https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Username222&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-22T15:53:52Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_25&diff=50547 2012 AMC 12B Problems/Problem 25 2013-01-13T03:50:10Z <p>Username222: </p> <hr /> <div>== Problem 25 ==<br /> <br /> Let &lt;math&gt;S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}&lt;/math&gt;. <br /> Let &lt;math&gt;T&lt;/math&gt; be the set of all right triangles whose vertices are in &lt;math&gt;S&lt;/math&gt;. For every right triangle &lt;math&gt;t=\triangle{ABC}&lt;/math&gt; with vertices &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; in counter-clockwise order and right angle at &lt;math&gt;A&lt;/math&gt;, let &lt;math&gt;f(t)=\tan(\angle{CBA})&lt;/math&gt;. What is &lt;cmath&gt;\prod_{t\in T} f(t)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24} &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==Solution 1==<br /> <br /> <br /> Four points in a rectangular arrangement allow 4 possible right angle triangles. These four congruent triangles will form two pairs of different vertice labelling - two ABC and two ACB. These pairs will multiple to equal 1 due to the fact that &lt;math&gt;\tan x \tan(90^{\circ}-x) = \tan x \cot x = 1.&lt;/math&gt;<br /> <br /> Due to the missing point (0,0) in the grid then in the rectangular arrangement (0,0), (x,0), (x,y), (0,y) there is only one triangle which will not cancel out to zero with a reflected version.<br /> <br /> <br /> So we need to consider all triangles of the form A(x,y), B(0,y), C(x,0). For these triangles &lt;math&gt;\text{tan B} = \frac{y}{x}&lt;/math&gt;.<br /> Multiplying them all together gives: &lt;math&gt;\frac{1}{1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \frac{3}{1} \cdot \frac{3}{2} \cdot \frac{3}{3} \cdot \frac{3}{4} \cdot \frac{4}{1} \cdot \frac{4}{2} \cdot \frac{4}{3} \cdot \frac{4}{4} \cdot \frac{5}{1} \cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} = \boxed{\textbf{(E)} \ \frac{625}{24}. } &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Consider reflections. For any right triangle &lt;math&gt;ABC&lt;/math&gt; with the right labeling described in the problem, any reflection &lt;math&gt;A'B'C'&lt;/math&gt; labeled that way will give us &lt;math&gt;\tan CBA \cdot \tan C'B'A' = 1&lt;/math&gt;. First we consider the reflection about the line &lt;math&gt;y=2.5&lt;/math&gt;. Only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex at &lt;math&gt;(0,5)&lt;/math&gt; do not reflect to a traingle &lt;math&gt;\subseteq S&lt;/math&gt;. Within those triangles, consider a reflection about the line &lt;math&gt;y=5-x&lt;/math&gt;. Then only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex on the line &lt;math&gt;y=0&lt;/math&gt; do not reflect to a triangle &lt;math&gt;\subseteq S&lt;/math&gt;. So we only need to look at right triangles that have vertices &lt;math&gt;(0,5), (*,0), (*,*)&lt;/math&gt;. There are three cases:<br /> <br /> Case 1: &lt;math&gt;A=(0,5)&lt;/math&gt;. Then &lt;math&gt;B=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Case 2: &lt;math&gt;B=(0,5)&lt;/math&gt;. Then we look for &lt;math&gt;A=(x,y)&lt;/math&gt; such that &lt;math&gt;\angle BAC=90^{\circ}&lt;/math&gt; and that &lt;math&gt;C=(*,0)&lt;/math&gt;. They are: &lt;math&gt;(A=(x,5), C=(x,0))&lt;/math&gt;, &lt;math&gt;(A=(2,4), C=(1,0))&lt;/math&gt; and &lt;math&gt;(A=(4,1), C=(3,0))&lt;/math&gt;. The product of their values of &lt;math&gt;\tan \angle CBA&lt;/math&gt; is &lt;math&gt;\frac{5}{1}\cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{625}{144}&lt;/math&gt;.<br /> <br /> Case 3: &lt;math&gt;C=(0,5)&lt;/math&gt;. Then &lt;math&gt;A=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Therefore &lt;math&gt;\boxed{\textbf{(B)} \ \frac{625}{144}}&lt;/math&gt; is the answer.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=24|after=}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_24&diff=50546 2012 AMC 12B Problems/Problem 24 2013-01-13T03:49:49Z <p>Username222: </p> <hr /> <div>== Problem 24 ==<br /> <br /> Define the function &lt;math&gt;f_1&lt;/math&gt; on the positive integers by setting &lt;math&gt;f_1(1)=1&lt;/math&gt; and if &lt;math&gt;n=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}&lt;/math&gt; is the prime factorization of &lt;math&gt;n&gt;1&lt;/math&gt;, then &lt;cmath&gt;f_1(n)=(p_1+1)^{e_1-1}(p_2+1)^{e_2-1}\cdots (p_k+1)^{e_k-1}.&lt;/cmath&gt;<br /> For every &lt;math&gt;m\ge 2&lt;/math&gt;, let &lt;math&gt;f_m(n)=f_1(f_{m-1}(n))&lt;/math&gt;. For how many &lt;math&gt;N&lt;/math&gt; in the range &lt;math&gt;1\le N\le 400&lt;/math&gt; is the sequence &lt;math&gt;(f_1(N),f_2(N),f_3(N),\dots )&lt;/math&gt; unbounded?<br /> <br /> '''Note:''' A sequence of positive numbers is unbounded if for every integer &lt;math&gt;B&lt;/math&gt;, there is a member of the sequence greater than &lt;math&gt;B&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 19 &lt;/math&gt;<br /> <br /> <br /> == Solution ==<br /> <br /> First of all, notice that for any odd prime &lt;math&gt;p&lt;/math&gt;, the largest prime that divides &lt;math&gt;p+1&lt;/math&gt; is no larger than &lt;math&gt;\frac{p+1}{2}&lt;/math&gt;, therefore eventually the factorization of &lt;math&gt;f_k(N)&lt;/math&gt; does not contain any prime larger than &lt;math&gt;3&lt;/math&gt;. Also, note that &lt;math&gt;f_2(2^m) = f_1(3^{m-1})=2^{2m-4}&lt;/math&gt;, when &lt;math&gt;m=4&lt;/math&gt; it stays the same but when &lt;math&gt;m\geq 5&lt;/math&gt; it grows indefinitely. Therefore any number &lt;math&gt;N&lt;/math&gt; that is divisible by &lt;math&gt;2^5&lt;/math&gt; or any number &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;f_k(N)&lt;/math&gt; is divisible by &lt;math&gt;2^5&lt;/math&gt; makes the sequence &lt;math&gt;(f_1(N),f_2(N),f_3(N),\dots )&lt;/math&gt; unbounded. There are &lt;math&gt;12&lt;/math&gt; multiples of &lt;math&gt;2^5&lt;/math&gt; within &lt;math&gt;400&lt;/math&gt;. &lt;math&gt;2^4 5^2=400&lt;/math&gt; also works: &lt;math&gt;f_2(2^4 5^2) = f_1(3^4 \cdot 2) = 2^6&lt;/math&gt;.<br /> <br /> Now let's look at the other cases. Any first power of prime in a prime factorization will not contribute the unboundedness because &lt;math&gt;f_1(p^1)=(p+1)^0=1&lt;/math&gt;. At least a square of prime is to contribute. So we test primes that are less than &lt;math&gt;\sqrt{400}=20&lt;/math&gt;:<br /> <br /> &lt;math&gt;f_1(3^4)=4^3=2^6&lt;/math&gt; works, therefore any number &lt;math&gt;\leq 400&lt;/math&gt; that are divisible by &lt;math&gt;3^4&lt;/math&gt; works: there are &lt;math&gt;4&lt;/math&gt; of them.<br /> <br /> &lt;math&gt;3^3 \cdot Q^2&lt;/math&gt; could also work if &lt;math&gt;Q^2&lt;/math&gt; satisfies &lt;math&gt;2~|~f_1(Q^2)&lt;/math&gt;, but &lt;math&gt;3^3 \cdot 5^2 &gt; 400&lt;/math&gt;.<br /> <br /> &lt;math&gt;f_1(5^3)=6^2 = 2^2 3^2&lt;/math&gt; does not work.<br /> <br /> &lt;math&gt;f_1(7^3)=8^2=2^6&lt;/math&gt; works. There are no other multiples of &lt;math&gt;7^3&lt;/math&gt; within &lt;math&gt;400&lt;/math&gt;.<br /> <br /> &lt;math&gt;7^2 \cdot Q^2&lt;/math&gt; could also work if &lt;math&gt;4~|~f_1(Q^2)&lt;/math&gt;, but &lt;math&gt;7^2 \cdot 3^2 &gt; 400&lt;/math&gt; already.<br /> <br /> For number that are only divisible by &lt;math&gt;p=11, 13, 17, 19&lt;/math&gt;, they don't work because none of these primes are such that &lt;math&gt;p+1&lt;/math&gt; could be a multiple of &lt;math&gt;2^5&lt;/math&gt; nor a multiple of &lt;math&gt;3^4&lt;/math&gt;.<br /> <br /> In conclusion, there are &lt;math&gt;12+1+4+1=18&lt;/math&gt; number of &lt;math&gt;N&lt;/math&gt;'s ... &lt;math&gt;\framebox{D}&lt;/math&gt;.<br /> <br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=23|num-a=25}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_23&diff=50545 2012 AMC 12B Problems/Problem 23 2013-01-13T03:49:26Z <p>Username222: </p> <hr /> <div>== Problem 23 ==<br /> <br /> Consider all polynomials of a complex variable, &lt;math&gt;P(z)=4z^4+az^3+bz^2+cz+d&lt;/math&gt;, where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are integers, &lt;math&gt;0\le d\le c\le b\le a\le 4&lt;/math&gt;, and the polynomial has a zero &lt;math&gt;z_0&lt;/math&gt; with &lt;math&gt;|z_0|=1.&lt;/math&gt; What is the sum of all values &lt;math&gt;P(1)&lt;/math&gt; over all the polynomials with these properties?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120 &lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution (doubtful) ==<br /> <br /> Since &lt;math&gt;z_0&lt;/math&gt; is a root of &lt;math&gt;P&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; has integer coefficients, &lt;math&gt;z_0&lt;/math&gt; must be algebraic. Since &lt;math&gt;z_0&lt;/math&gt; is algebraic and lies on the unit circle, &lt;math&gt;z_0&lt;/math&gt; must be a root of unity (Comment: this is not true. See this link: [http://math.stackexchange.com/questions/4323/are-all-algebraic-integers-with-absolute-value-1-roots-of-unity]). Since &lt;math&gt;P&lt;/math&gt; has degree 4, it seems reasonable (and we will assume this only temporarily) that &lt;math&gt;z_0&lt;/math&gt; must be a 2nd, 3rd, or 4th root of unity. These are among the set &lt;math&gt;\{\pm1,\pm i,(-1\pm i\sqrt{3})/2\}&lt;/math&gt;. Since complex roots of polynomials come in conjugate pairs, we have that &lt;math&gt;P&lt;/math&gt; has one (or more) of the following factors: &lt;math&gt;z+1&lt;/math&gt;, &lt;math&gt;z-1&lt;/math&gt;, &lt;math&gt;z^2+1&lt;/math&gt;, or &lt;math&gt;z^2+z+1&lt;/math&gt;. If &lt;math&gt;z=1&lt;/math&gt; then &lt;math&gt;a+b+c+d+4=0&lt;/math&gt;; a contradiction since &lt;math&gt;a,b,c,d&lt;/math&gt; are non-negative. On the other hand, suppose &lt;math&gt;z=-1&lt;/math&gt;. Then &lt;math&gt;(a+c)-(b+d)=4&lt;/math&gt;. This implies &lt;math&gt;a+b=8,7,6,5,4&lt;/math&gt; while &lt;math&gt;b+d=4,3,2,1,0&lt;/math&gt; correspondingly. After listing cases, the only such valid &lt;math&gt;a,b,c,d&lt;/math&gt; are &lt;math&gt;4,4,4,0&lt;/math&gt;, &lt;math&gt;4,3,3,0&lt;/math&gt;, &lt;math&gt;4,2,2,0&lt;/math&gt;, &lt;math&gt;4,1,1,0&lt;/math&gt;, and &lt;math&gt;4,0,0,0&lt;/math&gt;. <br /> <br /> Now suppose &lt;math&gt;z=i&lt;/math&gt;. Then &lt;math&gt;4=(a-c)i+(b-d)&lt;/math&gt; whereupon &lt;math&gt;a=c&lt;/math&gt; and &lt;math&gt;b-d=4&lt;/math&gt;. But then &lt;math&gt;a=b=c&lt;/math&gt; and &lt;math&gt;d=a-4&lt;/math&gt;. This gives only the cases &lt;math&gt;a,b,c,d&lt;/math&gt; equals &lt;math&gt;4,4,4,0&lt;/math&gt;, which we have already counted in a previous case.<br /> <br /> Suppose &lt;math&gt;z=-i&lt;/math&gt;. Then &lt;math&gt;4=i(c-a)+(b-d)&lt;/math&gt; so that &lt;math&gt;a=c&lt;/math&gt; and &lt;math&gt;b=4+d&lt;/math&gt;. This only gives rise to &lt;math&gt;a,b,c,d&lt;/math&gt; equal &lt;math&gt;4,4,4,0&lt;/math&gt; which we have previously counted. <br /> <br /> Finally suppose &lt;math&gt;z^2+z+1&lt;/math&gt; divides &lt;math&gt;P&lt;/math&gt;. Using polynomial division ((or that &lt;math&gt;z^3=1&lt;/math&gt; to make the same deductions) we ultimately obtain that &lt;math&gt;b=4+c&lt;/math&gt;. This can only happen if &lt;math&gt;a,b,c,d&lt;/math&gt; is &lt;math&gt;4,4,0,0&lt;/math&gt;. <br /> <br /> Hence we've the polynomials<br /> &lt;cmath&gt;4x^4+4x^3+4x^2+4x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+3x^2+3x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+2x^2+2x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+x^2+x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+4x^2&lt;/cmath&gt;<br /> However, by inspection &lt;math&gt;4x^4+4x^3+4x^2+4x+4&lt;/math&gt; has roots on the unit circle, because &lt;math&gt;x^4+x^3+x^2+x+1=(x^5-1)/(x-1)&lt;/math&gt; which brings the sum to 92 (choice B). Note that this polynomial has a 5th root of unity as a root. We will show that we were \textit{almost} correct in our initial assumption; that is that &lt;math&gt;z_0&lt;/math&gt; is at most a 5th root of unity, and that the last polynomial we obtained is the last polynomial with the given properties. Suppose that &lt;math&gt;z_0&lt;/math&gt; in an &lt;math&gt;n&lt;/math&gt;th root of unity where &lt;math&gt;n&gt;5&lt;/math&gt;, and &lt;math&gt;z_0&lt;/math&gt; is not a 3rd or 4th root of unity. (Note that 1st and 2nd roots of unity are themselves 4th roots of unity). If &lt;math&gt;n&lt;/math&gt; is prime, then \textit{every} &lt;math&gt;n&lt;/math&gt;th root of unity except 1 must satisfy our polynomial, but since &lt;math&gt;n&gt;5&lt;/math&gt; and the degree of our polynomial is 4, this is impossible. Suppose &lt;math&gt;n&lt;/math&gt; is composite. If it has a prime factor &lt;math&gt;p&lt;/math&gt; greater than 5 then again every &lt;math&gt;p&lt;/math&gt;th root of unity must satisfy our polynomial and we arrive at the same contradiction. Therefore suppose &lt;math&gt;n&lt;/math&gt; is divisible only by 2,3,or 5. Since by hypothesis &lt;math&gt;z_0&lt;/math&gt; is not a 2nd or 3rd root of unity, &lt;math&gt;z_0&lt;/math&gt; must be a 5th root of unity. Since 5 is prime, every 5th root of unity except 1 must satisfy our polynomial. That is, the other 4 complex 5th roots of unity must satisfy &lt;math&gt;P(z_0)=0&lt;/math&gt;. But &lt;math&gt;(x^5-1)/(x-1)&lt;/math&gt; has exactly all 5th roots of unity excluding 1, and &lt;math&gt;(x^5-1)/(x-1)=x^4+x^3+x^2+x+1&lt;/math&gt;. Thus this must divide &lt;math&gt;P&lt;/math&gt; which implies &lt;math&gt;P(x)=4(x^4+x^3+x^2+x+1)&lt;/math&gt;. This completes the proof.<br /> <br /> <br /> == Solution ==<br /> First, assume that &lt;math&gt;z_0\in \mathbb{R}&lt;/math&gt;, so &lt;math&gt;z_0=1&lt;/math&gt; or &lt;math&gt;-1&lt;/math&gt;. &lt;math&gt;1&lt;/math&gt; does not work because &lt;math&gt;P(1)\geq 4&lt;/math&gt;. Assume that &lt;math&gt;z_0=-1&lt;/math&gt;. Then &lt;math&gt;0=P(-1)=4-a+b-c+d&lt;/math&gt;, we have &lt;math&gt;4+b+d=a+c\leq 4+b&lt;/math&gt;, so &lt;math&gt;d=0&lt;/math&gt;. Also, &lt;math&gt;a=4&lt;/math&gt; has to be true since &lt;math&gt;4+b=a+c \leq a+b&lt;/math&gt;. Now &lt;math&gt;4+b=4+c&lt;/math&gt; gives &lt;math&gt;b=c&lt;/math&gt;, therefore the only possible choices for &lt;math&gt;(a,b,c,d)&lt;/math&gt; are &lt;math&gt;(4,t,t,0)&lt;/math&gt;. In these cases, &lt;math&gt;P(-1)=4-4+t-t+0=0&lt;/math&gt;. The sum of &lt;math&gt;P(-1)&lt;/math&gt; over these cases is &lt;math&gt;\sum_{t=0}^{4} (4+4+t+t) = 40+20=60&lt;/math&gt;.<br /> <br /> Second, assume that &lt;math&gt;z_0\in \mathbb{C} \backslash \mathbb{R}&lt;/math&gt;, so &lt;math&gt;z_0=x_0+iy_0&lt;/math&gt; for some real &lt;math&gt;x_0, y_0&lt;/math&gt;, &lt;math&gt;|x_0|&lt;1&lt;/math&gt;. By conjugate roots theorem we have that &lt;math&gt;P(z_0)=P(z_0^{*})=0&lt;/math&gt;, therefore &lt;math&gt;(z-z_0)(z-z_0^{*}) = (z^2 - 2x_0*z + 1)&lt;/math&gt; is a factor of &lt;math&gt;P(z)&lt;/math&gt;, and we may assume that <br /> <br /> &lt;cmath&gt;P(z) = (z^2-2x_0 z + 1)(4z^2 + pz + d)&lt;/cmath&gt;<br /> <br /> for some real &lt;math&gt;p&lt;/math&gt;. Expanding this polynomial and comparing the coefficients, we have the following equations:<br /> <br /> &lt;cmath&gt;p-8x_0 = a&lt;/cmath&gt;<br /> &lt;cmath&gt;d+4-2px_0 = b&lt;/cmath&gt;<br /> &lt;cmath&gt;p-2dx_0 = c&lt;/cmath&gt;<br /> <br /> From the first and the third we may deduce that &lt;math&gt;2x_0 = \frac{a-c}{d-4}&lt;/math&gt; and that &lt;math&gt;p=\frac{da-4c}{d-4}&lt;/math&gt;, if &lt;math&gt;d\neq 4&lt;/math&gt; (we will consider &lt;math&gt;d=4&lt;/math&gt; by the end). Let &lt;math&gt;k=2px_0=\frac{(a-c)(da-4c)}{(4-d)^2}&lt;/math&gt;. From the second equation, we know that &lt;math&gt;k=d+4-b&lt;/math&gt; is non-negative.<br /> <br /> Consider the following cases:<br /> <br /> Case 1: &lt;math&gt;a=c&lt;/math&gt;. Then &lt;math&gt;k=0&lt;/math&gt;, &lt;math&gt;b=d+4&lt;/math&gt;, so &lt;math&gt;a=b=c=4&lt;/math&gt;, &lt;math&gt;d=0&lt;/math&gt;. However, this has already been found (i.e. the form of &lt;math&gt;(4,t,t,0)&lt;/math&gt;).<br /> <br /> Case 2: &lt;math&gt;a&gt;c\geq 0&lt;/math&gt;. Then since &lt;math&gt;k\geq 0&lt;/math&gt;, we have &lt;math&gt;da-4c\geq 0&lt;/math&gt;. However, &lt;math&gt;da \leq 4c&lt;/math&gt;, therefore &lt;math&gt;da-4c=0&lt;/math&gt;. This is true only when &lt;math&gt;d=c&lt;/math&gt;. Also, we get &lt;math&gt;k=0&lt;/math&gt; again. In this case, &lt;math&gt;b=d+4&lt;/math&gt;, so &lt;math&gt;a=b=4&lt;/math&gt;, &lt;math&gt;c=d=0&lt;/math&gt;, &lt;math&gt;x_0=-1/2&lt;/math&gt;. &lt;math&gt;P(z)&lt;/math&gt; has a root &lt;math&gt;z_0=e^{i2\pi/3}&lt;/math&gt;. &lt;math&gt;P(1)=12&lt;/math&gt;.<br /> <br /> Last case: &lt;math&gt;d=4&lt;/math&gt;. We have &lt;math&gt;a=b=c=d=4&lt;/math&gt; and that &lt;math&gt;P(z)&lt;/math&gt; has a root &lt;math&gt;z_0=e^{i2\pi/5}&lt;/math&gt;. &lt;math&gt;P(1)=20&lt;/math&gt;.<br /> <br /> Therefore the desired sum is &lt;math&gt;60+12+20=92 ...\framebox{B}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=22|num-a=24}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_25&diff=50544 2012 AMC 10B Problems/Problem 25 2013-01-13T03:48:51Z <p>Username222: /* See Also */</p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #22]] and [[2012 AMC 10B Problems|2012 AMC 10B #25]]}}<br /> <br /> A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?<br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632));<br /> draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544));<br /> draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767));<br /> draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0));<br /> draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509));<br /> draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632));<br /> draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544));<br /> draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080));<br /> draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0));<br /> dot((0,0));<br /> dot((22,0));<br /> label(&quot;$A$&quot;,(0,0),WNW);<br /> label(&quot;$B$&quot;,(22,0),E);<br /> filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black);<br /> filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);<br /> filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black);<br /> filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black);<br /> filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);<br /> filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black);<br /> filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black);<br /> filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black);<br /> filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black);<br /> filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white);<br /> filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black);<br /> filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black);<br /> filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black);<br /> filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black);<br /> filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white);<br /> filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black);<br /> filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black);<br /> filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black);<br /> filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black);<br /> filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black);<br /> filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);<br /> filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black);<br /> filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632));<br /> draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544));<br /> draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767));<br /> draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0));<br /> draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509));<br /> draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632));<br /> draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544));<br /> draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080));<br /> draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0));<br /> dot((0,0));<br /> dot((22,0));<br /> label(&quot;$A$&quot;,(0,0),WNW);<br /> label(&quot;$B$&quot;,(22,0),E);<br /> filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,red);<br /> filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);<br /> filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,blue);<br /> filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black);<br /> filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);<br /> filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,blue);<br /> filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black);<br /> filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,green);<br /> filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,blue);<br /> filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white);<br /> filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,green);<br /> filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black);<br /> filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black);<br /> filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,green);<br /> filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white);<br /> filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,orange);<br /> filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,green);<br /> filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black);<br /> filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,orange);<br /> filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,red);<br /> filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);<br /> filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,blue);<br /> filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);&lt;/asy&gt;<br /> <br /> There is &lt;math&gt;1&lt;/math&gt; way to get to any of the red arrows. From the first red arrow, there are &lt;math&gt;2&lt;/math&gt; ways to get to each of the first and the second blue arrows; from the second red arrow, there are &lt;math&gt;3&lt;/math&gt; ways to get to each of the first and the second blue arrows. So there are in total &lt;math&gt;5&lt;/math&gt; ways to get to each of the blue arrows.<br /> <br /> From each of the first and second blue arrows, there are respectively &lt;math&gt;4&lt;/math&gt; ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively &lt;math&gt;8&lt;/math&gt; ways to get to each of the first and the second green arrows. Therefore there are in total &lt;math&gt;5 \cdot (4+4+8+8) = 120&lt;/math&gt; ways to get to each of the green arrows.<br /> <br /> Finally, from each of the first and second green arrows, there is respectively &lt;math&gt;2&lt;/math&gt; way to get to the first orange arrow; from each of the third and the fourth green arrows, there are &lt;math&gt;3&lt;/math&gt; ways to get to the first orange arrow. Therefore there are &lt;math&gt;120 \cdot (2+2+3+3) = 1200&lt;/math&gt; ways to get to each of the orange arrows, hence &lt;math&gt;2400&lt;/math&gt; ways to get to the point &lt;math&gt;B&lt;/math&gt;. &lt;math&gt;\framebox{E}&lt;/math&gt;<br /> <br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=24|after=}}<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=21|num-a=23}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_25&diff=50543 2012 AMC 10B Problems/Problem 25 2013-01-13T03:48:38Z <p>Username222: /* See Also */</p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #22]] and [[2012 AMC 10B Problems|2012 AMC 10B #25]]}}<br /> <br /> A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?<br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632));<br /> draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544));<br /> draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767));<br /> draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0));<br /> draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509));<br /> draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632));<br /> draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544));<br /> draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080));<br /> draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0));<br /> dot((0,0));<br /> dot((22,0));<br /> label(&quot;$A$&quot;,(0,0),WNW);<br /> label(&quot;$B$&quot;,(22,0),E);<br /> filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black);<br /> filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);<br /> filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black);<br /> filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black);<br /> filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);<br /> filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black);<br /> filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black);<br /> filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black);<br /> filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black);<br /> filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white);<br /> filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black);<br /> filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black);<br /> filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black);<br /> filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black);<br /> filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white);<br /> filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black);<br /> filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black);<br /> filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black);<br /> filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black);<br /> filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black);<br /> filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);<br /> filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black);<br /> filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632));<br /> draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544));<br /> draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767));<br /> draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0));<br /> draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509));<br /> draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632));<br /> draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544));<br /> draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080));<br /> draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0));<br /> dot((0,0));<br /> dot((22,0));<br /> label(&quot;$A$&quot;,(0,0),WNW);<br /> label(&quot;$B$&quot;,(22,0),E);<br /> filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,red);<br /> filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);<br /> filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,blue);<br /> filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black);<br /> filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);<br /> filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,blue);<br /> filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black);<br /> filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,green);<br /> filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,blue);<br /> filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white);<br /> filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,green);<br /> filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black);<br /> filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black);<br /> filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,green);<br /> filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white);<br /> filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,orange);<br /> filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,green);<br /> filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black);<br /> filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,orange);<br /> filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,red);<br /> filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);<br /> filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,blue);<br /> filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);&lt;/asy&gt;<br /> <br /> There is &lt;math&gt;1&lt;/math&gt; way to get to any of the red arrows. From the first red arrow, there are &lt;math&gt;2&lt;/math&gt; ways to get to each of the first and the second blue arrows; from the second red arrow, there are &lt;math&gt;3&lt;/math&gt; ways to get to each of the first and the second blue arrows. So there are in total &lt;math&gt;5&lt;/math&gt; ways to get to each of the blue arrows.<br /> <br /> From each of the first and second blue arrows, there are respectively &lt;math&gt;4&lt;/math&gt; ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively &lt;math&gt;8&lt;/math&gt; ways to get to each of the first and the second green arrows. Therefore there are in total &lt;math&gt;5 \cdot (4+4+8+8) = 120&lt;/math&gt; ways to get to each of the green arrows.<br /> <br /> Finally, from each of the first and second green arrows, there is respectively &lt;math&gt;2&lt;/math&gt; way to get to the first orange arrow; from each of the third and the fourth green arrows, there are &lt;math&gt;3&lt;/math&gt; ways to get to the first orange arrow. Therefore there are &lt;math&gt;120 \cdot (2+2+3+3) = 1200&lt;/math&gt; ways to get to each of the orange arrows, hence &lt;math&gt;2400&lt;/math&gt; ways to get to the point &lt;math&gt;B&lt;/math&gt;. &lt;math&gt;\framebox{E}&lt;/math&gt;<br /> <br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=24|after=}}<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=21|after=23}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_25&diff=50542 2012 AMC 10B Problems/Problem 25 2013-01-13T03:48:00Z <p>Username222: /* See Also */</p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #22]] and [[2012 AMC 10B Problems|2012 AMC 10B #25]]}}<br /> <br /> A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?<br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632));<br /> draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544));<br /> draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767));<br /> draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0));<br /> draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509));<br /> draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632));<br /> draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544));<br /> draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080));<br /> draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0));<br /> dot((0,0));<br /> dot((22,0));<br /> label(&quot;$A$&quot;,(0,0),WNW);<br /> label(&quot;$B$&quot;,(22,0),E);<br /> filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black);<br /> filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);<br /> filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black);<br /> filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black);<br /> filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);<br /> filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black);<br /> filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black);<br /> filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black);<br /> filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black);<br /> filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white);<br /> filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black);<br /> filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black);<br /> filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black);<br /> filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black);<br /> filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white);<br /> filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black);<br /> filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black);<br /> filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black);<br /> filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black);<br /> filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black);<br /> filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);<br /> filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black);<br /> filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632));<br /> draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544));<br /> draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767));<br /> draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0));<br /> draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509));<br /> draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632));<br /> draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544));<br /> draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080));<br /> draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0));<br /> dot((0,0));<br /> dot((22,0));<br /> label(&quot;$A$&quot;,(0,0),WNW);<br /> label(&quot;$B$&quot;,(22,0),E);<br /> filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,red);<br /> filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);<br /> filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,blue);<br /> filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black);<br /> filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);<br /> filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,blue);<br /> filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black);<br /> filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,green);<br /> filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,blue);<br /> filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white);<br /> filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,green);<br /> filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black);<br /> filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black);<br /> filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,green);<br /> filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white);<br /> filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,orange);<br /> filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,green);<br /> filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black);<br /> filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,orange);<br /> filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,red);<br /> filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);<br /> filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,blue);<br /> filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);&lt;/asy&gt;<br /> <br /> There is &lt;math&gt;1&lt;/math&gt; way to get to any of the red arrows. From the first red arrow, there are &lt;math&gt;2&lt;/math&gt; ways to get to each of the first and the second blue arrows; from the second red arrow, there are &lt;math&gt;3&lt;/math&gt; ways to get to each of the first and the second blue arrows. So there are in total &lt;math&gt;5&lt;/math&gt; ways to get to each of the blue arrows.<br /> <br /> From each of the first and second blue arrows, there are respectively &lt;math&gt;4&lt;/math&gt; ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively &lt;math&gt;8&lt;/math&gt; ways to get to each of the first and the second green arrows. Therefore there are in total &lt;math&gt;5 \cdot (4+4+8+8) = 120&lt;/math&gt; ways to get to each of the green arrows.<br /> <br /> Finally, from each of the first and second green arrows, there is respectively &lt;math&gt;2&lt;/math&gt; way to get to the first orange arrow; from each of the third and the fourth green arrows, there are &lt;math&gt;3&lt;/math&gt; ways to get to the first orange arrow. Therefore there are &lt;math&gt;120 \cdot (2+2+3+3) = 1200&lt;/math&gt; ways to get to each of the orange arrows, hence &lt;math&gt;2400&lt;/math&gt; ways to get to the point &lt;math&gt;B&lt;/math&gt;. &lt;math&gt;\framebox{E}&lt;/math&gt;<br /> <br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=21|after=23}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_25&diff=50541 2012 AMC 12B Problems/Problem 25 2013-01-13T03:45:38Z <p>Username222: /* Solution 1 */</p> <hr /> <div>== Problem 25 ==<br /> <br /> Let &lt;math&gt;S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}&lt;/math&gt;. <br /> Let &lt;math&gt;T&lt;/math&gt; be the set of all right triangles whose vertices are in &lt;math&gt;S&lt;/math&gt;. For every right triangle &lt;math&gt;t=\triangle{ABC}&lt;/math&gt; with vertices &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; in counter-clockwise order and right angle at &lt;math&gt;A&lt;/math&gt;, let &lt;math&gt;f(t)=\tan(\angle{CBA})&lt;/math&gt;. What is &lt;cmath&gt;\prod_{t\in T} f(t)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24} &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==Solution 1==<br /> <br /> <br /> Four points in a rectangular arrangement allow 4 possible right angle triangles. These four congruent triangles will form two pairs of different vertice labelling - two ABC and two ACB. These pairs will multiple to equal 1 due to the fact that &lt;math&gt;\tan x \tan(90^{\circ}-x) = \tan x \cot x = 1.&lt;/math&gt;<br /> <br /> Due to the missing point (0,0) in the grid then in the rectangular arrangement (0,0), (x,0), (x,y), (0,y) there is only one triangle which will not cancel out to zero with a reflected version.<br /> <br /> <br /> So we need to consider all triangles of the form A(x,y), B(0,y), C(x,0). For these triangles &lt;math&gt;\text{tan B} = \frac{y}{x}&lt;/math&gt;.<br /> Multiplying them all together gives: &lt;math&gt;\frac{1}{1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \frac{3}{1} \cdot \frac{3}{2} \cdot \frac{3}{3} \cdot \frac{3}{4} \cdot \frac{4}{1} \cdot \frac{4}{2} \cdot \frac{4}{3} \cdot \frac{4}{4} \cdot \frac{5}{1} \cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} = \boxed{\textbf{(E)} \ \frac{625}{24}. } &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Consider reflections. For any right triangle &lt;math&gt;ABC&lt;/math&gt; with the right labeling described in the problem, any reflection &lt;math&gt;A'B'C'&lt;/math&gt; labeled that way will give us &lt;math&gt;\tan CBA \cdot \tan C'B'A' = 1&lt;/math&gt;. First we consider the reflection about the line &lt;math&gt;y=2.5&lt;/math&gt;. Only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex at &lt;math&gt;(0,5)&lt;/math&gt; do not reflect to a traingle &lt;math&gt;\subseteq S&lt;/math&gt;. Within those triangles, consider a reflection about the line &lt;math&gt;y=5-x&lt;/math&gt;. Then only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex on the line &lt;math&gt;y=0&lt;/math&gt; do not reflect to a triangle &lt;math&gt;\subseteq S&lt;/math&gt;. So we only need to look at right triangles that have vertices &lt;math&gt;(0,5), (*,0), (*,*)&lt;/math&gt;. There are three cases:<br /> <br /> Case 1: &lt;math&gt;A=(0,5)&lt;/math&gt;. Then &lt;math&gt;B=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Case 2: &lt;math&gt;B=(0,5)&lt;/math&gt;. Then we look for &lt;math&gt;A=(x,y)&lt;/math&gt; such that &lt;math&gt;\angle BAC=90^{\circ}&lt;/math&gt; and that &lt;math&gt;C=(*,0)&lt;/math&gt;. They are: &lt;math&gt;(A=(x,5), C=(x,0))&lt;/math&gt;, &lt;math&gt;(A=(2,4), C=(1,0))&lt;/math&gt; and &lt;math&gt;(A=(4,1), C=(3,0))&lt;/math&gt;. The product of their values of &lt;math&gt;\tan \angle CBA&lt;/math&gt; is &lt;math&gt;\frac{5}{1}\cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{625}{144}&lt;/math&gt;.<br /> <br /> Case 3: &lt;math&gt;C=(0,5)&lt;/math&gt;. Then &lt;math&gt;A=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Therefore &lt;math&gt;\boxed{\textbf{(B)} \ \frac{625}{144}}&lt;/math&gt; is the answer.</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_25&diff=50540 2012 AMC 12B Problems/Problem 25 2013-01-13T03:45:24Z <p>Username222: /* Solution 1 */</p> <hr /> <div>== Problem 25 ==<br /> <br /> Let &lt;math&gt;S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}&lt;/math&gt;. <br /> Let &lt;math&gt;T&lt;/math&gt; be the set of all right triangles whose vertices are in &lt;math&gt;S&lt;/math&gt;. For every right triangle &lt;math&gt;t=\triangle{ABC}&lt;/math&gt; with vertices &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; in counter-clockwise order and right angle at &lt;math&gt;A&lt;/math&gt;, let &lt;math&gt;f(t)=\tan(\angle{CBA})&lt;/math&gt;. What is &lt;cmath&gt;\prod_{t\in T} f(t)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24} &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==Solution 1==<br /> <br /> <br /> Four points in a rectangular arrangement allow 4 possible right angle triangles. These four congruent triangles will form two pairs of different vertice labelling - two ABC and two ACB. These pairs will multiple to equal 1 due to the fact that &lt;math&gt;\tan x \tan(90^{\circ}-x) = \tan x \cot x = 1.&lt;/math&gt;<br /> <br /> Due to the missing point (0,0) in the grid then in the rectangular arrangement (0,0), (x,0), (x,y), (0,y) there is only one triangle which will not cancel out to zero with a reflected version.<br /> <br /> <br /> So we need to consider all triangles of the form A(x,y), B(0,y), C(x,0). For these triangle &lt;math&gt;\text{tan B} = \frac{y}{x}&lt;/math&gt;.<br /> Multiplying them all together gives: &lt;math&gt;\frac{1}{1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \frac{3}{1} \cdot \frac{3}{2} \cdot \frac{3}{3} \cdot \frac{3}{4} \cdot \frac{4}{1} \cdot \frac{4}{2} \cdot \frac{4}{3} \cdot \frac{4}{4} \cdot \frac{5}{1} \cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} = \boxed{\textbf{(E)} \ \frac{625}{24}. } &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Consider reflections. For any right triangle &lt;math&gt;ABC&lt;/math&gt; with the right labeling described in the problem, any reflection &lt;math&gt;A'B'C'&lt;/math&gt; labeled that way will give us &lt;math&gt;\tan CBA \cdot \tan C'B'A' = 1&lt;/math&gt;. First we consider the reflection about the line &lt;math&gt;y=2.5&lt;/math&gt;. Only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex at &lt;math&gt;(0,5)&lt;/math&gt; do not reflect to a traingle &lt;math&gt;\subseteq S&lt;/math&gt;. Within those triangles, consider a reflection about the line &lt;math&gt;y=5-x&lt;/math&gt;. Then only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex on the line &lt;math&gt;y=0&lt;/math&gt; do not reflect to a triangle &lt;math&gt;\subseteq S&lt;/math&gt;. So we only need to look at right triangles that have vertices &lt;math&gt;(0,5), (*,0), (*,*)&lt;/math&gt;. There are three cases:<br /> <br /> Case 1: &lt;math&gt;A=(0,5)&lt;/math&gt;. Then &lt;math&gt;B=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Case 2: &lt;math&gt;B=(0,5)&lt;/math&gt;. Then we look for &lt;math&gt;A=(x,y)&lt;/math&gt; such that &lt;math&gt;\angle BAC=90^{\circ}&lt;/math&gt; and that &lt;math&gt;C=(*,0)&lt;/math&gt;. They are: &lt;math&gt;(A=(x,5), C=(x,0))&lt;/math&gt;, &lt;math&gt;(A=(2,4), C=(1,0))&lt;/math&gt; and &lt;math&gt;(A=(4,1), C=(3,0))&lt;/math&gt;. The product of their values of &lt;math&gt;\tan \angle CBA&lt;/math&gt; is &lt;math&gt;\frac{5}{1}\cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{625}{144}&lt;/math&gt;.<br /> <br /> Case 3: &lt;math&gt;C=(0,5)&lt;/math&gt;. Then &lt;math&gt;A=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Therefore &lt;math&gt;\boxed{\textbf{(B)} \ \frac{625}{144}}&lt;/math&gt; is the answer.</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_25&diff=50539 2012 AMC 12B Problems/Problem 25 2013-01-13T03:44:29Z <p>Username222: /* Solution 2 */</p> <hr /> <div>== Problem 25 ==<br /> <br /> Let &lt;math&gt;S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}&lt;/math&gt;. <br /> Let &lt;math&gt;T&lt;/math&gt; be the set of all right triangles whose vertices are in &lt;math&gt;S&lt;/math&gt;. For every right triangle &lt;math&gt;t=\triangle{ABC}&lt;/math&gt; with vertices &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; in counter-clockwise order and right angle at &lt;math&gt;A&lt;/math&gt;, let &lt;math&gt;f(t)=\tan(\angle{CBA})&lt;/math&gt;. What is &lt;cmath&gt;\prod_{t\in T} f(t)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24} &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==Solution 1==<br /> <br /> <br /> Four points in a rectangular arrangement allow 4 possible right angle triangles. These four congruent triangles will form two pairs of different vertice labelling - two ABC and two ACB. These pairs will multiple to equal 1 due to the fact that &lt;math&gt;\tan x \tan(90^{\circ}-x) = \tan x \cot x = 1.&lt;/math&gt;<br /> <br /> Due to the missing point (0,0) in the grid then in the rectangular arrangement (0,0), (x,0), (x,y), (0,y) there is only one triangle which will not cancel out to zero with a reflected version.<br /> <br /> <br /> So we need to consider all triangles of the form A(x,y), B(0,y), C(x,0). For these triangle &lt;math&gt;tan B = \frac{y}{x}&lt;/math&gt;.<br /> Multiplying them all together gives: &lt;math&gt;\frac{1}{1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \frac{3}{1} \cdot \frac{3}{2} \cdot \frac{3}{3} \cdot \frac{3}{4} \cdot \frac{4}{1} \cdot \frac{4}{2} \cdot \frac{4}{3} \cdot \frac{4}{4} \cdot \frac{5}{1} \cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} = \boxed{\textbf{(E)} \ \frac{625}{24}. } &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Consider reflections. For any right triangle &lt;math&gt;ABC&lt;/math&gt; with the right labeling described in the problem, any reflection &lt;math&gt;A'B'C'&lt;/math&gt; labeled that way will give us &lt;math&gt;\tan CBA \cdot \tan C'B'A' = 1&lt;/math&gt;. First we consider the reflection about the line &lt;math&gt;y=2.5&lt;/math&gt;. Only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex at &lt;math&gt;(0,5)&lt;/math&gt; do not reflect to a traingle &lt;math&gt;\subseteq S&lt;/math&gt;. Within those triangles, consider a reflection about the line &lt;math&gt;y=5-x&lt;/math&gt;. Then only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex on the line &lt;math&gt;y=0&lt;/math&gt; do not reflect to a triangle &lt;math&gt;\subseteq S&lt;/math&gt;. So we only need to look at right triangles that have vertices &lt;math&gt;(0,5), (*,0), (*,*)&lt;/math&gt;. There are three cases:<br /> <br /> Case 1: &lt;math&gt;A=(0,5)&lt;/math&gt;. Then &lt;math&gt;B=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Case 2: &lt;math&gt;B=(0,5)&lt;/math&gt;. Then we look for &lt;math&gt;A=(x,y)&lt;/math&gt; such that &lt;math&gt;\angle BAC=90^{\circ}&lt;/math&gt; and that &lt;math&gt;C=(*,0)&lt;/math&gt;. They are: &lt;math&gt;(A=(x,5), C=(x,0))&lt;/math&gt;, &lt;math&gt;(A=(2,4), C=(1,0))&lt;/math&gt; and &lt;math&gt;(A=(4,1), C=(3,0))&lt;/math&gt;. The product of their values of &lt;math&gt;\tan \angle CBA&lt;/math&gt; is &lt;math&gt;\frac{5}{1}\cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{625}{144}&lt;/math&gt;.<br /> <br /> Case 3: &lt;math&gt;C=(0,5)&lt;/math&gt;. Then &lt;math&gt;A=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Therefore &lt;math&gt;\boxed{\textbf{(B)} \ \frac{625}{144}}&lt;/math&gt; is the answer.</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_25&diff=50538 2012 AMC 12B Problems/Problem 25 2013-01-13T03:44:18Z <p>Username222: /* Solution 2 */</p> <hr /> <div>== Problem 25 ==<br /> <br /> Let &lt;math&gt;S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}&lt;/math&gt;. <br /> Let &lt;math&gt;T&lt;/math&gt; be the set of all right triangles whose vertices are in &lt;math&gt;S&lt;/math&gt;. For every right triangle &lt;math&gt;t=\triangle{ABC}&lt;/math&gt; with vertices &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; in counter-clockwise order and right angle at &lt;math&gt;A&lt;/math&gt;, let &lt;math&gt;f(t)=\tan(\angle{CBA})&lt;/math&gt;. What is &lt;cmath&gt;\prod_{t\in T} f(t)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24} &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==Solution 1==<br /> <br /> <br /> Four points in a rectangular arrangement allow 4 possible right angle triangles. These four congruent triangles will form two pairs of different vertice labelling - two ABC and two ACB. These pairs will multiple to equal 1 due to the fact that &lt;math&gt;\tan x \tan(90^{\circ}-x) = \tan x \cot x = 1.&lt;/math&gt;<br /> <br /> Due to the missing point (0,0) in the grid then in the rectangular arrangement (0,0), (x,0), (x,y), (0,y) there is only one triangle which will not cancel out to zero with a reflected version.<br /> <br /> <br /> So we need to consider all triangles of the form A(x,y), B(0,y), C(x,0). For these triangle &lt;math&gt;tan B = \frac{y}{x}&lt;/math&gt;.<br /> Multiplying them all together gives: &lt;math&gt;\frac{1}{1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \frac{3}{1} \cdot \frac{3}{2} \cdot \frac{3}{3} \cdot \frac{3}{4} \cdot \frac{4}{1} \cdot \frac{4}{2} \cdot \frac{4}{3} \cdot \frac{4}{4} \cdot \frac{5}{1} \cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} = \boxed{\textbf{(E)} \ \frac{625}{24}. } &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Consider reflections. For any right triangle &lt;math&gt;ABC&lt;/math&gt; with the right labeling described in the problem, any reflection &lt;math&gt;A'B'C'&lt;/math&gt; labeled that way will give us &lt;math&gt;\tan CBA \cdot \tan C'B'A' = 1&lt;/math&gt;. First we consider the reflection about the line &lt;math&gt;y=2.5&lt;/math&gt;. Only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex at &lt;math&gt;(0,5)&lt;/math&gt; do not reflect to a traingle &lt;math&gt;\subseteq S&lt;/math&gt;. Within those triangles, consider a reflection about the line &lt;math&gt;y=5-x&lt;/math&gt;. Then only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex on the line &lt;math&gt;y=0&lt;/math&gt; do not reflect to a triangle &lt;math&gt;\subseteq S&lt;/math&gt;. So we only need to look at right triangles that have vertices &lt;math&gt;(0,5), (*,0), (*,*)&lt;/math&gt;. There are three cases:<br /> <br /> Case 1: &lt;math&gt;A=(0,5)&lt;/math&gt;. Then &lt;math&gt;B=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Case 2: &lt;math&gt;B=(0,5)&lt;/math&gt;. Then we look for &lt;math&gt;A=(x,y)&lt;/math&gt; such that &lt;math&gt;\angle BAC=90^{\circ}&lt;/math&gt; and that &lt;math&gt;C=(*,0)&lt;/math&gt;. They are: &lt;math&gt;(A=(x,5), C=(x,0))&lt;/math&gt;, &lt;math&gt;(A=(2,4), C=(1,0))&lt;/math&gt; and &lt;math&gt;(A=(4,1), C=(3,0))&lt;/math&gt;. The product of their values of &lt;math&gt;\tan \angle CBA&lt;/math&gt; is &lt;math&gt;\frac{5}{1}\cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{625}{144}&lt;/math&gt;.<br /> <br /> Case 3: &lt;math&gt;C=(0,5)&lt;/math&gt;. Then &lt;math&gt;A=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Therefore &lt;math&gt;\boxed{\textbf{(B)} \frac{625}{144}}&lt;/math&gt; is the answer.</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_25&diff=50537 2012 AMC 12B Problems/Problem 25 2013-01-13T03:43:49Z <p>Username222: /* Solution 2 */</p> <hr /> <div>== Problem 25 ==<br /> <br /> Let &lt;math&gt;S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}&lt;/math&gt;. <br /> Let &lt;math&gt;T&lt;/math&gt; be the set of all right triangles whose vertices are in &lt;math&gt;S&lt;/math&gt;. For every right triangle &lt;math&gt;t=\triangle{ABC}&lt;/math&gt; with vertices &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; in counter-clockwise order and right angle at &lt;math&gt;A&lt;/math&gt;, let &lt;math&gt;f(t)=\tan(\angle{CBA})&lt;/math&gt;. What is &lt;cmath&gt;\prod_{t\in T} f(t)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24} &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==Solution 1==<br /> <br /> <br /> Four points in a rectangular arrangement allow 4 possible right angle triangles. These four congruent triangles will form two pairs of different vertice labelling - two ABC and two ACB. These pairs will multiple to equal 1 due to the fact that &lt;math&gt;\tan x \tan(90^{\circ}-x) = \tan x \cot x = 1.&lt;/math&gt;<br /> <br /> Due to the missing point (0,0) in the grid then in the rectangular arrangement (0,0), (x,0), (x,y), (0,y) there is only one triangle which will not cancel out to zero with a reflected version.<br /> <br /> <br /> So we need to consider all triangles of the form A(x,y), B(0,y), C(x,0). For these triangle &lt;math&gt;tan B = \frac{y}{x}&lt;/math&gt;.<br /> Multiplying them all together gives: &lt;math&gt;\frac{1}{1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \frac{3}{1} \cdot \frac{3}{2} \cdot \frac{3}{3} \cdot \frac{3}{4} \cdot \frac{4}{1} \cdot \frac{4}{2} \cdot \frac{4}{3} \cdot \frac{4}{4} \cdot \frac{5}{1} \cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} = \boxed{\textbf{(E)} \ \frac{625}{24}. } &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Consider reflections. For any right triangle &lt;math&gt;ABC&lt;/math&gt; with the right labeling described in the problem, any reflection &lt;math&gt;A'B'C'&lt;/math&gt; labeled that way will give us &lt;math&gt;\tan CBA \cdot \tan C'B'A' = 1&lt;/math&gt;. First we consider the reflection about the line &lt;math&gt;y=2.5&lt;/math&gt;. Only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex at &lt;math&gt;(0,5)&lt;/math&gt; do not reflect to a traingle &lt;math&gt;\subseteq S&lt;/math&gt;. Within those triangles, consider a reflection about the line &lt;math&gt;y=5-x&lt;/math&gt;. Then only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex on the line &lt;math&gt;y=0&lt;/math&gt; do not reflect to a triangle &lt;math&gt;\subseteq S&lt;/math&gt;. So we only need to look at right triangles that have vertices &lt;math&gt;(0,5), (*,0), (*,*)&lt;/math&gt;. There are three cases:<br /> <br /> Case 1: &lt;math&gt;A=(0,5)&lt;/math&gt;. Then &lt;math&gt;B=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Case 2: &lt;math&gt;B=(0,5)&lt;/math&gt;. Then we look for &lt;math&gt;A=(x,y)&lt;/math&gt; such that &lt;math&gt;\angle BAC=90^{\circ}&lt;/math&gt; and that &lt;math&gt;C=(*,0)&lt;/math&gt;. They are: &lt;math&gt;(A=(x,5), C=(x,0))&lt;/math&gt;, &lt;math&gt;(A=(2,4), C=(1,0))&lt;/math&gt; and &lt;math&gt;(A=(4,1), C=(3,0))&lt;/math&gt;. The product of their values of &lt;math&gt;\tan \angle CBA&lt;/math&gt; is &lt;math&gt;\frac{5}{1}\cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{625}{144}&lt;/math&gt;.<br /> <br /> Case 3: &lt;math&gt;C=(0,5)&lt;/math&gt;. Then &lt;math&gt;A=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Therefore &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt; is the answer.</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_25&diff=50536 2012 AMC 12B Problems/Problem 25 2013-01-13T03:43:09Z <p>Username222: /* Solution 1 */</p> <hr /> <div>== Problem 25 ==<br /> <br /> Let &lt;math&gt;S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}&lt;/math&gt;. <br /> Let &lt;math&gt;T&lt;/math&gt; be the set of all right triangles whose vertices are in &lt;math&gt;S&lt;/math&gt;. For every right triangle &lt;math&gt;t=\triangle{ABC}&lt;/math&gt; with vertices &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; in counter-clockwise order and right angle at &lt;math&gt;A&lt;/math&gt;, let &lt;math&gt;f(t)=\tan(\angle{CBA})&lt;/math&gt;. What is &lt;cmath&gt;\prod_{t\in T} f(t)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24} &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==Solution 1==<br /> <br /> <br /> Four points in a rectangular arrangement allow 4 possible right angle triangles. These four congruent triangles will form two pairs of different vertice labelling - two ABC and two ACB. These pairs will multiple to equal 1 due to the fact that &lt;math&gt;\tan x \tan(90^{\circ}-x) = \tan x \cot x = 1.&lt;/math&gt;<br /> <br /> Due to the missing point (0,0) in the grid then in the rectangular arrangement (0,0), (x,0), (x,y), (0,y) there is only one triangle which will not cancel out to zero with a reflected version.<br /> <br /> <br /> So we need to consider all triangles of the form A(x,y), B(0,y), C(x,0). For these triangle &lt;math&gt;tan B = \frac{y}{x}&lt;/math&gt;.<br /> Multiplying them all together gives: &lt;math&gt;\frac{1}{1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \frac{3}{1} \cdot \frac{3}{2} \cdot \frac{3}{3} \cdot \frac{3}{4} \cdot \frac{4}{1} \cdot \frac{4}{2} \cdot \frac{4}{3} \cdot \frac{4}{4} \cdot \frac{5}{1} \cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} = \boxed{\textbf{(E)} \ \frac{625}{24}. } &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Consider reflections. For any right triangle &lt;math&gt;ABC&lt;/math&gt; with the right labeling described in the problem, any reflection &lt;math&gt;A'B'C'&lt;/math&gt; labeled that way will give us &lt;math&gt;\tan CBA \cdot \tan C'B'A' = 1&lt;/math&gt;. First we consider the reflection about the line &lt;math&gt;y=2.5&lt;/math&gt;. Only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex at &lt;math&gt;(0,5)&lt;/math&gt; do not reflect to a traingle &lt;math&gt;\subseteq S&lt;/math&gt;. Within those triangles, consider a reflection about the line &lt;math&gt;y=5-x&lt;/math&gt;. Then only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex on the line &lt;math&gt;y=0&lt;/math&gt; do not reflect to a triangle &lt;math&gt;\subseteq S&lt;/math&gt;. So we only need to look at right triangles that have vertices &lt;math&gt;(0,5), (*,0), (*,*)&lt;/math&gt;. There are three cases:<br /> <br /> Case 1: &lt;math&gt;A=(0,5)&lt;/math&gt;. Then &lt;math&gt;B=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Case 2: &lt;math&gt;B=(0,5)&lt;/math&gt;. Then we look for &lt;math&gt;A=(x,y)&lt;/math&gt; such that &lt;math&gt;\angle BAC=90^{\circ}&lt;/math&gt; and that &lt;math&gt;C=(*,0)&lt;/math&gt;. They are: &lt;math&gt;(A=(x,5), C=(x,0))&lt;/math&gt;, &lt;math&gt;(A=(2,4), C=(1,0))&lt;/math&gt; and &lt;math&gt;(A=(4,1), C=(3,0))&lt;/math&gt;. The product of their values of &lt;math&gt;\tan \angle CBA&lt;/math&gt; is &lt;math&gt;\frac{5}{1}\cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{625}{144}&lt;/math&gt;.<br /> <br /> Case 3: &lt;math&gt;C=(0,5)&lt;/math&gt;. Then &lt;math&gt;A=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Therefore &lt;math&gt;\framebox{B}&lt;/math&gt; is the answer.</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_4&diff=50535 2012 AMC 12B Problems/Problem 4 2013-01-13T03:41:50Z <p>Username222: Rewrote solution.</p> <hr /> <div>==Problem==<br /> <br /> Suppose that the euro is worth 1.3 dollars. If Diana has 500 dollars and Etienne has 400 euros, by what percent is the value of Etienne's money greater that the value of Diana's money?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6.5\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The ratio &lt;math&gt;\frac{400 \text{ euros}}{500 \text{ dollars}}&lt;/math&gt; can be simplified using simple dimensional analysis:<br /> <br /> &lt;math&gt;\frac{400 \text{ euros}}{500 \text{ dollars}} \cdot \frac{1.3 \text{ dollars}}{1 \text{ euro}} = \frac{520}{500} = 1.04&lt;/math&gt;, which means a value that is greater by &lt;math&gt;\boxed{ \textbf{(B)} \ 4 }&lt;/math&gt; percent.<br /> <br /> <br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=3|num-a=5}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_4&diff=50534 2012 AMC 12B Problems/Problem 4 2013-01-13T03:29:23Z <p>Username222: /* Problem */</p> <hr /> <div>==Problem==<br /> <br /> Suppose that the euro is worth 1.3 dollars. If Diana has 500 dollars and Etienne has 400 euros, by what percent is the value of Etienne's money greater that the value of Diana's money?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6.5\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> So convert everything to dollars; 400(euros) x 1.3 = 520 dollars. now, 520 divided by 500 = 1.04, which means a value that is 4% greater; B.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=3|num-a=5}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_3&diff=50533 2012 AMC 12B Problems/Problem 3 2013-01-13T03:28:58Z <p>Username222: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> For a science project, Sammy observed a chipmunk and squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> If &lt;math&gt;x&lt;/math&gt; is the number of holes that the chipmunk dug, then &lt;math&gt;3x=4(x-4)&lt;/math&gt;, so &lt;math&gt;3x=4x-16&lt;/math&gt;, and &lt;math&gt;x=16&lt;/math&gt;. The number of acorns hidden by the chipmunk is equal to &lt;math&gt;3x = \boxed{\textbf{(D)}\ 48}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=2|num-a=4}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_3&diff=50532 2012 AMC 12B Problems/Problem 3 2013-01-13T03:28:40Z <p>Username222: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> For a science project, Sammy observed a chipmunk and squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> If &lt;math&gt;x&lt;/math&gt; is the number of holes that the chipmunk dug, then &lt;math&gt;3x=4(x-4)&lt;/math&gt;, so &lt;math&gt;3x=4x-16&lt;/math&gt;, &lt;math&gt;x=16&lt;/math&gt;. The number of acorns hidden by the chipmunk is equal to &lt;math&gt;x = \boxed{\textbf{(D)}\ 48}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=2|num-a=4}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_3&diff=50531 2012 AMC 12B Problems/Problem 3 2013-01-13T03:26:26Z <p>Username222: /* Problem */</p> <hr /> <div>==Problem==<br /> <br /> For a science project, Sammy observed a chipmunk and squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> If x is the number of holes that the chipmunk dug then, 3x=4*(x-4). so 3x=4x-16. x=16, and to find number of acorns hid by chipmunk, multiply by three, to get 48; D.<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=2|num-a=4}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_2&diff=50530 2012 AMC 12B Problems/Problem 2 2013-01-13T03:25:17Z <p>Username222: /* Solution */</p> <hr /> <div>== Problem==<br /> A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200&lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution==<br /> If the radius is &lt;math&gt;5&lt;/math&gt;, then the width is &lt;math&gt;10&lt;/math&gt;, hence the length is &lt;math&gt;20&lt;/math&gt;. &lt;math&gt;10\times20= \boxed{\textbf{(E)}\ 200}.&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=1|num-a=3}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_1&diff=50529 2012 AMC 12B Problems/Problem 1 2013-01-13T03:23:36Z <p>Username222: /* See Also */</p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #1]] and [[2012 AMC 10B Problems|2012 AMC 10B #1]]}}<br /> <br /> == Problem ==<br /> <br /> Each third-grade classroom at Pearl Creek Elementary has 18 students and 2 pet rabbits. How many more students than rabbits are there in all 4 of the third-grade classrooms?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Multiplying 18 and 2 by 4 we get 72 and 8 students and rabbits respectively. Subtracting 8 from 72 we get <br /> &lt;math&gt;\boxed{\textbf{(C)}\ 64}&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> <br /> In each class, there are &lt;math&gt;18-2=16&lt;/math&gt; more students than rabbits. So for all classrooms, the difference between students and rabbits is &lt;math&gt;16 \times 4 = \boxed{\textbf{(C)}\ 64}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|before=|num-a=2}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_25&diff=50528 2012 AMC 10B Problems/Problem 25 2013-01-13T03:23:23Z <p>Username222: /* See Also */</p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #22]] and [[2012 AMC 10B Problems|2012 AMC 10B #25]]}}<br /> <br /> A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?<br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632));<br /> draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544));<br /> draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767));<br /> draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0));<br /> draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509));<br /> draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632));<br /> draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544));<br /> draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080));<br /> draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0));<br /> dot((0,0));<br /> dot((22,0));<br /> label(&quot;$A$&quot;,(0,0),WNW);<br /> label(&quot;$B$&quot;,(22,0),E);<br /> filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black);<br /> filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);<br /> filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black);<br /> filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black);<br /> filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);<br /> filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black);<br /> filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black);<br /> filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black);<br /> filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black);<br /> filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white);<br /> filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black);<br /> filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black);<br /> filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black);<br /> filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black);<br /> filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white);<br /> filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black);<br /> filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black);<br /> filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black);<br /> filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black);<br /> filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black);<br /> filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);<br /> filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black);<br /> filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632));<br /> draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544));<br /> draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767));<br /> draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0));<br /> draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509));<br /> draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632));<br /> draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544));<br /> draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080));<br /> draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0));<br /> dot((0,0));<br /> dot((22,0));<br /> label(&quot;$A$&quot;,(0,0),WNW);<br /> label(&quot;$B$&quot;,(22,0),E);<br /> filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,red);<br /> filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);<br /> filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,blue);<br /> filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black);<br /> filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);<br /> filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,blue);<br /> filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black);<br /> filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,green);<br /> filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,blue);<br /> filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white);<br /> filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,green);<br /> filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black);<br /> filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black);<br /> filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,green);<br /> filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white);<br /> filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,orange);<br /> filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,green);<br /> filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black);<br /> filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,orange);<br /> filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,red);<br /> filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);<br /> filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,blue);<br /> filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);&lt;/asy&gt;<br /> <br /> There is &lt;math&gt;1&lt;/math&gt; way to get to any of the red arrows. From the first red arrow, there are &lt;math&gt;2&lt;/math&gt; ways to get to each of the first and the second blue arrows; from the second red arrow, there are &lt;math&gt;3&lt;/math&gt; ways to get to each of the first and the second blue arrows. So there are in total &lt;math&gt;5&lt;/math&gt; ways to get to each of the blue arrows.<br /> <br /> From each of the first and second blue arrows, there are respectively &lt;math&gt;4&lt;/math&gt; ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively &lt;math&gt;8&lt;/math&gt; ways to get to each of the first and the second green arrows. Therefore there are in total &lt;math&gt;5 \cdot (4+4+8+8) = 120&lt;/math&gt; ways to get to each of the green arrows.<br /> <br /> Finally, from each of the first and second green arrows, there is respectively &lt;math&gt;2&lt;/math&gt; way to get to the first orange arrow; from each of the third and the fourth green arrows, there are &lt;math&gt;3&lt;/math&gt; ways to get to the first orange arrow. Therefore there are &lt;math&gt;120 \cdot (2+2+3+3) = 1200&lt;/math&gt; ways to get to each of the orange arrows, hence &lt;math&gt;2400&lt;/math&gt; ways to get to the point &lt;math&gt;B&lt;/math&gt;. &lt;math&gt;\framebox{E}&lt;/math&gt;<br /> <br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=24|after=}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_25&diff=50527 2012 AMC 10B Problems/Problem 25 2013-01-13T03:23:07Z <p>Username222: </p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #22]] and [[2012 AMC 10B Problems|2012 AMC 10B #25]]}}<br /> <br /> A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?<br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632));<br /> draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544));<br /> draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767));<br /> draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0));<br /> draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509));<br /> draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632));<br /> draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544));<br /> draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080));<br /> draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0));<br /> dot((0,0));<br /> dot((22,0));<br /> label(&quot;$A$&quot;,(0,0),WNW);<br /> label(&quot;$B$&quot;,(22,0),E);<br /> filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black);<br /> filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);<br /> filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black);<br /> filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black);<br /> filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);<br /> filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black);<br /> filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black);<br /> filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black);<br /> filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black);<br /> filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white);<br /> filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black);<br /> filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black);<br /> filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black);<br /> filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black);<br /> filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white);<br /> filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black);<br /> filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black);<br /> filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black);<br /> filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black);<br /> filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black);<br /> filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);<br /> filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black);<br /> filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632));<br /> draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544));<br /> draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767));<br /> draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0));<br /> draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509));<br /> draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632));<br /> draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544));<br /> draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080));<br /> draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0));<br /> dot((0,0));<br /> dot((22,0));<br /> label(&quot;$A$&quot;,(0,0),WNW);<br /> label(&quot;$B$&quot;,(22,0),E);<br /> filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,red);<br /> filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);<br /> filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,blue);<br /> filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black);<br /> filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);<br /> filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,blue);<br /> filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black);<br /> filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,green);<br /> filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,blue);<br /> filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white);<br /> filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,green);<br /> filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black);<br /> filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black);<br /> filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,green);<br /> filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white);<br /> filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,orange);<br /> filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,green);<br /> filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black);<br /> filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,orange);<br /> filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,red);<br /> filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);<br /> filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,blue);<br /> filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);&lt;/asy&gt;<br /> <br /> There is &lt;math&gt;1&lt;/math&gt; way to get to any of the red arrows. From the first red arrow, there are &lt;math&gt;2&lt;/math&gt; ways to get to each of the first and the second blue arrows; from the second red arrow, there are &lt;math&gt;3&lt;/math&gt; ways to get to each of the first and the second blue arrows. So there are in total &lt;math&gt;5&lt;/math&gt; ways to get to each of the blue arrows.<br /> <br /> From each of the first and second blue arrows, there are respectively &lt;math&gt;4&lt;/math&gt; ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively &lt;math&gt;8&lt;/math&gt; ways to get to each of the first and the second green arrows. Therefore there are in total &lt;math&gt;5 \cdot (4+4+8+8) = 120&lt;/math&gt; ways to get to each of the green arrows.<br /> <br /> Finally, from each of the first and second green arrows, there is respectively &lt;math&gt;2&lt;/math&gt; way to get to the first orange arrow; from each of the third and the fourth green arrows, there are &lt;math&gt;3&lt;/math&gt; ways to get to the first orange arrow. Therefore there are &lt;math&gt;120 \cdot (2+2+3+3) = 1200&lt;/math&gt; ways to get to each of the orange arrows, hence &lt;math&gt;2400&lt;/math&gt; ways to get to the point &lt;math&gt;B&lt;/math&gt;. &lt;math&gt;\framebox{E}&lt;/math&gt;<br /> <br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=24|num-a=26}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_21&diff=50526 2012 AMC 12B Problems/Problem 21 2013-01-13T03:22:50Z <p>Username222: </p> <hr /> <div>==Problem==<br /> <br /> Square &lt;math&gt;AXYZ&lt;/math&gt; is inscribed in equiangular hexagon &lt;math&gt;ABCDEF&lt;/math&gt; with &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt; on &lt;math&gt;\overline{DE}&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; on &lt;math&gt;\overline{EF}&lt;/math&gt;. Suppose that &lt;math&gt;AB=40&lt;/math&gt;, and &lt;math&gt;EF=41(\sqrt{3}-1)&lt;/math&gt;. What is the side-length of the square?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3}\qquad\textbf{(C)}\ 20\sqrt{3}+16&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(D)}\ 20\sqrt{2}+13\sqrt{3} <br /> \qquad\textbf{(E)}\ 21\sqrt{6}&lt;/math&gt;<br /> <br /> ==Solution (Long)==<br /> <br /> Extend &lt;math&gt;AF&lt;/math&gt; and &lt;math&gt;YE&lt;/math&gt; so that they meet at &lt;math&gt;G&lt;/math&gt;. Then &lt;math&gt;\angle FEG=\angle GFE=60^{\circ}&lt;/math&gt;, so &lt;math&gt;\angle FGE=60^{\circ}&lt;/math&gt; and therefore &lt;math&gt;AB&lt;/math&gt; is parallel to &lt;math&gt;YE&lt;/math&gt;. Also, since &lt;math&gt;AX&lt;/math&gt; is parallel and equal to &lt;math&gt;YZ&lt;/math&gt;, we get &lt;math&gt;\angle BAX = \angle ZYE&lt;/math&gt;, hence &lt;math&gt;\triangle ABX&lt;/math&gt; is congruent to &lt;math&gt;\triangle YEZ&lt;/math&gt;. We now get &lt;math&gt;YE=AB=40&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;a_1=EY=40&lt;/math&gt;, &lt;math&gt;a_2=AF&lt;/math&gt;, and &lt;math&gt;a_3=EF&lt;/math&gt;.<br /> <br /> Drop a perpendicular line from &lt;math&gt;A&lt;/math&gt; to the line of &lt;math&gt;EF&lt;/math&gt; that meets line &lt;math&gt;EF&lt;/math&gt; at &lt;math&gt;K&lt;/math&gt;, and a perpendicular line from &lt;math&gt;Y&lt;/math&gt; to the line of &lt;math&gt;EF&lt;/math&gt; that meets &lt;math&gt;EF&lt;/math&gt; at &lt;math&gt;L&lt;/math&gt;, then &lt;math&gt;\triangle AKZ&lt;/math&gt; is congruent to &lt;math&gt;\triangle ZLY&lt;/math&gt; since &lt;math&gt;\angle YZL&lt;/math&gt; is complementary to &lt;math&gt;\angle KZA&lt;/math&gt;. Then we have the following equations:<br /> <br /> &lt;cmath&gt;\frac{\sqrt{3}}{2}a_2 = AK=ZL = ZE+\frac{1}{2} a_1&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{\sqrt{3}}{2}a_1 = YL =ZK = ZF+\frac{1}{2} a_2&lt;/cmath&gt;<br /> <br /> The sum of these two yields that <br /> <br /> &lt;cmath&gt;\frac{\sqrt{3}}{2}(a_1+a_2) = \frac{1}{2} (a_1+a_2) + ZE+ZF = \frac{1}{2} (a_1+a_2) + EF&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{\sqrt{3}-1}{2}(a_1+a_2) = 41(\sqrt{3}-1)&lt;/cmath&gt;<br /> &lt;cmath&gt;a_1+a_2=82&lt;/cmath&gt;<br /> &lt;cmath&gt;a_2=82-40=42.&lt;/cmath&gt;<br /> <br /> So, we can now use the law of cosines in &lt;math&gt;\triangle AGY&lt;/math&gt;:<br /> <br /> &lt;cmath&gt; 2AZ^2 = AY^2 = AG^2 + YG^2 - 2AG\cdot YG \cdot \cos 60^{\circ}&lt;/cmath&gt;<br /> &lt;cmath&gt; = (a_2+a_3)^2 + (a_1+a_3)^2 - (a_2+a_3)(a_1+a_3)&lt;/cmath&gt;<br /> &lt;cmath&gt; = (41\sqrt{3}+1)^2 + (41\sqrt{3}-1)^2 - (41\sqrt{3}+1)(41\sqrt{3}-1)&lt;/cmath&gt;<br /> &lt;cmath&gt; = 6 \cdot 41^2 + 2 - 3 \cdot 41^2 + 1 = 3 (41^2 + 1) = 3\cdot 1682&lt;/cmath&gt;<br /> &lt;cmath&gt; AZ^2 = 3 \cdot 841 = 3 \cdot 29^2&lt;/cmath&gt;<br /> <br /> Therefore &lt;math&gt;AZ = 29\sqrt{3} ... \framebox{A}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=20|num-a=22}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_20&diff=50525 2012 AMC 12B Problems/Problem 20 2013-01-13T03:22:34Z <p>Username222: </p> <hr /> <div>== Problem 20 ==<br /> <br /> A trapezoid has side lengths 3, 5, 7, and 11. The sums of all the possible areas of the trapezoid can be written in the form of &lt;math&gt;r_1\sqrt{n_1}+r_2\sqrt{n_2}+r_3&lt;/math&gt;, where &lt;math&gt;r_1&lt;/math&gt;, &lt;math&gt;r_2&lt;/math&gt;, and &lt;math&gt;r_3&lt;/math&gt; are rational numbers and &lt;math&gt;n_1&lt;/math&gt; and &lt;math&gt;n_2&lt;/math&gt; are positive integers not divisible by the square of any prime. What is the greatest integer less than or equal to &lt;math&gt;r_1+r_2+r_3+n_1+n_2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 57\qquad\textbf{(B)}\ 59\qquad\textbf{(C)}\ 61\qquad\textbf{(D)}\ 63\qquad\textbf{(E)}\ 65&lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> ==Solution==<br /> Name the trapezoid &lt;math&gt;ABCD&lt;/math&gt;, where &lt;math&gt;AB&lt;/math&gt; is parallel to &lt;math&gt;CD&lt;/math&gt;, &lt;math&gt;AB&lt;CD&lt;/math&gt;, and &lt;math&gt;AD&lt;BC&lt;/math&gt;. Draw a line through &lt;math&gt;B&lt;/math&gt; parallel to &lt;math&gt;AD&lt;/math&gt;, crossing the side &lt;math&gt;CD&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt;. Then &lt;math&gt;BE=AD&lt;/math&gt;, &lt;math&gt;EC=DC-DE=DC-AB&lt;/math&gt;. One needs to guarantee that &lt;math&gt;BE+EC&gt;BC&lt;/math&gt;, so there are only three possible trapezoids:<br /> <br /> &lt;cmath&gt;AB=3, BC=7, CD=11, DA=5, CE=8&lt;/cmath&gt;<br /> &lt;cmath&gt;AB=5, BC=7, CD=11, DA=3, CE=6&lt;/cmath&gt;<br /> &lt;cmath&gt;AB=7, BC=5, CD=11, DA=3, CE=4&lt;/cmath&gt;<br /> <br /> <br /> In the first case, &lt;math&gt;\cos(\angle BCD) = (8^2+7^2-5^2)/(2\cdot 7\cdot 8) = 11/14&lt;/math&gt;, so &lt;math&gt;\sin (\angle BCD) = \sqrt{1-121/196} = 5\sqrt{3}/14&lt;/math&gt;. Therefore the area of this trapezoid is &lt;math&gt;\frac{1}{2} (3+11) \cdot 7 \cdot 5\sqrt{3}/14 = \frac{35}{2}\sqrt{3}&lt;/math&gt;.<br /> <br /> In the first case, &lt;math&gt;\cos(\angle BCD) = (6^2+7^2-3^2)/(2\cdot 6\cdot 7) = 19/21&lt;/math&gt;, so &lt;math&gt;\sin (\angle BCD) = \sqrt{1-361/441} = 4\sqrt{5}/21&lt;/math&gt;. Therefore the area of this trapezoid is &lt;math&gt;\frac{1}{2} (5+11) \cdot 7 \cdot 4\sqrt{5}/21 =\frac{32}{3}\sqrt{5}&lt;/math&gt;.<br /> <br /> In the first case, &lt;math&gt;\angle BCD = 90^{\circ}&lt;/math&gt;, therefore the area of this trapezoid is &lt;math&gt;\frac{1}{2} (7+11) \cdot 3 = 27&lt;/math&gt;.<br /> <br /> So &lt;math&gt;r_1 + r_2 + r_3 + n_1 + n_2 = 17.5 + 10.666... + 27 + 3 + 5&lt;/math&gt;, which is rounded down to &lt;math&gt;63... \framebox{D}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=19|num-a=21}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_19&diff=50524 2012 AMC 12B Problems/Problem 19 2013-01-13T03:22:12Z <p>Username222: </p> <hr /> <div>==Solution==<br /> <br /> [[File:2012_AMC-12B-19‎.jpg]]<br /> <br /> Observe the diagram above. Each dot represents one of the six vertices of the regular octahedron. Three dots have been placed exactly x units from the (0,0,0) corner of the unit cube. The other three dots have been placed exactly x units from the (1,1,1) corner of the unit cube. A red square has been drawn connecting four of the dots to provide perspective regarding the shape of the octahedron. Observe that the three dots that are near (0,0,0) are each (x)(&lt;math&gt;\sqrt{2}&lt;/math&gt;) from each other. The same is true for the three dots that are near (1,1,1). There is a unique x for which the rectangle drawn in red becomes a square. This will occur when the distance from (x,0,0) to (1,1-x, 1) is (x)(&lt;math&gt;\sqrt{2}&lt;/math&gt;).<br /> <br /> <br /> Using the distance formula we find the distance between the two points to be: &lt;math&gt;\sqrt{{(1-x)^2} + {(1-x)^2} + 1}&lt;/math&gt; = &lt;math&gt;\sqrt{2x^2 - 4x +3}&lt;/math&gt;. Equating this to (x)(&lt;math&gt;\sqrt{2}&lt;/math&gt;) and squaring both sides, we have the equation:<br /> <br /> &lt;math&gt;2{x^2}&lt;/math&gt; - &lt;math&gt;4x + 3&lt;/math&gt; = &lt;math&gt;2{x^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;-4x + 3 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;x&lt;/math&gt; = &lt;math&gt;\frac{3} {4}&lt;/math&gt;.<br /> <br /> <br /> Since the length of each side is (x)(&lt;math&gt;\sqrt{2}&lt;/math&gt;), we have a final result of &lt;math&gt;\frac{3 \sqrt{2}}{4}&lt;/math&gt;. Thus, Answer choice A is correct.<br /> <br /> <br /> (If someone can draw a better diagram with the points labeled P1,P2, etc., I would appreciate it).<br /> <br /> --[[User:Jm314|Jm314]] 14:55, 26 February 2012 (EST)<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=18|num-a=20}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_18&diff=50523 2012 AMC 12B Problems/Problem 18 2013-01-13T03:21:55Z <p>Username222: </p> <hr /> <div>== Problem 18 ==<br /> <br /> Let &lt;math&gt;(a_1,a_2, \dots ,a_{10})&lt;/math&gt; be a list of the first 10 positive integers such that for each &lt;math&gt;2 \le i \le 10&lt;/math&gt; either &lt;math&gt;a_i+1&lt;/math&gt; or &lt;math&gt;a_i-1&lt;/math&gt; or both appear somewhere before &lt;math&gt;a_i&lt;/math&gt; in the list. How many such lists are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ 362,880&lt;/math&gt;<br /> <br /> == Solution ==<br /> Let &lt;math&gt;1\leq k\leq 10&lt;/math&gt;. Assume that &lt;math&gt;a_1=k&lt;/math&gt;. If &lt;math&gt;k&lt;10&lt;/math&gt;, the first number appear after &lt;math&gt;k&lt;/math&gt; that is greater than &lt;math&gt;k&lt;/math&gt; must be &lt;math&gt;k+1&lt;/math&gt;, otherwise if it is any number &lt;math&gt;x&lt;/math&gt; larger than &lt;math&gt;k+1&lt;/math&gt;, there will be neither &lt;math&gt;x-1&lt;/math&gt; nor &lt;math&gt;x+1&lt;/math&gt; appearing before &lt;math&gt;x&lt;/math&gt;. Similarly, one can conclude that if &lt;math&gt;k+1&lt;10&lt;/math&gt;, the first number appear after &lt;math&gt;k+1&lt;/math&gt; that is larger than &lt;math&gt;k+1&lt;/math&gt; must be &lt;math&gt;k+2&lt;/math&gt;, and so forth.<br /> <br /> On the other hand, if &lt;math&gt;k&gt;1&lt;/math&gt;, the first number appear after &lt;math&gt;k&lt;/math&gt; that is less than &lt;math&gt;k&lt;/math&gt; must be &lt;math&gt;k-1&lt;/math&gt;, and then &lt;math&gt;k-2&lt;/math&gt;, and so forth.<br /> <br /> To count the number of possibilities when &lt;math&gt;a_1=k&lt;/math&gt; is given, we set up &lt;math&gt;9&lt;/math&gt; spots after &lt;math&gt;k&lt;/math&gt;, and assign &lt;math&gt;k-1&lt;/math&gt; of them to the numbers less than &lt;math&gt;k&lt;/math&gt; and the rest to the numbers greater than &lt;math&gt;k&lt;/math&gt;. The number of ways in doing so is &lt;math&gt;9&lt;/math&gt; choose &lt;math&gt;k-1&lt;/math&gt;.<br /> <br /> Therefore, when summing up the cases from &lt;math&gt;k=1&lt;/math&gt; to &lt;math&gt;10&lt;/math&gt;, we get<br /> <br /> &lt;cmath&gt;\binom{9}{0} + \binom{9}{1} + \cdots + \binom{9}{9} = 2^9=512 ...... \framebox{B}&lt;/cmath&gt;<br /> <br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=17|num-a=19}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_17&diff=50522 2012 AMC 12B Problems/Problem 17 2013-01-13T03:21:38Z <p>Username222: </p> <hr /> <div>==Problem==<br /> <br /> Square &lt;math&gt;PQRS&lt;/math&gt; lies in the first quadrant. Points &lt;math&gt;(3,0), (5,0), (7,0),&lt;/math&gt; and &lt;math&gt;(13,0)&lt;/math&gt; lie on lines &lt;math&gt;SP, RQ, PQ&lt;/math&gt;, and &lt;math&gt;SR&lt;/math&gt;, respectively. What is the sum of the coordinates of the center of the square &lt;math&gt;PQRS&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 6.2\qquad\textbf{(C)}\ 6.4\qquad\textbf{(D)}\ 6.6\qquad\textbf{(E)}\ 6.8 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Let the four points be labeled &lt;math&gt;P_1&lt;/math&gt;, &lt;math&gt;P_2&lt;/math&gt;, &lt;math&gt;P_3&lt;/math&gt;, and &lt;math&gt;P_4&lt;/math&gt;, respectively. Let the lines that go through each point be labeled &lt;math&gt;L_1&lt;/math&gt;, &lt;math&gt;L_2&lt;/math&gt;, &lt;math&gt;L_3&lt;/math&gt;, and &lt;math&gt;L_4&lt;/math&gt;, respectively. Since &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; go through &lt;math&gt;SP&lt;/math&gt; and &lt;math&gt;RQ&lt;/math&gt;, respectively, and &lt;math&gt;SP&lt;/math&gt; and &lt;math&gt;RQ&lt;/math&gt; are opposite sides of the square, we can say that &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; are parallel with slope &lt;math&gt;m&lt;/math&gt;. Similarly, &lt;math&gt;L_3&lt;/math&gt; and &lt;math&gt;L_4&lt;/math&gt; have slope &lt;math&gt;-\frac{1}{m}&lt;/math&gt;. Also, note that since square &lt;math&gt;PQRS&lt;/math&gt; lies in the first quadrant, &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; must have a positive slope. Using the point-slope form, we can now find the equations of all four lines: &lt;math&gt;L_1: y = m(x-3)&lt;/math&gt;, &lt;math&gt;L_2: y = m(x-5)&lt;/math&gt;, &lt;math&gt;L_3: y = -\frac{1}{m}(x-7)&lt;/math&gt;, &lt;math&gt;L_4: y = -\frac{1}{m}(x-13)&lt;/math&gt;.<br /> <br /> <br /> Since &lt;math&gt;PQRS&lt;/math&gt; is a square, it follows that &lt;math&gt;\Delta x&lt;/math&gt; between points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; is equal to &lt;math&gt;\Delta y&lt;/math&gt; between points &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;. Our approach will be to find &lt;math&gt;\Delta x&lt;/math&gt; and &lt;math&gt;\Delta y&lt;/math&gt; in terms of &lt;math&gt;m&lt;/math&gt; and equate the two to solve for &lt;math&gt;m&lt;/math&gt;. &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_3&lt;/math&gt; intersect at point &lt;math&gt;P&lt;/math&gt;. Setting the equations for &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_3&lt;/math&gt; equal to each other and solving for &lt;math&gt;x&lt;/math&gt;, we find that they intersect at &lt;math&gt;x = \frac{3m^2 + 7}{m^2 + 1}&lt;/math&gt;. &lt;math&gt;L_2&lt;/math&gt; and &lt;math&gt;L_3&lt;/math&gt; intersect at point &lt;math&gt;Q&lt;/math&gt;. Intersecting the two equations, the &lt;math&gt;x&lt;/math&gt;-coordinate of point &lt;math&gt;Q&lt;/math&gt; is found to be &lt;math&gt;x = \frac{5m^2 + 7}{m^2 + 1}&lt;/math&gt;. Subtracting the two, we get &lt;math&gt;\Delta x = \frac{2m^2}{m^2 + 1}&lt;/math&gt;. Substituting the &lt;math&gt;x&lt;/math&gt;-coordinate for point &lt;math&gt;Q&lt;/math&gt; found above into the equation for &lt;math&gt;L_2&lt;/math&gt;, we find that the &lt;math&gt;y&lt;/math&gt;-coordinate of point &lt;math&gt;Q&lt;/math&gt; is &lt;math&gt;y = \frac{2m}{m^2+1}&lt;/math&gt;. &lt;math&gt;L_2&lt;/math&gt; and &lt;math&gt;L_4&lt;/math&gt; intersect at point &lt;math&gt;R&lt;/math&gt;. Intersecting the two equations, the &lt;math&gt;y&lt;/math&gt;-coordinate of point &lt;math&gt;R&lt;/math&gt; is found to be &lt;math&gt;y = \frac{8m}{m^2 + 1}&lt;/math&gt;. Subtracting the two, we get &lt;math&gt;\Delta y = \frac{6m}{m^2 + 1}&lt;/math&gt;. Equating &lt;math&gt;\Delta x&lt;/math&gt; and &lt;math&gt;\Delta y&lt;/math&gt;, we get &lt;math&gt;2m^2 = 6m&lt;/math&gt; which gives us &lt;math&gt;m = 3&lt;/math&gt;. Finally, note that the line which goes though the midpoint of &lt;math&gt;P_1&lt;/math&gt; and &lt;math&gt;P_2&lt;/math&gt; with slope &lt;math&gt;3&lt;/math&gt; and the line which goes through the midpoint of &lt;math&gt;P_3&lt;/math&gt; and &lt;math&gt;P_4&lt;/math&gt; with slope &lt;math&gt;-\frac{1}{3}&lt;/math&gt; must intersect at at the center of the square. The equation of the line going through &lt;math&gt;(4,0)&lt;/math&gt; is given by &lt;math&gt;y = 3(x-4)&lt;/math&gt; and the equation of the line going through &lt;math&gt;(10,0)&lt;/math&gt; is &lt;math&gt;y = -\frac{1}{3}(x-10)&lt;/math&gt;. Equating the two, we find that they intersect at &lt;math&gt;(4.6, 1.8)&lt;/math&gt;. Adding the &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;-coordinates, we get &lt;math&gt;6.4&lt;/math&gt;. Thus, answer choice &lt;math&gt;\boxed{\textbf{(C)}}&lt;/math&gt; is correct.<br /> <br /> ==Solution 2==<br /> <br /> Note that the center of the square lies along a line that has an &lt;math&gt;x-&lt;/math&gt;intercept of &lt;math&gt;\frac{3+5}{2}=4&lt;/math&gt;, and also along another line with &lt;math&gt;x-&lt;/math&gt;intercept &lt;math&gt;\frac{7+13}{2}=10&lt;/math&gt;. Since these 2 lines are parallel to the sides of the square, they are perpindicular (since the sides of a square are). Let &lt;math&gt;m&lt;/math&gt; be the slope of the first line. Then &lt;math&gt;-\frac{1}{m}&lt;/math&gt; is the slope of the second line. We may use the point-slope form for the equation of a line to write &lt;math&gt;l_1:y=m(x-4)&lt;/math&gt; and &lt;math&gt;l_2:y=-\frac{1}{m}(x-10)&lt;/math&gt;. We easily calculate the intersection of these lines using substitution or elimination to obtain &lt;math&gt;\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)&lt;/math&gt; as the center or the square. Let &lt;math&gt;\theta&lt;/math&gt; denote the (acute) angle formed by &lt;math&gt;l_1&lt;/math&gt; and the &lt;math&gt;x-&lt;/math&gt;axis. Note that &lt;math&gt;\tan\theta=m&lt;/math&gt;. Let &lt;math&gt;s&lt;/math&gt; denote the side length of the square. Then &lt;math&gt;\sin\theta=s/2&lt;/math&gt;. On the other hand the acute angle formed by &lt;math&gt;l_2&lt;/math&gt; and the &lt;math&gt;x-&lt;/math&gt;axis is &lt;math&gt;90-\theta&lt;/math&gt; so that &lt;math&gt;\cos\theta=\sin(90-\theta)=s/6&lt;/math&gt;. Using &lt;math&gt;\cos\theta=\sqrt{1-\sin^2\theta}&lt;/math&gt; (for acute &lt;math&gt;\theta&lt;/math&gt;) we have &lt;math&gt;\frac{s}{6}=\sqrt{1-\left(\frac{s}{2}\right)^2}&lt;/math&gt; where upon &lt;math&gt;s=\frac{3\sqrt{10}}{5}&lt;/math&gt;. Then &lt;math&gt;m=\tan\theta=3&lt;/math&gt;. Substituting into &lt;math&gt;\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)&lt;/math&gt; we obtain &lt;math&gt;\left(\frac{23}{5},\frac{9}{5}\right)&lt;/math&gt; so that the sum of the coordinates is &lt;math&gt;\frac{32}{5}=6.4&lt;/math&gt;. Hence the answer is &lt;math&gt;\framebox{C}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Fast)==<br /> Suppose<br /> <br /> &lt;cmath&gt;SP: y=m(x-3)&lt;/cmath&gt;<br /> &lt;cmath&gt;RQ: y=m(x-5)&lt;/cmath&gt;<br /> &lt;cmath&gt;PQ: -my=x-7&lt;/cmath&gt;<br /> &lt;cmath&gt;SR: -my=x-13&lt;/cmath&gt;<br /> <br /> where &lt;math&gt;m &gt;0&lt;/math&gt;.<br /> <br /> Recall that the distance between two parallel lines &lt;math&gt;Ax+By+C=0&lt;/math&gt; and &lt;math&gt;Ax+By+C_1=0&lt;/math&gt; is &lt;math&gt;|C-C_1|/\sqrt{A^2+B^2}&lt;/math&gt;, we have distance between &lt;math&gt;SP&lt;/math&gt; and &lt;math&gt;RQ&lt;/math&gt; equals to &lt;math&gt;2m/\sqrt{1+m^2}&lt;/math&gt;, and the distance between &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;SR&lt;/math&gt; equals to &lt;math&gt;6/\sqrt{1+m^2}&lt;/math&gt;. Equating them, we get &lt;math&gt;m=3&lt;/math&gt;.<br /> <br /> Then, the center of the square is just the intersection between the following two &quot;mid&quot; lines:<br /> <br /> &lt;cmath&gt;L_1: y=3(x-4)&lt;/cmath&gt;<br /> &lt;cmath&gt;L_2: -3y = x-6&lt;/cmath&gt;<br /> <br /> The solution is &lt;math&gt;(4.6,1.8)&lt;/math&gt;, so we get the answer &lt;math&gt;4.6+1.8=6.4&lt;/math&gt;. &lt;math&gt;\framebox{C}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=16|num-a=18}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_16&diff=50521 2012 AMC 12B Problems/Problem 16 2013-01-13T03:21:22Z <p>Username222: </p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #16]] and [[2012 AMC 10B Problems|2012 AMC 10B #24]]}}<br /> <br /> == Problem==<br /> Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 &lt;/math&gt;<br /> <br /> <br /> <br /> <br /> <br /> == Solution==<br /> Let the ordered triple &lt;math&gt;(a,b,c)&lt;/math&gt; denote that &lt;math&gt;a&lt;/math&gt; songs are liked by Amy and Beth, &lt;math&gt;b&lt;/math&gt; songs by Beth and Jo, and &lt;math&gt;c&lt;/math&gt; songs by Jo and Amy. We claim that the only possible triples are &lt;math&gt;(1,1,1), (2,1,1), (1,2,1)(1,1,2)&lt;/math&gt;. <br /> <br /> To show this, observe these are all valid conditions. Second, note that none of &lt;math&gt;a,b,c&lt;/math&gt; can be bigger than 3. Suppose otherwise, that &lt;math&gt;a = 3&lt;/math&gt;. Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to be at least 1. In fact, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to equal 1, otherwise there will be a song liked by all three. Suppose &lt;math&gt;b = 1&lt;/math&gt;. Then we must have &lt;math&gt;c=0&lt;/math&gt; since no song is liked by all three girls, a contradiction.<br /> <br /> '''Case 1''':How many ways are there for &lt;math&gt;(a,b,c)&lt;/math&gt; to equal &lt;math&gt;(1,1,1)&lt;/math&gt;? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So &lt;math&gt;(a,b,c)=(1,1,1)&lt;/math&gt; in &lt;math&gt;4\cdot3\cdot2\cdot4 = 96&lt;/math&gt; ways.<br /> <br /> '''Case 2''':To find the number of ways for &lt;math&gt;(a,b,c) = (2,1,1)&lt;/math&gt;, observe there are &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are &lt;math&gt;6\cdot2\cdot3=36&lt;/math&gt; ways for the girls to like the songs.<br /> <br /> That gives a total of &lt;math&gt;96 + 36 = \boxed{132}&lt;/math&gt; ways for the girls to like the song, so the answer is &lt;math&gt;(\textrm{\textbf{B}})&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=15|num-a=17}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_15&diff=50520 2012 AMC 12B Problems/Problem 15 2013-01-13T03:21:10Z <p>Username222: </p> <hr /> <div>==Problem==<br /> <br /> Jesse cuts a circular disk of radius 12, along 2 radii to form 2 sectors, one with a central angle of 120. He makes two circular cones using each sector to form the lateral surface of each cone. What is the ratio of the volume of the smaller cone to the larger cone?<br /> <br /> <br /> <br /> ==Solution==<br /> <br /> If the original radius is &lt;math&gt;12&lt;/math&gt;, then the circumference is &lt;math&gt;24\pi&lt;/math&gt;; since arcs are defined by the central angles, the smaller arc, &lt;math&gt;120&lt;/math&gt; degree angle, is half the size of the larger sector. so the smaller arc is &lt;math&gt;8\pi&lt;/math&gt;, and the larger is &lt;math&gt;16\pi&lt;/math&gt;. Those two arc lengths become the two circumferences of the new cones; so the radius of the smaller cone is &lt;math&gt;4&lt;/math&gt; and the larger cone is &lt;math&gt;8&lt;/math&gt;. Using the Pythagorean theorem, the height of the larger cone is &lt;math&gt;4\cdot\sqrt{5}&lt;/math&gt; and the smaller cone is &lt;math&gt;8\cdot\sqrt{2}&lt;/math&gt;, and now for volume just square the radii and multiply by &lt;math&gt;\tfrac{1}{3}&lt;/math&gt; of the height to get the volume of each cone: &lt;math&gt;128\cdot\sqrt{2}&lt;/math&gt; and &lt;math&gt;256\cdot\sqrt{5}&lt;/math&gt; [both multiplied by three as ratio come out the same. now divide the volumes by each other to get the final ratio of &lt;math&gt;\boxed{\textbf{(C) } \frac{\sqrt{10}}{10}}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=14|num-a=16}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_14&diff=50519 2012 AMC 12B Problems/Problem 14 2013-01-13T03:20:53Z <p>Username222: </p> <hr /> <div>== Problem==<br /> Bernardo and Silvia play the following game. An integer between &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;999&lt;/math&gt; inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she addes &lt;math&gt;50&lt;/math&gt; to it and passes the result to Bernardo. The winner is the last person who produces a number less than &lt;math&gt;1000&lt;/math&gt;. Let &lt;math&gt;N&lt;/math&gt; be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 &lt;/math&gt;<br /> <br /> == Solution==<br /> <br /> === Solution 1 ===<br /> <br /> The last number that Bernado says has to be between 950 and 999. Note that 1-&gt;2-&gt;52-&gt;104-&gt;154-&gt;308-&gt;358-&gt;716-&gt;776 contains 4 doubling actions. Thus, we have &lt;math&gt;x \rightarrow 2x \rightarrow 2x+50 \rightarrow 4x+100 \rightarrow 4x+150 \rightarrow 8x+300 \rightarrow 8x+350 \rightarrow 16x+700&lt;/math&gt;. <br /> <br /> Thus, &lt;math&gt;950&lt;16x+700&lt;1000&lt;/math&gt;. Then, &lt;math&gt;16x&gt;250 \implies x \geq 16&lt;/math&gt;. If &lt;math&gt;x=16&lt;/math&gt;, we have &lt;math&gt;16x+700=956&lt;/math&gt;. Working backwards from 956, <br /> <br /> &lt;math&gt;956 \rightarrow 478 \rightarrow 428 \rightarrow 214 \rightarrow 164 \rightarrow 82 \rightarrow 32 \rightarrow 16&lt;/math&gt;. <br /> <br /> So the starting number is 16, and our answer is &lt;math&gt;1+6=\boxed{7}&lt;/math&gt;, which is A.<br /> <br /> === Solution 2 ===<br /> <br /> Work backwards. The last number Bernardo produces must be in the range &lt;math&gt;[950,999]&lt;/math&gt;. That means that before this, Silvia must produce a number in the range &lt;math&gt;[475,499]&lt;/math&gt;. Before this, Bernardo must produce a number in the range &lt;math&gt;[425,449]&lt;/math&gt;. Before this, Silvia must produce a number in the range &lt;math&gt;[213,224]&lt;/math&gt;. Before this, Bernardo must produce a number in the range &lt;math&gt;[163,174]&lt;/math&gt;. Before this, Silvia must produce a number in the range &lt;math&gt;[82,87]&lt;/math&gt;. Before this, Bernardo must produce a number in the range &lt;math&gt;[32,37]&lt;/math&gt;. Before this, Silvia must produce a number in the range &lt;math&gt;[16,18]&lt;/math&gt;. Bernardo could not have added 50 to any number before this to obtain a number in the range &lt;math&gt;[16,18]&lt;/math&gt;, hence the minimum &lt;math&gt;N&lt;/math&gt; is 16 with the sum of digits being &lt;math&gt;\boxed{7}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=13|num-a=15}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_13&diff=50518 2012 AMC 12B Problems/Problem 13 2013-01-13T03:20:41Z <p>Username222: </p> <hr /> <div>==Problem==<br /> <br /> Two parabolas have equations &lt;math&gt;y= x^2 + ax +b&lt;/math&gt; and &lt;math&gt;y= x^2 + cx +d&lt;/math&gt;, where &lt;math&gt;a, b, c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are integers, each chosen independently by rolling a fair six-sided die. What is the probability that the parabolas will have a least one point in common?<br /> <br /> ==Solution 1==<br /> <br /> Set the two equations equal to each other: &lt;math&gt;x^2 + ax + b = x^2 + cx + d&lt;/math&gt;. Now remove the x squared and get x's on one side: &lt;math&gt;ax-cx=d-b&lt;/math&gt;. Now factor &lt;math&gt;x&lt;/math&gt;: &lt;math&gt;x(a-c)=d-b&lt;/math&gt;. If a cannot equal &lt;math&gt;c&lt;/math&gt;, then there is always a solution, but if &lt;math&gt;a=c&lt;/math&gt;, a &lt;math&gt;1&lt;/math&gt; in &lt;math&gt;6&lt;/math&gt; chance, leaving a &lt;math&gt;1080&lt;/math&gt; out &lt;math&gt;1296&lt;/math&gt;, always having at least one point in common. And if &lt;math&gt;a=c&lt;/math&gt;, then the only way for that to work, is if &lt;math&gt;d=b&lt;/math&gt;, a &lt;math&gt;1&lt;/math&gt; in &lt;math&gt;36&lt;/math&gt; chance, however, this can occur &lt;math&gt;6&lt;/math&gt; ways, so a &lt;math&gt;1&lt;/math&gt; in &lt;math&gt;6&lt;/math&gt; chance of this happening. So adding one sixth to &lt;math&gt;\frac{1080}{1296}&lt;/math&gt;, we get the simplified fraction of &lt;math&gt;\frac{31}{36}&lt;/math&gt;; answer &lt;math&gt;(D)&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Proceed as above to obtain &lt;math&gt;x(a-c)=d-b&lt;/math&gt;. The probability that the parabolas have at least 1 point in common is 1 minus the probability that they do not intersect. The equation &lt;math&gt;x(a-c)=d-b&lt;/math&gt; has no solution if and only if &lt;math&gt;a=c&lt;/math&gt; and &lt;math&gt;d\neq b&lt;/math&gt;. The probability that &lt;math&gt;a=c&lt;/math&gt; is &lt;math&gt;\frac{1}{6}&lt;/math&gt; while the probability that &lt;math&gt;d\neq b&lt;/math&gt; is &lt;math&gt;\frac{5}{6}&lt;/math&gt;. Thus we have &lt;math&gt;1-\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)=\frac{31}{36}&lt;/math&gt; for the probability that the parabolas intersect.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=12|num-a=14}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_12&diff=50517 2012 AMC 12B Problems/Problem 12 2013-01-13T03:20:26Z <p>Username222: </p> <hr /> <div>==Solution 1==<br /> <br /> There are 20 Choose 2 selections however, we count these twice therefore<br /> <br /> 2* 20 C 2 = 380. The wording of the question implies D not E.<br /> <br /> MAA decided to accept both D and E, however.<br /> <br /> ==Solution 2==<br /> <br /> Consider the 20 term sequence of 0's and 1's. Keeping all other terms 1, a sequence of &lt;math&gt;k&gt;0&lt;/math&gt; consecutive 0's can be placed in &lt;math&gt;21-k&lt;/math&gt; locations. That is, there are 20 strings with 1 zero, 19 strings with 2 consecutive zeros, 18 strings with 3 consecutive zeros, ..., 1 string with 20 consecutive zeros. Hence there are &lt;math&gt;20+19+\cdots+1=\binom{21}{2}&lt;/math&gt; strings with consecutive zeros. The same argument shows there are &lt;math&gt;\binom{21}{2}&lt;/math&gt; strings with consecutive 1's. This yields &lt;math&gt;2\binom{21}{2}&lt;/math&gt; strings in all. However, we have counted twice those strings in which all the 1's and all the 0's are consecutive. These are the cases &lt;math&gt;01111...&lt;/math&gt;, &lt;math&gt;00111...&lt;/math&gt;, &lt;math&gt;000111...&lt;/math&gt;, ..., &lt;math&gt;000...0001&lt;/math&gt; (of which there are 19) as well as the cases &lt;math&gt;10000...&lt;/math&gt;, &lt;math&gt;11000...&lt;/math&gt;, &lt;math&gt;111000...&lt;/math&gt;, ..., &lt;math&gt;111...110&lt;/math&gt; (of which there are 19 as well). This yields &lt;math&gt;2\binom{21}{2}-2\cdot19=382&lt;/math&gt; so that the answer is &lt;math&gt;\framebox{E}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=11|num-a=13}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_11&diff=50516 2012 AMC 12B Problems/Problem 11 2013-01-13T03:20:12Z <p>Username222: </p> <hr /> <div>==Problem==<br /> <br /> In the equation, 132 (base A) + 43 (base B) = 69 (base A+B), A and B are consecutive integers. What is A+B?<br /> <br /> <br /> <br /> ==Solution==<br /> <br /> Change the equation to base 10: &lt;cmath&gt;A^2 + 3A +2 + 4B +3= 6A + 6B + 9&lt;/cmath&gt; &lt;cmath&gt; A^2 - 3A - 2B - 4=0&lt;/cmath&gt; <br /> <br /> Either &lt;math&gt;B = A + 1&lt;/math&gt; or &lt;math&gt;B = A - 1&lt;/math&gt;, so either &lt;math&gt;A^2 - 5A - 6, B = A + 1&lt;/math&gt; or &lt;math&gt;A^2 - 5A - 2, B = A - 1&lt;/math&gt;. The second case has no integer roots, and the first can be re-expressed as &lt;math&gt;(A-6)(A+1) = 0, B = A + 1&lt;/math&gt;. Since A must be positive, &lt;math&gt;A = 6, B = 7&lt;/math&gt; and &lt;math&gt;A+B = 13&lt;/math&gt;; C.<br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=10|num-a=12}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_10&diff=50515 2012 AMC 12B Problems/Problem 10 2013-01-13T03:20:00Z <p>Username222: </p> <hr /> <div>==Problem==<br /> <br /> What is the area of the polygon whose vertices are the points of intersection of the curves x^2 + y^2 =25 and (x-4)^2 + 9y^2 = 81.<br /> <br /> <br /> <br /> ==Solution==<br /> <br /> The first curve is a circle with radius 5 centered at the origin, and the second curve is an ellipse with center (4,0) and end points of (-5,0) and (13,0). Finding points of intersection, we get (-5,0) (4,3) and (4,-3), forming a triangle with height of 9 and base of 6. So 9x6x0.5 =27 ; B.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=9|num-a=11}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_9&diff=50514 2012 AMC 12B Problems/Problem 9 2013-01-13T03:19:44Z <p>Username222: </p> <hr /> <div>==Problem==<br /> <br /> It takes Clea 60 seconds to walk down an escalator when it is not moving, and 24 seconds when it is moving. How seconds would it take Clea to ride the escalator down when she is not walking?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52 &lt;/math&gt;<br /> <br /> ==Solution==<br /> She walks at a rate of &lt;math&gt;x&lt;/math&gt; units per second to travel a distance &lt;math&gt;y&lt;/math&gt;. As &lt;math&gt;vt=d&lt;/math&gt;, we find &lt;math&gt;60x=y&lt;/math&gt; and &lt;math&gt;24*(x+k)=y&lt;/math&gt;, where &lt;math&gt;k&lt;/math&gt; is the speed of the escalator. Setting the two equations equal to each other, &lt;math&gt;60x=24x+24k&lt;/math&gt;, which means that &lt;math&gt;k=1.5x&lt;/math&gt;. Now we divide &lt;math&gt;60&lt;/math&gt; by &lt;math&gt;1.5&lt;/math&gt; because you add the speed of the escalator but remove the walking, leaving the final answer that it takes to ride the escalator alone as &lt;math&gt;\boxed{\textbf{(B)}\ 40}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=8|num-a=10}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_8&diff=50513 2012 AMC 12B Problems/Problem 8 2013-01-13T03:19:25Z <p>Username222: /* Solution */</p> <hr /> <div>== Problem 8 ==<br /> <br /> A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304 &lt;/math&gt;<br /> <br /> ==Solution==<br /> We can count the number of possible foods for each day and then multiply to enumerate the number of combinations.<br /> <br /> On Friday, we have one possibility: cake.<br /> <br /> On Saturday, we have three possibilities: pie, ice cream, or pudding. This is the end of the week.<br /> <br /> On Thursday, we have three possibilities: pie, ice cream, or pudding. We can't have cake because we have to have cake the following day, which is the Friday with the birthday party.<br /> <br /> On Wednesday, we have three possibilities: cake, plus the two things that were not eaten on Thursday.<br /> <br /> Similarly, on Tuesday, we have three possibilities: the three things that were not eaten on Wednesday.<br /> <br /> Likewise on Monday: three possibilities, the three things that were not eaten on Tuesday.<br /> <br /> On Sunday, it is tempting to think there are four possibilities, but remember that cake must be served on Friday. This serves to limit the number of foods we can eat on Sunday, with the result being that there are three possibilities: The three things that were not eaten on Monday.<br /> <br /> So the number of menus is &lt;math&gt;3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 1 \cdot 3 = 729.&lt;/math&gt;<br /> The answer is &lt;math&gt; A.&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=7|num-a=9}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_8&diff=50512 2012 AMC 12B Problems/Problem 8 2013-01-13T03:18:43Z <p>Username222: </p> <hr /> <div>== Problem 8 ==<br /> <br /> A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304 &lt;/math&gt;<br /> <br /> ==Solution==<br /> We can count the number of possible foods for each day and then multiply to enumerate the number of combinations.<br /> <br /> On Friday, we have one possibility: cake.<br /> <br /> On Saturday, we have three possibilities: pie, ice cream, or pudding. This is the end of the week.<br /> <br /> On Thursday, we have three possibilities: pie, ice cream, or pudding. We can't have cake because we have to have cake the following day, which is the Friday with the birthday party.<br /> <br /> On Wednesday, we have three possibilities: cake, plus the two things that were not eaten on Thursday.<br /> <br /> Similarly, on Tuesday, we have three possibilities: the three things that were not eaten on Wednesday.<br /> <br /> Likewise on Monday: three possibilities, the three things that were not eaten on Tuesday.<br /> <br /> On Sunday, it is tempting to think there are four possibilities, but remember that cake must be served on Friday. This serves to limit the number of foods we can eat on Sunday, with the result being that there are three possibilities: The three things that were not eaten on Monday.<br /> <br /> So the number of menus is &lt;math&gt;3*3*3*3*3*1*3 = 729.&lt;/math&gt;<br /> The answer is &lt;math&gt; A.&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=7|num-a=9}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_7&diff=50511 2012 AMC 12B Problems/Problem 7 2013-01-13T02:51:09Z <p>Username222: </p> <hr /> <div>== Problem ==<br /> <br /> Small lights are hung on a string 6 inches apart in the order red, red, green, green, green, red, red, green, green, green, and so on continuing this pattern of 2 red lights followed by 3 green lights. How many feet separate the 3rd red light and the 21st red light?<br /> <br /> '''Note:''' 1 foot is equal to 12 inches.<br /> <br /> &lt;math&gt; \textbf{(A)}\ 18\qquad\textbf{(B)}\ 18.5\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 20.5\qquad\textbf{(E)}\ 22.5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> We know the repeating section is made of 2 red lights and 3 green lights. The 3rd red light would appear in the 2nd section of this pattern, and the 21st red light would appear in the 11th section. There would then be a total of 44 lights in between the 3rd and 21st red light, translating to 45 6-inch gaps. Since it wants the answer in feet, so the answer is &lt;math&gt;\frac{45*6}{12} \rightarrow \boxed{\textbf{(E)}\ 22.5}&lt;/math&gt; since it wants the answer in feet.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=6|num-a=8}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_6&diff=50510 2012 AMC 12B Problems/Problem 6 2013-01-13T02:50:56Z <p>Username222: </p> <hr /> <div>== Problem==<br /> In order to estimate the value of &lt;math&gt;x-y&lt;/math&gt; where &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are real numbers with &lt;math&gt;x&gt;y&gt;0&lt;/math&gt;, Xiaoli rounded &lt;math&gt;x&lt;/math&gt; up by a small amount, rounded &lt;math&gt;y&lt;/math&gt; down by the same amount, and then subtracted her rounded values. Which of the following statements is necessarily correct?<br /> <br /> '''(A)''' Her estimate is larger than &lt;math&gt;x-y&lt;/math&gt;.<br /> <br /> '''(B)''' Her estimate is smaller than &lt;math&gt;x-y&lt;/math&gt;.<br /> <br /> '''(C)''' Her estimate equals &lt;math&gt;x-y&lt;/math&gt;.<br /> <br /> '''(D)''' Her estimate equals &lt;math&gt;y-x&lt;/math&gt;.<br /> <br /> '''(E)''' Her estimate is &lt;math&gt;0&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> The original expression &lt;math&gt;x-y&lt;/math&gt; now becomes &lt;math&gt;(x+k) - (y-k)=(x-y)+2k&gt;x-y&lt;/math&gt;, where &lt;math&gt;k&lt;/math&gt; is a positive constant, hence the answer is '''(A)'''.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=5|num-a=7}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_5&diff=50509 2012 AMC 12B Problems/Problem 5 2013-01-13T02:50:42Z <p>Username222: </p> <hr /> <div>==Problem==<br /> <br /> Two integers have a sum of 26. when two more integers are added to the first two, the sum is 41. Finally, when two more integers are added to the sum of the previous 4 integers, the sum is 57. What is the minimum number of even integers among the 6 integers?<br /> <br /> <br /> <br /> ==Solution==<br /> <br /> So, x+y=26, x could equal 15, and y could equal 11, so no even integers required here. 41-26=15. a+b=15, a could equal 9 and b could equal 6, so one even integer is required here. 57-41=16. m+n=16, m could equal 9 and n could equal 7, so no even integers required here, meaning only 1 even integer is required; A.<br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=4|num-a=6}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_4&diff=50508 2012 AMC 12B Problems/Problem 4 2013-01-13T02:50:28Z <p>Username222: </p> <hr /> <div>==Problem==<br /> <br /> Suppose that the euro is worth 1.3 dollars. If Diana has 500 dollars and Etienne has 400 euros, by what percent is the value of Etienne's money greater that the value of Diana's money?<br /> <br /> <br /> <br /> <br /> ==Solution==<br /> <br /> So convert everything to dollars; 400(euros) x 1.3 = 520 dollars. now, 520 divided by 500 = 1.04, which means a value that is 4% greater; B.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=3|num-a=5}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_1&diff=50507 2012 AMC 12B Problems/Problem 1 2013-01-13T02:50:07Z <p>Username222: /* See Also */</p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #1]] and [[2012 AMC 10B Problems|2012 AMC 10B #1]]}}<br /> <br /> == Problem ==<br /> <br /> Each third-grade classroom at Pearl Creek Elementary has 18 students and 2 pet rabbits. How many more students than rabbits are there in all 4 of the third-grade classrooms?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Multiplying 18 and 2 by 4 we get 72 and 8 students and rabbits respectively. Subtracting 8 from 72 we get <br /> &lt;math&gt;\boxed{\textbf{(C)}\ 64}&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> <br /> In each class, there are &lt;math&gt;18-2=16&lt;/math&gt; more students than rabbits. So for all classrooms, the difference between students and rabbits is &lt;math&gt;16 \times 4 = \boxed{\textbf{(C)}\ 64}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|before=First Problem|num-a=2}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_1&diff=50506 2012 AMC 12B Problems/Problem 1 2013-01-13T02:49:42Z <p>Username222: </p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #1]] and [[2012 AMC 10B Problems|2012 AMC 10B #1]]}}<br /> <br /> == Problem ==<br /> <br /> Each third-grade classroom at Pearl Creek Elementary has 18 students and 2 pet rabbits. How many more students than rabbits are there in all 4 of the third-grade classrooms?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Multiplying 18 and 2 by 4 we get 72 and 8 students and rabbits respectively. Subtracting 8 from 72 we get <br /> &lt;math&gt;\boxed{\textbf{(C)}\ 64}&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> <br /> In each class, there are &lt;math&gt;18-2=16&lt;/math&gt; more students than rabbits. So for all classrooms, the difference between students and rabbits is &lt;math&gt;16 \times 4 = \boxed{\textbf{(C)}\ 64}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=First Problem|num-a=2}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_2&diff=50505 2012 AMC 12B Problems/Problem 2 2013-01-13T02:48:24Z <p>Username222: /* See Also */</p> <hr /> <div>== Problem==<br /> A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200&lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution==<br /> If the radius is &lt;math&gt;5&lt;/math&gt;, then the width is &lt;math&gt;10&lt;/math&gt;, hence the length is &lt;math&gt;20&lt;/math&gt;. &lt;math&gt;10\times20=200&lt;/math&gt;, &lt;math&gt;\boxed{\text{E}}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=1|num-a=3}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_3&diff=50504 2012 AMC 12B Problems/Problem 3 2013-01-13T02:48:14Z <p>Username222: </p> <hr /> <div>==Problem==<br /> <br /> For a science project, Sammy observed a chipmunk and squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?<br /> <br /> <br /> ==Solution==<br /> <br /> If x is the number of holes that the chipmunk dug then, 3x=4*(x-4). so 3x=4x-16. x=16, and to find number of acorns hid by chipmunk, multiply by three, to get 48; D.<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=2|num-a=4}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_2&diff=50503 2012 AMC 12B Problems/Problem 2 2013-01-13T02:47:45Z <p>Username222: </p> <hr /> <div>== Problem==<br /> A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200&lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution==<br /> If the radius is &lt;math&gt;5&lt;/math&gt;, then the width is &lt;math&gt;10&lt;/math&gt;, hence the length is &lt;math&gt;20&lt;/math&gt;. &lt;math&gt;10\times20=200&lt;/math&gt;, &lt;math&gt;\boxed{\text{E}}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=2|num-a=2}}</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=Talk:2012_AMC_12B_Problems/Problem_25&diff=50502 Talk:2012 AMC 12B Problems/Problem 25 2013-01-13T02:46:07Z <p>Username222: /* Answer... */</p> <hr /> <div>==Answer==<br /> <br /> Solution 1 gives an answer of E, while Solution 2 gives the answer as B. The answer key page lists the answer as B. Interestingly enough, I found E using the same logic as whoever posted this answer. What is the right answer?<br /> --[[User:Username222|Username222]] 21:43, 12 January 2013 (EST)</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=User:Username222&diff=50501 User:Username222 2013-01-13T02:43:24Z <p>Username222: Created page with &quot;Hi! You have reached my page.&quot;</p> <hr /> <div>Hi! You have reached my page.</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=Talk:2012_AMC_12B_Problems/Problem_25&diff=50500 Talk:2012 AMC 12B Problems/Problem 25 2013-01-13T02:43:10Z <p>Username222: Created page with &quot;==Answer...== The answers given in solution 1 is different from that given in solution 2. What's going on? --~~~~&quot;</p> <hr /> <div>==Answer...==<br /> <br /> The answers given in solution 1 is different from that given in solution 2. What's going on?<br /> --[[User:Username222|Username222]] 21:43, 12 January 2013 (EST)</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_25&diff=50499 2012 AMC 12B Problems/Problem 25 2013-01-13T02:41:33Z <p>Username222: /* Solution 1 */ adding LaTEX</p> <hr /> <div>== Problem 25 ==<br /> <br /> Let &lt;math&gt;S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}&lt;/math&gt;. <br /> Let &lt;math&gt;T&lt;/math&gt; be the set of all right triangles whose vertices are in &lt;math&gt;S&lt;/math&gt;. For every right triangle &lt;math&gt;t=\triangle{ABC}&lt;/math&gt; with vertices &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; in counter-clockwise order and right angle at &lt;math&gt;A&lt;/math&gt;, let &lt;math&gt;f(t)=\tan(\angle{CBA})&lt;/math&gt;. What is &lt;cmath&gt;\prod_{t\in T} f(t)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24} &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==Solution 1==<br /> <br /> <br /> Four points in a rectangular arrangement allow 4 possible right angle triangles. These four congruent triangles will form two pairs of different vertice labelling - two ABC and two ACB. These pairs will multiple to equal 1 due to the fact that &lt;math&gt;\tan x \tan(90^{\circ}-x) = \tan x \cot x = 1.&lt;/math&gt;<br /> <br /> Due to the missing point (0,0) in the grid then in the rectangular arrangement (0,0), (x,0), (x,y), (0,y) there is only one triangle which will not cancel out to zero with a reflected version.<br /> <br /> <br /> So we need to consider all triangles of the form A(x,y), B(0,y), C(x,0). For these triangle &lt;math&gt;tan B = \frac{y}{x}&lt;/math&gt;.<br /> Multiplying them all together gives: &lt;math&gt;\frac{1}{1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \frac{3}{1} \cdot \frac{3}{2} \cdot \frac{3}{3} \cdot \frac{3}{4} \cdot \frac{4}{1} \cdot \frac{4}{2} \cdot \frac{4}{3} \cdot \frac{4}{4} \cdot \frac{5}{1} \cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} = \frac{625}{24}. &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Consider reflections. For any right triangle &lt;math&gt;ABC&lt;/math&gt; with the right labeling described in the problem, any reflection &lt;math&gt;A'B'C'&lt;/math&gt; labeled that way will give us &lt;math&gt;\tan CBA \cdot \tan C'B'A' = 1&lt;/math&gt;. First we consider the reflection about the line &lt;math&gt;y=2.5&lt;/math&gt;. Only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex at &lt;math&gt;(0,5)&lt;/math&gt; do not reflect to a traingle &lt;math&gt;\subseteq S&lt;/math&gt;. Within those triangles, consider a reflection about the line &lt;math&gt;y=5-x&lt;/math&gt;. Then only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex on the line &lt;math&gt;y=0&lt;/math&gt; do not reflect to a triangle &lt;math&gt;\subseteq S&lt;/math&gt;. So we only need to look at right triangles that have vertices &lt;math&gt;(0,5), (*,0), (*,*)&lt;/math&gt;. There are three cases:<br /> <br /> Case 1: &lt;math&gt;A=(0,5)&lt;/math&gt;. Then &lt;math&gt;B=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Case 2: &lt;math&gt;B=(0,5)&lt;/math&gt;. Then we look for &lt;math&gt;A=(x,y)&lt;/math&gt; such that &lt;math&gt;\angle BAC=90^{\circ}&lt;/math&gt; and that &lt;math&gt;C=(*,0)&lt;/math&gt;. They are: &lt;math&gt;(A=(x,5), C=(x,0))&lt;/math&gt;, &lt;math&gt;(A=(2,4), C=(1,0))&lt;/math&gt; and &lt;math&gt;(A=(4,1), C=(3,0))&lt;/math&gt;. The product of their values of &lt;math&gt;\tan \angle CBA&lt;/math&gt; is &lt;math&gt;\frac{5}{1}\cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{625}{144}&lt;/math&gt;.<br /> <br /> Case 3: &lt;math&gt;C=(0,5)&lt;/math&gt;. Then &lt;math&gt;A=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Therefore &lt;math&gt;\framebox{B}&lt;/math&gt; is the answer.</div> Username222 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_25&diff=50498 2012 AMC 12B Problems/Problem 25 2013-01-13T02:39:58Z <p>Username222: /* Solution 1 */ In the process of LaTEXizing.</p> <hr /> <div>== Problem 25 ==<br /> <br /> Let &lt;math&gt;S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}&lt;/math&gt;. <br /> Let &lt;math&gt;T&lt;/math&gt; be the set of all right triangles whose vertices are in &lt;math&gt;S&lt;/math&gt;. For every right triangle &lt;math&gt;t=\triangle{ABC}&lt;/math&gt; with vertices &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; in counter-clockwise order and right angle at &lt;math&gt;A&lt;/math&gt;, let &lt;math&gt;f(t)=\tan(\angle{CBA})&lt;/math&gt;. What is &lt;cmath&gt;\prod_{t\in T} f(t)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24} &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==Solution 1==<br /> <br /> <br /> Four points in a rectangular arrangement allow 4 possible right angle triangles. These four congruent triangles will form two pairs of different vertice labelling - two ABC and two ACB. These pairs will multiple to equal 1 due to the fact that &lt;math&gt;\tan x \tan(90^{\circ}-x) = \tan x \cot x = 1.&lt;/math&gt;<br /> <br /> Due to the missing point (0,0) in the grid then in the rectangular arrangement (0,0), (x,0), (x,y), (0,y) there is only one triangle which will not cancel out to zero with a reflected version.<br /> <br /> <br /> So we need to consider all triangles of the form A(x,y), B(0,y), C(x,0). For these triangle tanB = y/x.<br /> Multiplying them all together gives: &lt;math&gt;\frac{1}{1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot * \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot * \frac{3}{1} \cdot \frac{3}{2} \cdot \frac{3}{3} \cdot \frac{3}{4} \cdot * \frac{4}{1} \cdot \frac{4}{2} \cdot \frac{4}{3} \cdot \frac{4}{4} \cdot * \frac{5}{1} \cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} = \frac{625}{24} &lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Consider reflections. For any right triangle &lt;math&gt;ABC&lt;/math&gt; with the right labeling described in the problem, any reflection &lt;math&gt;A'B'C'&lt;/math&gt; labeled that way will give us &lt;math&gt;\tan CBA \cdot \tan C'B'A' = 1&lt;/math&gt;. First we consider the reflection about the line &lt;math&gt;y=2.5&lt;/math&gt;. Only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex at &lt;math&gt;(0,5)&lt;/math&gt; do not reflect to a traingle &lt;math&gt;\subseteq S&lt;/math&gt;. Within those triangles, consider a reflection about the line &lt;math&gt;y=5-x&lt;/math&gt;. Then only those triangles &lt;math&gt;\subseteq S&lt;/math&gt; that have one vertex on the line &lt;math&gt;y=0&lt;/math&gt; do not reflect to a triangle &lt;math&gt;\subseteq S&lt;/math&gt;. So we only need to look at right triangles that have vertices &lt;math&gt;(0,5), (*,0), (*,*)&lt;/math&gt;. There are three cases:<br /> <br /> Case 1: &lt;math&gt;A=(0,5)&lt;/math&gt;. Then &lt;math&gt;B=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Case 2: &lt;math&gt;B=(0,5)&lt;/math&gt;. Then we look for &lt;math&gt;A=(x,y)&lt;/math&gt; such that &lt;math&gt;\angle BAC=90^{\circ}&lt;/math&gt; and that &lt;math&gt;C=(*,0)&lt;/math&gt;. They are: &lt;math&gt;(A=(x,5), C=(x,0))&lt;/math&gt;, &lt;math&gt;(A=(2,4), C=(1,0))&lt;/math&gt; and &lt;math&gt;(A=(4,1), C=(3,0))&lt;/math&gt;. The product of their values of &lt;math&gt;\tan \angle CBA&lt;/math&gt; is &lt;math&gt;\frac{5}{1}\cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{625}{144}&lt;/math&gt;.<br /> <br /> Case 3: &lt;math&gt;C=(0,5)&lt;/math&gt;. Then &lt;math&gt;A=(*,0)&lt;/math&gt; is impossible.<br /> <br /> Therefore &lt;math&gt;\framebox{B}&lt;/math&gt; is the answer.</div> Username222