https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Nukelauncher&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-24T13:00:13Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems&diff=102784 2019 AMC 10B Problems 2019-02-15T03:43:02Z <p>Nukelauncher: /* Problem 8 */</p> <hr /> <div>{{AMC10 Problems|year=2019|ab=B}}<br /> <br /> ==Problem 1==<br /> <br /> Alicia had two containers. The first was &lt;math&gt;\tfrac{5}{6}&lt;/math&gt; full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; full of water. What is the ratio of the volume of the first container to the volume of the second container?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{5}{8} \qquad \textbf{(B) } \frac{4}{5} \qquad \textbf{(C) } \frac{7}{8} \qquad \textbf{(D) } \frac{9}{10} \qquad \textbf{(E) } \frac{11}{12}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Consider the statement, &quot;If &lt;math&gt;n&lt;/math&gt; is not prime, then &lt;math&gt;n-2&lt;/math&gt; is prime.&quot; Which of the following values of &lt;math&gt;n&lt;/math&gt; is a counterexample to this statement.<br /> <br /> &lt;math&gt;\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> In a high school with &lt;math&gt;500&lt;/math&gt; students, &lt;math&gt;40\%&lt;/math&gt; of the seniors play a musical instrument, while &lt;math&gt;30\%&lt;/math&gt; of the non-seniors do not play a musical instrument. In all, &lt;math&gt;46.8\%&lt;/math&gt; of the students do not play a musical instrument. How many non-seniors play a musical instrument?<br /> <br /> &lt;math&gt;\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> All lines with equation &lt;math&gt;ax+by=c&lt;/math&gt; such that &lt;math&gt;a,b,c&lt;/math&gt; form an arithmetic progression pass through a common point. What are the coordinates of that point?<br /> <br /> &lt;math&gt;\textbf{(A) } (-1,2)<br /> \qquad\textbf{(B) } (0,1)<br /> \qquad\textbf{(C) } (1,-2)<br /> \qquad\textbf{(D) } (1,0)<br /> \qquad\textbf{(E) } (1,2)&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; lies in the first quadrant. Points &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; are reflected across the line &lt;math&gt;y=x&lt;/math&gt; to points &lt;math&gt;A'&lt;/math&gt;, &lt;math&gt;B'&lt;/math&gt;, and &lt;math&gt;C'&lt;/math&gt;, respectively. Assume that none of the vertices of the triangle lie on the line &lt;math&gt;y=x&lt;/math&gt;. Which of the following statements is &lt;i&gt;&lt;u&gt;not&lt;/u&gt;&lt;/i&gt; always true?<br /> <br /> &lt;math&gt;\textbf{(A) }&lt;/math&gt; Triangle &lt;math&gt;A'B'C'&lt;/math&gt; lies in the first quadrant.<br /> <br /> &lt;math&gt;\textbf{(B) }&lt;/math&gt; Triangles &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;A'B'C'&lt;/math&gt; have the same area.<br /> <br /> &lt;math&gt;\textbf{(C) }&lt;/math&gt; The slope of line &lt;math&gt;AA'&lt;/math&gt; is &lt;math&gt;-1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(D) }&lt;/math&gt; The slopes of lines &lt;math&gt;AA'&lt;/math&gt; and &lt;math&gt;CC'&lt;/math&gt; are the same.<br /> <br /> &lt;math&gt;\textbf{(E) }&lt;/math&gt; Lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;A'B'&lt;/math&gt; are perpendicular to each other.<br /> <br /> [[2019 AMC 10B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> There is a real &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;(n+1)! + (n+2)! = n! \cdot 440&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either 12 pieces of red candy, 14 pieces of green candy, 15 pieces of blue candy, or &lt;math&gt;n&lt;/math&gt; pieces of purple candy. A piece of purple candy costs 20 cents. What is the smallest possible value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> The figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length 2 and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region?<br /> <br /> &lt;asy&gt;<br /> pen white = gray(1);<br /> pen gray = gray(0.5);<br /> draw((0,0)--(2sqrt(3),0)--(2sqrt(3),2sqrt(3))--(0,2sqrt(3))--cycle);<br /> fill((0,0)--(2sqrt(3),0)--(2sqrt(3),2sqrt(3))--(0,2sqrt(3))--cycle, gray);<br /> draw((sqrt(3)-1,0)--(sqrt(3),sqrt(3))--(sqrt(3)+1,0)--cycle);<br /> fill((sqrt(3)-1,0)--(sqrt(3),sqrt(3))--(sqrt(3)+1,0)--cycle, white);<br /> draw((sqrt(3)-1,2sqrt(3))--(sqrt(3),sqrt(3))--(sqrt(3)+1,2sqrt(3))--cycle);<br /> fill((sqrt(3)-1,2sqrt(3))--(sqrt(3),sqrt(3))--(sqrt(3)+1,2sqrt(3))--cycle, white);<br /> draw((0,sqrt(3)-1)--(sqrt(3),sqrt(3))--(0,sqrt(3)+1)--cycle);<br /> fill((0,sqrt(3)-1)--(sqrt(3),sqrt(3))--(0,sqrt(3)+1)--cycle, white);<br /> draw((2sqrt(3),sqrt(3)-1)--(sqrt(3),sqrt(3))--(2sqrt(3),sqrt(3)+1)--cycle);<br /> fill((2sqrt(3),sqrt(3)-1)--(sqrt(3),sqrt(3))--(2sqrt(3),sqrt(3)+1)--cycle, white);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 4 \qquad \textbf{(B) } 12 - 4\sqrt{3} \qquad \textbf{(C) } 3\sqrt{3}\qquad \textbf{(D) } 4\sqrt{3} \qquad \textbf{(E) } 16 - \sqrt{3}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> The function &lt;math&gt;f&lt;/math&gt; is defined by &lt;cmath&gt;f(x) = \lfloor|x|\rfloor - |\lfloor x \rfloor|&lt;/cmath&gt;for all real numbers &lt;math&gt;x&lt;/math&gt;, where &lt;math&gt;\lfloor r \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to the real number &lt;math&gt;r&lt;/math&gt;. What is the range of &lt;math&gt;f&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \{-1, 0\} \qquad\textbf{(B) } \text{The set of nonpositive integers} \qquad\textbf{(C) } \{-1, 0, 1\} \qquad\textbf{(D) } \{0\} \qquad\textbf{(E) } \text{The set of nonnegative integers} &lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> In a given plane, points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are &lt;math&gt;10&lt;/math&gt; units apart. How many points &lt;math&gt;C&lt;/math&gt; are there in the plane such that the perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;50&lt;/math&gt; units and the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;100&lt;/math&gt; square units?<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar 1 the ratio of blue to green marbles is 9:1, and the ratio of blue to green marbles in Jar 2 is 8:1. There are 95 green marbles in all. How many more blue marbles are in Jar 1 than in Jar 2?<br /> <br /> &lt;math&gt;\textbf{(A) } 5<br /> \qquad\textbf{(B) } 10<br /> \qquad\textbf{(C) } 25<br /> \qquad\textbf{(D) } 45<br /> \qquad\textbf{(E) } 50&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than &lt;math&gt;2019&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 11<br /> \qquad\textbf{(B) } 14<br /> \qquad\textbf{(C) } 22<br /> \qquad\textbf{(D) } 23<br /> \qquad\textbf{(E) } 27&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> What is the sum of all real numbers &lt;math&gt;x&lt;/math&gt; for which the median of the numbers &lt;math&gt;4,6,8,17,&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; is equal to the mean of those five numbers?<br /> <br /> &lt;math&gt;\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> The base-ten representation for &lt;math&gt;19!&lt;/math&gt; is &lt;math&gt;121,6T5,100,40M,832,H00&lt;/math&gt;, where &lt;math&gt;T&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, and &lt;math&gt;H&lt;/math&gt; denote digits that are not given. What is &lt;math&gt;T+M+H&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }3 \qquad\textbf{(B) }8 \qquad\textbf{(C) }12 \qquad\textbf{(D) }14 \qquad\textbf{(E) } 17 &lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Two right triangles, &lt;math&gt;T_1&lt;/math&gt; and &lt;math&gt;T_2&lt;/math&gt;, have areas of 1 and 2, respectively. One side length of one triangle is congruent to a different side length in the other, and another side length of the first triangle is congruent to yet another side length in the other. What is the product of the third side lengths of &lt;math&gt;T_1&lt;/math&gt; and &lt;math&gt;T_2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{28}{3} \qquad\textbf{(B) }10\qquad\textbf{(C) } \frac{32}{3} \qquad\textbf{(D) } \frac{34}{3} \qquad\textbf{(E) }12&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; with a right angle at &lt;math&gt;C,&lt;/math&gt; point &lt;math&gt;D&lt;/math&gt; lies in the interior of &lt;math&gt;\overline{AB}&lt;/math&gt; and point &lt;math&gt;E&lt;/math&gt; lies in the interior of &lt;math&gt;\overline{BC}&lt;/math&gt; so that &lt;math&gt;AC=CD,&lt;/math&gt; &lt;math&gt;DE=EB,&lt;/math&gt; and the ratio &lt;math&gt;AC:DE=4:3.&lt;/math&gt; What is the ratio &lt;math&gt;AD:DB?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 2:3<br /> \qquad\textbf{(B) } 2:\sqrt{5}<br /> \qquad\textbf{(C) } 1:1<br /> \qquad\textbf{(D) } 3:\sqrt{5}<br /> \qquad\textbf{(E) } 3:2&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> A red ball and a green ball are randomly and independently tossed into bins numbered with positive integers so that for each ball, the probability that it is tossed into bin &lt;math&gt;k&lt;/math&gt; is &lt;math&gt;2^{-k}&lt;/math&gt; for &lt;math&gt;k=1,2,3,\ldots.&lt;/math&gt; What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Henry decides one morning to do a workout, and he walks &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; of the way from his home to his gym. The gym is &lt;math&gt;2&lt;/math&gt; kilometers away from Henry's home. At that point, he changes his mind and walks &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; of the distance toward either the gym or home from the point where he last changed his mind, he will get very close to walking back and forth between a point &lt;math&gt;A&lt;/math&gt; kilometers from home and a point &lt;math&gt;B&lt;/math&gt; kilometers from home. What is &lt;math&gt;|A-B|&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{2}{3} \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 1\frac{1}{5} \qquad \textbf{(D) } 1\frac{1}{4} \qquad \textbf{(E) } 1\frac{1}{2}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of all positive integer divisors of &lt;math&gt;100,000.&lt;/math&gt; How many numbers are the product of two distinct elements of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> As shown in the figure, line segment &lt;math&gt;\overline{AD}&lt;/math&gt; is trisected by points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;AB=BC=CD=2.&lt;/math&gt; Three semicircles of radius &lt;math&gt;1,&lt;/math&gt; &lt;math&gt;\overarc{AEB},\overarc{BFC},&lt;/math&gt; and &lt;math&gt;\overarc{CGD},&lt;/math&gt; have their diameters on &lt;math&gt;\overline{AD},&lt;/math&gt; and are tangent to line &lt;math&gt;EG&lt;/math&gt; at &lt;math&gt;E,F,&lt;/math&gt; and &lt;math&gt;G,&lt;/math&gt; respectively. A circle of radius &lt;math&gt;2&lt;/math&gt; has its center on &lt;math&gt;F. &lt;/math&gt; The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form<br /> &lt;cmath&gt;\frac{a}{b}\cdot\pi-\sqrt{c}+d,&lt;/cmath&gt;where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive integers and &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime. What is &lt;math&gt;a+b+c+d&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> filldraw(circle((0,0),2), gray(0.7));<br /> filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0));<br /> dot((-3,-1));<br /> label(&quot;$A$&quot;,(-3,-1),S);<br /> dot((-2,0));<br /> label(&quot;$E$&quot;,(-2,0),NW);<br /> dot((-1,-1));<br /> label(&quot;$B$&quot;,(-1,-1),S);<br /> dot((0,0));<br /> label(&quot;$F$&quot;,(0,0),N);<br /> dot((1,-1));<br /> label(&quot;$C$&quot;,(1,-1), S);<br /> dot((2,0));<br /> label(&quot;$G$&quot;, (2,0),NE);<br /> dot((3,-1));<br /> label(&quot;$D$&quot;, (3,-1), S);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> Debra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{36} \qquad \textbf{(B) } \frac{1}{24} \qquad \textbf{(C) } \frac{1}{18} \qquad \textbf{(D) } \frac{1}{12} \qquad \textbf{(E) } \frac{1}{6}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> Raashan, Sylvia, and Ted play the following game. Each starts with &lt;math&gt; \$1&lt;/math&gt;. A bell rings every &lt;math&gt;15&lt;/math&gt; seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives &lt;math&gt;\$1&lt;/math&gt; to that player. What is the probability that after the bell has rung &lt;math&gt;2019&lt;/math&gt; times, each player will have &lt;math&gt;\$1&lt;/math&gt;? (For example, Raashan and Ted may each decide to give &lt;math&gt;\$1&lt;/math&gt; to Sylvia, and Sylvia may decide to give her her dollar to Ted, at which point Raashan will have &lt;math&gt;\$0&lt;/math&gt;, Sylvia will have &lt;math&gt;\$2&lt;/math&gt;, and Ted will have &lt;math&gt;\$1&lt;/math&gt;, and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their &lt;math&gt; \$1&lt;/math&gt; to, and the holdings will be the same at the end of the second round.)<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{7} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{1}{2} \qquad\textbf{(E) } \frac{2}{3}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Points &lt;math&gt;A(6,13)&lt;/math&gt; and &lt;math&gt;B(12,11)&lt;/math&gt; lie on circle &lt;math&gt;\omega&lt;/math&gt; in the plane. Suppose that the tangent lines to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at a point on the &lt;math&gt;x&lt;/math&gt;-axis. What is the area of &lt;math&gt;\omega&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br /> \frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Define a sequence recursively by &lt;math&gt;x_0=5&lt;/math&gt; and<br /> &lt;cmath&gt;x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}&lt;/cmath&gt;for all nonnegative integers &lt;math&gt;n.&lt;/math&gt; Let &lt;math&gt;m&lt;/math&gt; be the least positive integer such that<br /> &lt;cmath&gt;x_m\leq 4+\frac{1}{2^{20}}.&lt;/cmath&gt;In which of the following intervals does &lt;math&gt;m&lt;/math&gt; lie?<br /> <br /> &lt;math&gt;\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty]&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> How many sequences of &lt;math&gt;0&lt;/math&gt;s and &lt;math&gt;1&lt;/math&gt;s of length &lt;math&gt;19&lt;/math&gt; are there that begin with a &lt;math&gt;0&lt;/math&gt;, end with a &lt;math&gt;0&lt;/math&gt;, contain no two consecutive &lt;math&gt;0&lt;/math&gt;s, and contain no three consecutive &lt;math&gt;1&lt;/math&gt;s?<br /> <br /> &lt;math&gt;\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2019|ab=B|before=[[2019 AMC 10A Problems]]|after=[[2020 AMC 10A Problems]]}}<br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_25&diff=101589 2019 AMC 10A Problems/Problem 25 2019-02-09T23:26:42Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #25]] and [[2019 AMC 12A Problems|2019 AMC 12A #24]]}}<br /> <br /> ==Problem==<br /> <br /> For how many integers &lt;math&gt;n&lt;/math&gt; between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;50&lt;/math&gt;, inclusive, is &lt;cmath&gt;\frac{(n^2-1)!}{(n!)^n}&lt;/cmath&gt; an integer? (Recall that &lt;math&gt;0! = 1&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35&lt;/math&gt;<br /> <br /> ==Solution==<br /> The main insight is that &lt;cmath&gt;\frac{(n^2)!}{(n!)^{n+1}}&lt;/cmath&gt; is always an integer. This is true because it is precisely the number of ways to split up &lt;math&gt;n^2&lt;/math&gt; objects into &lt;math&gt;n&lt;/math&gt; unordered groups of size &lt;math&gt;n&lt;/math&gt;, which has to be an integer. Thus, &lt;cmath&gt;\frac{(n^2-1)!}{(n!)^n}&lt;/cmath&gt; is not an integer iff &lt;math&gt;n^2 \nmid n!&lt;/math&gt;. This happens only for &lt;math&gt;n=4&lt;/math&gt; or &lt;math&gt;n&lt;/math&gt; prime. Thus, there are &lt;math&gt;16&lt;/math&gt; integers that do not work, which means the answer is &lt;math&gt;\boxed{\mathbf{(D)}\ 34}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=24|after=Last Problem}}<br /> {{AMC12 box|year=2019|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO_Problems/Problem_6&diff=94107 2018 USAJMO Problems/Problem 6 2018-04-21T17:55:15Z <p>Nukelauncher: </p> <hr /> <div>==Problem 6==<br /> Karl starts with &lt;math&gt;n&lt;/math&gt; cards labeled &lt;math&gt;1,2,3,\dots,n&lt;/math&gt; lined up in a random order on his desk. He calls a pair &lt;math&gt;(a,b)&lt;/math&gt; of these cards swapped if &lt;math&gt;a&gt;b&lt;/math&gt; and the card labeled &lt;math&gt;a&lt;/math&gt; is to the left of the card labeled &lt;math&gt;b&lt;/math&gt;. For instance, in the sequence of cards &lt;math&gt;3,1,4,2&lt;/math&gt;, there are three swapped pairs of cards, &lt;math&gt;(3,1)&lt;/math&gt;, &lt;math&gt;(3,2)&lt;/math&gt;, and &lt;math&gt;(4,2)&lt;/math&gt;.<br /> <br /> He picks up the card labeled 1 and inserts it back into the sequence in the opposite position: if the card labeled 1 had &lt;math&gt;i&lt;/math&gt; card to its left, then it now has &lt;math&gt;i&lt;/math&gt; cards to its right. He then picks up the card labeled &lt;math&gt;2&lt;/math&gt; and reinserts it in the same manner, and so on until he has picked up and put back each of the cards &lt;math&gt;1,2,\dots,n&lt;/math&gt; exactly once in that order. (For example, the process starting at &lt;math&gt;3,1,4,2&lt;/math&gt; would be &lt;math&gt;3,1,4,2\to 3,4,1,2\to 2,3,4,1\to 2,4,3,1\to 2,3,4,1&lt;/math&gt;.)<br /> <br /> Show that, no matter what lineup of cards Karl started with, his final lineup has the same number of swapped pairs as the starting lineup.<br /> <br /> <br /> ==Solution==<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2018|num-b=5|aftertext=|after=Last Problem}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_6&diff=94106 2017 USAJMO Problems/Problem 6 2018-04-21T17:55:03Z <p>Nukelauncher: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;P_1, \ldots, P_{2n}&lt;/math&gt; be &lt;math&gt;2n&lt;/math&gt; distinct points on the unit circle &lt;math&gt;x^2 + y^2 = 1&lt;/math&gt; other than &lt;math&gt;(1,0)&lt;/math&gt;. Each point is colored either red or blue, with exactly &lt;math&gt;n&lt;/math&gt; of them red and exactly &lt;math&gt;n&lt;/math&gt; of them blue. Let &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; be any ordering of the red points. Let &lt;math&gt;B_1&lt;/math&gt; be the nearest blue point to &lt;math&gt;R_1&lt;/math&gt; traveling counterclockwise around the circle starting from &lt;math&gt;R_1&lt;/math&gt;. Then let &lt;math&gt;B_2&lt;/math&gt; be the nearest of the remaining blue points to &lt;math&gt;R_2&lt;/math&gt; traveling counterclockwise around the circle from &lt;math&gt;R_2&lt;/math&gt;, and so on, until we have labeled all the blue points &lt;math&gt;B_1, \ldots, B_n&lt;/math&gt;. Show that the number of counterclockwise arcs of the form &lt;math&gt;R_i \rightarrow B_i&lt;/math&gt; that contain the point &lt;math&gt;(1,0)&lt;/math&gt; is independent of the way we chose the ordering &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; of the red points. <br /> <br /> ==Solution==<br /> I define a sequence to be, starting at &lt;math&gt;(1,0)&lt;/math&gt; and tracing the circle counterclockwise, and writing the color of the points in that order - either R or B. For example, possible sequences include &lt;math&gt;RB&lt;/math&gt;, &lt;math&gt;RBBR&lt;/math&gt;, &lt;math&gt;BBRRRB&lt;/math&gt;, &lt;math&gt;BRBRRBBR&lt;/math&gt;, etc.<br /> Note that choosing an &lt;math&gt;R_1&lt;/math&gt; is equivalent to choosing an &lt;math&gt;R&lt;/math&gt; in a sequence, and &lt;math&gt;B_1&lt;/math&gt; is defined as the &lt;math&gt;B&lt;/math&gt; closest to &lt;math&gt;R_1&lt;/math&gt; when moving rightwards. If no &lt;math&gt;B&lt;/math&gt;s exist to the right of &lt;math&gt;R_1&lt;/math&gt;, start from the far left. For example, if I have the above example &lt;math&gt;RBBR&lt;/math&gt;, and I define the 2nd &lt;math&gt;R&lt;/math&gt; to be &lt;math&gt;R_1&lt;/math&gt;, then the first &lt;math&gt;B&lt;/math&gt; will be &lt;math&gt;B_1&lt;/math&gt;. Because no &lt;math&gt;R&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt; can be named twice, I can simply remove &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;B_1&lt;/math&gt; from my sequence when I choose them. I define this to be a move. Hence, a possible move sequence of &lt;math&gt;BBRRRB&lt;/math&gt; is:<br /> &lt;math&gt;BBR_1RRB_1\implies B_2BRR_2\implies B_3R_3&lt;/math&gt;<br /> ---------<br /> Note that, if, in a move, &lt;math&gt;B_n&lt;/math&gt; appears to the left of &lt;math&gt;R_n&lt;/math&gt;, then &lt;math&gt;\stackrel{\frown}{R_nB_n}&lt;/math&gt; intersects &lt;math&gt;(1,0)&lt;/math&gt;<br /> <br /> Now, I define a commencing &lt;math&gt;B&lt;/math&gt; to be a &lt;math&gt;B&lt;/math&gt; which appears to the left of all &lt;math&gt;R&lt;/math&gt;s, and a terminating &lt;math&gt;R&lt;/math&gt; to be a &lt;math&gt;R&lt;/math&gt; which appears to the right of all &lt;math&gt;B&lt;/math&gt;s. Let the amount of commencing &lt;math&gt;B&lt;/math&gt;s be &lt;math&gt;j&lt;/math&gt;, and the amount of terminating &lt;math&gt;R&lt;/math&gt;s be &lt;math&gt;k&lt;/math&gt;, I claim that the number of arcs which cross &lt;math&gt;(1,0)&lt;/math&gt; is constant, and it is equal to &lt;math&gt;\text{max}(j,k)&lt;/math&gt;. I will show this with induction.<br /> <br /> Base case is when &lt;math&gt;n=1&lt;/math&gt;. In this case, there are only two possible sequences - &lt;math&gt;RB&lt;/math&gt; and &lt;math&gt;BR&lt;/math&gt;. In the first case, &lt;math&gt;\stackrel{\frown}{R_1B_1}&lt;/math&gt; does not cross &lt;math&gt;(1, 0)&lt;/math&gt;, but both &lt;math&gt;j&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; are &lt;math&gt;0&lt;/math&gt;, so &lt;math&gt;\text{max}(j,k)=0&lt;/math&gt;. In the second example, &lt;math&gt;j=1&lt;/math&gt;, &lt;math&gt;k=1&lt;/math&gt;, so &lt;math&gt;\text{max}(j,k)=1&lt;/math&gt;. &lt;math&gt;\stackrel{\frown}{R_1B_1}&lt;/math&gt; crosses &lt;math&gt;(1,0)&lt;/math&gt; since &lt;math&gt;B_1&lt;/math&gt; appears to the left of &lt;math&gt;R_1&lt;/math&gt;, so there is one arc which intersects. Hence, the base case is proved.<br /> <br /> For the inductive step, suppose that for a positive number &lt;math&gt;n&lt;/math&gt;, the number of arcs which cross &lt;math&gt;(1,0)&lt;/math&gt; is constant, and given by &lt;math&gt;\text{max}(j, k)&lt;/math&gt; for any configuration. Now, I will show it for &lt;math&gt;n+1&lt;/math&gt;.<br /> <br /> Suppose I first choose &lt;math&gt;R_1&lt;/math&gt; such that &lt;math&gt;B_1&lt;/math&gt; is to the right of &lt;math&gt;R_1&lt;/math&gt; in the sequence. This implies that &lt;math&gt;\stackrel{\frown}{R_1B_1}&lt;/math&gt; does not cross &lt;math&gt;(1,0)&lt;/math&gt;. But, neither &lt;math&gt;R_1&lt;/math&gt; nor &lt;math&gt;B_1&lt;/math&gt; is a commencing &lt;math&gt;B&lt;/math&gt; or terminating &lt;math&gt;R&lt;/math&gt;. These numbers remain constant, and now after this move we have a sequence of length &lt;math&gt;2n&lt;/math&gt;. Hence, by assumption, the total amount of arcs is &lt;math&gt;0+\text{max}(j,k)=\text{max}(j,k)&lt;/math&gt;.<br /> <br /> Now suppose that &lt;math&gt;R_1&lt;/math&gt; appears to the right of &lt;math&gt;B_1&lt;/math&gt;, but &lt;math&gt;B_1&lt;/math&gt; is not a commencing &lt;math&gt;B&lt;/math&gt;. This implies that there are no commencing &lt;math&gt;B&lt;/math&gt;s in the series, because there are no &lt;math&gt;B&lt;/math&gt;s to the left of &lt;math&gt;B_1&lt;/math&gt;, so &lt;math&gt;j=0&lt;/math&gt;. Note that this arc does intersect &lt;math&gt;(1,0)&lt;/math&gt;, and &lt;math&gt;R_1&lt;/math&gt; must be a terminating &lt;math&gt;R&lt;/math&gt;. &lt;math&gt;R_1&lt;/math&gt; must be a terminating &lt;math&gt;R&lt;/math&gt; because there are no &lt;math&gt;B&lt;/math&gt;s to the right of &lt;math&gt;R_1&lt;/math&gt;, or else that &lt;math&gt;B&lt;/math&gt; would be &lt;math&gt;B_1&lt;/math&gt;. The &lt;math&gt;2n&lt;/math&gt; length sequence that remains has &lt;math&gt;0&lt;/math&gt; commencing &lt;math&gt;B&lt;/math&gt;s and &lt;math&gt;k-1&lt;/math&gt; terminating &lt;math&gt;R&lt;/math&gt;s. Hence, by assumption, the total amount of arcs is &lt;math&gt;1+\text{max}(0,k-1)=1+k-1=k=\text{max}(j,k)&lt;/math&gt;.<br /> <br /> Finally, suppose that &lt;math&gt;R_1&lt;/math&gt; appears to the right of &lt;math&gt;B_1&lt;/math&gt;, and &lt;math&gt;B_1&lt;/math&gt; is a commencing &lt;math&gt;B&lt;/math&gt;. We know that this arc will cross &lt;math&gt;(1,0)&lt;/math&gt;. Analogous to the previous case, &lt;math&gt;R_1&lt;/math&gt; is a terminating &lt;math&gt;R&lt;/math&gt;, so the &lt;math&gt;2n&lt;/math&gt; length sequence which remains has &lt;math&gt;j-1&lt;/math&gt; commencing &lt;math&gt;B&lt;/math&gt;s and &lt;math&gt;k-1&lt;/math&gt; terminating &lt;math&gt;R&lt;/math&gt;s. Hence, by assumption, the total amount of arcs is &lt;math&gt;1+\text{max}(j-1,k-1)=1+\text{max}(j,k)-1=\text{max}(j,k)&lt;/math&gt;.<br /> <br /> There are no more possible cases, hence the induction is complete, and the number of arcs which intersect &lt;math&gt;(1,0)&lt;/math&gt; is indeed a constant which is given by &lt;math&gt;\text{max}(j,k)&lt;/math&gt;.<br /> <br /> -william122<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=5|aftertext=|after=Last Problem}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO_Problems/Problem_5&diff=94105 2018 USAJMO Problems/Problem 5 2018-04-21T17:53:40Z <p>Nukelauncher: </p> <hr /> <div>==Problem 5==<br /> Let &lt;math&gt;p&lt;/math&gt; be a prime, and let &lt;math&gt;a_1, \dots, a_p&lt;/math&gt; be integers. Show that there exists an integer &lt;math&gt;k&lt;/math&gt; such that the numbers &lt;cmath&gt;a_1 + k, a_2 + 2k, \dots, a_p + pk&lt;/cmath&gt;produce at least &lt;math&gt;\tfrac{1}{2} p&lt;/math&gt; distinct remainders upon division by &lt;math&gt;p&lt;/math&gt;.<br /> <br /> <br /> ==Solution==<br /> <br /> &lt;math&gt;\textbf{Lemma: }&lt;/math&gt; For fixed &lt;math&gt;i\neq j,&lt;/math&gt; where &lt;math&gt;i, j\in\{1, 2, ..., p\},&lt;/math&gt; the statement &lt;math&gt;a_i + ik\equiv a_j + jk\text{ (mod } p\text{)}&lt;/math&gt; holds for exactly one &lt;math&gt;k\in {1, 2, ..., p}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Proof: }&lt;/math&gt; Notice that the left side minus the right side is congruent to &lt;math&gt;(a_i - a_j) + (i - j)k&lt;/math&gt; modulo &lt;math&gt;p.&lt;/math&gt; For this difference to equal &lt;math&gt;0,&lt;/math&gt; there is a unique solution for &lt;math&gt;k&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; given by &lt;math&gt;k\equiv (a_j - a_i)(i - j)^{-1}\text{ (mod } p\text{)},&lt;/math&gt; where we have used the fact that every nonzero residue modulo &lt;math&gt;p&lt;/math&gt; has a unique multiplicative inverse. Therefore, there is exactly one &lt;math&gt;k\in {1, 2, ..., p}&lt;/math&gt; that satisfies &lt;math&gt;a_i + ik\equiv a_j + jk\text{ (mod } p\text{)}&lt;/math&gt; for any fixed &lt;math&gt;i\neq j. \textbf{ End Lemma}&lt;/math&gt;<br /> <br /> Suppose that you have &lt;math&gt;p&lt;/math&gt; graphs &lt;math&gt;G_1, G_2, ..., G_p,&lt;/math&gt; and graph &lt;math&gt;G_k&lt;/math&gt; consists of the vertices &lt;math&gt;(i, k)&lt;/math&gt; for all &lt;math&gt;1\le i\le p.&lt;/math&gt; Within any graph &lt;math&gt;G_k,&lt;/math&gt; vertices &lt;math&gt;(i_1, k)&lt;/math&gt; and &lt;math&gt;(i_2, k)&lt;/math&gt; are connected by an edge if and only if &lt;math&gt;a_{i_1} + i_1k\equiv a_{i_2} + i_2k\text{ (mod } p\text{)}.&lt;/math&gt; Notice that the number of disconnected components of any graph &lt;math&gt;G_k&lt;/math&gt; equals the number of distinct remainders when divided by &lt;math&gt;p&lt;/math&gt; given by the numbers &lt;math&gt;a_1 + k, a_2 + 2k, ..., a_p + pk.&lt;/math&gt;<br /> <br /> These &lt;math&gt;p&lt;/math&gt; graphs together have exactly one edge for every unordered pair of elements of &lt;math&gt;\{1, 2, ..., p\},&lt;/math&gt; so they have a total of exactly &lt;math&gt;\frac{p(p-1)}{2}&lt;/math&gt; edges. Therefore, there exists at least one graph &lt;math&gt;G_k&lt;/math&gt; that has strictly fewer than &lt;math&gt;\frac{p}{2}&lt;/math&gt; edges, meaning that it has more than &lt;math&gt;\frac{p}{2}&lt;/math&gt; disconnected components. Therefore, the collection of numbers &lt;math&gt;\{a_i + ik: 1\le i\le p\}&lt;/math&gt; for this particular value of &lt;math&gt;k&lt;/math&gt; has at least &lt;math&gt;\frac{p}{2}&lt;/math&gt; distinct remainders modulo &lt;math&gt;p.&lt;/math&gt; This completes the proof.<br /> <br /> (sujaykazi)<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2018|num-b=4|num-a=6}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO_Problems/Problem_4&diff=94104 2018 USAJMO Problems/Problem 4 2018-04-21T17:53:08Z <p>Nukelauncher: </p> <hr /> <div>==Problem 4==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in a circle of radius &lt;math&gt;2&lt;/math&gt; with &lt;math&gt;\angle ABC \geq 90^\circ&lt;/math&gt;, and &lt;math&gt;x&lt;/math&gt; is a real number satisfying the equation &lt;math&gt;x^4 + ax^3 + bx^2 + cx + 1 = 0&lt;/math&gt;, where &lt;math&gt;a=BC,b=CA,c=AB&lt;/math&gt;. Find all possible values of &lt;math&gt;x&lt;/math&gt;.<br /> <br /> <br /> ==Solution==<br /> Notice that<br /> &lt;cmath&gt;x^4 + ax^3 + bx^2 + cx + 1 = \left(x^2 + \frac{a}{2}x\right)^2 + \left(\frac{c}{2}x + 1\right)^2 + \left(b - \frac{a^2}{4} - \frac{c^2}{4}\right)x^2.&lt;/cmath&gt;<br /> Thus, if &lt;math&gt;b &gt; \frac{a^2}{4} + \frac{c^2}{4},&lt;/math&gt; then the expression above is strictly greater than &lt;math&gt;0&lt;/math&gt; for all &lt;math&gt;x,&lt;/math&gt; meaning that &lt;math&gt;x&lt;/math&gt; cannot satisfy the equation &lt;math&gt;x^4 + ax^3 + bx^2 + cx + 1 = 0.&lt;/math&gt; It follows that &lt;math&gt;b\le\frac{a^2}{4} + \frac{c^2}{4}.&lt;/math&gt;<br /> <br /> Since &lt;math&gt;\angle ABC\ge 90^{\circ},&lt;/math&gt; we have &lt;math&gt;b^2\ge a^2 + c^2.&lt;/math&gt; From this and the above we have &lt;math&gt;4b\le a^2 + c^2\le b^2,&lt;/math&gt; so &lt;math&gt;4b\le b^2.&lt;/math&gt; This is true for positive values of &lt;math&gt;b&lt;/math&gt; if and only if &lt;math&gt;b\ge 4.&lt;/math&gt; However, since &lt;math&gt;\triangle ABC&lt;/math&gt; is inscribed in a circle of radius &lt;math&gt;2,&lt;/math&gt; all of its side lengths must be at most the diameter of the circle, so &lt;math&gt;b\le 4.&lt;/math&gt; It follows that &lt;math&gt;b=4.&lt;/math&gt;<br /> <br /> We know that &lt;math&gt;4b\le a^2 + c^2\le b^2.&lt;/math&gt; Since &lt;math&gt;4b = b^2 = 16,&lt;/math&gt; we have &lt;math&gt;4b = a^2 + c^2 = b^2 = 16.&lt;/math&gt;<br /> <br /> The equation &lt;math&gt;x^4 + ax^3 + bx^2 + cx + 1 = 0&lt;/math&gt; can be rewritten as &lt;math&gt;\left(x^2 + \frac{a}{2}x\right)^2 + \left(\frac{c}{2}x + 1\right)^2 = 0,&lt;/math&gt; since &lt;math&gt;b = \frac{a^2}{4} + \frac{c^2}{4}.&lt;/math&gt; This has a real solution if and only if the two separate terms have zeroes in common. The zeroes of &lt;math&gt;\left(x^2 + \frac{a}{2}x\right)^2&lt;/math&gt; are &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;-\frac{a}{2},&lt;/math&gt; and the zero of &lt;math&gt;\left(\frac{c}{2}x + 1\right)^2 = 0&lt;/math&gt; is &lt;math&gt;-\frac{2}{c}.&lt;/math&gt; Clearly we cannot have &lt;math&gt;0=-\frac{2}{c},&lt;/math&gt; so the only other possibility is &lt;math&gt;-\frac{a}{2} = -\frac{2}{c},&lt;/math&gt; which means that &lt;math&gt;ac = 4.&lt;/math&gt;<br /> <br /> We have a system of equations: &lt;math&gt;ac = 4&lt;/math&gt; and &lt;math&gt;a^2 + c^2 = 16.&lt;/math&gt; Solving this system gives &lt;math&gt;(a, c) = \left(\sqrt{6}+\sqrt{2}, \sqrt{6}-\sqrt{2}\right), \left(\sqrt{6}-\sqrt{2}, \sqrt{6}+\sqrt{2}\right).&lt;/math&gt; Each of these gives solutions for &lt;math&gt;x&lt;/math&gt; as &lt;math&gt;-\frac{\sqrt{6}+\sqrt{2}}{2}&lt;/math&gt; and &lt;math&gt;-\frac{\sqrt{6}-\sqrt{2}}{2},&lt;/math&gt; respectively. Now that we know that any valid value of &lt;math&gt;x&lt;/math&gt; must be one of these two, we will verify that both of these values of &lt;math&gt;x&lt;/math&gt; are valid.<br /> <br /> First, consider a right triangle &lt;math&gt;ABC,&lt;/math&gt; inscribed in a circle of radius &lt;math&gt;2,&lt;/math&gt; with side lengths &lt;math&gt;a = \sqrt{6}+\sqrt{2}, b = 4, c = \sqrt{6}-\sqrt{2}.&lt;/math&gt; This generates the polynomial equation<br /> &lt;cmath&gt;x^4 + \left(\sqrt{6}+\sqrt{2}\right)x^3 + 4x^2 + \left(\sqrt{6}-\sqrt{2}\right)x + 1 = \left(\frac{\sqrt{6}-\sqrt{2}}{2}x + 1\right)^2\left(\left(\frac{\sqrt{6}+\sqrt{2}}{2}x\right)^2+1\right) = 0.&lt;/cmath&gt;<br /> This is satisfied by &lt;math&gt;x=-\frac{\sqrt{6}+\sqrt{2}}{2}.&lt;/math&gt;<br /> <br /> Second, consider a right triangle &lt;math&gt;ABC,&lt;/math&gt; inscribed in a circle of radius &lt;math&gt;2,&lt;/math&gt; with side lengths &lt;math&gt;a = \sqrt{6}-\sqrt{2}, b = 4, c = \sqrt{6}+\sqrt{2}.&lt;/math&gt; This generates the polynomial equation<br /> &lt;cmath&gt;x^4 + \left(\sqrt{6}-\sqrt{2}\right)x^3 + 4x^2 + \left(\sqrt{6}+\sqrt{2}\right)x + 1 = \left(\frac{\sqrt{6}+\sqrt{2}}{2}x + 1\right)^2\left(\left(\frac{\sqrt{6}-\sqrt{2}}{2}x\right)^2+1\right) = 0.&lt;/cmath&gt;<br /> This is satisfied by &lt;math&gt;x=-\frac{\sqrt{6}-\sqrt{2}}{2}.&lt;/math&gt;<br /> <br /> It follows that the possible values of &lt;math&gt;x&lt;/math&gt; are &lt;math&gt;-\frac{\sqrt{6}+\sqrt{2}}{2}&lt;/math&gt; and &lt;math&gt;-\frac{\sqrt{6}-\sqrt{2}}{2}.&lt;/math&gt;<br /> <br /> Fun fact: these solutions correspond to a &lt;math&gt;15&lt;/math&gt;-&lt;math&gt;75&lt;/math&gt;-&lt;math&gt;90&lt;/math&gt; triangle.<br /> <br /> (sujaykazi)<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2018|num-b=3|num-a=5}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO_Problems/Problem_3&diff=94103 2018 USAJMO Problems/Problem 3 2018-04-21T17:52:38Z <p>Nukelauncher: </p> <hr /> <div>== Problem ==<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABCD&lt;/math&gt; be a quadrilateral inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;\overline{AC} \perp \overline{BD}&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; be the reflections of &lt;math&gt;D&lt;/math&gt; over lines &lt;math&gt;BA&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, respectively, and let &lt;math&gt;P&lt;/math&gt; be the intersection of lines &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;EF&lt;/math&gt;. Suppose that the circumcircle of &lt;math&gt;\triangle EPD&lt;/math&gt; meets &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, and the circumcircle of &lt;math&gt;\triangle FPD&lt;/math&gt; meets &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;. Show that &lt;math&gt;EQ = FR&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(3cm); <br /> real labelscalefactor = 1.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); <br /> <br /> draw((0.5922362871684147,0.8057643453026269)--(0.8660254037844386,-0.5)--(0.06064145095757084,-0.9981596137020172)--(-0.9987724554622847,0.04953364724950819)--cycle, linewidth(2) + rvwvcq); <br /> /* draw figures */<br /> draw(circle((0,0), 1), linewidth(2) + wrwrwr); <br /> draw((0.8660254037844386,-0.5)--(-0.9987724554622847,0.04953364724950819), linewidth(2) + wrwrwr); <br /> draw((0.5922362871684147,0.8057643453026269)--(0.8660254037844386,-0.5), linewidth(2) + rvwvcq); <br /> draw((0.8660254037844386,-0.5)--(0.06064145095757084,-0.9981596137020172), linewidth(2) + rvwvcq); <br /> draw((0.06064145095757084,-0.9981596137020172)--(-0.9987724554622847,0.04953364724950819), linewidth(2) + rvwvcq); <br /> draw((-0.9987724554622847,0.04953364724950819)--(0.5922362871684147,0.8057643453026269), linewidth(2) + rvwvcq); <br /> draw((0.5922362871684147,0.8057643453026269)--(0.06064145095757084,-0.9981596137020172), linewidth(2) + wrwrwr); <br /> draw((0.5249726058304045,-2.4139334560841545)--(2.3530139989292476,0.7523271151020314), linewidth(2) + wrwrwr); <br /> draw((-0.9987724554622847,0.04953364724950819)--(1.5189031419104242,-0.6923952683993904), linewidth(2) + wrwrwr); <br /> draw((-0.9987724554622847,0.04953364724950819)--(0.5249726058304045,-2.4139334560841545), linewidth(2) + wrwrwr); <br /> draw((-0.9987724554622847,0.04953364724950819)--(2.3530139989292476,0.7523271151020314), linewidth(2) + wrwrwr); <br /> draw((0.8660254037844386,-0.5)--(2.3530139989292476,0.7523271151020314), linewidth(2) + wrwrwr); <br /> draw((0.8660254037844386,-0.5)--(0.5249726058304045,-2.4139334560841545), linewidth(2) + wrwrwr); <br /> draw(circle((0.5922362871684147,0.8057643453026269), 1.7615883990890795), linewidth(2) + linetype(&quot;4 4&quot;) + wrwrwr); <br /> draw(circle((0.06064145095757076,-0.9981596137020177), 1.4899728165839203), linewidth(2) + linetype(&quot;4 4&quot;) + wrwrwr); <br /> draw((0.5249726058304045,-2.4139334560841545)--(0.9854301844182564,0.17008042696736536), linewidth(2) + wrwrwr); <br /> draw((2.3530139989292476,0.7523271151020314)--(0.3454211217688861,-0.9384477868458769), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0,0),dotstyle); <br /> label(&quot;$O$&quot;, (0.03388760411534265,0.08889671794036069), NE * 0.5); <br /> dot((0.8660254037844386,-0.5),dotstyle); <br /> label(&quot;$B$&quot;, (0.9043736119372844,-0.4159851665963708), S * 3.5 * labelscalefactor); <br /> dot((-0.9987724554622847,0.04953364724950819),dotstyle); <br /> label(&quot;$D$&quot;, (-0.9671713048798903,0.13242101833145825), W * 3.5 * labelscalefactor); <br /> dot((0.5922362871684147,0.8057643453026269),dotstyle); <br /> label(&quot;$A$&quot;, (0.625818089434263,0.8897438451365556), NE * labelscalefactor); <br /> dot((0.06064145095757084,-0.9981596137020172),linewidth(4pt) + dotstyle); <br /> label(&quot;$C$&quot;, (0.09482162466287856,-0.9295719112113219), S * 3); <br /> dot((2.3530139989292476,0.7523271151020314),dotstyle); <br /> label(&quot;$E$&quot;, (2.3841998252345853,0.8375146846672384), NE * labelscalefactor); <br /> dot((0.5249726058304045,-2.4139334560841545),dotstyle); <br /> label(&quot;$F$&quot;, (0.5561792088085077,-2.3310543838046627), S * 3.5 * labelscalefactor); <br /> dot((1.5189031419104242,-0.6923952683993904),linewidth(4pt) + dotstyle); <br /> label(&quot;$P$&quot;, (1.5572381178037407,-0.6249018084736391), SE * 3); <br /> dot((0.3454211217688861,-0.9384477868458769),linewidth(4pt) + dotstyle); <br /> label(&quot;$Q$&quot;, (0.38208200724411934,-0.8686378906637853), S * 3); <br /> dot((0.9854301844182564,0.17008042696736536),linewidth(4pt) + dotstyle); <br /> label(&quot;$R$&quot;, (1.0175367929541368,0.23687933927009236), NE * labelscalefactor);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> First we have that &lt;math&gt;BE=BD=BF&lt;/math&gt; by the definition of a reflection. Let &lt;math&gt;\angle DEB = \alpha&lt;/math&gt; and &lt;math&gt;\angle DFB = \beta.&lt;/math&gt; Since &lt;math&gt;\triangle DBE&lt;/math&gt; is isosceles we have &lt;math&gt;\angle BDE = \alpha.&lt;/math&gt; Also, we see that &lt;math&gt;\angle BDE = \angle CAB = \angle CDB = \alpha,&lt;/math&gt; using similar triangles and the property of cyclic quadrilaterals. Similarly, &lt;cmath&gt;\angle DFB = \angle FDB = \angle ACB = \angle ADB = \beta.&lt;/cmath&gt; Now, from &lt;math&gt;BE=BD=BF&lt;/math&gt; we know that &lt;math&gt;B&lt;/math&gt; is the circumcenter of &lt;math&gt;\triangle DEF.&lt;/math&gt; Using the properties of the circumcenter and some elementary angle chasing, we find that &lt;cmath&gt;\angle DPE = 90^{\circ} + \beta - \alpha.&lt;/cmath&gt; <br /> <br /> Now, we claim that &lt;math&gt;Q&lt;/math&gt; is the intersection of ray &lt;math&gt;\overrightarrow{EB}&lt;/math&gt; and the circumcircle of &lt;math&gt;ABCD.&lt;/math&gt; To prove this, we just need to show that &lt;math&gt;DEPQ&lt;/math&gt; is cyclic by this definition of &lt;math&gt;Q.&lt;/math&gt; We have that &lt;cmath&gt;\angle DQE = \angle DCB = \angle DCA + \angle ACB = (90^{\circ}-\alpha)+\beta.&lt;/cmath&gt; We also have from before that &lt;cmath&gt;\angle DPE = 90+\beta-\alpha,&lt;/cmath&gt; so &lt;math&gt;\angle DQE=\angle DPE&lt;/math&gt; and this proves the claim.<br /> <br /> We can use a similar proof to show that &lt;math&gt;F, B, R&lt;/math&gt; are collinear.<br /> <br /> Now, &lt;math&gt;DP&lt;/math&gt; is the radical axis of the circumcircles of &lt;math&gt;\triangle EDP&lt;/math&gt; and &lt;math&gt;\triangle FDP.&lt;/math&gt; Since &lt;math&gt;B&lt;/math&gt; lies on &lt;math&gt;DP,&lt;/math&gt; and &lt;math&gt;E, Q&lt;/math&gt; lie on the circumcircle of &lt;math&gt;\triangle EPD&lt;/math&gt; and &lt;math&gt;F, R&lt;/math&gt; lie on the circumcircle of &lt;math&gt;\triangle FPD,&lt;/math&gt; we have that &lt;cmath&gt;BF \cdot BR = BE \cdot BQ.&lt;/cmath&gt; However, &lt;math&gt;BF=BE,&lt;/math&gt; so &lt;math&gt;BR=BQ.&lt;/math&gt; Since &lt;math&gt;E, B, Q&lt;/math&gt; are collinear and so are &lt;math&gt;F, B, R&lt;/math&gt; we can add these &lt;math&gt;2&lt;/math&gt; equations to get &lt;cmath&gt;EQ=BE+BQ=BF+BR=FR,&lt;/cmath&gt; which completes the proof.<br /> <br /> ~nukelauncher<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2018|num-b=2|num-a=4}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO_Problems/Problem_3&diff=94056 2018 USAJMO Problems/Problem 3 2018-04-20T16:59:00Z <p>Nukelauncher: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABCD&lt;/math&gt; be a quadrilateral inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;\overline{AC} \perp \overline{BD}&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; be the reflections of &lt;math&gt;D&lt;/math&gt; over lines &lt;math&gt;BA&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, respectively, and let &lt;math&gt;P&lt;/math&gt; be the intersection of lines &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;EF&lt;/math&gt;. Suppose that the circumcircle of &lt;math&gt;\triangle EPD&lt;/math&gt; meets &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, and the circumcircle of &lt;math&gt;\triangle FPD&lt;/math&gt; meets &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;. Show that &lt;math&gt;EQ = FR&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(3cm); <br /> real labelscalefactor = 1.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); <br /> <br /> draw((0.5922362871684147,0.8057643453026269)--(0.8660254037844386,-0.5)--(0.06064145095757084,-0.9981596137020172)--(-0.9987724554622847,0.04953364724950819)--cycle, linewidth(2) + rvwvcq); <br /> /* draw figures */<br /> draw(circle((0,0), 1), linewidth(2) + wrwrwr); <br /> draw((0.8660254037844386,-0.5)--(-0.9987724554622847,0.04953364724950819), linewidth(2) + wrwrwr); <br /> draw((0.5922362871684147,0.8057643453026269)--(0.8660254037844386,-0.5), linewidth(2) + rvwvcq); <br /> draw((0.8660254037844386,-0.5)--(0.06064145095757084,-0.9981596137020172), linewidth(2) + rvwvcq); <br /> draw((0.06064145095757084,-0.9981596137020172)--(-0.9987724554622847,0.04953364724950819), linewidth(2) + rvwvcq); <br /> draw((-0.9987724554622847,0.04953364724950819)--(0.5922362871684147,0.8057643453026269), linewidth(2) + rvwvcq); <br /> draw((0.5922362871684147,0.8057643453026269)--(0.06064145095757084,-0.9981596137020172), linewidth(2) + wrwrwr); <br /> draw((0.5249726058304045,-2.4139334560841545)--(2.3530139989292476,0.7523271151020314), linewidth(2) + wrwrwr); <br /> draw((-0.9987724554622847,0.04953364724950819)--(1.5189031419104242,-0.6923952683993904), linewidth(2) + wrwrwr); <br /> draw((-0.9987724554622847,0.04953364724950819)--(0.5249726058304045,-2.4139334560841545), linewidth(2) + wrwrwr); <br /> draw((-0.9987724554622847,0.04953364724950819)--(2.3530139989292476,0.7523271151020314), linewidth(2) + wrwrwr); <br /> draw((0.8660254037844386,-0.5)--(2.3530139989292476,0.7523271151020314), linewidth(2) + wrwrwr); <br /> draw((0.8660254037844386,-0.5)--(0.5249726058304045,-2.4139334560841545), linewidth(2) + wrwrwr); <br /> draw(circle((0.5922362871684147,0.8057643453026269), 1.7615883990890795), linewidth(2) + linetype(&quot;4 4&quot;) + wrwrwr); <br /> draw(circle((0.06064145095757076,-0.9981596137020177), 1.4899728165839203), linewidth(2) + linetype(&quot;4 4&quot;) + wrwrwr); <br /> draw((0.5249726058304045,-2.4139334560841545)--(0.9854301844182564,0.17008042696736536), linewidth(2) + wrwrwr); <br /> draw((2.3530139989292476,0.7523271151020314)--(0.3454211217688861,-0.9384477868458769), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0,0),dotstyle); <br /> label(&quot;$O$&quot;, (0.03388760411534265,0.08889671794036069), NE * 0.5); <br /> dot((0.8660254037844386,-0.5),dotstyle); <br /> label(&quot;$B$&quot;, (0.9043736119372844,-0.4159851665963708), S * 3.5 * labelscalefactor); <br /> dot((-0.9987724554622847,0.04953364724950819),dotstyle); <br /> label(&quot;$D$&quot;, (-0.9671713048798903,0.13242101833145825), W * 3.5 * labelscalefactor); <br /> dot((0.5922362871684147,0.8057643453026269),dotstyle); <br /> label(&quot;$A$&quot;, (0.625818089434263,0.8897438451365556), NE * labelscalefactor); <br /> dot((0.06064145095757084,-0.9981596137020172),linewidth(4pt) + dotstyle); <br /> label(&quot;$C$&quot;, (0.09482162466287856,-0.9295719112113219), S * 3); <br /> dot((2.3530139989292476,0.7523271151020314),dotstyle); <br /> label(&quot;$E$&quot;, (2.3841998252345853,0.8375146846672384), NE * labelscalefactor); <br /> dot((0.5249726058304045,-2.4139334560841545),dotstyle); <br /> label(&quot;$F$&quot;, (0.5561792088085077,-2.3310543838046627), S * 3.5 * labelscalefactor); <br /> dot((1.5189031419104242,-0.6923952683993904),linewidth(4pt) + dotstyle); <br /> label(&quot;$P$&quot;, (1.5572381178037407,-0.6249018084736391), SE * 3); <br /> dot((0.3454211217688861,-0.9384477868458769),linewidth(4pt) + dotstyle); <br /> label(&quot;$Q$&quot;, (0.38208200724411934,-0.8686378906637853), S * 3); <br /> dot((0.9854301844182564,0.17008042696736536),linewidth(4pt) + dotstyle); <br /> label(&quot;$R$&quot;, (1.0175367929541368,0.23687933927009236), NE * labelscalefactor);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> First we have that &lt;math&gt;BE=BD=BF&lt;/math&gt; by the definition of a reflection. Let &lt;math&gt;\angle DEB = \alpha&lt;/math&gt; and &lt;math&gt;\angle DFB = \beta.&lt;/math&gt; Since &lt;math&gt;\triangle DBE&lt;/math&gt; is isosceles we have &lt;math&gt;\angle BDE = \alpha.&lt;/math&gt; Also, we see that &lt;math&gt;\angle BDE = \angle CAB = \angle CDB = \alpha,&lt;/math&gt; using similar triangles and the property of cyclic quadrilaterals. Similarly, &lt;cmath&gt;\angle DFB = \angle FDB = \angle ACB = \angle ADB = \beta.&lt;/cmath&gt; Now, from &lt;math&gt;BE=BD=BF&lt;/math&gt; we know that &lt;math&gt;B&lt;/math&gt; is the circumcenter of &lt;math&gt;\triangle DEF.&lt;/math&gt; Using the properties of the circumcenter and some elementary angle chasing, we find that &lt;cmath&gt;\angle DPE = 90^{\circ} + \beta - \alpha.&lt;/cmath&gt; <br /> <br /> Now, we claim that &lt;math&gt;Q&lt;/math&gt; is the intersection of ray &lt;math&gt;\overrightarrow{EB}&lt;/math&gt; and the circumcircle of &lt;math&gt;ABCD.&lt;/math&gt; To prove this, we just need to show that &lt;math&gt;DEPQ&lt;/math&gt; is cyclic by this definition of &lt;math&gt;Q.&lt;/math&gt; We have that &lt;cmath&gt;\angle DQE = \angle DCB = \angle DCA + \angle ACB = (90^{\circ}-\alpha)+\beta.&lt;/cmath&gt; We also have from before that &lt;cmath&gt;\angle DPE = 90+\beta-\alpha,&lt;/cmath&gt; so &lt;math&gt;\angle DQE=\angle DPE&lt;/math&gt; and this proves the claim.<br /> <br /> We can use a similar proof to show that &lt;math&gt;F, B, R&lt;/math&gt; are collinear.<br /> <br /> Now, &lt;math&gt;DP&lt;/math&gt; is the radical axis of the circumcircles of &lt;math&gt;\triangle EDP&lt;/math&gt; and &lt;math&gt;\triangle FDP.&lt;/math&gt; Since &lt;math&gt;B&lt;/math&gt; lies on &lt;math&gt;DP,&lt;/math&gt; and &lt;math&gt;E, Q&lt;/math&gt; lie on the circumcircle of &lt;math&gt;\triangle EPD&lt;/math&gt; and &lt;math&gt;F, R&lt;/math&gt; lie on the circumcircle of &lt;math&gt;\triangle FPD,&lt;/math&gt; we have that &lt;cmath&gt;BF \cdot BR = BE \cdot BQ.&lt;/cmath&gt; However, &lt;math&gt;BF=BE,&lt;/math&gt; so &lt;math&gt;BR=BQ.&lt;/math&gt; Since &lt;math&gt;E, B, Q&lt;/math&gt; are collinear and so are &lt;math&gt;F, B, R&lt;/math&gt; we can add these &lt;math&gt;2&lt;/math&gt; equations to get &lt;cmath&gt;EQ=BE+BQ=BF+BR=FR,&lt;/cmath&gt; which completes the proof.<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2018|num-b=2|num-a=4}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO_Problems/Problem_3&diff=94054 2018 USAJMO Problems/Problem 3 2018-04-20T16:48:25Z <p>Nukelauncher: </p> <hr /> <div>== Problem ==<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABCD&lt;/math&gt; be a quadrilateral inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;\overline{AC} \perp \overline{BD}&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; be the reflections of &lt;math&gt;D&lt;/math&gt; over lines &lt;math&gt;BA&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, respectively, and let &lt;math&gt;P&lt;/math&gt; be the intersection of lines &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;EF&lt;/math&gt;. Suppose that the circumcircle of &lt;math&gt;\triangle EPD&lt;/math&gt; meets &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, and the circumcircle of &lt;math&gt;\triangle FPD&lt;/math&gt; meets &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;. Show that &lt;math&gt;EQ = FR&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(3cm); <br /> real labelscalefactor = 1.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); <br /> <br /> draw((0.5922362871684147,0.8057643453026269)--(0.8660254037844386,-0.5)--(0.06064145095757084,-0.9981596137020172)--(-0.9987724554622847,0.04953364724950819)--cycle, linewidth(2) + rvwvcq); <br /> /* draw figures */<br /> draw(circle((0,0), 1), linewidth(2) + wrwrwr); <br /> draw((0.8660254037844386,-0.5)--(-0.9987724554622847,0.04953364724950819), linewidth(2) + wrwrwr); <br /> draw((0.5922362871684147,0.8057643453026269)--(0.8660254037844386,-0.5), linewidth(2) + rvwvcq); <br /> draw((0.8660254037844386,-0.5)--(0.06064145095757084,-0.9981596137020172), linewidth(2) + rvwvcq); <br /> draw((0.06064145095757084,-0.9981596137020172)--(-0.9987724554622847,0.04953364724950819), linewidth(2) + rvwvcq); <br /> draw((-0.9987724554622847,0.04953364724950819)--(0.5922362871684147,0.8057643453026269), linewidth(2) + rvwvcq); <br /> draw((0.5922362871684147,0.8057643453026269)--(0.06064145095757084,-0.9981596137020172), linewidth(2) + wrwrwr); <br /> draw((0.5249726058304045,-2.4139334560841545)--(2.3530139989292476,0.7523271151020314), linewidth(2) + wrwrwr); <br /> draw((-0.9987724554622847,0.04953364724950819)--(1.5189031419104242,-0.6923952683993904), linewidth(2) + wrwrwr); <br /> draw((-0.9987724554622847,0.04953364724950819)--(0.5249726058304045,-2.4139334560841545), linewidth(2) + wrwrwr); <br /> draw((-0.9987724554622847,0.04953364724950819)--(2.3530139989292476,0.7523271151020314), linewidth(2) + wrwrwr); <br /> draw((0.8660254037844386,-0.5)--(2.3530139989292476,0.7523271151020314), linewidth(2) + wrwrwr); <br /> draw((0.8660254037844386,-0.5)--(0.5249726058304045,-2.4139334560841545), linewidth(2) + wrwrwr); <br /> draw(circle((0.5922362871684147,0.8057643453026269), 1.7615883990890795), linewidth(2) + linetype(&quot;4 4&quot;) + wrwrwr); <br /> draw(circle((0.06064145095757076,-0.9981596137020177), 1.4899728165839203), linewidth(2) + linetype(&quot;4 4&quot;) + wrwrwr); <br /> draw((0.5249726058304045,-2.4139334560841545)--(0.9854301844182564,0.17008042696736536), linewidth(2) + wrwrwr); <br /> draw((2.3530139989292476,0.7523271151020314)--(0.3454211217688861,-0.9384477868458769), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0,0),dotstyle); <br /> label(&quot;$O$&quot;, (0.03388760411534265,0.08889671794036069), NE * 0.5); <br /> dot((0.8660254037844386,-0.5),dotstyle); <br /> label(&quot;$B$&quot;, (0.9043736119372844,-0.4159851665963708), S * 3.5 * labelscalefactor); <br /> dot((-0.9987724554622847,0.04953364724950819),dotstyle); <br /> label(&quot;$D$&quot;, (-0.9671713048798903,0.13242101833145825), W * 3.5 * labelscalefactor); <br /> dot((0.5922362871684147,0.8057643453026269),dotstyle); <br /> label(&quot;$A$&quot;, (0.625818089434263,0.8897438451365556), NE * labelscalefactor); <br /> dot((0.06064145095757084,-0.9981596137020172),linewidth(4pt) + dotstyle); <br /> label(&quot;$C$&quot;, (0.09482162466287856,-0.9295719112113219), S * 3); <br /> dot((2.3530139989292476,0.7523271151020314),dotstyle); <br /> label(&quot;$E$&quot;, (2.3841998252345853,0.8375146846672384), NE * labelscalefactor); <br /> dot((0.5249726058304045,-2.4139334560841545),dotstyle); <br /> label(&quot;$F$&quot;, (0.5561792088085077,-2.3310543838046627), S * 3.5 * labelscalefactor); <br /> dot((1.5189031419104242,-0.6923952683993904),linewidth(4pt) + dotstyle); <br /> label(&quot;$P$&quot;, (1.5572381178037407,-0.6249018084736391), SE * 3); <br /> dot((0.3454211217688861,-0.9384477868458769),linewidth(4pt) + dotstyle); <br /> label(&quot;$Q$&quot;, (0.38208200724411934,-0.8686378906637853), S * 3); <br /> dot((0.9854301844182564,0.17008042696736536),linewidth(4pt) + dotstyle); <br /> label(&quot;$R$&quot;, (1.0175367929541368,0.23687933927009236), NE * labelscalefactor);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> First we have that &lt;math&gt;BE=BD=BF&lt;/math&gt; by the definition of a reflection. Let &lt;math&gt;\angle DEB = \alpha&lt;/math&gt; and &lt;math&gt;\angle DFB = \beta.&lt;/math&gt; Since &lt;math&gt;\triangle DBE.&lt;/math&gt; is isosceles we have &lt;math&gt;\angle BDE = \alpha.&lt;/math&gt; Also, we see that &lt;math&gt;\angle BDE = \angle CAB = \angle CDB = \alpha,&lt;/math&gt; using similar triangles and the property of cyclic quadrilaterals. Similarly, &lt;cmath&gt;\angle DFB = \angle FDB = \angle ACB = \angle ADB = \beta.&lt;/cmath&gt; Now, from &lt;math&gt;BE=BD=BF&lt;/math&gt; we know that &lt;math&gt;B&lt;/math&gt; is the circumcenter of &lt;math&gt;\triangle DEF.&lt;/math&gt; Using the properties of the circumcenter and some elementary angle chasing, we find that &lt;cmath&gt;\angle DPE = 90^{\circ} + \beta - \alpha.&lt;/cmath&gt; <br /> <br /> Now, we claim that &lt;math&gt;Q&lt;/math&gt; is the intersection of ray &lt;math&gt;\overrightarrow{EB}&lt;/math&gt; and the circumcircle of &lt;math&gt;ABCD.&lt;/math&gt; To prove this, we just need to show that &lt;math&gt;DEPQ&lt;/math&gt; is cyclic by this definition of &lt;math&gt;Q.&lt;/math&gt; We have that &lt;cmath&gt;\angle DQE = \angle DCB = \angle DCA + \angle ACB = (90^{\circ}-\alpha)+\beta.&lt;/cmath&gt; We also have from before that &lt;cmath&gt;\angle DPE = 90+\beta-\alpha,&lt;/cmath&gt; so &lt;math&gt;\angle DQE=\angle DPE&lt;/math&gt; and this proves the claim.<br /> <br /> We can use a similar proof to show that &lt;math&gt;F, B, R&lt;/math&gt; are collinear.<br /> <br /> Now, &lt;math&gt;DP&lt;/math&gt; is the radical axis of the circumcircles of &lt;math&gt;\triangle EDP&lt;/math&gt; and &lt;math&gt;\triangle FDP.&lt;/math&gt; Since &lt;math&gt;B&lt;/math&gt; lies on &lt;math&gt;DP,&lt;/math&gt; and &lt;math&gt;E, Q&lt;/math&gt; lie on the circumcircle of &lt;math&gt;\triangle EPD&lt;/math&gt; and &lt;math&gt;F, R&lt;/math&gt; lie on the circumcircle of &lt;math&gt;\triangle FPD,&lt;/math&gt; we have that &lt;cmath&gt;BF \cdot BR = BE \cdot BQ.&lt;/cmath&gt; However, &lt;math&gt;BF=BE,&lt;/math&gt; so &lt;math&gt;BR=BQ.&lt;/math&gt; Since &lt;math&gt;E, B, Q&lt;/math&gt; are collinear and so are &lt;math&gt;F, B, R&lt;/math&gt; we can add these &lt;math&gt;2&lt;/math&gt; equations to get &lt;cmath&gt;EQ=BE+BQ=BF+BR=FR,&lt;/cmath&gt; which completes the proof.<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2018|num-b=2|num-a=4}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO_Problems/Problem_3&diff=94051 2018 USAJMO Problems/Problem 3 2018-04-20T05:36:26Z <p>Nukelauncher: Created page with &quot;== Problem == (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABCD&lt;/math&gt; be a quadrilateral inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;\overline{AC} \perp \overline{BD}&lt;/math&gt;. Let &lt;math&gt;...&quot;</p> <hr /> <div>== Problem ==<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABCD&lt;/math&gt; be a quadrilateral inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;\overline{AC} \perp \overline{BD}&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; be the reflections of &lt;math&gt;D&lt;/math&gt; over lines &lt;math&gt;BA&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, respectively, and let &lt;math&gt;P&lt;/math&gt; be the intersection of lines &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;EF&lt;/math&gt;. Suppose that the circumcircle of &lt;math&gt;\triangle EPD&lt;/math&gt; meets &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, and the circumcircle of &lt;math&gt;\triangle FPD&lt;/math&gt; meets &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;. Show that &lt;math&gt;EQ = FR&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(3cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); <br /> <br /> draw((0.5922362871684147,0.8057643453026269)--(0.8660254037844386,-0.5)--(0.06064145095757084,-0.9981596137020172)--(-0.9987724554622847,0.04953364724950819)--cycle, linewidth(2) + rvwvcq); <br /> /* draw figures */<br /> draw(circle((0,0), 1), linewidth(2) + wrwrwr); <br /> draw((0.8660254037844386,-0.5)--(-0.9987724554622847,0.04953364724950819), linewidth(2) + wrwrwr); <br /> draw((0.5922362871684147,0.8057643453026269)--(0.8660254037844386,-0.5), linewidth(2) + rvwvcq); <br /> draw((0.8660254037844386,-0.5)--(0.06064145095757084,-0.9981596137020172), linewidth(2) + rvwvcq); <br /> draw((0.06064145095757084,-0.9981596137020172)--(-0.9987724554622847,0.04953364724950819), linewidth(2) + rvwvcq); <br /> draw((-0.9987724554622847,0.04953364724950819)--(0.5922362871684147,0.8057643453026269), linewidth(2) + rvwvcq); <br /> draw((0.5922362871684147,0.8057643453026269)--(0.06064145095757084,-0.9981596137020172), linewidth(2) + wrwrwr); <br /> draw((0.5249726058304045,-2.4139334560841545)--(2.3530139989292476,0.7523271151020314), linewidth(2) + wrwrwr); <br /> draw((-0.9987724554622847,0.04953364724950819)--(1.5189031419104242,-0.6923952683993904), linewidth(2) + wrwrwr); <br /> draw((-0.9987724554622847,0.04953364724950819)--(0.5249726058304045,-2.4139334560841545), linewidth(2) + wrwrwr); <br /> draw((-0.9987724554622847,0.04953364724950819)--(2.3530139989292476,0.7523271151020314), linewidth(2) + wrwrwr); <br /> draw((0.8660254037844386,-0.5)--(2.3530139989292476,0.7523271151020314), linewidth(2) + wrwrwr); <br /> draw((0.8660254037844386,-0.5)--(0.5249726058304045,-2.4139334560841545), linewidth(2) + wrwrwr); <br /> draw(circle((0.5922362871684147,0.8057643453026269), 1.7615883990890795), linewidth(2) + linetype(&quot;4 4&quot;) + wrwrwr); <br /> draw(circle((0.06064145095757076,-0.9981596137020177), 1.4899728165839203), linewidth(2) + linetype(&quot;4 4&quot;) + wrwrwr); <br /> draw((0.5249726058304045,-2.4139334560841545)--(0.9854301844182564,0.17008042696736536), linewidth(2) + wrwrwr); <br /> draw((2.3530139989292476,0.7523271151020314)--(0.3454211217688861,-0.9384477868458769), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0,0),dotstyle); <br /> label(&quot;$O$&quot;, (0.03388760411534265,0.08889671794036069), NE * labelscalefactor); <br /> dot((0.8660254037844386,-0.5),dotstyle); <br /> label(&quot;$B$&quot;, (0.9043736119372844,-0.4159851665963708), NE * labelscalefactor); <br /> dot((-0.9987724554622847,0.04953364724950819),dotstyle); <br /> label(&quot;$D$&quot;, (-0.9671713048798903,0.13242101833145825), NE * labelscalefactor); <br /> dot((0.5922362871684147,0.8057643453026269),dotstyle); <br /> label(&quot;$A$&quot;, (0.625818089434263,0.8897438451365556), NE * labelscalefactor); <br /> dot((0.06064145095757084,-0.9981596137020172),linewidth(4pt) + dotstyle); <br /> label(&quot;$C$&quot;, (0.09482162466287856,-0.9295719112113219), NE * labelscalefactor); <br /> dot((2.3530139989292476,0.7523271151020314),dotstyle); <br /> label(&quot;$E$&quot;, (2.3841998252345853,0.8375146846672384), NE * labelscalefactor); <br /> dot((0.5249726058304045,-2.4139334560841545),dotstyle); <br /> label(&quot;$F$&quot;, (0.5561792088085077,-2.3310543838046627), NE * labelscalefactor); <br /> dot((1.5189031419104242,-0.6923952683993904),linewidth(4pt) + dotstyle); <br /> label(&quot;$P$&quot;, (1.5572381178037407,-0.6249018084736391), NE * labelscalefactor); <br /> dot((0.3454211217688861,-0.9384477868458769),linewidth(4pt) + dotstyle); <br /> label(&quot;$Q$&quot;, (0.38208200724411934,-0.8686378906637853), NE * labelscalefactor); <br /> dot((0.9854301844182564,0.17008042696736536),linewidth(4pt) + dotstyle); <br /> label(&quot;$R$&quot;, (1.0175367929541368,0.23687933927009236), NE * labelscalefactor);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2018|num-b=2|num-a=4}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO_Problems/Problem_2&diff=94050 2018 USAJMO Problems/Problem 2 2018-04-20T05:32:36Z <p>Nukelauncher: Created page with &quot;== Problem == Let &lt;math&gt;a,b,c&lt;/math&gt; be positive real numbers such that &lt;math&gt;a+b+c=4\sqrt{abc}&lt;/math&gt;. Prove that &lt;cmath&gt;2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.&lt;/cm...&quot;</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;a,b,c&lt;/math&gt; be positive real numbers such that &lt;math&gt;a+b+c=4\sqrt{abc}&lt;/math&gt;. Prove that &lt;cmath&gt;2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.&lt;/cmath&gt;<br /> <br /> ==Solution 1==<br /> WLOG let &lt;math&gt;a \leq b \leq c.&lt;/math&gt;<br /> Add &lt;math&gt;2(ab+bc+ca)&lt;/math&gt; to both sides of the inequality and factor to get: &lt;cmath&gt;4(a(a+b+c)+bc) \geq (a+b+c)^2&lt;/cmath&gt; &lt;cmath&gt;\frac{4a\sqrt{abc}+bc}{2} \geq 2\sqrt{a^2b^2c^2}&lt;/cmath&gt;<br /> <br /> The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete.<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2018|num-b=1|num-a=3}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO_Problems/Problem_1&diff=94049 2018 USAJMO Problems/Problem 1 2018-04-20T05:24:33Z <p>Nukelauncher: /* Problem */</p> <hr /> <div>== Problem ==<br /> For each positive integer &lt;math&gt;n&lt;/math&gt;, find the number of &lt;math&gt;n&lt;/math&gt;-digit positive integers that satisfy both of the following conditions:<br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; no two consecutive digits are equal, and<br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; the last digit is a prime.<br /> <br /> ==Solution 1==<br /> The answer is &lt;math&gt;\boxed{\frac{2}{5}\left(9^n+(-1)^{n+1}\right)}&lt;/math&gt;.<br /> <br /> Suppose &lt;math&gt;a_n&lt;/math&gt; denotes the number of &lt;math&gt;n&lt;/math&gt;-digit numbers that satisfy the condition. We claim &lt;math&gt;a_n=4\cdot 9^{n-1}-a_{n-1}&lt;/math&gt;, with &lt;math&gt;a_1=4&lt;/math&gt;.<br /> [i]Proof.[/i] It is trivial to show that &lt;math&gt;a_1=4&lt;/math&gt;. Now, we can do casework on whether or not the tens digit of the &lt;math&gt;n&lt;/math&gt;-digit integer is prime. If the tens digit is prime, we can choose the digits before the units digit in &lt;math&gt;a_{n-1}&lt;/math&gt; ways and choose the units digit in &lt;math&gt;3&lt;/math&gt; ways, since it must be prime and not equal to the tens digit. Therefore, there are &lt;math&gt;3a_{n-1}&lt;/math&gt; ways in this case.<br /> <br /> If the tens digit is not prime, we can use complementary counting. First we consider the number of &lt;math&gt;(n-1)&lt;/math&gt;-digit integers that do not have consecutive digits. There are &lt;math&gt;9&lt;/math&gt; ways to choose the first digit and &lt;math&gt;9&lt;/math&gt; ways to choose the remaining digits. Thus, there are &lt;math&gt;9^{n-1}&lt;/math&gt; integers that satisfy this. Therefore, the number of those &lt;math&gt;(n-1)&lt;/math&gt;-digit integers whose units digit is not prime is &lt;math&gt;9^{n-1}-a_{n-1}&lt;/math&gt;. It is easy to see that there are &lt;math&gt;4&lt;/math&gt; ways to choose the units digit, so there are &lt;math&gt;4\left(9^{n-1}-a_{n-1}\right)&lt;/math&gt; numbers in this case. It follows that &lt;cmath&gt;a_n=3a_{n-1}+4\left(9^{n-1}-a_{n-1}\right)=4\cdot 9^{n-1}-a_{n-1},&lt;/cmath&gt;<br /> and our claim has been proven.<br /> <br /> Then, we can use induction to show that &lt;math&gt;a_n=\frac{2}{5}\left(9^n+(-1)^{n+1}\right)&lt;/math&gt;. It is easy to see that our base case is true, as &lt;math&gt;a_1=4&lt;/math&gt;. Then, &lt;cmath&gt;a_{n+1}=4\cdot 9^n-a_n=4\cdot 9^n-\frac{2}{5}\left(9^n+(-1)^{n+1}\right)=4\cdot 9^n-\frac{2}{5}\cdot 9^n-\frac{2}{5}(-1)^{n+1},&lt;/cmath&gt;<br /> which is equal to &lt;cmath&gt;a_{n+1}=\left(4-\frac{2}{5}\right)\cdot 9^n-\frac{2}{5}\cdot\frac{(-1)^{n+2}}{(-1)}=\frac{18}{5}\cdot 9^n+\frac{2}{5}\cdot(-1)^{n+2}=\frac{2}{5}\left(9^{n+1}+(-1)^{n+2}\right),&lt;/cmath&gt;<br /> as desired. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> Solution by TheUltimate123.<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2018|beforetext=|before=First Problem|num-a=2}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO_Problems&diff=94048 2018 USAJMO Problems 2018-04-20T05:23:00Z <p>Nukelauncher: /* Problem 2 */</p> <hr /> <div>==Day 1==<br /> <br /> Note: For any geometry problem whose statement begins with an asterisk (&lt;math&gt;*&lt;/math&gt;), the first page of the solution must be a large, in-scale, clearly labeled diagram. Failure to meet this requirement will result in an automatic 1-point deduction. <br /> <br /> ===Problem 1===<br /> For each positive integer &lt;math&gt;n&lt;/math&gt;, find the number of &lt;math&gt;n&lt;/math&gt;-digit positive integers that satisfy both of the following conditions:<br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; no two consecutive digits are equal, and<br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; the last digit is a prime.<br /> <br /> [[2018 USAJMO Problems/Problem 1|Solution]]<br /> <br /> ===Problem 2===<br /> Let &lt;math&gt;a,b,c&lt;/math&gt; be positive real numbers such that &lt;math&gt;a+b+c=4\sqrt{abc}&lt;/math&gt;. Prove that &lt;cmath&gt;2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.&lt;/cmath&gt;<br /> [[2018 USAJMO Problems/Problem 2|Solution]]<br /> <br /> ===Problem 3===<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABCD&lt;/math&gt; be a quadrilateral inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;\overline{AC} \perp \overline{BD}&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; be the reflections of &lt;math&gt;D&lt;/math&gt; over lines &lt;math&gt;BA&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, respectively, and let &lt;math&gt;P&lt;/math&gt; be the intersection of lines &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;EF&lt;/math&gt;. Suppose that the circumcircle of &lt;math&gt;\triangle EPD&lt;/math&gt; meets &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, and the circumcircle of &lt;math&gt;\triangle FPD&lt;/math&gt; meets &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;. Show that &lt;math&gt;EQ = FR&lt;/math&gt;.<br /> <br /> [[2018 USAJMO Problems/Problem 3|Solution]]<br /> <br /> ==Day 2==<br /> <br /> Note: For any geometry problem whose statement begins with an asterisk (&lt;math&gt;*&lt;/math&gt;), the first page of the solution must be a large, in-scale, clearly labeled diagram. Failure to meet this requirement will result in an automatic 1-point deduction. <br /> <br /> ===Problem 4===<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in a circle of radius 2 with &lt;math&gt;\angle ABC \geq 90^\circ&lt;/math&gt;, and &lt;math&gt;x&lt;/math&gt; is a real number satisfying the equation &lt;math&gt;x^4 + ax^3 + bx^2 + cx + 1 = 0&lt;/math&gt;, where &lt;math&gt;a=BC,b=CA,c=AB&lt;/math&gt;. Find all possible values of &lt;math&gt;x&lt;/math&gt;.<br /> <br /> [[2018 USAJMO Problems/Problem 4|Solution]]<br /> <br /> ===Problem 5===<br /> Let &lt;math&gt;p&lt;/math&gt; be a prime, and let &lt;math&gt;a_1, \dots, a_p&lt;/math&gt; be integers. Show that there exists an integer &lt;math&gt;k&lt;/math&gt; such that the numbers<br /> &lt;cmath&gt;a_1 + k, a_2 + 2k, \dots, a_p + pk&lt;/cmath&gt;produce at least &lt;math&gt;\tfrac{1}{2} p&lt;/math&gt; distinct remainders upon division by &lt;math&gt;p&lt;/math&gt;.<br /> <br /> [[2018 USAJMO Problems/Problem 5|Solution]]<br /> <br /> ===Problem 6===<br /> Karl starts with &lt;math&gt;n&lt;/math&gt; cards labeled &lt;math&gt;1,2,3,\dots,n&lt;/math&gt; lined up in a random order on his desk. He calls a pair &lt;math&gt;(a,b)&lt;/math&gt; of these cards swapped if &lt;math&gt;a&gt;b&lt;/math&gt; and the card labeled &lt;math&gt;a&lt;/math&gt; is to the left of the card labeled &lt;math&gt;b&lt;/math&gt;. For instance, in the sequence of cards &lt;math&gt;3,1,4,2&lt;/math&gt;, there are three swapped pairs of cards, &lt;math&gt;(3,1)&lt;/math&gt;, &lt;math&gt;(3,2)&lt;/math&gt;, and &lt;math&gt;(4,2)&lt;/math&gt;.<br /> <br /> He picks up the card labeled 1 and inserts it back into the sequence in the opposite position: if the card labeled 1 had &lt;math&gt;i&lt;/math&gt; card to its left, then it now has &lt;math&gt;i&lt;/math&gt; cards to its right. He then picks up the card labeled &lt;math&gt;2&lt;/math&gt; and reinserts it in the same manner, and so on until he has picked up and put back each of the cards &lt;math&gt;1,2,\dots,n&lt;/math&gt; exactly once in that order. (For example, the process starting at &lt;math&gt;3,1,4,2&lt;/math&gt; would be &lt;math&gt;3,1,4,2\to 3,4,1,2\to 2,3,4,1\to 2,4,3,1\to 2,3,4,1&lt;/math&gt;.)<br /> <br /> Show that, no matter what lineup of cards Karl started with, his final lineup has the same number of swapped pairs as the starting lineup.<br /> <br /> [[2018 USAJMO Problems/Problem 6|Solution]]<br /> <br /> {{MAA Notice}}<br /> <br /> {{USAJMO newbox|year= 2018 |before=[[2017 USAJMO]]|after=[[2019 USAJMO]]}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO_Problems/Problem_1&diff=94047 2018 USAJMO Problems/Problem 1 2018-04-20T05:22:25Z <p>Nukelauncher: /* Problem */</p> <hr /> <div>== Problem ==<br /> <br /> For each positive integer &lt;math&gt;n&lt;/math&gt;, find the number of &lt;math&gt;n&lt;/math&gt;-digit positive integers that satisfy both of the following conditions:<br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; no two consecutive digits are equal, and<br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; the last digit is a prime.<br /> <br /> ==Solution 1==<br /> The answer is &lt;math&gt;\boxed{\frac{2}{5}\left(9^n+(-1)^{n+1}\right)}&lt;/math&gt;.<br /> <br /> Suppose &lt;math&gt;a_n&lt;/math&gt; denotes the number of &lt;math&gt;n&lt;/math&gt;-digit numbers that satisfy the condition. We claim &lt;math&gt;a_n=4\cdot 9^{n-1}-a_{n-1}&lt;/math&gt;, with &lt;math&gt;a_1=4&lt;/math&gt;.<br /> [i]Proof.[/i] It is trivial to show that &lt;math&gt;a_1=4&lt;/math&gt;. Now, we can do casework on whether or not the tens digit of the &lt;math&gt;n&lt;/math&gt;-digit integer is prime. If the tens digit is prime, we can choose the digits before the units digit in &lt;math&gt;a_{n-1}&lt;/math&gt; ways and choose the units digit in &lt;math&gt;3&lt;/math&gt; ways, since it must be prime and not equal to the tens digit. Therefore, there are &lt;math&gt;3a_{n-1}&lt;/math&gt; ways in this case.<br /> <br /> If the tens digit is not prime, we can use complementary counting. First we consider the number of &lt;math&gt;(n-1)&lt;/math&gt;-digit integers that do not have consecutive digits. There are &lt;math&gt;9&lt;/math&gt; ways to choose the first digit and &lt;math&gt;9&lt;/math&gt; ways to choose the remaining digits. Thus, there are &lt;math&gt;9^{n-1}&lt;/math&gt; integers that satisfy this. Therefore, the number of those &lt;math&gt;(n-1)&lt;/math&gt;-digit integers whose units digit is not prime is &lt;math&gt;9^{n-1}-a_{n-1}&lt;/math&gt;. It is easy to see that there are &lt;math&gt;4&lt;/math&gt; ways to choose the units digit, so there are &lt;math&gt;4\left(9^{n-1}-a_{n-1}\right)&lt;/math&gt; numbers in this case. It follows that &lt;cmath&gt;a_n=3a_{n-1}+4\left(9^{n-1}-a_{n-1}\right)=4\cdot 9^{n-1}-a_{n-1},&lt;/cmath&gt;<br /> and our claim has been proven.<br /> <br /> Then, we can use induction to show that &lt;math&gt;a_n=\frac{2}{5}\left(9^n+(-1)^{n+1}\right)&lt;/math&gt;. It is easy to see that our base case is true, as &lt;math&gt;a_1=4&lt;/math&gt;. Then, &lt;cmath&gt;a_{n+1}=4\cdot 9^n-a_n=4\cdot 9^n-\frac{2}{5}\left(9^n+(-1)^{n+1}\right)=4\cdot 9^n-\frac{2}{5}\cdot 9^n-\frac{2}{5}(-1)^{n+1},&lt;/cmath&gt;<br /> which is equal to &lt;cmath&gt;a_{n+1}=\left(4-\frac{2}{5}\right)\cdot 9^n-\frac{2}{5}\cdot\frac{(-1)^{n+2}}{(-1)}=\frac{18}{5}\cdot 9^n+\frac{2}{5}\cdot(-1)^{n+2}=\frac{2}{5}\left(9^{n+1}+(-1)^{n+2}\right),&lt;/cmath&gt;<br /> as desired. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> Solution by TheUltimate123.<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2018|beforetext=|before=First Problem|num-a=2}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO_Problems&diff=94046 2018 USAJMO Problems 2018-04-20T05:21:56Z <p>Nukelauncher: Created page with &quot;==Day 1== Note: For any geometry problem whose statement begins with an asterisk (&lt;math&gt;*&lt;/math&gt;), the first page of the solution must be a large, in-scale, clearly labeled d...&quot;</p> <hr /> <div>==Day 1==<br /> <br /> Note: For any geometry problem whose statement begins with an asterisk (&lt;math&gt;*&lt;/math&gt;), the first page of the solution must be a large, in-scale, clearly labeled diagram. Failure to meet this requirement will result in an automatic 1-point deduction. <br /> <br /> ===Problem 1===<br /> For each positive integer &lt;math&gt;n&lt;/math&gt;, find the number of &lt;math&gt;n&lt;/math&gt;-digit positive integers that satisfy both of the following conditions:<br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; no two consecutive digits are equal, and<br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; the last digit is a prime.<br /> <br /> [[2018 USAJMO Problems/Problem 1|Solution]]<br /> <br /> ===Problem 2===<br /> Let &lt;math&gt;a,b,c&lt;/math&gt; be positive real numbers such that &lt;math&gt;a+b+c=4\sqrt{abc}&lt;/math&gt;. Prove that &lt;cmath&gt;2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.&lt;/cmath&gt;<br /> [[2017 USAJMO Problems/Problem 2|Solution]]<br /> <br /> ===Problem 3===<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABCD&lt;/math&gt; be a quadrilateral inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;\overline{AC} \perp \overline{BD}&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; be the reflections of &lt;math&gt;D&lt;/math&gt; over lines &lt;math&gt;BA&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, respectively, and let &lt;math&gt;P&lt;/math&gt; be the intersection of lines &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;EF&lt;/math&gt;. Suppose that the circumcircle of &lt;math&gt;\triangle EPD&lt;/math&gt; meets &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, and the circumcircle of &lt;math&gt;\triangle FPD&lt;/math&gt; meets &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;. Show that &lt;math&gt;EQ = FR&lt;/math&gt;.<br /> <br /> [[2018 USAJMO Problems/Problem 3|Solution]]<br /> <br /> ==Day 2==<br /> <br /> Note: For any geometry problem whose statement begins with an asterisk (&lt;math&gt;*&lt;/math&gt;), the first page of the solution must be a large, in-scale, clearly labeled diagram. Failure to meet this requirement will result in an automatic 1-point deduction. <br /> <br /> ===Problem 4===<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in a circle of radius 2 with &lt;math&gt;\angle ABC \geq 90^\circ&lt;/math&gt;, and &lt;math&gt;x&lt;/math&gt; is a real number satisfying the equation &lt;math&gt;x^4 + ax^3 + bx^2 + cx + 1 = 0&lt;/math&gt;, where &lt;math&gt;a=BC,b=CA,c=AB&lt;/math&gt;. Find all possible values of &lt;math&gt;x&lt;/math&gt;.<br /> <br /> [[2018 USAJMO Problems/Problem 4|Solution]]<br /> <br /> ===Problem 5===<br /> Let &lt;math&gt;p&lt;/math&gt; be a prime, and let &lt;math&gt;a_1, \dots, a_p&lt;/math&gt; be integers. Show that there exists an integer &lt;math&gt;k&lt;/math&gt; such that the numbers<br /> &lt;cmath&gt;a_1 + k, a_2 + 2k, \dots, a_p + pk&lt;/cmath&gt;produce at least &lt;math&gt;\tfrac{1}{2} p&lt;/math&gt; distinct remainders upon division by &lt;math&gt;p&lt;/math&gt;.<br /> <br /> [[2018 USAJMO Problems/Problem 5|Solution]]<br /> <br /> ===Problem 6===<br /> Karl starts with &lt;math&gt;n&lt;/math&gt; cards labeled &lt;math&gt;1,2,3,\dots,n&lt;/math&gt; lined up in a random order on his desk. He calls a pair &lt;math&gt;(a,b)&lt;/math&gt; of these cards swapped if &lt;math&gt;a&gt;b&lt;/math&gt; and the card labeled &lt;math&gt;a&lt;/math&gt; is to the left of the card labeled &lt;math&gt;b&lt;/math&gt;. For instance, in the sequence of cards &lt;math&gt;3,1,4,2&lt;/math&gt;, there are three swapped pairs of cards, &lt;math&gt;(3,1)&lt;/math&gt;, &lt;math&gt;(3,2)&lt;/math&gt;, and &lt;math&gt;(4,2)&lt;/math&gt;.<br /> <br /> He picks up the card labeled 1 and inserts it back into the sequence in the opposite position: if the card labeled 1 had &lt;math&gt;i&lt;/math&gt; card to its left, then it now has &lt;math&gt;i&lt;/math&gt; cards to its right. He then picks up the card labeled &lt;math&gt;2&lt;/math&gt; and reinserts it in the same manner, and so on until he has picked up and put back each of the cards &lt;math&gt;1,2,\dots,n&lt;/math&gt; exactly once in that order. (For example, the process starting at &lt;math&gt;3,1,4,2&lt;/math&gt; would be &lt;math&gt;3,1,4,2\to 3,4,1,2\to 2,3,4,1\to 2,4,3,1\to 2,3,4,1&lt;/math&gt;.)<br /> <br /> Show that, no matter what lineup of cards Karl started with, his final lineup has the same number of swapped pairs as the starting lineup.<br /> <br /> [[2018 USAJMO Problems/Problem 6|Solution]]<br /> <br /> {{MAA Notice}}<br /> <br /> {{USAJMO newbox|year= 2018 |before=[[2017 USAJMO]]|after=[[2019 USAJMO]]}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems&diff=92977 2018 AIME I Problems 2018-03-08T03:12:30Z <p>Nukelauncher: /* Problem 10 */</p> <hr /> <div>{{AIME Problems|year=2018|n=I}}<br /> <br /> ==Problem 1==<br /> Let &lt;math&gt;S&lt;/math&gt; be the number of ordered pairs of integers &lt;math&gt;(a,b)&lt;/math&gt; with &lt;math&gt;1 \leq a \leq 100&lt;/math&gt; and &lt;math&gt;b \geq 0&lt;/math&gt; such that the polynomial &lt;math&gt;x^2+ax+b&lt;/math&gt; can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when &lt;math&gt;S&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 1 | Solution]]<br /> <br /> ==Problem 2==<br /> The number &lt;math&gt;n&lt;/math&gt; can be written in base &lt;math&gt;14&lt;/math&gt; as &lt;math&gt;\underline{a}\text{ }\underline{b}\text{ }\underline{c}&lt;/math&gt;, can be written in base &lt;math&gt;15&lt;/math&gt; as &lt;math&gt;\underline{a}\text{ }\underline{c}\text{ }\underline{b}&lt;/math&gt;, and can be written in base &lt;math&gt;6&lt;/math&gt; as &lt;math&gt;\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }&lt;/math&gt;, where &lt;math&gt;a &gt; 0&lt;/math&gt;. Find the base-&lt;math&gt;10&lt;/math&gt; representation of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 2 | Solution]]<br /> <br /> ==Problem 3==<br /> Kathy has &lt;math&gt;5&lt;/math&gt; red cards and &lt;math&gt;5&lt;/math&gt; green cards. She shuffles the &lt;math&gt;10&lt;/math&gt; cards and lays out &lt;math&gt;5&lt;/math&gt; of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders &lt;math&gt;RRGGG, GGGGR,&lt;/math&gt; or &lt;math&gt;RRRRR&lt;/math&gt; will make Kathy happy, but &lt;math&gt;RRRGR&lt;/math&gt; will not. The probability that Kathy will be happy is &lt;math&gt; \dfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 3 | Solution]]<br /> <br /> ==Problem 4==<br /> In &lt;math&gt;\triangle ABC, AB = AC = 10&lt;/math&gt; and &lt;math&gt;BC = 12&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; lies strictly between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; and point &lt;math&gt;E&lt;/math&gt; lies strictly between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; on &lt;math&gt;\overline{AC}&lt;/math&gt;) so that &lt;math&gt;AD = DE = EC&lt;/math&gt;. Then &lt;math&gt;AD&lt;/math&gt; can be expressed in the form &lt;math&gt;\dfrac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 4 | Solution]]<br /> <br /> ==Problem 5==<br /> For each ordered pair of real numbers &lt;math&gt;(x,y)&lt;/math&gt; satisfying<br /> &lt;cmath&gt; \log_2(2x+y) = \log_4(x^2+xy+7y^2) &lt;/cmath&gt;there is a real number &lt;math&gt;K&lt;/math&gt; such that<br /> &lt;cmath&gt; \log_3(3x+y) = \log_9(3x^2+4xy+Ky^2). &lt;/cmath&gt;Find the product of all possible values of &lt;math&gt;K&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 5 | Solution]]<br /> <br /> ==Problem 6==<br /> Let &lt;math&gt;N&lt;/math&gt; be the number of complex numbers &lt;math&gt;z&lt;/math&gt; with the properties that &lt;math&gt;|z|=1&lt;/math&gt; and &lt;math&gt;z^{6!}-z^{5!}&lt;/math&gt; is a real number. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 6 | Solution]]<br /> <br /> ==Problem 7==<br /> A right hexagonal prism has height &lt;math&gt;2&lt;/math&gt;. The bases are regular hexagons with side length &lt;math&gt;1&lt;/math&gt;. Any &lt;math&gt;3&lt;/math&gt; of the &lt;math&gt;12&lt;/math&gt; vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles).<br /> <br /> [[2018 AIME I Problems/Problem 7 | Solution]]<br /> <br /> ==Problem 8==<br /> Let &lt;math&gt;ABCDEF&lt;/math&gt; be an equiangular hexagon such that &lt;math&gt;AB=6, BC=8, CD=10&lt;/math&gt;, and &lt;math&gt;DE=12&lt;/math&gt;. Denote &lt;math&gt;d&lt;/math&gt; the diameter of the largest circle that fits inside the hexagon. Find &lt;math&gt;d^2&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 8 | Solution]]<br /> <br /> ==Problem 9==<br /> Find the number of four-element subsets of &lt;math&gt;\{1,2,3,4,\dots, 20\}&lt;/math&gt; with the property that two distinct elements of a subset have a sum of &lt;math&gt;16&lt;/math&gt;, and two distinct elements of a subset have a sum of &lt;math&gt;24&lt;/math&gt;. For example, &lt;math&gt;\{3,5,13,19\}&lt;/math&gt; and &lt;math&gt;\{6,10,20,18\}&lt;/math&gt; are two such subsets.<br /> <br /> <br /> [[2018 AIME I Problems/Problem 9 | Solution]]<br /> <br /> ==Problem 10==<br /> The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point &lt;math&gt;A&lt;/math&gt;. At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug only walks in a clockwise direction. For example, the bug could travel along the path &lt;math&gt;AJABCHCHIJA&lt;/math&gt;, which has &lt;math&gt;10&lt;/math&gt; steps. Let &lt;math&gt;n&lt;/math&gt; be the number of paths with &lt;math&gt;15&lt;/math&gt; steps that begin and end at point &lt;math&gt;A&lt;/math&gt;. Find the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> <br /> draw(unitcircle);<br /> draw(scale(2) * unitcircle);<br /> for(int d = 90; d &lt; 360 + 90; d += 72){<br /> draw(2 * dir(d) -- dir(d));<br /> }<br /> <br /> dot(1 * dir( 90), linewidth(5));<br /> dot(1 * dir(162), linewidth(5));<br /> dot(1 * dir(234), linewidth(5));<br /> dot(1 * dir(306), linewidth(5));<br /> dot(1 * dir(378), linewidth(5));<br /> dot(2 * dir(378), linewidth(5));<br /> dot(2 * dir(306), linewidth(5));<br /> dot(2 * dir(234), linewidth(5));<br /> dot(2 * dir(162), linewidth(5));<br /> dot(2 * dir( 90), linewidth(5));<br /> <br /> label(&quot;$A$&quot;, 1 * dir( 90), -dir( 90));<br /> label(&quot;$B$&quot;, 1 * dir(162), -dir(162));<br /> label(&quot;$C$&quot;, 1 * dir(234), -dir(234));<br /> label(&quot;$D$&quot;, 1 * dir(306), -dir(306));<br /> label(&quot;$E$&quot;, 1 * dir(378), -dir(378));<br /> label(&quot;$F$&quot;, 2 * dir(378), dir(378));<br /> label(&quot;$G$&quot;, 2 * dir(306), dir(306));<br /> label(&quot;$H$&quot;, 2 * dir(234), dir(234));<br /> label(&quot;$I$&quot;, 2 * dir(162), dir(162));<br /> label(&quot;$J$&quot;, 2 * dir( 90), dir( 90));<br /> &lt;/asy&gt;<br /> <br /> [[2018 AIME I Problems/Problem 10 | Solution]]<br /> <br /> ==Problem 11==<br /> Find the least positive integer &lt;math&gt;n&lt;/math&gt; such that when &lt;math&gt;3^n&lt;/math&gt; is written in base &lt;math&gt;143&lt;/math&gt;, its two right-most digits in base &lt;math&gt;143&lt;/math&gt; are &lt;math&gt;01&lt;/math&gt;.<br /> <br /> <br /> [[2018 AIME I Problems/Problem 11 | Solution]]<br /> <br /> ==Problem 12==<br /> For every subset &lt;math&gt;T&lt;/math&gt; of &lt;math&gt;U = \{ 1,2,3,\ldots,18 \}&lt;/math&gt;, let &lt;math&gt;s(T)&lt;/math&gt; be the sum of the elements of &lt;math&gt;T&lt;/math&gt;, with &lt;math&gt;s(\emptyset)&lt;/math&gt; defined to be &lt;math&gt;0&lt;/math&gt;. If &lt;math&gt;T&lt;/math&gt; is chosen at random among all subsets of &lt;math&gt;U&lt;/math&gt;, the probability that &lt;math&gt;s(T)&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 12 | Solution]]<br /> <br /> ==Problem 13==<br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; have side lengths &lt;math&gt;AB=30&lt;/math&gt;, &lt;math&gt;BC=32&lt;/math&gt;, and &lt;math&gt;AC=34&lt;/math&gt;. Point &lt;math&gt;X&lt;/math&gt; lies in the interior of &lt;math&gt;\overline{BC}&lt;/math&gt;, and points &lt;math&gt;I_1&lt;/math&gt; and &lt;math&gt;I_2&lt;/math&gt; are the incenters of &lt;math&gt;\triangle ABX&lt;/math&gt; and &lt;math&gt;\triangle ACX&lt;/math&gt;, respectively. Find the minimum possible area of &lt;math&gt;\triangle AI_1I_2&lt;/math&gt; as &lt;math&gt;X&lt;/math&gt; varies along &lt;math&gt;\overline{BC}&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 13 | Solution]]<br /> <br /> ==Problem 14==<br /> Let &lt;math&gt;SP_1P_2P_3EP_4P_5&lt;/math&gt; be a heptagon. A frog starts jumping at vertex &lt;math&gt;S&lt;/math&gt;. From any vertex of the heptagon except &lt;math&gt;E&lt;/math&gt;, the frog may jump to either fo the two adjacent vertices. When it reaches vertex &lt;math&gt;E&lt;/math&gt;, the frog stops and stays there. Find the number of distinct sequences of jumps of no more than &lt;math&gt;12&lt;/math&gt; jumps that end at &lt;math&gt;E&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 14 | Solution]]<br /> <br /> ==Problem 15==<br /> David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, &lt;math&gt;A,\text{ }B,\text{ }C&lt;/math&gt;, which can each be inscribed in a circle with radius &lt;math&gt;1&lt;/math&gt;. Let &lt;math&gt;\varphi_A&lt;/math&gt; denote the measure of the acute angle made by the diagonals of quadrilateral $$A$$, and define &lt;math&gt;\varphi_B&lt;/math&gt; and &lt;math&gt;\varphi_C&lt;/math&gt; similarly. Suppose that &lt;math&gt;\sin\varphi_A=\frac{2}{3}&lt;/math&gt;, &lt;math&gt;\sin\varphi_B=\frac{3}{5}&lt;/math&gt;, and &lt;math&gt;\sin\varphi_C=\frac{6}{7}&lt;/math&gt;. All three quadrilaterals have the same area &lt;math&gt;K&lt;/math&gt;, which can be written in the form &lt;math&gt;\dfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 15 | Solution]]<br /> <br /> {{AIME box|year=2018|n=I|before=[[2017 AIME II]]|after=[[2018 AIME II]]}}<br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_12&diff=88912 2017 AMC 8 Problems/Problem 12 2017-12-15T02:05:40Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 12==<br /> The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?<br /> <br /> &lt;math&gt;\textbf{(A) }2\text{ and }19\qquad\textbf{(B) }20\text{ and }39\qquad\textbf{(C) }40\text{ and }59\qquad\textbf{(D) }60\text{ and }79\qquad\textbf{(E) }80\text{ and }124&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The &lt;math&gt;\text{lcm}(4,5,6)&lt;/math&gt; is 60. Since &lt;math&gt;60+1=61&lt;/math&gt;, and that is in the range of &lt;math&gt;\boxed{\textbf{(D)}\ \text{60 and 79}}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=11|num-a=13}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_12&diff=88911 2017 AMC 8 Problems/Problem 12 2017-12-15T02:05:23Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 12==<br /> The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?<br /> <br /> &lt;math&gt;\textbf{(A) }2\text{ and }19\qquad\textbf{(B) }20\text{ and }39\qquad\textbf{(C) }40\text{ and }59\qquad\textbf{(D) }60\text{ and }79\qquad\textbf{(E) }80\text{ and }124&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The &lt;math&gt;\lcm(4,5,6)&lt;/math&gt; is 60. Since &lt;math&gt;60+1=61&lt;/math&gt;, and that is in the range of &lt;math&gt;\boxed{\textbf{(D)}\ \text{60 and 79}}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=11|num-a=13}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_19&diff=88701 2017 AMC 8 Problems/Problem 19 2017-11-28T06:32:07Z <p>Nukelauncher: /* Solution 2 */</p> <hr /> <div>==Problem 19==<br /> For any positive integer &lt;math&gt;M&lt;/math&gt;, the notation &lt;math&gt;M!&lt;/math&gt; denotes the product of the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;M&lt;/math&gt;. What is the largest integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;5^n&lt;/math&gt; is a factor of the sum &lt;math&gt;98!+99!+100!&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27&lt;/math&gt; <br /> <br /> ==Solution 1==<br /> Factoring out &lt;math&gt;98!&lt;/math&gt;, we have &lt;math&gt;98!(10,000)&lt;/math&gt;. Next, &lt;math&gt;98!&lt;/math&gt; has &lt;math&gt;\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;. Now &lt;math&gt;10,000&lt;/math&gt; has &lt;math&gt;4&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;, so there are a total of &lt;math&gt;22 + 4 = \boxed{\textbf{(D)}\ 26}&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The number of &lt;math&gt;5&lt;/math&gt;'s in the factorization of &lt;math&gt;98! + 99! + 100!&lt;/math&gt; is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by &lt;math&gt;5&lt;/math&gt;, until you can't divide by &lt;math&gt;5&lt;/math&gt; anymore. Factorizing &lt;math&gt;98! + 99! + 100!&lt;/math&gt;, you get &lt;math&gt;98!(1+99+9900)=98!(10000)&lt;/math&gt;. To find the number of trailing zeroes in 98!, we do &lt;math&gt;\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22&lt;/math&gt;. Now since &lt;math&gt;10000&lt;/math&gt; has 4 zeroes, we add &lt;math&gt;22 + 4&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(D)}\ 26}&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=18|num-a=20}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_20&diff=88584 2017 AMC 8 Problems/Problem 20 2017-11-23T01:55:39Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 20==<br /> <br /> An integer between &lt;math&gt;1000&lt;/math&gt; and &lt;math&gt;9999&lt;/math&gt;, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{14}{75}\qquad\textbf{(B) }\frac{56}{225}\qquad\textbf{(C) }\frac{107}{400}\qquad\textbf{(D) }\frac{7}{25}\qquad\textbf{(E) }\frac{9}{25}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> There are &lt;math&gt;5&lt;/math&gt; options for the last digit, as the integer must be odd. The first digit now has &lt;math&gt;8&lt;/math&gt; options left (it can't be &lt;math&gt;0&lt;/math&gt; or the same as the last digit). The second digit also has &lt;math&gt;8&lt;/math&gt; options left (it can't be the same as the first or last digit). Finally, the third digit has &lt;math&gt;7&lt;/math&gt; options (it can't be the same as the three digits that are already chosen). <br /> <br /> Since there are &lt;math&gt;9000&lt;/math&gt; total integers, out answer is &lt;cmath&gt;\frac{8 \cdot 8 \cdot 7 \cdot 5}{9000} = \boxed{\textbf{(B)}\ \frac{56}{225}}.&lt;/cmath&gt;<br /> <br /> ~nukelauncher<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=19|num-a=21}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_16&diff=88583 2017 AMC 8 Problems/Problem 16 2017-11-23T01:54:57Z <p>Nukelauncher: /* Solution 2 */</p> <hr /> <div>==Problem 16==<br /> <br /> In the figure below, choose point &lt;math&gt;D&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt; so that &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\triangle ABD&lt;/math&gt; have equal perimeters. What is the area of &lt;math&gt;\triangle ABD&lt;/math&gt;?<br /> &lt;asy&gt;draw((0,0)--(4,0)--(0,3)--(0,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$B$&quot;, (4,0), ESE);<br /> label(&quot;$C$&quot;, (0, 3), N);<br /> label(&quot;$3$&quot;, (0, 1.5), W);<br /> label(&quot;$4$&quot;, (2, 0), S);<br /> label(&quot;$5$&quot;, (2, 1.5), NE);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }\frac{3}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\frac{12}{5}\qquad\textbf{(E) }\frac{5}{2}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Essentially, we see that if we draw a line from point A to imaginary point D, that line would apply to both triangles. Let us say that x is the length of the line from B to D. So, the perimeter of &lt;math&gt;\triangle{ABD}&lt;/math&gt; would be &lt;math&gt;\overline{AD} + 4 + x&lt;/math&gt;, while the perimeter of &lt;math&gt;\triangle{ACD}&lt;/math&gt; would be &lt;math&gt;\overline{AD} + 3 + (5 - x)&lt;/math&gt;. Notice that we can find out &lt;math&gt;x&lt;/math&gt; from these two equations. We can find out that &lt;math&gt;x = 2&lt;/math&gt;, so that means that the area of &lt;math&gt;\triangle{ABD} = \frac{2 \cdot 6}{5} = \boxed{\textbf{(D) } \frac{12}{5}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We know that the perimeters of the two small triangles are &lt;math&gt;3+CD+AD&lt;/math&gt; and &lt;math&gt;4+BD+AD&lt;/math&gt;. Setting both equal and using &lt;math&gt;BD+CD = 5&lt;/math&gt;, we have &lt;math&gt;BD = 2&lt;/math&gt; and &lt;math&gt;CD = 3&lt;/math&gt;. Now, we simply have to find the area of &lt;math&gt;\triangle ABD&lt;/math&gt;. Since &lt;math&gt;\frac{BD}{CD} = \frac{2}{3}&lt;/math&gt;, we must have &lt;math&gt;\frac{[ABD]}{[ACD]} = 2/3&lt;/math&gt;. Combining this with the fact that &lt;math&gt;[ABC] = [ABD] + [ACD] = \frac{3*4}{2} = 6&lt;/math&gt;, we get &lt;math&gt;[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} * 6 = \boxed{\textbf{(D) } \frac{12}{5}}&lt;/math&gt;<br /> <br /> ==Solution 3== <br /> <br /> Since point &lt;math&gt;D&lt;/math&gt; is on line &lt;math&gt;BC&lt;/math&gt;, it will split it into &lt;math&gt;CD&lt;/math&gt; and &lt;math&gt;DB&lt;/math&gt;. Let &lt;math&gt;CD = 5 - x&lt;/math&gt; and &lt;math&gt;DB = x&lt;/math&gt;. Triangle &lt;math&gt;CAD&lt;/math&gt; has side lengths &lt;math&gt;3, 5 - x, AD&lt;/math&gt; and triangle &lt;math&gt;DAB&lt;/math&gt; has side lengths &lt;math&gt;x, 4, AD&lt;/math&gt;. Since both perimeters are equal, we have the equation &lt;math&gt;3 + 5 - x + AD = 4 + x + AD&lt;/math&gt;. Eliminating &lt;math&gt;AD&lt;/math&gt; and solving the resulting linear equation gives &lt;math&gt;x = 2&lt;/math&gt;. Draw a perpendicular from point &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;AB&lt;/math&gt;. Call the point of intersection &lt;math&gt;F&lt;/math&gt;. Because angle &lt;math&gt;ABC&lt;/math&gt; is common to both triangles &lt;math&gt;DBF&lt;/math&gt; and &lt;math&gt;ABC&lt;/math&gt;, and both are right triangles, both are similar. The hypotenuse of triangle &lt;math&gt;DBF&lt;/math&gt; is 2, so the altitude must be &lt;math&gt;6/5&lt;/math&gt; Because &lt;math&gt;DBF&lt;/math&gt; and &lt;math&gt;ABD&lt;/math&gt; share the same altitude, the height of &lt;math&gt;ABD&lt;/math&gt; therefore must be &lt;math&gt;6/5&lt;/math&gt;. The base of &lt;math&gt;ABD&lt;/math&gt; is 4, so &lt;math&gt;[ABD] = \frac{1}{2} * 4 * \frac{6}{5} = \frac{12}{5} \implies \boxed{\textbf{(D) } \frac{12}{5}}&lt;/math&gt;<br /> <br /> ==Solution 4 (trig)==<br /> Using any preferred method, realize &lt;math&gt;BD = 2&lt;/math&gt;. Since we are given a 3-4-5 right triangle, we know the value of &lt;math&gt;\sin(\angle ABC) = \frac{3}{5}&lt;/math&gt;. Since we are given &lt;math&gt;AB = 4&lt;/math&gt;, apply the Sine Area Formula to get &lt;math&gt;\frac{1}{2} * 4 * 2 * \frac{3}{5} = \boxed{\textbf{(D) } \frac{12}{5}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=15|num-a=17}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_11&diff=88514 2017 AMC 8 Problems/Problem 11 2017-11-22T20:07:02Z <p>Nukelauncher: /* Problem 11 */</p> <hr /> <div>==Problem 11==<br /> A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?<br /> <br /> &lt;math&gt;\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369&lt;/math&gt;<br /> <br /> ==Solution==<br /> Since the number of tiles lying on both diagonals is &lt;math&gt;37&lt;/math&gt;, counting one tile twice, there are &lt;math&gt;37=2x-1\implies x=19&lt;/math&gt; tiles on each side. Hence, our answer is &lt;math&gt;19^2=361=\boxed{\textbf{(C)}\ 361}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_11&diff=88513 2017 AMC 8 Problems/Problem 11 2017-11-22T20:06:38Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 11==<br /> A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?<br /> &lt;math&gt;\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369&lt;/math&gt;<br /> <br /> ==Solution==<br /> Since the number of tiles lying on both diagonals is &lt;math&gt;37&lt;/math&gt;, counting one tile twice, there are &lt;math&gt;37=2x-1\implies x=19&lt;/math&gt; tiles on each side. Hence, our answer is &lt;math&gt;19^2=361=\boxed{\textbf{(C)}\ 361}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_11&diff=88512 2017 AMC 8 Problems/Problem 11 2017-11-22T20:06:14Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 11==<br /> A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?<br /> &lt;math&gt;\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369&lt;/math&gt;<br /> <br /> ==Solution==<br /> Since the number of tiles lying on both diagonals is &lt;math&gt;37&lt;/math&gt;, counting one tile twice, there are &lt;math&gt;37=2x-1\implies x=19&lt;/math&gt; tiles on each side. Hence, our answer is &lt;math&gt;19^2=361=\boxed{\textbf{C}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_15&diff=88511 2017 AMC 8 Problems/Problem 15 2017-11-22T20:05:24Z <p>Nukelauncher: Created page with &quot;==Problem 15== In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path allows only moves fr...&quot;</p> <hr /> <div>==Problem 15==<br /> <br /> In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path allows only moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture. <br /> &lt;asy&gt; fill((0.5, 4.5)--(1.5,4.5)--(1.5,2.5)--(0.5,2.5)--cycle,lightgray); fill((1.5,3.5)--(2.5,3.5)--(2.5,1.5)--(1.5,1.5)--cycle,lightgray); label(&quot;$8$&quot;, (1, 0)); label(&quot;$C$&quot;, (2, 0)); label(&quot;$8$&quot;, (3, 0)); label(&quot;$8$&quot;, (0, 1)); label(&quot;$C$&quot;, (1, 1)); label(&quot;$M$&quot;, (2, 1)); label(&quot;$C$&quot;, (3, 1)); label(&quot;$8$&quot;, (4, 1)); label(&quot;$C$&quot;, (0, 2)); label(&quot;$M$&quot;, (1, 2)); label(&quot;$A$&quot;, (2, 2)); label(&quot;$M$&quot;, (3, 2)); label(&quot;$C$&quot;, (4, 2)); label(&quot;$8$&quot;, (0, 3)); label(&quot;$C$&quot;, (1, 3)); label(&quot;$M$&quot;, (2, 3)); label(&quot;$C$&quot;, (3, 3)); label(&quot;$8$&quot;, (4, 3)); label(&quot;$8$&quot;, (1, 4)); label(&quot;$C$&quot;, (2, 4)); label(&quot;$8$&quot;, (3, 4));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Notice that the &lt;math&gt;A&lt;/math&gt; is adjacent to &lt;math&gt;4&lt;/math&gt; &lt;math&gt;M&lt;/math&gt;s, each &lt;math&gt;M&lt;/math&gt; is adjacent to &lt;math&gt;3&lt;/math&gt; &lt;math&gt;C&lt;/math&gt;s, and each &lt;math&gt;C&lt;/math&gt; is adjacent to &lt;math&gt;2&lt;/math&gt; &lt;math&gt;8&lt;/math&gt;s. Thus, the answer is &lt;math&gt;4\cdot 3\cdot 2 = \boxed{\textbf{(D)}\ 24}.&lt;/math&gt;<br /> <br /> ~nukelauncher<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=14|num-a=16}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_8&diff=88505 2017 AMC 8 Problems/Problem 8 2017-11-22T19:50:58Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 8==<br /> Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, &quot;My house number has two digits, and exactly three of the following four statements about it are true.&quot;<br /> <br /> (1) It is prime.<br /> <br /> (2) It is even.<br /> <br /> (3) It is divisible by 7.<br /> <br /> (4) One of its digits is 9.<br /> <br /> This information allows Malcolm to determine Isabella's house number. What is its units digit?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Notice that (1) cannot be true, as we quickly see that we cannot have &lt;math&gt;2&lt;/math&gt; of the &lt;math&gt;3&lt;/math&gt; remaining conditions be true without running into a contradiction. Thus, we must have (2), (3), and (4) true. By (2), the &lt;math&gt;2&lt;/math&gt;-digit number is even, and thus the digit in the tens place must be &lt;math&gt;9&lt;/math&gt;. The only even &lt;math&gt;2&lt;/math&gt;-digit number starting with &lt;math&gt;9&lt;/math&gt; and divisible by &lt;math&gt;7&lt;/math&gt; is &lt;math&gt;98&lt;/math&gt;, which has a units digit of &lt;math&gt;\boxed{\textbf{(D)}\ 8}.&lt;/math&gt;<br /> <br /> ~nukelauncher<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=7|num-a=9}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_7&diff=88504 2017 AMC 8 Problems/Problem 7 2017-11-22T19:50:43Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 7==<br /> <br /> Let &lt;math&gt;Z&lt;/math&gt; be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of &lt;math&gt;Z&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;Z = \overline{ABCABC} = 1001 \cdot \overline{ABC} = 7 \cdot 11 \cdot 13 \cdot \overline{ABC}.&lt;/math&gt; Clearly, &lt;math&gt;Z&lt;/math&gt; is divisible by &lt;math&gt;\boxed{\textbf{(A)}\ 11}&lt;/math&gt;.<br /> <br /> ~nukelauncher<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=6|num-a=8}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_19&diff=88503 2017 AMC 8 Problems/Problem 19 2017-11-22T19:49:46Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 19==<br /> For any positive integer &lt;math&gt;M&lt;/math&gt;, the notation &lt;math&gt;M!&lt;/math&gt; denotes the product of the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;M&lt;/math&gt;. What is the largest integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;5^n&lt;/math&gt; is a factor of the sum &lt;math&gt;98!+99!+100!&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27&lt;/math&gt; <br /> <br /> ==Solution==<br /> Factoring out &lt;math&gt;98!&lt;/math&gt;, we have &lt;math&gt;98!(10,000)&lt;/math&gt;. Next, &lt;math&gt;98!&lt;/math&gt; has &lt;math&gt;\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;. Now &lt;math&gt;10,000&lt;/math&gt; has &lt;math&gt;4&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;, so there are a total of &lt;math&gt;22 + 4 = \boxed{\textbf{(D)}\ 26}&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=18|num-a=20}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_19&diff=88502 2017 AMC 8 Problems/Problem 19 2017-11-22T19:49:33Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 19==<br /> For any positive integer &lt;math&gt;M&lt;/math&gt;, the notation &lt;math&gt;M!&lt;/math&gt; denotes the product of the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;M&lt;/math&gt;. What is the largest integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;5^n&lt;/math&gt; is a factor of the sum &lt;math&gt;98!+99!+100!&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27&lt;/math&gt; <br /> <br /> ==Solution==<br /> Factoring out &lt;math&gt;98!&lt;/math&gt;, we have &lt;math&gt;98!(10,000)&lt;/math&gt;. Next, &lt;math&gt;98!&lt;/math&gt; has &lt;math&gt;\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;. Now &lt;math&gt;10,000&lt;/math&gt; has &lt;math&gt;4&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;, so there are a total of &lt;math&gt;22 + 4 = \boxed{\textbf{(D)}\ 26&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=18|num-a=20}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_18&diff=88501 2017 AMC 8 Problems/Problem 18 2017-11-22T19:46:42Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 18==<br /> In the non-convex quadrilateral &lt;math&gt;ABCD&lt;/math&gt; shown below, &lt;math&gt;\angle BCD&lt;/math&gt; is a right angle, &lt;math&gt;AB=12&lt;/math&gt;, &lt;math&gt;BC=4&lt;/math&gt;, &lt;math&gt;CD=3&lt;/math&gt;, and &lt;math&gt;AD=13&lt;/math&gt;. &lt;asy&gt;draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label(&quot;$B$&quot;, (0, 0), SW); label(&quot;$A$&quot;, (12, 0), ESE); label(&quot;$C$&quot;, (2.4, 3.6), SE); label(&quot;$D$&quot;, (0, 5), N);&lt;/asy&gt; What is the area of quadrilateral &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }12\qquad\textbf{(B) }24\qquad\textbf{(C) }26\qquad\textbf{(D) }30\qquad\textbf{(E) }36&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can see a Pythagorean triple's two longer lengths: &lt;math&gt;12&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;. So &lt;math&gt;BD&lt;/math&gt; should be &lt;math&gt;5&lt;/math&gt;. This is certainly the case because &lt;math&gt;3^2 + 4^2 = 5^2&lt;/math&gt;, which is &lt;math&gt;BD&lt;/math&gt;. Thus the area of quadrialteral &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;30-6 = \boxed{\textbf{(B)}\ 24}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=17|num-a=19}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_18&diff=88499 2017 AMC 8 Problems/Problem 18 2017-11-22T19:46:26Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 18==<br /> In the non-convex quadrilateral &lt;math&gt;ABCD&lt;/math&gt; shown below, &lt;math&gt;\angle BCD&lt;/math&gt; is a right angle, &lt;math&gt;AB=12&lt;/math&gt;, &lt;math&gt;BC=4&lt;/math&gt;, &lt;math&gt;CD=3&lt;/math&gt;, and &lt;math&gt;AD=13&lt;/math&gt;. &lt;asy&gt;draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label(&quot;$B$&quot;, (0, 0), SW); label(&quot;$A$&quot;, (12, 0), ESE); label(&quot;$C$&quot;, (2.4, 3.6), SE); label(&quot;$D$&quot;, (0, 5), N);&lt;/asy&gt; What is the area of quadrilateral &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }12\qquad\textbf{(B) }24\qquad\textbf{(C) }26\qquad\textbf{(D) }30\qquad\textbf{(E) }36&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can see a Pythagorean triple's two longer lengths: &lt;math&gt;12&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;. So &lt;math&gt;BD&lt;/math&gt; should be &lt;math&gt;5&lt;/math&gt;. This is certainly the case because &lt;math&gt;3^2 + 4^2 = 5^2&lt;/math&gt;, which is &lt;math&gt;BD&lt;/math&gt;. Thus the area of quadrialteral ABCD is &lt;math&gt;30-6 = \boxed{\textbf{(B)}\ 24}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=17|num-a=19}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_16&diff=88498 2017 AMC 8 Problems/Problem 16 2017-11-22T19:44:22Z <p>Nukelauncher: /* Problem 16 */</p> <hr /> <div>==Problem 16==<br /> <br /> In the figure below, choose point &lt;math&gt;D&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt; so that &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\triangle ABD&lt;/math&gt; have equal perimeters. What is the area of &lt;math&gt;\triangle ABD&lt;/math&gt;?<br /> &lt;asy&gt;draw((0,0)--(4,0)--(0,3)--(0,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$B$&quot;, (4,0), ESE);<br /> label(&quot;$C$&quot;, (0, 3), N);<br /> label(&quot;$3$&quot;, (0, 1.5), W);<br /> label(&quot;$4$&quot;, (2, 0), S);<br /> label(&quot;$5$&quot;, (2, 1.5), NE);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }\frac{3}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\frac{12}{5}\qquad\textbf{(E) }\frac{5}{2}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> We know that the perimeters of the two small triangles are &lt;math&gt;3+CD+AD&lt;/math&gt; and &lt;math&gt;4+BD+AD.&lt;/math&gt; Setting both equal and using &lt;math&gt;BD+CD = 5,&lt;/math&gt; we have &lt;math&gt;BD = 2&lt;/math&gt; and &lt;math&gt;CD = 3.&lt;/math&gt; Now, we simply have to find the area of &lt;math&gt;\triangle ABD.&lt;/math&gt; We can use &lt;math&gt;AB&lt;/math&gt; as the base and the altitude from &lt;math&gt;D&lt;/math&gt;. Let's call the foot of the altitude &lt;math&gt;E.&lt;/math&gt; We have &lt;math&gt;\triangle BDE&lt;/math&gt; similar to &lt;math&gt;BAC.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=15|num-a=17}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_12&diff=88497 2017 AMC 8 Problems/Problem 12 2017-11-22T19:43:39Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 12==<br /> The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?<br /> <br /> &lt;math&gt;\textbf{(A) }2\text{ and }19\qquad\textbf{(B) }20\text{ and }39\qquad\textbf{(C) }40\text{ and }59\qquad\textbf{(D) }60\text{ and }79\qquad\textbf{(E) }80\text{ and }124&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Finding the LCM of &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, and &lt;math&gt;6&lt;/math&gt;; we find it is &lt;math&gt;60&lt;/math&gt;. Now, &lt;math&gt;60+1=61&lt;/math&gt;, and that is in the range of &lt;math&gt;\boxed{\textbf{(D)}\ \text{60 and 79}}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=11|after=13}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_22&diff=88496 2017 AMC 8 Problems/Problem 22 2017-11-22T19:42:15Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 22==<br /> <br /> In the right triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AC=12&lt;/math&gt;, &lt;math&gt;BC=5&lt;/math&gt;, and angle &lt;math&gt;C&lt;/math&gt; is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?<br /> &lt;asy&gt;<br /> draw((0,0)--(12,0)--(12,5)--(0,0));<br /> draw(arc((8.67,0),(12,0),(5.33,0)));<br /> label(&quot;$A$&quot;, (0,0), W);<br /> label(&quot;$C$&quot;, (12,0), E);<br /> label(&quot;$B$&quot;, (12,5), NE);<br /> label(&quot;$12$&quot;, (6, 0), S);<br /> label(&quot;$5$&quot;, (12, 2.5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can reflect triangle &lt;math&gt;ABC&lt;/math&gt; on line &lt;math&gt;AC.&lt;/math&gt; This forms the triangle &lt;math&gt;AB'C&lt;/math&gt; and a circle out of the semicircle. Let us call the center of the circle &lt;math&gt;O.&lt;/math&gt;<br /> <br /> We can see that Circle &lt;math&gt;O&lt;/math&gt; is the incircle of &lt;math&gt;AB'C.&lt;/math&gt; We can use the formula for finding the radius of the incircle to solve this problem. The area of &lt;math&gt;AB'C&lt;/math&gt; is &lt;math&gt;12\times5 = 60.&lt;/math&gt; The semiperimeter is &lt;math&gt;5+13 = 18.&lt;/math&gt; Simplifying &lt;math&gt;\dfrac{60}{18} = \dfrac{10}{3}.&lt;/math&gt; Our answer is therefore &lt;math&gt;\boxed{\textbf{(D)}\ \frac{10}{3}}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=21|num-a=23}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_22&diff=88495 2017 AMC 8 Problems/Problem 22 2017-11-22T19:41:41Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 22==<br /> <br /> In the right triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AC=12&lt;/math&gt;, &lt;math&gt;BC=5&lt;/math&gt;, and angle &lt;math&gt;C&lt;/math&gt; is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?<br /> &lt;asy&gt;<br /> draw((0,0)--(12,0)--(12,5)--(0,0));<br /> draw(arc((8.67,0),(12,0),(5.33,0)));<br /> label(&quot;$A$&quot;, (0,0), W);<br /> label(&quot;$C$&quot;, (12,0), E);<br /> label(&quot;$B$&quot;, (12,5), NE);<br /> label(&quot;$12$&quot;, (6, 0), S);<br /> label(&quot;$5$&quot;, (12, 2.5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can reflect triangle &lt;math&gt;ABC&lt;/math&gt; on line &lt;math&gt;AC.&lt;/math&gt; This forms the triangle &lt;math&gt;AB'C&lt;/math&gt; and a circle out of the semicircle. Let us call the center of the circle &lt;math&gt;O.&lt;/math&gt;<br /> <br /> We can see that Circle &lt;math&gt;O&lt;/math&gt; is the incircle of &lt;math&gt;AB'C.&lt;/math&gt; We can use the formula for finding the radius of the incircle to solve this problem. The are of &lt;math&gt;AB'C&lt;/math&gt; is &lt;math&gt;12\times5 = 60.&lt;/math&gt; The semiperimeter is &lt;math&gt;5+13 = 18.&lt;/math&gt; Simplifying &lt;math&gt;\dfrac{60}{18} = \dfrac{10}{3}.&lt;/math&gt; Our answer is therefore &lt;math&gt;\boxed{\textbf{(D)}\ \frac{10}{3}}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=21|num-a=23}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_22&diff=88494 2017 AMC 8 Problems/Problem 22 2017-11-22T19:41:31Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 22==<br /> <br /> In the right triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AC=12&lt;/math&gt;, &lt;math&gt;BC=5&lt;/math&gt;, and angle &lt;math&gt;C&lt;/math&gt; is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?<br /> &lt;asy&gt;<br /> draw((0,0)--(12,0)--(12,5)--(0,0));<br /> draw(arc((8.67,0),(12,0),(5.33,0)));<br /> label(&quot;$A$&quot;, (0,0), W);<br /> label(&quot;$C$&quot;, (12,0), E);<br /> label(&quot;$B$&quot;, (12,5), NE);<br /> label(&quot;$12$&quot;, (6, 0), S);<br /> label(&quot;$5$&quot;, (12, 2.5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can reflect triangle &lt;math&gt;ABC&lt;/math&gt; on line &lt;math&gt;AC.&lt;/math&gt; This forms the triangle &lt;math&gt;AB'C&lt;/math&gt; and a circle out of the semicircle. Let us call the center of the circle &lt;math&gt;O.&lt;/math&gt;<br /> <br /> We can see that Circle &lt;math&gt;O&lt;/math&gt; is the incircle of &lt;math&gt;AB'C.&lt;/math&gt; We can use the formula for finding the radius of the incircle to solve this problem. The are of &lt;math&gt;AB'C&lt;/math&gt; is &lt;math&gt;12\times5 = 60.&lt;/math&gt; The semiperimeter is &lt;math&gt;5+13 = 18.&lt;/math&gt; Simplifying &lt;math&gt;\dfrac{60}{18} = \dfrac{10}{3}.&lt;/math&gt; Our answer is therefore &lt;math&gt;\boxed{\textbf{D)}\ \frac{10}{3}}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=21|num-a=23}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_22&diff=88492 2017 AMC 8 Problems/Problem 22 2017-11-22T19:40:45Z <p>Nukelauncher: </p> <hr /> <div>==Problem 22==<br /> <br /> In the right triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AC=12&lt;/math&gt;, &lt;math&gt;BC=5&lt;/math&gt;, and angle &lt;math&gt;C&lt;/math&gt; is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?<br /> &lt;asy&gt;<br /> draw((0,0)--(12,0)--(12,5)--(0,0));<br /> draw(arc((8.67,0),(12,0),(5.33,0)));<br /> label(&quot;$A$&quot;, (0,0), W);<br /> label(&quot;$C$&quot;, (12,0), E);<br /> label(&quot;$B$&quot;, (12,5), NE);<br /> label(&quot;$12$&quot;, (6, 0), S);<br /> label(&quot;$5$&quot;, (12, 2.5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can reflect triangle &lt;math&gt;ABC&lt;/math&gt; on line &lt;math&gt;AC.&lt;/math&gt; This forms the triangle &lt;math&gt;AB'C&lt;/math&gt; and a circle out of the semicircle. Let us call the center of the circle &lt;math&gt;O.&lt;/math&gt;<br /> <br /> We can see that Circle &lt;math&gt;O&lt;/math&gt; is the incircle of &lt;math&gt;AB'C.&lt;/math&gt; We can use the formula for finding the radius of the incircle to solve this problem. The are of &lt;math&gt;AB'C&lt;/math&gt; is &lt;math&gt;12\times5 = 60.&lt;/math&gt; The semiperimeter is &lt;math&gt;5+13 = 18.&lt;/math&gt; Simplifying &lt;math&gt;\dfrac{60}{18} = \dfrac{10}{3}.&lt;/math&gt; Our answer is therefore &lt;math&gt;\text{D)}&lt;/math&gt; &lt;math&gt;\dfrac{10}{3}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=21|num-a=23}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_22&diff=88491 2017 AMC 8 Problems/Problem 22 2017-11-22T19:40:27Z <p>Nukelauncher: /* Solution: */</p> <hr /> <div>==Problem 22==<br /> <br /> In the right triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AC=12&lt;/math&gt;, &lt;math&gt;BC=5&lt;/math&gt;, and angle &lt;math&gt;C&lt;/math&gt; is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?<br /> &lt;asy&gt;<br /> draw((0,0)--(12,0)--(12,5)--(0,0));<br /> draw(arc((8.67,0),(12,0),(5.33,0)));<br /> label(&quot;$A$&quot;, (0,0), W);<br /> label(&quot;$C$&quot;, (12,0), E);<br /> label(&quot;$B$&quot;, (12,5), NE);<br /> label(&quot;$12$&quot;, (6, 0), S);<br /> label(&quot;$5$&quot;, (12, 2.5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can reflect triangle &lt;math&gt;ABC&lt;/math&gt; on line &lt;math&gt;AC.&lt;/math&gt; This forms the triangle &lt;math&gt;AB'C&lt;/math&gt; and a circle out of the semicircle. Let us call the center of the circle &lt;math&gt;O.&lt;/math&gt;<br /> <br /> We can see that Circle &lt;math&gt;O&lt;/math&gt; is the incircle of &lt;math&gt;AB'C.&lt;/math&gt; We can use the formula for finding the radius of the incircle to solve this problem. The are of &lt;math&gt;AB'C&lt;/math&gt; is &lt;math&gt;12\times5 = 60.&lt;/math&gt; The semiperimeter is &lt;math&gt;5+13 = 18.&lt;/math&gt; Simplifying &lt;math&gt;\dfrac{60}{18} = \dfrac{10}{3}.&lt;/math&gt; Our answer is therefore &lt;math&gt;\text{D)}&lt;/math&gt; &lt;math&gt;\dfrac{10}{3}.&lt;/math&gt;</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_22&diff=88490 2017 AMC 8 Problems/Problem 22 2017-11-22T19:40:20Z <p>Nukelauncher: /* Problem 22 */</p> <hr /> <div>==Problem 22==<br /> <br /> In the right triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AC=12&lt;/math&gt;, &lt;math&gt;BC=5&lt;/math&gt;, and angle &lt;math&gt;C&lt;/math&gt; is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?<br /> &lt;asy&gt;<br /> draw((0,0)--(12,0)--(12,5)--(0,0));<br /> draw(arc((8.67,0),(12,0),(5.33,0)));<br /> label(&quot;$A$&quot;, (0,0), W);<br /> label(&quot;$C$&quot;, (12,0), E);<br /> label(&quot;$B$&quot;, (12,5), NE);<br /> label(&quot;$12$&quot;, (6, 0), S);<br /> label(&quot;$5$&quot;, (12, 2.5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}&lt;/math&gt;<br /> <br /> ==Solution:==<br /> We can reflect triangle &lt;math&gt;ABC&lt;/math&gt; on line &lt;math&gt;AC.&lt;/math&gt; This forms the triangle &lt;math&gt;AB'C&lt;/math&gt; and a circle out of the semicircle. Let us call the center of the circle &lt;math&gt;O.&lt;/math&gt;<br /> <br /> We can see that Circle &lt;math&gt;O&lt;/math&gt; is the incircle of &lt;math&gt;AB'C.&lt;/math&gt; We can use the formula for finding the radius of the incircle to solve this problem. The are of &lt;math&gt;AB'C&lt;/math&gt; is &lt;math&gt;12\times5 = 60.&lt;/math&gt; The semiperimeter is &lt;math&gt;5+13 = 18.&lt;/math&gt; Simplifying &lt;math&gt;\dfrac{60}{18} = \dfrac{10}{3}.&lt;/math&gt; Our answer is therefore &lt;math&gt;\text{D)}&lt;/math&gt; &lt;math&gt;\dfrac{10}{3}.&lt;/math&gt;</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_17&diff=88488 2017 AMC 8 Problems/Problem 17 2017-11-22T19:39:08Z <p>Nukelauncher: </p> <hr /> <div>==Problem 17==<br /> <br /> Starting with some gold coins and some empty treasure chests, I tried to put &lt;math&gt;9 &lt;/math&gt; gold coins in each treasure chest, but that left &lt;math&gt;2&lt;/math&gt; treasure chests empty. So instead I put &lt;math&gt;6&lt;/math&gt; gold coins in each treasure chest, but then I had &lt;math&gt;3&lt;/math&gt; gold coins left over. How many gold coins did I have?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad\textbf{(D) }63\qquad\textbf{(E) }81&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can represent the amount of gold with &lt;math&gt;g&lt;/math&gt; and the amount of chests with &lt;math&gt;c&lt;/math&gt;. We can use the problem to make the following equations:<br /> &lt;cmath&gt;9c-18 = g&lt;/cmath&gt;<br /> &lt;cmath&gt;6c+3 = g&lt;/cmath&gt;<br /> <br /> Therefore, &lt;math&gt;6c+3 = 9c-18.&lt;/math&gt; This implies that &lt;math&gt;c = 7.&lt;/math&gt; We therefore have &lt;math&gt;g = 45.&lt;/math&gt; So, our answer is &lt;math&gt;\boxed{\textbf{C)}\ 45}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=16|num-a=18}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_17&diff=88487 2017 AMC 8 Problems/Problem 17 2017-11-22T19:38:45Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 17==<br /> <br /> Starting with some gold coins and some empty treasure chests, I tried to put &lt;math&gt;9 &lt;/math&gt; gold coins in each treasure chest, but that left &lt;math&gt;2&lt;/math&gt; treasure chests empty. So instead I put &lt;math&gt;6&lt;/math&gt; gold coins in each treasure chest, but then I had &lt;math&gt;3&lt;/math&gt; gold coins left over. How many gold coins did I have?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad\textbf{(D) }63\qquad\textbf{(E) }81&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can represent the amount of gold with &lt;math&gt;g&lt;/math&gt; and the amount of chests with &lt;math&gt;c&lt;/math&gt;. We can use the problem to make the following equations:<br /> &lt;cmath&gt;9c-18 = g&lt;/cmath&gt;<br /> &lt;cmath&gt;6c+3 = g&lt;/cmath&gt;<br /> <br /> Therefore, &lt;math&gt;6c+3 = 9c-18.&lt;/math&gt; This implies that &lt;math&gt;c = 7.&lt;/math&gt; We therefore have &lt;math&gt;g = 45.&lt;/math&gt; So, our answer is &lt;math&gt;\boxed{\textbf{C)}\ 45}.&lt;/math&gt;</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_17&diff=88486 2017 AMC 8 Problems/Problem 17 2017-11-22T19:38:05Z <p>Nukelauncher: /* Problem 17 */</p> <hr /> <div>==Problem 17==<br /> <br /> Starting with some gold coins and some empty treasure chests, I tried to put &lt;math&gt;9 &lt;/math&gt; gold coins in each treasure chest, but that left &lt;math&gt;2&lt;/math&gt; treasure chests empty. So instead I put &lt;math&gt;6&lt;/math&gt; gold coins in each treasure chest, but then I had &lt;math&gt;3&lt;/math&gt; gold coins left over. How many gold coins did I have?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad\textbf{(D) }63\qquad\textbf{(E) }81&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can represent the amount of gold with &lt;math&gt;g&lt;/math&gt; and the amount of chests with &lt;math&gt;c&lt;/math&gt;. We can use the problem to make the following equations:<br /> &lt;cmath&gt;9c-18 = g&lt;/cmath&gt;<br /> &lt;cmath&gt;6c+3 = g&lt;/cmath&gt;<br /> <br /> Therefore, &lt;math&gt;6c+3 = 9c-18.&lt;/math&gt; This implies that &lt;math&gt;c = 7.&lt;/math&gt; We therefore have &lt;math&gt;g = 45.&lt;/math&gt; So, our answer is &lt;math&gt;\text{C)}&lt;/math&gt; &lt;math&gt;45.&lt;/math&gt;</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_17&diff=88485 2017 AMC 8 Problems/Problem 17 2017-11-22T19:37:43Z <p>Nukelauncher: /* Solution: */</p> <hr /> <div>==Problem 17==<br /> <br /> Starting with some gold coins and some empty treasure chests, I tried to put &lt;math&gt;9 &lt;/math&gt; gold coins in each treasure chest, but that left &lt;math&gt;2&lt;/math&gt; treasure chests empty. So instead I put &lt;math&gt;6&lt;/math&gt; gold coins in each treasure chest, but then I had &lt;math&gt;3&lt;/math&gt; gold coins left over. How many gold coins did I have?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad\textbf{(D) }63\qquad\textbf{(E) }81&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 17|Solution<br /> ]]<br /> <br /> ==Solution==<br /> We can represent the amount of gold with &lt;math&gt;g&lt;/math&gt; and the amount of chests with &lt;math&gt;c&lt;/math&gt;. We can use the problem to make the following equations:<br /> &lt;cmath&gt;9c-18 = g&lt;/cmath&gt;<br /> &lt;cmath&gt;6c+3 = g&lt;/cmath&gt;<br /> <br /> Therefore, &lt;math&gt;6c+3 = 9c-18.&lt;/math&gt; This implies that &lt;math&gt;c = 7.&lt;/math&gt; We therefore have &lt;math&gt;g = 45.&lt;/math&gt; So, our answer is &lt;math&gt;\text{C)}&lt;/math&gt; &lt;math&gt;45.&lt;/math&gt;</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_8&diff=88484 2017 AMC 8 Problems/Problem 8 2017-11-22T19:36:47Z <p>Nukelauncher: </p> <hr /> <div>==Problem 8==<br /> Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, &quot;My house number has two digits, and exactly three of the following four statements about it are true.&quot;<br /> <br /> (1) It is prime.<br /> <br /> (2) It is even.<br /> <br /> (3) It is divisible by 7.<br /> <br /> (4) One of its digits is 9.<br /> <br /> This information allows Malcolm to determine Isabella's house number. What is its units digit?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Notice that (1) cannot be true, as we quickly see that we cannot have &lt;math&gt;2&lt;/math&gt; of the &lt;math&gt;3&lt;/math&gt; remaining conditions be true without running into a contradiction. Thus, we must have (2), (3), and (4) true. By (2), the &lt;math&gt;2&lt;/math&gt;-digit number is even, and thus the digit in the tens place must be &lt;math&gt;9&lt;/math&gt;. The only even &lt;math&gt;2&lt;/math&gt;-digit number starting with &lt;math&gt;9&lt;/math&gt; and divisible by &lt;math&gt;7&lt;/math&gt; is &lt;math&gt;98&lt;/math&gt;, which has a units digit of &lt;math&gt;\boxed{\textbf{(D)}\ 8}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=7|num-a=9}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_7&diff=88480 2017 AMC 8 Problems/Problem 7 2017-11-22T19:32:25Z <p>Nukelauncher: </p> <hr /> <div>==Problem 7==<br /> <br /> Let &lt;math&gt;Z&lt;/math&gt; be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of &lt;math&gt;Z&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;Z = \overline{ABCABC} = 1001 \cdot \overline{ABC} = 7 \cdot 11 \cdot 13 \cdot \overline{ABC}.&lt;/math&gt; Clearly, &lt;math&gt;Z&lt;/math&gt; is divisible by &lt;math&gt;\boxed{\textbf{(A)}\ 11}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=6|num-a=8}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_7&diff=88479 2017 AMC 8 Problems/Problem 7 2017-11-22T19:32:00Z <p>Nukelauncher: </p> <hr /> <div>==Problem 7==<br /> <br /> Let &lt;math&gt;Z&lt;/math&gt; be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of &lt;math&gt;Z&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;Z = \overline{ABCABC} = 1001 \cdot \overline{ABC} = 7 \cdot 11 \cdot 13 \cdot \overline{ABC}.&lt;/math&gt; Clearly, &lt;math&gt;Z&lt;/math&gt; is divisible by &lt;math&gt;\boxed{\textbf{(A)}\ 11}&lt;/math&gt;.</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_5&diff=88474 2017 AMC 8 Problems/Problem 5 2017-11-22T19:28:44Z <p>Nukelauncher: /* See Also */</p> <hr /> <div>==Problem 5==<br /> What is the value of the expression &lt;math&gt;\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360&lt;/math&gt; <br /> <br /> ==Solution==<br /> We evaluate both the top and bottom: &lt;math&gt;\frac{40320}{36}&lt;/math&gt;. This simplifies to &lt;math&gt;\boxed{\textbf{(B)}\ 1120}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_3&diff=88473 2017 AMC 8 Problems/Problem 3 2017-11-22T19:27:38Z <p>Nukelauncher: /* Solution */</p> <hr /> <div>==Problem 3==<br /> <br /> What is the value of the expression &lt;math&gt;\sqrt{16\sqrt{8\sqrt{4}}}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }4\sqrt{2}\qquad\textbf{(C) }8\qquad\textbf{(D) }8\sqrt{2}\qquad\textbf{(E) }16&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> &lt;math&gt;\sqrt{16\sqrt{8\sqrt{4}}} = \sqrt{16\sqrt{8\cdot 2}} = \sqrt{16\sqrt{16}} = \sqrt{16\cdot 4} = \sqrt{64} = \boxed{\textbf{(C)}\ 8}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=2|num-a=4}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_4&diff=88470 2017 AMC 8 Problems/Problem 4 2017-11-22T19:24:16Z <p>Nukelauncher: </p> <hr /> <div>==Problem 4==<br /> When &lt;math&gt;0.000315&lt;/math&gt; is multiplied by &lt;math&gt;7,928,564&lt;/math&gt; the product is closest to which of the following?<br /> <br /> &lt;math&gt;\textbf{(A) }210\qquad\textbf{(B) }240\qquad\textbf{(C) }2100\qquad\textbf{(D) }2400\qquad\textbf{(E) }24000&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can approximate &lt;math&gt;7,928,564&lt;/math&gt; to &lt;math&gt;8,000,000,&lt;/math&gt; and &lt;math&gt;0.000315&lt;/math&gt; to &lt;math&gt;0.0003.&lt;/math&gt; Multiplying the two yields &lt;math&gt;2400.&lt;/math&gt; This gives our answer to be &lt;math&gt;\boxed{\textbf{(D)}\ 2400}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=3|num-a=5}}<br /> <br /> {{MAA Notice}}</div> Nukelauncher https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_4&diff=88468 2017 AMC 8 Problems/Problem 4 2017-11-22T19:23:49Z <p>Nukelauncher: /* Solution: */</p> <hr /> <div>==Problem 4==<br /> When &lt;math&gt;0.000315&lt;/math&gt; is multiplied by &lt;math&gt;7,928,564&lt;/math&gt; the product is closest to which of the following?<br /> <br /> &lt;math&gt;\textbf{(A) }210\qquad\textbf{(B) }240\qquad\textbf{(C) }2100\qquad\textbf{(D) }2400\qquad\textbf{(E) }24000&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can approximate &lt;math&gt;7,928,564&lt;/math&gt; to &lt;math&gt;8,000,000,&lt;/math&gt; and &lt;math&gt;0.000315&lt;/math&gt; to &lt;math&gt;0.0003.&lt;/math&gt; Multiplying the two yields &lt;math&gt;2400.&lt;/math&gt; This gives our answer to be &lt;math&gt;\boxed{\textbf{(D)}\ 2400}.&lt;/math&gt;</div> Nukelauncher