https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathcool2009&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-23T02:20:29Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_15&diff=132011 2001 AIME I Problems/Problem 15 2020-08-18T02:31:03Z <p>Mathcool2009: </p> <hr /> <div>== Problem ==<br /> The numbers &lt;math&gt;1, 2, 3, 4, 5, 6, 7,&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; are randomly written on the faces of a regular [[octahedron]] so that each face contains a different number. The [[probability]] that no two consecutive numbers, where &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt; are considered to be consecutive, are written on faces that share an edge is &lt;math&gt;m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n.&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1===<br /> Choose one face of the octahedron randomly and label it with &lt;math&gt;1&lt;/math&gt;. There are three faces adjacent to this one, which we will call A-faces. There are three faces adjacent to two of the A-faces, which we will call B-faces, and one face adjacent to the three B-faces, which we will call the C-face.<br /> <br /> Clearly, the labels for the A-faces must come from the set &lt;math&gt;\{3,4,5,6,7\}&lt;/math&gt;, since these faces are all adjacent to &lt;math&gt;1&lt;/math&gt;. There are thus &lt;math&gt;5 \cdot 4 \cdot 3 = 60&lt;/math&gt; ways to assign the labels for the A-faces. <br /> <br /> The labels for the B-faces and C-face are the two remaining numbers from the above set, plus &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;. The number on the C-face must not be consecutive to any of the numbers on the B-faces. <br /> <br /> From here it is easiest to brute force the &lt;math&gt;10&lt;/math&gt; possibilities for the &lt;math&gt;4&lt;/math&gt; numbers on the B and C faces:<br /> <br /> *2348 (2678): 8(2) is the only one not adjacent to any of the others, so it goes on the C-face. 4(6) has only one B-face it can go to, while 2 and 3 (7 and 8) can be assigned randomly to the last two. 2 possibilities here.<br /> <br /> *2358 (2578): 5 cannot go on any of the B-faces, so it must be on the C-face. 3 and 8 (2 and 7) have only one allowable B-face, so just 1 possibility here.<br /> <br /> *2368 (2478): 6(4) cannot go on any of the B-faces, so it must be on the C-face. 3 and 8 (2 and 7) have only one allowable B-face, so 1 possibility here.<br /> <br /> *2458 (2568): All of the numbers have only one B-face they could go to. 2 and 4 (6 and 8) can go on the same, so one must go to the C-face. Only 2(8) is not consecutive with any of the others, so it goes on the C-face. 1 possibility.<br /> <br /> *2378: None of the numbers can go on the C-face because they will be consecutive with one of the B-face numbers. So this possibility is impossible. <br /> <br /> *2468: Both 4 and 6 cannot go on any B-face. They cannot both go on the C-face, so this possibility is impossible.<br /> <br /> There is a total of &lt;math&gt;10&lt;/math&gt; possibilities. There are &lt;math&gt;3!=6&lt;/math&gt; permutations (more like &quot;rotations&quot;) of each, so &lt;math&gt;60&lt;/math&gt; acceptable ways to fill in the rest of the octahedron given the &lt;math&gt;1&lt;/math&gt;. There are &lt;math&gt;7!=5040&lt;/math&gt; ways to randomly fill in the rest of the octahedron. So the probability is &lt;math&gt;\frac {60}{5040} = \frac {1}{84}&lt;/math&gt;. The answer is &lt;math&gt;\boxed{085}&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> Consider the cube formed from the face centers of the regular octahedron. Color the vertices in a checker board fashion. We seek the number of circuits traversing the cube entirely composed of diagonals. Notice for any vertex, it can be linked to at most one different-colored vertex, i.e. its opposite vertex. Thus, there are only two possible types of circuits (&lt;math&gt;B&lt;/math&gt; for black and &lt;math&gt;W&lt;/math&gt; for white).<br /> <br /> Type I: &lt;math&gt;BB-WWWW-BB&lt;/math&gt;. There are &lt;math&gt;4!&lt;/math&gt; ways to arrange the black vertices and consequently two of the white vertices and &lt;math&gt;2!&lt;/math&gt; ways to arrange the other two white vertices. Since the template has a period of &lt;math&gt;8&lt;/math&gt;, there are &lt;math&gt;4!\cdot 2!\cdot 8 = 384&lt;/math&gt; circuits of type I.<br /> <br /> Type II: &lt;math&gt;B-WW-BB-WW-B&lt;/math&gt;. There are &lt;math&gt;4!&lt;/math&gt; ways to arrange the black vertices and consequently the white vertices. Since the template has a period of &lt;math&gt;4&lt;/math&gt;, there are &lt;math&gt;4! \cdot 4 = 96&lt;/math&gt; circuits of type II.<br /> <br /> Thus, there are &lt;math&gt;384+96=480&lt;/math&gt; circuits satisfying the given condition, out of the &lt;math&gt;8!&lt;/math&gt; possible circuits. Therefore, the desired probability is &lt;math&gt;\frac{480}{8!} = \frac{1}{84}&lt;/math&gt;. The answer is &lt;math&gt;\boxed{085}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> As in the previous solution, consider the cube formed by taking each face of the octahedron as a vertex. Let one fixed vertex be A. Then each configuration (letting each vertex have a number value from 1-8) of A and the three vertices adjacent to A uniquely determine a configuration that satisfies the conditions, i.e. no two vertices have consecutive numbers. Thus, the number of desired configurations is equivalent to the number of ways of choosing the values of A and its three adjacent vertices. <br /> <br /> The value of A can be chosen in 8 ways, and the 3 vertices adjacent to A can be chosen in &lt;math&gt;5\cdot4\cdot3=60&lt;/math&gt; ways, since they aren't adjacent to each other, but they can't, after all, be consecutive values to A. For example, if A=1, then the next vertex can't be 1,2 or 8, so there are 5 choices. However, the next vertex also adjacent to A can be chosen in 4 ways; it can't be equal to 1,2,8, or the value of the previously chosen vertex. With the same reasoning, the last such vertex has 3 possible choices.<br /> <br /> The total number of ways to choose the values of the vertices of the cube independently is 8!, so our probability is thus &lt;math&gt;\frac{8\cdot60}{8!}=\frac{8\cdot5\cdot4\cdot3}{8!}=\frac{1}{84}&lt;/math&gt;, from which the answer is &lt;math&gt;\boxed{085}&lt;/math&gt;.<br /> <br /> , the average is one per choice of 3 to border the 1, ex 5,6,7 border 2 solutions, 3,5,7 border 0 solution<br /> <br /> === Solution 4 ===<br /> <br /> &lt;asy&gt;<br /> import three;<br /> draw((0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0));<br /> draw((1,1,0)--(1,1,1));<br /> draw((0,1,0)--(0,1,1));<br /> draw((0,0,0)--(1,0,0));<br /> draw((0,0,1)--(1,0,1));<br /> for(int i = 0; i &lt; 2; ++i) {<br /> for(int j = 0; j &lt; 2; ++j) {<br /> for(int k = 0; k &lt; 2; ++k) {<br /> dot((i,j,k));<br /> }<br /> }<br /> }<br /> // dot((0,0,1),blue);<br /> // dot((0,1,0),green);<br /> // dot((1,0,0),red);<br /> draw((0,0,0)--(1,1,0));<br /> draw((0,1,0)--(1,0,0));<br /> draw((0,0,1)--(1,1,1));<br /> draw((0,1,1)--(1,0,1));<br /> &lt;/asy&gt;<br /> <br /> The probability is equivalent to counting the number of Hamiltonian cycles in this 3D graph over &lt;math&gt;7!.&lt;/math&gt; This is because each Hamiltonian cycle corresponds to eight unique ways to label the faces. Label the vertices &lt;math&gt;AR,BR,CR,DR,AX,BX,CX,DX&lt;/math&gt; where vertices &lt;math&gt;ab&lt;/math&gt; and &lt;math&gt;cd&lt;/math&gt; are connected if &lt;math&gt;a=c&lt;/math&gt; or &lt;math&gt;b=d.&lt;/math&gt;<br /> <br /> Case 1: Four of the vertical edges are used. &lt;math&gt;6\cdot 2=12.&lt;/math&gt;<br /> <br /> Case 2: Two of the vertical edges are used. &lt;math&gt;4\cdot 3 \cdot 2\cdot 2=48.&lt;/math&gt;<br /> <br /> So, the probability is &lt;math&gt;\frac{60}{5040}=\frac{1}{84}.&lt;/math&gt; Therefore, our answer is &lt;math&gt;\boxed{85.}&lt;/math&gt;<br /> <br /> == Solution 5 ==<br /> As with some of the previous solutions, consider the cube formed by connecting the centroids of the faces on the octahedron. We choose a random vertex(hence fixing the diagram), giving us &lt;math&gt;7!&lt;/math&gt; ways as our denominator. WLOG, we color this start vertex red, and we color all &lt;math&gt;3&lt;/math&gt; vertices adjacent to it blue. We repeat this for the other vertices.<br /> <br /> Note that there is a 1-1 correspondence between the number of valid face numberings with rotational symmetry and the number of ways to move a particle to every vertex of the cube and returning to the start vertex with only diagonal moves. We can move along either short diagonals and long diagonals.<br /> <br /> Next, note that we can only move from the red vertices to the blue vertices and back with long diagonals(since short diagonals keep us on the color we are currently on). Thus, it is easy to check that the only possible sequences of long and short diagonals are cyclic permutations of &lt;math&gt;SSSLSSSL&lt;/math&gt; and &lt;math&gt;SLSLSLSL&lt;/math&gt;.<br /> <br /> Case 1: &lt;math&gt;SSSLSSSL&lt;/math&gt;.<br /> There are &lt;math&gt;4&lt;/math&gt; cyclic permutations, and we can traverse the entirety of the red tetrahedron in &lt;math&gt;4&lt;/math&gt; steps in &lt;math&gt;3!&lt;/math&gt; ways. Then, after the move to the blue tetrahedron from one of the non-starting red vertices, we realize that our first step cannot be to the blue vertex opposite the starting red vertex. Hence, there are &lt;math&gt;2&lt;/math&gt; possibilities for our first step. However, once we make our first move, the path is fixed. This leaves us with &lt;math&gt;4! \cdot 2&lt;/math&gt; ways.<br /> <br /> Case 2: &lt;math&gt;SLSLSLSL&lt;/math&gt;<br /> There are &lt;math&gt;2&lt;/math&gt; cyclic permutations in this case. After we choose one of the &lt;math&gt;3&lt;/math&gt; red vertices to go to from the starting vertex and move to the blue tetrahedron, we are once again left with &lt;math&gt;2&lt;/math&gt; choices to move to by similar logic to the first case. However, after we move back to the red tetrahedron, our choices are fixed. This leaves us with &lt;math&gt;2 \cdot 3 \cdot 2&lt;/math&gt; ways.<br /> <br /> Finally, summing and dividing by &lt;math&gt;7!&lt;/math&gt; gives us &lt;math&gt;\frac{60}{7!} = \frac{1}{84} \rightarrow \boxed{085}&lt;/math&gt;, as desired. - Spacesam<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=Characteristic_polynomial&diff=65842 Characteristic polynomial 2014-11-03T05:28:01Z <p>Mathcool2009: /* Proof 3 (Partial fractions) */ Fixed typos</p> <hr /> <div>The '''characteristic polynomial''' of a linear [[operator]] refers to the [[polynomial]] whose roots are the [[eigenvalue]]s of the operator. It carries much information about the operator. <br /> <br /> In the context of problem-solving, the characteristic polynomial is often used to find closed forms for the solutions of [[#Linear recurrences|linear recurrences]].<br /> <br /> == Definition ==<br /> Suppose &lt;math&gt;A&lt;/math&gt; is a &lt;math&gt;n \times n&lt;/math&gt; [[matrix]] (over a field &lt;math&gt;K&lt;/math&gt;). Then the characteristic polynomial of &lt;math&gt;A&lt;/math&gt; is defined as &lt;math&gt;P_A(t) = \det(tI - A)&lt;/math&gt;, which is a &lt;math&gt;n&lt;/math&gt;th degree polynomial in &lt;math&gt;t&lt;/math&gt;. Here, &lt;math&gt;I&lt;/math&gt; refers to the &lt;math&gt;n\times n&lt;/math&gt; [[identity matrix]].<br /> <br /> Written out, the characteristic polynomial is the [[determinant]]<br /> <br /> &lt;center&gt;&lt;cmath&gt;P_A(t) = \begin{vmatrix}t-a_{11} &amp; a_{12} &amp; \cdots &amp; a_{1n} \\ a_{21} &amp; t-a_{22} &amp; \cdots &amp; a_{2n} \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ a_{n1} &amp; a_{n2} &amp; \cdots &amp; t-a_{nn}\end{vmatrix}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> == Properties ==<br /> An [[eigenvector]] &lt;math&gt;\bold{v} \in K^n&lt;/math&gt; is a non-zero vector that satisfies the relation &lt;math&gt;A\bold{v} = \lambda\bold{v}&lt;/math&gt;, for some scalar &lt;math&gt;\lambda \in K&lt;/math&gt;. In other words, applying a linear operator to an eigenvector causes the eigenvector to dilate. The associated number &lt;math&gt;\lambda&lt;/math&gt; is called the [[eigenvalue]]. <br /> <br /> There are at most &lt;math&gt;n&lt;/math&gt; distinct eigenvalues, whose values are exactly the roots of the characteristic polynomial of the square matrix. To prove this, we use the fact that the determinant of a matrix is &lt;math&gt;0&lt;/math&gt; [[iff]] the column vectors of the matrix are [[linearly dependent]]. Observe that if &lt;math&gt;\lambda&lt;/math&gt; satisfies &lt;math&gt;\det(\lambda I-A) = 0&lt;/math&gt;, then the column vectors of the &lt;math&gt;n\times n&lt;/math&gt; matrix &lt;math&gt;\lambda I - A&lt;/math&gt; are linearly dependent. Indeed, if we define &lt;math&gt;T = \lambda I - A&lt;/math&gt; and let &lt;math&gt;\bold{T}^1, \bold{T}^2, \ldots, \bold{T}^n&lt;/math&gt; denote the column vectors of &lt;math&gt;T&lt;/math&gt;, this is equivalent to saying that there exists not all zero scalars &lt;math&gt;v_i \in K&lt;/math&gt; such that <br /> <br /> &lt;cmath&gt;v_1 \cdot \bold{T}^1 + v_2 \cdot \bold{T}^2 + \cdots + v_n \cdot \bold{T}^n = \bold{O}.&lt;/cmath&gt;<br /> <br /> Hence, there exists a non-zero vector &lt;math&gt;\bold{v} = (v_1, v_2, \ldots, v_n) \in K^n&lt;/math&gt; such that &lt;math&gt;(\lambda I - A) \bold{v} = \bold{O}&lt;/math&gt;. Distributing and re-arranging, &lt;math&gt;A\bold{v} = \lambda\bold{v}&lt;/math&gt;, as desired. In the other direction, if &lt;math&gt;A \bold{v} = \lambda \bold{v}&lt;/math&gt;, then &lt;math&gt;\lambda I \bold{v} - A \bold{v} = \bold{O}&lt;/math&gt;. But then, the column vectors of &lt;math&gt;\lambda I - A&lt;/math&gt; are linearly dependent, so it follows that &lt;math&gt;\det(\lambda I - A) = 0&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> Note that if &lt;math&gt;t = 0&lt;/math&gt;, then &lt;math&gt;P_A(0) = \det (tI - A) = \det (-A) = (-1)^n \det (A)&lt;/math&gt;. Hence, the characteristic polynomial encodes the determinant of the matrix. Also, the [[coefficient]] of the &lt;math&gt;t^{n-1}&lt;/math&gt; term of &lt;math&gt;P_A(t)&lt;/math&gt; gives the negative of the [[trace]] of the matrix (which follows from [[Vieta's formulas]]). <br /> <br /> By the [[Hamilton-Cayley Theorem]], the characteristic polynomial of a square matrix applied to the square matrix itself is zero, that is &lt;math&gt;P_A(A) = 0&lt;/math&gt;. The [[minimal polynomial]] of &lt;math&gt;A&lt;/math&gt; thus divides the characteristic polynomial &lt;math&gt;p_A&lt;/math&gt;. <br /> <br /> == Linear recurrences ==<br /> Let &lt;math&gt;x_1, x_2, \ldots, &lt;/math&gt; be a sequence of real numbers. Consider a monic [[homogenous]] [[linear recurrence]] of the form <br /> <br /> &lt;center&gt;&lt;cmath&gt;x_{n} = c_1x_{n-1} + c_2x_{n-2} + \cdots + c_kx_{n-k}, \quad (*)&lt;/cmath&gt;&lt;/center&gt; <br /> <br /> where &lt;math&gt;c_1, \ldots, c_k&lt;/math&gt; are real constants. The characteristic polynomial of this recurrence is defined as the polynomial <br /> <br /> &lt;center&gt;&lt;cmath&gt;P(x) = x^k - c_1x^{k-1} - c_2x^{k-2} - \cdots -c_{k-1}x - c_k.&lt;/cmath&gt;&lt;/center&gt; <br /> <br /> For example, let &lt;math&gt;F_n&lt;/math&gt; be the &lt;math&gt;n&lt;/math&gt;th [[Fibonacci number]] defined by &lt;math&gt;F_1 = F_2 = 1&lt;/math&gt;, and <br /> <br /> &lt;center&gt;&lt;cmath&gt;F_n = F_{n-1} + F_{n-2} \Longleftrightarrow F_n - F_{n-1} - F_{n-2} = 0.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> Then, its characteristic polynomial is &lt;math&gt;x^2 - x - 1&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> The roots of the polynomial can be used to write a closed form for the recurrence. If the roots of this polynomial &lt;math&gt;P&lt;/math&gt; are distinct, then suppose the roots are &lt;math&gt;r_1,r_2, \cdots, r_k&lt;/math&gt;. Then, there exists real constants &lt;math&gt;a_1, a_2, \cdots, a_k&lt;/math&gt; such that<br /> <br /> &lt;center&gt;&lt;cmath&gt;x_n = a_1 \cdot r_1^n + a_2 \cdot r_2^n + \cdots + a_k \cdot r_k^n&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> If we evaluate &lt;math&gt;k&lt;/math&gt; different values of &lt;math&gt;x_i&lt;/math&gt; (typically &lt;math&gt;x_0, x_1, \cdots, x_{k-1}&lt;/math&gt;), we can find a linear system in the &lt;math&gt;a_i&lt;/math&gt;s that can be solved for each constant. Refer to the [[#Introductory|introductory problems]] below to see an example of how to do this. In particular, for the Fibonacci numbers, this yields [[Binet's formula]]. &lt;br&gt;&lt;br&gt;<br /> <br /> If there are roots with multiplicity greater than &lt;math&gt;1&lt;/math&gt;, suppose &lt;math&gt;r_1 = r_2 = \cdots = r_{k_1}&lt;/math&gt;. Then we would replace the term &lt;math&gt;a_1 \cdot r^n + a_2 \cdot r_2^n + \cdots + a_{k_1} \cdot r_{k_1}^n&lt;/math&gt; with the expression <br /> <br /> &lt;center&gt;&lt;cmath&gt;(a_1 \cdot n^{k_1-1} + a_2 \cdot n^{k_1 - 2} + \cdots + a_{k_1-1} \cdot n + a_{k_1}) \cdot r^n.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> That is, there is now a polynomial in &lt;math&gt;n&lt;/math&gt; multiplied with the exponent. Note that there are still the same number of constants that we must solve for. For example, consider the recurrence relation &lt;math&gt;x_n = -2x_{n-1} -x_{n-2} \Longleftrightarrow x_n + 2x_{n-1} + x_{n-2} = 0&lt;/math&gt;. It’s characteristic polynomial, &lt;math&gt;x^2 + 2x + 1&lt;/math&gt;, has a &lt;math&gt;-1&lt;/math&gt; double root. Then, its closed form solution is of the type &lt;math&gt;x_n = (-1)^n(an + b)&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> Given a linear recurrence of the form &lt;math&gt;x_n - a_1x_{n-1} - \cdots - a_kx_{n-k} = f(n)&lt;/math&gt;, we often try to find a new sequence &lt;math&gt;y_n = x_n - g(n)&lt;/math&gt; such that &lt;math&gt;y_n&lt;/math&gt; is a homogenous linear recurrence. Then, we can find a closed form for &lt;math&gt;y_n&lt;/math&gt;, and then the answer is given by &lt;math&gt;x_n = y_n + g(n)&lt;/math&gt;. <br /> <br /> Of the following proofs, the second and third are more approachable. <br /> <br /> === Proof 1 (Linear Algebra) ===<br /> Note: The ideas expressed in this section can be transferred to the next section about differential equations. This requires some knowledge of [[linear algebra]] (upto the [[Spectral Theorem]]).<br /> <br /> Let &lt;math&gt;\bold{y}_{n-1} = \begin{pmatrix} x_{n-1} \\ x_{n-2} \\ \vdots \\ x_{n-k} \end{pmatrix}&lt;/math&gt;. Then, we can express our linear recurrence as the matrix <br /> <br /> &lt;center&gt;&lt;cmath&gt;A = \begin{pmatrix}a_1 &amp; a_2 &amp; a_3 &amp; \cdots &amp; a_{k-1} &amp; a_{k} \\ 1 &amp; 0 &amp; 0 &amp; \cdots &amp;0 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; \cdots &amp; 0 &amp; 0 \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \ddots &amp; \vdots \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 1 &amp; 0 \end{pmatrix},&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> so that &lt;math&gt;\bold{y}_n = A\bold{y}_{n-1}&lt;/math&gt; (try to verify this). The characteristic polynomial of &lt;math&gt;A&lt;/math&gt; is precisely &lt;math&gt;P_A(t) = \det (tI - A) = a_1 \cdot t^{k-1} + a_2 \cdot t^{k-2} \cdots + a_k = 0&lt;/math&gt;. This is not difficult to show via [[Laplace's expansion]] and induction, and is left as an exercise to the reader. <br /> <br /> If the roots of &lt;math&gt;P_A&lt;/math&gt; are distinct, then there exists a [[basis]] (of &lt;math&gt;\mathbb{R}^{k-1}&lt;/math&gt;) consisting of [[eigenvector]]s of &lt;math&gt;A&lt;/math&gt; (since eigenvectors of different eigenvalues are [[linearly independent]]). That is, applying a change of bases, we can write <br /> <br /> &lt;center&gt;&lt;cmath&gt;A = U^{-1} D U&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> for a matrix &lt;math&gt;U&lt;/math&gt; and a diagonal matrix &lt;math&gt;D&lt;/math&gt;. Then, &lt;math&gt;A^2 = U^{-1} D U U^{-1} D U = U^{-1} D^2 U&lt;/math&gt;, and in general, &lt;math&gt;A^n = U^{-1} D^n U&lt;/math&gt;. Thus, &lt;math&gt;\bold{y}_{n} = A^{n}\bold{y}_0 = U^{-1}D^{n+k}U\bold{y}_0&lt;/math&gt;. Here, &lt;math&gt;U^{-1}, U,&lt;/math&gt; and &lt;math&gt;\bold{y}_0&lt;/math&gt; are fixed (note that to find the values of &lt;math&gt;\bold{y}_0&lt;/math&gt;, we may need to trace the recurrence backwards. We only take the &lt;math&gt;0&lt;/math&gt;th index for simplicity). It follows that &lt;math&gt;y_{n}&lt;/math&gt; is a linear combination of the diagonal elements of &lt;math&gt;D&lt;/math&gt;, namely &lt;math&gt;r_1^n, r_2^n, \ldots, r_k^n&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> Suppose now that that the roots of &lt;math&gt;P_A&lt;/math&gt; are not distinct. Then, we can write the matrix in the [[Jordan normal form]]. For simplicity, first consider just a single root &lt;math&gt;r&lt;/math&gt; repeated &lt;math&gt;k&lt;/math&gt; times. The corresponding Jordan form of the matrix is given by (that is, the matrix is [[similar]] to the following):<br /> <br /> &lt;center&gt;&lt;cmath&gt;\begin{pmatrix} r &amp; 1 &amp; 0 &amp; \cdots &amp; 0 \\ 0 &amp; r &amp; 1 &amp; \cdots &amp; 0 \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; r \end{pmatrix}&lt;/cmath&gt;.&lt;/center&gt;<br /> <br /> Exponentiating this matrix to the &lt;math&gt;n&lt;/math&gt;th power will yield [[binomial coefficient]]s as follows<br /> <br /> &lt;center&gt;&lt;cmath&gt;\begin{pmatrix} r^n &amp; {n \choose 1}r^{n-1} &amp; {n \choose 2}r^{n-2} &amp; \cdots &amp; {n \choose k}r^{n-k} \\ 0 &amp; r &amp; {n \choose 1}r^{n-1} &amp; \cdots &amp; {n \choose {k-1}}r^{n-k+1} \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; r^n \end{pmatrix}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> We can treat the binomial coefficient as a polynomial in &lt;math&gt;n&lt;/math&gt;. Furthermore, we can scale away the powers of the eigenvalue (which is a constant); after taking the appropriate linear combinations of the binomial coefficients and a little bit of work, the result follows. &lt;br&gt;&lt;br&gt;<br /> <br /> See also a &lt;url&gt;viewtopic.php?t=290351 graph-theoretic&lt;/url&gt; approach.<br /> <br /> === Proof 2 (Induction) ===<br /> <br /> There are a couple of lower-level ways to prove this. One is by [[induction]], though the proof is not very revealing; we can explicitly check that a sequence &lt;math&gt;\{a_1 \cdot r_1^n + a_2 \cdot r_2^n + \cdots + a_k \cdot r_k^n\}&lt;/math&gt;, for real numbers &lt;math&gt;a_1, a_2, \ldots, a_k&lt;/math&gt;, satisfies the linear recurrence relation &lt;math&gt;(*)&lt;/math&gt;. If the two sequences are the same for the first &lt;math&gt;k&lt;/math&gt; values of the sequence, it follows by induction that the two sequences must be exactly the same.<br /> <br /> In particular, for &lt;math&gt;k=2&lt;/math&gt;, we can check this by using the identity <br /> <br /> &lt;center&gt;&lt;cmath&gt;ax^{n+1} + by^{n+1} = (x+y)(ax^n + by^n) - xy(ax^{n-1} + by^{n-1}).&lt;/cmath&gt;&lt;/center&gt; &lt;br&gt;<br /> <br /> It is also possible to reduce the recurrence to a [[telescoping sum]]. However, the details are slightly messy. <br /> <br /> === Proof 3 (Partial fractions) ===<br /> <br /> Another method uses [[partial fractions]] and [[generating function]]s, though not much knowledge of each is required for this proof. Let &lt;math&gt;G_n = a_1G_{n-1} + \cdots + a_kG_{n-k}&lt;/math&gt; be a linear recurrence. Consider the [[generating function]] given by <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = G_0 + G_1x + G_2x^2 + \cdots&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> Then, writing the following expressing out and carefully comparing coefficients (try it), <br /> <br /> &lt;center&gt;&lt;cmath&gt;a_kG(x) \cdot x^{k-1} + a_{k-1}G(x) \cdot x^{k-2} + \cdots + a_{2}G(x) \cdot x + a_{1}G(x) = \frac{G(x) + R(x)}x,&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> where &lt;math&gt;R(x)&lt;/math&gt; is a remainder polynomial with degree &lt;math&gt;\le k-1&lt;/math&gt;. Re-arranging, <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = \frac{R(x)}{a_k\cdot x^{k} + a_{k-1}\cdot x^{k-1} + \cdots + a_{1}x - 1}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> This polynomial in the denominator is the reverse of the characteristic polynomial of the recurrence. Hence, its roots are &lt;math&gt;1/r_1, 1/r_2, \ldots, 1/r_k&lt;/math&gt;, assuming that the roots are distinct. Using [[partial fraction]] decomposition (essentially the reverse of the process of adding fractions by combining denominators, except we now pull the denominators apart), we can write <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = \frac{c_1}{1 - r_1x} + \frac{c_2}{1-r_2x} + \cdots + \frac{c_k}{1-r_kx } &lt;/cmath&gt;&lt;/center&gt;<br /> <br /> for some constants &lt;math&gt;c_1, \ldots, c_k&lt;/math&gt;. Using the [[geometric series]] formula, we have &lt;math&gt;\frac{c}{1-y} = c + cy + cy^2 + \ldots&lt;/math&gt;. Thus, <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = (c_1 + \cdots + c_k) + (c_1r_1 + \cdots + c_kr_k)x + (c_1r_1^2 + \cdots + c_kr_k^2)x^2 + \cdots.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> Comparing coefficients with our original definition of &lt;math&gt;G(x)&lt;/math&gt; gives &lt;math&gt;G_n = c_1r_1^n + c_2r_2^n + \cdots + c_kr_k^n&lt;/math&gt;, as desired. &lt;br&gt;&lt;br&gt;<br /> <br /> The generalization to characteristic polynomials with multiple roots is not difficult from here, and is left to the reader. The partial fraction decomposition will have terms of the form &lt;math&gt;\frac{c_i}{(1-r_ix)^\alpha}&lt;/math&gt; for &lt;math&gt;\alpha &gt; 1&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> A note about this argument: all of the power series used here are defined formally, and so we do not actually need to worry whether or not they converge. See these [http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=250866 blog][http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=215833 posts] for further ideas.<br /> <br /> == Differential equations ==<br /> Given a monic linear [[homogenous]] [[differential equation]] of the form &lt;math&gt;D^ny +c_{n-1}D^{n-1}y + \cdots + c_1Dy + c_0y = 0&lt;/math&gt;, then the characteristic polynomial of the equation is the polynomial <br /> <br /> &lt;center&gt;&lt;cmath&gt;P(t) = t^n + c_{n-1}t^{n-1} + c_{n-2}t^{n-2} + \cdots + c_0.&lt;/cmath&gt;&lt;/center&gt; <br /> <br /> Here, &lt;math&gt;D = \frac{d}{dx}&lt;/math&gt; is short-hand for the differential operator. &lt;br&gt;&lt;br&gt;<br /> <br /> If the roots of the polynomial are distinct, say &lt;math&gt;r_1, r_2, \cdots, r_n&lt;/math&gt;, then the solutions of this differential equation are precisely the linear combinations &lt;math&gt;y(x) = a_1e^{r_1x} + a_2e^{r_2x} + \cdots + a_ne^{r_nx}&lt;/math&gt;. Similarly, if there is a set of roots with multiplicity greater than &lt;math&gt;1&lt;/math&gt;, say &lt;math&gt;r_1, r_2, \cdots, r_k&lt;/math&gt;, then we would replace the term &lt;math&gt;a_1e^{r_1x} + \cdots + a_ke^{r_kx}&lt;/math&gt; with the expression &lt;math&gt;(a_1x^{k-1} + a_2x^{k-2} + \cdots + a_{k-1}x + a_k)e^{r_1x}&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> In general, given a linear differential equation of the form &lt;math&gt;Ly = f&lt;/math&gt;, where &lt;math&gt;L = c_nD^n + c_{n-1}D^{n-1} + \cdots + c_0&lt;/math&gt; is a linear differential operator, then the set of solutions is given by the sum of any solution to the homogenous equation &lt;math&gt;Ly = 0&lt;/math&gt; and a specific solution to &lt;math&gt;Ly = f&lt;/math&gt;.<br /> <br /> === Proof ===<br /> We can apply an induction technique similar to the section on linear recurrences above. <br /> <br /> From linear algebra, we can use the following [[vector space]] decomposition theorem. Let &lt;math&gt;A: V \to V&lt;/math&gt; be a linear operator for a [[vector space]]s &lt;math&gt;V&lt;/math&gt; over a field &lt;math&gt;K&lt;/math&gt;. Suppose that there exists a polynomial &lt;math&gt;f(x) \in K[x]&lt;/math&gt; such that &lt;math&gt;f = gh&lt;/math&gt;, where &lt;math&gt;g&lt;/math&gt; and &lt;math&gt;h&lt;/math&gt; are non-zero polynomials such that &lt;math&gt;\text{gcd}\,(g,h) = 1&lt;/math&gt;, and such that &lt;math&gt;f(A) = O&lt;/math&gt;. Then &lt;math&gt;V = \ker g(A) \oplus \ker h(A)&lt;/math&gt;. This allows us to reduce the differential equation into finding the solutions to the equation &lt;math&gt;(D - \lambda I)^my = 0&lt;/math&gt;, which has a basis of functions &lt;math&gt;e^{\lambda t}, te^{\lambda t}, \ldots, t^{m-1}e^{\lambda t}&lt;/math&gt;.<br /> <br /> == Problems ==<br /> === Introductory ===<br /> *Prove [[Binet's formula]]. Find a similar closed form equation for the [[Lucas sequence]], defined with the starting terms &lt;math&gt;L_1 = 2, L_2 = 1&lt;/math&gt;, and satisfying the recursion &lt;math&gt;L_n = L_{n-1} + L_{n-2}&lt;/math&gt;. <br /> *Let &lt;math&gt;\{x_n\}&lt;/math&gt; denote the sequence defined by the recursion &lt;math&gt;x_0 = 3, x_1 = 1&lt;/math&gt;, and &lt;math&gt;x_n = 2x_{n-1} + 3x_{n-2}&lt;/math&gt;. Find a closed form expression for &lt;math&gt;x_n&lt;/math&gt;.<br /> *Given &lt;math&gt;a_0 = 1&lt;/math&gt;, &lt;math&gt;a_1 = 3&lt;/math&gt;, and the general relation &lt;math&gt;a_n^2 - a_{n - 1}a_{n + 1} = ( - 1)^n&lt;/math&gt; for &lt;math&gt;n \ge 1&lt;/math&gt;. Find a linear recurrence for &lt;math&gt;a_n&lt;/math&gt;. (AHSME 1958, Problem 40)<br /> <br /> === Intermediate ===<br /> *Let &lt;math&gt;S_n&lt;/math&gt; denote the number of ternary sequences (consisting of &lt;math&gt;0&lt;/math&gt;,&lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;2&lt;/math&gt;s) of length &lt;math&gt;n&lt;/math&gt;, such that they do not contain a substring of &quot;10&quot;, &quot;01&quot;, or &quot;11&quot;. Find a closed form expression for &lt;math&gt;S_n&lt;/math&gt;. <br /> *Let &lt;math&gt;\{X_n\}&lt;/math&gt; and &lt;math&gt;\{Y_n\}&lt;/math&gt; be sequences defined as follows: <br /> <br /> &lt;center&gt;&lt;cmath&gt;X_0 = Y_0 = X_1 = Y_1 = 1, X_{n+1} = X_n + 2X_{n-1}, Y_{n+1} = 3Y_n + 4Y_{n-1}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> :Let &lt;math&gt;k&lt;/math&gt; be the largest integer that satisfies all of the following conditions:<br /> <br /> ::(i) &lt;math&gt;|X_i - k| \le 2007&lt;/math&gt;, for some positive integer &lt;math&gt;i&lt;/math&gt;;<br /> ::(ii) &lt;math&gt;|Y_j - k| \le 2007&lt;/math&gt;, for some positive integer &lt;math&gt;j&lt;/math&gt;;<br /> ::(iii) &lt;math&gt;k &lt; 10^{2007}&lt;/math&gt;. <br /> <br /> :Find the remainder when &lt;math&gt;k&lt;/math&gt; is divided by &lt;math&gt;2007&lt;/math&gt;. (2007 [[iTest]], #47)<br /> <br /> *Let &lt;math&gt;a_{n}&lt;/math&gt;, &lt;math&gt;b_{n}&lt;/math&gt;, and &lt;math&gt;c_{n}&lt;/math&gt; be geometric sequences with different common ratios and let &lt;math&gt;a_{n}+b_{n}+c_{n}=d_{n}&lt;/math&gt; for all integers &lt;math&gt;n&lt;/math&gt;. If &lt;math&gt;d_{1}=1&lt;/math&gt;, &lt;math&gt;d_{2}=2&lt;/math&gt;, &lt;math&gt;d_{3}=3&lt;/math&gt;, &lt;math&gt;d_{4}=-7&lt;/math&gt;, &lt;math&gt;d_{5}=13&lt;/math&gt;, and &lt;math&gt;d_{6}=-16&lt;/math&gt;, find &lt;math&gt;d_{7}&lt;/math&gt;. ([[Mock AIME 1 2006-2007/Problem 13|Mock AIME 1 2006-2007, Problem 13]])<br /> <br /> *Find all possible values of &lt;math&gt;x_0&lt;/math&gt; and &lt;math&gt;x_1&lt;/math&gt; such that the sequence defined by:<br /> <br /> &lt;center&gt;&lt;cmath&gt;x_{n + 1} = \frac {x_{n - 1} x_n}{3x_{n - 1} - 2x_n}, \quad n \ge 1&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> :contains infinitely many natural numbers.<br /> <br /> *Show that &lt;math&gt;a^n + b^n + c^n&lt;/math&gt; can be written as a linear combination of the [[elementary symmetric polynomials]] &lt;math&gt;abc, ab+bc+ca, a+b+c&lt;/math&gt;. In general, prove [[Newton's sums]]. <br /> <br /> === Olympiad ===<br /> *Let &lt;math&gt;r&lt;/math&gt; be a real number, and let &lt;math&gt;x_n&lt;/math&gt; be a sequence such that &lt;math&gt;x_0 = 0, x_1 = 1&lt;/math&gt;, and &lt;math&gt;x_{n+2} = rx_{n+1} - x_n&lt;/math&gt; for &lt;math&gt;n \ge 0&lt;/math&gt;. For which values of &lt;math&gt;r&lt;/math&gt; does &lt;math&gt;x_1 + x_3 + \cdots + x_{2m-1} = x_m^2&lt;/math&gt; for all positive integers &lt;math&gt;m&lt;/math&gt;? ([[WOOT]])<br /> * Let &lt;math&gt;a(x,y)&lt;/math&gt; be the polynomial &lt;math&gt;x^2y + xy^2&lt;/math&gt;, and &lt;math&gt;b(x,y)&lt;/math&gt; the polynomial &lt;math&gt;x^2 + xy + y^2&lt;/math&gt;. Prove that we can find a polynomial &lt;math&gt;p_n(a, b)&lt;/math&gt; which is identically equal to &lt;math&gt;(x + y)^n + (-1)^n (x^n + y^n)&lt;/math&gt;. For example, &lt;math&gt;p_4(a, b) = 2b^2&lt;/math&gt;. ([[1976 Putnam Problems/Problem A2|1976 Putnam, Problem A2]]).<br /> *&lt;math&gt;a_1,a_2,a_3,b_1,b_2,b_3&lt;/math&gt; are distinct positive integers such that &lt;math&gt;(n + 1)a_1^n + na_2^n + (n - 1)a_3^n|(n + 1)b_1^n + nb_2^n + (n - 1)b_3^n&lt;/math&gt; holds for all positive integer &lt;math&gt;n&lt;/math&gt;. Prove that there exists &lt;math&gt;k\in N&lt;/math&gt; such that &lt;math&gt;b_i = ka_i&lt;/math&gt; for &lt;math&gt;i = 1,2,3&lt;/math&gt;. (2010 Chinese MO, Problem 6)<br /> <br /> == Hints/Solutions ==<br /> === Introductory ===<br /> * A proof of [[Binet's formula]] may be found in that link. The characteristic polynomial of the Lucas sequence is exactly the same. Hence, the only thing we have to change are the coefficients. <br /> * We will work out this problem in full detail. The recurrence relation is &lt;math&gt;x_n -2x_{n-1} -3x_{n-2}&lt;/math&gt;, and its characteristic polynomial is given by &lt;math&gt;x^2-2x-3&lt;/math&gt;. The roots of this polynomial are &lt;math&gt;-1&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;. Thus, a closed form solution is given by &lt;math&gt;x_n = a \cdot (-1)^n + b \cdot 3^n&lt;/math&gt;. For &lt;math&gt;n = 0&lt;/math&gt;, we get &lt;math&gt;x_0 = a + b = 3&lt;/math&gt;, and for &lt;math&gt;n = 1&lt;/math&gt;, we get &lt;math&gt;x_1 = -a + 3b = 1&lt;/math&gt;. Solving gives &lt;math&gt;a = 2, b = 1&lt;/math&gt;. Thus, our answer is &lt;math&gt;x_n = 2 \cdot (-1)^n + 3^n&lt;/math&gt;. <br /> *&lt;url&gt;viewtopic.php?t=282744 Discussion&lt;/url&gt; (1958 AHSME, 40)<br /> <br /> === Intermediate ===<br /> *&lt;url&gt;viewtopic.php?t=335138 Discussion&lt;/url&gt;<br /> *Answer (2007 iTest 47): The answer is &lt;math&gt;k = 7468&lt;/math&gt; (before taking &lt;math&gt;\mod{2007}&lt;/math&gt;).<br /> *&lt;url&gt;viewtopic.php?t=287441 Discussion&lt;/url&gt; <br /> *Hint (Newton’s Sum): Let us work backwards. Suppose &lt;math&gt;a,b,c&lt;/math&gt; are the roots of the characteristic polynomial of a linear recurrence. Then apply [[Vieta's sums]].<br /> <br /> === Olympiad ===<br /> *Hint (WOOT): substitute the closed form solution. It is true for all &lt;math&gt;r&lt;/math&gt;. <br /> *Hint (1976 Putnam A2): do the even and odd cases separately.<br /> *&lt;url&gt;viewtopic.php?search_id=804457492&amp;t=327474 Discussion&lt;/url&gt; (2010 Chinese MO, 6)<br /> <br /> [[Category:Linear algebra]]</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=Characteristic_polynomial&diff=65841 Characteristic polynomial 2014-11-03T05:23:37Z <p>Mathcool2009: /* Proof 3 (Partial fractions) */ Whoops</p> <hr /> <div>The '''characteristic polynomial''' of a linear [[operator]] refers to the [[polynomial]] whose roots are the [[eigenvalue]]s of the operator. It carries much information about the operator. <br /> <br /> In the context of problem-solving, the characteristic polynomial is often used to find closed forms for the solutions of [[#Linear recurrences|linear recurrences]].<br /> <br /> == Definition ==<br /> Suppose &lt;math&gt;A&lt;/math&gt; is a &lt;math&gt;n \times n&lt;/math&gt; [[matrix]] (over a field &lt;math&gt;K&lt;/math&gt;). Then the characteristic polynomial of &lt;math&gt;A&lt;/math&gt; is defined as &lt;math&gt;P_A(t) = \det(tI - A)&lt;/math&gt;, which is a &lt;math&gt;n&lt;/math&gt;th degree polynomial in &lt;math&gt;t&lt;/math&gt;. Here, &lt;math&gt;I&lt;/math&gt; refers to the &lt;math&gt;n\times n&lt;/math&gt; [[identity matrix]].<br /> <br /> Written out, the characteristic polynomial is the [[determinant]]<br /> <br /> &lt;center&gt;&lt;cmath&gt;P_A(t) = \begin{vmatrix}t-a_{11} &amp; a_{12} &amp; \cdots &amp; a_{1n} \\ a_{21} &amp; t-a_{22} &amp; \cdots &amp; a_{2n} \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ a_{n1} &amp; a_{n2} &amp; \cdots &amp; t-a_{nn}\end{vmatrix}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> == Properties ==<br /> An [[eigenvector]] &lt;math&gt;\bold{v} \in K^n&lt;/math&gt; is a non-zero vector that satisfies the relation &lt;math&gt;A\bold{v} = \lambda\bold{v}&lt;/math&gt;, for some scalar &lt;math&gt;\lambda \in K&lt;/math&gt;. In other words, applying a linear operator to an eigenvector causes the eigenvector to dilate. The associated number &lt;math&gt;\lambda&lt;/math&gt; is called the [[eigenvalue]]. <br /> <br /> There are at most &lt;math&gt;n&lt;/math&gt; distinct eigenvalues, whose values are exactly the roots of the characteristic polynomial of the square matrix. To prove this, we use the fact that the determinant of a matrix is &lt;math&gt;0&lt;/math&gt; [[iff]] the column vectors of the matrix are [[linearly dependent]]. Observe that if &lt;math&gt;\lambda&lt;/math&gt; satisfies &lt;math&gt;\det(\lambda I-A) = 0&lt;/math&gt;, then the column vectors of the &lt;math&gt;n\times n&lt;/math&gt; matrix &lt;math&gt;\lambda I - A&lt;/math&gt; are linearly dependent. Indeed, if we define &lt;math&gt;T = \lambda I - A&lt;/math&gt; and let &lt;math&gt;\bold{T}^1, \bold{T}^2, \ldots, \bold{T}^n&lt;/math&gt; denote the column vectors of &lt;math&gt;T&lt;/math&gt;, this is equivalent to saying that there exists not all zero scalars &lt;math&gt;v_i \in K&lt;/math&gt; such that <br /> <br /> &lt;cmath&gt;v_1 \cdot \bold{T}^1 + v_2 \cdot \bold{T}^2 + \cdots + v_n \cdot \bold{T}^n = \bold{O}.&lt;/cmath&gt;<br /> <br /> Hence, there exists a non-zero vector &lt;math&gt;\bold{v} = (v_1, v_2, \ldots, v_n) \in K^n&lt;/math&gt; such that &lt;math&gt;(\lambda I - A) \bold{v} = \bold{O}&lt;/math&gt;. Distributing and re-arranging, &lt;math&gt;A\bold{v} = \lambda\bold{v}&lt;/math&gt;, as desired. In the other direction, if &lt;math&gt;A \bold{v} = \lambda \bold{v}&lt;/math&gt;, then &lt;math&gt;\lambda I \bold{v} - A \bold{v} = \bold{O}&lt;/math&gt;. But then, the column vectors of &lt;math&gt;\lambda I - A&lt;/math&gt; are linearly dependent, so it follows that &lt;math&gt;\det(\lambda I - A) = 0&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> Note that if &lt;math&gt;t = 0&lt;/math&gt;, then &lt;math&gt;P_A(0) = \det (tI - A) = \det (-A) = (-1)^n \det (A)&lt;/math&gt;. Hence, the characteristic polynomial encodes the determinant of the matrix. Also, the [[coefficient]] of the &lt;math&gt;t^{n-1}&lt;/math&gt; term of &lt;math&gt;P_A(t)&lt;/math&gt; gives the negative of the [[trace]] of the matrix (which follows from [[Vieta's formulas]]). <br /> <br /> By the [[Hamilton-Cayley Theorem]], the characteristic polynomial of a square matrix applied to the square matrix itself is zero, that is &lt;math&gt;P_A(A) = 0&lt;/math&gt;. The [[minimal polynomial]] of &lt;math&gt;A&lt;/math&gt; thus divides the characteristic polynomial &lt;math&gt;p_A&lt;/math&gt;. <br /> <br /> == Linear recurrences ==<br /> Let &lt;math&gt;x_1, x_2, \ldots, &lt;/math&gt; be a sequence of real numbers. Consider a monic [[homogenous]] [[linear recurrence]] of the form <br /> <br /> &lt;center&gt;&lt;cmath&gt;x_{n} = c_1x_{n-1} + c_2x_{n-2} + \cdots + c_kx_{n-k}, \quad (*)&lt;/cmath&gt;&lt;/center&gt; <br /> <br /> where &lt;math&gt;c_1, \ldots, c_k&lt;/math&gt; are real constants. The characteristic polynomial of this recurrence is defined as the polynomial <br /> <br /> &lt;center&gt;&lt;cmath&gt;P(x) = x^k - c_1x^{k-1} - c_2x^{k-2} - \cdots -c_{k-1}x - c_k.&lt;/cmath&gt;&lt;/center&gt; <br /> <br /> For example, let &lt;math&gt;F_n&lt;/math&gt; be the &lt;math&gt;n&lt;/math&gt;th [[Fibonacci number]] defined by &lt;math&gt;F_1 = F_2 = 1&lt;/math&gt;, and <br /> <br /> &lt;center&gt;&lt;cmath&gt;F_n = F_{n-1} + F_{n-2} \Longleftrightarrow F_n - F_{n-1} - F_{n-2} = 0.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> Then, its characteristic polynomial is &lt;math&gt;x^2 - x - 1&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> The roots of the polynomial can be used to write a closed form for the recurrence. If the roots of this polynomial &lt;math&gt;P&lt;/math&gt; are distinct, then suppose the roots are &lt;math&gt;r_1,r_2, \cdots, r_k&lt;/math&gt;. Then, there exists real constants &lt;math&gt;a_1, a_2, \cdots, a_k&lt;/math&gt; such that<br /> <br /> &lt;center&gt;&lt;cmath&gt;x_n = a_1 \cdot r_1^n + a_2 \cdot r_2^n + \cdots + a_k \cdot r_k^n&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> If we evaluate &lt;math&gt;k&lt;/math&gt; different values of &lt;math&gt;x_i&lt;/math&gt; (typically &lt;math&gt;x_0, x_1, \cdots, x_{k-1}&lt;/math&gt;), we can find a linear system in the &lt;math&gt;a_i&lt;/math&gt;s that can be solved for each constant. Refer to the [[#Introductory|introductory problems]] below to see an example of how to do this. In particular, for the Fibonacci numbers, this yields [[Binet's formula]]. &lt;br&gt;&lt;br&gt;<br /> <br /> If there are roots with multiplicity greater than &lt;math&gt;1&lt;/math&gt;, suppose &lt;math&gt;r_1 = r_2 = \cdots = r_{k_1}&lt;/math&gt;. Then we would replace the term &lt;math&gt;a_1 \cdot r^n + a_2 \cdot r_2^n + \cdots + a_{k_1} \cdot r_{k_1}^n&lt;/math&gt; with the expression <br /> <br /> &lt;center&gt;&lt;cmath&gt;(a_1 \cdot n^{k_1-1} + a_2 \cdot n^{k_1 - 2} + \cdots + a_{k_1-1} \cdot n + a_{k_1}) \cdot r^n.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> That is, there is now a polynomial in &lt;math&gt;n&lt;/math&gt; multiplied with the exponent. Note that there are still the same number of constants that we must solve for. For example, consider the recurrence relation &lt;math&gt;x_n = -2x_{n-1} -x_{n-2} \Longleftrightarrow x_n + 2x_{n-1} + x_{n-2} = 0&lt;/math&gt;. It’s characteristic polynomial, &lt;math&gt;x^2 + 2x + 1&lt;/math&gt;, has a &lt;math&gt;-1&lt;/math&gt; double root. Then, its closed form solution is of the type &lt;math&gt;x_n = (-1)^n(an + b)&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> Given a linear recurrence of the form &lt;math&gt;x_n - a_1x_{n-1} - \cdots - a_kx_{n-k} = f(n)&lt;/math&gt;, we often try to find a new sequence &lt;math&gt;y_n = x_n - g(n)&lt;/math&gt; such that &lt;math&gt;y_n&lt;/math&gt; is a homogenous linear recurrence. Then, we can find a closed form for &lt;math&gt;y_n&lt;/math&gt;, and then the answer is given by &lt;math&gt;x_n = y_n + g(n)&lt;/math&gt;. <br /> <br /> Of the following proofs, the second and third are more approachable. <br /> <br /> === Proof 1 (Linear Algebra) ===<br /> Note: The ideas expressed in this section can be transferred to the next section about differential equations. This requires some knowledge of [[linear algebra]] (upto the [[Spectral Theorem]]).<br /> <br /> Let &lt;math&gt;\bold{y}_{n-1} = \begin{pmatrix} x_{n-1} \\ x_{n-2} \\ \vdots \\ x_{n-k} \end{pmatrix}&lt;/math&gt;. Then, we can express our linear recurrence as the matrix <br /> <br /> &lt;center&gt;&lt;cmath&gt;A = \begin{pmatrix}a_1 &amp; a_2 &amp; a_3 &amp; \cdots &amp; a_{k-1} &amp; a_{k} \\ 1 &amp; 0 &amp; 0 &amp; \cdots &amp;0 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; \cdots &amp; 0 &amp; 0 \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \ddots &amp; \vdots \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 1 &amp; 0 \end{pmatrix},&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> so that &lt;math&gt;\bold{y}_n = A\bold{y}_{n-1}&lt;/math&gt; (try to verify this). The characteristic polynomial of &lt;math&gt;A&lt;/math&gt; is precisely &lt;math&gt;P_A(t) = \det (tI - A) = a_1 \cdot t^{k-1} + a_2 \cdot t^{k-2} \cdots + a_k = 0&lt;/math&gt;. This is not difficult to show via [[Laplace's expansion]] and induction, and is left as an exercise to the reader. <br /> <br /> If the roots of &lt;math&gt;P_A&lt;/math&gt; are distinct, then there exists a [[basis]] (of &lt;math&gt;\mathbb{R}^{k-1}&lt;/math&gt;) consisting of [[eigenvector]]s of &lt;math&gt;A&lt;/math&gt; (since eigenvectors of different eigenvalues are [[linearly independent]]). That is, applying a change of bases, we can write <br /> <br /> &lt;center&gt;&lt;cmath&gt;A = U^{-1} D U&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> for a matrix &lt;math&gt;U&lt;/math&gt; and a diagonal matrix &lt;math&gt;D&lt;/math&gt;. Then, &lt;math&gt;A^2 = U^{-1} D U U^{-1} D U = U^{-1} D^2 U&lt;/math&gt;, and in general, &lt;math&gt;A^n = U^{-1} D^n U&lt;/math&gt;. Thus, &lt;math&gt;\bold{y}_{n} = A^{n}\bold{y}_0 = U^{-1}D^{n+k}U\bold{y}_0&lt;/math&gt;. Here, &lt;math&gt;U^{-1}, U,&lt;/math&gt; and &lt;math&gt;\bold{y}_0&lt;/math&gt; are fixed (note that to find the values of &lt;math&gt;\bold{y}_0&lt;/math&gt;, we may need to trace the recurrence backwards. We only take the &lt;math&gt;0&lt;/math&gt;th index for simplicity). It follows that &lt;math&gt;y_{n}&lt;/math&gt; is a linear combination of the diagonal elements of &lt;math&gt;D&lt;/math&gt;, namely &lt;math&gt;r_1^n, r_2^n, \ldots, r_k^n&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> Suppose now that that the roots of &lt;math&gt;P_A&lt;/math&gt; are not distinct. Then, we can write the matrix in the [[Jordan normal form]]. For simplicity, first consider just a single root &lt;math&gt;r&lt;/math&gt; repeated &lt;math&gt;k&lt;/math&gt; times. The corresponding Jordan form of the matrix is given by (that is, the matrix is [[similar]] to the following):<br /> <br /> &lt;center&gt;&lt;cmath&gt;\begin{pmatrix} r &amp; 1 &amp; 0 &amp; \cdots &amp; 0 \\ 0 &amp; r &amp; 1 &amp; \cdots &amp; 0 \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; r \end{pmatrix}&lt;/cmath&gt;.&lt;/center&gt;<br /> <br /> Exponentiating this matrix to the &lt;math&gt;n&lt;/math&gt;th power will yield [[binomial coefficient]]s as follows<br /> <br /> &lt;center&gt;&lt;cmath&gt;\begin{pmatrix} r^n &amp; {n \choose 1}r^{n-1} &amp; {n \choose 2}r^{n-2} &amp; \cdots &amp; {n \choose k}r^{n-k} \\ 0 &amp; r &amp; {n \choose 1}r^{n-1} &amp; \cdots &amp; {n \choose {k-1}}r^{n-k+1} \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; r^n \end{pmatrix}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> We can treat the binomial coefficient as a polynomial in &lt;math&gt;n&lt;/math&gt;. Furthermore, we can scale away the powers of the eigenvalue (which is a constant); after taking the appropriate linear combinations of the binomial coefficients and a little bit of work, the result follows. &lt;br&gt;&lt;br&gt;<br /> <br /> See also a &lt;url&gt;viewtopic.php?t=290351 graph-theoretic&lt;/url&gt; approach.<br /> <br /> === Proof 2 (Induction) ===<br /> <br /> There are a couple of lower-level ways to prove this. One is by [[induction]], though the proof is not very revealing; we can explicitly check that a sequence &lt;math&gt;\{a_1 \cdot r_1^n + a_2 \cdot r_2^n + \cdots + a_k \cdot r_k^n\}&lt;/math&gt;, for real numbers &lt;math&gt;a_1, a_2, \ldots, a_k&lt;/math&gt;, satisfies the linear recurrence relation &lt;math&gt;(*)&lt;/math&gt;. If the two sequences are the same for the first &lt;math&gt;k&lt;/math&gt; values of the sequence, it follows by induction that the two sequences must be exactly the same.<br /> <br /> In particular, for &lt;math&gt;k=2&lt;/math&gt;, we can check this by using the identity <br /> <br /> &lt;center&gt;&lt;cmath&gt;ax^{n+1} + by^{n+1} = (x+y)(ax^n + by^n) - xy(ax^{n-1} + by^{n-1}).&lt;/cmath&gt;&lt;/center&gt; &lt;br&gt;<br /> <br /> It is also possible to reduce the recurrence to a [[telescoping sum]]. However, the details are slightly messy. <br /> <br /> === Proof 3 (Partial fractions) ===<br /> <br /> Another method uses [[partial fractions]] and [[generating function]]s, though not much knowledge of each is required for this proof. Let &lt;math&gt;G_n = a_1G_{n-1} + \cdots + a_kG_{n-k}&lt;/math&gt; be a linear recurrence. Consider the [[generating function]] given by <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = G_0 + G_1x + G_2x^2 + \cdots&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> Then, writing the following expressing out and carefully comparing coefficients (try it), <br /> <br /> &lt;center&gt;&lt;cmath&gt;a_kG(x) \cdot x^{k-1} + a_{k-1}G(x) \cdot x^{k-2} + \cdots + a_{2}G(x) \cdot x + a_{1}G(x) = \frac{G(x) - R(x)}x,&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> where &lt;math&gt;R(x)&lt;/math&gt; is a remainder polynomial with degree &lt;math&gt;&lt; k-1&lt;/math&gt;. Re-arranging, <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = \frac{xR(x)}{a_k\cdot x^{k} + a_{k-1}\cdot x^{k-1} + \cdots + a_{1}x - 1}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> This polynomial in the denominator is the reverse of the characteristic polynomial of the recurrence. Hence, its roots are &lt;math&gt;1/r_1, 1/r_2, \ldots, 1/r_k&lt;/math&gt;, assuming that the roots are distinct. Using [[partial fraction]] decomposition (essentially the reverse of the process of adding fractions by combining denominators, except we now pull the denominators apart), we can write <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = \frac{c_1}{1 - r_1x} + \frac{c_2}{1-r_2x} + \cdots + \frac{c_k}{1-r_kx } &lt;/cmath&gt;&lt;/center&gt;<br /> <br /> for some constants &lt;math&gt;c_1, \ldots, c_k&lt;/math&gt;. Using the [[geometric series]] formula, we have &lt;math&gt;\frac{c}{1-y} = c + cy + cy^2 + \ldots&lt;/math&gt;. Thus, <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = (c_1 + \cdots + c_k) + (c_1r_1 + \cdots + c_kr_k)x + (c_1r_1^2 + \cdots + c_kr_k^2)x^2 + \cdots.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> Comparing coefficients with our original definition of &lt;math&gt;G(x)&lt;/math&gt; gives &lt;math&gt;G_n = c_1r_1^n + c_2r_2^n + \cdots + c_kr_k^n&lt;/math&gt;, as desired. &lt;br&gt;&lt;br&gt;<br /> <br /> The generalization to characteristic polynomials with multiple roots is not difficult from here, and is left to the reader. The partial fraction decomposition will have terms of the form &lt;math&gt;\frac{c_i}{(1-r_ix)^\alpha}&lt;/math&gt; for &lt;math&gt;\alpha &gt; 1&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> A note about this argument: all of the power series used here are defined formally, and so we do not actually need to worry whether or not they converge. See these [http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=250866 blog][http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=215833 posts] for further ideas.<br /> <br /> == Differential equations ==<br /> Given a monic linear [[homogenous]] [[differential equation]] of the form &lt;math&gt;D^ny +c_{n-1}D^{n-1}y + \cdots + c_1Dy + c_0y = 0&lt;/math&gt;, then the characteristic polynomial of the equation is the polynomial <br /> <br /> &lt;center&gt;&lt;cmath&gt;P(t) = t^n + c_{n-1}t^{n-1} + c_{n-2}t^{n-2} + \cdots + c_0.&lt;/cmath&gt;&lt;/center&gt; <br /> <br /> Here, &lt;math&gt;D = \frac{d}{dx}&lt;/math&gt; is short-hand for the differential operator. &lt;br&gt;&lt;br&gt;<br /> <br /> If the roots of the polynomial are distinct, say &lt;math&gt;r_1, r_2, \cdots, r_n&lt;/math&gt;, then the solutions of this differential equation are precisely the linear combinations &lt;math&gt;y(x) = a_1e^{r_1x} + a_2e^{r_2x} + \cdots + a_ne^{r_nx}&lt;/math&gt;. Similarly, if there is a set of roots with multiplicity greater than &lt;math&gt;1&lt;/math&gt;, say &lt;math&gt;r_1, r_2, \cdots, r_k&lt;/math&gt;, then we would replace the term &lt;math&gt;a_1e^{r_1x} + \cdots + a_ke^{r_kx}&lt;/math&gt; with the expression &lt;math&gt;(a_1x^{k-1} + a_2x^{k-2} + \cdots + a_{k-1}x + a_k)e^{r_1x}&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> In general, given a linear differential equation of the form &lt;math&gt;Ly = f&lt;/math&gt;, where &lt;math&gt;L = c_nD^n + c_{n-1}D^{n-1} + \cdots + c_0&lt;/math&gt; is a linear differential operator, then the set of solutions is given by the sum of any solution to the homogenous equation &lt;math&gt;Ly = 0&lt;/math&gt; and a specific solution to &lt;math&gt;Ly = f&lt;/math&gt;.<br /> <br /> === Proof ===<br /> We can apply an induction technique similar to the section on linear recurrences above. <br /> <br /> From linear algebra, we can use the following [[vector space]] decomposition theorem. Let &lt;math&gt;A: V \to V&lt;/math&gt; be a linear operator for a [[vector space]]s &lt;math&gt;V&lt;/math&gt; over a field &lt;math&gt;K&lt;/math&gt;. Suppose that there exists a polynomial &lt;math&gt;f(x) \in K[x]&lt;/math&gt; such that &lt;math&gt;f = gh&lt;/math&gt;, where &lt;math&gt;g&lt;/math&gt; and &lt;math&gt;h&lt;/math&gt; are non-zero polynomials such that &lt;math&gt;\text{gcd}\,(g,h) = 1&lt;/math&gt;, and such that &lt;math&gt;f(A) = O&lt;/math&gt;. Then &lt;math&gt;V = \ker g(A) \oplus \ker h(A)&lt;/math&gt;. This allows us to reduce the differential equation into finding the solutions to the equation &lt;math&gt;(D - \lambda I)^my = 0&lt;/math&gt;, which has a basis of functions &lt;math&gt;e^{\lambda t}, te^{\lambda t}, \ldots, t^{m-1}e^{\lambda t}&lt;/math&gt;.<br /> <br /> == Problems ==<br /> === Introductory ===<br /> *Prove [[Binet's formula]]. Find a similar closed form equation for the [[Lucas sequence]], defined with the starting terms &lt;math&gt;L_1 = 2, L_2 = 1&lt;/math&gt;, and satisfying the recursion &lt;math&gt;L_n = L_{n-1} + L_{n-2}&lt;/math&gt;. <br /> *Let &lt;math&gt;\{x_n\}&lt;/math&gt; denote the sequence defined by the recursion &lt;math&gt;x_0 = 3, x_1 = 1&lt;/math&gt;, and &lt;math&gt;x_n = 2x_{n-1} + 3x_{n-2}&lt;/math&gt;. Find a closed form expression for &lt;math&gt;x_n&lt;/math&gt;.<br /> *Given &lt;math&gt;a_0 = 1&lt;/math&gt;, &lt;math&gt;a_1 = 3&lt;/math&gt;, and the general relation &lt;math&gt;a_n^2 - a_{n - 1}a_{n + 1} = ( - 1)^n&lt;/math&gt; for &lt;math&gt;n \ge 1&lt;/math&gt;. Find a linear recurrence for &lt;math&gt;a_n&lt;/math&gt;. (AHSME 1958, Problem 40)<br /> <br /> === Intermediate ===<br /> *Let &lt;math&gt;S_n&lt;/math&gt; denote the number of ternary sequences (consisting of &lt;math&gt;0&lt;/math&gt;,&lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;2&lt;/math&gt;s) of length &lt;math&gt;n&lt;/math&gt;, such that they do not contain a substring of &quot;10&quot;, &quot;01&quot;, or &quot;11&quot;. Find a closed form expression for &lt;math&gt;S_n&lt;/math&gt;. <br /> *Let &lt;math&gt;\{X_n\}&lt;/math&gt; and &lt;math&gt;\{Y_n\}&lt;/math&gt; be sequences defined as follows: <br /> <br /> &lt;center&gt;&lt;cmath&gt;X_0 = Y_0 = X_1 = Y_1 = 1, X_{n+1} = X_n + 2X_{n-1}, Y_{n+1} = 3Y_n + 4Y_{n-1}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> :Let &lt;math&gt;k&lt;/math&gt; be the largest integer that satisfies all of the following conditions:<br /> <br /> ::(i) &lt;math&gt;|X_i - k| \le 2007&lt;/math&gt;, for some positive integer &lt;math&gt;i&lt;/math&gt;;<br /> ::(ii) &lt;math&gt;|Y_j - k| \le 2007&lt;/math&gt;, for some positive integer &lt;math&gt;j&lt;/math&gt;;<br /> ::(iii) &lt;math&gt;k &lt; 10^{2007}&lt;/math&gt;. <br /> <br /> :Find the remainder when &lt;math&gt;k&lt;/math&gt; is divided by &lt;math&gt;2007&lt;/math&gt;. (2007 [[iTest]], #47)<br /> <br /> *Let &lt;math&gt;a_{n}&lt;/math&gt;, &lt;math&gt;b_{n}&lt;/math&gt;, and &lt;math&gt;c_{n}&lt;/math&gt; be geometric sequences with different common ratios and let &lt;math&gt;a_{n}+b_{n}+c_{n}=d_{n}&lt;/math&gt; for all integers &lt;math&gt;n&lt;/math&gt;. If &lt;math&gt;d_{1}=1&lt;/math&gt;, &lt;math&gt;d_{2}=2&lt;/math&gt;, &lt;math&gt;d_{3}=3&lt;/math&gt;, &lt;math&gt;d_{4}=-7&lt;/math&gt;, &lt;math&gt;d_{5}=13&lt;/math&gt;, and &lt;math&gt;d_{6}=-16&lt;/math&gt;, find &lt;math&gt;d_{7}&lt;/math&gt;. ([[Mock AIME 1 2006-2007/Problem 13|Mock AIME 1 2006-2007, Problem 13]])<br /> <br /> *Find all possible values of &lt;math&gt;x_0&lt;/math&gt; and &lt;math&gt;x_1&lt;/math&gt; such that the sequence defined by:<br /> <br /> &lt;center&gt;&lt;cmath&gt;x_{n + 1} = \frac {x_{n - 1} x_n}{3x_{n - 1} - 2x_n}, \quad n \ge 1&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> :contains infinitely many natural numbers.<br /> <br /> *Show that &lt;math&gt;a^n + b^n + c^n&lt;/math&gt; can be written as a linear combination of the [[elementary symmetric polynomials]] &lt;math&gt;abc, ab+bc+ca, a+b+c&lt;/math&gt;. In general, prove [[Newton's sums]]. <br /> <br /> === Olympiad ===<br /> *Let &lt;math&gt;r&lt;/math&gt; be a real number, and let &lt;math&gt;x_n&lt;/math&gt; be a sequence such that &lt;math&gt;x_0 = 0, x_1 = 1&lt;/math&gt;, and &lt;math&gt;x_{n+2} = rx_{n+1} - x_n&lt;/math&gt; for &lt;math&gt;n \ge 0&lt;/math&gt;. For which values of &lt;math&gt;r&lt;/math&gt; does &lt;math&gt;x_1 + x_3 + \cdots + x_{2m-1} = x_m^2&lt;/math&gt; for all positive integers &lt;math&gt;m&lt;/math&gt;? ([[WOOT]])<br /> * Let &lt;math&gt;a(x,y)&lt;/math&gt; be the polynomial &lt;math&gt;x^2y + xy^2&lt;/math&gt;, and &lt;math&gt;b(x,y)&lt;/math&gt; the polynomial &lt;math&gt;x^2 + xy + y^2&lt;/math&gt;. Prove that we can find a polynomial &lt;math&gt;p_n(a, b)&lt;/math&gt; which is identically equal to &lt;math&gt;(x + y)^n + (-1)^n (x^n + y^n)&lt;/math&gt;. For example, &lt;math&gt;p_4(a, b) = 2b^2&lt;/math&gt;. ([[1976 Putnam Problems/Problem A2|1976 Putnam, Problem A2]]).<br /> *&lt;math&gt;a_1,a_2,a_3,b_1,b_2,b_3&lt;/math&gt; are distinct positive integers such that &lt;math&gt;(n + 1)a_1^n + na_2^n + (n - 1)a_3^n|(n + 1)b_1^n + nb_2^n + (n - 1)b_3^n&lt;/math&gt; holds for all positive integer &lt;math&gt;n&lt;/math&gt;. Prove that there exists &lt;math&gt;k\in N&lt;/math&gt; such that &lt;math&gt;b_i = ka_i&lt;/math&gt; for &lt;math&gt;i = 1,2,3&lt;/math&gt;. (2010 Chinese MO, Problem 6)<br /> <br /> == Hints/Solutions ==<br /> === Introductory ===<br /> * A proof of [[Binet's formula]] may be found in that link. The characteristic polynomial of the Lucas sequence is exactly the same. Hence, the only thing we have to change are the coefficients. <br /> * We will work out this problem in full detail. The recurrence relation is &lt;math&gt;x_n -2x_{n-1} -3x_{n-2}&lt;/math&gt;, and its characteristic polynomial is given by &lt;math&gt;x^2-2x-3&lt;/math&gt;. The roots of this polynomial are &lt;math&gt;-1&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;. Thus, a closed form solution is given by &lt;math&gt;x_n = a \cdot (-1)^n + b \cdot 3^n&lt;/math&gt;. For &lt;math&gt;n = 0&lt;/math&gt;, we get &lt;math&gt;x_0 = a + b = 3&lt;/math&gt;, and for &lt;math&gt;n = 1&lt;/math&gt;, we get &lt;math&gt;x_1 = -a + 3b = 1&lt;/math&gt;. Solving gives &lt;math&gt;a = 2, b = 1&lt;/math&gt;. Thus, our answer is &lt;math&gt;x_n = 2 \cdot (-1)^n + 3^n&lt;/math&gt;. <br /> *&lt;url&gt;viewtopic.php?t=282744 Discussion&lt;/url&gt; (1958 AHSME, 40)<br /> <br /> === Intermediate ===<br /> *&lt;url&gt;viewtopic.php?t=335138 Discussion&lt;/url&gt;<br /> *Answer (2007 iTest 47): The answer is &lt;math&gt;k = 7468&lt;/math&gt; (before taking &lt;math&gt;\mod{2007}&lt;/math&gt;).<br /> *&lt;url&gt;viewtopic.php?t=287441 Discussion&lt;/url&gt; <br /> *Hint (Newton’s Sum): Let us work backwards. Suppose &lt;math&gt;a,b,c&lt;/math&gt; are the roots of the characteristic polynomial of a linear recurrence. Then apply [[Vieta's sums]].<br /> <br /> === Olympiad ===<br /> *Hint (WOOT): substitute the closed form solution. It is true for all &lt;math&gt;r&lt;/math&gt;. <br /> *Hint (1976 Putnam A2): do the even and odd cases separately.<br /> *&lt;url&gt;viewtopic.php?search_id=804457492&amp;t=327474 Discussion&lt;/url&gt; (2010 Chinese MO, 6)<br /> <br /> [[Category:Linear algebra]]</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=Characteristic_polynomial&diff=65840 Characteristic polynomial 2014-11-03T05:22:26Z <p>Mathcool2009: /* Proof 3 (Partial fractions) */ Fixed order</p> <hr /> <div>The '''characteristic polynomial''' of a linear [[operator]] refers to the [[polynomial]] whose roots are the [[eigenvalue]]s of the operator. It carries much information about the operator. <br /> <br /> In the context of problem-solving, the characteristic polynomial is often used to find closed forms for the solutions of [[#Linear recurrences|linear recurrences]].<br /> <br /> == Definition ==<br /> Suppose &lt;math&gt;A&lt;/math&gt; is a &lt;math&gt;n \times n&lt;/math&gt; [[matrix]] (over a field &lt;math&gt;K&lt;/math&gt;). Then the characteristic polynomial of &lt;math&gt;A&lt;/math&gt; is defined as &lt;math&gt;P_A(t) = \det(tI - A)&lt;/math&gt;, which is a &lt;math&gt;n&lt;/math&gt;th degree polynomial in &lt;math&gt;t&lt;/math&gt;. Here, &lt;math&gt;I&lt;/math&gt; refers to the &lt;math&gt;n\times n&lt;/math&gt; [[identity matrix]].<br /> <br /> Written out, the characteristic polynomial is the [[determinant]]<br /> <br /> &lt;center&gt;&lt;cmath&gt;P_A(t) = \begin{vmatrix}t-a_{11} &amp; a_{12} &amp; \cdots &amp; a_{1n} \\ a_{21} &amp; t-a_{22} &amp; \cdots &amp; a_{2n} \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ a_{n1} &amp; a_{n2} &amp; \cdots &amp; t-a_{nn}\end{vmatrix}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> == Properties ==<br /> An [[eigenvector]] &lt;math&gt;\bold{v} \in K^n&lt;/math&gt; is a non-zero vector that satisfies the relation &lt;math&gt;A\bold{v} = \lambda\bold{v}&lt;/math&gt;, for some scalar &lt;math&gt;\lambda \in K&lt;/math&gt;. In other words, applying a linear operator to an eigenvector causes the eigenvector to dilate. The associated number &lt;math&gt;\lambda&lt;/math&gt; is called the [[eigenvalue]]. <br /> <br /> There are at most &lt;math&gt;n&lt;/math&gt; distinct eigenvalues, whose values are exactly the roots of the characteristic polynomial of the square matrix. To prove this, we use the fact that the determinant of a matrix is &lt;math&gt;0&lt;/math&gt; [[iff]] the column vectors of the matrix are [[linearly dependent]]. Observe that if &lt;math&gt;\lambda&lt;/math&gt; satisfies &lt;math&gt;\det(\lambda I-A) = 0&lt;/math&gt;, then the column vectors of the &lt;math&gt;n\times n&lt;/math&gt; matrix &lt;math&gt;\lambda I - A&lt;/math&gt; are linearly dependent. Indeed, if we define &lt;math&gt;T = \lambda I - A&lt;/math&gt; and let &lt;math&gt;\bold{T}^1, \bold{T}^2, \ldots, \bold{T}^n&lt;/math&gt; denote the column vectors of &lt;math&gt;T&lt;/math&gt;, this is equivalent to saying that there exists not all zero scalars &lt;math&gt;v_i \in K&lt;/math&gt; such that <br /> <br /> &lt;cmath&gt;v_1 \cdot \bold{T}^1 + v_2 \cdot \bold{T}^2 + \cdots + v_n \cdot \bold{T}^n = \bold{O}.&lt;/cmath&gt;<br /> <br /> Hence, there exists a non-zero vector &lt;math&gt;\bold{v} = (v_1, v_2, \ldots, v_n) \in K^n&lt;/math&gt; such that &lt;math&gt;(\lambda I - A) \bold{v} = \bold{O}&lt;/math&gt;. Distributing and re-arranging, &lt;math&gt;A\bold{v} = \lambda\bold{v}&lt;/math&gt;, as desired. In the other direction, if &lt;math&gt;A \bold{v} = \lambda \bold{v}&lt;/math&gt;, then &lt;math&gt;\lambda I \bold{v} - A \bold{v} = \bold{O}&lt;/math&gt;. But then, the column vectors of &lt;math&gt;\lambda I - A&lt;/math&gt; are linearly dependent, so it follows that &lt;math&gt;\det(\lambda I - A) = 0&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> Note that if &lt;math&gt;t = 0&lt;/math&gt;, then &lt;math&gt;P_A(0) = \det (tI - A) = \det (-A) = (-1)^n \det (A)&lt;/math&gt;. Hence, the characteristic polynomial encodes the determinant of the matrix. Also, the [[coefficient]] of the &lt;math&gt;t^{n-1}&lt;/math&gt; term of &lt;math&gt;P_A(t)&lt;/math&gt; gives the negative of the [[trace]] of the matrix (which follows from [[Vieta's formulas]]). <br /> <br /> By the [[Hamilton-Cayley Theorem]], the characteristic polynomial of a square matrix applied to the square matrix itself is zero, that is &lt;math&gt;P_A(A) = 0&lt;/math&gt;. The [[minimal polynomial]] of &lt;math&gt;A&lt;/math&gt; thus divides the characteristic polynomial &lt;math&gt;p_A&lt;/math&gt;. <br /> <br /> == Linear recurrences ==<br /> Let &lt;math&gt;x_1, x_2, \ldots, &lt;/math&gt; be a sequence of real numbers. Consider a monic [[homogenous]] [[linear recurrence]] of the form <br /> <br /> &lt;center&gt;&lt;cmath&gt;x_{n} = c_1x_{n-1} + c_2x_{n-2} + \cdots + c_kx_{n-k}, \quad (*)&lt;/cmath&gt;&lt;/center&gt; <br /> <br /> where &lt;math&gt;c_1, \ldots, c_k&lt;/math&gt; are real constants. The characteristic polynomial of this recurrence is defined as the polynomial <br /> <br /> &lt;center&gt;&lt;cmath&gt;P(x) = x^k - c_1x^{k-1} - c_2x^{k-2} - \cdots -c_{k-1}x - c_k.&lt;/cmath&gt;&lt;/center&gt; <br /> <br /> For example, let &lt;math&gt;F_n&lt;/math&gt; be the &lt;math&gt;n&lt;/math&gt;th [[Fibonacci number]] defined by &lt;math&gt;F_1 = F_2 = 1&lt;/math&gt;, and <br /> <br /> &lt;center&gt;&lt;cmath&gt;F_n = F_{n-1} + F_{n-2} \Longleftrightarrow F_n - F_{n-1} - F_{n-2} = 0.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> Then, its characteristic polynomial is &lt;math&gt;x^2 - x - 1&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> The roots of the polynomial can be used to write a closed form for the recurrence. If the roots of this polynomial &lt;math&gt;P&lt;/math&gt; are distinct, then suppose the roots are &lt;math&gt;r_1,r_2, \cdots, r_k&lt;/math&gt;. Then, there exists real constants &lt;math&gt;a_1, a_2, \cdots, a_k&lt;/math&gt; such that<br /> <br /> &lt;center&gt;&lt;cmath&gt;x_n = a_1 \cdot r_1^n + a_2 \cdot r_2^n + \cdots + a_k \cdot r_k^n&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> If we evaluate &lt;math&gt;k&lt;/math&gt; different values of &lt;math&gt;x_i&lt;/math&gt; (typically &lt;math&gt;x_0, x_1, \cdots, x_{k-1}&lt;/math&gt;), we can find a linear system in the &lt;math&gt;a_i&lt;/math&gt;s that can be solved for each constant. Refer to the [[#Introductory|introductory problems]] below to see an example of how to do this. In particular, for the Fibonacci numbers, this yields [[Binet's formula]]. &lt;br&gt;&lt;br&gt;<br /> <br /> If there are roots with multiplicity greater than &lt;math&gt;1&lt;/math&gt;, suppose &lt;math&gt;r_1 = r_2 = \cdots = r_{k_1}&lt;/math&gt;. Then we would replace the term &lt;math&gt;a_1 \cdot r^n + a_2 \cdot r_2^n + \cdots + a_{k_1} \cdot r_{k_1}^n&lt;/math&gt; with the expression <br /> <br /> &lt;center&gt;&lt;cmath&gt;(a_1 \cdot n^{k_1-1} + a_2 \cdot n^{k_1 - 2} + \cdots + a_{k_1-1} \cdot n + a_{k_1}) \cdot r^n.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> That is, there is now a polynomial in &lt;math&gt;n&lt;/math&gt; multiplied with the exponent. Note that there are still the same number of constants that we must solve for. For example, consider the recurrence relation &lt;math&gt;x_n = -2x_{n-1} -x_{n-2} \Longleftrightarrow x_n + 2x_{n-1} + x_{n-2} = 0&lt;/math&gt;. It’s characteristic polynomial, &lt;math&gt;x^2 + 2x + 1&lt;/math&gt;, has a &lt;math&gt;-1&lt;/math&gt; double root. Then, its closed form solution is of the type &lt;math&gt;x_n = (-1)^n(an + b)&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> Given a linear recurrence of the form &lt;math&gt;x_n - a_1x_{n-1} - \cdots - a_kx_{n-k} = f(n)&lt;/math&gt;, we often try to find a new sequence &lt;math&gt;y_n = x_n - g(n)&lt;/math&gt; such that &lt;math&gt;y_n&lt;/math&gt; is a homogenous linear recurrence. Then, we can find a closed form for &lt;math&gt;y_n&lt;/math&gt;, and then the answer is given by &lt;math&gt;x_n = y_n + g(n)&lt;/math&gt;. <br /> <br /> Of the following proofs, the second and third are more approachable. <br /> <br /> === Proof 1 (Linear Algebra) ===<br /> Note: The ideas expressed in this section can be transferred to the next section about differential equations. This requires some knowledge of [[linear algebra]] (upto the [[Spectral Theorem]]).<br /> <br /> Let &lt;math&gt;\bold{y}_{n-1} = \begin{pmatrix} x_{n-1} \\ x_{n-2} \\ \vdots \\ x_{n-k} \end{pmatrix}&lt;/math&gt;. Then, we can express our linear recurrence as the matrix <br /> <br /> &lt;center&gt;&lt;cmath&gt;A = \begin{pmatrix}a_1 &amp; a_2 &amp; a_3 &amp; \cdots &amp; a_{k-1} &amp; a_{k} \\ 1 &amp; 0 &amp; 0 &amp; \cdots &amp;0 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; \cdots &amp; 0 &amp; 0 \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \ddots &amp; \vdots \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 1 &amp; 0 \end{pmatrix},&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> so that &lt;math&gt;\bold{y}_n = A\bold{y}_{n-1}&lt;/math&gt; (try to verify this). The characteristic polynomial of &lt;math&gt;A&lt;/math&gt; is precisely &lt;math&gt;P_A(t) = \det (tI - A) = a_1 \cdot t^{k-1} + a_2 \cdot t^{k-2} \cdots + a_k = 0&lt;/math&gt;. This is not difficult to show via [[Laplace's expansion]] and induction, and is left as an exercise to the reader. <br /> <br /> If the roots of &lt;math&gt;P_A&lt;/math&gt; are distinct, then there exists a [[basis]] (of &lt;math&gt;\mathbb{R}^{k-1}&lt;/math&gt;) consisting of [[eigenvector]]s of &lt;math&gt;A&lt;/math&gt; (since eigenvectors of different eigenvalues are [[linearly independent]]). That is, applying a change of bases, we can write <br /> <br /> &lt;center&gt;&lt;cmath&gt;A = U^{-1} D U&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> for a matrix &lt;math&gt;U&lt;/math&gt; and a diagonal matrix &lt;math&gt;D&lt;/math&gt;. Then, &lt;math&gt;A^2 = U^{-1} D U U^{-1} D U = U^{-1} D^2 U&lt;/math&gt;, and in general, &lt;math&gt;A^n = U^{-1} D^n U&lt;/math&gt;. Thus, &lt;math&gt;\bold{y}_{n} = A^{n}\bold{y}_0 = U^{-1}D^{n+k}U\bold{y}_0&lt;/math&gt;. Here, &lt;math&gt;U^{-1}, U,&lt;/math&gt; and &lt;math&gt;\bold{y}_0&lt;/math&gt; are fixed (note that to find the values of &lt;math&gt;\bold{y}_0&lt;/math&gt;, we may need to trace the recurrence backwards. We only take the &lt;math&gt;0&lt;/math&gt;th index for simplicity). It follows that &lt;math&gt;y_{n}&lt;/math&gt; is a linear combination of the diagonal elements of &lt;math&gt;D&lt;/math&gt;, namely &lt;math&gt;r_1^n, r_2^n, \ldots, r_k^n&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> Suppose now that that the roots of &lt;math&gt;P_A&lt;/math&gt; are not distinct. Then, we can write the matrix in the [[Jordan normal form]]. For simplicity, first consider just a single root &lt;math&gt;r&lt;/math&gt; repeated &lt;math&gt;k&lt;/math&gt; times. The corresponding Jordan form of the matrix is given by (that is, the matrix is [[similar]] to the following):<br /> <br /> &lt;center&gt;&lt;cmath&gt;\begin{pmatrix} r &amp; 1 &amp; 0 &amp; \cdots &amp; 0 \\ 0 &amp; r &amp; 1 &amp; \cdots &amp; 0 \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; r \end{pmatrix}&lt;/cmath&gt;.&lt;/center&gt;<br /> <br /> Exponentiating this matrix to the &lt;math&gt;n&lt;/math&gt;th power will yield [[binomial coefficient]]s as follows<br /> <br /> &lt;center&gt;&lt;cmath&gt;\begin{pmatrix} r^n &amp; {n \choose 1}r^{n-1} &amp; {n \choose 2}r^{n-2} &amp; \cdots &amp; {n \choose k}r^{n-k} \\ 0 &amp; r &amp; {n \choose 1}r^{n-1} &amp; \cdots &amp; {n \choose {k-1}}r^{n-k+1} \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; r^n \end{pmatrix}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> We can treat the binomial coefficient as a polynomial in &lt;math&gt;n&lt;/math&gt;. Furthermore, we can scale away the powers of the eigenvalue (which is a constant); after taking the appropriate linear combinations of the binomial coefficients and a little bit of work, the result follows. &lt;br&gt;&lt;br&gt;<br /> <br /> See also a &lt;url&gt;viewtopic.php?t=290351 graph-theoretic&lt;/url&gt; approach.<br /> <br /> === Proof 2 (Induction) ===<br /> <br /> There are a couple of lower-level ways to prove this. One is by [[induction]], though the proof is not very revealing; we can explicitly check that a sequence &lt;math&gt;\{a_1 \cdot r_1^n + a_2 \cdot r_2^n + \cdots + a_k \cdot r_k^n\}&lt;/math&gt;, for real numbers &lt;math&gt;a_1, a_2, \ldots, a_k&lt;/math&gt;, satisfies the linear recurrence relation &lt;math&gt;(*)&lt;/math&gt;. If the two sequences are the same for the first &lt;math&gt;k&lt;/math&gt; values of the sequence, it follows by induction that the two sequences must be exactly the same.<br /> <br /> In particular, for &lt;math&gt;k=2&lt;/math&gt;, we can check this by using the identity <br /> <br /> &lt;center&gt;&lt;cmath&gt;ax^{n+1} + by^{n+1} = (x+y)(ax^n + by^n) - xy(ax^{n-1} + by^{n-1}).&lt;/cmath&gt;&lt;/center&gt; &lt;br&gt;<br /> <br /> It is also possible to reduce the recurrence to a [[telescoping sum]]. However, the details are slightly messy. <br /> <br /> === Proof 3 (Partial fractions) ===<br /> <br /> Another method uses [[partial fractions]] and [[generating function]]s, though not much knowledge of each is required for this proof. Let &lt;math&gt;G_n = a_1G_{n-1} + \cdots + a_kG_{n-k}&lt;/math&gt; be a linear recurrence. Consider the [[generating function]] given by <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = G_0 + G_1x + G_2x^2 + \cdots&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> Then, writing the following expressing out and carefully comparing coefficients (try it), <br /> <br /> &lt;center&gt;&lt;cmath&gt;a_kG(x) \cdot x^{k-1} + a_{k-1}G(x) \cdot x^{k-2} + \cdots + a_{2}G(x) \cdot x + a_{1}G(x) = \frac{R(x) - G(x)}x,&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> where &lt;math&gt;R(x)&lt;/math&gt; is a remainder polynomial with degree &lt;math&gt;&lt; k-1&lt;/math&gt;. Re-arranging, <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = \frac{xR(x)}{a_k\cdot x^{k} + a_{k-1}\cdot x^{k-1} + \cdots + a_{1}x + 1}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> This polynomial in the denominator is the reverse of the characteristic polynomial of the recurrence. Hence, its roots are &lt;math&gt;1/r_1, 1/r_2, \ldots, 1/r_k&lt;/math&gt;, assuming that the roots are distinct. Using [[partial fraction]] decomposition (essentially the reverse of the process of adding fractions by combining denominators, except we now pull the denominators apart), we can write <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = \frac{c_1}{1 - r_1x} + \frac{c_2}{1-r_2x} + \cdots + \frac{c_k}{1-r_kx } &lt;/cmath&gt;&lt;/center&gt;<br /> <br /> for some constants &lt;math&gt;c_1, \ldots, c_k&lt;/math&gt;. Using the [[geometric series]] formula, we have &lt;math&gt;\frac{c}{1-y} = c + cy + cy^2 + \ldots&lt;/math&gt;. Thus, <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = (c_1 + \cdots + c_k) + (c_1r_1 + \cdots + c_kr_k)x + (c_1r_1^2 + \cdots + c_kr_k^2)x^2 + \cdots.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> Comparing coefficients with our original definition of &lt;math&gt;G(x)&lt;/math&gt; gives &lt;math&gt;G_n = c_1r_1^n + c_2r_2^n + \cdots + c_kr_k^n&lt;/math&gt;, as desired. &lt;br&gt;&lt;br&gt;<br /> <br /> The generalization to characteristic polynomials with multiple roots is not difficult from here, and is left to the reader. The partial fraction decomposition will have terms of the form &lt;math&gt;\frac{c_i}{(1-r_ix)^\alpha}&lt;/math&gt; for &lt;math&gt;\alpha &gt; 1&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> A note about this argument: all of the power series used here are defined formally, and so we do not actually need to worry whether or not they converge. See these [http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=250866 blog][http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=215833 posts] for further ideas.<br /> <br /> == Differential equations ==<br /> Given a monic linear [[homogenous]] [[differential equation]] of the form &lt;math&gt;D^ny +c_{n-1}D^{n-1}y + \cdots + c_1Dy + c_0y = 0&lt;/math&gt;, then the characteristic polynomial of the equation is the polynomial <br /> <br /> &lt;center&gt;&lt;cmath&gt;P(t) = t^n + c_{n-1}t^{n-1} + c_{n-2}t^{n-2} + \cdots + c_0.&lt;/cmath&gt;&lt;/center&gt; <br /> <br /> Here, &lt;math&gt;D = \frac{d}{dx}&lt;/math&gt; is short-hand for the differential operator. &lt;br&gt;&lt;br&gt;<br /> <br /> If the roots of the polynomial are distinct, say &lt;math&gt;r_1, r_2, \cdots, r_n&lt;/math&gt;, then the solutions of this differential equation are precisely the linear combinations &lt;math&gt;y(x) = a_1e^{r_1x} + a_2e^{r_2x} + \cdots + a_ne^{r_nx}&lt;/math&gt;. Similarly, if there is a set of roots with multiplicity greater than &lt;math&gt;1&lt;/math&gt;, say &lt;math&gt;r_1, r_2, \cdots, r_k&lt;/math&gt;, then we would replace the term &lt;math&gt;a_1e^{r_1x} + \cdots + a_ke^{r_kx}&lt;/math&gt; with the expression &lt;math&gt;(a_1x^{k-1} + a_2x^{k-2} + \cdots + a_{k-1}x + a_k)e^{r_1x}&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> In general, given a linear differential equation of the form &lt;math&gt;Ly = f&lt;/math&gt;, where &lt;math&gt;L = c_nD^n + c_{n-1}D^{n-1} + \cdots + c_0&lt;/math&gt; is a linear differential operator, then the set of solutions is given by the sum of any solution to the homogenous equation &lt;math&gt;Ly = 0&lt;/math&gt; and a specific solution to &lt;math&gt;Ly = f&lt;/math&gt;.<br /> <br /> === Proof ===<br /> We can apply an induction technique similar to the section on linear recurrences above. <br /> <br /> From linear algebra, we can use the following [[vector space]] decomposition theorem. Let &lt;math&gt;A: V \to V&lt;/math&gt; be a linear operator for a [[vector space]]s &lt;math&gt;V&lt;/math&gt; over a field &lt;math&gt;K&lt;/math&gt;. Suppose that there exists a polynomial &lt;math&gt;f(x) \in K[x]&lt;/math&gt; such that &lt;math&gt;f = gh&lt;/math&gt;, where &lt;math&gt;g&lt;/math&gt; and &lt;math&gt;h&lt;/math&gt; are non-zero polynomials such that &lt;math&gt;\text{gcd}\,(g,h) = 1&lt;/math&gt;, and such that &lt;math&gt;f(A) = O&lt;/math&gt;. Then &lt;math&gt;V = \ker g(A) \oplus \ker h(A)&lt;/math&gt;. This allows us to reduce the differential equation into finding the solutions to the equation &lt;math&gt;(D - \lambda I)^my = 0&lt;/math&gt;, which has a basis of functions &lt;math&gt;e^{\lambda t}, te^{\lambda t}, \ldots, t^{m-1}e^{\lambda t}&lt;/math&gt;.<br /> <br /> == Problems ==<br /> === Introductory ===<br /> *Prove [[Binet's formula]]. Find a similar closed form equation for the [[Lucas sequence]], defined with the starting terms &lt;math&gt;L_1 = 2, L_2 = 1&lt;/math&gt;, and satisfying the recursion &lt;math&gt;L_n = L_{n-1} + L_{n-2}&lt;/math&gt;. <br /> *Let &lt;math&gt;\{x_n\}&lt;/math&gt; denote the sequence defined by the recursion &lt;math&gt;x_0 = 3, x_1 = 1&lt;/math&gt;, and &lt;math&gt;x_n = 2x_{n-1} + 3x_{n-2}&lt;/math&gt;. Find a closed form expression for &lt;math&gt;x_n&lt;/math&gt;.<br /> *Given &lt;math&gt;a_0 = 1&lt;/math&gt;, &lt;math&gt;a_1 = 3&lt;/math&gt;, and the general relation &lt;math&gt;a_n^2 - a_{n - 1}a_{n + 1} = ( - 1)^n&lt;/math&gt; for &lt;math&gt;n \ge 1&lt;/math&gt;. Find a linear recurrence for &lt;math&gt;a_n&lt;/math&gt;. (AHSME 1958, Problem 40)<br /> <br /> === Intermediate ===<br /> *Let &lt;math&gt;S_n&lt;/math&gt; denote the number of ternary sequences (consisting of &lt;math&gt;0&lt;/math&gt;,&lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;2&lt;/math&gt;s) of length &lt;math&gt;n&lt;/math&gt;, such that they do not contain a substring of &quot;10&quot;, &quot;01&quot;, or &quot;11&quot;. Find a closed form expression for &lt;math&gt;S_n&lt;/math&gt;. <br /> *Let &lt;math&gt;\{X_n\}&lt;/math&gt; and &lt;math&gt;\{Y_n\}&lt;/math&gt; be sequences defined as follows: <br /> <br /> &lt;center&gt;&lt;cmath&gt;X_0 = Y_0 = X_1 = Y_1 = 1, X_{n+1} = X_n + 2X_{n-1}, Y_{n+1} = 3Y_n + 4Y_{n-1}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> :Let &lt;math&gt;k&lt;/math&gt; be the largest integer that satisfies all of the following conditions:<br /> <br /> ::(i) &lt;math&gt;|X_i - k| \le 2007&lt;/math&gt;, for some positive integer &lt;math&gt;i&lt;/math&gt;;<br /> ::(ii) &lt;math&gt;|Y_j - k| \le 2007&lt;/math&gt;, for some positive integer &lt;math&gt;j&lt;/math&gt;;<br /> ::(iii) &lt;math&gt;k &lt; 10^{2007}&lt;/math&gt;. <br /> <br /> :Find the remainder when &lt;math&gt;k&lt;/math&gt; is divided by &lt;math&gt;2007&lt;/math&gt;. (2007 [[iTest]], #47)<br /> <br /> *Let &lt;math&gt;a_{n}&lt;/math&gt;, &lt;math&gt;b_{n}&lt;/math&gt;, and &lt;math&gt;c_{n}&lt;/math&gt; be geometric sequences with different common ratios and let &lt;math&gt;a_{n}+b_{n}+c_{n}=d_{n}&lt;/math&gt; for all integers &lt;math&gt;n&lt;/math&gt;. If &lt;math&gt;d_{1}=1&lt;/math&gt;, &lt;math&gt;d_{2}=2&lt;/math&gt;, &lt;math&gt;d_{3}=3&lt;/math&gt;, &lt;math&gt;d_{4}=-7&lt;/math&gt;, &lt;math&gt;d_{5}=13&lt;/math&gt;, and &lt;math&gt;d_{6}=-16&lt;/math&gt;, find &lt;math&gt;d_{7}&lt;/math&gt;. ([[Mock AIME 1 2006-2007/Problem 13|Mock AIME 1 2006-2007, Problem 13]])<br /> <br /> *Find all possible values of &lt;math&gt;x_0&lt;/math&gt; and &lt;math&gt;x_1&lt;/math&gt; such that the sequence defined by:<br /> <br /> &lt;center&gt;&lt;cmath&gt;x_{n + 1} = \frac {x_{n - 1} x_n}{3x_{n - 1} - 2x_n}, \quad n \ge 1&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> :contains infinitely many natural numbers.<br /> <br /> *Show that &lt;math&gt;a^n + b^n + c^n&lt;/math&gt; can be written as a linear combination of the [[elementary symmetric polynomials]] &lt;math&gt;abc, ab+bc+ca, a+b+c&lt;/math&gt;. In general, prove [[Newton's sums]]. <br /> <br /> === Olympiad ===<br /> *Let &lt;math&gt;r&lt;/math&gt; be a real number, and let &lt;math&gt;x_n&lt;/math&gt; be a sequence such that &lt;math&gt;x_0 = 0, x_1 = 1&lt;/math&gt;, and &lt;math&gt;x_{n+2} = rx_{n+1} - x_n&lt;/math&gt; for &lt;math&gt;n \ge 0&lt;/math&gt;. For which values of &lt;math&gt;r&lt;/math&gt; does &lt;math&gt;x_1 + x_3 + \cdots + x_{2m-1} = x_m^2&lt;/math&gt; for all positive integers &lt;math&gt;m&lt;/math&gt;? ([[WOOT]])<br /> * Let &lt;math&gt;a(x,y)&lt;/math&gt; be the polynomial &lt;math&gt;x^2y + xy^2&lt;/math&gt;, and &lt;math&gt;b(x,y)&lt;/math&gt; the polynomial &lt;math&gt;x^2 + xy + y^2&lt;/math&gt;. Prove that we can find a polynomial &lt;math&gt;p_n(a, b)&lt;/math&gt; which is identically equal to &lt;math&gt;(x + y)^n + (-1)^n (x^n + y^n)&lt;/math&gt;. For example, &lt;math&gt;p_4(a, b) = 2b^2&lt;/math&gt;. ([[1976 Putnam Problems/Problem A2|1976 Putnam, Problem A2]]).<br /> *&lt;math&gt;a_1,a_2,a_3,b_1,b_2,b_3&lt;/math&gt; are distinct positive integers such that &lt;math&gt;(n + 1)a_1^n + na_2^n + (n - 1)a_3^n|(n + 1)b_1^n + nb_2^n + (n - 1)b_3^n&lt;/math&gt; holds for all positive integer &lt;math&gt;n&lt;/math&gt;. Prove that there exists &lt;math&gt;k\in N&lt;/math&gt; such that &lt;math&gt;b_i = ka_i&lt;/math&gt; for &lt;math&gt;i = 1,2,3&lt;/math&gt;. (2010 Chinese MO, Problem 6)<br /> <br /> == Hints/Solutions ==<br /> === Introductory ===<br /> * A proof of [[Binet's formula]] may be found in that link. The characteristic polynomial of the Lucas sequence is exactly the same. Hence, the only thing we have to change are the coefficients. <br /> * We will work out this problem in full detail. The recurrence relation is &lt;math&gt;x_n -2x_{n-1} -3x_{n-2}&lt;/math&gt;, and its characteristic polynomial is given by &lt;math&gt;x^2-2x-3&lt;/math&gt;. The roots of this polynomial are &lt;math&gt;-1&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;. Thus, a closed form solution is given by &lt;math&gt;x_n = a \cdot (-1)^n + b \cdot 3^n&lt;/math&gt;. For &lt;math&gt;n = 0&lt;/math&gt;, we get &lt;math&gt;x_0 = a + b = 3&lt;/math&gt;, and for &lt;math&gt;n = 1&lt;/math&gt;, we get &lt;math&gt;x_1 = -a + 3b = 1&lt;/math&gt;. Solving gives &lt;math&gt;a = 2, b = 1&lt;/math&gt;. Thus, our answer is &lt;math&gt;x_n = 2 \cdot (-1)^n + 3^n&lt;/math&gt;. <br /> *&lt;url&gt;viewtopic.php?t=282744 Discussion&lt;/url&gt; (1958 AHSME, 40)<br /> <br /> === Intermediate ===<br /> *&lt;url&gt;viewtopic.php?t=335138 Discussion&lt;/url&gt;<br /> *Answer (2007 iTest 47): The answer is &lt;math&gt;k = 7468&lt;/math&gt; (before taking &lt;math&gt;\mod{2007}&lt;/math&gt;).<br /> *&lt;url&gt;viewtopic.php?t=287441 Discussion&lt;/url&gt; <br /> *Hint (Newton’s Sum): Let us work backwards. Suppose &lt;math&gt;a,b,c&lt;/math&gt; are the roots of the characteristic polynomial of a linear recurrence. Then apply [[Vieta's sums]].<br /> <br /> === Olympiad ===<br /> *Hint (WOOT): substitute the closed form solution. It is true for all &lt;math&gt;r&lt;/math&gt;. <br /> *Hint (1976 Putnam A2): do the even and odd cases separately.<br /> *&lt;url&gt;viewtopic.php?search_id=804457492&amp;t=327474 Discussion&lt;/url&gt; (2010 Chinese MO, 6)<br /> <br /> [[Category:Linear algebra]]</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_8_Problems&diff=63221 2005 AMC 8 Problems 2014-08-22T19:04:09Z <p>Mathcool2009: /* Problem 11 */</p> <hr /> <div>==Problem 1==<br /> Connie multiplies a number by 2 and gets 60 as her answer. However, she should<br /> have divided the number by 2 to get the correct answer. What is the correct<br /> answer?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 7.5\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 240 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Karl bought five folders from Pay-A-Lot at a cost of &lt;math&gt; \textdollar 2.50 &lt;/math&gt; each.<br /> Pay-A-Lot had a 20%-off sale the following day. How much could<br /> Karl have saved on the purchase by waiting a day?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \textdollar 1.00 \qquad\textbf{(B)}\ \textdollar 2.00 \qquad\textbf{(C)}\ \textdollar 2.50\qquad\textbf{(D)}\ \textdollar 2.75 \qquad\textbf{(E)}\ \textdollar 5.00 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> What is the minimum number of small squares that must be colored black so that a line of symmetry lies on the diagonal &lt;math&gt; \overline{BD}&lt;/math&gt; of square &lt;math&gt; ABCD&lt;/math&gt;?<br /> &lt;asy&gt;defaultpen(linewidth(1));<br /> for ( int x = 0; x &amp;lt; 5; ++x )<br /> {<br /> draw((0,x)--(4,x));<br /> draw((x,0)--(x,4));<br /> }<br /> <br /> fill((1,0)--(2,0)--(2,1)--(1,1)--cycle);<br /> fill((0,3)--(1,3)--(1,4)--(0,4)--cycle);<br /> fill((2,3)--(4,3)--(4,4)--(2,4)--cycle);<br /> fill((3,1)--(4,1)--(4,2)--(3,2)--cycle);<br /> label(&quot;$A$&quot;, (0, 4), NW);<br /> label(&quot;$B$&quot;, (4, 4), NE);<br /> label(&quot;$C$&quot;, (4, 0), SE);<br /> label(&quot;$D$&quot;, (0, 0), SW);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are 6.1 cm, 8.2 cm and 9.7 cm. What is the area of the square in square centimeters?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 24\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Soda is sold in packs of 6, 12 and 24 cans. What is the minimum number of packs needed to buy exactly 90 cans of soda?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 15 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Suppose &lt;math&gt;d&lt;/math&gt; is a digit. For how many values of &lt;math&gt;d&lt;/math&gt; is &lt;math&gt;2.00d5 &gt; 2.005&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 10 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Bill walks &lt;math&gt;\tfrac12&lt;/math&gt; mile south, then &lt;math&gt;\tfrac34&lt;/math&gt; mile east, and finally &lt;math&gt;\tfrac12&lt;/math&gt; mile south. How many miles is he, in a direct line, from his starting point?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 1\tfrac14\qquad\textbf{(C)}\ 1\tfrac12\qquad\textbf{(D)}\ 1\tfrac34\qquad\textbf{(E)}\ 2 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Suppose m and n are positive odd integers. Which of the following must also be an odd integer?<br /> <br /> &lt;math&gt; \textbf{(A)}\ m+3n\qquad\textbf{(B)}\ 3m-n\qquad\textbf{(C)}\ 3m^2 + 3n^2\qquad\textbf{(D)}\ (nm + 3)^2\qquad\textbf{(E)}\ 3mn &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> In quadrilateral &lt;math&gt; ABCD&lt;/math&gt;, sides &lt;math&gt; \overline{AB}&lt;/math&gt; and &lt;math&gt; \overline{BC}&lt;/math&gt; both have length 10, sides &lt;math&gt; \overline{CD}&lt;/math&gt; and &lt;math&gt; \overline{DA}&lt;/math&gt; both have length 17, and the measure of angle &lt;math&gt; ADC&lt;/math&gt; is &lt;math&gt; 60^\circ&lt;/math&gt;. What is the length of diagonal &lt;math&gt; \overline{AC}&lt;/math&gt;?<br /> &lt;asy&gt;draw((0,0)--(17,0));<br /> draw(rotate(301, (17,0))*(0,0)--(17,0));<br /> picture p;<br /> draw(p, (0,0)--(0,10));<br /> draw(p, rotate(115, (0,10))*(0,0)--(0,10));<br /> add(rotate(3)*p);<br /> <br /> draw((0,0)--(8.25,14.5), linetype(&quot;8 8&quot;));<br /> <br /> label(&quot;$A$&quot;, (8.25, 14.5), N);<br /> label(&quot;$B$&quot;, (-0.25, 10), W);<br /> label(&quot;$C$&quot;, (0,0), SW);<br /> label(&quot;$D$&quot;, (17, 0), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 15.5\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18.5 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Joe had walked half way from home to school when he realized he was late. He ran the rest of the way to school. He ran 3 times as fast as he walked. Joe took 6 minutes to walk half way to school. How many minutes did it take Joe to get from home to school?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 7\qquad\textbf{(B)}\ 7.3\qquad\textbf{(C)}\ 7.7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 8.3 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its &amp;#36;90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up &amp;#36;90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up &amp;#36;90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?<br /> <br /> &lt;math&gt; \textbf{(A)}\ - \textdollar1.06\qquad\textbf{(B)}\ - \textdollar 0.53\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \textdollar 0.53\qquad\textbf{(E)}\ \textdollar 1.06 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> The area of polygon &lt;math&gt; ABCDEF&lt;/math&gt; is 52 with &lt;math&gt; AB\equal{}8&lt;/math&gt;, &lt;math&gt; BC\equal{}9&lt;/math&gt; and &lt;math&gt; FA\equal{}5&lt;/math&gt;. What is &lt;math&gt; DE\plus{}EF&lt;/math&gt;?<br /> &lt;asy&gt;pair a=(0,9), b=(8,9), c=(8,0), d=(4,0), e=(4,4), f=(0,4);<br /> draw(a--b--c--d--e--f--cycle);<br /> draw(shift(0,-.25)*a--shift(.25,-.25)*a--shift(.25,0)*a);<br /> draw(shift(-.25,0)*b--shift(-.25,-.25)*b--shift(0,-.25)*b);<br /> draw(shift(-.25,0)*c--shift(-.25,.25)*c--shift(0,.25)*c);<br /> draw(shift(.25,0)*d--shift(.25,.25)*d--shift(0,.25)*d);<br /> draw(shift(.25,0)*f--shift(.25,.25)*f--shift(0,.25)*f);<br /> label(&quot;$A$&quot;, a, NW);<br /> label(&quot;$B$&quot;, b, NE);<br /> label(&quot;$C$&quot;, c, SE);<br /> label(&quot;$D$&quot;, d, SW);<br /> label(&quot;$E$&quot;, e, SW);<br /> label(&quot;$F$&quot;, f, SW);<br /> label(&quot;5&quot;, (0,6.5), W);<br /> label(&quot;8&quot;, (4,9), N);<br /> label(&quot;9&quot;, (8, 4.5), E);&lt;/asy&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 /> [[2005 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> The Little Twelve Basketball League has two divisions, with six teams in each division. Each team plays each of the other teams in its own division twice and every team in the other division once. How many games are scheduled?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 80\qquad\textbf{(B)}\ 96\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 192 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> How many different isosceles triangles have integer side lengths and perimeter 23?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> A five-legged Martian has a drawer full of socks, each of which is red, white or blue, and there are at least five socks of each color. The Martian pulls out one sock at a time without looking. How many socks must the Martian remove from the drawer to be certain there will be 5 socks of the same color?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> The results of a cross-country team's training run are graphed below. Which student has the greatest average speed?<br /> &lt;asy&gt;<br /> for ( int i = 1; i &lt;= 7; ++i )<br /> {<br /> draw((i,0)--(i,6));<br /> }<br /> <br /> for ( int i = 1; i &lt;= 5; ++i )<br /> {<br /> draw((0,i)--(8,i));<br /> }<br /> draw((-0.5,0)--(8,0), linewidth(1));<br /> draw((0,-0.5)--(0,6), linewidth(1));<br /> label(&quot;$O$&quot;, (0,0), SW);<br /> label(scale(.85)*rotate(90)*&quot;distance&quot;, (0, 3), W);<br /> label(scale(.85)*&quot;time&quot;, (4, 0), S);<br /> dot((1.25, 4.5));<br /> label(scale(.85)*&quot;Evelyn&quot;, (1.25, 4.8), N);<br /> dot((2.5, 2.2));<br /> label(scale(.85)*&quot;Briana&quot;, (2.5, 2.2), S);<br /> dot((4.25,5.2));<br /> label(scale(.85)*&quot;Carla&quot;, (4.25, 5.2), SE);<br /> dot((5.6, 2.8));<br /> label(scale(.85)*&quot;Debra&quot;, (5.6, 2.8), N);<br /> dot((6.8, 1.4));<br /> label(scale(.85)*&quot;Angela&quot;, (6.8, 1.4), E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{Angela}\qquad\textbf{(B)}\ \text{Briana}\qquad\textbf{(C)}\ \text{Carla}\qquad\textbf{(D)}\ \text{Debra}\qquad\textbf{(E)}\ \text{Evelyn} &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> How many three-digit numbers are divisible by 13?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 7\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ 69\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 77&lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> What is the perimeter of trapezoid &lt;math&gt; ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;size(3inch, 1.5inch);<br /> pair a=(0,0), b=(18,24), c=(68,24), d=(75,0), f=(68,0), e=(18,0);<br /> draw(a--b--c--d--cycle);<br /> draw(b--e);<br /> draw(shift(0,2)*e--shift(2,2)*e--shift(2,0)*e);<br /> label(&quot;30&quot;, (9,12), W);<br /> label(&quot;50&quot;, (43,24), N);<br /> label(&quot;25&quot;, (71.5, 12), E);<br /> label(&quot;24&quot;, (18, 12), E);<br /> label(&quot;$A$&quot;, a, SW);<br /> label(&quot;$B$&quot;, b, N);<br /> label(&quot;$C$&quot;, c, N);<br /> label(&quot;$D$&quot;, d, SE);<br /> label(&quot;$E$&quot;, e, S);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 180\qquad\textbf{(B)}\ 188\qquad\textbf{(C)}\ 196\qquad\textbf{(D)}\ 200\qquad\textbf{(E)}\ 204 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise.<br /> In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> How many distinct triangles can be drawn using three of the dots below as vertices?<br /> <br /> &lt;asy&gt;dot(origin^^(1,0)^^(2,0)^^(0,1)^^(1,1)^^(2,1));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24 &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> A company sells detergent in three different sized boxes: small (S), medium (M) and large (L). The medium size costs 50% more than the small size and contains 20% less detergent than the large size. The large size contains twice as much detergent as the small size and costs 30% more than the medium size. Rank the three sizes from best to worst buy.<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{SML}\qquad\textbf{(B)}\ \text{LMS}\qquad\textbf{(C)}\ \text{MSL}\qquad\textbf{(D)}\ \text{LSM}\qquad\textbf{(E)}\ \text{MLS} &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Isosceles right triangle &lt;math&gt; ABC&lt;/math&gt; encloses a semicircle of area &lt;math&gt; 2\pi&lt;/math&gt;. The circle has its center &lt;math&gt; O&lt;/math&gt; on hypotenuse &lt;math&gt; \overline{AB}&lt;/math&gt; and is tangent to sides &lt;math&gt; \overline{AC}&lt;/math&gt; and &lt;math&gt; \overline{BC}&lt;/math&gt;. What is the area of triangle &lt;math&gt; ABC&lt;/math&gt;?<br /> <br /> &lt;asy&gt;pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2);<br /> draw(circle(o, 2));<br /> clip(a--b--c--cycle);<br /> draw(a--b--c--cycle);<br /> dot(o);<br /> label(&quot;$C$&quot;, c, NW);<br /> label(&quot;$A$&quot;, a, NE);<br /> label(&quot;$B$&quot;, b, SW);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 3\pi\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 4\pi &lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A certain calculator has only two keys [+1] and [x2]. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed &quot;9&quot; and you pressed [+1], it would display &quot;10.&quot; If you then pressed [x2], it would display &quot;20.&quot; Starting with the display &quot;1,&quot; what is the fewest number of keystrokes you would need to reach &quot;200&quot;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?<br /> <br /> &lt;asy&gt;pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2);<br /> draw(a--d--b--c--cycle);<br /> draw(circle(o, 2.5));&lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A)}\ \frac{2}{\sqrt{\pi}} \qquad \textbf{(B)}\ \frac{1\plus{}\sqrt{2}}{2} \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ \sqrt{\pi}&lt;/math&gt;<br /> <br /> [[2005 AMC 8 Problems/Problem 25|Solution]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2005|before=[[2004 AMC 8 Problems|2004 AMC 8]]|after=[[2006 AMC 8 Problems|2006 AMC 8]]}}<br /> * [[AMC 8]]<br /> * [[AMC 8 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2014_USAJMO_Problems/Problem_4&diff=63069 2014 USAJMO Problems/Problem 4 2014-08-14T21:00:15Z <p>Mathcool2009: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;b\geq 2&lt;/math&gt; be an integer, and let &lt;math&gt;s_b(n)&lt;/math&gt; denote the sum of the digits of &lt;math&gt;n&lt;/math&gt; when it is written in base &lt;math&gt;b&lt;/math&gt;. Show that there are infinitely many positive integers that cannot be represented in the form &lt;math&gt;n+s_b(n)&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is a positive integer.<br /> ==Solution==<br /> Define &lt;math&gt;S(n) = n + s_b(n)&lt;/math&gt;, and call a number ''unrepresentable'' if it cannot equal &lt;math&gt;S(n)&lt;/math&gt; for a positive integer &lt;math&gt;n&lt;/math&gt;.<br /> We claim that in the interval &lt;math&gt;(b^p, b^{p+1}]&lt;/math&gt; there exists an unrepresentable number, for every positive integer &lt;math&gt;p&lt;/math&gt;.<br /> <br /> If &lt;math&gt;b^{p+1}&lt;/math&gt; is unrepresentable, we're done. Otherwise, time for our lemma:<br /> <br /> Lemma: Define the function &lt;math&gt;f(p)&lt;/math&gt; to equal the number of integer x less than &lt;math&gt;b^p&lt;/math&gt; such that &lt;math&gt;S(x) \ge b^p&lt;/math&gt;. If &lt;math&gt;b^{p+1} = S(y)&lt;/math&gt; for some y, then &lt;math&gt;f(p+1) &gt; f(p)&lt;/math&gt;.<br /> <br /> Proof: Let &lt;math&gt;F(p)&lt;/math&gt; be the set of integers x less than &lt;math&gt;b^p&lt;/math&gt; such that &lt;math&gt;S(x) \ge b^p&lt;/math&gt;. Then for every integer in &lt;math&gt;F(p)&lt;/math&gt;, append the digit &lt;math&gt;(b-1)&lt;/math&gt; to the front of it to create a valid integer in &lt;math&gt;F(p+1)&lt;/math&gt;. Also, notice that &lt;math&gt;(b-1) \cdot b^p \le y &lt; b^{p+1}&lt;/math&gt;. Removing the digit &lt;math&gt;(b-1)&lt;/math&gt; from the front of y creates a number that is not in &lt;math&gt;F(p)&lt;/math&gt;. Hence, &lt;math&gt;F(p) \rightarrow F(p+1)&lt;/math&gt;, but there exists an element of &lt;math&gt;F(p+1)&lt;/math&gt; not corresponding with &lt;math&gt;F(p)&lt;/math&gt;, so &lt;math&gt;f(p+1) &gt; f(p)&lt;/math&gt;.<br /> <br /> Note that our lemma combined with the Pigeonhole Principle essentially proves the claim. Therefore, because there are infinitely many intervals containing an unrepresentable number, there are infinitely many unrepresentable numbers.</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2014_USAJMO_Problems/Problem_4&diff=63068 2014 USAJMO Problems/Problem 4 2014-08-14T20:41:49Z <p>Mathcool2009: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;b\geq 2&lt;/math&gt; be an integer, and let &lt;math&gt;s_b(n)&lt;/math&gt; denote the sum of the digits of &lt;math&gt;n&lt;/math&gt; when it is written in base &lt;math&gt;b&lt;/math&gt;. Show that there are infinitely many positive integers that cannot be represented in the form &lt;math&gt;n+s_b(n)&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is a positive integer.<br /> ==Solution==<br /> Define &lt;math&gt;S(n) = n + s_b(n)&lt;/math&gt;, and call a number ''unrepresentable'' if it cannot equal &lt;math&gt;S(n)&lt;/math&gt; for a positive integer &lt;math&gt;n&lt;/math&gt;.<br /> We claim that in the interval &lt;math&gt;(b^p, b^{p+1}]&lt;/math&gt; there exists an unrepresentable number, for every positive integer &lt;math&gt;p&lt;/math&gt;.<br /> <br /> If &lt;math&gt;b^{p+1}&lt;/math&gt; is unrepresentable, we're done. Otherwise, time for our lemma:<br /> <br /> Lemma: Define the function &lt;math&gt;f(p)&lt;/math&gt; to equal the number of integer x less than &lt;math&gt;b^p&lt;/math&gt; such that &lt;math&gt;S(x) \ge b^p&lt;/math&gt;. If &lt;math&gt;b^{p+1} = S(y)&lt;/math&gt; for some y, then &lt;math&gt;f(p+1) &gt; f(p)&lt;/math&gt;.<br /> <br /> Proof: Let &lt;math&gt;F(p)&lt;/math&gt; be the set of integers x less than &lt;math&gt;b^p&lt;/math&gt; such that &lt;math&gt;S(x) &gt; b^p&lt;/math&gt;. Then for every integer in &lt;math&gt;F(p)&lt;/math&gt;, append the digit &lt;math&gt;(b-1)&lt;/math&gt; to the front of it to create a valid integer in &lt;math&gt;F(p+1)&lt;/math&gt;. Also, notice that &lt;math&gt;(b-1) \cdot b^p \le y &lt; b^{p+1}&lt;/math&gt;. Removing the digit &lt;math&gt;(b-1)&lt;/math&gt; from the front of y creates a number that is not in &lt;math&gt;F(p)&lt;/math&gt;. Hence, &lt;math&gt;F(p) \rightarrow F(p+1)&lt;/math&gt;, but there exists an element of &lt;math&gt;F(p+1)&lt;/math&gt; not corresponding with &lt;math&gt;F(p)&lt;/math&gt;, so &lt;math&gt;f(p+1) &gt; f(p)&lt;/math&gt;.<br /> <br /> Note that our lemma combined with the Pigeonhole Principle essentially proves the claim. Therefore, because there are infinitely many intervals containing an unrepresentable number, there are infinitely many unrepresentable numbers.</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2014_USAJMO_Problems/Problem_4&diff=63067 2014 USAJMO Problems/Problem 4 2014-08-14T20:33:30Z <p>Mathcool2009: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;b\geq 2&lt;/math&gt; be an integer, and let &lt;math&gt;s_b(n)&lt;/math&gt; denote the sum of the digits of &lt;math&gt;n&lt;/math&gt; when it is written in base &lt;math&gt;b&lt;/math&gt;. Show that there are infinitely many positive integers that cannot be represented in the form &lt;math&gt;n+s_b(n)&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is a positive integer.<br /> ==Solution==<br /> Define &lt;math&gt;S(n) = n + s_b(n)&lt;/math&gt;, and call a number ''unrepresentable'' if it cannot equal &lt;math&gt;S(n)&lt;/math&gt; for a positive integer &lt;math&gt;n&lt;/math&gt;.<br /> We claim that in the interval &lt;math&gt;(b^p, b^{p+1}]&lt;/math&gt; there exists an unrepresentable number, for every positive integer &lt;math&gt;p&lt;/math&gt;.<br /> <br /> If &lt;math&gt;b^{p+1}&lt;/math&gt; is unrepresentable, we're done. Otherwise, time for our lemma:<br /> <br /> Lemma: Define the function &lt;math&gt;f(p)&lt;/math&gt; to equal the number of integer x less than &lt;math&gt;b^p&lt;/math&gt; such that &lt;math&gt;S(x) \ge b^p&lt;/math&gt;. If &lt;math&gt;b^{p+1} = S(y)&lt;/math&gt; for some y, then &lt;math&gt;f(p+1) &gt; f(p)&lt;/math&gt;.<br /> <br /> Proof: Let &lt;math&gt;F(p)&lt;/math&gt; be the set of integers x less than &lt;math&gt;b^p&lt;/math&gt; such that &lt;math&gt;S(x) \ge b^p&lt;/math&gt;. Then for every integer in &lt;math&gt;F(p)&lt;/math&gt;, append the digit &lt;math&gt;(b-1)&lt;/math&gt; to the front of it to create a valid integer in &lt;math&gt;F(p+1)&lt;/math&gt;. Also, notice that &lt;math&gt;(b-1) \cdot b^p \le y &lt; b^{p+1}&lt;/math&gt;. Removing the digit &lt;math&gt;(b-1)&lt;/math&gt; from the front of y creates a number that is not in &lt;math&gt;F(p)&lt;/math&gt;. Hence, &lt;math&gt;F(p) \rightarrow F(p+1)&lt;/math&gt;, but there exists an element of &lt;math&gt;F(p+1)&lt;/math&gt; not corresponding with &lt;math&gt;F(p)&lt;/math&gt;, so &lt;math&gt;f(p+1) &gt; f(p)&lt;/math&gt;.<br /> <br /> Note that our lemma combined with the Pigeonhole Principle essentially proves the claim. Therefore, because there are infinitely many intervals containing an unrepresentable number, there are infinitely many unrepresentable numbers.</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2014_USAJMO_Problems/Problem_4&diff=63063 2014 USAJMO Problems/Problem 4 2014-08-14T18:29:24Z <p>Mathcool2009: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;b\geq 2&lt;/math&gt; be an integer, and let &lt;math&gt;s_b(n)&lt;/math&gt; denote the sum of the digits of &lt;math&gt;n&lt;/math&gt; when it is written in base &lt;math&gt;b&lt;/math&gt;. Show that there are infinitely many positive integers that cannot be represented in the form &lt;math&gt;n+s_b(n)&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is a positive integer.<br /> ==Solution==<br /> Define &lt;math&gt;S(n) = n + s_b(n)&lt;/math&gt;, and call a number ''unrepresentable'' if it cannot equal &lt;math&gt;S(n)&lt;/math&gt; for a positive integer &lt;math&gt;n&lt;/math&gt;.<br /> We claim that in the interval &lt;math&gt;[b^p, b^{p+1})&lt;/math&gt; there exists an unrepresentable number, for every positive integer &lt;math&gt;p&lt;/math&gt;.<br /> <br /> If &lt;math&gt;b^{p+1}&lt;/math&gt; is unrepresentable, we're done. Otherwise, time for our lemma:<br /> <br /> Lemma: Define the function &lt;math&gt;f(p)&lt;/math&gt; to equal the number of integer x less than &lt;math&gt;b^p&lt;/math&gt; such that &lt;math&gt;S(x) &gt; b^p&lt;/math&gt;. If &lt;math&gt;b^{p+1} = S(y)&lt;/math&gt; for some y, then &lt;math&gt;f(p+1) &gt; f(p)&lt;/math&gt;.<br /> <br /> Proof: Let &lt;math&gt;F(p)&lt;/math&gt; be the set of integers x less than &lt;math&gt;b^p&lt;/math&gt; such that &lt;math&gt;S(x) \ge b^p&lt;/math&gt;. Then for every integer in &lt;math&gt;F(p)&lt;/math&gt;, append the digit &lt;math&gt;(b-1)&lt;/math&gt; to the front of it to create a valid integer in &lt;math&gt;F(p+1)&lt;/math&gt;. Also, notice that &lt;math&gt;(b-1) \cdot b^p \le y &lt; b^{p+1}&lt;/math&gt;. Removing the digit &lt;math&gt;(b-1)&lt;/math&gt; from the front of y creates a number that is not in &lt;math&gt;F(p)&lt;/math&gt;. Hence, &lt;math&gt;F(p) \rightarrow F(p+1)&lt;/math&gt;, but there exists an element of &lt;math&gt;F(p+1)&lt;/math&gt; not corresponding with &lt;math&gt;F(p)&lt;/math&gt;, so &lt;math&gt;f(p+1) &gt; f(p)&lt;/math&gt;.<br /> <br /> Note that our lemma combined with the Pigeonhole Principle essentially proves the claim. Therefore, because there are infinitely many intervals containing an unrepresentable number, there are infinitely many unrepresentable numbers.</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2014_USAJMO_Problems/Problem_4&diff=63062 2014 USAJMO Problems/Problem 4 2014-08-14T18:28:39Z <p>Mathcool2009: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;b\geq 2&lt;/math&gt; be an integer, and let &lt;math&gt;s_b(n)&lt;/math&gt; denote the sum of the digits of &lt;math&gt;n&lt;/math&gt; when it is written in base &lt;math&gt;b&lt;/math&gt;. Show that there are infinitely many positive integers that cannot be represented in the form &lt;math&gt;n+s_b(n)&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is a positive integer.<br /> ==Solution==<br /> Define &lt;math&gt;S(n) = n + s_b(n)&lt;/math&gt;, and call a number ''unrepresentable'' if it cannot equal &lt;math&gt;S(n)&lt;/math&gt; for a positive integer &lt;math&gt;n&lt;/math&gt;.<br /> We claim that in the interval &lt;math&gt;(b^p, b^{p+1}]&lt;/math&gt; there exists an unrepresentable number, for every positive integer &lt;math&gt;p&lt;/math&gt;.<br /> <br /> If &lt;math&gt;b^{p+1}&lt;/math&gt; is unrepresentable, we're done. Otherwise, time for our lemma:<br /> <br /> Lemma: Define the function &lt;math&gt;f(p)&lt;/math&gt; to equal the number of integer x less than &lt;math&gt;b^p&lt;/math&gt; such that &lt;math&gt;S(x) &gt; b^p&lt;/math&gt;. If &lt;math&gt;b^{p+1} = S(y)&lt;/math&gt; for some y, then &lt;math&gt;f(p+1) &gt; f(p)&lt;/math&gt;.<br /> <br /> Proof: Let &lt;math&gt;F(p)&lt;/math&gt; be the set of integers x less than &lt;math&gt;b^p&lt;/math&gt; such that &lt;math&gt;S(x) \ge b^p&lt;/math&gt;. Then for every integer in &lt;math&gt;F(p)&lt;/math&gt;, append the digit &lt;math&gt;(b-1)&lt;/math&gt; to the front of it to create a valid integer in &lt;math&gt;F(p+1)&lt;/math&gt;. Also, notice that &lt;math&gt;(b-1) \cdot b^p \le y &lt; b^{p+1}&lt;/math&gt;. Removing the digit &lt;math&gt;(b-1)&lt;/math&gt; from the front of y creates a number that is not in &lt;math&gt;F(p)&lt;/math&gt;. Hence, &lt;math&gt;F(p) \rightarrow F(p+1)&lt;/math&gt;, but there exists an element of &lt;math&gt;F(p+1)&lt;/math&gt; not corresponding with &lt;math&gt;F(p)&lt;/math&gt;, so &lt;math&gt;f(p+1) &gt; f(p)&lt;/math&gt;.<br /> <br /> Note that our lemma combined with the Pigeonhole Principle essentially proves the claim. Therefore, because there are infinitely many intervals containing an unrepresentable number, there are infinitely many unrepresentable numbers.</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2013_USAJMO_Problems/Problem_1&diff=62695 2013 USAJMO Problems/Problem 1 2014-07-27T01:41:00Z <p>Mathcool2009: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Are there integers &lt;math&gt; a &lt;/math&gt; and &lt;math&gt; b &lt;/math&gt; such that &lt;math&gt; a^5b+3 &lt;/math&gt; and &lt;math&gt; ab^5+3 &lt;/math&gt; are both perfect cubes of integers?<br /> <br /> ==Solution==<br /> <br /> No, such integers do not exist. This shall be proven by contradiction, by showing that if &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube then &lt;math&gt;ab^5+3&lt;/math&gt; cannot be.<br /> <br /> Remark that perfect cubes are always congruent to &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, or &lt;math&gt;-1&lt;/math&gt; modulo &lt;math&gt;9&lt;/math&gt;. Therefore, if &lt;math&gt;a^5b+3\equiv 0,1,\text{ or} -1\pmod{9}&lt;/math&gt;, then &lt;math&gt;a^5b\equiv 5,6,\text{ or }7\pmod{9}&lt;/math&gt;.<br /> <br /> If &lt;math&gt;a^5b\equiv 6\pmod 9&lt;/math&gt;, then note that &lt;math&gt;3|b&lt;/math&gt;. (This is because if &lt;math&gt;3|a&lt;/math&gt; then &lt;math&gt;a^5b\equiv 0\pmod 9&lt;/math&gt;.) Therefore &lt;math&gt;ab^5\equiv 0\pmod 9&lt;/math&gt; and &lt;math&gt;ab^5+3\equiv 3\pmod 9&lt;/math&gt;, contradiction.<br /> <br /> Otherwise, either &lt;math&gt;a^5b\equiv 5\pmod 9&lt;/math&gt; or &lt;math&gt;a^5b\equiv 7\pmod 9&lt;/math&gt;. Note that since &lt;math&gt;a^6b^6&lt;/math&gt; is a perfect sixth power, and since neither &lt;math&gt;a&lt;/math&gt; nor &lt;math&gt;b&lt;/math&gt; contains a factor of &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;a^6b^6\equiv 1\pmod 9&lt;/math&gt;. If &lt;math&gt;a^5b\equiv 5\pmod 9&lt;/math&gt;, then &lt;cmath&gt;a^6b^6\equiv (a^5b)(ab^5)\equiv 5ab^5\equiv 1\pmod 9\implies ab^5\equiv 2\pmod 9.&lt;/cmath&gt; Similarly, if &lt;math&gt;a^5b\equiv 7\pmod 9&lt;/math&gt;, then &lt;cmath&gt;a^6b^6\equiv (a^5b)(ab^5)\equiv 7ab^5\equiv 1\pmod 9\implies ab^5\equiv 4\pmod 9.&lt;/cmath&gt; Therefore &lt;math&gt;ab^5+3\equiv 5,7\pmod 9&lt;/math&gt;, contradiction.<br /> <br /> Therefore no such integers exist.<br /> <br /> ==Solution 2==<br /> We shall prove that such integers do not exist via contradiction.<br /> Suppose that &lt;math&gt;a^5b + 3 = x^3&lt;/math&gt; and &lt;math&gt;ab^5 + 3 = y^3&lt;/math&gt; for integers x and y. Rearranging terms gives &lt;math&gt;a^5b = x^3 - 3&lt;/math&gt; and &lt;math&gt;ab^5 = y^3 - 3&lt;/math&gt;. Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a = &lt;math&gt;(x^3 - 3)^\frac{5}{24} (y^3 - 3)^\frac{-1}{24}&lt;/math&gt; and b = &lt;math&gt;(y^3 - 3)^\frac{5}{24} (x^3 - 3)^\frac{-1}{24}&lt;/math&gt;. Consider a prime p in the prime factorization of &lt;math&gt;x^3 - 3&lt;/math&gt; and &lt;math&gt;y^3 - 3&lt;/math&gt;. If it has power &lt;math&gt;r_1&lt;/math&gt; in &lt;math&gt;x^3 - 3&lt;/math&gt; and power &lt;math&gt;r_2&lt;/math&gt; in &lt;math&gt;y^3 - 3&lt;/math&gt;, then &lt;math&gt;5r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; is a multiple of 24 and &lt;math&gt;5r_2&lt;/math&gt; - &lt;math&gt;r_1&lt;/math&gt; also is a multiple of 24. <br /> <br /> Adding and subtracting the divisions gives that &lt;math&gt;r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; divides 12. (actually, &lt;math&gt;r_1 - r_2&lt;/math&gt; is a multiple of 4, as you can verify if &lt;math&gt;\{^{5r_1 - r_2 = 24}_{5r_2 - r_1 = 48}&lt;/math&gt;. So the rest of the proof is invalid.) Because &lt;math&gt;5r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; also divides 12, &lt;math&gt;4r_1&lt;/math&gt; divides 12 and thus &lt;math&gt;r_1&lt;/math&gt; divides 3. Repeating this trick for all primes in &lt;math&gt;x^3 - 3&lt;/math&gt;, we see that &lt;math&gt;x^3 - 3&lt;/math&gt; is a perfect cube, say &lt;math&gt;q^3&lt;/math&gt;. Then &lt;math&gt;x^3 - q^3 = 3,&lt;/math&gt; and &lt;math&gt;(x-q)(x^2 + xq + q^2) = 3&lt;/math&gt;, so that &lt;math&gt;x - q = 1&lt;/math&gt; and &lt;math&gt;x^2 + xq + q^2 = 3&lt;/math&gt;. Clearly, this system of equations has no integer solutions for &lt;math&gt;x&lt;/math&gt; or &lt;math&gt;q&lt;/math&gt;, a contradiction, hence completing the proof.<br /> <br /> Therefore no such integers exist.<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_15&diff=62512 2001 AIME I Problems/Problem 15 2014-07-15T22:52:31Z <p>Mathcool2009: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> The numbers &lt;math&gt;1, 2, 3, 4, 5, 6, 7,&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; are randomly written on the faces of a regular [[octahedron]] so that each face contains a different number. The [[probability]] that no two consecutive numbers, where &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt; are considered to be consecutive, are written on faces that share an edge is &lt;math&gt;m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n.&lt;/math&gt;<br /> <br /> == Solution 1==<br /> Choose one face of the octahedron randomly and label it with &lt;math&gt;1&lt;/math&gt;. There are three faces adjacent to this one, which we will call A-faces. There are three faces adjacent to two of the A-faces, which we will call B-faces, and one face adjacent to the three B-faces, which we will call the C-face.<br /> <br /> Clearly, the labels for the A-faces must come from the set &lt;math&gt;\{3,4,5,6,7\}&lt;/math&gt;, since these faces are all adjacent to &lt;math&gt;1&lt;/math&gt;. There are thus &lt;math&gt;5 \cdot 4 \cdot 3 = 60&lt;/math&gt; ways to assign the labels for the A-faces. <br /> <br /> The labels for the B-faces and C-face are the two remaining numbers from the above set, plus &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;. The number on the C-face must not be consecutive to any of the numbers on the B-faces. <br /> <br /> From here it is easiest to brute force the &lt;math&gt;10&lt;/math&gt; possibilities for the &lt;math&gt;4&lt;/math&gt; numbers on the B and C faces:<br /> <br /> *2348 (2678): 8(2) is the only one not adjacent to any of the others, so it goes on the C-face. 4(6) has only one B-face it can go to, while 2 and 3 (7 and 8) can be assigned randomly to the last two. 2 possibilities here.<br /> <br /> *2358 (2578): 5 cannot go on any of the B-faces, so it must be on the C-face. 3 and 8 (2 and 7) have only one allowable B-face, so just 1 possibility here.<br /> <br /> *2368 (2478): 6(4) cannot go on any of the B-faces, so it must be on the C-face. 3 and 8 (2 and 7) have only one allowable B-face, so 1 possibility here.<br /> <br /> *2458 (2568): All of the numbers have only one B-face they could go to. 2 and 4 (6 and 8) can go on the same, so one must go to the C-face. Only 2(8) is not consecutive with any of the others, so it goes on the C-face. 1 possibility.<br /> <br /> *2378: None of the numbers can go on the C-face because they will be consecutive with one of the B-face numbers. So this possibility is impossible. <br /> <br /> *2468: Both 4 and 6 cannot go on any B-face. They cannot both go on the C-face, so this possibility is impossible.<br /> <br /> There is a total of &lt;math&gt;10&lt;/math&gt; possibilities. There are &lt;math&gt;3!=6&lt;/math&gt; permutations (more like &quot;rotations&quot;) of each, so &lt;math&gt;60&lt;/math&gt; acceptable ways to fill in the rest of the octahedron given the &lt;math&gt;1&lt;/math&gt;. There are &lt;math&gt;7!=5040&lt;/math&gt; ways to randomly fill in the rest of the octahedron. So the probability is &lt;math&gt;\frac {60}{5040} = \frac {1}{84}&lt;/math&gt;. The answer is &lt;math&gt;\boxed{085}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> Consider the cube formed from the face centers of the regular octahedron. Color the vertices in a checker board fashion. We seek the number of circuits traversing the cube entirely composed of diagonals. Notice for any vertex, it can be linked to at most one different-colored vertex, i.e. its opposite vertex. Thus, there are only two possible types of circuits (&lt;math&gt;B&lt;/math&gt; for black and &lt;math&gt;W&lt;/math&gt; for white).<br /> <br /> Type I: &lt;math&gt;BB-WWWW-BB&lt;/math&gt;. There are &lt;math&gt;4!&lt;/math&gt; ways to arrange the black vertices and consequently two of the white vertices and &lt;math&gt;2!&lt;/math&gt; ways to arrange the other two white vertices. Since the template has a period of &lt;math&gt;8&lt;/math&gt;, there are &lt;math&gt;4!\cdot 2!\cdot 8 = 384&lt;/math&gt; circuits of type I.<br /> <br /> Type II: &lt;math&gt;B-WW-BB-WW-B&lt;/math&gt;. There are &lt;math&gt;4!&lt;/math&gt; ways to arrange the black vertices and consequently the white vertices. Since the template has a period of &lt;math&gt;4&lt;/math&gt;, there are &lt;math&gt;4! \cdot 4 = 96&lt;/math&gt; circuits of type II.<br /> <br /> Thus, there are &lt;math&gt;384+96=480&lt;/math&gt; circuits satisfying the given condition, out of the &lt;math&gt;8!&lt;/math&gt; possible circuits. Therefore, the desired probability is &lt;math&gt;\frac{480}{8!} = \frac{1}{84}&lt;/math&gt;. The answer is &lt;math&gt;\boxed{085}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> As in the previous solution, consider the cube formed by taking each face of the octahedron as a vertex. Let one fixed vertex be A. Then each configuration (letting each vertex have a number value from 1-8) of A and the three vertices adjacent to A uniquely determine a configuration that satisfies the conditions, i.e. no two vertices have consecutive numbers. Thus, the number of desired configurations is equivalent to the number of ways of choosing the values of A and its three adjacent vertices. <br /> <br /> The value of A can be chosen in 8 ways, and the 3 vertices adjacent to A can be chosen in &lt;math&gt;5\cdot4\cdot3=60&lt;/math&gt; ways, since they aren't adjacent to each other, but they can't, after all, be consecutive values to A. For example, if A=1, then the next vertex can't be 1,2 or 8, so there are 5 choices. However, the next vertex also adjacent to A can be chosen in 4 ways; it can't be equal to 1,2,8, or the value of the previously chosen vertex. With the same reasoning, the last such vertex has 3 possible choices.<br /> <br /> The total number of ways to choose the values of the vertices of the cube independently is 8!, so our probability is thus &lt;math&gt;\frac{8\cdot60}{8!}=\frac{8\cdot5\cdot4\cdot3}{8!}=\frac{1}{84}&lt;/math&gt;, from which the answer is &lt;math&gt;\boxed{085}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_15&diff=62511 2001 AIME I Problems/Problem 15 2014-07-15T22:46:56Z <p>Mathcool2009: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> The numbers &lt;math&gt;1, 2, 3, 4, 5, 6, 7,&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; are randomly written on the faces of a regular [[octahedron]] so that each face contains a different number. The [[probability]] that no two consecutive numbers, where &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt; are considered to be consecutive, are written on faces that share an edge is &lt;math&gt;m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n.&lt;/math&gt;<br /> <br /> == Solution 1==<br /> Choose one face of the octahedron randomly and label it with &lt;math&gt;1&lt;/math&gt;. There are three faces adjacent to this one, which we will call A-faces. There are three faces adjacent to two of the A-faces, which we will call B-faces, and one face adjacent to the three B-faces, which we will call the C-face.<br /> <br /> Clearly, the labels for the A-faces must come from the set &lt;math&gt;\{3,4,5,6,7\}&lt;/math&gt;, since these faces are all adjacent to &lt;math&gt;1&lt;/math&gt;. There are thus &lt;math&gt;5 \cdot 4 \cdot 3 = 60&lt;/math&gt; ways to assign the labels for the A-faces. <br /> <br /> The labels for the B-faces and C-face are the two remaining numbers from the above set, plus &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;. The number on the C-face must not be consecutive to any of the numbers on the B-faces. <br /> <br /> From here it is easiest to brute force the &lt;math&gt;10&lt;/math&gt; possibilities for the &lt;math&gt;4&lt;/math&gt; numbers on the B and C faces:<br /> <br /> *2348 (2678): 8(2) is the only one not adjacent to any of the others, so it goes on the C-face. 4(6) has only one B-face it can go to, while 2 and 3 (7 and 8) can be assigned randomly to the last two. 2 possibilities here.<br /> <br /> *2358 (2578): 5 cannot go on any of the B-faces, so it must be on the C-face. 3 and 8 (2 and 7) have only one allowable B-face, so just 1 possibility here.<br /> <br /> *2368 (2478): 6(4) cannot go on any of the B-faces, so it must be on the C-face. 3 and 8 (2 and 7) have only one allowable B-face, so 1 possibility here.<br /> <br /> *2458 (2568): All of the numbers have only one B-face they could go to. 2 and 4 (6 and 8) can go on the same, so one must go to the C-face. Only 2(8) is not consecutive with any of the others, so it goes on the C-face. 1 possibility.<br /> <br /> *2378: None of the numbers can go on the C-face because they will be consecutive with one of the B-face numbers. So this possibility is impossible. <br /> <br /> *2468: Both 4 and 6 cannot go on any B-face. They cannot both go on the C-face, so this possibility is impossible.<br /> <br /> There is a total of &lt;math&gt;10&lt;/math&gt; possibilities. There are &lt;math&gt;3!=6&lt;/math&gt; permutations of each, so &lt;math&gt;60&lt;/math&gt; acceptable ways to fill in the rest of the octahedron given the &lt;math&gt;1&lt;/math&gt;. There are &lt;math&gt;7!=5040&lt;/math&gt; ways to randomly fill in the rest of the octahedron. So the probability is &lt;math&gt;\frac {60}{5040} = \frac {1}{84}&lt;/math&gt;. The answer is &lt;math&gt;\boxed{085}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> Consider the cube formed from the face centers of the regular octahedron. Color the vertices in a checker board fashion. We seek the number of circuits traversing the cube entirely composed of diagonals. Notice for any vertex, it can be linked to at most one different-colored vertex, i.e. its opposite vertex. Thus, there are only two possible types of circuits (&lt;math&gt;B&lt;/math&gt; for black and &lt;math&gt;W&lt;/math&gt; for white).<br /> <br /> Type I: &lt;math&gt;BB-WWWW-BB&lt;/math&gt;. There are &lt;math&gt;4!&lt;/math&gt; ways to arrange the black vertices and consequently two of the white vertices and &lt;math&gt;2!&lt;/math&gt; ways to arrange the other two white vertices. Since the template has a period of &lt;math&gt;8&lt;/math&gt;, there are &lt;math&gt;4!\cdot 2!\cdot 8 = 384&lt;/math&gt; circuits of type I.<br /> <br /> Type II: &lt;math&gt;B-WW-BB-WW-B&lt;/math&gt;. There are &lt;math&gt;4!&lt;/math&gt; ways to arrange the black vertices and consequently the white vertices. Since the template has a period of &lt;math&gt;4&lt;/math&gt;, there are &lt;math&gt;4! \cdot 4 = 96&lt;/math&gt; circuits of type II.<br /> <br /> Thus, there are &lt;math&gt;384+96=480&lt;/math&gt; circuits satisfying the given condition, out of the &lt;math&gt;8!&lt;/math&gt; possible circuits. Therefore, the desired probability is &lt;math&gt;\frac{480}{8!} = \frac{1}{84}&lt;/math&gt;. The answer is &lt;math&gt;\boxed{085}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> As in the previous solution, consider the cube formed by taking each face of the octahedron as a vertex. Let one fixed vertex be A. Then each configuration (letting each vertex have a number value from 1-8) of A and the three vertices adjacent to A uniquely determine a configuration that satisfies the conditions, i.e. no two vertices have consecutive numbers. Thus, the number of desired configurations is equivalent to the number of ways of choosing the values of A and its three adjacent vertices. <br /> <br /> The value of A can be chosen in 8 ways, and the 3 vertices adjacent to A can be chosen in &lt;math&gt;5\cdot4\cdot3=60&lt;/math&gt; ways, since they aren't adjacent to each other, but they can't, after all, be consecutive values to A. For example, if A=1, then the next vertex can't be 1,2 or 8, so there are 5 choices. However, the next vertex also adjacent to A can be chosen in 4 ways; it can't be equal to 1,2,8, or the value of the previously chosen vertex. With the same reasoning, the last such vertex has 3 possible choices.<br /> <br /> The total number of ways to choose the values of the vertices of the cube independently is 8!, so our probability is thus &lt;math&gt;\frac{8\cdot60}{8!}=\frac{8\cdot5\cdot4\cdot3}{8!}=\frac{1}{84}&lt;/math&gt;, from which the answer is &lt;math&gt;\boxed{085}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_5&diff=62510 2001 AIME I Problems/Problem 5 2014-07-15T22:32:44Z <p>Mathcool2009: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> An [[equilateral triangle]] is inscribed in the [[ellipse]] whose equation is &lt;math&gt;x^2+4y^2=4&lt;/math&gt;. One vertex of the triangle is &lt;math&gt;(0,1)&lt;/math&gt;, one altitude is contained in the y-axis, and the length of each side is &lt;math&gt;\sqrt{\frac mn}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solution ==<br /> &lt;center&gt;&lt;asy&gt;<br /> pointpen = black; pathpen = black + linewidth(0.7);<br /> path e = xscale(2)*unitcircle; real x = -8/13*3^.5;<br /> D((-3,0)--(3,0)); D((0,-2)--(0,2)); /* axes */<br /> D(e); D(D((0,1))--(x,x*3^.5+1)--(-x,x*3^.5+1)--cycle);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> === Solution 1 ===<br /> Denote the vertices of the triangle &lt;math&gt;A,B,&lt;/math&gt; and &lt;math&gt;C,&lt;/math&gt; where &lt;math&gt;B&lt;/math&gt; is in [[quadrant]] 4 and &lt;math&gt;C&lt;/math&gt; is in quadrant &lt;math&gt;3.&lt;/math&gt;<br /> <br /> Note that the slope of &lt;math&gt;\overline{AC}&lt;/math&gt; is &lt;math&gt;\tan 60^\circ = \sqrt {3}.&lt;/math&gt; Hence, the equation of the line containing &lt;math&gt;\overline{AC}&lt;/math&gt; is<br /> &lt;cmath&gt;<br /> y = x\sqrt {3} + 1.<br /> &lt;/cmath&gt;<br /> This will intersect the ellipse when<br /> &lt;cmath&gt;<br /> \begin{eqnarray*}4 = x^{2} + 4y^{2} &amp; = &amp; x^{2} + 4(x\sqrt {3} + 1)^{2} \\<br /> &amp; = &amp; x^{2} + 4(3x^{2} + 2x\sqrt {3} + 1) \implies x = \frac { - 8\sqrt {3}}{13}. \end{eqnarray*}<br /> &lt;/cmath&gt;<br /> Since the triangle is symmetric with respect to the y-axis, the coordinates of &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are now &lt;math&gt;\left(\frac {8\sqrt {3}}{13},y_{0}\right)&lt;/math&gt; and &lt;math&gt;\left(\frac { - 8\sqrt {3}}{13},y_{0}\right),&lt;/math&gt; respectively, for some value of &lt;math&gt;y_{0}.&lt;/math&gt;<br /> <br /> Since we're going to use the distance formula, the value of &lt;math&gt;y_{0}&lt;/math&gt; is irrelevant. Our answer is<br /> &lt;cmath&gt;<br /> BC = \sqrt {2\left(\frac {8\sqrt {3}}{13}\right)^{2}} = \sqrt {\frac {768}{169}}\implies m + n = \boxed{937}.<br /> &lt;/cmath&gt;<br /> <br /> === Solution 2 ===<br /> Solving for &lt;math&gt;y&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt; gives &lt;math&gt;y=\sqrt{4-x^2}/2&lt;/math&gt;, so the two other points of the triangle are &lt;math&gt;(x,\sqrt{4-x^2}/2)&lt;/math&gt; and &lt;math&gt;(-x,\sqrt{4-x^2}/2)&lt;/math&gt;, which are a distance of &lt;math&gt;2x&lt;/math&gt; apart. Thus &lt;math&gt;2x&lt;/math&gt; equals the distance between &lt;math&gt;(x,\sqrt{4-x^2}/2)&lt;/math&gt; and &lt;math&gt;(0,1)&lt;/math&gt;, so by the distance formula we have <br /> <br /> &lt;cmath&gt;2x=\sqrt{x^2+(1-\sqrt{4-x^2}/2)^2}.&lt;/cmath&gt; <br /> <br /> Squaring both sides and simplifying through algebra yields &lt;math&gt;x^2=192/169&lt;/math&gt;, so &lt;math&gt;2x=\sqrt{768/169}&lt;/math&gt; and the answer is &lt;math&gt;\boxed{937}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Since the altitude goes along the &lt;math&gt;y&lt;/math&gt; axis, this means that the base is a horizontal line, which means that the endpoints of the base are &lt;math&gt;(x,y)&lt;/math&gt; and &lt;math&gt;(-x,y)&lt;/math&gt;, and WLOG, we can say that &lt;math&gt;x&lt;/math&gt; is positive.<br /> <br /> Now, since all sides of an equilateral triangle are the same, we can do this (distance from one of the endpoints of the base to the vertex and the length of the base):<br /> <br /> &lt;math&gt;\sqrt{x^2 + (y-1)^2} = 2x&lt;/math&gt;<br /> <br /> Square both sides,<br /> <br /> &lt;math&gt;x^2 + (y-1)^2 = 4x^2&lt;/math&gt;<br /> &lt;math&gt;(y-1)^2 = 3x^2&lt;/math&gt;<br /> <br /> Now, with the equation of the ellipse:<br /> &lt;math&gt;x^2 + 4y^2 = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2 = 4-4y^2&lt;/math&gt;<br /> <br /> &lt;math&gt;3x^2 = 12-12y^2&lt;/math&gt;<br /> <br /> Substituting, <br /> <br /> &lt;math&gt;12-12y^2 = y^2 - 2y +1&lt;/math&gt;<br /> <br /> Moving stuff around and solving:<br /> <br /> &lt;math&gt;y = \frac{-11}{3}, 1&lt;/math&gt;<br /> <br /> The second is found to be extraneous, so, when we go back and figure out &lt;math&gt;x&lt;/math&gt; and then &lt;math&gt;2x&lt;/math&gt; (which is the side length), we find it to be:<br /> <br /> &lt;math&gt;\sqrt{\frac{768}{169}}&lt;/math&gt; <br /> <br /> and so we get the desired answer of &lt;math&gt;\boxed{937}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_6&diff=62509 2001 AIME I Problems/Problem 6 2014-07-15T22:28:20Z <p>Mathcool2009: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A fair die is rolled four times. The [[probability]] that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form &lt;math&gt;\frac{m}{n}&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are [[relatively prime]] [[positive]] [[integer]]s. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solutions ==<br /> === Solution 1 ===<br /> Recast the problem entirely as a block-walking problem. Call the respective dice &lt;math&gt;a, b, c, d&lt;/math&gt;. In the diagram below, the lowest &lt;math&gt;y&lt;/math&gt;-coordinate at each of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; corresponds to the value of the roll.<br /> <br /> [[File:AIME01IN6.png]]<br /> <br /> The red path corresponds to the sequence of rolls &lt;math&gt;2, 3, 5, 5&lt;/math&gt;. This establishes a one-to-one correspondence between valid dice roll sequences and block walking paths. <br /> <br /> The solution to this problem is therefore &lt;math&gt;\dfrac{\binom{9}{4}}{6^4} = \boxed{\dfrac{7}{72}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> If we take any [[combination]] of four numbers, there is only one way to order them in a non-decreasing order. It suffices now to find the number of combinations for four numbers from &lt;math&gt;\{1,2,3,4,5,6\}&lt;/math&gt;. We can visualize this as taking the four dice and splitting them into 6 slots (each slot representing one of {1,2,3,4,5,6}), or dividing them amongst 5 separators. Thus, there are &lt;math&gt;{9\choose4} = 126&lt;/math&gt; outcomes of four dice. The solution is therefore &lt;math&gt;\frac{126}{6^4} = \frac{7}{72}&lt;/math&gt;, and &lt;math&gt;7 + 72 = \boxed{079}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Call the dice rolls &lt;math&gt;a, b, c, d&lt;/math&gt;. The difference between the &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; distinguishes the number of possible rolls there are. <br /> <br /> *If &lt;math&gt;a - d = 0&lt;/math&gt;, then the values of &lt;math&gt;b,\ c&lt;/math&gt; are set, and so there are &lt;math&gt;6&lt;/math&gt; values for &lt;math&gt;a,\ d&lt;/math&gt;.<br /> *If &lt;math&gt;a - d = 1&lt;/math&gt;, then there are &lt;math&gt;{3\choose2} = 3&lt;/math&gt; ways to arrange for values of &lt;math&gt;b,\ c&lt;/math&gt;, but only &lt;math&gt;5&lt;/math&gt; values for &lt;math&gt;a,\ d&lt;/math&gt;.<br /> *If &lt;math&gt;a - d = 2&lt;/math&gt;, then there are &lt;math&gt;{4\choose2} = 6&lt;/math&gt; ways to arrange &lt;math&gt;b, c&lt;/math&gt;, and there are only &lt;math&gt;6 - 2 = 4&lt;/math&gt; values for &lt;math&gt;a, d&lt;/math&gt;.<br /> <br /> Continuing, we see that the sum is equal to &lt;math&gt;\sum_{i = 0}^{5}{i+2\choose2}(6 - i) = 6 + 15 + 24 + 30 + 30 + 21 = 126&lt;/math&gt;. The requested probability is &lt;math&gt;\frac{126}{6^4} = \frac{7}{72}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> The dice rolls can be in the form <br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;<br /> ABCD&lt;br /&gt;<br /> AABC&lt;br /&gt;<br /> AABB&lt;br /&gt;<br /> AAAB&lt;br /&gt;<br /> AAAA<br /> &lt;/div&gt;<br /> <br /> where A, B, C, D are some possible value of the dice rolls. (These forms are not keeping track of whether or not the dice are in ascending order, just the possible outcomes.) <br /> <br /> #Now, for the first case, there are &lt;math&gt;{6\choose4} = 15&lt;/math&gt; ways for this. We do not have to consider the order because the combination counts only one of the permutations; we can say that it counts the correct (ascending order) permutation. <br /> #Second case: &lt;math&gt;{6\choose3} = 20&lt;/math&gt; ways to pick 3 numbers, &lt;math&gt;{3\choose1}&lt;/math&gt; ways to pick 1 of those 3 to duplicate. A total of 60 for this case. <br /> #Third case: &lt;math&gt;{6\choose2}=15&lt;/math&gt; ways to pick 2 numbers. We will duplicate both, so nothing else in this case matters. <br /> #Fourth case: &lt;math&gt;{6\choose2} = 15&lt;/math&gt; ways to pick 2 numbers. We pick one to duplicate with &lt;math&gt;{2\choose1} = 2&lt;/math&gt;, so there are a total of 30 in this case. <br /> #Fifth case: &lt;math&gt;{6\choose1} = 6&lt;/math&gt;; all get duplicated so nothing else matters. <br /> <br /> There are a total of &lt;math&gt;6^4&lt;/math&gt; possible dice rolls. <br /> <br /> Thus, <br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;\frac{m}{n} = \frac{15 + 60 + 15 + 30 + 6}{6^4} = \frac{126}{6^4} = \frac{7}{72}&lt;/math&gt;&lt;/div&gt;<br /> <br /> === Solution 5 ===<br /> Consider the number of possible dice roll combinations which work after &lt;math&gt;1&lt;/math&gt; roll, after &lt;math&gt;2&lt;/math&gt; rolls, and so on. There is 6 possible rolls for the first dice. If the number rolled is a 1, then there are 6 further values that are possible for the second dice; if the number rolled is a 2, then there are 5 further values that are possible for the second dice, and so on. <br /> <br /> Suppose we generalize this as a function, say &lt;math&gt;f(l,r)&lt;/math&gt; return the number of possible combinations after &lt;math&gt;l&lt;/math&gt; rolls and &lt;math&gt;r&lt;/math&gt; being the beginning value of the first roll. It becomes clear that from above, &lt;math&gt;f(1,r) = 1&lt;/math&gt;; every value of &lt;math&gt;l&lt;/math&gt; after that is equal to the sum of the number of combinations of &lt;math&gt;l - 1&lt;/math&gt; rolls that have a starting value of at least &lt;math&gt;r&lt;/math&gt;. If we slowly count through and add up all the possible combinations we get &lt;math&gt;\frac{7}{72}&lt;/math&gt; possibilities.<br /> <br /> === Solution 6 ===<br /> In a manner similar to the above solution, instead consider breaking it down into two sets of two dice rolls. The first [[subset]] must have a maximum value which is &lt;math&gt;\le&lt;/math&gt; the minimum value of the second subset. <br /> <br /> *If the first subset ends in a 1, there is &lt;math&gt;1&lt;/math&gt; such subset and there are &lt;math&gt;6 + 5 + 4 + 3 + 2 + 1 = \frac{6}{2}(6 + 1) = 21&lt;/math&gt; ways of making the second subset.<br /> *If the first subset ends in a 2, there is &lt;math&gt;2&lt;/math&gt; such subsets and there are &lt;math&gt;5 + \ldots + 1 = \frac{5}{2}(5 + 1) = 15&lt;/math&gt; ways of making the second subset.<br /> <br /> Thus, the number of combinations is &lt;math&gt;\sum_{i=1}^6 i \cdot \left(\frac{(7 - i)(8 - i)}{2}\right) = 21 + 30 + 30 + 24 + 15 + 6 = 126&lt;/math&gt;, and the probability again is &lt;math&gt;\frac7{72}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_6&diff=62508 2001 AIME I Problems/Problem 6 2014-07-15T22:27:44Z <p>Mathcool2009: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A fair die is rolled four times. The [[probability]] that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form &lt;math&gt;\frac{m}{n}&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are [[relatively prime]] [[positive]] [[integer]]s. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solutions ==<br /> === Solution 1 ===<br /> Recast the problem entirely as a block-walking problem. Call the respective dice &lt;math&gt;a, b, c, d&lt;/math&gt;. In the diagram below, the lowest &lt;math&gt;y&lt;/math&gt;-coordinate at each of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; corresponds to the value of the roll.<br /> <br /> [[File:AIME01IN6.png]]<br /> <br /> The red path corresponds to the sequence of rolls &lt;math&gt;2, 3, 5, 5&lt;/math&gt;. This establishes a one-to-one correspondence between valid dice roll sequences and block walking paths. <br /> <br /> The solution to this problem is therefore &lt;math&gt;\dfrac{\binom{9}{4}}{6^4} = \boxed{\dfrac{7}{72}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> If we take any [[combination]] of four numbers, there is only one way to order them in a non-decreasing order. It suffices now to find the number of combinations for four numbers from &lt;math&gt;\{1,2,3,4,5,6\}&lt;/math&gt;. We can visualize this as taking the four dice and splitting them into 6 slots (each slot representing one possible roll), or dividing them amongst 5 separators. Thus, there are &lt;math&gt;{9\choose4} = 126&lt;/math&gt; outcomes of four dice. The solution is therefore &lt;math&gt;\frac{126}{6^4} = \frac{7}{72}&lt;/math&gt;, and &lt;math&gt;7 + 72 = \boxed{079}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Call the dice rolls &lt;math&gt;a, b, c, d&lt;/math&gt;. The difference between the &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; distinguishes the number of possible rolls there are. <br /> <br /> *If &lt;math&gt;a - d = 0&lt;/math&gt;, then the values of &lt;math&gt;b,\ c&lt;/math&gt; are set, and so there are &lt;math&gt;6&lt;/math&gt; values for &lt;math&gt;a,\ d&lt;/math&gt;.<br /> *If &lt;math&gt;a - d = 1&lt;/math&gt;, then there are &lt;math&gt;{3\choose2} = 3&lt;/math&gt; ways to arrange for values of &lt;math&gt;b,\ c&lt;/math&gt;, but only &lt;math&gt;5&lt;/math&gt; values for &lt;math&gt;a,\ d&lt;/math&gt;.<br /> *If &lt;math&gt;a - d = 2&lt;/math&gt;, then there are &lt;math&gt;{4\choose2} = 6&lt;/math&gt; ways to arrange &lt;math&gt;b, c&lt;/math&gt;, and there are only &lt;math&gt;6 - 2 = 4&lt;/math&gt; values for &lt;math&gt;a, d&lt;/math&gt;.<br /> <br /> Continuing, we see that the sum is equal to &lt;math&gt;\sum_{i = 0}^{5}{i+2\choose2}(6 - i) = 6 + 15 + 24 + 30 + 30 + 21 = 126&lt;/math&gt;. The requested probability is &lt;math&gt;\frac{126}{6^4} = \frac{7}{72}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> The dice rolls can be in the form <br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;<br /> ABCD&lt;br /&gt;<br /> AABC&lt;br /&gt;<br /> AABB&lt;br /&gt;<br /> AAAB&lt;br /&gt;<br /> AAAA<br /> &lt;/div&gt;<br /> <br /> where A, B, C, D are some possible value of the dice rolls. (These forms are not keeping track of whether or not the dice are in ascending order, just the possible outcomes.) <br /> <br /> #Now, for the first case, there are &lt;math&gt;{6\choose4} = 15&lt;/math&gt; ways for this. We do not have to consider the order because the combination counts only one of the permutations; we can say that it counts the correct (ascending order) permutation. <br /> #Second case: &lt;math&gt;{6\choose3} = 20&lt;/math&gt; ways to pick 3 numbers, &lt;math&gt;{3\choose1}&lt;/math&gt; ways to pick 1 of those 3 to duplicate. A total of 60 for this case. <br /> #Third case: &lt;math&gt;{6\choose2}=15&lt;/math&gt; ways to pick 2 numbers. We will duplicate both, so nothing else in this case matters. <br /> #Fourth case: &lt;math&gt;{6\choose2} = 15&lt;/math&gt; ways to pick 2 numbers. We pick one to duplicate with &lt;math&gt;{2\choose1} = 2&lt;/math&gt;, so there are a total of 30 in this case. <br /> #Fifth case: &lt;math&gt;{6\choose1} = 6&lt;/math&gt;; all get duplicated so nothing else matters. <br /> <br /> There are a total of &lt;math&gt;6^4&lt;/math&gt; possible dice rolls. <br /> <br /> Thus, <br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;\frac{m}{n} = \frac{15 + 60 + 15 + 30 + 6}{6^4} = \frac{126}{6^4} = \frac{7}{72}&lt;/math&gt;&lt;/div&gt;<br /> <br /> === Solution 5 ===<br /> Consider the number of possible dice roll combinations which work after &lt;math&gt;1&lt;/math&gt; roll, after &lt;math&gt;2&lt;/math&gt; rolls, and so on. There is 6 possible rolls for the first dice. If the number rolled is a 1, then there are 6 further values that are possible for the second dice; if the number rolled is a 2, then there are 5 further values that are possible for the second dice, and so on. <br /> <br /> Suppose we generalize this as a function, say &lt;math&gt;f(l,r)&lt;/math&gt; return the number of possible combinations after &lt;math&gt;l&lt;/math&gt; rolls and &lt;math&gt;r&lt;/math&gt; being the beginning value of the first roll. It becomes clear that from above, &lt;math&gt;f(1,r) = 1&lt;/math&gt;; every value of &lt;math&gt;l&lt;/math&gt; after that is equal to the sum of the number of combinations of &lt;math&gt;l - 1&lt;/math&gt; rolls that have a starting value of at least &lt;math&gt;r&lt;/math&gt;. If we slowly count through and add up all the possible combinations we get &lt;math&gt;\frac{7}{72}&lt;/math&gt; possibilities.<br /> <br /> === Solution 6 ===<br /> In a manner similar to the above solution, instead consider breaking it down into two sets of two dice rolls. The first [[subset]] must have a maximum value which is &lt;math&gt;\le&lt;/math&gt; the minimum value of the second subset. <br /> <br /> *If the first subset ends in a 1, there is &lt;math&gt;1&lt;/math&gt; such subset and there are &lt;math&gt;6 + 5 + 4 + 3 + 2 + 1 = \frac{6}{2}(6 + 1) = 21&lt;/math&gt; ways of making the second subset.<br /> *If the first subset ends in a 2, there is &lt;math&gt;2&lt;/math&gt; such subsets and there are &lt;math&gt;5 + \ldots + 1 = \frac{5}{2}(5 + 1) = 15&lt;/math&gt; ways of making the second subset.<br /> <br /> Thus, the number of combinations is &lt;math&gt;\sum_{i=1}^6 i \cdot \left(\frac{(7 - i)(8 - i)}{2}\right) = 21 + 30 + 30 + 24 + 15 + 6 = 126&lt;/math&gt;, and the probability again is &lt;math&gt;\frac7{72}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_14&diff=62507 2001 AIME I Problems/Problem 14 2014-07-15T22:08:40Z <p>Mathcool2009: </p> <hr /> <div>== Problem ==<br /> A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible?<br /> <br /> == Solution ==<br /> Let &lt;math&gt;0&lt;/math&gt; represent a house that does not receive mail and &lt;math&gt;1&lt;/math&gt; represent a house that does receive mail. This problem is now asking for the number of &lt;math&gt;19&lt;/math&gt;-digit strings of &lt;math&gt;0&lt;/math&gt;'s and &lt;math&gt;1&lt;/math&gt;'s such that there are no two consecutive &lt;math&gt;1&lt;/math&gt;'s and no three consecutive &lt;math&gt;0&lt;/math&gt;'s. <br /> <br /> The last two digits of any &lt;math&gt;n&lt;/math&gt;-digit string can't be &lt;math&gt;11&lt;/math&gt;, so the only possibilities are &lt;math&gt;00&lt;/math&gt;, &lt;math&gt;01&lt;/math&gt;, and &lt;math&gt;10&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;a_n&lt;/math&gt; be the number of &lt;math&gt;n&lt;/math&gt;-digit strings ending in &lt;math&gt;00&lt;/math&gt;, &lt;math&gt;b_n&lt;/math&gt; be the number of &lt;math&gt;n&lt;/math&gt;-digit strings ending in &lt;math&gt;01&lt;/math&gt;, and &lt;math&gt;c_n&lt;/math&gt; be the number of &lt;math&gt;n&lt;/math&gt;-digit strings ending in &lt;math&gt;10&lt;/math&gt;. <br /> <br /> If an &lt;math&gt;n&lt;/math&gt;-digit string ends in &lt;math&gt;00&lt;/math&gt;, then the previous digit must be a &lt;math&gt;1&lt;/math&gt;, and the last two digits of the &lt;math&gt;n-1&lt;/math&gt; digits substring will be &lt;math&gt;10&lt;/math&gt;. So <br /> &lt;cmath&gt;a_{n} = c_{n-1}.&lt;/cmath&gt;<br /> <br /> If an &lt;math&gt;n&lt;/math&gt;-digit string ends in &lt;math&gt;01&lt;/math&gt;, then the previous digit can be either a &lt;math&gt;0&lt;/math&gt; or a &lt;math&gt;1&lt;/math&gt;, and the last two digits of the &lt;math&gt;n-1&lt;/math&gt; digits substring can be either &lt;math&gt;00&lt;/math&gt; or &lt;math&gt;10&lt;/math&gt;. So <br /> &lt;cmath&gt;b_{n} = a_{n-1} + c_{n-1}.&lt;/cmath&gt;<br /> <br /> If an &lt;math&gt;n&lt;/math&gt;-digit string ends in &lt;math&gt;10&lt;/math&gt;, then the previous digit must be a &lt;math&gt;0&lt;/math&gt;, and the last two digits of the &lt;math&gt;n-1&lt;/math&gt; digits substring will be &lt;math&gt;01&lt;/math&gt;. So <br /> &lt;cmath&gt;c_{n} = b_{n-1}.&lt;/cmath&gt;<br /> <br /> Clearly, &lt;math&gt;a_2=b_2=c_2=1&lt;/math&gt;. Using the [[recursion|recursive]] equations and initial values: <br /> &lt;cmath&gt;\begin{tabular}[t]{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}<br /> \multicolumn{19}{c}{}\\\hline<br /> n&amp;2&amp;3&amp;4&amp;5&amp;6&amp;7&amp;8&amp;9&amp;10&amp;11&amp;12&amp;13&amp;14&amp;15&amp;16&amp;17&amp;18&amp;19\\\hline<br /> a_n&amp;1&amp;1&amp;1&amp;2&amp;2&amp;3&amp;4&amp;5&amp;7&amp;9&amp;12&amp;16&amp;21&amp;28&amp;37&amp;49&amp;65&amp;86\\\hline<br /> b_n&amp;1&amp;2&amp;2&amp;3&amp;4&amp;5&amp;7&amp;9&amp;12&amp;16&amp;21&amp;28&amp;37&amp;49&amp;65&amp;86&amp;114&amp;151\\\hline<br /> c_n&amp;1&amp;1&amp;2&amp;2&amp;3&amp;4&amp;5&amp;7&amp;9&amp;12&amp;16&amp;21&amp;28&amp;37&amp;49&amp;65&amp;86&amp;114\\\hline<br /> \end{tabular}&lt;/cmath&gt;<br /> <br /> Therefore, the number of &lt;math&gt;19&lt;/math&gt;-digit strings is &lt;math&gt;a_{19}+b_{19}+c_{19} = 86+151+114 = \boxed{351}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=13|num-a=15|t=384210}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_10&diff=62496 2000 AIME II Problems/Problem 10 2014-07-14T21:52:10Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> A [[circle]] is [[inscribe]]d in [[quadrilateral]] &lt;math&gt;ABCD&lt;/math&gt;, [[tangent]] to &lt;math&gt;\overline{AB}&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt; and to &lt;math&gt;\overline{CD}&lt;/math&gt; at &lt;math&gt;Q&lt;/math&gt;. Given that &lt;math&gt;AP=19&lt;/math&gt;, &lt;math&gt;PB=26&lt;/math&gt;, &lt;math&gt;CQ=37&lt;/math&gt;, and &lt;math&gt;QD=23&lt;/math&gt;, find the [[Perfect square|square]] of the [[radius]] of the circle.<br /> <br /> == Solution ==<br /> Call the [[center]] of the circle &lt;math&gt;O&lt;/math&gt;. By drawing the lines from &lt;math&gt;O&lt;/math&gt; tangent to the sides and from &lt;math&gt;O&lt;/math&gt; to the vertices of the quadrilateral, four pairs of congruent [[right triangle]]s are formed.<br /> <br /> Thus, &lt;math&gt;\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180&lt;/math&gt;, or &lt;math&gt;(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=180&lt;/math&gt;.<br /> <br /> Take the &lt;math&gt;\tan&lt;/math&gt; of both sides and use the identity for &lt;math&gt;\tan(A+B)&lt;/math&gt; to get &lt;math&gt;\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0&lt;/math&gt;.<br /> <br /> Use the identity for &lt;math&gt;\tan(A+B)&lt;/math&gt; again to get &lt;math&gt;\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}=0&lt;/math&gt;.<br /> <br /> Solving gives &lt;math&gt;r^2=\boxed{647}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=II|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_14&diff=62468 1998 AIME Problems/Problem 14 2014-07-07T22:04:40Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> An &lt;math&gt;m\times n\times p&lt;/math&gt; rectangular box has half the volume of an &lt;math&gt;(m + 2)\times(n + 2)\times(p + 2)&lt;/math&gt; rectangular box, where &lt;math&gt;m, n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are integers, and &lt;math&gt;m\le n\le p.&lt;/math&gt; What is the largest possible value of &lt;math&gt;p&lt;/math&gt;?<br /> <br /> == Solution ==<br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;2mnp = (m+2)(n+2)(p+2)&lt;/math&gt;&lt;/div&gt;<br /> <br /> Let’s solve for &lt;math&gt;p&lt;/math&gt;:<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4}&lt;/math&gt;&lt;/div&gt;<br /> <br /> For the denominator, we will use a factoring trick (colloquially known as [[Simon's Favorite Factoring Trick|SFFT]]), which states that &lt;math&gt;xy + ax + ay + a^2 = (x+a)(y+a)&lt;/math&gt;. <br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;p = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}&lt;/math&gt;&lt;/div&gt;<br /> <br /> Clearly, we want to minimize the denominator, so &lt;math&gt;(m-2)(n-2) - 8 = 1 \Longrightarrow (m-2)(n-2) = 9&lt;/math&gt;. The possible pairs of factors of &lt;math&gt;9&lt;/math&gt; are &lt;math&gt;(1,9)(3,3)&lt;/math&gt;. These give &lt;math&gt;m = 3, n = 11&lt;/math&gt; and &lt;math&gt;m = 5, n = 5&lt;/math&gt; respectively. Substituting into the numerator, we see that the first pair gives &lt;math&gt;130&lt;/math&gt;, while the second pair gives &lt;math&gt;98&lt;/math&gt;. We can quickly test for the denominator assuming other values to convince ourselves that &lt;math&gt;1&lt;/math&gt; is the best possible value for the denominator. Hence, the solution is &lt;math&gt;130&lt;/math&gt;.<br /> <br /> <br /> Proof that setting the denominator &lt;math&gt;(m - 2)(n - 2) - 8&lt;/math&gt; to &lt;math&gt;1&lt;/math&gt; is optimal: Suppose &lt;math&gt;(m - 2)(n - 2) = 9&lt;/math&gt;, and suppose for the sake of contradiction that there exist &lt;math&gt;m', n'&lt;/math&gt; such that &lt;math&gt;(m' - 2)(n' - 2) = 8 + d&lt;/math&gt; for some &lt;math&gt;d &gt; 1&lt;/math&gt; and such that<br /> &lt;cmath&gt;\frac{2(m + 2)(n + 2)}{(m - 2)(n - 2) - 8} &lt; \frac{2(m' + 2)(n' + 2)}{(m' - 2)(n' - 2) - 8}.&lt;/cmath&gt;<br /> This implies that<br /> &lt;cmath&gt;d(m + 2)(n + 2) &lt; (m' + 2)(n' + 2),&lt;/cmath&gt;<br /> and<br /> &lt;cmath&gt;d((m - 2)(n - 2) + 4(m + n)) &lt; (m' - 2)(n' - 2) + 4(m' + n').&lt;/cmath&gt;<br /> Substituting gives<br /> &lt;cmath&gt;d(9 + 4(m + n)) &lt; 8 + d + 4(m' + n'),&lt;/cmath&gt;<br /> which we rewrite as<br /> &lt;cmath&gt;d(8 + 4(m + n)) &lt; 24 + 4((m' - 2) + (n' - 2)).&lt;/cmath&gt;<br /> Next, note that for &lt;math&gt;p'&lt;/math&gt; to be positive, we must have &lt;math&gt;m' - 2&lt;/math&gt; and &lt;math&gt;n' - 2&lt;/math&gt; be positive, so <br /> &lt;cmath&gt;(m' - 2) + (n' - 2) \leq (m' - 2)(n' - 2) + 1 = 9 + d.&lt;/cmath&gt;<br /> So<br /> &lt;cmath&gt;d(8 + 4(m + n)) &lt; 8 + 4(9 + d)&lt;/cmath&gt; //shouldn't this be 24 + 4(9+d)? then the entire proof is fallacious (it results in 2(1 + 3 + 3) &lt; 15 which is true)<br /> &lt;cmath&gt;d(4 + 4(m + n)) &lt; 44&lt;/cmath&gt;<br /> &lt;cmath&gt;d(1 + m + n) &lt; 11&lt;/cmath&gt;<br /> Next, we must have that &lt;math&gt;m - 2&lt;/math&gt; and &lt;math&gt;n - 2&lt;/math&gt; are positive, so &lt;math&gt;3 \leq m&lt;/math&gt; and &lt;math&gt;3 \leq n&lt;/math&gt;. Also, &lt;math&gt;2 \leq d&lt;/math&gt; by how we defined &lt;math&gt;d&lt;/math&gt;. So<br /> &lt;cmath&gt;2(1 + 3 + 3) &lt; 11,&lt;/cmath&gt;<br /> a contradiction. We already showed above that there are some values of &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;(m - 2)(n - 2) = 9&lt;/math&gt; that work, so this proves that one of these pairs of values of &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; must yield the maximal value of &lt;math&gt;p&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1998|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_9&diff=62321 1991 AIME Problems/Problem 9 2014-06-23T18:27:00Z <p>Mathcool2009: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Suppose that &lt;math&gt;\sec x+\tan x=\frac{22}7&lt;/math&gt; and that &lt;math&gt;\csc x+\cot x=\frac mn,&lt;/math&gt; where &lt;math&gt;\frac mn&lt;/math&gt; is in lowest terms. Find &lt;math&gt;m+n^{}_{}.&lt;/math&gt;<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> Use the two [[Trigonometric identities#Pythagorean Identities|trigonometric Pythagorean identities]] &lt;math&gt;1 + \tan^2 x = \sec^2 x&lt;/math&gt; and &lt;math&gt;1 + \cot^2 x = \csc^2 x&lt;/math&gt;. <br /> <br /> If we square the given &lt;math&gt;\sec x = \frac{22}{7} - \tan x&lt;/math&gt;, we find that <br /> <br /> &lt;cmath&gt;\begin{align*}<br /> \sec^2 x &amp;= \left(\frac{22}7\right)^2 - 2\left(\frac{22}7\right)\tan x + \tan^2 x \\<br /> 1 &amp;= \left(\frac{22}7\right)^2 - \frac{44}7 \tan x \end{align*}&lt;/cmath&gt;<br /> <br /> This yields &lt;math&gt;\tan x = \frac{435}{308}&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;y = \frac mn&lt;/math&gt;. Then squaring, <br /> <br /> &lt;cmath&gt;\csc^2 x = (y - \cot x)^2 \Longrightarrow 1 = y^2 - 2y\cot x.&lt;/cmath&gt; <br /> <br /> Substituting &lt;math&gt;\cot x = \frac{1}{\tan x} = \frac{308}{435}&lt;/math&gt; yields a [[quadratic equation]]: &lt;math&gt;0 = 435y^2 - 616y - 435 = (15y - 29)(29y + 15)&lt;/math&gt;. It turns out that only the [[positive]] root will work, so the value of &lt;math&gt;y = \frac{29}{15}&lt;/math&gt; and &lt;math&gt;m + n = \boxed{044}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Recall that &lt;math&gt;\sec^2 x - \tan^2 x = 1&lt;/math&gt;, from which we find that &lt;math&gt;\sec x - \tan x = 7/22&lt;/math&gt;. Adding the equations<br /> <br /> &lt;cmath&gt;\begin{eqnarray*} \sec x + \tan x &amp; = &amp; 22/7 \\<br /> \sec x - \tan x &amp; = &amp; 7/22\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> together and dividing by 2 gives &lt;math&gt;\sec x = 533/308&lt;/math&gt;, and subtracting the equations and dividing by 2 gives &lt;math&gt;\tan x = 435/308&lt;/math&gt;. Hence, &lt;math&gt;\cos x = 308/533&lt;/math&gt; and &lt;math&gt;\sin x = \tan x \cos x = (435/308)(308/533) = 435/533&lt;/math&gt;. Thus, &lt;math&gt;\csc x = 533/435&lt;/math&gt; and &lt;math&gt;\cot x = 308/435&lt;/math&gt;. Finally,<br /> <br /> &lt;cmath&gt;\csc x + \cot x = \frac {841}{435} = \frac {29}{15},&lt;/cmath&gt;<br /> <br /> so &lt;math&gt;m + n = 044&lt;/math&gt;.<br /> <br /> === Solution 3 (least computation)===<br /> By the given,<br /> &lt;math&gt;\frac {1}{\cos x} + \frac {\sin x}{\cos x} = \frac {22}{7}&lt;/math&gt; and<br /> &lt;math&gt;\frac {1}{\sin x} + \frac {\cos x}{\sin x} = k&lt;/math&gt;.<br /> <br /> Multiplying the two, we have<br /> <br /> &lt;cmath&gt;\frac {1}{\sin x \cos x} + \frac {1}{\sin x} + \frac {1}{\cos x} + 1 = \frac {22}{7}k&lt;/cmath&gt;<br /> <br /> Subtracting both of the two given equations from this, and simpliyfing with the identity &lt;math&gt;\frac {\sin x}{\cos x} + \frac {\cos x}{\sin x} = \frac{\sin ^2 x + \cos ^2 x}{\sin x \cos x} = \frac {1}{\sin x \cos x}&lt;/math&gt;, we get <br /> <br /> &lt;cmath&gt;1 = \frac {22}{7}k - \frac {22}{7} - k.&lt;/cmath&gt;<br /> <br /> Solving yields &lt;math&gt;k = \frac {29}{15}&lt;/math&gt;, and &lt;math&gt;m+n = 044&lt;/math&gt;<br /> <br /> === Solution 4 ===<br /> Make the substitution &lt;math&gt;u = \tan \frac x2&lt;/math&gt; (a substitution commonly used in calculus). &lt;math&gt;\tan \frac x2 = \frac{\sin x}{1+\cos x}&lt;/math&gt;, so &lt;math&gt;\csc x + \cot x = \frac{1+\cos x}{\sin x} = \frac1u = \frac mn&lt;/math&gt;. &lt;math&gt;\sec x + \tan x = \frac{1 + \sin x}{\cos x}.&lt;/math&gt; Now note the following:<br /> <br /> &lt;cmath&gt;\begin{align*}\sin x &amp;= \frac{2u}{1+u^2}\\<br /> \cos x &amp;= \frac{1-u^2}{1+u^2}\end{align*}&lt;/cmath&gt;<br /> <br /> Plugging these into our equality gives:<br /> <br /> &lt;cmath&gt;\frac{1+\frac{2u}{1+u^2}}{\frac{1-u^2}{1+u^2}} = \frac{22}7&lt;/cmath&gt;<br /> <br /> This simplifies to &lt;math&gt;\frac{1+u}{1-u} = \frac{22}7&lt;/math&gt;, and solving for &lt;math&gt;u&lt;/math&gt; gives &lt;math&gt;u = \frac{15}{29}&lt;/math&gt;, and &lt;math&gt;\frac mn = \frac{29}{15}&lt;/math&gt;. Finally, &lt;math&gt;m+n = 044&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> We are given that &lt;math&gt;\frac{1+\sin x}{\cos x}=\frac{22}7\implies\frac{1+\sin x}{\cos x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin^2x}{\cos x(1-\sin x)}=\frac{\cos^2x}{\cos x(1-\sin x)}&lt;/math&gt;<br /> &lt;math&gt;=\frac{\cos x}{1-\sin x}&lt;/math&gt;, or equivalently, &lt;math&gt;\cos x=\frac{7+7\sin x}{22}=\frac{22-22\sin x}7\implies\sin x=\frac{22^2-7^2}{22^2+7^2}&lt;/math&gt;<br /> &lt;math&gt;\implies\cos x=\frac{2\cdot22\cdot7}{22^2+7^2}&lt;/math&gt;. Note that what we want is just &lt;math&gt;\frac{1+\cos x}{\sin x}=\frac{1+\frac{2\cdot22\cdot7}{22^2+7^2}}{\frac{22^2-7^2}{22^2+7^2}}=\frac{22^2+7^2+2\cdot22\cdot7}{22^2-7^2}=\frac{(22+7)^2}{(22-7)(22+7)}=\frac{22+7}{22-7}&lt;/math&gt;<br /> &lt;math&gt;=\frac{29}{15}\implies m+n=29+15=\boxed{044}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1991|num-b=8|num-a=10|t=8322}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_7&diff=62320 1991 AIME Problems/Problem 7 2014-06-23T18:18:40Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> Find &lt;math&gt;A^2_{}&lt;/math&gt;, where &lt;math&gt;A^{}_{}&lt;/math&gt; is the sum of the [[absolute value]]s of all roots of the following equation:<br /> &lt;div style=&quot;text-align:center&quot;&gt;&lt;math&gt;x = \sqrt{19} + \frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{x}}}}}}}}}<br /> &lt;/math&gt;&lt;/div&gt;<br /> <br /> == Solution ==<br /> Let &lt;math&gt;f(x) = \sqrt{19} + \frac{91}{x}&lt;/math&gt;. Then &lt;math&gt;x = f(f(f(f(f(x)))))&lt;/math&gt;, from which we realize that &lt;math&gt;f(x) = x&lt;/math&gt;. This is because if we expand the entire expression, we will get a fraction of the form &lt;math&gt;\frac{ax + b}{cx + d}&lt;/math&gt; on the right hand side, which makes the equation simplify to a quadratic. As this quadratic will have two roots, they must be the same roots as the quadratic &lt;math&gt;f(x)=x&lt;/math&gt;.<br /> <br /> The given finite expansion can then be easily seen to reduce to the [[quadratic equation]] &lt;math&gt;x_{}^{2}-\sqrt{19}x-91=0&lt;/math&gt;. The solutions are &lt;math&gt;x_{\pm}^{}=&lt;/math&gt;&lt;math&gt;\frac{\sqrt{19}\pm\sqrt{383}}{2}&lt;/math&gt;. Therefore, &lt;math&gt;A_{}^{}=\vert x_{+}\vert+\vert x_{-}\vert=\sqrt{383}&lt;/math&gt;. We conclude that &lt;math&gt;A_{}^{2}=383&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1991|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems&diff=62319 1991 AIME Problems 2014-06-23T18:07:07Z <p>Mathcool2009: /* Problem 11 */ I made the diagram more clear than before--the old diagram implies that the centers of the disks are on C, which is false.</p> <hr /> <div>{{AIME Problems|year=1991}}<br /> <br /> == Problem 1 ==<br /> Find &lt;math&gt;x^2+y^2_{}&lt;/math&gt; if &lt;math&gt;x_{}^{}&lt;/math&gt; and &lt;math&gt;y_{}^{}&lt;/math&gt; are positive integers such that<br /> &lt;center&gt;&lt;math&gt;xy_{}^{}+x+y = 71&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;x^2y+xy^2 = 880^{}_{}.&lt;/math&gt;&lt;/center&gt;<br /> <br /> [[1991 AIME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> Rectangle &lt;math&gt;ABCD_{}^{}&lt;/math&gt; has sides &lt;math&gt;\overline {AB}&lt;/math&gt; of length 4 and &lt;math&gt;\overline {CB}&lt;/math&gt; of length 3. Divide &lt;math&gt;\overline {AB}&lt;/math&gt; into 168 congruent segments with points &lt;math&gt;A_{}^{}=P_0, P_1, \ldots, P_{168}=B&lt;/math&gt;, and divide &lt;math&gt;\overline {CB}&lt;/math&gt; into 168 congruent segments with points &lt;math&gt;C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B&lt;/math&gt;. For &lt;math&gt;1_{}^{} \le k \le 167&lt;/math&gt;, draw the segments &lt;math&gt;\overline {P_kQ_k}&lt;/math&gt;. Repeat this construction on the sides &lt;math&gt;\overline {AD}&lt;/math&gt; and &lt;math&gt;\overline {CD}&lt;/math&gt;, and then draw the diagonal &lt;math&gt;\overline {AC}&lt;/math&gt;. Find the sum of the lengths of the 335 parallel segments drawn.<br /> <br /> [[1991 AIME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> Expanding &lt;math&gt;(1+0.2)^{1000}_{}&lt;/math&gt; by the binomial theorem and doing no further manipulation gives<br /> &lt;center&gt;&lt;math&gt;{1000 \choose 0}(0.2)^0+{1000 \choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;= A_0 + A_1 + A_2 + \cdots + A_{1000},&lt;/math&gt;&lt;/center&gt;<br /> where &lt;math&gt;A_k = {1000 \choose k}(0.2)^k&lt;/math&gt; for &lt;math&gt;k = 0,1,2,\ldots,1000&lt;/math&gt;. For which &lt;math&gt;k_{}^{}&lt;/math&gt; is &lt;math&gt;A_k^{}&lt;/math&gt; the largest?<br /> <br /> [[1991 AIME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> How many real numbers &lt;math&gt;x^{}_{}&lt;/math&gt; satisfy the equation &lt;math&gt;\frac{1}{5}\log_2 x = \sin (5\pi x)&lt;/math&gt;?<br /> <br /> [[1991 AIME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> Given a rational number, write it as a fraction in lowest terms and calculate the product of the resulting numerator and denominator. For how many rational numbers between 0 and 1 will be &lt;math&gt;20_{}^{}!&lt;/math&gt; the resulting product? <br /> <br /> [[1991 AIME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> Suppose &lt;math&gt;r^{}_{}&lt;/math&gt; is a real number for which<br /> &lt;center&gt;&lt;math&gt;<br /> \left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546.<br /> &lt;/math&gt;&lt;/center&gt;<br /> Find &lt;math&gt;\lfloor 100r \rfloor&lt;/math&gt;. (For real &lt;math&gt;x^{}_{}&lt;/math&gt;, &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is the greatest integer less than or equal to &lt;math&gt;x^{}_{}&lt;/math&gt;.)<br /> <br /> [[1991 AIME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Find &lt;math&gt;A^2_{}&lt;/math&gt;, where &lt;math&gt;A^{}_{}&lt;/math&gt; is the sum of the absolute values of all roots of the following equation:<br /> &lt;center&gt;&lt;math&gt;x = \sqrt{19} + \frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{x}}}}}}}}}<br /> &lt;/math&gt;&lt;/center&gt;<br /> <br /> [[1991 AIME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> For how many real numbers &lt;math&gt;a^{}_{}&lt;/math&gt; does the quadratic equation &lt;math&gt;x^2 + ax^{}_{} + 6a=0&lt;/math&gt; have only integer roots for &lt;math&gt;x^{}_{}&lt;/math&gt;?<br /> <br /> [[1991 AIME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Suppose that &lt;math&gt;\sec x+\tan x=\frac{22}7&lt;/math&gt; and that &lt;math&gt;\csc x+\cot x=\frac mn,&lt;/math&gt; where &lt;math&gt;\frac mn&lt;/math&gt; is in lowest terms. Find &lt;math&gt;m+n^{}_{}.&lt;/math&gt;<br /> <br /> [[1991 AIME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> Two three-letter strings, &lt;math&gt;aaa^{}_{}&lt;/math&gt; and &lt;math&gt;bbb^{}_{}&lt;/math&gt;, are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a 1/3 chance of being received incorrectly, as an &lt;math&gt;a^{}_{}&lt;/math&gt; when it should have been a &lt;math&gt;b^{}_{}&lt;/math&gt;, or as a &lt;math&gt;b^{}_{}&lt;/math&gt; when it should be an &lt;math&gt;a^{}_{}&lt;/math&gt;. However, whether a given letter is received correctly or incorrectly is independent of the reception of any other letter. Let &lt;math&gt;S_a^{}&lt;/math&gt; be the three-letter string received when &lt;math&gt;aaa^{}_{}&lt;/math&gt; is transmitted and let &lt;math&gt;S_b^{}&lt;/math&gt; be the three-letter string received when &lt;math&gt;bbb^{}_{}&lt;/math&gt; is transmitted. Let &lt;math&gt;p&lt;/math&gt; be the probability that &lt;math&gt;S_a^{}&lt;/math&gt; comes before &lt;math&gt;S_b^{}&lt;/math&gt; in alphabetical order. When &lt;math&gt;p&lt;/math&gt; is written as a fraction in lowest terms, what is its numerator?<br /> <br /> [[1991 AIME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> Twelve congruent disks are placed on a circle &lt;math&gt;C^{}_{}&lt;/math&gt; of radius 1 in such a way that the twelve disks cover &lt;math&gt;C^{}_{}&lt;/math&gt;, no two of the disks overlap, and so that each of the twelve disks is tangent to its two neighbors. The resulting arrangement of disks is shown in the figure below. The sum of the areas of the twelve disks can be written in the from &lt;math&gt;\pi(a-b\sqrt{c})&lt;/math&gt;, where &lt;math&gt;a,b,c^{}_{}&lt;/math&gt; are positive integers and &lt;math&gt;c^{}_{}&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;a+b+c^{}_{}&lt;/math&gt;.<br /> <br /> [[Image:AIME_1991_Problem_11.png|300px]]<br /> <br /> [[1991 AIME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> Rhombus &lt;math&gt;PQRS^{}_{}&lt;/math&gt; is inscribed in rectangle &lt;math&gt;ABCD^{}_{}&lt;/math&gt; so that vertices &lt;math&gt;P^{}_{}&lt;/math&gt;, &lt;math&gt;Q^{}_{}&lt;/math&gt;, &lt;math&gt;R^{}_{}&lt;/math&gt;, and &lt;math&gt;S^{}_{}&lt;/math&gt; are interior points on sides &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, and &lt;math&gt;\overline{DA}&lt;/math&gt;, respectively. It is given that &lt;math&gt;PB^{}_{}=15&lt;/math&gt;, &lt;math&gt;BQ^{}_{}=20&lt;/math&gt;, &lt;math&gt;PR^{}_{}=30&lt;/math&gt;, and &lt;math&gt;QS^{}_{}=40&lt;/math&gt;. Let &lt;math&gt;m/n^{}_{}&lt;/math&gt;, in lowest terms, denote the perimeter of &lt;math&gt;ABCD^{}_{}&lt;/math&gt;. Find &lt;math&gt;m+n^{}_{}&lt;/math&gt;.<br /> <br /> [[1991 AIME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> A drawer contains a mixture of red socks and blue socks, at most 1991 in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly &lt;math&gt;\frac{1}{2}&lt;/math&gt; that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data? <br /> <br /> [[1991 AIME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> A hexagon is inscribed in a circle. Five of the sides have length 81 and the sixth, denoted by &lt;math&gt;\overline{AB}&lt;/math&gt;, has length 31. Find the sum of the lengths of the three diagonals that can be drawn from &lt;math&gt;A_{}^{}&lt;/math&gt;.<br /> <br /> [[1991 AIME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> For positive integer &lt;math&gt;n_{}^{}&lt;/math&gt;, define &lt;math&gt;S_n^{}&lt;/math&gt; to be the minimum value of the sum<br /> &lt;center&gt;&lt;math&gt;\sum_{k=1}^n \sqrt{(2k-1)^2+a_k^2},&lt;/math&gt;&lt;/center&gt;<br /> where &lt;math&gt;a_1,a_2,\ldots,a_n^{}&lt;/math&gt; are positive real numbers whose sum is 17. There is a unique positive integer &lt;math&gt;n^{}_{}&lt;/math&gt; for which &lt;math&gt;S_n^{}&lt;/math&gt; is also an integer. Find this &lt;math&gt;n^{}_{}&lt;/math&gt;.<br /> <br /> [[1991 AIME Problems/Problem 15|Solution]]<br /> <br /> == See also ==<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:AIME Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_8&diff=62215 1988 AIME Problems/Problem 8 2014-06-07T05:43:28Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> The [[function]] &lt;math&gt;f&lt;/math&gt;, defined on the set of ordered pairs of positive [[integer]]s, satisfies the following properties:<br /> &lt;cmath&gt;<br /> \begin{eqnarray*} f(x,x) &amp; = &amp; x, \\<br /> f(x,y) &amp; = &amp; f(y,x), \quad \text{and} \\<br /> (x + y) f(x,y) &amp; = &amp; yf(x,x + y). \end{eqnarray*}<br /> &lt;/cmath&gt;<br /> Calculate &lt;math&gt;f(14,52)&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Since all of the function's properties contain a recursive definition except for the first one, we know that &lt;math&gt;f(x,x) = x&lt;/math&gt; in order to obtain an integer answer. So, we have to transform &lt;math&gt;f(14,52)&lt;/math&gt; to this form by exploiting the other properties. The second one doesn't help us immediately, so we will use the third one. <br /> <br /> Note that<br /> <br /> &lt;cmath&gt;f(14,52) = \frac {38}{38}f(14,14 + 38) = \frac {52}{38}f(14,38)&lt;/cmath&gt;<br /> <br /> Repeating the process several times,<br /> &lt;cmath&gt;<br /> \begin{eqnarray*}f(14,52) &amp; = &amp; \frac {38}{38}f(14,14 + 38) = \frac {52}{38}f(14,38) \\<br /> &amp; = &amp; \frac {52}{38}\times \frac {38}{24}f(14,14 + 24) = \frac {52}{24}f(14,24) \\<br /> &amp; = &amp; \frac {52}{10}f(10,14) \\<br /> &amp; = &amp; \frac {52}{10}\times \frac {14}{4}f(10,4) = \frac {91}{5}f(4,10) \\<br /> &amp; = &amp; \frac {91}{3}f(4,6) \\<br /> &amp; = &amp; 91f(2,4) \\<br /> &amp; = &amp; 91\times 2 f(2,2) = 364. \end{eqnarray*}<br /> &lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> Notice that &lt;math&gt;f(x,y) = \text{lcm}(x,y)&lt;/math&gt; satisfies all three properties: <br /> <br /> Clearly, &lt;math&gt;\text{lcm}(x,x) = x&lt;/math&gt; and &lt;math&gt;\text{lcm}(x,y) = \text{lcm}(y,x)&lt;/math&gt;. <br /> <br /> Using the identities &lt;math&gt;\text{gcd}(x,y) \cdot \text{lcm}(x,y) = xy&lt;/math&gt; and &lt;math&gt;\text{gcd}(x,x+y) = \text{gcd}(x,y)&lt;/math&gt;, we have: <br /> <br /> &lt;math&gt;y \cdot \text{lcm}(x,x+y) &lt;/math&gt; &lt;math&gt;= \dfrac{y \cdot x(x+y)}{\text{gcd}(x,x+y)} &lt;/math&gt; &lt;math&gt;= \dfrac{(x+y) \cdot xy}{\text{gcd}(x,y)} &lt;/math&gt; &lt;math&gt;= (x+y) \cdot \text{lcm}(x,y)&lt;/math&gt;. <br /> <br /> Hence, &lt;math&gt;f(x,y) = \text{lcm}(x,y)&lt;/math&gt; is a solution to the functional equation. <br /> <br /> Since this is an [[AIME]] problem, there is exactly one correct answer, and thus, exactly one possible value of &lt;math&gt;f(14,52)&lt;/math&gt;. <br /> <br /> Therefore, &lt;math&gt;f(14,52) = \text{lcm}(14,52) = \text{lcm}(2 \cdot 7,2^2 \cdot 13) = 2^2 \cdot 7 \cdot 13 = \boxed{364}&lt;/math&gt;. <br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_7&diff=62211 1988 AIME Problems/Problem 7 2014-06-06T05:41:22Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> In [[triangle]] &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;\tan \angle CAB = 22/7&lt;/math&gt;, and the [[altitude]] from &lt;math&gt;A&lt;/math&gt; divides &lt;math&gt;BC&lt;/math&gt; into [[segment]]s of length 3 and 17. What is the area of triangle &lt;math&gt;ABC&lt;/math&gt;?<br /> <br /> == Solution ==<br /> &lt;center&gt;[[Image:AIME_1988_Solution_07.png]]&lt;/center&gt;<br /> <br /> Let &lt;math&gt;D&lt;/math&gt; be the intersection of the [[altitude]] with &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;h&lt;/math&gt; be the length of the altitude. [[Without loss of generality]], let &lt;math&gt;BD = 17&lt;/math&gt; and &lt;math&gt;CD = 3&lt;/math&gt;. Then &lt;math&gt;\tan \angle DAB = \frac{17}{h}&lt;/math&gt; and &lt;math&gt;\tan \angle CAD = \frac{3}{h}&lt;/math&gt;. Using the [[Trigonometric_identities#Angle_Addition.2FSubtraction_Identities|tangent sum formula]],<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;<br /> &lt;math&gt;\begin{eqnarray*}<br /> \tan CAB &amp;=&amp; \tan (DAB + CAD)\\<br /> \frac{22}{7} &amp;=&amp; \frac{\tan DAB + \tan CAD}{1 - \tan DAB \cdot \tan CAD} \\<br /> &amp;=&amp; \frac{\frac{17}{h} + \frac{3}{h}}{1 - \left(\frac{17}{h}\right)\left(\frac{3}{h}\right)} \\<br /> \frac{22}{7} &amp;=&amp; \frac{20h}{h^2 - 51}\\<br /> 0 &amp;=&amp; 22h^2 - 140h - 22 \cdot 51\\<br /> 0 &amp;=&amp; (11h + 51)(h - 11)<br /> \end{eqnarray*}&lt;/math&gt;&lt;/div&gt;<br /> <br /> The postive value of &lt;math&gt;h = 11&lt;/math&gt;, so the area is &lt;math&gt;\frac{1}{2}(17 + 3)\cdot 11 = 110&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_6&diff=62210 1988 AIME Problems/Problem 6 2014-06-06T05:33:14Z <p>Mathcool2009: /* Solution 1 (specific) */</p> <hr /> <div>== Problem ==<br /> It is possible to place positive integers into the vacant twenty-one squares of the &lt;math&gt;5 \times 5&lt;/math&gt; square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (*).<br /> <br /> [[Image:1988_AIME-6.png]]<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 (specific) ===<br /> Let the coordinates of the square at the bottom left be &lt;math&gt;(0,0)&lt;/math&gt;, the square to the right &lt;math&gt;(1,0)&lt;/math&gt;, etc.<br /> <br /> Label the leftmost column (from bottom to top) &lt;math&gt;0, a, 2a, 3a, 4a&lt;/math&gt; and the bottom-most row (from left to right) &lt;math&gt;0, b, 2b, 3b, 4b&lt;/math&gt;. Our method will be to use the given numbers to set up equations to solve for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, and then calculate &lt;math&gt;(*)&lt;/math&gt;.<br /> <br /> &lt;math&gt;\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a &amp; &amp; &amp; * &amp; \\<br /> \hline 3a &amp; 74 &amp; &amp; &amp; \\<br /> \hline 2a &amp; &amp; &amp; &amp; 186 \\<br /> \hline a &amp; &amp; 103 &amp; &amp; \\<br /> \hline 0 &amp; b &amp; 2b &amp; 3b &amp; 4b \\<br /> \hline \end{tabular}&lt;/math&gt;<br /> <br /> We can compute the squares at the intersections of two existing numbers in terms of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;; two such equations will give us the values of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. On the fourth row from the bottom, the common difference is &lt;math&gt;74 - 3a&lt;/math&gt;, so the square at &lt;math&gt;(2,3)&lt;/math&gt; has a value of &lt;math&gt;148 - 3a&lt;/math&gt;. On the third column from the left, the common difference is &lt;math&gt;103 - 2b&lt;/math&gt;, so that square also has a value of &lt;math&gt;2b + 3(103 - 2b) = 309 - 4b&lt;/math&gt;. Equating, we get &lt;math&gt;148 - 3a = 309 - 4b \Longrightarrow 4b - 3a = 161&lt;/math&gt;.<br /> <br /> Now we compute the square &lt;math&gt;(2,2)&lt;/math&gt;. By rows, this value is simply the average of &lt;math&gt;2a&lt;/math&gt; and &lt;math&gt;186&lt;/math&gt;, so it is equal to &lt;math&gt;\frac{2a + 186}{2} = a + 93&lt;/math&gt;. By columns, the common difference is &lt;math&gt;103 - 2b&lt;/math&gt;, so our value is &lt;math&gt;206 - 2b&lt;/math&gt;. Equating, &lt;math&gt;a + 93 = 206 - 2b \Longrightarrow a + 2b = 113&lt;/math&gt;.<br /> <br /> Solving <br /> &lt;cmath&gt;\begin{eqnarray*}4b - 3a &amp;=&amp; 161\\<br /> a + 2b &amp;=&amp; 113&lt;/cmath&gt;<br /> <br /> gives &lt;math&gt;a = 13&lt;/math&gt;, &lt;math&gt;b = 50&lt;/math&gt;. Now it is simple to calculate &lt;math&gt;(4,3)&lt;/math&gt;. One way to do it is to see that &lt;math&gt;(2,2)&lt;/math&gt; has &lt;math&gt;206 - 2b = 106&lt;/math&gt; and &lt;math&gt;(4,2)&lt;/math&gt; has &lt;math&gt;186&lt;/math&gt;, so &lt;math&gt;(3,2)&lt;/math&gt; has &lt;math&gt;\frac{106 + 186}{2} = 146&lt;/math&gt;. Now, &lt;math&gt;(3,0)&lt;/math&gt; has &lt;math&gt;3b = 150&lt;/math&gt;, so &lt;math&gt;(3,2) = \frac{(3,0) + (3,4)}{2} \Longrightarrow (3,4) = * = 142&lt;/math&gt;.<br /> <br /> === Solution 2 (general) ===<br /> First, let &lt;math&gt;a =&lt;/math&gt; the number to be placed in the first column, fourth row. Let &lt;math&gt;b =&lt;/math&gt; the number to be placed in the second column, fifth row. We can determine the entire first column and fifth row in terms of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;:<br /> <br /> &lt;math&gt;\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a &amp; &amp; &amp; &amp; \\<br /> \hline 3a &amp; &amp; &amp; &amp; \\<br /> \hline 2a &amp; &amp; &amp; &amp; \\<br /> \hline a &amp; &amp; &amp; &amp; \\<br /> \hline 0 &amp; b &amp; 2b &amp; 3b &amp; 4b \\<br /> \hline \end{tabular}&lt;/math&gt;<br /> <br /> Next, let &lt;math&gt;a + b + c =&lt;/math&gt; the number to be placed in the second column, fourth row. We can determine the entire second column and fourth row in terms of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;:<br /> <br /> &lt;math&gt;\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a &amp; 4a + b + 4c &amp; &amp; &amp; \\<br /> \hline 3a &amp; 3a + b + 3c &amp; &amp; &amp; \\<br /> \hline 2a &amp; 2a + b + 2c &amp; &amp; &amp; \\<br /> \hline a &amp; a + b + c &amp; a + 2b + 2c &amp; a + 3b + 3c &amp; a + 4b + 4c \\<br /> \hline 0 &amp; b &amp; 2b &amp; 3b &amp; 4b \\<br /> \hline \end{tabular}&lt;/math&gt;<br /> <br /> We have now determined at least two values in each row and column. We can finish the table without introducing any more variables:<br /> <br /> &lt;math&gt;\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a &amp; 4a + b + 4c &amp; 4a + 2b + 8c &amp; 4a + 3b + 12c &amp; 4a + 4b + 16c \\<br /> \hline 3a &amp; 3a + b + 3c &amp; 3a + 2b + 6c &amp; 3a + 3b + 9c &amp; 3a + 4b + 12c \\<br /> \hline 2a &amp; 2a + b + 2c &amp; 2a + 2b + 4c &amp; 2a + 3b + 6c &amp; 2a + 4b + 8c \\<br /> \hline a &amp; a + b + c &amp; a + 2b + 2c &amp; a + 3b + 3c &amp; a + 4b + 4c \\<br /> \hline 0 &amp; b &amp; 2b &amp; 3b &amp; 4b \\<br /> \hline \end{tabular}&lt;/math&gt;<br /> <br /> We now have a system of equations.<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;3a + b + 3c = 74&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;2a + 4b + 8c = 186&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;a + 2b + 2c = 103&lt;/math&gt;&lt;/div&gt;<br /> <br /> Solving, we find that &lt;math&gt;(a,b,c) = (13,50, - 5)&lt;/math&gt;. The number in the square marked by the asterisk is &lt;math&gt;4a + 3b + 12c = \boxed{142}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_6&diff=62209 1988 AIME Problems/Problem 6 2014-06-06T05:32:27Z <p>Mathcool2009: /* Solution 1 (specific) */</p> <hr /> <div>== Problem ==<br /> It is possible to place positive integers into the vacant twenty-one squares of the &lt;math&gt;5 \times 5&lt;/math&gt; square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (*).<br /> <br /> [[Image:1988_AIME-6.png]]<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 (specific) ===<br /> Let the coordinates of the square at the bottom left be &lt;math&gt;(0,0)&lt;/math&gt;, the square to the right &lt;math&gt;(1,0)&lt;/math&gt;, etc.<br /> <br /> Label the leftmost column (from bottom to top) &lt;math&gt;0, a, 2a, 3a, 4a&lt;/math&gt; and the bottom-most row (from left to right) &lt;math&gt;0, b, 2b, 3b, 4b&lt;/math&gt;. Our method will be to use the given numbers to set up equations to solve for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, and then calculate &lt;math&gt;(*)&lt;/math&gt;.<br /> <br /> &lt;math&gt;\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a &amp; &amp; &amp; * &amp; \\<br /> \hline 3a &amp; 74 &amp; &amp; &amp; \\<br /> \hline 2a &amp; &amp; &amp; &amp; 186 \\<br /> \hline a &amp; &amp; 103 &amp; &amp; \\<br /> \hline 0 &amp; b &amp; 2b &amp; 3b &amp; 4b \\<br /> \hline \end{tabular}&lt;/math&gt;<br /> <br /> We can compute the squares at the intersections of two existing numbers in terms of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;; two such equations will give us the values of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. On the fourth row from the bottom, the common difference is &lt;math&gt;74 - 3a&lt;/math&gt;, so the square at &lt;math&gt;(2,3)&lt;/math&gt; has a value of &lt;math&gt;148 - 3a&lt;/math&gt;. On the third column from the left, the common difference is &lt;math&gt;103 - 2b&lt;/math&gt;, so that square also has a value of &lt;math&gt;2b + 3(103 - 2b) = 309 - 4b&lt;/math&gt;. Equating, we get &lt;math&gt;148 - 3a = 309 - 4b \Longrightarrow 4b - 3a = 161&lt;/math&gt;.<br /> <br /> Now we compute the square &lt;math&gt;(2,2)&lt;/math&gt;. By rows, this value is simply the average of &lt;math&gt;2a&lt;/math&gt; and &lt;math&gt;186&lt;/math&gt;, so it is equal to &lt;math&gt;\frac{2a + 186}{2} = a + 93&lt;/math&gt;. By columns, the common difference is &lt;math&gt;103 - 2b&lt;/math&gt;, so our value is &lt;math&gt;206 - 2b&lt;/math&gt;. Equating, &lt;math&gt;a + 93 = 206 - 2b \Longrightarrow a + 2b = 113&lt;/math&gt;.<br /> <br /> Solving <br /> &lt;cmath&gt;\begin{eqnarray*}4b - 3a &amp;=&amp; 161\\<br /> a + 2b &amp;=&amp; 113&lt;/cmath&gt;<br /> <br /> gives &lt;math&gt;a = 13&lt;/math&gt;, &lt;math&gt;b = 50&lt;/math&gt;. Now it is simple to calculate &lt;math&gt;(4,3)&lt;/math&gt;. One way to do it is to see that &lt;math&gt;(2,2)&lt;/math&gt; has &lt;math&gt;206 - 2b = 106&lt;/math&gt; and &lt;math&gt;(4,2)&lt;/math&gt; has &lt;math&gt;186&lt;/math&gt;, so &lt;math&gt;(3,2)&lt;/math&gt; has &lt;math&gt;\frac{106 + 186}{2} = 146&lt;/math&gt;. Now, &lt;math&gt;(3,0)&lt;/math&gt; has &lt;math&gt;3b = 150&lt;/math&gt;, so &lt;math&gt;(3,2) = \frac{(3,0) + (3,4)}{2} = \Longrightarrow (3,4) = * = 142&lt;/math&gt;.<br /> <br /> === Solution 2 (general) ===<br /> First, let &lt;math&gt;a =&lt;/math&gt; the number to be placed in the first column, fourth row. Let &lt;math&gt;b =&lt;/math&gt; the number to be placed in the second column, fifth row. We can determine the entire first column and fifth row in terms of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;:<br /> <br /> &lt;math&gt;\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a &amp; &amp; &amp; &amp; \\<br /> \hline 3a &amp; &amp; &amp; &amp; \\<br /> \hline 2a &amp; &amp; &amp; &amp; \\<br /> \hline a &amp; &amp; &amp; &amp; \\<br /> \hline 0 &amp; b &amp; 2b &amp; 3b &amp; 4b \\<br /> \hline \end{tabular}&lt;/math&gt;<br /> <br /> Next, let &lt;math&gt;a + b + c =&lt;/math&gt; the number to be placed in the second column, fourth row. We can determine the entire second column and fourth row in terms of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;:<br /> <br /> &lt;math&gt;\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a &amp; 4a + b + 4c &amp; &amp; &amp; \\<br /> \hline 3a &amp; 3a + b + 3c &amp; &amp; &amp; \\<br /> \hline 2a &amp; 2a + b + 2c &amp; &amp; &amp; \\<br /> \hline a &amp; a + b + c &amp; a + 2b + 2c &amp; a + 3b + 3c &amp; a + 4b + 4c \\<br /> \hline 0 &amp; b &amp; 2b &amp; 3b &amp; 4b \\<br /> \hline \end{tabular}&lt;/math&gt;<br /> <br /> We have now determined at least two values in each row and column. We can finish the table without introducing any more variables:<br /> <br /> &lt;math&gt;\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a &amp; 4a + b + 4c &amp; 4a + 2b + 8c &amp; 4a + 3b + 12c &amp; 4a + 4b + 16c \\<br /> \hline 3a &amp; 3a + b + 3c &amp; 3a + 2b + 6c &amp; 3a + 3b + 9c &amp; 3a + 4b + 12c \\<br /> \hline 2a &amp; 2a + b + 2c &amp; 2a + 2b + 4c &amp; 2a + 3b + 6c &amp; 2a + 4b + 8c \\<br /> \hline a &amp; a + b + c &amp; a + 2b + 2c &amp; a + 3b + 3c &amp; a + 4b + 4c \\<br /> \hline 0 &amp; b &amp; 2b &amp; 3b &amp; 4b \\<br /> \hline \end{tabular}&lt;/math&gt;<br /> <br /> We now have a system of equations.<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;3a + b + 3c = 74&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;2a + 4b + 8c = 186&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;a + 2b + 2c = 103&lt;/math&gt;&lt;/div&gt;<br /> <br /> Solving, we find that &lt;math&gt;(a,b,c) = (13,50, - 5)&lt;/math&gt;. The number in the square marked by the asterisk is &lt;math&gt;4a + 3b + 12c = \boxed{142}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_4&diff=62208 1988 AIME Problems/Problem 4 2014-06-06T05:23:38Z <p>Mathcool2009: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Suppose that &lt;math&gt;|x_i| &lt; 1&lt;/math&gt; for &lt;math&gt;i = 1, 2, \dots, n&lt;/math&gt;. Suppose further that<br /> &lt;math&gt;<br /> |x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n|.<br /> &lt;/math&gt;<br /> What is the smallest possible value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> Since &lt;math&gt;|x_i| &lt; 1&lt;/math&gt; then<br /> <br /> &lt;cmath&gt;|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n| &lt; n&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;n \ge 20&lt;/math&gt;. We now just need to find an example where &lt;math&gt;n = 20&lt;/math&gt;: suppose &lt;math&gt;x_{2k-1} = \frac{19}{20}&lt;/math&gt; and &lt;math&gt;x_{2k} = -\frac{19}{20}&lt;/math&gt;; then on the left hand side we have &lt;math&gt;\left|\frac{19}{20}\right| + \left|-\frac{19}{20}\right| + \dots + \left|-\frac{19}{20}\right| = 20\left(\frac{19}{20}\right) = 19&lt;/math&gt;. On the right hand side, we have &lt;math&gt;19 + \left|\frac{19}{20} - \frac{19}{20} + \dots - \frac{19}{20}\right| = 19 + 0 = 19&lt;/math&gt;, and so the equation can hold for &lt;math&gt;n = \boxed{020}&lt;/math&gt;.<br /> <br /> ===Solution 2 (???dubious)===<br /> <br /> Let &lt;math&gt;|x_1 + x_2 + \dots + x_n| = 0&lt;/math&gt; and &lt;math&gt;|x_1| + |x_2| + \dots + |x_n| = 19&lt;/math&gt;. Then the smallest value of &lt;math&gt;n = \boxed {20}&lt;/math&gt; because &lt;math&gt;|x_i| &lt; 1&lt;/math&gt;, and therefore &lt;math&gt;n &gt; 19&lt;/math&gt;.<br /> <br /> == See also ==<br /> *[[Triangle Inequality]]<br /> {{AIME box|year=1988|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_4&diff=62207 1988 AIME Problems/Problem 4 2014-06-06T05:22:32Z <p>Mathcool2009: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Suppose that &lt;math&gt;|x_i| &lt; 1&lt;/math&gt; for &lt;math&gt;i = 1, 2, \dots, n&lt;/math&gt;. Suppose further that<br /> &lt;math&gt;<br /> |x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n|.<br /> &lt;/math&gt;<br /> What is the smallest possible value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> Since &lt;math&gt;|x_i| &lt; 1&lt;/math&gt; then<br /> <br /> &lt;cmath&gt;|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n| &lt; n&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;n \ge 20&lt;/math&gt;. We now just need to find an example where &lt;math&gt;n = 20&lt;/math&gt;: suppose &lt;math&gt;x_{2k-1} = \frac{19}{20}&lt;/math&gt; and &lt;math&gt;x_{2k} = -\frac{19}{20}&lt;/math&gt;; then on the left hand side we have &lt;math&gt;\left|\frac{19}{20}\right| + \left|-\frac{19}{20}\right| + \dots + \left|-\frac{19}{20}\right| = 20\left(\frac{19}{20}\right) = 19&lt;/math&gt;. On the right hand side, we have &lt;math&gt;19 + \left|\frac{19}{20} - \frac{19}{20} + \dots - \frac{19}{20}\right| = 19 + 0 = 19&lt;/math&gt;, and so the equation can hold for &lt;math&gt;n = \boxed{020}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> <br /> Let &lt;math&gt;|x_1 + x_2 + \dots + x_n| = 0&lt;/math&gt; and &lt;math&gt;|x_1| + |x_2| + \dots + |x_n| = 19&lt;/math&gt;. Then the smallest value of &lt;math&gt;n = \boxed {20}&lt;/math&gt; because &lt;math&gt;|x_i| &lt; 1&lt;/math&gt;, and therefore &lt;math&gt;n &gt; 19&lt;/math&gt;.<br /> <br /> == See also ==<br /> *[[Triangle Inequality]]<br /> {{AIME box|year=1988|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_2&diff=62206 1988 AIME Problems/Problem 2 2014-06-06T05:19:53Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> For any positive integer &lt;math&gt;k&lt;/math&gt;, let &lt;math&gt;f_1(k)&lt;/math&gt; denote the square of the sum of the digits of &lt;math&gt;k&lt;/math&gt;. For &lt;math&gt;n \ge 2&lt;/math&gt;, let &lt;math&gt;f_n(k) = f_1(f_{n - 1}(k))&lt;/math&gt;. Find &lt;math&gt;f_{1988}(11)&lt;/math&gt;.<br /> <br /> == Solution ==<br /> We see that &lt;math&gt;f_{1}(11)=4&lt;/math&gt;<br /> <br /> &lt;math&gt;f_2(11) = f_1(4)=16&lt;/math&gt;<br /> <br /> &lt;math&gt;f_3(11) = f_1(16)=49&lt;/math&gt;<br /> <br /> &lt;math&gt;f_4(11) = f_1(49)=169&lt;/math&gt;<br /> <br /> &lt;math&gt;f_5(11) = f_1(169)=256&lt;/math&gt;<br /> <br /> &lt;math&gt;f_6(11) = f_1(256)=169&lt;/math&gt;<br /> <br /> Note that this revolves between the two numbers.<br /> &lt;math&gt;f_{1988}(169)=169\implies f_{1988}(11)=\boxed{169}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_1&diff=62205 1988 AIME Problems/Problem 1 2014-06-06T05:18:03Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> One commercially available ten-button lock may be opened by depressing -- in any order -- the correct five buttons. The sample shown below has &lt;math&gt;\{1,2,3,6,9\}&lt;/math&gt; as its [[combination]]. Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many additional combinations would this allow?<br /> <br /> [[Image:1988-1.png]]<br /> <br /> == Solution ==<br /> Currently there are &lt;math&gt;{10 \choose 5}&lt;/math&gt; possible combinations.<br /> With any integer &lt;math&gt;x&lt;/math&gt; from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;9&lt;/math&gt;, the number of ways to choose a set of &lt;math&gt;x&lt;/math&gt; buttons is &lt;math&gt;\sum^{9}_{k=1}{10 \choose k}&lt;/math&gt;.<br /> Now we can use the identity &lt;math&gt;\sum^{n}_{k=0}{n \choose k}=2^{n}&lt;/math&gt;.<br /> So the number of additional combinations is just<br /> &lt;math&gt;2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=\boxed{770}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1988|before=First Question|num-a=2}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_1&diff=62204 1988 AIME Problems/Problem 1 2014-06-06T05:17:34Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> One commercially available ten-button lock may be opened by depressing -- in any order -- the correct five buttons. The sample shown below has &lt;math&gt;\{1,2,3,6,9\}&lt;/math&gt; as its [[combination]]. Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many additional combinations would this allow?<br /> <br /> [[Image:1988-1.png]]<br /> <br /> == Solution ==<br /> Currently there are &lt;math&gt;{10 \choose 5}&lt;/math&gt; possible combinations.<br /> With any integer &lt;math&gt;x&lt;/math&gt; from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;9&lt;/math&gt;, the number of ways to choose a set of &lt;math&gt;x&lt;/math&gt; buttons is &lt;math&gt;\sum^{9}_{k=1}{10 \choose k}&lt;/math&gt;.<br /> Now we can use the identity &lt;math&gt;\sum^{n}_{k=0}{n \choose k}=2^{n}&lt;/math&gt;.<br /> So the number of ways is just<br /> &lt;math&gt;2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=\boxed{770}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1988|before=First Question|num-a=2}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_1&diff=62203 1988 AIME Problems/Problem 1 2014-06-06T05:17:02Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> One commercially available ten-button lock may be opened by depressing -- in any order -- the correct five buttons. The sample shown below has &lt;math&gt;\{1,2,3,6,9\}&lt;/math&gt; as its [[combination]]. Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many additional combinations would this allow?<br /> <br /> [[Image:1988-1.png]]<br /> <br /> == Solution ==<br /> Currently there are &lt;math&gt;{10 \choose 5}&lt;/math&gt; possible ways.<br /> With any integer &lt;math&gt;x&lt;/math&gt; from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;9&lt;/math&gt;, the number of ways to choose a set of &lt;math&gt;x&lt;/math&gt; buttons is &lt;math&gt;\sum^{9}_{k=1}{10 \choose k}&lt;/math&gt;.<br /> Now we can use the identity &lt;math&gt;\sum^{n}_{k=0}{n \choose k}=2^{n}&lt;/math&gt;.<br /> So the number of ways is just<br /> &lt;math&gt;2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=\boxed{770}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1988|before=First Question|num-a=2}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_13&diff=62110 1987 AIME Problems/Problem 13 2014-05-25T03:41:35Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> A given [[sequence]] &lt;math&gt;r_1, r_2, \dots, r_n&lt;/math&gt; of [[distinct]] [[real number]]s can be put in [[ascending]] order by means of one or more &quot;bubble passes&quot;. A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term, &lt;math&gt;r_n&lt;/math&gt;, with its current predecessor and exchanging them if and only if the last term is smaller. <br /> <br /> The example below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass. The numbers compared at each step are underlined.<br /> &lt;center&gt;&lt;math&gt;\underline{1 \quad 9} \quad 8 \quad 7&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;1 \quad {}\underline{9 \quad 8} \quad 7&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;1 \quad 8 \quad \underline{9 \quad 7}&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;1 \quad 8 \quad 7 \quad 9&lt;/math&gt;&lt;/center&gt;<br /> Suppose that &lt;math&gt;n = 40&lt;/math&gt;, and that the terms of the initial sequence &lt;math&gt;r_1, r_2, \dots, r_{40}&lt;/math&gt; are distinct from one another and are in random order. Let &lt;math&gt;p/q&lt;/math&gt;, in lowest terms, be the [[probability]] that the number that begins as &lt;math&gt;r_{20}&lt;/math&gt; will end up, after one bubble pass, in the &lt;math&gt;30^{\mbox{th}}&lt;/math&gt; place. Find &lt;math&gt;p + q&lt;/math&gt;.<br /> == Solution ==<br /> If any of &lt;math&gt;r_1, \ldots, r_{19}&lt;/math&gt; is larger than &lt;math&gt;r_{20}&lt;/math&gt;, one of these numbers will be compared with &lt;math&gt;r_{20}&lt;/math&gt; on the 19th step of the first bubble pass and &lt;math&gt;r_{20}&lt;/math&gt; will be moved back to the 19th position. Thus, &lt;math&gt;r_{20}&lt;/math&gt; must be the largest of the first 20 terms. In addition, &lt;math&gt;r_{20}&lt;/math&gt; must be larger than &lt;math&gt;r_{21}, r_{22}, \ldots, r_{30}&lt;/math&gt; but smaller than &lt;math&gt;r_{31}&lt;/math&gt; in order that it move right to the 30th position but then not continue moving right to the 31st.<br /> <br /> Thus, our problem can be restated: What is the probability that in a sequence of 31 distinct real numbers, the largest is in position 31 and the second-largest is in position 20 (the other 29 numbers are irrelevant)?<br /> <br /> This is much easier to solve: there are &lt;math&gt;31!&lt;/math&gt; ways to order the first thirty-one numbers and &lt;math&gt;29!&lt;/math&gt; ways to arrange them so that the largest number is in the 31st position and the second-largest is in the 20th. This gives us a desired probability of &lt;math&gt;\frac{29!}{31!} = \frac{1}{31\cdot 30} = \frac{1}{930}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{931}&lt;/math&gt;.<br /> <br /> == See also ==<br /> * [[Factorial]]<br /> {{AIME box|year=1987|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_13&diff=62109 1987 AIME Problems/Problem 13 2014-05-25T03:40:38Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> A given [[sequence]] &lt;math&gt;r_1, r_2, \dots, r_n&lt;/math&gt; of [[distinct]] [[real number]]s can be put in [[ascending]] order by means of one or more &quot;bubble passes&quot;. A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term, &lt;math&gt;r_n&lt;/math&gt;, with its current predecessor and exchanging them if and only if the last term is smaller. <br /> <br /> The example below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass. The numbers compared at each step are underlined.<br /> &lt;center&gt;&lt;math&gt;\underline{1 \quad 9} \quad 8 \quad 7&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;1 \quad {}\underline{9 \quad 8} \quad 7&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;1 \quad 8 \quad \underline{9 \quad 7}&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;1 \quad 8 \quad 7 \quad 9&lt;/math&gt;&lt;/center&gt;<br /> Suppose that &lt;math&gt;n = 40&lt;/math&gt;, and that the terms of the initial sequence &lt;math&gt;r_1, r_2, \dots, r_{40}&lt;/math&gt; are distinct from one another and are in random order. Let &lt;math&gt;p/q&lt;/math&gt;, in lowest terms, be the [[probability]] that the number that begins as &lt;math&gt;r_{20}&lt;/math&gt; will end up, after one bubble pass, in the &lt;math&gt;30^{\mbox{th}}&lt;/math&gt; place. Find &lt;math&gt;p + q&lt;/math&gt;.<br /> == Solution ==<br /> If any of &lt;math&gt;r_1, \ldots, r_{19}&lt;/math&gt; is larger than &lt;math&gt;r_{20}&lt;/math&gt;, one of these numbers will be compared with &lt;math&gt;r_{20}&lt;/math&gt; on the 19th step of the first bubble pass and &lt;math&gt;r_{20}&lt;/math&gt; will be moved back to the 19th position. Thus, &lt;math&gt;r_{20}&lt;/math&gt; must be the largest of the first 20 terms. In addition, &lt;math&gt;r_{20}&lt;/math&gt; must be larger than &lt;math&gt;r_{21}, r_{22}, \ldots, r_{30}&lt;/math&gt; but smaller than &lt;math&gt;r_{31}&lt;/math&gt; in order that it move right to the 30th position but then not continue moving right to the 31st.<br /> <br /> Thus, our problem can be restated: What is the probability that in a sequence of 31 distinct real numbers, the largest is in position 31 and the second-largest is in position 20 (the other 29 numbers are irrelevant)?<br /> <br /> This is much easier to solve: there are &lt;math&gt;31!&lt;/math&gt; ways to order the first thirty-one numbers and &lt;math&gt;29!&lt;/math&gt; ways to arrange them so that the largest number is in the 31st position and the second-largest is in the 20th. This gives us a desired probability of &lt;math&gt;\frac{29!}{31!} = \frac{1}{31\cdot 30} = \frac{1}{930}&lt;/math&gt;, so the answer is &lt;math&gt;931&lt;/math&gt;.<br /> <br /> == See also ==<br /> * [[Factorial]]<br /> {{AIME box|year=1987|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2014_USAJMO_Problems/Problem_1&diff=61823 2014 USAJMO Problems/Problem 1 2014-04-30T03:52:38Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; be real numbers greater than or equal to &lt;math&gt;1&lt;/math&gt;. Prove that &lt;cmath&gt;\min{\left (\frac{10a^2-5a+1}{b^2-5b+1},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )}\leq abc &lt;/cmath&gt;<br /> ==Solution==<br /> Since &lt;math&gt;(a-1)^5\geqslant 0&lt;/math&gt;,<br /> &lt;cmath&gt;a^5-5a^4+10a^3-10a^2+5a-1\geqslant 0&lt;/cmath&gt;<br /> or<br /> &lt;cmath&gt;10a^2-5a+1\leqslant a^3(a^2-5a+10)&lt;/cmath&gt;<br /> Since &lt;math&gt;a^2-5a+10=\left( a-\dfrac{5}{2}\right)^2 +\dfrac{15}{4}\geqslant 0&lt;/math&gt;,<br /> &lt;cmath&gt; \frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3 &lt;/cmath&gt;<br /> Also note that &lt;math&gt;10a^2-5a+1=10\left( a-\dfrac{1}{4}\right)^2+\dfrac{3}{8}\geqslant 0&lt;/math&gt;,<br /> We conclude<br /> &lt;cmath&gt;0\leqslant\frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3&lt;/cmath&gt;<br /> Similarly, <br /> &lt;cmath&gt;0\leqslant\frac{10b^2-5b+1}{b^2-5b+10}\leqslant b^3&lt;/cmath&gt;<br /> &lt;cmath&gt;0\leqslant\frac{10c^2-5c+1}{c^2-5c+10}\leqslant c^3&lt;/cmath&gt;<br /> So &lt;cmath&gt;\left(\frac{10a^2-5a+1}{a^2-5a+10}\right)\left(\frac{10b^2-5b+1}{b^2-5b+10}\right)\left(\frac{10c^2-5c+1}{c^2-5c+10}\right)\leqslant a^3b^3c^3&lt;/cmath&gt;<br /> or<br /> &lt;cmath&gt;\left(\frac{10a^2-5a+1}{b^2-5b+10}\right)\left(\frac{10b^2-5b+1}{c^2-5c+10}\right)\left(\frac{10c^2-5c+1}{a^2-5a+10}\right) \leqslant(abc)^3&lt;/cmath&gt;<br /> Therefore,<br /> &lt;cmath&gt; \min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )\leq abc. &lt;/cmath&gt;</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2014_USAJMO_Problems/Problem_1&diff=61822 2014 USAJMO Problems/Problem 1 2014-04-30T03:52:10Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; be real numbers greater than or equal to &lt;math&gt;1&lt;/math&gt;. Prove that &lt;cmath&gt;\min{\left (\frac{10a^2-5a+1}{b^2-5b+1},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )}\leq abc &lt;/cmath&gt;<br /> ==Solution==<br /> Since &lt;math&gt;(a-1)^5\geqslant 0&lt;/math&gt;,<br /> &lt;cmath&gt;a^5-5a^4+10a^3-10a^2+5a-1\geqslant 0&lt;/cmath&gt;<br /> or<br /> &lt;cmath&gt;10a^2-5a+1\leqslant a^3(a^2-5a+10)&lt;/cmath&gt;<br /> Since &lt;math&gt;a^2-5a+10=\left( a-\dfrac{5}{2}\right)^2 +\dfrac{15}{4}\geqslant 0&lt;/math&gt;,<br /> &lt;cmath&gt; \frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3 &lt;/cmath&gt;<br /> Also note that &lt;math&gt;10a^2-5a+1=10\left( a-\dfrac{1}{4}\right)+\dfrac{3}{8}\geqslant 0&lt;/math&gt;,<br /> We conclude<br /> &lt;cmath&gt;0\leqslant\frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3&lt;/cmath&gt;<br /> Similarly, <br /> &lt;cmath&gt;0\leqslant\frac{10b^2-5b+1}{b^2-5b+10}\leqslant b^3&lt;/cmath&gt;<br /> &lt;cmath&gt;0\leqslant\frac{10c^2-5c+1}{c^2-5c+10}\leqslant c^3&lt;/cmath&gt;<br /> So &lt;cmath&gt;\left(\frac{10a^2-5a+1}{a^2-5a+10}\right)\left(\frac{10b^2-5b+1}{b^2-5b+10}\right)\left(\frac{10c^2-5c+1}{c^2-5c+10}\right)\leqslant a^3b^3c^3&lt;/cmath&gt;<br /> or<br /> &lt;cmath&gt;\left(\frac{10a^2-5a+1}{b^2-5b+10}\right)\left(\frac{10b^2-5b+1}{c^2-5c+10}\right)\left(\frac{10c^2-5c+1}{a^2-5a+10}\right) \leqslant(abc)^3&lt;/cmath&gt;<br /> Therefore,<br /> &lt;cmath&gt; \min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )\leq abc. &lt;/cmath&gt;</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2003_USAMO_Problems/Problem_2&diff=61818 2003 USAMO Problems/Problem 2 2014-04-30T00:50:24Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> A convex polygon &lt;math&gt; \mathcal{P} &lt;/math&gt; in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon &lt;math&gt; \mathcal{P} &lt;/math&gt; are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.<br /> <br /> == Solution ==<br /> <br /> When &lt;math&gt; \mathcal{P} &lt;/math&gt; is a triangle, the problem is trivial. Otherwise, it is sufficient to prove that any two diagonals of the polygon cut each other into rational lengths. Let two diagonals which intersect at a point &lt;math&gt;P &lt;/math&gt; within the polygon be &lt;math&gt;AC &lt;/math&gt; and &lt;math&gt;BD &lt;/math&gt;. Since &lt;math&gt;ABCD &lt;/math&gt; is a convex quadrilateral with sides and diagonals of rational length, we consider it in isolation.<br /> <br /> By the [[Law of Cosines]], &lt;math&gt; \cos BAC = \frac{AB^2 + CA^2 - BC^2}{2AB \cdot CA} &lt;/math&gt;, which is rational. Similarly, &lt;math&gt;\cos CAD &lt;/math&gt; is rational, as well as &lt;math&gt;\cos BAD = \cos (BAC + CAD) = \cos BAC \cos CAD - \sin BAC \sin CAD &lt;/math&gt;. It follows that &lt;math&gt;\sin BAC \sin CAD &lt;/math&gt; is rational. Since &lt;math&gt;\sin^2 CAD = 1 - \cos^2 CAD &lt;/math&gt; is rational, this means that &lt;math&gt; \frac{\sin BAC \sin CAD}{\sin^2 CAD} = \frac{\sin BAC}{\sin CAD} = \frac{\sin BAP}{\sin PAD} &lt;/math&gt; is rational. This implies that &lt;math&gt; \frac{ AB \sin BAP}{ AD \sin PAD} = \frac{\frac{1}{2} AB \cdot AP \sin BAP}{\frac{1}{2} AP \cdot AD \sin PAD} = \frac{[BAP]}{[PAD]} = \frac{BP}{PD} &lt;/math&gt; is rational. Define &lt;math&gt;r&lt;/math&gt; to be equal to &lt;math&gt;\frac{BP}{PD} &lt;/math&gt;. We know that &lt;math&gt;\frac{BP}{PD} &lt;/math&gt; is rational; hence &lt;math&gt;r&lt;/math&gt; is rational. We also have &lt;math&gt;(1+r)PD = (1+\frac{BP}{PD} )PD = PD + BP = BD &lt;/math&gt;, which is, of course, rational. It follows that &lt;math&gt;BP &lt;/math&gt; and &lt;math&gt;PD &lt;/math&gt; both have rational length, as desired.<br /> <br /> <br /> {{alternate solutions}}<br /> <br /> == See also ==<br /> {{USAMO newbox|year=2003|num-b=1|num-a=3}}<br /> <br /> <br /> [[Category:Olympiad Geometry Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=2003_USAMO_Problems/Problem_2&diff=61817 2003 USAMO Problems/Problem 2 2014-04-30T00:47:50Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> A convex polygon &lt;math&gt; \mathcal{P} &lt;/math&gt; in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon &lt;math&gt; \mathcal{P} &lt;/math&gt; are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.<br /> <br /> == Solution ==<br /> <br /> When &lt;math&gt; \mathcal{P} &lt;/math&gt; is a triangle, the problem is trivial. Otherwise, it is sufficient to prove that any two diagonals of the polygon cut each other into rational lengths. Let two diagonals which intersect at a point &lt;math&gt;P &lt;/math&gt; within the polygon be &lt;math&gt;AC &lt;/math&gt; and &lt;math&gt;BD &lt;/math&gt;. Since &lt;math&gt;ABCD &lt;/math&gt; is a convex quadrilateral with sides and diagonals of rational length, we consider it in isolation.<br /> <br /> By the [[Law of Cosines]], &lt;math&gt; \cos BAC = \frac{AB^2 + CA^2 - BC^2}{2AB \cdot CA} &lt;/math&gt;, which is rational. Similarly, &lt;math&gt;\cos CAD &lt;/math&gt; is rational, as well as &lt;math&gt;\cos BAD = \cos (BAC + CAD) = \cos BAC \cos CAD - \sin BAC \sin CAD &lt;/math&gt;. It follows that &lt;math&gt;\sin BAC \sin CAD &lt;/math&gt; is rational. Since &lt;math&gt;\sin^2 CAD = 1 - \cos^2 CAD &lt;/math&gt; is rational, this means that &lt;math&gt; \frac{\sin BAC \sin CAD}{\sin^2 CAD} = \frac{\sin BAC}{\sin CAD} = \frac{\sin BAP}{\sin PAD} &lt;/math&gt; is rational. This implies that &lt;math&gt; \frac{ AB \sin BAP}{ AD \sin PAD} = \frac{\frac{1}{2} AB \cdot AP \sin BAP}{\frac{1}{2} AP \cdot AD \sin PAD} = \frac{[BAP]}{[PAD]} = \frac{BP}{PD} &lt;/math&gt; is rational. This means that for some rational number &lt;math&gt;r &lt;/math&gt;, &lt;math&gt;(1+r)PD = BP + PD = BD &lt;/math&gt;, which is, of course, rational. It follows that &lt;math&gt;BP &lt;/math&gt; and &lt;math&gt;PD &lt;/math&gt; both have rational length, as desired.<br /> <br /> <br /> {{alternate solutions}}<br /> <br /> == See also ==<br /> {{USAMO newbox|year=2003|num-b=1|num-a=3}}<br /> <br /> <br /> [[Category:Olympiad Geometry Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=Talk:2003_USAMO_Problems/Problem_1&diff=61813 Talk:2003 USAMO Problems/Problem 1 2014-04-30T00:34:19Z <p>Mathcool2009: </p> <hr /> <div>\\By the way Solution 2 is incorrect--subtracting b from a where both a and b have the same number of digits and have only odd digits does not necessarily result in a difference with all odd digits. Consider 73 minus 35. Both have only odd digits, and yet 73 - 35 = 38, which does not have only odd digits.//--[[User:Mathcool2009|Mathcool2009]] 20:28, 29 April 2014 (EDT)</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=Talk:2003_USAMO_Problems/Problem_1&diff=61812 Talk:2003 USAMO Problems/Problem 1 2014-04-30T00:28:36Z <p>Mathcool2009: Created page with &quot;\\By the way Solution 2 is incorrect--~~~~&quot;</p> <hr /> <div>\\By the way Solution 2 is incorrect--[[User:Mathcool2009|Mathcool2009]] 20:28, 29 April 2014 (EDT)</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_12&diff=58574 1986 AIME Problems/Problem 12 2014-01-03T19:36:10Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let the sum of a set of numbers be the sum of its elements. Let &lt;math&gt;S&lt;/math&gt; be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of &lt;math&gt;S&lt;/math&gt; have the same sum. What is the largest sum a set &lt;math&gt;S&lt;/math&gt; with these properties can have? <br /> == Solution ==<br /> The maximum is &lt;math&gt;\boxed{061}&lt;/math&gt;, attained when &lt;math&gt;S=\{ 15,14,13,11,8\}&lt;/math&gt;. We must now prove that no such set has sum at least 62. Suppose such a set &lt;math&gt;S&lt;/math&gt; existed. Then &lt;math&gt;S&lt;/math&gt; must have more than 4 elements, otherwise its sum would be at most &lt;math&gt;15+14+13+12=54&lt;/math&gt; if it had 4 elements.<br /> <br /> But also, &lt;math&gt;S&lt;/math&gt; can't have at least 6 elements. To see why, note that at least &lt;math&gt;1 + 6 + \dbinom{6}{2} + \dbinom{6}{3} + \dbinom{6}{4}=57&lt;/math&gt; of its subsets have at most four elements (the number of subsets with no elements plus the number of subsets with one element plus the number of subsets with two elements plus the number of subsets with three elements plus the number of subsets with four elements), so each of them have sum at most 54. By the Pigeonhole Principle, two of these subsets would have the same sum, a contradiction to the givens.<br /> <br /> <br /> Thus, &lt;math&gt;S&lt;/math&gt; must have exactly 5 elements. &lt;math&gt;S&lt;/math&gt; contains both 15 and 14 (otherwise its sum is at most &lt;math&gt;10+11+12+13+15=61&lt;/math&gt;). It follows that &lt;math&gt;S&lt;/math&gt; cannot contain both &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;a-1&lt;/math&gt; for any &lt;math&gt;a\leq 13&lt;/math&gt;, or the subsets &lt;math&gt;\{a,14\}&lt;/math&gt; and &lt;math&gt;\{a-1,15\}&lt;/math&gt; would have the same sum. So now &lt;math&gt;S&lt;/math&gt; must contain 13 (otherwise its sum is at most &lt;math&gt;15+14+12+10+8=59&lt;/math&gt;), and &lt;math&gt;S&lt;/math&gt; cannot contain 12, or the subsets &lt;math&gt;\{12,15\}&lt;/math&gt; and &lt;math&gt;\{13,14\}&lt;/math&gt; would have the same sum.<br /> <br /> Now the only way &lt;math&gt;S&lt;/math&gt; could have sum at least &lt;math&gt;62=15+14+13+11+9&lt;/math&gt; would be if &lt;math&gt;S=\{ 15,14,13,11,9\}&lt;/math&gt;. But the subsets &lt;math&gt;\{9,15\}&lt;/math&gt; and &lt;math&gt;\{11,13\}&lt;/math&gt; have the same sum, so this set does not work, a contradiction to the givens. Therefore no &lt;math&gt;S&lt;/math&gt; with sum at least 62 is possible and 61 is indeed the maximum.<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_12&diff=58573 1986 AIME Problems/Problem 12 2014-01-03T19:34:24Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let the sum of a set of numbers be the sum of its elements. Let &lt;math&gt;S&lt;/math&gt; be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of &lt;math&gt;S&lt;/math&gt; have the same sum. What is the largest sum a set &lt;math&gt;S&lt;/math&gt; with these properties can have? <br /> == Solution ==<br /> The maximum is &lt;math&gt;\boxed{061}&lt;/math&gt;, attained when &lt;math&gt;S=\{ 15,14,13,11,8\}&lt;/math&gt;. We must now prove that no such set has sum at least 62. Suppose such a set &lt;math&gt;S&lt;/math&gt; existed. Then &lt;math&gt;S&lt;/math&gt; must have more than 4 elements, otherwise its sum would be at most &lt;math&gt;15+14+13+12=54&lt;/math&gt; if it had 4 elements.<br /> <br /> But also, &lt;math&gt;S&lt;/math&gt; can't have at least 6 elements. To see why, note that &lt;math&gt;1 + 6 + \dbinom{6}{2} + \dbinom{6}{3} + \dbinom{6}{4}=57&lt;/math&gt; of its subsets have at most four elements (the number of subsets with no elements plus the number of subsets with one element plus the number of subsets with two elements plus the number of subsets with three elements plus the number of subsets with four elements), so each of them have sum at most 54. By the Pigeonhole Principle, two of these subsets would have the same sum, a contradiction to the givens.<br /> <br /> <br /> Thus, &lt;math&gt;S&lt;/math&gt; must have exactly 5 elements. &lt;math&gt;S&lt;/math&gt; contains both 15 and 14 (otherwise its sum is at most &lt;math&gt;10+11+12+13+15=61&lt;/math&gt;). It follows that &lt;math&gt;S&lt;/math&gt; cannot contain both &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;a-1&lt;/math&gt; for any &lt;math&gt;a\leq 13&lt;/math&gt;, or the subsets &lt;math&gt;\{a,14\}&lt;/math&gt; and &lt;math&gt;\{a-1,15\}&lt;/math&gt; would have the same sum. So now &lt;math&gt;S&lt;/math&gt; must contain 13 (otherwise its sum is at most &lt;math&gt;15+14+12+10+8=59&lt;/math&gt;), and &lt;math&gt;S&lt;/math&gt; cannot contain 12, or the subsets &lt;math&gt;\{12,15\}&lt;/math&gt; and &lt;math&gt;\{13,14\}&lt;/math&gt; would have the same sum.<br /> <br /> Now the only way &lt;math&gt;S&lt;/math&gt; could have sum at least &lt;math&gt;62=15+14+13+11+9&lt;/math&gt; would be if &lt;math&gt;S=\{ 15,14,13,11,9\}&lt;/math&gt;. But the subsets &lt;math&gt;\{9,15\}&lt;/math&gt; and &lt;math&gt;\{11,13\}&lt;/math&gt; have the same sum, so this set does not work, a contradiction to the givens. Therefore no &lt;math&gt;S&lt;/math&gt; with sum at least 62 is possible and 61 is indeed the maximum.<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_12&diff=58572 1986 AIME Problems/Problem 12 2014-01-03T19:34:09Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let the sum of a set of numbers be the sum of its elements. Let &lt;math&gt;S&lt;/math&gt; be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of &lt;math&gt;S&lt;/math&gt; have the same sum. What is the largest sum a set &lt;math&gt;S&lt;/math&gt; with these properties can have? <br /> == Solution ==<br /> The maximum is &lt;math&gt;61&lt;/math&gt;, attained when &lt;math&gt;S=\{ 15,14,13,11,8\}&lt;/math&gt;. We must now prove that no such set has sum at least 62. Suppose such a set &lt;math&gt;S&lt;/math&gt; existed. Then &lt;math&gt;S&lt;/math&gt; must have more than 4 elements, otherwise its sum would be at most &lt;math&gt;15+14+13+12=54&lt;/math&gt; if it had 4 elements.<br /> <br /> But also, &lt;math&gt;S&lt;/math&gt; can't have at least 6 elements. To see why, note that &lt;math&gt;1 + 6 + \dbinom{6}{2} + \dbinom{6}{3} + \dbinom{6}{4}=57&lt;/math&gt; of its subsets have at most four elements (the number of subsets with no elements plus the number of subsets with one element plus the number of subsets with two elements plus the number of subsets with three elements plus the number of subsets with four elements), so each of them have sum at most 54. By the Pigeonhole Principle, two of these subsets would have the same sum, a contradiction to the givens.<br /> <br /> <br /> Thus, &lt;math&gt;S&lt;/math&gt; must have exactly 5 elements. &lt;math&gt;S&lt;/math&gt; contains both 15 and 14 (otherwise its sum is at most &lt;math&gt;10+11+12+13+15=61&lt;/math&gt;). It follows that &lt;math&gt;S&lt;/math&gt; cannot contain both &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;a-1&lt;/math&gt; for any &lt;math&gt;a\leq 13&lt;/math&gt;, or the subsets &lt;math&gt;\{a,14\}&lt;/math&gt; and &lt;math&gt;\{a-1,15\}&lt;/math&gt; would have the same sum. So now &lt;math&gt;S&lt;/math&gt; must contain 13 (otherwise its sum is at most &lt;math&gt;15+14+12+10+8=59&lt;/math&gt;), and &lt;math&gt;S&lt;/math&gt; cannot contain 12, or the subsets &lt;math&gt;\{12,15\}&lt;/math&gt; and &lt;math&gt;\{13,14\}&lt;/math&gt; would have the same sum.<br /> <br /> Now the only way &lt;math&gt;S&lt;/math&gt; could have sum at least &lt;math&gt;62=15+14+13+11+9&lt;/math&gt; would be if &lt;math&gt;S=\{ 15,14,13,11,9\}&lt;/math&gt;. But the subsets &lt;math&gt;\{9,15\}&lt;/math&gt; and &lt;math&gt;\{11,13\}&lt;/math&gt; have the same sum, so this set does not work, a contradiction to the givens. Therefore no &lt;math&gt;S&lt;/math&gt; with sum at least 62 is possible and 61 is indeed the maximum.<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_12&diff=58571 1986 AIME Problems/Problem 12 2014-01-03T19:28:58Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let the sum of a set of numbers be the sum of its elements. Let &lt;math&gt;S&lt;/math&gt; be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of &lt;math&gt;S&lt;/math&gt; have the same sum. What is the largest sum a set &lt;math&gt;S&lt;/math&gt; with these properties can have? <br /> == Solution ==<br /> The maximum is &lt;math&gt;61&lt;/math&gt;, attained when &lt;math&gt;S=\{ 15,14,13,11,8\}&lt;/math&gt;. We must now prove that no such set has sum at least 62. Suppose such a set &lt;math&gt;S&lt;/math&gt; existed. Then &lt;math&gt;S&lt;/math&gt; must have more than 4 elements, otherwise its sum would be at most &lt;math&gt;15+14+13+12=54&lt;/math&gt; if it had 4 elements.<br /> <br /> But also, &lt;math&gt;S&lt;/math&gt; can't have at least 6 elements. To see why, note that &lt;math&gt;1 + 6 + \dbinom{6}{2} + \dbinom{6}{3} + \dbinom{6}{4}=57&lt;/math&gt; of its subsets have at most four elements (the number of subsets with no elements plus the number of subsets with one element plus the number of subsets with two elements plus the number of subsets with three elements plus the number of subsets with four elements), so each of them have sum at most 54. By the Pigeonhole Principle, two of these subsets would have the same sum, a contradiction to the givens.<br /> <br /> <br /> Thus, &lt;math&gt;S&lt;/math&gt; must have exactly 5 elements. &lt;math&gt;S&lt;/math&gt; contains both 15 and 14 (otherwise its sum is at most &lt;math&gt;10+11+12+13+15=61&lt;/math&gt;). It follows that &lt;math&gt;S&lt;/math&gt; cannot contain both &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;a-1&lt;/math&gt; for any &lt;math&gt;a\leq 13&lt;/math&gt;. So now &lt;math&gt;S&lt;/math&gt; must contain 13 (otherwise its sum is at most &lt;math&gt;15+14+12+10+8=59&lt;/math&gt;), and &lt;math&gt;S&lt;/math&gt; cannot contain 12.<br /> <br /> Now the only way &lt;math&gt;S&lt;/math&gt; could have sum at least &lt;math&gt;62=15+14+13+11+9&lt;/math&gt; would be if &lt;math&gt;S=\{ 15,14,13,11,9\}&lt;/math&gt;. But &lt;math&gt;15+9=13+11&lt;/math&gt; so this set does not work, a contradiction. Therefore 61 is indeed the maximum.<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_15&diff=58469 1985 AIME Problems/Problem 15 2014-01-01T03:50:12Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> Three 12 cm &lt;math&gt;\times&lt;/math&gt;12 cm [[square (geometry) | squares]] are each cut into two pieces &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, as shown in the first figure below, by joining the [[midpoint]]s of two adjacent sides. These six pieces are then attached to a [[regular polygon | regular]] [[hexagon]], as shown in the second figure, so as to fold into a [[polyhedron]]. What is the [[volume]] (in &lt;math&gt;\mathrm{cm}^3&lt;/math&gt;) of this polyhedron?<br /> <br /> [[Image:AIME 1985 Problem 15.png]]<br /> <br /> == Solution ==<br /> Note that gluing two of the given polyhedra together along a hexagonal face (rotated &lt;math&gt;60^\circ&lt;/math&gt; from each other) yields a [[cube (geometry) | cube]], so the volume is &lt;math&gt;\frac12 \cdot 12^3 = 864&lt;/math&gt;.<br /> <br /> Image: [[Media:AoPS_AIME_1985.png]]<br /> <br /> == See also ==<br /> {{AIME box|year=1985|num-b=14|after=Last Question}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=File:AoPS_AIME_1985.png&diff=58468 File:AoPS AIME 1985.png 2014-01-01T03:47:32Z <p>Mathcool2009: </p> <hr /> <div></div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_14&diff=58467 1985 AIME Problems/Problem 14 2014-01-01T03:16:42Z <p>Mathcool2009: /* Solution */</p> <hr /> <div>== Problem ==<br /> In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded 1 point, the loser got 0 points, and each of the two players earned 1/2 point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?<br /> == Solution ==<br /> Let us suppose for convenience that there were &lt;math&gt;n + 10&lt;/math&gt; players over all. Among the &lt;math&gt;n&lt;/math&gt; players not in the weakest 10 there were &lt;math&gt;n \choose 2&lt;/math&gt; games played and thus &lt;math&gt;n \choose 2&lt;/math&gt; points earned. By the givens, this means that these &lt;math&gt;n&lt;/math&gt; players also earned &lt;math&gt;n \choose 2&lt;/math&gt; points against our weakest 10. Now, the 10 weakest players playing amongst themselves played &lt;math&gt;{10 \choose 2} = 45&lt;/math&gt; games and so earned 45 points playing each other. Then they also earned 45 points playing against the stronger &lt;math&gt;n&lt;/math&gt; players. Since every point earned falls into one of these categories, It follows that the total number of points earned was &lt;math&gt;2{n \choose 2} + 90 = n^2 - n + 90&lt;/math&gt;. However, there was one point earned per game, and there were a total of &lt;math&gt;{n + 10 \choose 2} = \frac{(n + 10)(n + 9)}{2}&lt;/math&gt; games played and thus &lt;math&gt;\frac{(n + 10)(n + 9)}{2}&lt;/math&gt; points earned. So we have &lt;math&gt;n^2 -n + 90 = \frac{(n + 10)(n + 9)}{2}&lt;/math&gt; so &lt;math&gt;2n^2 - 2n + 180 = n^2 + 19n + 90&lt;/math&gt; and &lt;math&gt;n^2 -21n + 90 = 0&lt;/math&gt; and &lt;math&gt;n = 6&lt;/math&gt; or &lt;math&gt;n = 15&lt;/math&gt;. Now, note that the top &lt;math&gt;n&lt;/math&gt; players got &lt;math&gt;n(n - 1)&lt;/math&gt; points in total (by our previous calculation) for an average of &lt;math&gt;n - 1&lt;/math&gt;, while the bottom 10 got 90 points total, for an average of 9. Thus we must have &lt;math&gt;n &gt; 10&lt;/math&gt;, so &lt;math&gt;n = 15&lt;/math&gt; and the answer is &lt;math&gt;15 + 10 = \boxed{025}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1985|num-b=13|num-a=15}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_15&diff=58443 1984 AIME Problems/Problem 15 2013-12-30T03:00:33Z <p>Mathcool2009: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Determine &lt;math&gt;w^2+x^2+y^2+z^2&lt;/math&gt; if<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt; \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 &lt;/math&gt;&lt;/div&gt;<br /> <br /> == Solution 1 ==<br /> Rewrite the system of equations as &lt;math&gt; \frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1. &lt;/math&gt; This equation is satisfied when &lt;math&gt;t = 4,16,36,64&lt;/math&gt;, as then the equation is equivalent to the given equations.<br /> After clearing fractions, for each of the values &lt;math&gt;t=4,16,36,64&lt;/math&gt;, we have the [[equation]] &lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)&lt;/math&gt;. When &lt;math&gt;t=4,16,36,64&lt;/math&gt;, a &lt;math&gt;(t-4)(t-16)(t-36)(t-64)&lt;/math&gt; term can be subtracted from the right-hand side because it equals 0. Thus we have the following [[equation]] which holds for &lt;math&gt;t=4,16,36,64&lt;/math&gt;:<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)&lt;/math&gt;<br /> &lt;math&gt;=(t-1)(t-9)(t-25)(t-49)-(t-4)(t-16)(t-36)(t-64)&lt;/math&gt;<br /> &lt;/div&gt;<br /> <br /> Each side of this equation is a [[polynomial]] in &lt;math&gt;t&lt;/math&gt; of degree at most 3, and they are equal for 4 values of &lt;math&gt;t&lt;/math&gt; (when &lt;math&gt;t=4,16,36,64&lt;/math&gt;). Therefore, the polynomials must be equal for all &lt;math&gt;t&lt;/math&gt;.*<br /> <br /> Now we can plug in &lt;math&gt;t=1&lt;/math&gt; into the polynomial equation. Most terms drop, and we end up with<br /> <br /> &lt;cmath&gt;x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)&lt;/cmath&gt;<br /> <br /> so that<br /> <br /> &lt;cmath&gt;x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}&lt;/cmath&gt;<br /> <br /> Similarly, we can plug in &lt;math&gt;t=9,25,49&lt;/math&gt; and get<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> y^2&amp;=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\<br /> z^2&amp;=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}\\<br /> w^2&amp;=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}&lt;/cmath&gt;<br /> <br /> Now adding them up,<br /> <br /> &lt;cmath&gt;\begin{align*}z^2+w^2&amp;=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\<br /> x^2+y^2&amp;=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}\end{align*}&lt;/cmath&gt;<br /> <br /> with a sum of<br /> <br /> &lt;cmath&gt;\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.&lt;/cmath&gt;<br /> <br /> /*Lengthy proof that any two cubic polynomials in &lt;math&gt;t&lt;/math&gt; which are equal at 4 values of &lt;math&gt;t&lt;/math&gt; are themselves equivalent:<br /> Let the two polynomials be &lt;math&gt;A(t)&lt;/math&gt; and &lt;math&gt;B(t)&lt;/math&gt; and let them be equal at &lt;math&gt;t=a,b,c,d&lt;/math&gt;. Thus we have &lt;math&gt;A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0&lt;/math&gt;. Also the polynomial &lt;math&gt;A(t) - B(t)&lt;/math&gt; is cubic, but it equals 0 at 4 values of &lt;math&gt;t&lt;/math&gt;. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into &lt;math&gt;(t-a)(t-b)(t-c)(t-d)(&lt;/math&gt;some nonzero polynomial&lt;math&gt;)&lt;/math&gt; which would have a degree greater than or equal to 4, contradicting the statement that &lt;math&gt;A(t) - B(t)&lt;/math&gt; is cubic. Because &lt;math&gt;A(t) - B(t) = 0, A(t)&lt;/math&gt; and &lt;math&gt;B(t)&lt;/math&gt; are equivalent and must be equal for all &lt;math&gt;t&lt;/math&gt;.<br /> <br /> '''Post script for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes &lt;math&gt;x^2,y^2,z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; separately before adding them to obtain the final answer is appealing because it gives the individual values of &lt;math&gt;x^2,y^2,z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; which can be plugged into the given equations to check.<br /> <br /> == Solution 2 ==<br /> As in Solution 1, we have <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)&lt;/math&gt;<br /> &lt;math&gt;=(t-4)(t-16)(t-36)(t-64)&lt;/math&gt;<br /> &lt;/div&gt;<br /> Now the coefficient of &lt;math&gt;t^3&lt;/math&gt; on both sides must be equal. Therefore we have &lt;math&gt;1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_15&diff=58442 1984 AIME Problems/Problem 15 2013-12-30T02:58:39Z <p>Mathcool2009: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Determine &lt;math&gt;w^2+x^2+y^2+z^2&lt;/math&gt; if<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt; \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 &lt;/math&gt;&lt;/div&gt;<br /> <br /> == Solution 1 ==<br /> Rewrite the system of equations as &lt;math&gt; \frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1. &lt;/math&gt; This equation is satisfied when &lt;math&gt;t = 4,16,36,64&lt;/math&gt;, as then the equation is equivalent to the given equations.<br /> After clearing fractions, for each of the values &lt;math&gt;t=4,16,36,64&lt;/math&gt;, we have the [[equation]] &lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)&lt;/math&gt;. When &lt;math&gt;t=4,16,36,64&lt;/math&gt;, a &lt;math&gt;(t-4)(t-16)(t-36)(t-64)&lt;/math&gt; term can be subtracted from the right-hand side because it equals 0. Thus we have the following [[equation]] which holds for &lt;math&gt;t=4,16,36,64&lt;/math&gt;:<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)&lt;/math&gt;<br /> &lt;math&gt;=(t-1)(t-9)(t-25)(t-49)-(t-4)(t-16)(t-36)(t-64)&lt;/math&gt;<br /> &lt;/div&gt;<br /> <br /> Each side of this equation is a [[polynomial]] in &lt;math&gt;t&lt;/math&gt; of degree at most 3, and they are equal for 4 values of &lt;math&gt;t&lt;/math&gt; (when &lt;math&gt;t=4,16,36,64&lt;/math&gt;). Therefore, the polynomials must be equal.*<br /> <br /> Now we can plug in &lt;math&gt;t=1&lt;/math&gt; into the polynomial equation. Most terms drop, and we end up with<br /> <br /> &lt;cmath&gt;x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)&lt;/cmath&gt;<br /> <br /> so that<br /> <br /> &lt;cmath&gt;x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}&lt;/cmath&gt;<br /> <br /> Similarly, we can plug in &lt;math&gt;t=9,25,49&lt;/math&gt; and get<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> y^2&amp;=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\<br /> z^2&amp;=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}\\<br /> w^2&amp;=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}&lt;/cmath&gt;<br /> <br /> Now adding them up,<br /> <br /> &lt;cmath&gt;\begin{align*}z^2+w^2&amp;=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\<br /> x^2+y^2&amp;=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}\end{align*}&lt;/cmath&gt;<br /> <br /> with a sum of<br /> <br /> &lt;cmath&gt;\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.&lt;/cmath&gt;<br /> <br /> /*Lengthy proof that any two cubic polynomials in &lt;math&gt;t&lt;/math&gt; which are equal at 4 values of &lt;math&gt;t&lt;/math&gt; are themselves equivalent:<br /> Let the two polynomials be &lt;math&gt;A(t)&lt;/math&gt; and &lt;math&gt;B(t)&lt;/math&gt; and let them be equal at &lt;math&gt;t=a,b,c,d&lt;/math&gt;. Thus we have &lt;math&gt;A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0&lt;/math&gt;. Also the polynomial &lt;math&gt;A(t) - B(t)&lt;/math&gt; is cubic, but it equals 0 at 4 values of &lt;math&gt;t&lt;/math&gt;. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into &lt;math&gt;(t-a)(t-b)(t-c)(t-d)(&lt;/math&gt;some nonzero polynomial&lt;math&gt;)&lt;/math&gt; which would have a degree greater than or equal to 4, contradicting the statement that &lt;math&gt;A(t) - B(t)&lt;/math&gt; is cubic.<br /> <br /> '''Post script for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes &lt;math&gt;x^2,y^2,z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; separately before adding them to obtain the final answer is appealing because it gives the individual values of &lt;math&gt;x^2,y^2,z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; which can be plugged into the given equations to check.<br /> <br /> == Solution 2 ==<br /> As in Solution 1, we have <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)&lt;/math&gt;<br /> &lt;math&gt;=(t-4)(t-16)(t-36)(t-64)&lt;/math&gt;<br /> &lt;/div&gt;<br /> Now the coefficient of &lt;math&gt;t^3&lt;/math&gt; on both sides must be equal. Therefore we have &lt;math&gt;1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Mathcool2009 https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_15&diff=58441 1984 AIME Problems/Problem 15 2013-12-30T02:57:54Z <p>Mathcool2009: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Determine &lt;math&gt;w^2+x^2+y^2+z^2&lt;/math&gt; if<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt; \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 &lt;/math&gt;&lt;/div&gt;<br /> <br /> == Solution 1 ==<br /> Rewrite the system of equations as &lt;math&gt; \frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1. &lt;/math&gt; This equation is satisfied when &lt;math&gt;t = 4,16,36,64&lt;/math&gt;, as then the equation is equivalent to the given equations.<br /> After clearing fractions, for each of the values &lt;math&gt;t=4,16,36,64&lt;/math&gt;, we have the [[equation]] &lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)&lt;/math&gt;. When &lt;math&gt;t=4,16,36,64&lt;/math&gt;, a &lt;math&gt;(t-4)(t-16)(t-36)(t-64)&lt;/math&gt; term can be subtracted from the right-hand side because it equals 0. Thus we have the following [[equation]]:<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)&lt;/math&gt;<br /> &lt;math&gt;=(t-1)(t-9)(t-25)(t-49)-(t-4)(t-16)(t-36)(t-64)&lt;/math&gt;<br /> &lt;/div&gt;<br /> <br /> Each side of this equation is a [[polynomial]] in &lt;math&gt;t&lt;/math&gt; of degree at most 3, and they are equal for 4 values of &lt;math&gt;t&lt;/math&gt; (when &lt;math&gt;t=4,16,36,64&lt;/math&gt;). Therefore, the polynomials must be equal.*<br /> <br /> Now we can plug in &lt;math&gt;t=1&lt;/math&gt; into the polynomial equation. Most terms drop, and we end up with<br /> <br /> &lt;cmath&gt;x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)&lt;/cmath&gt;<br /> <br /> so that<br /> <br /> &lt;cmath&gt;x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}&lt;/cmath&gt;<br /> <br /> Similarly, we can plug in &lt;math&gt;t=9,25,49&lt;/math&gt; and get<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> y^2&amp;=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\<br /> z^2&amp;=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}\\<br /> w^2&amp;=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}&lt;/cmath&gt;<br /> <br /> Now adding them up,<br /> <br /> &lt;cmath&gt;\begin{align*}z^2+w^2&amp;=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\<br /> x^2+y^2&amp;=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}\end{align*}&lt;/cmath&gt;<br /> <br /> with a sum of<br /> <br /> &lt;cmath&gt;\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.&lt;/cmath&gt;<br /> <br /> /*Lengthy proof that any two cubic polynomials in &lt;math&gt;t&lt;/math&gt; which are equal at 4 values of &lt;math&gt;t&lt;/math&gt; are themselves equivalent:<br /> Let the two polynomials be &lt;math&gt;A(t)&lt;/math&gt; and &lt;math&gt;B(t)&lt;/math&gt; and let them be equal at &lt;math&gt;t=a,b,c,d&lt;/math&gt;. Thus we have &lt;math&gt;A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0&lt;/math&gt;. Also the polynomial &lt;math&gt;A(t) - B(t)&lt;/math&gt; is cubic, but it equals 0 at 4 values of &lt;math&gt;t&lt;/math&gt;. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into &lt;math&gt;(t-a)(t-b)(t-c)(t-d)(&lt;/math&gt;some nonzero polynomial&lt;math&gt;)&lt;/math&gt; which would have a degree greater than or equal to 4, contradicting the statement that &lt;math&gt;A(t) - B(t)&lt;/math&gt; is cubic.<br /> <br /> '''Post script for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes &lt;math&gt;x^2,y^2,z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; separately before adding them to obtain the final answer is appealing because it gives the individual values of &lt;math&gt;x^2,y^2,z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; which can be plugged into the given equations to check.<br /> <br /> == Solution 2 ==<br /> As in Solution 1, we have <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)&lt;/math&gt;<br /> &lt;math&gt;=(t-4)(t-16)(t-36)(t-64)&lt;/math&gt;<br /> &lt;/div&gt;<br /> Now the coefficient of &lt;math&gt;t^3&lt;/math&gt; on both sides must be equal. Therefore we have &lt;math&gt;1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Mathcool2009