https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Abe27342&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T10:08:29ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_2&diff=477482011 AMC 12A Problems/Problem 22012-07-27T16:53:01Z<p>Abe27342: /* Solution */</p>
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<div>== Problem ==<br />
There are <math>5</math> coins placed flat on a table according to the figure. What is the order of the coins from top to bottom?<br />
<asy><br />
size(100); defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
draw(arc((0,1), 1.2, 25, 214));<br />
draw(arc((.951,.309), 1.2, 0, 360));<br />
draw(arc((.588,-.809), 1.2, 132, 370));<br />
draw(arc((-.588,-.809), 1.2, 75, 300));<br />
draw(arc((-.951,.309), 1.2, 96, 228));<br />
label("$A$",(0,1),NW); label("$B$",(-1.1,.309),NW); label("$C$",(.951,.309),E); label("$D$",(-.588,-.809),W); label("$E$",(.588,-.809),S);</asy><br />
<math><br />
\textbf{(A)}\ (C, A, E, D, B) \qquad<br />
\textbf{(B)}\ (C, A, D, E, B) \qquad<br />
\textbf{(C)}\ (C, D, E, A, B) \qquad<br />
\textbf{(D)}\ (C, E, A, D, B) \qquad \\<br />
\textbf{(E)}\ (C, E, D, A, B) </math><br />
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== Solution ==<br />
By inspection, the answer is <math>\textbf{(E)}</math>.<br />
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(rigor?)<br />
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== See also ==<br />
{{AMC12 box|year=2011|num-b=1|num-a=3|ab=A}}<br />
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[[Category:Introductory Combinatorics Problems]]</div>Abe27342https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_6&diff=461162012 AIME II Problems/Problem 62012-04-04T19:31:13Z<p>Abe27342: </p>
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<div>== Problem 6 ==<br />
Let <math>z=a+bi</math> be the complex number with <math>\vert z \vert = 5</math> and <math>b > 0</math> such that the distance between <math>(1+2i)z^3</math> and <math>z^5</math> is maximized, and let <math>z^4 = c+di</math>. Find <math>c+d</math>.<br />
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== Solution ==<br />
Let's consider the maximization constraint first: we want to maximize the value of <math>|z^5 - (1+2i)z^3|</math><br />
Simplifying, we have<br />
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<math>|z^3| * |z^2 - (1+2i)|</math><br />
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<math>=|z|^3 * |z^2 - (1+2i)|</math><br />
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<math>=125|z^2 - (1+2i)|</math><br />
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Thus we only need to maximize the value of <math>|z^2 - (1+2i)|</math>.<br />
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To maximize this value, we must have that <math>z^2</math> is in the opposite direction of <math>1+2i</math>. The unit vector in the complex plane in the desired direction is <math>\frac{-1}{\sqrt{5}} + \frac{-2}{\sqrt{5}} i</math>. Furthermore, we know that the magnitude of <math>z^2</math> is <math>25</math>, because the magnitude of <math>z</math> is <math>5</math>. From this information, we can find that <math>z^2 = \sqrt{5} (-5 - 10i)</math><br />
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Squaring, we get <math>z^4 = 5 (25 - 100 + 100i) = -375 + 500i</math>. Finally, <math>c+d = -375 + 500 = 125</math><br />
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== See Also ==<br />
{{AIME box|year=2012|n=II|num-b=5|num-a=7}}</div>Abe27342https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_11&diff=382932011 AIME I Problems/Problem 112011-05-01T05:55:18Z<p>Abe27342: /* Solution */</p>
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<div>== Problem ==<br />
Let <math>R</math> be the set of all possible remainders when a number of the form <math>2^n</math>, <math>n</math> a nonnegative integer, is divided by <math>1000</math>. Let <math>S</math> be the sum of the elements in <math>R</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>.<br />
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== Solution ==<br />
Note that the cycle of remainders of <math>2^n</math> will start after <math>2^2</math> because remainders of <math>1, 2, 4</math> will not be possible after (the numbers following will always be congruent to 0 modulo 8). Now we have to find the order. Note that <math>2^{100}\equiv 1\mod 125</math>. The order is <math>100</math> starting with remainder <math>8</math>. All that is left is find <math>S</math> in mod <math>1000</math> after some computation.<br />
<cmath>S=2^0+2^1+2^2+2^3+2^4...+2^{99}\equiv 2^{100}-1\equiv 8-1\equiv \boxed{007}\mod 1000</cmath><br />
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== See also ==<br />
{{AIME box|year=2011|n=I|num-b=10|num-a=12}}</div>Abe27342https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_10&diff=382922011 AIME I Problems/Problem 102011-04-30T23:39:32Z<p>Abe27342: /* Solution */</p>
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<div>== Problem ==<br />
The probability that a set of three distinct vertices chosen at random from among the vertices of a regular n-gon determine an obtuse triangle is <math>\frac{93}{125}</math> . Find the sum of all possible values of <math>n</math>.<br />
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== Solution 1 ==<br />
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Inscribe the regular polygon inside a circle. A triangle inside this circle will be obtuse if and only if its three vertices lie on one side of a diameter of the circle.<br />
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Break up the problem into two cases: an even number of sides <math>2n</math>, or an odd number of sides <math>2n-1</math>. After looking at a few diagrams, it becomes apparent that there are exactly <math>n</math> points on one side of a diameter.<br />
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Case 1: <math>2n</math>-sided polygon. There are clearly <math> \binom{2n}{3}</math> different triangles total. To find triangles that meet the criteria, choose the left-most point. There are obviously <math>2n</math> choices for this point. From there, the other two points must be within the <math>n-1</math> points remaining on the same side of the diameter. So our desired probability is<br />
<math>\frac{2n\binom{n-1}{2}}{\binom{2n}{3}}</math><br />
<math>=\frac{n(n-1)(n-2)}{\frac{2n(2n-1)(2n-2)}{6}}</math><br />
<math>=\frac{6n(n-1)(n-2)}{2n(2n-1)(2n-2)}</math><br />
<math>=\frac{3(n-2)}{2(2n-1)}</math><br />
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so <math>\frac{93}{125}=\frac{3(n-2)}{2(2n-1)}</math><br />
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<math>186(2n-1)=375(n-2)</math>.<br />
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<math>372n-186=375n-750</math><br />
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<math>3n=564</math><br />
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<math>n=188</math> and so the polygon has <math>376</math> sides.<br />
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Case 2: <math>2n-1</math>-sided polygon. Similarly, <math>\bimon{2n-1}{3}</math> total triangles. Again choose the leftmost point, with <math>2n-1</math> choices. For the other two points, there are again <math>\binom{n-1}{2}</math> possibilities.<br />
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The probability is <math>\frac{(2n-1)\binom{n-1}{2}}{\binom{2n-1}{3}}</math><br />
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<math>=\frac{3(2n-1)(n-1)(n-2)}{(2n-1)(2n-2)(2n-3)}</math><br />
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<math>=\frac{3(n-2)}{2(2n-3)}</math><br />
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so <math>\frac{93}{125}=\frac{3(n-2)}{2(2n-3)}</math><br />
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<math>186(2n-3)=375(n-2)</math><br />
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<math>375n-750=372n-558</math><br />
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<math>3n=192</math><br />
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<math>n=64</math> and our polygon has <math>127</math> sides.<br />
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Adding, <math>127+376=\boxed{503}</math><br />
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== Incomplete Solution ==<br />
NOTE: This is not complete; but it can probably become a viable solution. If you have a different one, please put it under an alternate solution until it's been verified.<br />
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We use casework on the locations of the vertices, if we choose the locations of vertices <math>v_a, v_b, v_c</math> on the n-gon (where the vertices of the n-gon are <math>v_0, v_1, v_2, ... v_{n-1},</math> in clockwise order) to be the vertices of triangle ABC, in order, with the restriction that <math>a<b<c</math>.<br />
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By symmetry, we can assume W/O LOG that the location of vertex A is vertex <math>v_0</math>.<br />
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Now, vertex B can be any of <math>v_1, v_2, ... v_{n-2}</math>. We start in on casework.<br />
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Case 1: vertex B is at one of the locations <math>v_{n-2}, v_{n-3}, ... v_{\lfloor n/2 \rfloor +1}</math>. (The floor function is necessary for the cases in which n is odd.)<br />
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Now, since the clockwise arc from A to B measures more than 180 degrees; for every location of vertex C we can choose in the above restrictions, angle C will be an obtuse angle.<br />
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There are <math>\lfloor n/2 \rfloor - 1</math> choices for vertex B now (again, the floor function is necessary to satisfy both odd and even cases of n). If vertex B is placed at <math>v_m</math>, there are <math>n - m - 1</math> possible places for vertex C.<br />
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Summing over all these possibilities, we obtain that the number of obtuse triangles obtainable from this case is <math>\frac{(n- \lfloor n/2 \rfloor - 1)(n - \lfloor n/2 \rfloor)}{2}</math>.<br />
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Case 2: vertex B is at one of the locations not covered in the first case.<br />
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Note that this will result in the same number of obtuse triangles as case 1, but multiplied by 2. This is because fixing vertex B in <math>v_0</math>, then counting up the cases for vertices C, and again for vertices C and A, respectively, is combinatorially equivalent to fixing vertex A at <math>v_0</math>, then counting cases for vertex B, as every triangle obtained in this way can be rotated in the n-gon to place vertex A at <math>v_0</math>, and will not be congruent to any obtuse triangle obtained in case 1, as there will be a different side opposite the obtuse angle in this case.<br />
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Therefore, there are <math>\frac{3(n- \lfloor n/2 \rfloor - 1)(n - \lfloor n/2 \rfloor)}{2}</math> total obtuse triangles obtainable.<br />
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The total number of triangles obtainable is <math>1+2+3+...+(n-2) = \frac{(n-2)(n-1)}{2}</math>.<br />
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The ratio of obtuse triangles obtainable to all triangles obtainable is therefore<br />
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<math>\frac{\frac{3(n- \lfloor n/2 \rfloor - 1)(n - \lfloor n/2 \rfloor)}{2}}{\frac{(n-2)(n-1)}{2}} = \frac{3(n- \lfloor n/2 \rfloor - 1)(n - \lfloor n/2 \rfloor)}{(n-2)(n-1)} = \frac{93}{125}</math>.<br />
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So, <math> \frac{(n- \lfloor n/2 \rfloor - 1)(n - \lfloor n/2 \rfloor)}{(n-2)(n-1)} = \frac{31}{125}</math>.<br />
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Now, we have that <math>(n-2)(n-1)</math> is divisible by <math>125 = 5^3</math>. It is now much easier to perform trial-and-error on possible values of n, because we see that <math>n \equiv 1,2 mod 125</math>.<br />
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However, there is no solution to this that is less than 1000. I must have made an error somewhere; if someone could find and fix it, I would be much obliged.<br />
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== See also ==<br />
{{AIME box|year=2011|n=I|num-b=9|num-a=11}}</div>Abe27342