https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Pragmatictnt&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-13T20:33:26Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12A_Problems&diff=71004 2014 AMC 12A Problems 2015-07-05T21:37:21Z <p>Pragmatictnt: /* Problem 25 */</p> <hr /> <div>{{AMC12 Problems|year=2014|ab=A}}<br /> <br /> <br /> ==Problem 1==<br /> What is &lt;math&gt;10 \cdot \left(\tfrac{1}{2} + \tfrac{1}{5} + \tfrac{1}{10}\right)^{-1}?&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{25}{2}\qquad\textbf{(D)}\ \frac{170}{3}\qquad\textbf{(E)}\ 170&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> At the theater children get in for half price. The price for &lt;math&gt;5&lt;/math&gt; adult tickets and &lt;math&gt;4&lt;/math&gt; child tickets is &lt;math&gt;24.50&lt;/math&gt;. How much would &lt;math&gt;8&lt;/math&gt; adult tickets and &lt;math&gt;6&lt;/math&gt; child tickets cost?<br /> <br /> &lt;math&gt;\textbf{(A) }35\qquad<br /> \textbf{(B) }38.50\qquad<br /> \textbf{(C) }40\qquad<br /> \textbf{(D) }42\qquad<br /> \textbf{(E) }42.50&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 4==<br /> Suppose that &lt;math&gt;a&lt;/math&gt; cows give &lt;math&gt;b&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days. At this rate, how many gallons of milk will &lt;math&gt;d&lt;/math&gt; cows give in &lt;math&gt;e&lt;/math&gt; days?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 5==<br /> On an algebra quiz, &lt;math&gt;10\%&lt;/math&gt; of the students scored &lt;math&gt;70&lt;/math&gt; points, &lt;math&gt;35\%&lt;/math&gt; scored &lt;math&gt;80&lt;/math&gt; points, &lt;math&gt;30\%&lt;/math&gt; scored &lt;math&gt;90&lt;/math&gt; points, and the rest scored &lt;math&gt;100&lt;/math&gt; points. What is the difference between the mean and median score of the students' scores on this quiz?<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 /> [[2014 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> The difference between a two-digit number and the number obtained by reversing its digits is &lt;math&gt;5&lt;/math&gt; times the sum of the digits of either number. What is the sum of the two digit number and its reverse?<br /> <br /> &lt;math&gt;\textbf{(A) }44\qquad<br /> \textbf{(B) }55\qquad<br /> \textbf{(C) }77\qquad<br /> \textbf{(D) }99\qquad<br /> \textbf{(E) }110&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> The first three terms of a geometric progression are &lt;math&gt;\sqrt 3&lt;/math&gt;, &lt;math&gt;\sqrt3&lt;/math&gt;, and &lt;math&gt;\sqrt3&lt;/math&gt;. What is the fourth term?<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad<br /> \textbf{(B) }\sqrt3\qquad<br /> \textbf{(C) }\sqrt3\qquad<br /> \textbf{(D) }\sqrt3\qquad<br /> \textbf{(E) }\sqrt3\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> A customer who intends to purchase an appliance has three coupons, only one of which may be used:<br /> <br /> Coupon 1: &lt;math&gt;10\%&lt;/math&gt; off the listed price if the listed price is at least &lt;math&gt;50&lt;/math&gt;<br /> <br /> Coupon 2: &lt;math&gt;20&lt;/math&gt; dollars off the listed price if the listed price is at least &lt;math&gt;100&lt;/math&gt;<br /> <br /> Coupon 3: &lt;math&gt;18\%&lt;/math&gt; off the amount by which the listed price exceeds &lt;math&gt;100&lt;/math&gt;<br /> <br /> For which of the following listed prices will coupon &lt;math&gt;1&lt;/math&gt; offer a greater price reduction than either coupon &lt;math&gt;2&lt;/math&gt; or coupon &lt;math&gt;3&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }179.95\qquad<br /> \textbf{(B) }199.95\qquad<br /> \textbf{(C) }219.95\qquad<br /> \textbf{(D) }239.95\qquad<br /> \textbf{(E) }259.95\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 9==<br /> Five positive consecutive integers starting with &lt;math&gt;a&lt;/math&gt; have average &lt;math&gt;b&lt;/math&gt;. What is the average of &lt;math&gt;5&lt;/math&gt; consecutive integers that start with &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ a+3\qquad\textbf{(B)}\ a+4\qquad\textbf{(C)}\ a+5\qquad\textbf{(D)}\ a+6\qquad\textbf{(E)}\ a+7&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 10==<br /> Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length &lt;math&gt;1&lt;/math&gt;. The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles?<br /> <br /> &lt;math&gt;\textbf{(A) }\dfrac{\sqrt3}4\qquad<br /> \textbf{(B) }\dfrac{\sqrt3}3\qquad<br /> \textbf{(C) }\dfrac23\qquad<br /> \textbf{(D) }\dfrac{\sqrt2}2\qquad<br /> \textbf{(E) }\dfrac{\sqrt3}2&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> David drives from his home to the airport to catch a flight. He drives &lt;math&gt;35&lt;/math&gt; miles in the first hour, but realizes that he will be &lt;math&gt;1&lt;/math&gt; hour late if he continues at this speed. He increases his speed by &lt;math&gt;15&lt;/math&gt; miles per hour for the rest of the way to the airport and arrives &lt;math&gt;30&lt;/math&gt; minutes early. How many miles is the airport from his home?<br /> <br /> &lt;math&gt;\textbf{(A) }140\qquad<br /> \textbf{(B) }175\qquad<br /> \textbf{(C) }210\qquad<br /> \textbf{(D) }245\qquad<br /> \textbf{(E) }280\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 12==<br /> Two circles intersect at points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. The minor arcs &lt;math&gt;AB&lt;/math&gt; measure &lt;math&gt;30^\circ&lt;/math&gt; on one circle and &lt;math&gt;60^\circ&lt;/math&gt; on the other circle. What is the ratio of the area of the larger circle to the area of the smaller circle?<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad<br /> \textbf{(B) }1+\sqrt3\qquad<br /> \textbf{(C) }3\qquad<br /> \textbf{(D) }2+\sqrt3\qquad<br /> \textbf{(E) }4\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> A fancy bed and breakfast inn has &lt;math&gt;5&lt;/math&gt; rooms, each with a distinctive color-coded decor. One day &lt;math&gt;5&lt;/math&gt; friends arrive to spend the night. There are no other guests that night. The friends can room in any combination they wish, but with no more than &lt;math&gt;2&lt;/math&gt; friends per room. In how many ways can the innkeeper assign the guests to the rooms?<br /> <br /> &lt;math&gt;\textbf{(A) }2100\qquad<br /> \textbf{(B) }2220\qquad<br /> \textbf{(C) }3000\qquad<br /> \textbf{(D) }3120\qquad<br /> \textbf{(E) }3125\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> Let &lt;math&gt;a&lt;b&lt;c&lt;/math&gt; be three integers such that &lt;math&gt;a,b,c&lt;/math&gt; is an arithmetic progression and &lt;math&gt;a,c,b&lt;/math&gt; is a geometric progression. What is the smallest possible value of &lt;math&gt;c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }-2\qquad<br /> \textbf{(B) }1\qquad<br /> \textbf{(C) }2\qquad<br /> \textbf{(D) }4\qquad<br /> \textbf{(E) }6\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> A five-digit palindrome is a positive integer with respective digits &lt;math&gt;abcba&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; is non-zero. Let &lt;math&gt;S&lt;/math&gt; be the sum of all five-digit palindromes. What is the sum of the digits of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad<br /> \textbf{(B) }18\qquad<br /> \textbf{(C) }27\qquad<br /> \textbf{(D) }36\qquad<br /> \textbf{(E) }45\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> The product &lt;math&gt;(8)(888\ldots 8)&lt;/math&gt;, where the second factor has &lt;math&gt;k&lt;/math&gt; digits, is an integer whose digits have a sum of &lt;math&gt;1000&lt;/math&gt;. What is &lt;math&gt;k&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }901\qquad<br /> \textbf{(B) }911\qquad<br /> \textbf{(C) }919\qquad<br /> \textbf{(D) }991\qquad<br /> \textbf{(E) }999\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 17==<br /> A &lt;math&gt;4\times 4\times h&lt;/math&gt; rectangular box contains a sphere of radius &lt;math&gt;2&lt;/math&gt; and eight smaller spheres of radius &lt;math&gt;1&lt;/math&gt;. The smaller spheres are each tangent to three sides of the box, and the larger sphere is tangent to each of the smaller spheres. What is &lt;math&gt;h&lt;/math&gt;?<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> import graph3;<br /> import solids;<br /> real h=2+2*sqrt(7);<br /> currentprojection=orthographic((0.75,-5,h/2+1),target=(2,2,h/2));<br /> currentlight=light(4,-4,4);<br /> draw((0,0,0)--(4,0,0)--(4,4,0)--(0,4,0)--(0,0,0)^^(4,0,0)--(4,0,h)--(4,4,h)--(0,4,h)--(0,4,0));<br /> draw(shift((1,3,1))*unitsphere,gray(0.85));<br /> draw(shift((3,3,1))*unitsphere,gray(0.85));<br /> draw(shift((3,1,1))*unitsphere,gray(0.85));<br /> draw(shift((1,1,1))*unitsphere,gray(0.85));<br /> draw(shift((2,2,h/2))*scale(2,2,2)*unitsphere,gray(0.85));<br /> draw(shift((1,3,h-1))*unitsphere,gray(0.85));<br /> draw(shift((3,3,h-1))*unitsphere,gray(0.85));<br /> draw(shift((3,1,h-1))*unitsphere,gray(0.85));<br /> draw(shift((1,1,h-1))*unitsphere,gray(0.85));<br /> draw((0,0,0)--(0,0,h)--(4,0,h)^^(0,0,h)--(0,4,h));<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }2+2\sqrt 7\qquad<br /> \textbf{(B) }3+2\sqrt 5\qquad<br /> \textbf{(C) }4+2\sqrt 7\qquad<br /> \textbf{(D) }4\sqrt 5\qquad<br /> \textbf{(E) }4\sqrt 7\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> The domain of the function &lt;math&gt;f(x)=\log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))))&lt;/math&gt; is an interval of length &lt;math&gt;\tfrac 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. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }19\qquad<br /> \textbf{(B) }31\qquad<br /> \textbf{(C) }271\qquad<br /> \textbf{(D) }319\qquad<br /> \textbf{(E) }511\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> There are exactly &lt;math&gt;N&lt;/math&gt; distinct rational numbers &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;|k|&lt;200&lt;/math&gt; and &lt;cmath&gt;5x^2+kx+12=0&lt;/cmath&gt; has at least one integer solution for &lt;math&gt;x&lt;/math&gt;. What is &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }6\qquad<br /> \textbf{(B) }12\qquad<br /> \textbf{(C) }24\qquad<br /> \textbf{(D) }48\qquad<br /> \textbf{(E) }78\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> In &lt;math&gt;\triangle BAC&lt;/math&gt;, &lt;math&gt;\angle BAC=40^\circ&lt;/math&gt;, &lt;math&gt;AB=10&lt;/math&gt;, and &lt;math&gt;AC=6&lt;/math&gt;. Points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; lie on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt; respectively. What is the minimum possible value of &lt;math&gt;BE+DE+CD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }6\sqrt 3+3\qquad<br /> \textbf{(B) }\dfrac{27}2\qquad<br /> \textbf{(C) }8\sqrt 3\qquad<br /> \textbf{(D) }14\qquad<br /> \textbf{(E) }3\sqrt 3+9\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> For every real number &lt;math&gt;x&lt;/math&gt;, let &lt;math&gt;\lfloor x\rfloor&lt;/math&gt; denote the greatest integer not exceeding &lt;math&gt;x&lt;/math&gt;, and let &lt;cmath&gt;f(x)=\lfloor x\rfloor(2014^{x-\lfloor x\rfloor}-1).&lt;/cmath&gt; The set of all numbers &lt;math&gt;x&lt;/math&gt; such that &lt;math&gt;1\leq x&lt;2014&lt;/math&gt; and &lt;math&gt;f(x)\leq 1&lt;/math&gt; is a union of disjoint intervals. What is the sum of the lengths of those intervals?<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad<br /> \textbf{(B) }\dfrac{\log 2015}{\log 2014}\qquad<br /> \textbf{(C) }\dfrac{\log 2014}{\log 2013}\qquad<br /> \textbf{(D) }\dfrac{2014}{2013}\qquad<br /> \textbf{(E) }2014^{\frac1{2014}}\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> The number &lt;math&gt;5^{867}&lt;/math&gt; is between &lt;math&gt;2^{2013}&lt;/math&gt; and &lt;math&gt;2^{2014}&lt;/math&gt;. How many pairs of integers &lt;math&gt;(m,n)&lt;/math&gt; are there such that &lt;math&gt;1\leq m\leq 2012&lt;/math&gt; and &lt;cmath&gt;5^n&lt;2^m&lt;2^{m+2}&lt;5^{n+1}?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }278\qquad<br /> \textbf{(B) }279\qquad<br /> \textbf{(C) }280\qquad<br /> \textbf{(D) }281\qquad<br /> \textbf{(E) }282\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==Problem 23==<br /> The fraction &lt;cmath&gt;\dfrac1{99^2}=0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},&lt;/cmath&gt; where &lt;math&gt;n&lt;/math&gt; is the length of the period of the repeating decimal expansion. What is the sum &lt;math&gt;b_0+b_1+\cdots+b_{n-1}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }874\qquad<br /> \textbf{(B) }883\qquad<br /> \textbf{(C) }887\qquad<br /> \textbf{(D) }891\qquad<br /> \textbf{(E) }892\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> Let &lt;math&gt;f_0(x)=x+|x-100|-|x+100|&lt;/math&gt;, and for &lt;math&gt;n\geq 1&lt;/math&gt;, let &lt;math&gt;f_n(x)=|f_{n-1}(x)|-1&lt;/math&gt;. For how many values of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;f_{100}(x)=0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }299\qquad<br /> \textbf{(B) }300\qquad<br /> \textbf{(C) }301\qquad<br /> \textbf{(D) }302\qquad<br /> \textbf{(E) }303\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> The parabola &lt;math&gt;P&lt;/math&gt; has focus &lt;math&gt;(0,0)&lt;/math&gt; and goes through the points &lt;math&gt;(4,3)&lt;/math&gt; and &lt;math&gt;(-4,-3)&lt;/math&gt;. For how many points &lt;math&gt;(x,y)\in P&lt;/math&gt; with integer coordinates is it true that &lt;math&gt;|4x+3y|\leq 1000&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }38\qquad<br /> \textbf{(B) }40\qquad<br /> \textbf{(C) }42\qquad<br /> \textbf{(D) }44\qquad<br /> \textbf{(E) }46\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 12A Problems/Problem 25|Solution]]</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_25&diff=70956 2013 AMC 12B Problems/Problem 25 2015-07-02T02:13:05Z <p>Pragmatictnt: /* Problem */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;G&lt;/math&gt; be the set of polynomials of the form<br /> &lt;cmath&gt; P(z)=z^n+c_{n-1}z^{n-1}+\cdots+c_2z^2+c_1z+50, &lt;/cmath&gt;<br /> where &lt;math&gt; c_1,c_2,\cdots, c_{n-1} &lt;/math&gt; are integers and &lt;math&gt;P(z)&lt;/math&gt; has distinct roots of the form &lt;math&gt;a+ib&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; integers. How many polynomials are in &lt;math&gt;G&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 288\qquad\textbf{(B)}\ 528\qquad\textbf{(C)}\ 576\qquad\textbf{(D)}\ 992\qquad\textbf{(E)}\ 1056 &lt;/math&gt;<br /> <br /> ==Solution==<br /> If we factor into irreducible polynomials (in &lt;math&gt;\mathbb{Q}[x]&lt;/math&gt;), each factor &lt;math&gt;f_i&lt;/math&gt; has exponent &lt;math&gt;1&lt;/math&gt; in the factorization and degree at most &lt;math&gt;2&lt;/math&gt; (since the &lt;math&gt;a+bi&lt;/math&gt; with &lt;math&gt;b\ne0&lt;/math&gt; come in conjugate pairs with product &lt;math&gt;a^2+b^2&lt;/math&gt;). Clearly we want the product of constant terms of these polynomials to equal &lt;math&gt;50&lt;/math&gt;; for &lt;math&gt;d\mid 50&lt;/math&gt;, let &lt;math&gt;f(d)&lt;/math&gt; be the number of permitted &lt;math&gt;f_i&lt;/math&gt; with constant term &lt;math&gt;d&lt;/math&gt;. It's easy to compute &lt;math&gt;f(1)=2&lt;/math&gt;, &lt;math&gt;f(2)=3&lt;/math&gt;, &lt;math&gt;f(5)=5&lt;/math&gt;, &lt;math&gt;f(10)=5&lt;/math&gt;, &lt;math&gt;f(25)=6&lt;/math&gt;, &lt;math&gt;f(50)=7&lt;/math&gt;, and obviously &lt;math&gt;f(d) = 1&lt;/math&gt; for negative &lt;math&gt;d\mid 50&lt;/math&gt;.<br /> <br /> Note that by the distinctness condition, the only constant terms &lt;math&gt;d&lt;/math&gt; that can be repeated are those with &lt;math&gt;d^2\mid 50&lt;/math&gt; and &lt;math&gt;f(d)&gt;1&lt;/math&gt;, i.e. &lt;math&gt;+1&lt;/math&gt; and &lt;math&gt;+5&lt;/math&gt;. Also, the &lt;math&gt;+1&lt;/math&gt;s don't affect the product, so we can simply count the number of polynomials with no constant terms of &lt;math&gt;+1&lt;/math&gt; and multiply by &lt;math&gt;2^{f(1)} = 4&lt;/math&gt; at the end.<br /> <br /> We do casework on the (unique) even constant term &lt;math&gt;d\in\{\pm2,\pm10,\pm50\}&lt;/math&gt; in our product. For convenience, let &lt;math&gt;F(d)&lt;/math&gt; be the number of ways to get a product of &lt;math&gt;50/d&lt;/math&gt; without using &lt;math&gt;\pm 1&lt;/math&gt; (so only using &lt;math&gt;\pm5,\pm25&lt;/math&gt;) and recall &lt;math&gt;f(-1) = 1&lt;/math&gt;; then our final answer will be &lt;math&gt;2^{f(1)}\sum_{d\in\{2,10,50\}}(f(-d)+f(d))(F(-d)+F(d))&lt;/math&gt;. It's easy to compute &lt;math&gt;F(-50)=0&lt;/math&gt;, &lt;math&gt;F(50)=1&lt;/math&gt;, &lt;math&gt;F(-10)=f(5)=5&lt;/math&gt;, &lt;math&gt;F(10)=f(-5)=1&lt;/math&gt;, &lt;math&gt;F(-2)=f(-25)+f(-5)f(5)=6&lt;/math&gt;, &lt;math&gt;F(2)=f(25)+\binom{f(5)}{2}=16&lt;/math&gt;, so we get<br /> &lt;cmath&gt; 4 [ (1+3)(6+16) + (1+5)(1+5) + (1+7)(0+1) ] = 4 = \boxed{\textbf{(B) }528} &lt;/cmath&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=B|num-b=24|after=Last Question}}<br /> <br /> {{MAA Notice}}</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_17&diff=70742 2012 AMC 12B Problems/Problem 17 2015-06-12T05:21:37Z <p>Pragmatictnt: /* Solution 3 (Fast) */</p> <hr /> <div>==Problem==<br /> <br /> Square &lt;math&gt;PQRS&lt;/math&gt; lies in the first quadrant. Points &lt;math&gt;(3,0), (5,0), (7,0),&lt;/math&gt; and &lt;math&gt;(13,0)&lt;/math&gt; lie on lines &lt;math&gt;SP, RQ, PQ&lt;/math&gt;, and &lt;math&gt;SR&lt;/math&gt;, respectively. What is the sum of the coordinates of the center of the square &lt;math&gt;PQRS&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 6.2\qquad\textbf{(C)}\ 6.4\qquad\textbf{(D)}\ 6.6\qquad\textbf{(E)}\ 6.8 &lt;/math&gt;<br /> <br /> ==Solutions==<br /> ===Solution 1===<br /> {{image}}<br /> Let the four points be labeled &lt;math&gt;P_1&lt;/math&gt;, &lt;math&gt;P_2&lt;/math&gt;, &lt;math&gt;P_3&lt;/math&gt;, and &lt;math&gt;P_4&lt;/math&gt;, respectively. Let the lines that go through each point be labeled &lt;math&gt;L_1&lt;/math&gt;, &lt;math&gt;L_2&lt;/math&gt;, &lt;math&gt;L_3&lt;/math&gt;, and &lt;math&gt;L_4&lt;/math&gt;, respectively. Since &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; go through &lt;math&gt;SP&lt;/math&gt; and &lt;math&gt;RQ&lt;/math&gt;, respectively, and &lt;math&gt;SP&lt;/math&gt; and &lt;math&gt;RQ&lt;/math&gt; are opposite sides of the square, we can say that &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; are parallel with slope &lt;math&gt;m&lt;/math&gt;. Similarly, &lt;math&gt;L_3&lt;/math&gt; and &lt;math&gt;L_4&lt;/math&gt; have slope &lt;math&gt;-\frac{1}{m}&lt;/math&gt;. Also, note that since square &lt;math&gt;PQRS&lt;/math&gt; lies in the first quadrant, &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; must have a positive slope. Using the point-slope form, we can now find the equations of all four lines: &lt;math&gt;L_1: y = m(x-3)&lt;/math&gt;, &lt;math&gt;L_2: y = m(x-5)&lt;/math&gt;, &lt;math&gt;L_3: y = -\frac{1}{m}(x-7)&lt;/math&gt;, &lt;math&gt;L_4: y = -\frac{1}{m}(x-13)&lt;/math&gt;.<br /> <br /> <br /> Since &lt;math&gt;PQRS&lt;/math&gt; is a square, it follows that &lt;math&gt;\Delta x&lt;/math&gt; between points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; is equal to &lt;math&gt;\Delta y&lt;/math&gt; between points &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;. Our approach will be to find &lt;math&gt;\Delta x&lt;/math&gt; and &lt;math&gt;\Delta y&lt;/math&gt; in terms of &lt;math&gt;m&lt;/math&gt; and equate the two to solve for &lt;math&gt;m&lt;/math&gt;. &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_3&lt;/math&gt; intersect at point &lt;math&gt;P&lt;/math&gt;. Setting the equations for &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_3&lt;/math&gt; equal to each other and solving for &lt;math&gt;x&lt;/math&gt;, we find that they intersect at &lt;math&gt;x = \frac{3m^2 + 7}{m^2 + 1}&lt;/math&gt;. &lt;math&gt;L_2&lt;/math&gt; and &lt;math&gt;L_3&lt;/math&gt; intersect at point &lt;math&gt;Q&lt;/math&gt;. Intersecting the two equations, the &lt;math&gt;x&lt;/math&gt;-coordinate of point &lt;math&gt;Q&lt;/math&gt; is found to be &lt;math&gt;x = \frac{5m^2 + 7}{m^2 + 1}&lt;/math&gt;. Subtracting the two, we get &lt;math&gt;\Delta x = \frac{2m^2}{m^2 + 1}&lt;/math&gt;. Substituting the &lt;math&gt;x&lt;/math&gt;-coordinate for point &lt;math&gt;Q&lt;/math&gt; found above into the equation for &lt;math&gt;L_2&lt;/math&gt;, we find that the &lt;math&gt;y&lt;/math&gt;-coordinate of point &lt;math&gt;Q&lt;/math&gt; is &lt;math&gt;y = \frac{2m}{m^2+1}&lt;/math&gt;. &lt;math&gt;L_2&lt;/math&gt; and &lt;math&gt;L_4&lt;/math&gt; intersect at point &lt;math&gt;R&lt;/math&gt;. Intersecting the two equations, the &lt;math&gt;y&lt;/math&gt;-coordinate of point &lt;math&gt;R&lt;/math&gt; is found to be &lt;math&gt;y = \frac{8m}{m^2 + 1}&lt;/math&gt;. Subtracting the two, we get &lt;math&gt;\Delta y = \frac{6m}{m^2 + 1}&lt;/math&gt;. Equating &lt;math&gt;\Delta x&lt;/math&gt; and &lt;math&gt;\Delta y&lt;/math&gt;, we get &lt;math&gt;2m^2 = 6m&lt;/math&gt; which gives us &lt;math&gt;m = 3&lt;/math&gt;. Finally, note that the line which goes though the midpoint of &lt;math&gt;P_1&lt;/math&gt; and &lt;math&gt;P_2&lt;/math&gt; with slope &lt;math&gt;3&lt;/math&gt; and the line which goes through the midpoint of &lt;math&gt;P_3&lt;/math&gt; and &lt;math&gt;P_4&lt;/math&gt; with slope &lt;math&gt;-\frac{1}{3}&lt;/math&gt; must intersect at at the center of the square. The equation of the line going through &lt;math&gt;(4,0)&lt;/math&gt; is given by &lt;math&gt;y = 3(x-4)&lt;/math&gt; and the equation of the line going through &lt;math&gt;(10,0)&lt;/math&gt; is &lt;math&gt;y = -\frac{1}{3}(x-10)&lt;/math&gt;. Equating the two, we find that they intersect at &lt;math&gt;(4.6, 1.8)&lt;/math&gt;. Adding the &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;-coordinates, we get &lt;math&gt;6.4&lt;/math&gt;. Thus, answer choice &lt;math&gt;\boxed{\textbf{(C)}}&lt;/math&gt; is correct.<br /> <br /> ===Solution 2===<br /> <br /> Note that the center of the square lies along a line that has an &lt;math&gt;x-&lt;/math&gt;intercept of &lt;math&gt;\frac{3+5}{2}=4&lt;/math&gt;, and also along another line with &lt;math&gt;x-&lt;/math&gt;intercept &lt;math&gt;\frac{7+13}{2}=10&lt;/math&gt;. Since these 2 lines are parallel to the sides of the square, they are perpindicular (since the sides of a square are). Let &lt;math&gt;m&lt;/math&gt; be the slope of the first line. Then &lt;math&gt;-\frac{1}{m}&lt;/math&gt; is the slope of the second line. We may use the point-slope form for the equation of a line to write &lt;math&gt;l_1:y=m(x-4)&lt;/math&gt; and &lt;math&gt;l_2:y=-\frac{1}{m}(x-10)&lt;/math&gt;. We easily calculate the intersection of these lines using substitution or elimination to obtain &lt;math&gt;\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)&lt;/math&gt; as the center or the square. Let &lt;math&gt;\theta&lt;/math&gt; denote the (acute) angle formed by &lt;math&gt;l_1&lt;/math&gt; and the &lt;math&gt;x-&lt;/math&gt;axis. Note that &lt;math&gt;\tan\theta=m&lt;/math&gt;. Let &lt;math&gt;s&lt;/math&gt; denote the side length of the square. Then &lt;math&gt;\sin\theta=s/2&lt;/math&gt;. On the other hand the acute angle formed by &lt;math&gt;l_2&lt;/math&gt; and the &lt;math&gt;x-&lt;/math&gt;axis is &lt;math&gt;90-\theta&lt;/math&gt; so that &lt;math&gt;\cos\theta=\sin(90-\theta)=s/6&lt;/math&gt;. Using &lt;math&gt;\cos\theta=\sqrt{1-\sin^2\theta}&lt;/math&gt; (for acute &lt;math&gt;\theta&lt;/math&gt;) we have &lt;math&gt;\frac{s}{6}=\sqrt{1-\left(\frac{s}{2}\right)^2}&lt;/math&gt; where upon &lt;math&gt;s=\frac{3\sqrt{10}}{5}&lt;/math&gt;. Then &lt;math&gt;m=\tan\theta=3&lt;/math&gt;. Substituting into &lt;math&gt;\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)&lt;/math&gt; we obtain &lt;math&gt;\left(\frac{23}{5},\frac{9}{5}\right)&lt;/math&gt; so that the sum of the coordinates is &lt;math&gt;\frac{32}{5}=6.4&lt;/math&gt;. Hence the answer is &lt;math&gt;\framebox{C}&lt;/math&gt;.<br /> <br /> ===Solution 3 (Fast)===<br /> Suppose<br /> <br /> &lt;cmath&gt;SP: y=m(x-3)&lt;/cmath&gt;<br /> &lt;cmath&gt;RQ: y=m(x-5)&lt;/cmath&gt;<br /> &lt;cmath&gt;PQ: -my=x-7&lt;/cmath&gt;<br /> &lt;cmath&gt;SR: -my=x-13&lt;/cmath&gt;<br /> <br /> where &lt;math&gt;m &gt;0&lt;/math&gt;.<br /> <br /> Recall that the distance between two parallel lines &lt;math&gt;Ax+By+C=0&lt;/math&gt; and &lt;math&gt;Ax+By+C_1=0&lt;/math&gt; is &lt;math&gt;|C-C_1|/\sqrt{A^2+B^2}&lt;/math&gt;, we have distance between &lt;math&gt;SP&lt;/math&gt; and &lt;math&gt;RQ&lt;/math&gt; equals to &lt;math&gt;2m/\sqrt{1+m^2}&lt;/math&gt;, and the distance between &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;SR&lt;/math&gt; equals to &lt;math&gt;6/\sqrt{1+m^2}&lt;/math&gt;. Equating them, we get &lt;math&gt;m=3&lt;/math&gt;.<br /> <br /> Then, the center of the square is just the intersection between the following two &quot;mid&quot; lines:<br /> <br /> &lt;cmath&gt;L_1: y=3(x-4)&lt;/cmath&gt;<br /> &lt;cmath&gt;L_2: -3y = x-10&lt;/cmath&gt;<br /> <br /> The solution is &lt;math&gt;(4.6,1.8)&lt;/math&gt;, so we get the answer &lt;math&gt;4.6+1.8=6.4&lt;/math&gt;. &lt;math&gt;\framebox{C}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=16|num-a=18}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_17&diff=70741 2012 AMC 12B Problems/Problem 17 2015-06-12T04:56:39Z <p>Pragmatictnt: /* Solution 3 (Fast) */</p> <hr /> <div>==Problem==<br /> <br /> Square &lt;math&gt;PQRS&lt;/math&gt; lies in the first quadrant. Points &lt;math&gt;(3,0), (5,0), (7,0),&lt;/math&gt; and &lt;math&gt;(13,0)&lt;/math&gt; lie on lines &lt;math&gt;SP, RQ, PQ&lt;/math&gt;, and &lt;math&gt;SR&lt;/math&gt;, respectively. What is the sum of the coordinates of the center of the square &lt;math&gt;PQRS&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 6.2\qquad\textbf{(C)}\ 6.4\qquad\textbf{(D)}\ 6.6\qquad\textbf{(E)}\ 6.8 &lt;/math&gt;<br /> <br /> ==Solutions==<br /> ===Solution 1===<br /> {{image}}<br /> Let the four points be labeled &lt;math&gt;P_1&lt;/math&gt;, &lt;math&gt;P_2&lt;/math&gt;, &lt;math&gt;P_3&lt;/math&gt;, and &lt;math&gt;P_4&lt;/math&gt;, respectively. Let the lines that go through each point be labeled &lt;math&gt;L_1&lt;/math&gt;, &lt;math&gt;L_2&lt;/math&gt;, &lt;math&gt;L_3&lt;/math&gt;, and &lt;math&gt;L_4&lt;/math&gt;, respectively. Since &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; go through &lt;math&gt;SP&lt;/math&gt; and &lt;math&gt;RQ&lt;/math&gt;, respectively, and &lt;math&gt;SP&lt;/math&gt; and &lt;math&gt;RQ&lt;/math&gt; are opposite sides of the square, we can say that &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; are parallel with slope &lt;math&gt;m&lt;/math&gt;. Similarly, &lt;math&gt;L_3&lt;/math&gt; and &lt;math&gt;L_4&lt;/math&gt; have slope &lt;math&gt;-\frac{1}{m}&lt;/math&gt;. Also, note that since square &lt;math&gt;PQRS&lt;/math&gt; lies in the first quadrant, &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; must have a positive slope. Using the point-slope form, we can now find the equations of all four lines: &lt;math&gt;L_1: y = m(x-3)&lt;/math&gt;, &lt;math&gt;L_2: y = m(x-5)&lt;/math&gt;, &lt;math&gt;L_3: y = -\frac{1}{m}(x-7)&lt;/math&gt;, &lt;math&gt;L_4: y = -\frac{1}{m}(x-13)&lt;/math&gt;.<br /> <br /> <br /> Since &lt;math&gt;PQRS&lt;/math&gt; is a square, it follows that &lt;math&gt;\Delta x&lt;/math&gt; between points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; is equal to &lt;math&gt;\Delta y&lt;/math&gt; between points &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;. Our approach will be to find &lt;math&gt;\Delta x&lt;/math&gt; and &lt;math&gt;\Delta y&lt;/math&gt; in terms of &lt;math&gt;m&lt;/math&gt; and equate the two to solve for &lt;math&gt;m&lt;/math&gt;. &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_3&lt;/math&gt; intersect at point &lt;math&gt;P&lt;/math&gt;. Setting the equations for &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_3&lt;/math&gt; equal to each other and solving for &lt;math&gt;x&lt;/math&gt;, we find that they intersect at &lt;math&gt;x = \frac{3m^2 + 7}{m^2 + 1}&lt;/math&gt;. &lt;math&gt;L_2&lt;/math&gt; and &lt;math&gt;L_3&lt;/math&gt; intersect at point &lt;math&gt;Q&lt;/math&gt;. Intersecting the two equations, the &lt;math&gt;x&lt;/math&gt;-coordinate of point &lt;math&gt;Q&lt;/math&gt; is found to be &lt;math&gt;x = \frac{5m^2 + 7}{m^2 + 1}&lt;/math&gt;. Subtracting the two, we get &lt;math&gt;\Delta x = \frac{2m^2}{m^2 + 1}&lt;/math&gt;. Substituting the &lt;math&gt;x&lt;/math&gt;-coordinate for point &lt;math&gt;Q&lt;/math&gt; found above into the equation for &lt;math&gt;L_2&lt;/math&gt;, we find that the &lt;math&gt;y&lt;/math&gt;-coordinate of point &lt;math&gt;Q&lt;/math&gt; is &lt;math&gt;y = \frac{2m}{m^2+1}&lt;/math&gt;. &lt;math&gt;L_2&lt;/math&gt; and &lt;math&gt;L_4&lt;/math&gt; intersect at point &lt;math&gt;R&lt;/math&gt;. Intersecting the two equations, the &lt;math&gt;y&lt;/math&gt;-coordinate of point &lt;math&gt;R&lt;/math&gt; is found to be &lt;math&gt;y = \frac{8m}{m^2 + 1}&lt;/math&gt;. Subtracting the two, we get &lt;math&gt;\Delta y = \frac{6m}{m^2 + 1}&lt;/math&gt;. Equating &lt;math&gt;\Delta x&lt;/math&gt; and &lt;math&gt;\Delta y&lt;/math&gt;, we get &lt;math&gt;2m^2 = 6m&lt;/math&gt; which gives us &lt;math&gt;m = 3&lt;/math&gt;. Finally, note that the line which goes though the midpoint of &lt;math&gt;P_1&lt;/math&gt; and &lt;math&gt;P_2&lt;/math&gt; with slope &lt;math&gt;3&lt;/math&gt; and the line which goes through the midpoint of &lt;math&gt;P_3&lt;/math&gt; and &lt;math&gt;P_4&lt;/math&gt; with slope &lt;math&gt;-\frac{1}{3}&lt;/math&gt; must intersect at at the center of the square. The equation of the line going through &lt;math&gt;(4,0)&lt;/math&gt; is given by &lt;math&gt;y = 3(x-4)&lt;/math&gt; and the equation of the line going through &lt;math&gt;(10,0)&lt;/math&gt; is &lt;math&gt;y = -\frac{1}{3}(x-10)&lt;/math&gt;. Equating the two, we find that they intersect at &lt;math&gt;(4.6, 1.8)&lt;/math&gt;. Adding the &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;-coordinates, we get &lt;math&gt;6.4&lt;/math&gt;. Thus, answer choice &lt;math&gt;\boxed{\textbf{(C)}}&lt;/math&gt; is correct.<br /> <br /> ===Solution 2===<br /> <br /> Note that the center of the square lies along a line that has an &lt;math&gt;x-&lt;/math&gt;intercept of &lt;math&gt;\frac{3+5}{2}=4&lt;/math&gt;, and also along another line with &lt;math&gt;x-&lt;/math&gt;intercept &lt;math&gt;\frac{7+13}{2}=10&lt;/math&gt;. Since these 2 lines are parallel to the sides of the square, they are perpindicular (since the sides of a square are). Let &lt;math&gt;m&lt;/math&gt; be the slope of the first line. Then &lt;math&gt;-\frac{1}{m}&lt;/math&gt; is the slope of the second line. We may use the point-slope form for the equation of a line to write &lt;math&gt;l_1:y=m(x-4)&lt;/math&gt; and &lt;math&gt;l_2:y=-\frac{1}{m}(x-10)&lt;/math&gt;. We easily calculate the intersection of these lines using substitution or elimination to obtain &lt;math&gt;\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)&lt;/math&gt; as the center or the square. Let &lt;math&gt;\theta&lt;/math&gt; denote the (acute) angle formed by &lt;math&gt;l_1&lt;/math&gt; and the &lt;math&gt;x-&lt;/math&gt;axis. Note that &lt;math&gt;\tan\theta=m&lt;/math&gt;. Let &lt;math&gt;s&lt;/math&gt; denote the side length of the square. Then &lt;math&gt;\sin\theta=s/2&lt;/math&gt;. On the other hand the acute angle formed by &lt;math&gt;l_2&lt;/math&gt; and the &lt;math&gt;x-&lt;/math&gt;axis is &lt;math&gt;90-\theta&lt;/math&gt; so that &lt;math&gt;\cos\theta=\sin(90-\theta)=s/6&lt;/math&gt;. Using &lt;math&gt;\cos\theta=\sqrt{1-\sin^2\theta}&lt;/math&gt; (for acute &lt;math&gt;\theta&lt;/math&gt;) we have &lt;math&gt;\frac{s}{6}=\sqrt{1-\left(\frac{s}{2}\right)^2}&lt;/math&gt; where upon &lt;math&gt;s=\frac{3\sqrt{10}}{5}&lt;/math&gt;. Then &lt;math&gt;m=\tan\theta=3&lt;/math&gt;. Substituting into &lt;math&gt;\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)&lt;/math&gt; we obtain &lt;math&gt;\left(\frac{23}{5},\frac{9}{5}\right)&lt;/math&gt; so that the sum of the coordinates is &lt;math&gt;\frac{32}{5}=6.4&lt;/math&gt;. Hence the answer is &lt;math&gt;\framebox{C}&lt;/math&gt;.<br /> <br /> ===Solution 3 (Fast)===<br /> Suppose<br /> <br /> &lt;cmath&gt;SP: y=m(x-3)&lt;/cmath&gt;<br /> &lt;cmath&gt;RQ: y=m(x-5)&lt;/cmath&gt;<br /> &lt;cmath&gt;PQ: -my=x-7&lt;/cmath&gt;<br /> &lt;cmath&gt;SR: -my=x-13&lt;/cmath&gt;<br /> <br /> where &lt;math&gt;m &gt;0&lt;/math&gt;.<br /> <br /> Recall that the distance between two parallel lines &lt;math&gt;Ax+By+C=0&lt;/math&gt; and &lt;math&gt;Ax+By+C_1=0&lt;/math&gt; is &lt;math&gt;|C-C_1|/\sqrt{A^2+B^2}&lt;/math&gt;, we have distance between &lt;math&gt;SP&lt;/math&gt; and &lt;math&gt;RQ&lt;/math&gt; equals to &lt;math&gt;2m/\sqrt{1+m^2}&lt;/math&gt;, and the distance between &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;SR&lt;/math&gt; equals to &lt;math&gt;6m/\sqrt{1+m^2}&lt;/math&gt;. Equating them, we get &lt;math&gt;m=3&lt;/math&gt;.<br /> <br /> Then, the center of the square is just the intersection between the following two &quot;mid&quot; lines:<br /> <br /> &lt;cmath&gt;L_1: y=3(x-4)&lt;/cmath&gt;<br /> &lt;cmath&gt;L_2: -3y = x-10&lt;/cmath&gt;<br /> <br /> The solution is &lt;math&gt;(4.6,1.8)&lt;/math&gt;, so we get the answer &lt;math&gt;4.6+1.8=6.4&lt;/math&gt;. &lt;math&gt;\framebox{C}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=16|num-a=18}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12A_Problems/Problem_25&diff=70677 2012 AMC 12A Problems/Problem 25 2015-06-07T23:59:06Z <p>Pragmatictnt: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;f(x)=|2\{x\}-1|&lt;/math&gt; where &lt;math&gt;\{x\}&lt;/math&gt; denotes the fractional part of &lt;math&gt;x&lt;/math&gt;. The number &lt;math&gt;n&lt;/math&gt; is the smallest positive integer such that the equation &lt;cmath&gt;nf(xf(x))=x&lt;/cmath&gt; has at least &lt;math&gt;2012&lt;/math&gt; real solutions. What is &lt;math&gt;n&lt;/math&gt;? '''Note:''' the fractional part of &lt;math&gt;x&lt;/math&gt; is a real number &lt;math&gt;y=\{x\}&lt;/math&gt; such that &lt;math&gt;0\le y&lt;1&lt;/math&gt; and &lt;math&gt;x-y&lt;/math&gt; is an integer.<br /> <br /> &lt;math&gt; \textbf{(A)}\ 30\qquad\textbf{(B)}\ 31\qquad\textbf{(C)}\ 32\qquad\textbf{(D)}\ 62\qquad\textbf{(E)}\ 64 &lt;/math&gt;<br /> <br /> == Solution ==<br /> Our goal is to determine how many times intersects the line &lt;math&gt;y=x&lt;/math&gt;. We begin by analyzing the behavior of &lt;math&gt;\{x\}&lt;/math&gt;. It increases linearly with a slope of one, then when it reaches the next integer, it repeats itself. We can deduce that the function is like a sawtooth wave, with a period of one. We then analyze the function &lt;math&gt;f(x)=|2\{x\}-1|&lt;/math&gt;. The slope of the teeth is multiplied by 2 to get 2, and the function is moved one unit downward. The function can then be described as starting at -1, moving upward with a slope of 2 to get to 1, and then repeating itself, still with a period of 1. The absolute value of the function is then taken. This results in all the negative segments becoming flipped in the Y direction. The positive slope starting at -1 of the function ranging from &lt;math&gt;u&lt;/math&gt; to &lt;math&gt;u.5&lt;/math&gt;, where u is any arbitrary integer, is now a negative slope starting at positive 1. The function now looks like the letter V repeated within every square in the first row. It is now that we address the goal of this, which is to determine how many times the function intersects the line &lt;math&gt;y=x&lt;/math&gt;. Since there are two line segments per box, the function has two chances to intersect the line &lt;math&gt;y=x&lt;/math&gt; for every integer. If the height of the function is higher than &lt;math&gt;y=x&lt;/math&gt; for every integer on an interval, then every chance within that interval intersects the line. Returning to analyzing the function, we note that it is multiplied by &lt;math&gt;x&lt;/math&gt;, and then fed into &lt;math&gt;f(x)&lt;/math&gt;. Since &lt;math&gt;f(x)&lt;/math&gt; is a periodic function, we can model it as multiplying the function's frequency by &lt;math&gt;x&lt;/math&gt;. This gives us &lt;math&gt;2x&lt;/math&gt; chances for every integer, which is then multiplied by 2 once more to get &lt;math&gt;4x&lt;/math&gt; chances for every integer. The amplitude of this function is initially 1, and then it is multiplied by n, to give an amplitude of n. The function intersects the line &lt;math&gt;y=x&lt;/math&gt; for every chance in the interval of &lt;math&gt;0\leq x \leq n&lt;/math&gt;, since the function is n units high. The function ceases to intersect &lt;math&gt;y=x&lt;/math&gt; when &lt;math&gt;n &lt; x&lt;/math&gt;, since the height of the function is lower than &lt;math&gt;y=x&lt;/math&gt;. The number of times the function intersects &lt;math&gt;y=x&lt;/math&gt; is then therefore equal to &lt;math&gt;\int_0^n \! 4x \, \mathrm{d}x&lt;/math&gt;. It is easy to see that this is equal to &lt;math&gt;2n^2&lt;/math&gt;. The problem then simplifies to the algebraic expression &lt;math&gt;2n^2=2012&lt;/math&gt;, which simplifies to, &lt;math&gt;n^2=1006&lt;/math&gt;, and then to &lt;math&gt;n=\sqrt{1006}&lt;/math&gt;, which rounds up to 32. &lt;math&gt;\boxed{\text{C}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2012|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_23&diff=70142 2010 AMC 12A Problems/Problem 23 2015-05-08T01:29:19Z <p>Pragmatictnt: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> The number obtained from the last two nonzero digits of &lt;math&gt;90!&lt;/math&gt; is equal to &lt;math&gt;n&lt;/math&gt;. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68&lt;/math&gt;<br /> <br /> == Hints and Method of Attack ==<br /> Let &lt;math&gt;P&lt;/math&gt; be the result of dividing &lt;math&gt;90!&lt;/math&gt; by tens such that &lt;math&gt;P&lt;/math&gt; is not divisible by 10. We want to consider &lt;math&gt;P \mod 100&lt;/math&gt;. But because 100 is not prime, and because &lt;math&gt;P&lt;/math&gt; is obviously divisible by 4 (if in doubt, look at the answer choices), we only need to consider &lt;math&gt;P \mod 25&lt;/math&gt;.<br /> <br /> However, 25 is a very particular number. &lt;math&gt;1 * 2 * 3 * 4 \equiv -1 \mod 25&lt;/math&gt;, and so is &lt;math&gt;6 * 7 * 8 * 9&lt;/math&gt;. How can we group terms to take advantage of this fact?<br /> <br /> There might be a problem when you cancel out the 10s from &lt;math&gt;90!&lt;/math&gt;. One method is to cancel out a factor of 2 from an existing number along with a factor of 5. But this might prove cumbersome, as the grouping method will not be as effective. Instead, take advantage of ''inverses'' in modular arithmetic. Just leave the negative powers of 2 in a &quot;storage base,&quot; and take care of the other terms first. Then, use Fermat's Little Theorem to solve for the power of 2.<br /> <br /> == Solution ==<br /> <br /> We will use the fact that for any integer &lt;math&gt;n&lt;/math&gt;,<br /> &lt;cmath&gt;\begin{align*}(5n+1)(5n+2)(5n+3)(5n+4)&amp;=[(5n+4)(5n+1)][(5n+2)(5n+3)]\\<br /> &amp;=(25n^2+25n+4)(25n^2+25n+6)\equiv 4\cdot 6\\<br /> &amp;=24\pmod{25}\equiv -1\pmod{25}.\end{align*}&lt;/cmath&gt;<br /> <br /> First, we find that the number of factors of &lt;math&gt;10&lt;/math&gt; in &lt;math&gt;90!&lt;/math&gt; is equal to &lt;math&gt;\left\lfloor \frac{90}5\right\rfloor+\left\lfloor\frac{90}{25}\right\rfloor=18+3=21&lt;/math&gt;. Let &lt;math&gt;N=\frac{90!}{10^{21}}&lt;/math&gt;. The &lt;math&gt;n&lt;/math&gt; we want is therefore the last two digits of &lt;math&gt;N&lt;/math&gt;, or &lt;math&gt;N\pmod{100}&lt;/math&gt;. If instead we find &lt;math&gt;N\pmod{25}&lt;/math&gt;, we know that &lt;math&gt;N\pmod{100}&lt;/math&gt;, what we are looking for, could be &lt;math&gt;N\pmod{25}&lt;/math&gt;, &lt;math&gt;N\pmod{25}+25&lt;/math&gt;, &lt;math&gt;N\pmod{25}+50&lt;/math&gt;, or &lt;math&gt;N\pmod{25}+75&lt;/math&gt;. Only one of these numbers will be a multiple of four, and whichever one that is will be the answer, because &lt;math&gt;N\pmod{100}&lt;/math&gt; has to be a multiple of 4.<br /> <br /> If we divide &lt;math&gt;N&lt;/math&gt; by &lt;math&gt;5^{21}&lt;/math&gt; by taking out all the factors of &lt;math&gt;5&lt;/math&gt; in &lt;math&gt;N&lt;/math&gt;, we can write &lt;math&gt;N&lt;/math&gt; as &lt;math&gt;\frac M{2^{21}}&lt;/math&gt; where<br /> &lt;cmath&gt;M=1\cdot 2\cdot 3\cdot 4\cdot 1\cdot 6\cdot 7\cdot 8\cdot 9\cdot 2\cdots 89\cdot 18,&lt;/cmath&gt;<br /> where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form &lt;math&gt;5n&lt;/math&gt; is replaced by &lt;math&gt;n&lt;/math&gt;, and every number in the form &lt;math&gt;25n&lt;/math&gt; is replaced by &lt;math&gt;n&lt;/math&gt;.<br /> <br /> The number &lt;math&gt;M&lt;/math&gt; can be grouped as follows:<br /> <br /> &lt;cmath&gt;\begin{align*}M= &amp;(1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots(86\cdot 87\cdot 88\cdot 89)\\<br /> &amp;\cdot (1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots (16\cdot 17\cdot 18) \\<br /> &amp;\cdot (1\cdot 2\cdot 3).\end{align*}&lt;/cmath&gt;<br /> <br /> Where the first line is composed of the numbers in &lt;math&gt;90!&lt;/math&gt; that aren't multiples of five, the second line is the multiples of five '''and not 25''' after they have been divided by five, and the third line is multiples of 25 after they have been divided by 25.<br /> <br /> Using the identity at the beginning of the solution, we can reduce &lt;math&gt;M&lt;/math&gt; to<br /> <br /> &lt;cmath&gt;\begin{align*}M&amp;\equiv(-1)^{18} \cdot (-1)^3(16\cdot 17\cdot 18) \cdot (1\cdot 2\cdot 3) \\<br /> &amp;= 1\cdot -21\cdot 6\\<br /> &amp;= -1\pmod{25} =24\pmod{25}.\end{align*}&lt;/cmath&gt;<br /> <br /> Using the fact that &lt;math&gt;2^{10}=1024\equiv -1\pmod{25}&lt;/math&gt; (or simply the fact that &lt;math&gt;2^{21}=2097152&lt;/math&gt; if you have your powers of 2 memorized), we can deduce that &lt;math&gt;2^{21}\equiv 2\pmod{25}&lt;/math&gt;. Therefore &lt;math&gt;N=\frac M{2^{21}}\equiv \frac {24}2\pmod{25}=12\pmod{25}&lt;/math&gt;.<br /> <br /> Finally, combining with the fact that &lt;math&gt;N\equiv 0\pmod 4&lt;/math&gt; yields &lt;math&gt;n=\boxed{\textbf{(A)}\ 12}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;P&lt;/math&gt; be &lt;math&gt;90!&lt;/math&gt; after we truncate its zeros. Notice that &lt;math&gt;90!&lt;/math&gt; has exactly (floored) &lt;math&gt;\lfloor\frac{90}{5}\rfloor + \lfloor\frac{90}{25}\rfloor = 21&lt;/math&gt; factors of 5; thus, &lt;cmath&gt;P = 2^{-21}*5^{-21}*90!.&lt;/cmath&gt; We shall consider &lt;math&gt;P&lt;/math&gt; modulo 4 and 25, to determine its residue modulo 100. It is easy to prove that &lt;math&gt;P&lt;/math&gt; is divisible by 4 (consider the number of 2s dividing &lt;math&gt;90!&lt;/math&gt; minus the number of 5s dividing &lt;math&gt;90!&lt;/math&gt;), and so we only need to consider &lt;math&gt;P&lt;/math&gt; modulo 25.<br /> <br /> Now, notice that for integers &lt;math&gt;a, n&lt;/math&gt; we have&lt;cmath&gt;(5n + a)(5n - a) \equiv -a^2 \mod 25.&lt;/cmath&gt;<br /> <br /> Thus, for integral a: &lt;cmath&gt;(10a + 1)(10a + 2)(10a + 3)(10a + 4)(10a + 6)(10a + 7)(10a + 8)(10a + 9) \equiv (-1)(-4)(-9)(-16) \equiv 576 \equiv 1 \mod 25.&lt;/cmath&gt; Using this process, we can essentially remove all the numbers which had not formerly been a multiple of 5 in &lt;math&gt;90!&lt;/math&gt; from consideration.<br /> <br /> Now, we consider the remnants of the 5, 10, 15, 20, ..., 90 not yet eliminated. The 10, 20, 30, ..., 90 becomes 1, 2, 3, 4, 1, 6, 7, 8, 9, whose product is 1 mod 25. Also, the 5, 5, 15, 25, ..., 85 becomes 1, 1, 3, 1, 7, 9, 11, 13, 3, 17 and &lt;math&gt;2^{-12}&lt;/math&gt;. We deduce that from multiplying out the 1, 1, 3, 1, 7, ..., 17 is equivalent to 2 modulo 25, and so we need to compute &lt;math&gt;2^{-11}&lt;/math&gt;. But this is simply by Fermat's Little Theorem &lt;math&gt;2^9 = 512 \equiv 12 \mod 25&lt;/math&gt;. Because 12 is also a multiple of 4, we can utilize the Chinese Remainder Theorem to show that &lt;math&gt;P = 12 \mod 100&lt;/math&gt; and so the answer is &lt;math&gt;\boxed{12}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=22|num-a=24|ab=A}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_23&diff=70141 2010 AMC 12A Problems/Problem 23 2015-05-08T01:28:02Z <p>Pragmatictnt: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> The number obtained from the last two nonzero digits of &lt;math&gt;90!&lt;/math&gt; is equal to &lt;math&gt;n&lt;/math&gt;. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68&lt;/math&gt;<br /> <br /> == Hints and Method of Attack ==<br /> Let &lt;math&gt;P&lt;/math&gt; be the result of dividing &lt;math&gt;90!&lt;/math&gt; by tens such that &lt;math&gt;P&lt;/math&gt; is not divisible by 10. We want to consider &lt;math&gt;P \mod 100&lt;/math&gt;. But because 100 is not prime, and because &lt;math&gt;P&lt;/math&gt; is obviously divisible by 4 (if in doubt, look at the answer choices), we only need to consider &lt;math&gt;P \mod 25&lt;/math&gt;.<br /> <br /> However, 25 is a very particular number. &lt;math&gt;1 * 2 * 3 * 4 \equiv -1 \mod 25&lt;/math&gt;, and so is &lt;math&gt;6 * 7 * 8 * 9&lt;/math&gt;. How can we group terms to take advantage of this fact?<br /> <br /> There might be a problem when you cancel out the 10s from &lt;math&gt;90!&lt;/math&gt;. One method is to cancel out a factor of 2 from an existing number along with a factor of 5. But this might prove cumbersome, as the grouping method will not be as effective. Instead, take advantage of ''inverses'' in modular arithmetic. Just leave the negative powers of 2 in a &quot;storage base,&quot; and take care of the other terms first. Then, use Fermat's Little Theorem to solve for the power of 2.<br /> <br /> == Solution ==<br /> <br /> We will use the fact that for any integer &lt;math&gt;n&lt;/math&gt;,<br /> &lt;cmath&gt;\begin{align*}(5n+1)(5n+2)(5n+3)(5n+4)&amp;=[(5n+4)(5n+1)][(5n+2)(5n+3)]\\<br /> &amp;=(25n^2+25n+4)(25n^2+25n+6)\equiv 4\cdot 6\\<br /> &amp;=24\pmod{25}\equiv -1\pmod{25}.\end{align*}&lt;/cmath&gt;<br /> <br /> First, we find that the number of factors of &lt;math&gt;10&lt;/math&gt; in &lt;math&gt;90!&lt;/math&gt; is equal to &lt;math&gt;\left\lfloor \frac{90}5\right\rfloor+\left\lfloor\frac{90}{25}\right\rfloor=18+3=21&lt;/math&gt;. Let &lt;math&gt;N=\frac{90!}{10^{21}}&lt;/math&gt;. The &lt;math&gt;n&lt;/math&gt; we want is therefore the last two digits of &lt;math&gt;N&lt;/math&gt;, or &lt;math&gt;N\pmod{100}&lt;/math&gt;. If instead we find &lt;math&gt;N\pmod{25}&lt;/math&gt;, we know that &lt;math&gt;N\pmod{100}&lt;/math&gt;, what we are looking for, could be &lt;math&gt;N\pmod{25}&lt;/math&gt;, &lt;math&gt;N\pmod{25}+25&lt;/math&gt;, &lt;math&gt;N\pmod{25}+50&lt;/math&gt;, or &lt;math&gt;N\pmod{25}+75&lt;/math&gt;. Only one of these numbers will be a multiple of four, and whichever one that is will be the answer, because &lt;math&gt;N\pmod{100}&lt;/math&gt; has to be a multiple of 4.<br /> <br /> If we divide &lt;math&gt;N&lt;/math&gt; by &lt;math&gt;5^{21}&lt;/math&gt; by taking out all the factors of &lt;math&gt;5&lt;/math&gt; in &lt;math&gt;N&lt;/math&gt;, we can write &lt;math&gt;N&lt;/math&gt; as &lt;math&gt;\frac M{2^{21}}&lt;/math&gt; where<br /> &lt;cmath&gt;M=1\cdot 2\cdot 3\cdot 4\cdot 1\cdot 6\cdot 7\cdot 8\cdot 9\cdot 2\cdots 89\cdot 18,&lt;/cmath&gt;<br /> where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form &lt;math&gt;5n&lt;/math&gt; is replaced by &lt;math&gt;n&lt;/math&gt;, and every number in the form &lt;math&gt;25n&lt;/math&gt; is replaced by &lt;math&gt;n&lt;/math&gt;.<br /> <br /> The number &lt;math&gt;M&lt;/math&gt; can be grouped as follows:<br /> <br /> &lt;cmath&gt;\begin{align*}M= &amp;(1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots(86\cdot 87\cdot 88\cdot 89)\\<br /> &amp;\cdot (1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots (16\cdot 17\cdot 18) \\<br /> &amp;\cdot (1\cdot 2\cdot 3).\end{align*}&lt;/cmath&gt;<br /> <br /> Where the first line is composed of the numbers in &lt;math&gt;90!&lt;/math&gt; that aren't multiples of five, the second line is the multiples of five '''and not 25''' after they have been divided by five, and the third line is multiples of 25 after they have been divided by 25.<br /> <br /> Using the identity at the beginning of the solution, we can reduce &lt;math&gt;M&lt;/math&gt; to<br /> <br /> &lt;cmath&gt;\begin{align*}M&amp;\equiv(-1)^{18} \cdot (-1)^3(16\cdot 17\cdot 18) \cdot (1\cdot 2\cdot 3) \\<br /> &amp;= 1\cdot -21\cdot 6\\<br /> &amp;= -1\pmod{25} =24\pmod{25}.\end{align*}&lt;/cmath&gt;<br /> <br /> Using the fact that &lt;math&gt;2^{10}=1024\equiv -1\pmod{25}&lt;/math&gt; (or simply the fact that &lt;math&gt;2^{21}=2097152&lt;/math&gt; if you have your powers of 2 memorized), we can deduce that &lt;math&gt;2^{21}\equiv 2\pmod{25}&lt;/math&gt;. Therefore &lt;math&gt;N=\frac M{2^{21}}\equiv \frac {24}2\pmod{25}=12\pmod{25}&lt;/math&gt;.<br /> <br /> Finally, combining with the fact that &lt;math&gt;N\equiv 0\pmod 4&lt;/math&gt; yields &lt;math&gt;n=\boxed{\textbf{(A)}\ 12}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;P&lt;/math&gt; be &lt;math&gt;90!&lt;/math&gt; after we truncate its zeros. Notice that &lt;math&gt;90!&lt;/math&gt; has exactly (floored) &lt;math&gt;\lfloor{\frac{90}{5}} + \rfloor{\frac{90}{25}} = 21&lt;/math&gt; factors of 5; thus, &lt;cmath&gt;P = 2^{-21}*5^{-21}*90!.&lt;/cmath&gt; We shall consider &lt;math&gt;P&lt;/math&gt; modulo 4 and 25, to determine its residue modulo 100. It is easy to prove that &lt;math&gt;P&lt;/math&gt; is divisible by 4 (consider the number of 2s dividing &lt;math&gt;90!&lt;/math&gt; minus the number of 5s dividing &lt;math&gt;90!&lt;/math&gt;), and so we only need to consider &lt;math&gt;P&lt;/math&gt; modulo 25.<br /> <br /> Now, notice that for integers &lt;math&gt;a, n&lt;/math&gt; we have&lt;cmath&gt;(5n + a)(5n - a) \equiv -a^2 \mod 25.&lt;/cmath&gt;<br /> <br /> Thus, for integral a: &lt;cmath&gt;(10a + 1)(10a + 2)(10a + 3)(10a + 4)(10a + 6)(10a + 7)(10a + 8)(10a + 9) \equiv (-1)(-4)(-9)(-16) \equiv 576 \equiv 1 \mod 25.&lt;/cmath&gt; Using this process, we can essentially remove all the numbers which had not formerly been a multiple of 5 in &lt;math&gt;90!&lt;/math&gt; from consideration.<br /> <br /> Now, we consider the remnants of the 5, 10, 15, 20, ..., 90 not yet eliminated. The 10, 20, 30, ..., 90 becomes 1, 2, 3, 4, 1, 6, 7, 8, 9, whose product is 1 mod 25. Also, the 5, 5, 15, 25, ..., 85 becomes 1, 1, 3, 1, 7, 9, 11, 13, 3, 17 and &lt;math&gt;2^{-12}&lt;/math&gt;. We deduce that from multiplying out the 1, 1, 3, 1, 7, ..., 17 is equivalent to 2 modulo 25, and so we need to compute &lt;math&gt;2^{-11}&lt;/math&gt;. But this is simply by Fermat's Little Theorem &lt;math&gt;2^9 = 512 \equiv 12 \mod 25&lt;/math&gt;. Because 12 is also a multiple of 4, we can utilize the Chinese Remainder Theorem to show that &lt;math&gt;P = 12 \mod 100&lt;/math&gt; and so the answer is &lt;math&gt;\boxed{12}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=22|num-a=24|ab=A}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_25&diff=70126 2009 AMC 12B Problems/Problem 25 2015-05-06T18:37:40Z <p>Pragmatictnt: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> The set &lt;math&gt;G&lt;/math&gt; is defined by the points &lt;math&gt;(x,y)&lt;/math&gt; with integer coordinates, &lt;math&gt;3\le|x|\le7&lt;/math&gt;, &lt;math&gt;3\le|y|\le7&lt;/math&gt;. How many squares of side at least &lt;math&gt;6&lt;/math&gt; have their four vertices in &lt;math&gt;G&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> defaultpen(black+0.75bp+fontsize(8pt));<br /> size(5cm);<br /> path p = scale(.15)*unitcircle;<br /> draw((-8,0)--(8.5,0),Arrow(HookHead,1mm));<br /> draw((0,-8)--(0,8.5),Arrow(HookHead,1mm));<br /> int i,j;<br /> for (i=-7;i&lt;8;++i) {<br /> for (j=-7;j&lt;8;++j) {<br /> if (((-7 &lt;= i) &amp;&amp; (i &lt;= -3)) || ((3 &lt;= i) &amp;&amp; (i&lt;= 7))) { if (((-7 &lt;= j) &amp;&amp; (j &lt;= -3)) || ((3 &lt;= j) &amp;&amp; (j&lt;= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp);<br /> draw((-3,-.2)--(-3,.2),black+0.5bp);<br /> draw((3,-.2)--(3,.2),black+0.5bp);<br /> draw((7,-.2)--(7,.2),black+0.5bp);<br /> draw((-.2,-7)--(.2,-7),black+0.5bp);<br /> draw((-.2,-3)--(.2,-3),black+0.5bp);<br /> draw((-.2,3)--(.2,3),black+0.5bp);<br /> draw((-.2,7)--(.2,7),black+0.5bp);<br /> label(&quot;$-7$&quot;,(-7,0),S);<br /> label(&quot;$-3$&quot;,(-3,0),S);<br /> label(&quot;$3$&quot;,(3,0),S);<br /> label(&quot;$7$&quot;,(7,0),S);<br /> label(&quot;$-7$&quot;,(0,-7),W);<br /> label(&quot;$-3$&quot;,(0,-3),W);<br /> label(&quot;$3$&quot;,(0,3),W);<br /> label(&quot;$7$&quot;,(0,7),W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 125\qquad \textbf{(B)}\ 150\qquad \textbf{(C)}\ 175\qquad \textbf{(D)}\ 200\qquad \textbf{(E)}\ 225&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> We need to find a reasonably easy way to count the squares. <br /> <br /> First, obviously the maximum distance between two points in the same quadrant is &lt;math&gt;4\sqrt 2 &lt; 6&lt;/math&gt;, hence each square has exactly one vertex in each quadrant.<br /> <br /> Given any square, we can circumscribe another axes-parallel square around it. In the picture below, the original square is red and the circumscribed one is blue.<br /> <br /> &lt;asy&gt;<br /> defaultpen(black+0.75bp+fontsize(8pt));<br /> size(7.5cm);<br /> path p = scale(.15)*unitcircle;<br /> draw((-8,0)--(8.5,0),Arrow(HookHead,1mm));<br /> draw((0,-8)--(0,8.5),Arrow(HookHead,1mm));<br /> int i,j;<br /> for (i=-7;i&lt;8;++i) {<br /> for (j=-7;j&lt;8;++j) {<br /> if (((-7 &lt;= i) &amp;&amp; (i &lt;= -3)) || ((3 &lt;= i) &amp;&amp; (i&lt;= 7))) { if (((-7 &lt;= j) &amp;&amp; (j &lt;= -3)) || ((3 &lt;= j) &amp;&amp; (j&lt;= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp);<br /> draw((-3,-.2)--(-3,.2),black+0.5bp);<br /> draw((3,-.2)--(3,.2),black+0.5bp);<br /> draw((7,-.2)--(7,.2),black+0.5bp);<br /> draw((-.2,-7)--(.2,-7),black+0.5bp);<br /> draw((-.2,-3)--(.2,-3),black+0.5bp);<br /> draw((-.2,3)--(.2,3),black+0.5bp);<br /> draw((-.2,7)--(.2,7),black+0.5bp);<br /> label(&quot;$-7$&quot;,(-7,0),S);<br /> label(&quot;$-3$&quot;,(-3,0),S);<br /> label(&quot;$3$&quot;,(3,0),S);<br /> label(&quot;$7$&quot;,(7,0),S);<br /> label(&quot;$-7$&quot;,(0,-7),W);<br /> label(&quot;$-3$&quot;,(0,-3),W);<br /> label(&quot;$3$&quot;,(0,3),W);<br /> label(&quot;$7$&quot;,(0,7),W);<br /> draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red );<br /> draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, dashed + blue );<br /> &lt;/asy&gt;<br /> <br /> Let's now consider the opposite direction. Assume that we picked the blue square, how many different red squares do share it?<br /> <br /> Answering this question is not as simple as it may seem. Consider the picture below. It shows all three red squares that share the same blue square. In addition, the picture shows a green square that is not valid, as two of its vertices are in bad locations.<br /> <br /> &lt;asy&gt;<br /> defaultpen(black+0.75bp+fontsize(8pt));<br /> size(7.5cm);<br /> path p = scale(.15)*unitcircle;<br /> draw((-8,0)--(8.5,0),Arrow(HookHead,1mm));<br /> draw((0,-8)--(0,8.5),Arrow(HookHead,1mm));<br /> int i,j;<br /> for (i=-7;i&lt;8;++i) {<br /> for (j=-7;j&lt;8;++j) {<br /> if (((-7 &lt;= i) &amp;&amp; (i &lt;= -3)) || ((3 &lt;= i) &amp;&amp; (i&lt;= 7))) { if (((-7 &lt;= j) &amp;&amp; (j &lt;= -3)) || ((3 &lt;= j) &amp;&amp; (j&lt;= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp);<br /> draw((-3,-.2)--(-3,.2),black+0.5bp);<br /> draw((3,-.2)--(3,.2),black+0.5bp);<br /> draw((7,-.2)--(7,.2),black+0.5bp);<br /> draw((-.2,-7)--(.2,-7),black+0.5bp);<br /> draw((-.2,-3)--(.2,-3),black+0.5bp);<br /> draw((-.2,3)--(.2,3),black+0.5bp);<br /> draw((-.2,7)--(.2,7),black+0.5bp);<br /> label(&quot;$-7$&quot;,(-7,0),S);<br /> label(&quot;$-3$&quot;,(-3,0),S);<br /> label(&quot;$3$&quot;,(3,0),S);<br /> label(&quot;$7$&quot;,(7,0),S);<br /> label(&quot;$-7$&quot;,(0,-7),W);<br /> label(&quot;$-3$&quot;,(0,-3),W);<br /> label(&quot;$3$&quot;,(0,3),W);<br /> label(&quot;$7$&quot;,(0,7),W);<br /> draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, red );<br /> draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red );<br /> draw( (5,-3) -- (-2,-5) -- (-4,2) -- (3,4) -- cycle, dashed + green );<br /> draw( (5,-4) -- (-3,-5) -- (-4,3) -- (4,4) -- cycle, red );<br /> draw( scale(1.05)*((5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle), dashed + blue );<br /> &lt;/asy&gt;<br /> <br /> The size of the blue square can range from &lt;math&gt;6\times 6&lt;/math&gt; to &lt;math&gt;14\times 14&lt;/math&gt;, and for the intermediate sizes there is more than one valid placement. We will now examine the cases one after another. Also, we can use symmetry to reduce the number of cases.<br /> <br /> size upper_right solutions symmetries total<br /> 6 (3,3) 1 1 1<br /> <br /> 7 (3,3) 1 4 4<br /> <br /> 8 (3,3) 1 4 4<br /> 8 (3,4) 1 4 4<br /> 8 (4,4) 3 1 3<br /> <br /> 9 (3,3) 1 4 4<br /> 9 (3,4) 1 8 8<br /> 9 (4,4) 3 4 12<br /> <br /> 10 (3,3) 1 4 4<br /> 10 (3,4) 1 8 8<br /> 10 (3,5) 1 4 4<br /> 10 (4,4) 3 4 12<br /> 10 (4,5) 3 4 12<br /> 10 (5,5) 5 1 5<br /> <br /> 11 (4,4) 3 4 12<br /> 11 (4,5) 3 8 24<br /> 11 (5,5) 5 4 20<br /> <br /> 12 (5,5) 5 4 20<br /> 12 (5,6) 5 4 20<br /> 12 (6,6) 7 1 7<br /> <br /> 13 (6,6) 7 4 28<br /> <br /> 14 (7,7) 9 1 9<br /> <br /> Summing the last column, we get that the answer is &lt;math&gt;\boxed{225}&lt;/math&gt;.<br /> == Solution 2==<br /> <br /> This is based on a clever bijection given in [http://answers.yahoo.com/question/index?qid=20110101175843AAhdbwT this page].<br /> <br /> Consider any square &lt;math&gt;ABCD&lt;/math&gt; where all four vertices are in &lt;math&gt;G&lt;/math&gt;, and the side length is at least &lt;math&gt;6&lt;/math&gt;, so the four vertices must lie in distinct quadrants (Same proof as in solution 1). Without loss of generality, assume that &lt;math&gt;A,B,C,D&lt;/math&gt; are in the first, second, third, fourth quadrant. Then we consider the following mapping:<br /> <br /> &lt;cmath&gt;A \to A' = A&lt;/cmath&gt;<br /> &lt;cmath&gt;B \to B' = B + (10,0)&lt;/cmath&gt;<br /> &lt;cmath&gt;C \to C' = C + (10,10)&lt;/cmath&gt;<br /> &lt;cmath&gt;D \to D' = D + (0,10)&lt;/cmath&gt;<br /> <br /> Then the new points &lt;math&gt;A'&lt;/math&gt;, &lt;math&gt;B'&lt;/math&gt;, &lt;math&gt;C'&lt;/math&gt;, &lt;math&gt;D'&lt;/math&gt; are either being the same point or forming a square in &lt;math&gt;G_1 = G \cap \{ x&gt;0, y&gt;0 \}&lt;/math&gt;, a 5x5 grid. <br /> <br /> Conversely, for any point in &lt;math&gt;G_1&lt;/math&gt;, it can be reversed to a square &lt;math&gt;ABCD&lt;/math&gt;; however, for any square in &lt;math&gt;G_1&lt;/math&gt;, there are four possible squares &lt;math&gt;ABCD&lt;/math&gt; that were mapped to them. Therefore the number of possible squares &lt;math&gt;ABCD&lt;/math&gt; is equal to &lt;math&gt;25 + 4N&lt;/math&gt;, where &lt;math&gt;N&lt;/math&gt; is the number of squares inscribed in &lt;math&gt;G_1&lt;/math&gt;.<br /> <br /> Moreover by the same idea in solution 1, each square (with sides parallel or slanted to the axes) in a &lt;math&gt;G_1&lt;/math&gt; can be inscribed in a square in &lt;math&gt;G_1&lt;/math&gt;, with sides parallel to one of the axes, call it &quot;standard square&quot;. Noticing that each standard square of side length &lt;math&gt;a&lt;/math&gt; corresponds to &lt;math&gt;a&lt;/math&gt; inscribed squares, and that there are &lt;math&gt;(5-a)^2&lt;/math&gt; number of standard squares of side length &lt;math&gt;a&lt;/math&gt;, we have<br /> <br /> &lt;cmath&gt;N = \sum_{a=1}^{4} a(5-a)^2 = 1\cdot 16 + 2\cdot 9 + 3 \cdot 4 + 4 \cdot 1 = 16+18+12+4=50&lt;/cmath&gt;<br /> <br /> So the answer is &lt;math&gt;25+4\cdot 50 = 225&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=B|num-b=24|after=Last Question}}<br /> {{MAA Notice}}</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_24&diff=70110 2009 AMC 12A Problems/Problem 24 2015-05-03T01:50:38Z <p>Pragmatictnt: /* Solution */</p> <hr /> <div>== Problem ==<br /> The ''tower function of twos'' is defined recursively as follows: &lt;math&gt;T(1) = 2&lt;/math&gt; and &lt;math&gt;T(n + 1) = 2^{T(n)}&lt;/math&gt; for &lt;math&gt;n\ge1&lt;/math&gt;. Let &lt;math&gt;A = (T(2009))^{T(2009)}&lt;/math&gt; and &lt;math&gt;B = (T(2009))^A&lt;/math&gt;. What is the largest integer &lt;math&gt;k&lt;/math&gt; such that<br /> <br /> &lt;center&gt;&lt;cmath&gt;\underbrace{\log_2\log_2\log_2\ldots\log_2B}_{k\text{ times}}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> is defined?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2009\qquad \textbf{(B)}\ 2010\qquad \textbf{(C)}\ 2011\qquad \textbf{(D)}\ 2012\qquad \textbf{(E)}\ 2013&lt;/math&gt; <br /> <br /> == Solution ==<br /> We just look at the last three logarithms for the moment, and use the fact that &lt;math&gt;\log_2 T(k) = T(k - 1)&lt;/math&gt;. We wish to find:<br /> &lt;center&gt;<br /> &lt;cmath&gt;\log_2\log_2\log_2\left({T(2009)^{\left({T(2009)}^{T(2009)}\right)}}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;= \log_2(T(2009)\log_2(T(2009)\log_2 T(2009)))&lt;/cmath&gt;<br /> &lt;cmath&gt;= \log_2(T(2009)\log_2(T(2009)T(2008)))&lt;/cmath&gt;<br /> &lt;cmath&gt;= \log_2(T(2009)(T(2008) + T(2007)))&lt;/cmath&gt;<br /> &lt;/center&gt;<br /> <br /> <br /> Now we realize that &lt;math&gt;T(n - 1)&lt;/math&gt; is much smaller than &lt;math&gt;T(n)&lt;/math&gt;. So we approximate this, remembering we have rounded down, as:<br /> <br /> &lt;center&gt;&lt;cmath&gt;\log_2(T(2009)) = T(2008)&lt;/cmath&gt;&lt;/center&gt;<br /> We have used &lt;math&gt;3&lt;/math&gt; logarithms so far. Applying &lt;math&gt;2007&lt;/math&gt; more to the left of our expression, we get &lt;math&gt;T(1) = 2&lt;/math&gt;. Then we can apply the logarithm &lt;math&gt;2&lt;/math&gt; more times, until we get to &lt;math&gt;0&lt;/math&gt;. So our answer is approximately &lt;math&gt;3 + 2007 + 2 = 2012&lt;/math&gt;. But we rounded down, so that means that after &lt;math&gt;2012&lt;/math&gt; logarithms we get a number slightly greater than &lt;math&gt;0&lt;/math&gt;, so we can apply logarithms one more time. We can be sure it is small enough so that the logarithm can only be applied &lt;math&gt;1&lt;/math&gt; more time since &lt;math&gt;2012 + 1 = 2013&lt;/math&gt; is the largest answer choice. So the answer is &lt;math&gt;\mathbf{(E)}&lt;/math&gt;.<br /> <br /> == Alternative Solution ==<br /> Let &lt;math&gt;L_k(x)=\log_2\log_2...\log_2(x)&lt;/math&gt; where there are &lt;math&gt;k&lt;/math&gt; &lt;math&gt;\log_2&lt;/math&gt;'s. &lt;math&gt;L_k(B)&lt;/math&gt; is defined iff &lt;math&gt;L_{k-1}(B) &gt; 0&lt;/math&gt; iff &lt;math&gt;L_{k-2}(B) &gt; 1&lt;/math&gt;. Note &lt;math&gt;\log_2 T(k) = T(k - 1)&lt;/math&gt;, so &lt;math&gt;L_{k-2}(T(k-2))=1&lt;/math&gt;. Thus, we seek the largest &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;B &gt; T(k-2)&lt;/math&gt;. Now note that<br /> <br /> &lt;math&gt;T(2009)^{T(2009)^{T(2009)}} &gt; 2^{2^{T(2009)}} = T(2011)&lt;/math&gt;<br /> <br /> so &lt;math&gt;k=2013&lt;/math&gt; satisfies the inequality. Since it is the largest choice, it is the answer.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2009|ab=A|num-b=23|num-a=25}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_23&diff=70109 2008 AMC 12B Problems/Problem 23 2015-05-02T20:32:49Z <p>Pragmatictnt: /* Solution 2 */</p> <hr /> <div>==Problem 23==<br /> The sum of the base-&lt;math&gt;10&lt;/math&gt; logarithms of the divisors of &lt;math&gt;10^n&lt;/math&gt; is &lt;math&gt;792&lt;/math&gt;. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)}\ 11\qquad \text{(B)}\ 12\qquad \text{(C)}\ 13\qquad \text{(D)}\ 14\qquad \text{(E)}\ 15&lt;/math&gt;<br /> <br /> __TOC__<br /> ==Solution==<br /> === Solution 1 ===<br /> Every factor of &lt;math&gt;10^n&lt;/math&gt; will be of the form &lt;math&gt;2^a \times 5^b , a\leq n , b\leq n&lt;/math&gt;. Using the logarithmic property &lt;math&gt;\log(a \times b) = \log(a)+\log(b)&lt;/math&gt;, it suffices to count the total number of 2's and 5's running through all possible &lt;math&gt;(a,b)&lt;/math&gt;. For every factor &lt;math&gt;2^a \times 5^b&lt;/math&gt;, there will be another &lt;math&gt;2^b \times 5^a&lt;/math&gt;, so it suffices to count the total number of 2's occurring in all factors (because of this symmetry, the number of 5's will be equal). And since &lt;math&gt;\log(2)+\log(5) = \log(10) = 1&lt;/math&gt;, the final sum will be the total number of 2's occurring in all factors of &lt;math&gt;10^n&lt;/math&gt;.<br /> <br /> There are &lt;math&gt;n+1&lt;/math&gt; choices for the exponent of 5 in each factor, and for each of those choices, there are &lt;math&gt;n+1&lt;/math&gt; factors (each corresponding to a different exponent of 2), yielding &lt;math&gt;0+1+2+3...+n = \frac{n(n+1)}{2}&lt;/math&gt; total 2's. The total number of 2's is therefore &lt;math&gt;\frac{n \cdot(n+1)^2}{2} = \frac{n^3+2n^2+n}{2}&lt;/math&gt;. Plugging in our answer choices into this formula yields 11 (answer choice &lt;math&gt;\mathrm{(A)}&lt;/math&gt;) as the correct answer.<br /> <br /> <br /> ===Solution 2===<br /> <br /> We are given &lt;cmath&gt; \log_{10}d_1 + \log_{10}d_2 + \ldots + \log_{10}d_k = 792 &lt;/cmath&gt; The property &lt;math&gt;\log(ab) = \log(a)+\log(b)&lt;/math&gt; now gives &lt;cmath&gt; \log_{10}(d_1 d_2\cdot\ldots d_k) = 792 &lt;/cmath&gt; The product of the divisors is (from elementary number theory) &lt;math&gt;a^{d(n)/2}&lt;/math&gt; where &lt;math&gt;d(n)&lt;/math&gt; is the number of divisors. Note that &lt;math&gt;10^n = 2^n\cdot 5^n&lt;/math&gt;, so &lt;math&gt;d(n) = (n + 1)^2&lt;/math&gt;. Substituting these values with &lt;math&gt;a = 10^n&lt;/math&gt; in our equation above, we get &lt;math&gt;n(n + 1)^2 = 1584&lt;/math&gt;, from whence we immediately obtain &lt;math&gt;\framebox[1.2\width]{(A)}&lt;/math&gt; as the correct answer.<br /> <br /> === Solution 3 ===<br /> For every divisor &lt;math&gt;d&lt;/math&gt; of &lt;math&gt;10^n&lt;/math&gt;, &lt;math&gt;d \le \sqrt{10^n}&lt;/math&gt;, we have &lt;math&gt;\log d + \log \frac{10^n}{d} = \log 10^n = n&lt;/math&gt;. There are &lt;math&gt;\left \lfloor \frac{(n+1)^2}{2} \right \rfloor&lt;/math&gt; divisors of &lt;math&gt;10^n = 2^n \times 5^n&lt;/math&gt; that are &lt;math&gt;\le \sqrt{10^n}&lt;/math&gt;. After casework on the parity of &lt;math&gt;n&lt;/math&gt;, we find that the answer is given by &lt;math&gt;n \times \frac{(n+1)^2}{2} = 792 \Longrightarrow n = 11\ \mathrm{(A)}&lt;/math&gt;. <br /> <br /> ==See Also==<br /> {{AMC12 box|year=2008|ab=B|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_20&diff=70108 2008 AMC 12A Problems/Problem 20 2015-05-02T20:26:09Z <p>Pragmatictnt: /* Solution */</p> <hr /> <div>==Problem==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has &lt;math&gt;AC=3&lt;/math&gt;, &lt;math&gt;BC=4&lt;/math&gt;, and &lt;math&gt;AB=5&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; is on &lt;math&gt;\overline{AB}&lt;/math&gt;, and &lt;math&gt;\overline{CD}&lt;/math&gt; bisects the right angle. The inscribed circles of &lt;math&gt;\triangle ADC&lt;/math&gt; and &lt;math&gt;\triangle BCD&lt;/math&gt; have radii &lt;math&gt;r_a&lt;/math&gt; and &lt;math&gt;r_b&lt;/math&gt;, respectively. What is &lt;math&gt;r_a/r_b&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac{1}{28}\left(10-\sqrt{2}\right)\qquad\mathrm{(B)}\ \frac{3}{56}\left(10-\sqrt{2}\right)\qquad\mathrm{(C)}\ \frac{1}{14}\left(10-\sqrt{2}\right)\qquad\mathrm{(D)}\ \frac{5}{56}\left(10-\sqrt{2}\right)\\\mathrm{(E)}\ \frac{3}{28}\left(10-\sqrt{2}\right)&lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;center&gt;&lt;asy&gt;<br /> import olympiad;<br /> size(300);<br /> defaultpen(0.8);<br /> pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429);<br /> pair O=incenter(A,C,D), P=incenter(B,C,D);<br /> picture p = new picture; <br /> draw(p,Circle(C,0.2)); draw(p,Circle(B,0.2));<br /> clip(p,B--C--D--cycle);<br /> add(p);<br /> draw(A--B--C--D--C--cycle);<br /> draw(incircle(A,C,D));<br /> draw(incircle(B,C,D));<br /> dot(O);dot(P);<br /> label(&quot;$$A$$&quot;,A,W);<br /> label(&quot;$$B$$&quot;,B,E);<br /> label(&quot;$$C$$&quot;,C,W);<br /> label(&quot;$$D$$&quot;,D,NE);<br /> label(&quot;$$O_A$$&quot;,O,W);<br /> label(&quot;$$O_B$$&quot;,P,W);<br /> label(&quot;$$3$$&quot;,(A+C)/2,W);<br /> label(&quot;$$4$$&quot;,(B+C)/2,S);<br /> label(&quot;$$\frac{15}{7}$$&quot;,(A+D)/2,NE);<br /> label(&quot;$$\frac{20}{7}$$&quot;,(B+D)/2,NE);<br /> label(&quot;$$45^{\circ}$$&quot;,(.2,.1),E);<br /> label(&quot;$$\sin \theta = \frac{3}{5}$$&quot;,B-(.2,-.1),W);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> By the [[Angle Bisector Theorem]], <br /> &lt;cmath&gt;\frac{BD}{4} = \frac{5-BD}{3} \Longrightarrow BD = \frac{20}7&lt;/cmath&gt;<br /> By [[Law of Sines]] on &lt;math&gt;\triangle BCD&lt;/math&gt;, <br /> &lt;cmath&gt;\frac{BD}{\sin 45^{\circ}} = \frac{CD}{\sin \angle B} \Longrightarrow \frac{20/7}{\sqrt{2}/2} = \frac{CD}{3/5} \Longrightarrow CD=\frac{12\sqrt{2}}{7}&lt;/cmath&gt;<br /> Since the area of a triangle satisfies &lt;math&gt;[\triangle]=rs&lt;/math&gt;, where &lt;math&gt;r = &lt;/math&gt; the [[inradius]] and &lt;math&gt;s =&lt;/math&gt; the [[semiperimeter]], we have <br /> &lt;cmath&gt;\frac{r_A}{r_B} = \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A}&lt;/cmath&gt;<br /> &lt;!--Using any of various formulas for triangle area, we find the area &lt;math&gt;[BCD]&lt;/math&gt; to be <br /> &lt;cmath&gt;[BCD] = \frac{1}{2} (\sin \angle CBD) \cdot (BD) \cdot (CD) = \frac 12 \cdot \frac 35 \cdot \frac{20}{7} \cdot 4 = \frac{24}{7}&lt;/cmath&gt;<br /> and <br /> &lt;cmath&gt;[ACD] = [ABC] - [BCD] = \frac 12 (3)(4) - \frac{24}{7} = \frac{18}{7}&lt;/cmath&gt;--&gt;<br /> Since &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\triangle BCD&lt;/math&gt; share the [[altitude]] (to &lt;math&gt;\overline{AB}&lt;/math&gt;), their areas are the ratio of their bases, or &lt;cmath&gt;\frac{[ACD]}{[BCD]} = \frac{AD}{BD} = \frac{3}{4}&lt;/cmath&gt;<br /> The semiperimeters are &lt;math&gt;s_A = \left(3 + \frac{15}{7} + \frac{12\sqrt{2}}{7}\right)\left/\right.2 = \frac{18+6\sqrt{2}}{7}&lt;/math&gt; and &lt;math&gt;s_B = \frac{24+ 6\sqrt{2}}{7}&lt;/math&gt;. Thus,<br /> &lt;cmath&gt;\begin{align*}<br /> \frac{r_A}{r_B} &amp;= \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A} = \frac{3}{4} \cdot \frac{(24+ 6\sqrt{2})/7}{(18+6\sqrt{2})/7} \\<br /> &amp;= \frac{3(4+\sqrt{2})}{4(3+\sqrt{2})} \cdot \left(\frac{3-\sqrt{2}}{3-\sqrt{2}}\right) = \frac{3}{28}(10-\sqrt{2}) \Rightarrow \mathrm{(E)}\qquad \blacksquare \end{align*}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2008|num-b=19|num-a=21|ab=A}}<br /> {{MAA Notice}}</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_20&diff=70107 2007 AMC 12B Problems/Problem 20 2015-05-02T20:22:11Z <p>Pragmatictnt: </p> <hr /> <div>==Problem==<br /> The parallelogram bounded by the lines &lt;math&gt;y=ax+c&lt;/math&gt;, &lt;math&gt;y=ax+d&lt;/math&gt;, &lt;math&gt;y=bx+c&lt;/math&gt;, and &lt;math&gt;y=bx+d&lt;/math&gt; has area &lt;math&gt;18&lt;/math&gt;. The parallelogram bounded by the lines &lt;math&gt;y=ax+c&lt;/math&gt;, &lt;math&gt;y=ax-d&lt;/math&gt;, &lt;math&gt;y=bx+c&lt;/math&gt;, and &lt;math&gt;y=bx-d&lt;/math&gt; has area &lt;math&gt;72&lt;/math&gt;. Given that &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; are positive integers, what is the smallest possible value of &lt;math&gt;a+b+c+d&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17&lt;/math&gt;<br /> <br /> ==Solution==<br /> {{incomplete|solution}}<br /> &lt;!-- &lt;center&gt;&lt;asy&gt;<br /> pathpen = linewidth(0.7);<br /> real a = 3, b = 1, c = 9, d = 3; <br /> D((0,c) -- ((d-c)/(a-b),(a*d-b*c)/(a-b)) -- (0,d) -- ((c-d)/(a-b),(b*c-a*d)/(a-b)) -- cycle);<br /> D((0,c) -- ((-d-c)/(a-b),(-a*d-b*c)/(a-b)) -- (0,-d) -- ((c+d)/(a-b),-(-a*d-b*c)/(a-b)) -- cycle);<br /> &lt;/asy&gt;&lt;/center&gt; --&gt;<br /> Plotting the parallelogram on the coordinate plane, the 4 corners are at &lt;math&gt;(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)&lt;/math&gt;. Because &lt;math&gt;72= 4\cdot 18&lt;/math&gt;, we have that &lt;math&gt;4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)&lt;/math&gt; or that &lt;math&gt;2(c-d)=c+d&lt;/math&gt;, which gives &lt;math&gt;c=3d&lt;/math&gt; (consider a [[homothety]], or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by &lt;math&gt;4\times&lt;/math&gt;, it follows that the stretch along the diagonal, or the ratio of side lengths, is &lt;math&gt;2\times&lt;/math&gt;). The area of triangular half of the parallelogram on the right side of the y-axis is given by &lt;math&gt;9 = \frac{1}{2} (c-d)\left(\frac{d-c}{a-b}\right)&lt;/math&gt;, so substituting &lt;math&gt;c = 3d&lt;/math&gt;:<br /> &lt;center&gt;&lt;cmath&gt;<br /> \frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) = 9 \quad \Longrightarrow \quad 2d^2 = 9(a-b)&lt;/cmath&gt;&lt;/center&gt;<br /> Thus &lt;math&gt;3|d&lt;/math&gt;, and we verify that &lt;math&gt;d = 3&lt;/math&gt;, &lt;math&gt;a-b = 2 \Longrightarrow a = 3, b = 1&lt;/math&gt; will give us a minimum value for &lt;math&gt;a+b+c+d&lt;/math&gt;. Then &lt;math&gt;a+b+c+d = 3 + 1 + 9 + 3 = 16\ \mathbf{(D)}&lt;/math&gt;.<br /> ==Solution 2==<br /> {{incomplete|solution}}<br /> The key to this solution is that area is invariant under translation. By suitably shifting the plane, the problem is mapped to the lines &lt;math&gt;c,d,(b-a)x+c,(b-a)x+d&lt;/math&gt; and &lt;math&gt;c,-d,(b-a)x+c,(b-a)x-d&lt;/math&gt;. Now, the area of the parallelogram contained by is the former is equal to the area of a rectangle with sides &lt;math&gt;d-c&lt;/math&gt; and &lt;math&gt;\frac{d-c}{b-a}&lt;/math&gt;, &lt;math&gt;\frac{(d-c)^2}{b-a}=18&lt;/math&gt;, and the area contained by the latter is &lt;math&gt;\frac{(c+d)^2}{b-a}=72&lt;/math&gt;. Thus, &lt;math&gt;d=3c&lt;/math&gt; and &lt;math&gt;b-a&lt;/math&gt; must be even if the former quantity is to equal &lt;math&gt;18&lt;/math&gt;. &lt;math&gt;c^2=18(b-a)&lt;/math&gt; so &lt;math&gt;c&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;. Putting this all together, the minimal solution for &lt;math&gt;(a,b,c,d)=(3,1,3,9)&lt;/math&gt;, so the sum is &lt;math&gt; \boxed{\textbf{(D)} 16} &lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC12 box|year=2007|ab=B|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_23&diff=70106 2007 AMC 12B Problems/Problem 23 2015-05-02T20:20:19Z <p>Pragmatictnt: /* Solution #2 */</p> <hr /> <div>==Problem 23==<br /> How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to &lt;math&gt;3&lt;/math&gt; times their perimeters?<br /> <br /> &lt;math&gt;\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be the two legs of the triangle.<br /> <br /> We have &lt;math&gt;\frac{1}{2}ab = 3(a+b+c)&lt;/math&gt;.<br /> <br /> Then &lt;math&gt;ab=6\cdot (a+b+\sqrt {a^2 + b^2})&lt;/math&gt;.<br /> <br /> We can complete the square under the root, and we get, &lt;math&gt;ab=6\cdot (a+b+\sqrt {(a+b)^2 - 2ab})&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;ab=p&lt;/math&gt; and &lt;math&gt;a+b=s&lt;/math&gt;, we have &lt;math&gt;p=6\cdot (s+ \sqrt {s^2 - 2p})&lt;/math&gt;.<br /> <br /> After rearranging, squaring both sides, and simplifying, we have &lt;math&gt;p=12s-72&lt;/math&gt;.<br /> <br /> <br /> Putting back &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, and after factoring using &lt;math&gt;SFFT&lt;/math&gt;, we've got &lt;math&gt;(a-12)\cdot (b-12)=72&lt;/math&gt;.<br /> <br /> <br /> Factoring 72, we get 6 pairs of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).&lt;/math&gt;<br /> <br /> <br /> And this gives us &lt;math&gt;6&lt;/math&gt; solutions &lt;math&gt;\Rightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Solution #2==<br /> We will proceed by using the fact that &lt;math&gt;[ABC] = r\cdot s&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the radius of the incircle and &lt;math&gt;s&lt;/math&gt; is the semiperimeter (&lt;math&gt;s = \frac{p}{2}&lt;/math&gt;).<br /> <br /> We are given &lt;math&gt;[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6&lt;/math&gt;.<br /> <br /> The incircle of &lt;math&gt;ABC&lt;/math&gt; breaks the triangle's sides into segments such that &lt;math&gt;AB = x + y&lt;/math&gt;, &lt;math&gt;BC = x + z&lt;/math&gt; and &lt;math&gt;AC = y + z&lt;/math&gt;. Since ABC is a right triangle, one of &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; is equal to its radius, 6. Let's assume &lt;math&gt;z = 6&lt;/math&gt;.<br /> <br /> The side lengths then become &lt;math&gt;AB = x + y&lt;/math&gt;, &lt;math&gt;BC = x + 6&lt;/math&gt; and &lt;math&gt;AC = y + 6&lt;/math&gt;. Plugging into Pythagorean's theorem:<br /> <br /> &lt;math&gt;(x + y)^2 = (x+6)^2 + (y + 6)^2&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2 + 2xy + y^2 = x^2 + 12x + 36 + y^2 + 12y + 36&lt;/math&gt;<br /> <br /> &lt;math&gt;2xy - 12x - 12y = 72&lt;/math&gt;<br /> <br /> &lt;math&gt;xy - 6x - 6y = 36&lt;/math&gt;<br /> <br /> &lt;math&gt;(x - 6)(y - 6) - 36 = 36&lt;/math&gt;<br /> <br /> &lt;math&gt;(x - 6)(y - 6) = 72&lt;/math&gt;<br /> <br /> We can factor &lt;math&gt;72&lt;/math&gt; to arrive with &lt;math&gt;6&lt;/math&gt; pairs of solutions: &lt;math&gt;(7, 78), (8,42), (9, 30), (10, 24), (12, 18),&lt;/math&gt; and &lt;math&gt;(14, 15) \Rightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2007|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_24&diff=70105 2007 AMC 12B Problems/Problem 24 2015-05-02T20:19:30Z <p>Pragmatictnt: /* Solution 3 */</p> <hr /> <div>==Problem 24==<br /> How many pairs of positive integers &lt;math&gt;(a,b)&lt;/math&gt; are there such that &lt;math&gt;\gcd(a,b)=1&lt;/math&gt; and &lt;cmath&gt;\frac{a}{b}+\frac{14b}{9a}&lt;/cmath&gt; is an integer?<br /> <br /> &lt;math&gt;\mathrm {(A)} 4&lt;/math&gt; &lt;math&gt;\mathrm {(B)} 6&lt;/math&gt; &lt;math&gt;\mathrm {(C)} 9&lt;/math&gt; &lt;math&gt;\mathrm {(D)} 12&lt;/math&gt; &lt;math&gt;\mathrm {(E)} \text{infinitely many}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Combining the fraction, &lt;math&gt;\frac{9a^2 + 14b^2}{9ab}&lt;/math&gt; must be an integer.<br /> <br /> Since the denominator contains a factor of &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b&lt;/math&gt;<br /> <br /> Since &lt;math&gt;b = 3n&lt;/math&gt; for some positive integer &lt;math&gt;n&lt;/math&gt;, we can rewrite the fraction(divide by &lt;math&gt;9&lt;/math&gt; on both top and bottom) as &lt;math&gt;\frac{a^2 + 14n^2}{3an}&lt;/math&gt;<br /> <br /> Since the denominator now contains a factor of &lt;math&gt;n&lt;/math&gt;, we get &lt;math&gt;n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2&lt;/math&gt;.<br /> <br /> But since &lt;math&gt;1=gcd(a,b)=gcd(a,3n)=gcd(a,n)&lt;/math&gt;, we must have &lt;math&gt;n=1&lt;/math&gt;, and thus &lt;math&gt;b=3&lt;/math&gt;.<br /> <br /> For &lt;math&gt;b=3&lt;/math&gt; the original fraction simplifies to &lt;math&gt;\frac{a^2 + 14}{3a}&lt;/math&gt;.<br /> <br /> For that to be an integer, &lt;math&gt;a&lt;/math&gt; must divide &lt;math&gt;14&lt;/math&gt;, and therefore we must have &lt;math&gt;a\in\{1,2,7,14\}&lt;/math&gt;. Each of these values does indeed yield an integer.<br /> <br /> Thus there are four solutions: &lt;math&gt;(1,3)&lt;/math&gt;, &lt;math&gt;(2,3)&lt;/math&gt;, &lt;math&gt;(7,3)&lt;/math&gt;, &lt;math&gt;(14,3)&lt;/math&gt; and the answer is &lt;math&gt;\mathrm {(A)}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let's assume that &lt;math&gt;\frac{a}{b} + \frac{14b}{9a} = m&lt;/math&gt; We get--<br /> <br /> &lt;math&gt;9a^2 - 9mab + 14b^2 = 0&lt;/math&gt;<br /> <br /> Factoring this, we get &lt;math&gt;4&lt;/math&gt; equations-<br /> <br /> &lt;math&gt;(3a-2b)(3a-7b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(3a-b)(3a-14b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(a-2b)(9a-7b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(a-b)(9a-14b) = 0&lt;/math&gt;<br /> <br /> (It's all negative, because if we had positive signs, &lt;math&gt;a&lt;/math&gt; would be the opposite sign of &lt;math&gt;b&lt;/math&gt;)<br /> <br /> Now we look at these, and see that-<br /> <br /> &lt;math&gt;3a=2b&lt;/math&gt; <br /> <br /> &lt;math&gt;3a=b&lt;/math&gt;<br /> <br /> &lt;math&gt;3a=7b&lt;/math&gt; <br /> <br /> &lt;math&gt;3a=14b&lt;/math&gt;<br /> <br /> &lt;math&gt;a=2b&lt;/math&gt; <br /> <br /> &lt;math&gt;9a=7b&lt;/math&gt;<br /> <br /> &lt;math&gt;a=b&lt;/math&gt; <br /> <br /> &lt;math&gt;9a=14b&lt;/math&gt;<br /> <br /> This gives us &lt;math&gt;8&lt;/math&gt; solutions, but we note that the middle term needs to give you back &lt;math&gt;9m&lt;/math&gt;.<br /> <br /> For example, in the case <br /> <br /> &lt;math&gt;(a-2b)(9a-7b)&lt;/math&gt;, the middle term is &lt;math&gt;-25ab&lt;/math&gt;, which is not equal by &lt;math&gt;-9m&lt;/math&gt; for whatever integar &lt;math&gt;m&lt;/math&gt;.<br /> <br /> Similar reason for the fourth equation. This elimnates the last four solutions out of the above eight listed, giving us 4 solutions total &lt;math&gt;\mathrm {(A)}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;u = \frac{a}{b}&lt;/math&gt;. Then the given equation becomes &lt;math&gt;u + \frac{14}{9u} = \frac{9u^2 + 14}{9u}&lt;/math&gt;.<br /> <br /> Let's set this equal to some value, &lt;math&gt;k \Rightarrow \frac{9u^2 + 14}{9u} = k&lt;/math&gt;.<br /> <br /> Clearing the denominator and simplifying, we get a quadratic in terms of &lt;math&gt;u&lt;/math&gt;:<br /> <br /> &lt;math&gt;9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}&lt;/math&gt;<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers, &lt;math&gt;u&lt;/math&gt; is a rational number. This means that &lt;math&gt;\sqrt{(9k)^2 - 504}&lt;/math&gt; is an integer.<br /> <br /> Let &lt;math&gt;\sqrt{(9k)^2 - 504} = x&lt;/math&gt;. Squaring and rearranging yields:<br /> <br /> &lt;math&gt;(9k)^2 - x^2 = 504&lt;/math&gt;<br /> <br /> &lt;math&gt;(9k+x)(9k-x) = 504&lt;/math&gt;.<br /> <br /> In order for both &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;a&lt;/math&gt; to be an integer, &lt;math&gt;9k + x&lt;/math&gt; and &lt;math&gt;9k - x&lt;/math&gt; must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let &lt;math&gt;9k + x = 2m&lt;/math&gt; and &lt;math&gt;9k - x = 2n&lt;/math&gt;.<br /> <br /> Then:<br /> <br /> &lt;math&gt;2m \cdot 2n = 504&lt;/math&gt;<br /> <br /> &lt;math&gt;mn = 126&lt;/math&gt;.<br /> <br /> Factoring 126, we get &lt;math&gt;6&lt;/math&gt; pairs of numbers: &lt;math&gt;(1,126), (2,63), (3,42), (6,21), (7,18),&lt;/math&gt; and &lt;math&gt;(9,14)&lt;/math&gt;.<br /> <br /> Looking back at our equations for &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, we can solve for &lt;math&gt;k = \frac{2m + 2n}{18} = \frac{m+n}{9}&lt;/math&gt;. Since &lt;math&gt;k&lt;/math&gt; is an integer, there are only &lt;math&gt;2&lt;/math&gt; pairs of &lt;math&gt;(m,n)&lt;/math&gt; that work: &lt;math&gt;(3,42)&lt;/math&gt; and &lt;math&gt;(6,21)&lt;/math&gt;. This means that there are &lt;math&gt;2&lt;/math&gt; values of &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;u&lt;/math&gt; is an integer. But looking back at &lt;math&gt;u&lt;/math&gt; in terms of &lt;math&gt;k&lt;/math&gt;, we have &lt;math&gt;\pm&lt;/math&gt;, meaning that there are &lt;math&gt;2&lt;/math&gt; values of &lt;math&gt;u&lt;/math&gt; for every &lt;math&gt;k&lt;/math&gt;. Thus, the answer is &lt;math&gt;2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_24&diff=70104 2007 AMC 12B Problems/Problem 24 2015-05-02T20:17:54Z <p>Pragmatictnt: /* Solution 2 */</p> <hr /> <div>==Problem 24==<br /> How many pairs of positive integers &lt;math&gt;(a,b)&lt;/math&gt; are there such that &lt;math&gt;\gcd(a,b)=1&lt;/math&gt; and &lt;cmath&gt;\frac{a}{b}+\frac{14b}{9a}&lt;/cmath&gt; is an integer?<br /> <br /> &lt;math&gt;\mathrm {(A)} 4&lt;/math&gt; &lt;math&gt;\mathrm {(B)} 6&lt;/math&gt; &lt;math&gt;\mathrm {(C)} 9&lt;/math&gt; &lt;math&gt;\mathrm {(D)} 12&lt;/math&gt; &lt;math&gt;\mathrm {(E)} \text{infinitely many}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Combining the fraction, &lt;math&gt;\frac{9a^2 + 14b^2}{9ab}&lt;/math&gt; must be an integer.<br /> <br /> Since the denominator contains a factor of &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b&lt;/math&gt;<br /> <br /> Since &lt;math&gt;b = 3n&lt;/math&gt; for some positive integer &lt;math&gt;n&lt;/math&gt;, we can rewrite the fraction(divide by &lt;math&gt;9&lt;/math&gt; on both top and bottom) as &lt;math&gt;\frac{a^2 + 14n^2}{3an}&lt;/math&gt;<br /> <br /> Since the denominator now contains a factor of &lt;math&gt;n&lt;/math&gt;, we get &lt;math&gt;n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2&lt;/math&gt;.<br /> <br /> But since &lt;math&gt;1=gcd(a,b)=gcd(a,3n)=gcd(a,n)&lt;/math&gt;, we must have &lt;math&gt;n=1&lt;/math&gt;, and thus &lt;math&gt;b=3&lt;/math&gt;.<br /> <br /> For &lt;math&gt;b=3&lt;/math&gt; the original fraction simplifies to &lt;math&gt;\frac{a^2 + 14}{3a}&lt;/math&gt;.<br /> <br /> For that to be an integer, &lt;math&gt;a&lt;/math&gt; must divide &lt;math&gt;14&lt;/math&gt;, and therefore we must have &lt;math&gt;a\in\{1,2,7,14\}&lt;/math&gt;. Each of these values does indeed yield an integer.<br /> <br /> Thus there are four solutions: &lt;math&gt;(1,3)&lt;/math&gt;, &lt;math&gt;(2,3)&lt;/math&gt;, &lt;math&gt;(7,3)&lt;/math&gt;, &lt;math&gt;(14,3)&lt;/math&gt; and the answer is &lt;math&gt;\mathrm {(A)}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let's assume that &lt;math&gt;\frac{a}{b} + \frac{14b}{9a} = m&lt;/math&gt; We get--<br /> <br /> &lt;math&gt;9a^2 - 9mab + 14b^2 = 0&lt;/math&gt;<br /> <br /> Factoring this, we get &lt;math&gt;4&lt;/math&gt; equations-<br /> <br /> &lt;math&gt;(3a-2b)(3a-7b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(3a-b)(3a-14b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(a-2b)(9a-7b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(a-b)(9a-14b) = 0&lt;/math&gt;<br /> <br /> (It's all negative, because if we had positive signs, &lt;math&gt;a&lt;/math&gt; would be the opposite sign of &lt;math&gt;b&lt;/math&gt;)<br /> <br /> Now we look at these, and see that-<br /> <br /> &lt;math&gt;3a=2b&lt;/math&gt; <br /> <br /> &lt;math&gt;3a=b&lt;/math&gt;<br /> <br /> &lt;math&gt;3a=7b&lt;/math&gt; <br /> <br /> &lt;math&gt;3a=14b&lt;/math&gt;<br /> <br /> &lt;math&gt;a=2b&lt;/math&gt; <br /> <br /> &lt;math&gt;9a=7b&lt;/math&gt;<br /> <br /> &lt;math&gt;a=b&lt;/math&gt; <br /> <br /> &lt;math&gt;9a=14b&lt;/math&gt;<br /> <br /> This gives us &lt;math&gt;8&lt;/math&gt; solutions, but we note that the middle term needs to give you back &lt;math&gt;9m&lt;/math&gt;.<br /> <br /> For example, in the case <br /> <br /> &lt;math&gt;(a-2b)(9a-7b)&lt;/math&gt;, the middle term is &lt;math&gt;-25ab&lt;/math&gt;, which is not equal by &lt;math&gt;-9m&lt;/math&gt; for whatever integar &lt;math&gt;m&lt;/math&gt;.<br /> <br /> Similar reason for the fourth equation. This elimnates the last four solutions out of the above eight listed, giving us 4 solutions total &lt;math&gt;\mathrm {(A)}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;u = \frac{a}{b}&lt;/math&gt;. Then the given equation becomes &lt;math&gt;u + \frac{14}{9u} = \frac{9u^2 + 14}{9u}&lt;/math&gt;.<br /> <br /> Let's set this equal to some value, &lt;math&gt;k \Rightarrow \frac{9u^2 + 14}{9u} = k&lt;/math&gt;.<br /> <br /> Clearing the denominator and simplifying, we get a quadratic in terms of &lt;math&gt;u&lt;/math&gt;:<br /> <br /> &lt;math&gt;9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}&lt;/math&gt;<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers, &lt;math&gt;u&lt;/math&gt; is a rational number. This means that &lt;math&gt;\sqrt{(9k)^2 - 504}&lt;/math&gt; is an integer.<br /> <br /> Let &lt;math&gt;\sqrt{(9k)^2 - 504} = x&lt;/math&gt;. Squaring and rearranging yields:<br /> <br /> &lt;math&gt;(9k)^2 - x^2 = 504&lt;/math&gt;<br /> <br /> &lt;math&gt;(9k+x)(9k-x) = 504&lt;/math&gt;.<br /> <br /> In order for both &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;a&lt;/math&gt; to be an integer, &lt;math&gt;9k + x&lt;/math&gt; and &lt;math&gt;9k - x&lt;/math&gt; must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let &lt;math&gt;9k + x = 2m&lt;/math&gt; and &lt;math&gt;9k - x = 2n&lt;/math&gt;.<br /> <br /> Then:<br /> <br /> &lt;math&gt;2m \cdot 2n = 504&lt;/math&gt;<br /> <br /> &lt;math&gt;mn = 126&lt;/math&gt;.<br /> <br /> Factoring 126, we get &lt;math&gt;6&lt;/math&gt; pairs of numbers: &lt;math&gt;(1,126), (2,63), (3,42), (6,21), (7,18) &amp; (9,14)&lt;/math&gt;.<br /> <br /> Looking back at our equations for &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, we can solve for &lt;math&gt;k = \frac{2m + 2n}{18} = \frac{m+n}{9}&lt;/math&gt;. Since &lt;math&gt;k&lt;/math&gt; is an integer, there are only &lt;math&gt;2&lt;/math&gt; pairs of &lt;math&gt;(m,n)&lt;/math&gt; that work: &lt;math&gt;(3,42) &amp; (6,21)&lt;/math&gt;. This means that there are &lt;math&gt;2&lt;/math&gt; values of &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;u&lt;/math&gt; is an integer. But looking back at &lt;math&gt;u&lt;/math&gt; in terms of &lt;math&gt;k&lt;/math&gt;, we have &lt;math&gt;\pm&lt;/math&gt;, meaning that there are &lt;math&gt;2&lt;/math&gt; values of &lt;math&gt;u&lt;/math&gt; for every &lt;math&gt;k&lt;/math&gt;. Thus, the answer is &lt;math&gt;2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_25&diff=70103 2007 AMC 12B Problems/Problem 25 2015-05-02T20:17:06Z <p>Pragmatictnt: /* Solution */</p> <hr /> <div>==Problem==<br /> Points &lt;math&gt;A,B,C,D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; are located in 3-dimensional space with &lt;math&gt;AB=BC=CD=DE=EA=2&lt;/math&gt; and &lt;math&gt;\angle ABC=\angle CDE=\angle DEA=90^o&lt;/math&gt;. The plane of &lt;math&gt;\triangle ABC&lt;/math&gt; is parallel to &lt;math&gt;\overline{DE}&lt;/math&gt;. What is the area of &lt;math&gt;\triangle BDE&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm {(A)} \sqrt{2}\qquad \mathrm {(B)} \sqrt{3}\qquad \mathrm {(C)} 2\qquad \mathrm {(D)} \sqrt{5}\qquad \mathrm {(E)} \sqrt{6}&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;A=(0,0,0)&lt;/math&gt;, and &lt;math&gt;B=(2,0,0)&lt;/math&gt;. Since &lt;math&gt;EA=2&lt;/math&gt;, we could let &lt;math&gt;C=(2,0,2)&lt;/math&gt;, &lt;math&gt;D=(2,2,2)&lt;/math&gt;, and &lt;math&gt;E=(2,2,0)&lt;/math&gt;. Now to get back to &lt;math&gt;A&lt;/math&gt; we need another vertex &lt;math&gt;F=(0,2,0)&lt;/math&gt;. Now if we look at this configuration as if it was two dimensions, we would see a square missing a side if we don't draw &lt;math&gt;FA&lt;/math&gt;. Now we can bend these three sides into an equilateral triangle, and the coordinates change: &lt;math&gt;A=(0,0,0)&lt;/math&gt;, &lt;math&gt;B=(2,0,0)&lt;/math&gt;, &lt;math&gt;C=(2,0,2)&lt;/math&gt;, &lt;math&gt;D=(1,\sqrt{3},2)&lt;/math&gt;, and &lt;math&gt;E=(1,\sqrt{3},0)&lt;/math&gt;. Checking for all the requirements, they are all satisfied. Now we find the area of triangle &lt;math&gt;BDE&lt;/math&gt;. It is a &lt;math&gt;2-2-2\sqrt{2}&lt;/math&gt; triangle, which is an isosceles right triangle. Thus the area of it is &lt;math&gt;\frac{2\cdot2}{2}=2\Rightarrow \mathrm{(C)}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC12 box|year=2007|ab=B|num-b=24|after=Last Problem}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Pragmatictnt https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_23&diff=70102 2008 AMC 12A Problems/Problem 23 2015-05-02T17:17:37Z <p>Pragmatictnt: /* Solution */</p> <hr /> <div>==Problem==<br /> The solutions of the equation &lt;math&gt;z^4+4z^3i-6z^2-4zi-i=0&lt;/math&gt; are the vertices of a convex polygon in the complex plane. What is the area of the polygon?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2^{\frac{5}{8}}\qquad\mathrm{(B)}\ 2^{\frac{3}{4}}\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 2^{\frac{5}{4}}\qquad\mathrm{(E)}\ 2^{\frac{3}{2}}&lt;/math&gt;<br /> <br /> ==Solution==<br /> Looking at the coefficients, we are immediately reminded of the binomial expansion of &lt;math&gt;{\left(x+1\right)}^{4}&lt;/math&gt;.<br /> <br /> Modifying this slightly, we can write the given equation as:<br /> &lt;cmath&gt; {\left(z+i\right)}^{4}=1+i=2^{\frac{1}{2}}\cdot \text{cis}\, \frac {\pi}{4} &lt;/cmath&gt;<br /> We can apply a translation of &lt;math&gt;-i&lt;/math&gt; and a rotation of &lt;math&gt;-\frac{\pi}{4}&lt;/math&gt; (both operations preserve area) to simplify the problem:<br /> &lt;cmath&gt;z^{4}=2^{\frac{1}{2}}&lt;/cmath&gt;<br /> <br /> Because the roots of this equation are created by rotating &lt;math&gt;\frac{\pi}{2}&lt;/math&gt; radians successively about the origin, the quadrilateral is a square.<br /> <br /> We know that half the diagonal length of the square is &lt;math&gt;{\left(2^{\frac{1}{2}}\right)}^{\frac{1}{4}}=2^{\frac{1}{8}}&lt;/math&gt;<br /> <br /> Therefore, the area of the square is<br /> &lt;math&gt; \frac{{\left( 2 \cdot 2^{\frac{1}{8}}\right)}^2}{2}=\frac{2^{\frac{9}{4}}}{2}=2^{\frac{5}{4}} \Rightarrow D. &lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2008|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Pragmatictnt