https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Laffytaffy&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T23:55:30ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_16&diff=1997872018 AMC 10B Problems/Problem 162023-10-18T03:19:09Z<p>Laffytaffy: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>a_1,a_2,\dots,a_{2018}</math> be a strictly increasing sequence of positive integers such that <cmath>a_1+a_2+\cdots+a_{2018}=2018^{2018}.</cmath><br />
What is the remainder when <math>a_1^3+a_2^3+\cdots+a_{2018}^3</math> is divided by <math>6</math>?<br />
<br />
<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math><br />
<br />
==Solution 1==<br />
<br />
Note that, because of fermat's little theorem, for any integer <math>a</math> relatively prime to 6, we have <math>a^2 \equiv 1 \pmod{6} \implies a^3 \equiv a \pmod{6}</math>. This is also trivially true if <math>a</math> is not relatively prime to 6 because <math>a \equiv 0 \pmod{6}</math>. Therefore,<br />
<br />
<cmath>a_1+a_2+\cdots+a_{2018} \equiv a_1^3+a_2^3+\cdots+a_{2018}^3 \pmod{6}.</cmath><br />
<br />
Therefore the answer is congruent to <math>2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ (E)4}</math> because <math>2^n \pmod{6}</math> alternates with <math>2</math> and <math>4</math> when <math>n</math> increases.<br />
<br />
~Dolphindesigner<br />
<br />
==Solution 2==<br />
<br />
Note that <math>\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k</math><br />
<br />
Note that <math><br />
a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2({2018}^{2018}-a_1)+3a_2^2({2018}^{2018}-a_2)+\cdots+3a_{2018}^2({2018}^{2018}-a_{2018})<br />
\equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6<br />
</math><br />
Therefore, <math>-2(a_1^3+a_2^3+\cdots+a_{2018}^3)\equiv \left(2018^{2018}\right)^3\equiv\left( 2^{2018}\right)^3\equiv 4^3\equiv 4\pmod{6}</math>.<br />
<br />
Thus, <math>a_1^3+a_2^3+\cdots+a_{2018}^3\equiv 1\pmod 3</math>. However, since cubing preserves parity, and the sum of the individual terms is even, the sum of the cubes is also even, and our answer is <math>\boxed{\text{(E) }4}</math><br />
<br />
==Solution 3 (Partial Proof)==<br />
First, we can assume that the problem will have a consistent answer for all possible values of <math>a_1</math>. For the purpose of this solution, we will assume that <math>a_1 = 1</math>.<br />
<br />
We first note that <math>1^3+2^3+...+n^3 = (1+2+...+n)^2</math>. So what we are trying to find is what <math>\left(2018^{2018}\right)^2=\left(2018^{4036}\right)</math> mod <math>6</math>. We start by noting that <math>2018</math> is congruent to <math>2 \pmod{6}</math>. So we are trying to find <math>\left(2^{4036}\right) \pmod{6}</math>. Instead of trying to do this with some number theory skills, we could just look for a pattern. We start with small powers of <math>2</math> and see that <math>2^1</math> is <math>2</math> mod <math>6</math>, <math>2^2</math> is <math>4</math> mod <math>6</math>, <math>2^3</math> is <math>2</math> mod <math>6</math>, <math>2^4</math> is <math>4</math> mod <math>6</math>, and so on... So we see that since <math>\left(2^{4036}\right)</math> has an even power, it must be congruent to <math>4 \pmod{6}</math>, thus giving our answer <math>\boxed{\text{(E) }4}</math>. You can prove this pattern using mods. But I thought this was easier.<br />
<br />
-TheMagician<br />
<br />
==Solution 4 (Lazy solution)==<br />
First, we can assume that the problem will have a consistent answer for all possible values of <math>a_1</math>. For the purpose of this solution, assume <math>a_1, a_2, ... a_{2017}</math> are multiples of 6 and find <math>2018^{2018} \pmod{6}</math> (which happens to be <math>4</math>). Then <math>{a_1}^3 + ... + {a_{2018}}^3</math> is congruent to <math>64 \pmod{6}</math> or just <math>\boxed{\textbf{(E)} 4}</math>. <br />
<br />
-Patrick4President<br />
<br />
~minor edit made by CatachuKetchup~<br />
<br />
==Solution 5 (Even Lazier Solution)==<br />
Due to the large amounts of variables in the problem, and the fact that the test is only 75 minutes, you can assume that the answer is probably just <math>2018^{2018} \pmod{6} </math>, which is <math>\boxed{\textbf{(E)} 4}</math>.<br />
<br />
~ Zeeshan12 [Be warned that this technique is not recommended for all problems and you should use it as a last resort]<br />
<br />
==Algebraic Insight into Given Property==<br />
Mods is a good way to prove <math>a^3 \equiv a \pmod6</math>: residues are simply <math>3, \pm 2, \pm 1, 0</math>. Only <math>2</math> and <math>3</math> are necessary to check.<br />
Another way is to observe that <math>a^3-a</math> factors into <math>(a-1)a(a+1)</math>. Any <math>k</math> consecutive numbers must be a multiple of <math>k</math>, so <math>a^3-a</math> is both divisible by <math>2</math> and <math>3</math>. This provides an algebraic method for proving <math>a^3 \equiv a \pmod6</math> for all <math>a</math>.<br />
<br />
==Video Solution 1==<br />
With Modular Arithmetic Intro<br />
https://www.youtube.com/watch?v=wbv3TArroSs<br />
<br />
~IceMatrix<br />
<br />
==Video Solution 2==<br />
https://www.youtube.com/watch?v=SRjZ6B5DR74<br />
<br />
~bunny1<br />
<br />
== Video Solution 3 by OmegaLearn==<br />
https://youtu.be/4_x1sgcQCp4?t=112<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Laffytaffyhttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_21&diff=1464252021 AMC 10B Problems/Problem 212021-02-13T15:33:55Z<p>Laffytaffy: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
A square piece of paper has side length <math>1</math> and vertices <math>A,B,C,</math> and <math>D</math> in that order. As shown in the figure, the paper is folded so that vertex <math>C</math> meets edge <math>\overline{AD}</math> at point <math>C'</math>, and edge <math>\overline{AB}</math> at point <math>E</math>. Suppose that <math>C'D = \frac{1}{3}</math>. What is the perimeter of triangle <math>\bigtriangleup AEC' ?</math><br />
<br />
<math>\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{136} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}</math><br />
<asy><br />
/* Made by samrocksnature */<br />
pair A=(0,1);<br />
pair CC=(0.666666666666,1);<br />
pair D=(1,1);<br />
pair F=(1,0.62);<br />
pair C=(1,0);<br />
pair B=(0,0);<br />
pair G=(0,0.25);<br />
pair H=(-0.13,0.41);<br />
pair E=(0,0.5);<br />
dot(A^^CC^^D^^C^^B^^E);<br />
draw(E--A--D--F);<br />
draw(G--B--C--F, dashed);<br />
fill(E--CC--F--G--H--E--CC--cycle, gray);<br />
draw(E--CC--F--G--H--E--CC);<br />
label("A",A,NW);<br />
label("B",B,SW);<br />
label("C",C,SE);<br />
label("D",D,NE);<br />
label("E",E,NW);<br />
label("C'",CC,N);<br />
</asy><br />
<br />
==Solution 1==<br />
<br />
We can set the point on <math>CD</math> where the fold occurs as point <math>F</math>. Then, we can set <math>FD</math> as <math>x</math>, and <math>CF</math> as <math>1-x</math> because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for <math>x</math>, we get, <br />
<br />
<cmath>x^2 + (\frac{1}{3})^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}</cmath><br />
<br />
We know this is a 3-4-5 triangle because the side lengths are <math>\frac{3}{9}, \frac{4}{9}, \frac{5}{9}</math>. We also know that <math>EAC'</math> is similar to <math>C'DF</math> because angle <math>C'</math> is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of <math>C'DF * \frac{AC'}{DF}</math>. Thats just <math>\frac{4}{3} * \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} * \frac{3}{2} = 2</math>. Therefore, the final answer is <math>\boxed{A}</math><br />
<br />
~Tony_Li2007<br />
<br />
==Solution 2==<br />
Let line we're reflecting over be <math>\ell</math>, and let the points where it hits <math>AB</math> and <math>CD</math>, be <math>M</math> and <math>N</math>, respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line <math>\ell</math>. The segment <math>CC'</math> has slope <math>\frac{0 - 1}{1 - 2/3} = -3</math>, implying line <math>\ell</math> has a slope of <math>\frac{1}{3}</math>. Also, the midpoint of segment <math>CC'</math> is <math>\left( \frac{5}{6}, \frac{1}{2} \right)</math>, so line <math>\ell</math> passes through this point. Then, we get the equation of line <math>\ell</math> is simply <math>y = \frac{1}{3} x + \frac{2}{9}</math>. Then, if the point where <math>B</math> is reflected over line <math>\ell</math> is <math>B'</math>, then we get <math>BB'</math> is the line <math>y = -3x</math>. The intersection of <math>\ell</math> and segment <math>BB'</math> is <math>\left( - \frac{1}{15}, \frac{1}{5} \right)</math>. So, we get <math>B' = \left(- \frac{2}{15}, \frac{2}{5} \right)</math>. Then, line segment <math>B'C'</math> has equation <math>y = \frac{3}{4} x + \frac{1}{2}</math>, so the point <math>E</math> is the <math>y</math>-intercept, or <math>\left(0, \frac{1}{2} \right)</math>. This implies that <math>AE = \frac{1}{2}, AC' = \frac{2}{3}</math>, and by the Pythagorean Theorem, <math>EC' = \frac{5}{6}</math> (or you could notice <math>\triangle AEC'</math> is a <math>3-4-5</math> right triangle). Then, the perimeter is <math>\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2</math>, so our answer is <math>\boxed{\textbf{(A)} ~2}</math>. ~rocketsri<br />
<br />
==Solution 3 (Fakesolve):==<br />
Assume that E is the midpoint of <math>\overline{AB}</math>. Then, <math>\overline{AE}=\frac{1}{2}</math> and since <math>C'D=\frac{1}{3}</math>, <math>\overline{AC'}=\frac{2}{3}</math>. By the Pythagorean Theorem, <math>\overline{EC'}=\frac{5}{6}</math>. It easily follows that our desired perimeter is <math>2 \rightarrow \boxed{A}</math> ~samrocksnature<br />
<br />
==Solution 4==<br />
As described in Solution 1, we can find that <math>DF=\frac{4}{9}</math>, and <math>C'F = \frac{5}{9}.</math> <br />
<br />
<br />
Then, we can find we can find the length of <math>\overline{AE}</math> by expressing the length of <math>\overline{EF}</math> in two different ways, in terms of <math>AE</math>. If let <math>AE = a</math>, by the Pythagorean Theorem we have that <math>EC = \sqrt{a^2 + \left(\frac{2}{3}\right)^2} = \sqrt{a^2 + \frac{4}{9}}.</math> Therefore, since we know that <math>\angle EC'F</math> is right, by Pythagoras again we have that <math>EF = \sqrt{\left(\sqrt{a^2+\frac{4}{9}}\right)^2 + \left(\frac{5}{9}\right)^2} = \sqrt{a^2 + \frac{61}{81}}.</math><br />
<br />
<br />
Another way we can express <math>EF</math> is by using Pythagoras on <math>\triangle XEF</math>, where <math>X</math> is the foot of the perpendicular from <math>F</math> to <math>\overline{AE}.</math> We see that <math>ADFX</math> is a rectangle, so we know that <math>AD = 1 = FX</math>. Secondly, since <math>FD = \frac{4}{9}, EX = a - \frac{4}{9}</math>. Therefore, through the Pythagorean Theorem, we find that <math>EF = \sqrt{\left(a-\frac{4}{9}\right)^2 + 1^2} = \sqrt{a^2 - \frac{8}{9}a +\frac{97}{81}}.</math><br />
<br />
Since we have found two expressions for the same length, we have the equation <math>\sqrt{a^2 + \frac{61}{81}} = \sqrt{a^2 - \frac{8}{9}a +\frac{97}{81}}.</math> Solving this, we find that <math>a=\frac{1}{2}</math>.<br />
<br />
Finally, we see that the perimeter of <math>\triangle AEC'</math> is <math>\frac{1}{2} + \frac{2}{3} + \sqrt{\left(\frac{1}{2}\right)^2 + \frac{4}{9}},</math> which we can simplify to be <math>2</math>. Thus, the answer is <math>\boxed{\textbf{(A)} ~2}.</math> ~laffytaffy<br />
<br />
== Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) ==<br />
https://youtu.be/cagzLmdbqYQ<br />
<br />
~ pi_is_3.14<br />
<br />
{{AMC10 box|year=2021|ab=B|num-b=20|num-a=22}}</div>Laffytaffyhttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_21&diff=1464242021 AMC 10B Problems/Problem 212021-02-13T15:32:22Z<p>Laffytaffy: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
A square piece of paper has side length <math>1</math> and vertices <math>A,B,C,</math> and <math>D</math> in that order. As shown in the figure, the paper is folded so that vertex <math>C</math> meets edge <math>\overline{AD}</math> at point <math>C'</math>, and edge <math>\overline{AB}</math> at point <math>E</math>. Suppose that <math>C'D = \frac{1}{3}</math>. What is the perimeter of triangle <math>\bigtriangleup AEC' ?</math><br />
<br />
<math>\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{136} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}</math><br />
<asy><br />
/* Made by samrocksnature */<br />
pair A=(0,1);<br />
pair CC=(0.666666666666,1);<br />
pair D=(1,1);<br />
pair F=(1,0.62);<br />
pair C=(1,0);<br />
pair B=(0,0);<br />
pair G=(0,0.25);<br />
pair H=(-0.13,0.41);<br />
pair E=(0,0.5);<br />
dot(A^^CC^^D^^C^^B^^E);<br />
draw(E--A--D--F);<br />
draw(G--B--C--F, dashed);<br />
fill(E--CC--F--G--H--E--CC--cycle, gray);<br />
draw(E--CC--F--G--H--E--CC);<br />
label("A",A,NW);<br />
label("B",B,SW);<br />
label("C",C,SE);<br />
label("D",D,NE);<br />
label("E",E,NW);<br />
label("C'",CC,N);<br />
</asy><br />
<br />
==Solution 1==<br />
<br />
We can set the point on <math>CD</math> where the fold occurs as point <math>F</math>. Then, we can set <math>FD</math> as <math>x</math>, and <math>CF</math> as <math>1-x</math> because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for <math>x</math>, we get, <br />
<br />
<cmath>x^2 + (\frac{1}{3})^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}</cmath><br />
<br />
We know this is a 3-4-5 triangle because the side lengths are <math>\frac{3}{9}, \frac{4}{9}, \frac{5}{9}</math>. We also know that <math>EAC'</math> is similar to <math>C'DF</math> because angle <math>C'</math> is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of <math>C'DF * \frac{AC'}{DF}</math>. Thats just <math>\frac{4}{3} * \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} * \frac{3}{2} = 2</math>. Therefore, the final answer is <math>\boxed{A}</math><br />
<br />
~Tony_Li2007<br />
<br />
==Solution 2==<br />
Let line we're reflecting over be <math>\ell</math>, and let the points where it hits <math>AB</math> and <math>CD</math>, be <math>M</math> and <math>N</math>, respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line <math>\ell</math>. The segment <math>CC'</math> has slope <math>\frac{0 - 1}{1 - 2/3} = -3</math>, implying line <math>\ell</math> has a slope of <math>\frac{1}{3}</math>. Also, the midpoint of segment <math>CC'</math> is <math>\left( \frac{5}{6}, \frac{1}{2} \right)</math>, so line <math>\ell</math> passes through this point. Then, we get the equation of line <math>\ell</math> is simply <math>y = \frac{1}{3} x + \frac{2}{9}</math>. Then, if the point where <math>B</math> is reflected over line <math>\ell</math> is <math>B'</math>, then we get <math>BB'</math> is the line <math>y = -3x</math>. The intersection of <math>\ell</math> and segment <math>BB'</math> is <math>\left( - \frac{1}{15}, \frac{1}{5} \right)</math>. So, we get <math>B' = \left(- \frac{2}{15}, \frac{2}{5} \right)</math>. Then, line segment <math>B'C'</math> has equation <math>y = \frac{3}{4} x + \frac{1}{2}</math>, so the point <math>E</math> is the <math>y</math>-intercept, or <math>\left(0, \frac{1}{2} \right)</math>. This implies that <math>AE = \frac{1}{2}, AC' = \frac{2}{3}</math>, and by the Pythagorean Theorem, <math>EC' = \frac{5}{6}</math> (or you could notice <math>\triangle AEC'</math> is a <math>3-4-5</math> right triangle). Then, the perimeter is <math>\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2</math>, so our answer is <math>\boxed{\textbf{(A)} ~2}</math>. ~rocketsri<br />
<br />
==Solution 3 (Fakesolve):==<br />
Assume that E is the midpoint of <math>\overline{AB}</math>. Then, <math>\overline{AE}=\frac{1}{2}</math> and since <math>C'D=\frac{1}{3}</math>, <math>\overline{AC'}=\frac{2}{3}</math>. By the Pythagorean Theorem, <math>\overline{EC'}=\frac{5}{6}</math>. It easily follows that our desired perimeter is <math>2 \rightarrow \boxed{A}</math> ~samrocksnature<br />
<br />
==Solution 4==<br />
As described in Solution 1, we can find that <math>DF=\frac{4}{9}</math>, and <math>C'F = \frac{5}{9}.</math> <br />
<br />
<br />
Then, we can find we can find the length of <math>\overline{AE}</math> by expressing the length of <math>\overline{EF}</math> in two different ways, in terms of <math>AE</math>. If let <math>AE = a</math>, by the Pythagorean Theorem we have that <math>EC = \sqrt{a^2 + \left(\frac{2}{3}\right)^2} = \sqrt{a^2 + \frac{4}{9}}.</math> Therefore, since we know that <math>\angle EC'F</math> is right, by Pythagoras again we have that <math>EF = \sqrt{\left(\sqrt{a^2+\frac{4}{9}}\right)^2 + \left(\frac{5}{9}\right)^2} = \sqrt{a^2 + \frac{61}{81}}.</math><br />
<br />
<br />
Another way we can express <math>EF</math> is by using Pythagoras on <math>\triangle XEF</math>, where <math>X</math> is the foot of the perpendicular from <math>F</math> to <math>\overline{AE}.</math> We see that <math>ADFX</math> is a rectangle, so we know that <math>AD = 1 = FX</math>. Secondly, since <math>FD = \frac{4}{9}, EX = a - \frac{4}{9}</math>. Therefore, through the Pythagorean Theorem, we find that <math>EF = \sqrt{\left(a-\frac{4}{9}\right)^2 + 1^2} = \sqrt{a^2 - \frac{8}{9}a +\frac{97}{81}}.</math><br />
<br />
Since we have found two expressions for the same length, we have the equation <math>\sqrt{a^2 + \frac{61}{81}} = \sqrt{a^2 - \frac{8}{9}a +\frac{97}{81}}.</math> Solving this, we find that <math>a=\frac{1}{2}</math>.<br />
<br />
Finally, we see that the perimeter of <math>\triangle AEC'</math> is <math>\frac{1}{2} + \frac{2}{3} + \sqrt{\left(\frac{1}{2}\right)^2 + \frac{4}{9}},</math> which we can simplify to be <math>2</math>. Thus, the answer is <math>\boxed{\textbf{(A)} ~2}.</math><br />
<br />
== Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) ==<br />
https://youtu.be/cagzLmdbqYQ<br />
<br />
~ pi_is_3.14<br />
<br />
{{AMC10 box|year=2021|ab=B|num-b=20|num-a=22}}</div>Laffytaffyhttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_21&diff=1464212021 AMC 10B Problems/Problem 212021-02-13T15:08:48Z<p>Laffytaffy: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
A square piece of paper has side length <math>1</math> and vertices <math>A,B,C,</math> and <math>D</math> in that order. As shown in the figure, the paper is folded so that vertex <math>C</math> meets edge <math>\overline{AD}</math> at point <math>C'</math>, and edge <math>\overline{AB}</math> at point <math>E</math>. Suppose that <math>C'D = \frac{1}{3}</math>. What is the perimeter of triangle <math>\bigtriangleup AEC' ?</math><br />
<br />
<math>\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{136} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}</math><br />
<asy><br />
/* Made by samrocksnature */<br />
pair A=(0,1);<br />
pair CC=(0.666666666666,1);<br />
pair D=(1,1);<br />
pair F=(1,0.62);<br />
pair C=(1,0);<br />
pair B=(0,0);<br />
pair G=(0,0.25);<br />
pair H=(-0.13,0.41);<br />
pair E=(0,0.5);<br />
dot(A^^CC^^D^^C^^B^^E);<br />
draw(E--A--D--F);<br />
draw(G--B--C--F, dashed);<br />
fill(E--CC--F--G--H--E--CC--cycle, gray);<br />
draw(E--CC--F--G--H--E--CC);<br />
label("A",A,NW);<br />
label("B",B,SW);<br />
label("C",C,SE);<br />
label("D",D,NE);<br />
label("E",E,NW);<br />
label("C'",CC,N);<br />
</asy><br />
<br />
==Solution 1==<br />
<br />
We can set the point on <math>CD</math> where the fold occurs as point <math>F</math>. Then, we can set <math>FD</math> as <math>x</math>, and <math>CF</math> as <math>1-x</math> because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for <math>x</math>, we get, <br />
<br />
<cmath>x^2 + (\frac{1}{3})^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}</cmath><br />
<br />
We know this is a 3-4-5 triangle because the side lengths are <math>\frac{3}{9}, \frac{4}{9}, \frac{5}{9}</math>. We also know that <math>EAC'</math> is similar to <math>C'DF</math> because angle <math>C'</math> is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of <math>C'DF * \frac{AC'}{DF}</math>. Thats just <math>\frac{4}{3} * \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} * \frac{3}{2} = 2</math>. Therefore, the final answer is <math>\boxed{A}</math><br />
<br />
~Tony_Li2007<br />
<br />
==Solution 2==<br />
Let line we're reflecting over be <math>\ell</math>, and let the points where it hits <math>AB</math> and <math>CD</math>, be <math>M</math> and <math>N</math>, respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line <math>\ell</math>. The segment <math>CC'</math> has slope <math>\frac{0 - 1}{1 - 2/3} = -3</math>, implying line <math>\ell</math> has a slope of <math>\frac{1}{3}</math>. Also, the midpoint of segment <math>CC'</math> is <math>\left( \frac{5}{6}, \frac{1}{2} \right)</math>, so line <math>\ell</math> passes through this point. Then, we get the equation of line <math>\ell</math> is simply <math>y = \frac{1}{3} x + \frac{2}{9}</math>. Then, if the point where <math>B</math> is reflected over line <math>\ell</math> is <math>B'</math>, then we get <math>BB'</math> is the line <math>y = -3x</math>. The intersection of <math>\ell</math> and segment <math>BB'</math> is <math>\left( - \frac{1}{15}, \frac{1}{5} \right)</math>. So, we get <math>B' = \left(- \frac{2}{15}, \frac{2}{5} \right)</math>. Then, line segment <math>B'C'</math> has equation <math>y = \frac{3}{4} x + \frac{1}{2}</math>, so the point <math>E</math> is the <math>y</math>-intercept, or <math>\left(0, \frac{1}{2} \right)</math>. This implies that <math>AE = \frac{1}{2}, AC' = \frac{2}{3}</math>, and by the Pythagorean Theorem, <math>EC' = \frac{5}{6}</math> (or you could notice <math>\triangle AEC'</math> is a <math>3-4-5</math> right triangle). Then, the perimeter is <math>\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2</math>, so our answer is <math>\boxed{\textbf{(A)} ~2}</math>. ~rocketsri<br />
<br />
==Solution 3 (Fakesolve):==<br />
Assume that E is the midpoint of <math>\overline{AB}</math>. Then, <math>\overline{AE}=\frac{1}{2}</math> and since <math>C'D=\frac{1}{3}</math>, <math>\overline{AC'}=\frac{2}{3}</math>. By the Pythagorean Theorem, <math>\overline{EC'}=\frac{5}{6}</math>. It easily follows that our desired perimeter is <math>2 \rightarrow \boxed{A}</math> ~samrocksnature<br />
<br />
==Solution 4==<br />
As described in Solution 1, we can find that <math>DF=\frac{4}{9}</math>, and <math>C'F = \frac{5}{9}.</math> <br />
<br />
<br />
Then, we can find we can find the length of <math>AE</math> by expressing the length of <math>EF</math> in two different ways, in terms of <math>AE</math>. If let <math>AE = a</math>, by the Pythagorean Theorem we have that <math>EC = \sqrt{a^2 + \left(\frac{2}{3}\right)^2} = \sqrt{a^2 + \frac{4}{9}}.</math> Therefore, since we know that <math>\angle EC'F</math> is right, by Pythagoras again we have that <math>EF = \sqrt{\left(\sqrt{a^2+\frac{4}{9}\right)^2}</math><br />
<br />
== Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) ==<br />
https://youtu.be/cagzLmdbqYQ<br />
<br />
~ pi_is_3.14<br />
<br />
{{AMC10 box|year=2021|ab=B|num-b=20|num-a=22}}</div>Laffytaffyhttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_21&diff=1464192021 AMC 10B Problems/Problem 212021-02-13T15:05:58Z<p>Laffytaffy: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
A square piece of paper has side length <math>1</math> and vertices <math>A,B,C,</math> and <math>D</math> in that order. As shown in the figure, the paper is folded so that vertex <math>C</math> meets edge <math>\overline{AD}</math> at point <math>C'</math>, and edge <math>\overline{AB}</math> at point <math>E</math>. Suppose that <math>C'D = \frac{1}{3}</math>. What is the perimeter of triangle <math>\bigtriangleup AEC' ?</math><br />
<br />
<math>\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{136} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}</math><br />
<asy><br />
/* Made by samrocksnature */<br />
pair A=(0,1);<br />
pair CC=(0.666666666666,1);<br />
pair D=(1,1);<br />
pair F=(1,0.62);<br />
pair C=(1,0);<br />
pair B=(0,0);<br />
pair G=(0,0.25);<br />
pair H=(-0.13,0.41);<br />
pair E=(0,0.5);<br />
dot(A^^CC^^D^^C^^B^^E);<br />
draw(E--A--D--F);<br />
draw(G--B--C--F, dashed);<br />
fill(E--CC--F--G--H--E--CC--cycle, gray);<br />
draw(E--CC--F--G--H--E--CC);<br />
label("A",A,NW);<br />
label("B",B,SW);<br />
label("C",C,SE);<br />
label("D",D,NE);<br />
label("E",E,NW);<br />
label("C'",CC,N);<br />
</asy><br />
<br />
==Solution 1==<br />
<br />
We can set the point on <math>CD</math> where the fold occurs as point <math>F</math>. Then, we can set <math>FD</math> as <math>x</math>, and <math>CF</math> as <math>1-x</math> because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for <math>x</math>, we get, <br />
<br />
<cmath>x^2 + (\frac{1}{3})^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}</cmath><br />
<br />
We know this is a 3-4-5 triangle because the side lengths are <math>\frac{3}{9}, \frac{4}{9}, \frac{5}{9}</math>. We also know that <math>EAC'</math> is similar to <math>C'DF</math> because angle <math>C'</math> is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of <math>C'DF * \frac{AC'}{DF}</math>. Thats just <math>\frac{4}{3} * \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} * \frac{3}{2} = 2</math>. Therefore, the final answer is <math>\boxed{A}</math><br />
<br />
~Tony_Li2007<br />
<br />
==Solution 2==<br />
Let line we're reflecting over be <math>\ell</math>, and let the points where it hits <math>AB</math> and <math>CD</math>, be <math>M</math> and <math>N</math>, respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line <math>\ell</math>. The segment <math>CC'</math> has slope <math>\frac{0 - 1}{1 - 2/3} = -3</math>, implying line <math>\ell</math> has a slope of <math>\frac{1}{3}</math>. Also, the midpoint of segment <math>CC'</math> is <math>\left( \frac{5}{6}, \frac{1}{2} \right)</math>, so line <math>\ell</math> passes through this point. Then, we get the equation of line <math>\ell</math> is simply <math>y = \frac{1}{3} x + \frac{2}{9}</math>. Then, if the point where <math>B</math> is reflected over line <math>\ell</math> is <math>B'</math>, then we get <math>BB'</math> is the line <math>y = -3x</math>. The intersection of <math>\ell</math> and segment <math>BB'</math> is <math>\left( - \frac{1}{15}, \frac{1}{5} \right)</math>. So, we get <math>B' = \left(- \frac{2}{15}, \frac{2}{5} \right)</math>. Then, line segment <math>B'C'</math> has equation <math>y = \frac{3}{4} x + \frac{1}{2}</math>, so the point <math>E</math> is the <math>y</math>-intercept, or <math>\left(0, \frac{1}{2} \right)</math>. This implies that <math>AE = \frac{1}{2}, AC' = \frac{2}{3}</math>, and by the Pythagorean Theorem, <math>EC' = \frac{5}{6}</math> (or you could notice <math>\triangle AEC'</math> is a <math>3-4-5</math> right triangle). Then, the perimeter is <math>\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2</math>, so our answer is <math>\boxed{\textbf{(A)} ~2}</math>. ~rocketsri<br />
<br />
==Solution 3 (Fakesolve):==<br />
Assume that E is the midpoint of <math>\overline{AB}</math>. Then, <math>\overline{AE}=\frac{1}{2}</math> and since <math>C'D=\frac{1}{3}</math>, <math>\overline{AC'}=\frac{2}{3}</math>. By the Pythagorean Theorem, <math>\overline{EC'}=\frac{5}{6}</math>. It easily follows that our desired perimeter is <math>2 \rightarrow \boxed{A}</math> ~samrocksnature<br />
<br />
==Solution 4==<br />
As described in Solution 1, we can find that <math>DF=\frac{4}{9}</math>, and <math>C'F = \frac{5}{9}.</math> <br />
<br />
<br />
Then, we can find we can find the length of <math>AE</math> by expressing the length of <math>EF</math> in two different ways, in terms of <math>AE</math>. If let <math>AE = a</math>, by the Pythagorean Theorem we have that <math>EC = \sqrt{a^2 + \left(\frac{2}{3}\right)^2} = \sqrt{a^2 + \frac{4}{9}}.</math><br />
<br />
== Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) ==<br />
https://youtu.be/cagzLmdbqYQ<br />
<br />
~ pi_is_3.14<br />
<br />
{{AMC10 box|year=2021|ab=B|num-b=20|num-a=22}}</div>Laffytaffyhttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_21&diff=1464182021 AMC 10B Problems/Problem 212021-02-13T15:04:25Z<p>Laffytaffy: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
A square piece of paper has side length <math>1</math> and vertices <math>A,B,C,</math> and <math>D</math> in that order. As shown in the figure, the paper is folded so that vertex <math>C</math> meets edge <math>\overline{AD}</math> at point <math>C'</math>, and edge <math>\overline{AB}</math> at point <math>E</math>. Suppose that <math>C'D = \frac{1}{3}</math>. What is the perimeter of triangle <math>\bigtriangleup AEC' ?</math><br />
<br />
<math>\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{136} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}</math><br />
<asy><br />
/* Made by samrocksnature */<br />
pair A=(0,1);<br />
pair CC=(0.666666666666,1);<br />
pair D=(1,1);<br />
pair F=(1,0.62);<br />
pair C=(1,0);<br />
pair B=(0,0);<br />
pair G=(0,0.25);<br />
pair H=(-0.13,0.41);<br />
pair E=(0,0.5);<br />
dot(A^^CC^^D^^C^^B^^E);<br />
draw(E--A--D--F);<br />
draw(G--B--C--F, dashed);<br />
fill(E--CC--F--G--H--E--CC--cycle, gray);<br />
draw(E--CC--F--G--H--E--CC);<br />
label("A",A,NW);<br />
label("B",B,SW);<br />
label("C",C,SE);<br />
label("D",D,NE);<br />
label("E",E,NW);<br />
label("C'",CC,N);<br />
</asy><br />
<br />
==Solution 1==<br />
<br />
We can set the point on <math>CD</math> where the fold occurs as point <math>F</math>. Then, we can set <math>FD</math> as <math>x</math>, and <math>CF</math> as <math>1-x</math> because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for <math>x</math>, we get, <br />
<br />
<cmath>x^2 + (\frac{1}{3})^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}</cmath><br />
<br />
We know this is a 3-4-5 triangle because the side lengths are <math>\frac{3}{9}, \frac{4}{9}, \frac{5}{9}</math>. We also know that <math>EAC'</math> is similar to <math>C'DF</math> because angle <math>C'</math> is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of <math>C'DF * \frac{AC'}{DF}</math>. Thats just <math>\frac{4}{3} * \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} * \frac{3}{2} = 2</math>. Therefore, the final answer is <math>\boxed{A}</math><br />
<br />
~Tony_Li2007<br />
<br />
==Solution 2==<br />
Let line we're reflecting over be <math>\ell</math>, and let the points where it hits <math>AB</math> and <math>CD</math>, be <math>M</math> and <math>N</math>, respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line <math>\ell</math>. The segment <math>CC'</math> has slope <math>\frac{0 - 1}{1 - 2/3} = -3</math>, implying line <math>\ell</math> has a slope of <math>\frac{1}{3}</math>. Also, the midpoint of segment <math>CC'</math> is <math>\left( \frac{5}{6}, \frac{1}{2} \right)</math>, so line <math>\ell</math> passes through this point. Then, we get the equation of line <math>\ell</math> is simply <math>y = \frac{1}{3} x + \frac{2}{9}</math>. Then, if the point where <math>B</math> is reflected over line <math>\ell</math> is <math>B'</math>, then we get <math>BB'</math> is the line <math>y = -3x</math>. The intersection of <math>\ell</math> and segment <math>BB'</math> is <math>\left( - \frac{1}{15}, \frac{1}{5} \right)</math>. So, we get <math>B' = \left(- \frac{2}{15}, \frac{2}{5} \right)</math>. Then, line segment <math>B'C'</math> has equation <math>y = \frac{3}{4} x + \frac{1}{2}</math>, so the point <math>E</math> is the <math>y</math>-intercept, or <math>\left(0, \frac{1}{2} \right)</math>. This implies that <math>AE = \frac{1}{2}, AC' = \frac{2}{3}</math>, and by the Pythagorean Theorem, <math>EC' = \frac{5}{6}</math> (or you could notice <math>\triangle AEC'</math> is a <math>3-4-5</math> right triangle). Then, the perimeter is <math>\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2</math>, so our answer is <math>\boxed{\textbf{(A)} ~2}</math>. ~rocketsri<br />
<br />
==Solution 3 (Fakesolve):==<br />
Assume that E is the midpoint of <math>\overline{AB}</math>. Then, <math>\overline{AE}=\frac{1}{2}</math> and since <math>C'D=\frac{1}{3}</math>, <math>\overline{AC'}=\frac{2}{3}</math>. By the Pythagorean Theorem, <math>\overline{EC'}=\frac{5}{6}</math>. It easily follows that our desired perimeter is <math>2 \rightarrow \boxed{A}</math> ~samrocksnature<br />
<br />
==Solution 4==<br />
As described in Solution 1, we can find that <math>DF=\frac{4}{9}</math>, and <math>C'F = \frac{5}{9}.</math> <br />
<br />
Then, we can find we can find the length of <math>AE</math> by expressing the length of <math>EF</math> in two different ways in terms of <math>AE</math>. If let <math>AE = a</math>, by the Pythagorean Theorem we have that <math>EC = \sqrt{a^2 + \left(\frac{2}{3}\right)^2}</math><br />
<br />
== Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) ==<br />
https://youtu.be/cagzLmdbqYQ<br />
<br />
~ pi_is_3.14<br />
<br />
{{AMC10 box|year=2021|ab=B|num-b=20|num-a=22}}</div>Laffytaffyhttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_21&diff=1464162021 AMC 10B Problems/Problem 212021-02-13T14:57:15Z<p>Laffytaffy: </p>
<hr />
<div>==Problem==<br />
A square piece of paper has side length <math>1</math> and vertices <math>A,B,C,</math> and <math>D</math> in that order. As shown in the figure, the paper is folded so that vertex <math>C</math> meets edge <math>\overline{AD}</math> at point <math>C'</math>, and edge <math>\overline{AB}</math> at point <math>E</math>. Suppose that <math>C'D = \frac{1}{3}</math>. What is the perimeter of triangle <math>\bigtriangleup AEC' ?</math><br />
<br />
<math>\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{136} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}</math><br />
<asy><br />
/* Made by samrocksnature */<br />
pair A=(0,1);<br />
pair CC=(0.666666666666,1);<br />
pair D=(1,1);<br />
pair F=(1,0.62);<br />
pair C=(1,0);<br />
pair B=(0,0);<br />
pair G=(0,0.25);<br />
pair H=(-0.13,0.41);<br />
pair E=(0,0.5);<br />
dot(A^^CC^^D^^C^^B^^E);<br />
draw(E--A--D--F);<br />
draw(G--B--C--F, dashed);<br />
fill(E--CC--F--G--H--E--CC--cycle, gray);<br />
draw(E--CC--F--G--H--E--CC);<br />
label("A",A,NW);<br />
label("B",B,SW);<br />
label("C",C,SE);<br />
label("D",D,NE);<br />
label("E",E,NW);<br />
label("C'",CC,N);<br />
</asy><br />
<br />
==Solution 1==<br />
<br />
We can set the point on <math>CD</math> where the fold occurs as point <math>F</math>. Then, we can set <math>FD</math> as <math>x</math>, and <math>CF</math> as <math>1-x</math> because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for <math>x</math>, we get, <br />
<br />
<cmath>x^2 + (\frac{1}{3})^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}</cmath><br />
<br />
We know this is a 3-4-5 triangle because the side lengths are <math>\frac{3}{9}, \frac{4}{9}, \frac{5}{9}</math>. We also know that <math>EAC'</math> is similar to <math>C'DF</math> because angle <math>C'</math> is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of <math>C'DF * \frac{AC'}{DF}</math>. Thats just <math>\frac{4}{3} * \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} * \frac{3}{2} = 2</math>. Therefore, the final answer is <math>\boxed{A}</math><br />
<br />
~Tony_Li2007<br />
<br />
==Solution 2==<br />
Let line we're reflecting over be <math>\ell</math>, and let the points where it hits <math>AB</math> and <math>CD</math>, be <math>M</math> and <math>N</math>, respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line <math>\ell</math>. The segment <math>CC'</math> has slope <math>\frac{0 - 1}{1 - 2/3} = -3</math>, implying line <math>\ell</math> has a slope of <math>\frac{1}{3}</math>. Also, the midpoint of segment <math>CC'</math> is <math>\left( \frac{5}{6}, \frac{1}{2} \right)</math>, so line <math>\ell</math> passes through this point. Then, we get the equation of line <math>\ell</math> is simply <math>y = \frac{1}{3} x + \frac{2}{9}</math>. Then, if the point where <math>B</math> is reflected over line <math>\ell</math> is <math>B'</math>, then we get <math>BB'</math> is the line <math>y = -3x</math>. The intersection of <math>\ell</math> and segment <math>BB'</math> is <math>\left( - \frac{1}{15}, \frac{1}{5} \right)</math>. So, we get <math>B' = \left(- \frac{2}{15}, \frac{2}{5} \right)</math>. Then, line segment <math>B'C'</math> has equation <math>y = \frac{3}{4} x + \frac{1}{2}</math>, so the point <math>E</math> is the <math>y</math>-intercept, or <math>\left(0, \frac{1}{2} \right)</math>. This implies that <math>AE = \frac{1}{2}, AC' = \frac{2}{3}</math>, and by the Pythagorean Theorem, <math>EC' = \frac{5}{6}</math> (or you could notice <math>\triangle AEC'</math> is a <math>3-4-5</math> right triangle). Then, the perimeter is <math>\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2</math>, so our answer is <math>\boxed{\textbf{(A)} ~2}</math>. ~rocketsri<br />
<br />
==Solution 3 (Fakesolve):==<br />
Assume that E is the midpoint of <math>\overline{AB}</math>. Then, <math>\overline{AE}=\frac{1}{2}</math> and since <math>C'D=\frac{1}{3}</math>, <math>\overline{AC'}=\frac{2}{3}</math>. By the Pythagorean Theorem, <math>\overline{EC'}=\frac{5}{6}</math>. It easily follows that our desired perimeter is <math>2 \rightarrow \boxed{A}</math> ~samrocksnature<br />
<br />
==Solution 4==<br />
<br />
== Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) ==<br />
https://youtu.be/cagzLmdbqYQ<br />
<br />
~ pi_is_3.14<br />
<br />
{{AMC10 box|year=2021|ab=B|num-b=20|num-a=22}}</div>Laffytaffyhttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_16&diff=1401952001 AMC 12 Problems/Problem 162020-12-22T16:07:44Z<p>Laffytaffy: /* Best Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?<br />
<br />
<math><br />
\text{(A) }8!<br />
\qquad<br />
\text{(B) }2^8 \cdot 8!<br />
\qquad<br />
\text{(C) }(8!)^2<br />
\qquad<br />
\text{(D) }\frac {16!}{2^8}<br />
\qquad<br />
\text{(E) }16!<br />
</math><br />
==Best Solution==<br />
Pretty clearly, if we choose each sock-shoe pair individually, they will automatically align themselves.<br />
<br />
<math>\binom{16}{2}\binom{14}{2}\cdots\binom{2}{2} = \frac{16 \cdot 15 \cdots 2}{2 \cdot 2 \cdots 2} = \frac{16!}{2^8}</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
<br />
Let the spider try to put on all <math>16</math> things in a random order. Each of the <math>16!</math> permutations is equally probable. For any fixed leg, the probability that he will first put on the sock and only then the shoe is clearly <math>\frac{1}{2}</math>. Then the probability that he will correctly put things on all legs is <math>\frac{1}{2^{8}}</math>. Therefore the number of correct permutations must be <math>\boxed{\frac {16!}{2^8}}</math>.<br />
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=== Solution 2 ===<br />
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Each dressing sequence can be uniquely described by a sequence containing two <math>1</math>s, two <math>2</math>s, ..., and two <math>8</math>s -- the first occurrence of number <math>x</math> means that the spider puts the sock onto leg <math>x</math>, the second occurrence of <math>x</math> means he puts the shoe onto leg <math>x</math>. <br />
If the numbers were all unique, the answer would be <math> 16! </math>. However, since 8 terms appear twice, the answer is <math> \frac{16!}{(2!)^8} = \boxed{\dfrac {16!}{2^8}}</math>.<br />
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=== Solution 3 ===<br />
You can put all <math>8</math> socks on first for <math>8!</math> ways and then all <math>8</math> shoes on next for <math>8!</math> more ways. This is not the only possibility, so the lower bound is <math>(8!)^2</math>. You can choose all <math>16</math> in a random fashion, but some combinations would violate the rules, so the upper bound is <math>16!</math>. <math>\text{(C)}</math> & <math>\text{(E)}</math> are the lower and upper bounds, so the answer is in between them, <math>\boxed{\frac {16!}{2^8}}</math>.<br />
== Also See ==<br />
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{{AMC12 box|year=2001|num-b=15|num-a=17}}<br />
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