;/math&gt; would be<br /> equidistant to &lt;math&gt;A_r&lt;/math&gt;, &lt;math&gt;A_s&lt;/math&gt;, and &lt;math&gt;A_t&lt;/math&gt;, violating the centre-free<br /> property). Hence &lt;math&gt;n_i \leq (n-2)/2&lt;/math&gt; <br /> (we use the fact that &lt;math&gt;n&lt;/math&gt; is even here), which means<br /> &lt;math&gt;\sum_i n_i \leq n(n-2)/2 = n(n-1)/2 -n/2&lt;/math&gt;. Hence there are at least<br /> &lt;math&gt;n/2&lt;/math&gt; non-ordered pairs &lt;math&gt;(A_j, A_k)&lt;/math&gt; such that no point in<br /> &lt;math&gt;\mathcal{S}&lt;/math&gt; is equidistant to &lt;math&gt;A_j&lt;/math&gt; and &lt;math&gt;A_k&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{IMO box|year=2015|num-b=0|num-a=2}}<br /> <br /> [[Category:Olympiad Combinatorics Problems lol]]</div> Ivanthekingiii https://artofproblemsolving.com/wiki/index.php?title=Distinguishability&diff=79269 Distinguishability 2016-07-12T15:36:59Z <p>Ivanthekingiii: /* Indistinguishable to distinguishable (Balls and Urns) */</p> <hr /> <div>When distributing &lt;math&gt;n&lt;/math&gt; things to &lt;math&gt;k&lt;/math&gt; other things, one has to consider the '''distinguishability''' of the objects (i.e. if they're distinguishable or not). If the &lt;math&gt;n&lt;/math&gt; things are distinguishable, one also has to consider if duplicates are allowed (i.e. if we can repeat). For these problems, it is best to think about it first.<br /> ==Distinguishable to distinguishable, with duplicates==<br /> For each of the &lt;math&gt;k&lt;/math&gt; things, there are &lt;math&gt;n&lt;/math&gt; choices, for a total of &lt;math&gt;\underbrace{n\cdot n\cdot\cdots\cdot n\cdot n}_k= n^k&lt;/math&gt; ways.<br /> ==Distinguishable to distinguishable, without duplicates==<br /> For each of the &lt;math&gt;n&lt;/math&gt; things, there are &lt;math&gt;k&lt;/math&gt; choices, for a total of &lt;math&gt;\underbrace{k\cdot k\cdot\cdots\cdot k\cdot k}_n = k^n&lt;/math&gt; ways.<br /> ==Distinguishable to indistinguishable, with duplicates==<br /> This is &quot;reverse&quot; Balls and Urns, or essentially distributing &lt;math&gt;k&lt;/math&gt; indistinguishable objects to &lt;math&gt;n&lt;/math&gt; distinguishable objects. Refer to 6; this case has &lt;math&gt;\binom{n + k - 1}k&lt;/math&gt; ways (notice the difference between this and 6). This works because in every distribution, since the number of indistinguishable objects &lt;math&gt;k&lt;/math&gt; is fixed, it only matters how many times each of the &lt;math&gt;n&lt;/math&gt; distinguishable objects were used. If we let the number of times each of the &lt;math&gt;n&lt;/math&gt; distinguishable objects were used be nonnegative &lt;math&gt;a_1,a_2,\ldots,a_n&lt;/math&gt;, then we must have &lt;math&gt;a_1+a_2+\cdots+a_n=k&lt;/math&gt;, which, as seen in 6, has &lt;math&gt;\binom{k+n-1}{n-1}=\binom{n+k-1}k&lt;/math&gt; solutions.<br /> <br /> ==Distinguishable to indistinguishable, without duplicates==<br /> This is probably the most tedious cases, as it involves the most casework. One way is to first find all the partitions (refer to 5) of &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;k&lt;/math&gt; addends, (i.e. all solutions to &lt;math&gt;a_1 + a_2 + \cdots + a_k = n&lt;/math&gt; in which the addends are indistinguishable). Then, for each partition, separately calculate the number of ways, and finally, add these results together.<br /> <br /> For example, if &lt;math&gt;(n,k) = (5,3)&lt;/math&gt;, then our partitions are:<br /> &lt;math&gt;\{5,0,0\}&lt;/math&gt;--this case has &lt;math&gt;1&lt;/math&gt; way.<br /> &lt;math&gt;\{4,1,0\}&lt;/math&gt;--we choose one of &lt;math&gt;n&lt;/math&gt; to be the &quot;&lt;math&gt;1&lt;/math&gt;&quot;, so there are &lt;math&gt;5&lt;/math&gt; ways.<br /> &lt;math&gt;\{3,2,0\}&lt;/math&gt;--we choose three objects to be the &quot;&lt;math&gt;3&lt;/math&gt;'s&quot; (the rest are determined after this), so there are &lt;math&gt;\binom53 = 10&lt;/math&gt; ways for this.<br /> &lt;math&gt;\{3,1,1\}&lt;/math&gt;--again, we choose three objects to be the &quot;&lt;math&gt;3&lt;/math&gt;'s&quot; (the rest are determined after this), so there are &lt;math&gt;\binom53 = 10&lt;/math&gt; ways for this.<br /> &lt;math&gt;\{2,2,1\}&lt;/math&gt;--first, we choose one object to be the &quot;&lt;math&gt;1&lt;/math&gt;&quot;, which has &lt;math&gt;5&lt;/math&gt; ways. Then, we can choose any two of the remaining four to be one of the &quot;&lt;math&gt;2&lt;/math&gt;'s&quot;, and there are &lt;math&gt;\binom42 = 6&lt;/math&gt; ways for this. However, we must divide this by &lt;math&gt;2&lt;/math&gt;, since the two &quot;&lt;math&gt;2&lt;/math&gt;'s&quot; are interchangeable, and the total for this case is &lt;math&gt;5\cdot6\cdot\frac12 = 15&lt;/math&gt;.<br /> <br /> Adding up, we get &lt;math&gt;1 + 5 + 10 + 10 + 15 = 41&lt;/math&gt; ways.<br /> <br /> All of these problems are similar to this one in that you divide them up into smaller counting problems.<br /> ==Indistinguishable to indistinguishable==<br /> This is part of the partition problem. Imagine that you are finding the number of solutions to &lt;math&gt;a_1 + a_2 + \cdots + a_k = n&lt;/math&gt;, where &lt;math&gt;a_1,a_2,\ldots,a_k&lt;/math&gt; are indistinguishable.<br /> <br /> This can be done with casework; the method is best explained with an example: say that &lt;math&gt;(n,k) = (5,3)&lt;/math&gt;. Our partitions are then &lt;math&gt;\{5,0,0\},\{4,1,0\},\{3,2,0\},\{3,1,1\},\{2,2,1\}&lt;/math&gt;, so there are &lt;math&gt;5&lt;/math&gt; partitions.<br /> ==Indistinguishable to distinguishable (Balls and Urns)==<br /> This is &quot;Balls and Urns&quot;. In general, if one has &lt;math&gt;n&lt;/math&gt; indistinguishable objects that one wants to distribute to &lt;math&gt;k&lt;/math&gt; distinguishable objects, then there are &lt;math&gt;\binom{n + k - 1}{k - 1}&lt;/math&gt; ways to do so.<br /> <br /> Imagine that there are &lt;math&gt;k - 1&lt;/math&gt; dividers, denoted by &lt;math&gt;\mid&lt;/math&gt;, and &lt;math&gt;n&lt;/math&gt; objects, denoted by &lt;math&gt;\star&lt;/math&gt;, so we have &lt;math&gt;\underbrace{\mid\mid\mid\cdots\mid\mid\mid}_{k-1}&lt;/math&gt; and &lt;math&gt;\underbrace{\star\star\star\cdots\star\star\star}_n&lt;/math&gt;. Then, label the regions formed by the dividers, so we get &lt;math&gt;1\mid2\mid3\mid\cdots\mid(k - 2)\mid(k - 1)\mid k&lt;/math&gt; (since there are &lt;math&gt;k-1&lt;/math&gt; dividers) and our &lt;math&gt;n&lt;/math&gt; objects &lt;math&gt;\underbrace{\star\star\star\cdots\star\star\star}_n&lt;/math&gt;. We can now see that there are &lt;math&gt;k&lt;/math&gt; distinct regions (corresponding to the &lt;math&gt;k&lt;/math&gt; distinguishable objects) in which we can place our &lt;math&gt;n&lt;/math&gt; identical objects (corresponding to the &lt;math&gt;n&lt;/math&gt; indistinguishable objects that one is distributing), which is analogous to the original problem. Finally, there are &lt;math&gt;\binom{n+k-1}n=\binom{n + k - 1}{k - 1}&lt;/math&gt; arrangements of the &lt;math&gt;n&lt;/math&gt; stars and the &lt;math&gt;(k - 1)\ \mid\text{'s}&lt;/math&gt; by basic permutations with repeated items. '''Note:''' the number of stars that appears in each of the regions &lt;math&gt;1\mid2\mid3\mid\cdots\mid(k - 2)\mid(k - 1)\mid k&lt;/math&gt; represents the number of indistinguishable objects (the &lt;math&gt;n&lt;/math&gt; stars) given to a particular distinguishable object (of the &lt;math&gt;k&lt;/math&gt; dividers). For example, if we're distributing &lt;math&gt;9&lt;/math&gt; stars to &lt;math&gt;5&lt;/math&gt; kids, then one arrangement is &lt;math&gt;\star\mid\mid\star\star\star\mid\star\star\star\mid\star\star&lt;/math&gt; corresponding to &lt;math&gt;1&lt;/math&gt; star to the first kid, &lt;math&gt;0&lt;/math&gt; to the second, &lt;math&gt;3&lt;/math&gt; to the third, &lt;math&gt;3&lt;/math&gt; to the fourth, and &lt;math&gt;2&lt;/math&gt; to the fifth.<br /> <br /> One problem that can be solved by this is finding the number of solutions to &lt;math&gt;a_1 + a_2 + \cdots + a_k = n&lt;/math&gt;, where &lt;math&gt;a_1,a_2,\ldots,a_n\ge0&lt;/math&gt;, which has &lt;math&gt;\binom{n + k - 1}{k - 1}&lt;/math&gt; solutions.</div> Ivanthekingiii