https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=ZekromReshiram&feedformat=atom AoPS Wiki - User contributions [en] 2021-09-28T08:37:51Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_9&diff=80884 1998 AIME Problems/Problem 9 2016-10-31T03:26:36Z <p>ZekromReshiram: </p> <hr /> <div>== Problem ==<br /> Two mathematicians take a morning coffee break each day. They arrive at the cafeteria independently, at random times between 9 a.m. and 10 a.m., and stay for exactly &lt;math&gt;m&lt;/math&gt; minutes. The [[probability]] that either one arrives while the other is in the cafeteria is &lt;math&gt;40 \%,&lt;/math&gt; and &lt;math&gt;m = a - b\sqrt {c},&lt;/math&gt; where &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are [[positive]] [[integer]]s, and &lt;math&gt;c&lt;/math&gt; is not divisible by the square of any [[prime]]. Find &lt;math&gt;a + b + c.&lt;/math&gt;<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Let the two mathematicians be &lt;math&gt;M_1&lt;/math&gt; and &lt;math&gt;M_2&lt;/math&gt;. Consider plotting the times that they are on break on a [[coordinate plane]] with one axis being the time &lt;math&gt;M_1&lt;/math&gt; arrives and the second axis being the time &lt;math&gt;M_2&lt;/math&gt; arrives (in minutes past 9 a.m.). The two mathematicians meet each other when &lt;math&gt;|M_1-M_2| \leq m&lt;/math&gt;. Also because the mathematicians arrive between 9 and 10, &lt;math&gt;0 \leq M_1,M_2 \leq 60&lt;/math&gt;. Therefore, &lt;math&gt;60\times 60&lt;/math&gt; square represents the possible arrival times of the mathematicians, while the shaded region represents the arrival times where they meet.<br /> &lt;asy&gt;<br /> import graph;<br /> size(180);<br /> real m=60-12*sqrt(15);<br /> draw((0,0)--(60,0)--(60,60)--(0,60)--cycle);<br /> fill((m,0)--(60,60-m)--(60,60)--(60-m,60)--(0,m)--(0,0)--cycle,lightgray);<br /> draw((m,0)--(60,60-m)--(60,60)--(60-m,60)--(0,m)--(0,0)--cycle);<br /> xaxis(&quot;$M_1$&quot;,-10,80);<br /> yaxis(&quot;$M_2$&quot;,-10,80);<br /> label(rotate(45)*&quot;$M_1-M_2\le m$&quot;,((m+60)/2,(60-m)/2),NW,fontsize(9));<br /> label(rotate(45)*&quot;$M_1-M_2\ge -m$&quot;,((60-m)/2,(m+60)/2),SE,fontsize(9));<br /> label(&quot;$m$&quot;,(m,0),S);<br /> label(&quot;$m$&quot;,(0,m),W);<br /> label(&quot;$60$&quot;,(60,0),S);<br /> label(&quot;$60$&quot;,(0,60),W);<br /> &lt;/asy&gt;<br /> It's easier to compete the area of the unshaded region over the area of the total region, which is the probability that the mathematicians do not meet:<br /> &lt;div style=&quot;text-align:center;&quot;&gt;<br /> &lt;math&gt;\frac{(60-m)^2}{60^2} = .6&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;(60-m)^2 = 36\cdot 60&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;60 - m = 12\sqrt{15}&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;\Rightarrow m = 60-12\sqrt{15}&lt;/math&gt;<br /> &lt;/div&gt;<br /> So the answer is &lt;math&gt;60 + 12 + 15 = 087&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Case 1:<br /> <br /> [[Image:AIME_1998-9.png]]<br /> Case 2:<br /> <br /> [[Image:AIME_1998-9b.png]]<br /> <br /> We draw a [[number line]] representing the time interval. If mathematician &lt;math&gt;M_1&lt;/math&gt; comes in at the center of the time period, then the two mathematicions will meet if &lt;math&gt;M_2&lt;/math&gt; comes in somewhere between &lt;math&gt;m&lt;/math&gt; minutes before and after &lt;math&gt;M_1&lt;/math&gt; comes (a total range of &lt;math&gt;2m&lt;/math&gt; minutes). However, if &lt;math&gt;M_1&lt;/math&gt; comes into the cafeteria in the first or last &lt;math&gt;m&lt;/math&gt; minutes, then the range in which &lt;math&gt;M_2&lt;/math&gt; is reduced to somewhere in between &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;2m&lt;/math&gt;.<br /> <br /> We know try to find the [[weighted average]] of the chance that the two meet. In the central &lt;math&gt;60-2m&lt;/math&gt; minutes, &lt;math&gt;M_1&lt;/math&gt; and &lt;math&gt;M_2&lt;/math&gt; have to enter the cafeteria within &lt;math&gt;m&lt;/math&gt; minutes of each other; so if we fix point &lt;math&gt;M_1&lt;/math&gt; then &lt;math&gt;M_2&lt;/math&gt; has a &lt;math&gt;\frac{2m}{60} = \frac{m}{30}&lt;/math&gt; probability of meeting.<br /> <br /> In the first and last &lt;math&gt;2m&lt;/math&gt; minutes, the probability that the two meet ranges from &lt;math&gt;\frac{m}{60}&lt;/math&gt; to &lt;math&gt;\frac{2m}{60}&lt;/math&gt;, depending upon the location of &lt;math&gt;M_1&lt;/math&gt; with respect to the endpoints. Intuitively, the average probability will occur at &lt;math&gt;\frac{\frac{3}{2}m}{60} = \frac{m}{40}&lt;/math&gt;.<br /> <br /> So the weighted average is:<br /> :&lt;math&gt;\frac{\frac{m}{30}(60-2m) + \frac{m}{40}(2m)}{60} = \frac{40}{100}&lt;/math&gt;<br /> :&lt;math&gt;0 = \frac{m^2}{60} - 2m + 24&lt;/math&gt;<br /> :&lt;math&gt;0 = m^2 - 120m + 1440&lt;/math&gt;<br /> <br /> Solving this [[quadratic equation|quadratic]], we get two roots, &lt;math&gt;60 \pm 12\sqrt{15}&lt;/math&gt;. However, &lt;math&gt;m &lt; 60&lt;/math&gt;, so we discard the greater root; and thus our answer &lt;math&gt;60 + 12 + 15 = 087&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1998|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> ZekromReshiram https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems&diff=71198 2003 AIME II Problems 2015-07-17T00:37:30Z <p>ZekromReshiram: </p> <hr /> <div>{{AIME Problems|year=2003|n=II}}<br /> <br /> == Problem 1 ==<br /> The product &lt;math&gt;N&lt;/math&gt; of three positive integers is 6 times their sum, and one of the integers is the sum of the other two. Find the sum of all possible values of &lt;math&gt;N&lt;/math&gt;.<br /> <br /> [[2003 AIME II Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> Let &lt;math&gt;N&lt;/math&gt; be the greatest integer multiple of 8, whose digits are all different. What is the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000.<br /> <br /> [[2003 AIME II Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> Define a &lt;math&gt;good~word&lt;/math&gt; as a sequence of letters that consists only of the letters &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; - some of these letters may not appear in the sequence - and in which &lt;math&gt;A&lt;/math&gt; is never immediately followed by &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; is never immediately followed by &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; is never immediately followed by &lt;math&gt;A&lt;/math&gt;. How many seven-letter good words are there?<br /> <br /> [[2003 AIME II Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is &lt;math&gt;m/n&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2003 AIME II Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> A cylindrical log has diameter &lt;math&gt;12&lt;/math&gt; inches. A wedge is cut from the log by making two planar cuts that go entirely through the log. The first is perpendicular to the axis of the cylinder, and the plane of the second cut forms a &lt;math&gt;45^\circ&lt;/math&gt; angle with the plane of the first cut. The intersection of these two planes has exactly one point in common with the log. The number of cubic inches in the wedge can be expressed as &lt;math&gt;n\pi&lt;/math&gt;, where n is a positive integer. Find &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2003 AIME II Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> In triangle &lt;math&gt;ABC,&lt;/math&gt; &lt;math&gt;AB = 13,&lt;/math&gt; &lt;math&gt;BC = 14,&lt;/math&gt; &lt;math&gt;AC = 15,&lt;/math&gt; and point &lt;math&gt;G&lt;/math&gt; is the intersection of the medians. Points &lt;math&gt;A',&lt;/math&gt; &lt;math&gt;B',&lt;/math&gt; and &lt;math&gt;C',&lt;/math&gt; are the images of &lt;math&gt;A,&lt;/math&gt; &lt;math&gt;B,&lt;/math&gt; and &lt;math&gt;C,&lt;/math&gt; respectively, after a &lt;math&gt;180^\circ&lt;/math&gt; rotation about &lt;math&gt;G.&lt;/math&gt; What is the area of the union of the two regions enclosed by the triangles &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;A'B'C'?&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Find the area of rhombus &lt;math&gt;ABCD&lt;/math&gt; given that the radii of the circles circumscribed around triangles &lt;math&gt;ABD&lt;/math&gt; and &lt;math&gt;ACD&lt;/math&gt; are &lt;math&gt;12.5&lt;/math&gt; and &lt;math&gt;25&lt;/math&gt;, respectively.<br /> <br /> [[2003 AIME II Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Find the eighth term of the sequence &lt;math&gt;1440,&lt;/math&gt; &lt;math&gt;1716,&lt;/math&gt; &lt;math&gt;1848,\ldots,&lt;/math&gt; whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.<br /> <br /> [[2003 AIME II Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Consider the polynomials &lt;math&gt;P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x&lt;/math&gt; and &lt;math&gt;Q(x) = x^{4} - x^{3} - x^{2} - 1.&lt;/math&gt; Given that &lt;math&gt;z_{1},z_{2},z_{3},&lt;/math&gt; and &lt;math&gt;z_{4}&lt;/math&gt; are the roots of &lt;math&gt;Q(x) = 0,&lt;/math&gt; find &lt;math&gt;P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> Two positive integers differ by &lt;math&gt;60.&lt;/math&gt; The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?<br /> <br /> [[2003 AIME II Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is a right triangle with &lt;math&gt;AC = 7,&lt;/math&gt; &lt;math&gt;BC = 24,&lt;/math&gt; and right angle at &lt;math&gt;C.&lt;/math&gt; Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;AB,&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; is on the same side of line &lt;math&gt;AB&lt;/math&gt; as &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;AD = BD = 15.&lt;/math&gt; Given that the area of triangle &lt;math&gt;CDM&lt;/math&gt; may be expressed as &lt;math&gt;\frac {m\sqrt {n}}{p},&lt;/math&gt; where &lt;math&gt;m,&lt;/math&gt; &lt;math&gt;n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are relatively prime, and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime, find &lt;math&gt;m + n + p.&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The members of a distinguished committee were choosing a president, and each member gave one vote to one of the &lt;math&gt;27&lt;/math&gt; candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least &lt;math&gt;1&lt;/math&gt; than the number of votes for that candidate. What is the smallest possible number of members of the committee?<br /> <br /> [[2003 AIME II Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is &lt;math&gt;m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m + n.&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Let &lt;math&gt;A = (0,0)&lt;/math&gt; and &lt;math&gt;B = (b,2)&lt;/math&gt; be points on the coordinate plane. Let &lt;math&gt;ABCDEF&lt;/math&gt; be a convex equilateral hexagon such that &lt;math&gt;\angle FAB = 120^\circ,&lt;/math&gt; &lt;math&gt;\overline{AB}\parallel \overline{DE},&lt;/math&gt; &lt;math&gt;\overline{BC}\parallel \overline{EF,}&lt;/math&gt; &lt;math&gt;\overline{CD}\parallel \overline{FA},&lt;/math&gt; and the y-coordinates of its vertices are distinct elements of the set &lt;math&gt;\{0,2,4,6,8,10\}.&lt;/math&gt; The area of the hexagon can be written in the form &lt;math&gt;m\sqrt {n},&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers and n is not divisible by the square of any prime. Find &lt;math&gt;m + n.&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> Let<br /> &lt;center&gt;&lt;math&gt;P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).&lt;/math&gt;&lt;/center&gt;<br /> Let &lt;math&gt;z_{1},z_{2},\ldots,z_{r}&lt;/math&gt; be the distinct zeros of &lt;math&gt;P(x),&lt;/math&gt; and let &lt;math&gt;z_{k}^{2} = a_{k} + b_{k}i&lt;/math&gt; for &lt;math&gt;k = 1,2,\ldots,r,&lt;/math&gt; where &lt;math&gt;i = \sqrt { - 1},&lt;/math&gt; and &lt;math&gt;a_{k}&lt;/math&gt; and &lt;math&gt;b_{k}&lt;/math&gt; are real numbers. Let<br /> &lt;center&gt;&lt;math&gt;\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},&lt;/math&gt;&lt;/center&gt;<br /> where &lt;math&gt;m,&lt;/math&gt; &lt;math&gt;n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m + n + p.&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 15|Solution]]<br /> <br /> == See also ==<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> ZekromReshiram https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_20&diff=67625 2011 AMC 12B Problems/Problem 20 2015-02-05T19:35:56Z <p>ZekromReshiram: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has &lt;math&gt;AB = 13, BC = 14&lt;/math&gt;, and &lt;math&gt;AC = 15&lt;/math&gt;. The points &lt;math&gt;D, E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; are the midpoints of &lt;math&gt;\overline{AB}, \overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt; respectively. Let &lt;math&gt;X \neq E&lt;/math&gt; be the intersection of the circumcircles of &lt;math&gt;\triangle BDE&lt;/math&gt; and &lt;math&gt;\triangle CEF&lt;/math&gt;. What is &lt;math&gt;XA + XB + XC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 14\sqrt{3} \qquad \textbf{(C)}\ \frac{195}{8} \qquad \textbf{(D)}\ \frac{129\sqrt{7}}{14} \qquad \textbf{(E)}\ \frac{69\sqrt{2}}{4}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Answer: (C)<br /> <br /> Let us also consider the circumcircle of &lt;math&gt;\triangle ADF&lt;/math&gt;.<br /> <br /> Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt; which is &lt;math&gt;P&lt;/math&gt;, Also, since &lt;math&gt;m\angle ADP = m\angle AFP = 90^\circ&lt;/math&gt;. &lt;math&gt;ADPF&lt;/math&gt; is cyclic, similarly, &lt;math&gt;BDPE&lt;/math&gt; and &lt;math&gt;CEPF&lt;/math&gt; are also cyclic. With this, we know that the circumcircles of &lt;math&gt;\triangle ADF&lt;/math&gt;, &lt;math&gt;\triangle BDE&lt;/math&gt; and &lt;math&gt;\triangle CEF&lt;/math&gt; all intercept at &lt;math&gt;P&lt;/math&gt;, so &lt;math&gt;P&lt;/math&gt; is &lt;math&gt;X&lt;/math&gt;.<br /> <br /> The question now becomes calculate the sum of distance from each vertices to the circumcenter. <br /> <br /> We can calculate the distances with coordinate geometry. (Note that &lt;math&gt;XA = XB = XC&lt;/math&gt; because &lt;math&gt;X&lt;/math&gt; is the circumcenter.)<br /> <br /> Let &lt;math&gt;A = (5,12)&lt;/math&gt;, &lt;math&gt;B = (0,0)&lt;/math&gt;, &lt;math&gt;C = (14, 0)&lt;/math&gt;, &lt;math&gt;X= (x_0, y_0)&lt;/math&gt;<br /> <br /> Then &lt;math&gt;X&lt;/math&gt; is on the line &lt;math&gt;x = 7&lt;/math&gt; and also the line with slope &lt;math&gt;-\frac{5}{12}&lt;/math&gt; and passes through &lt;math&gt;(2.5, 6)&lt;/math&gt;. <br /> <br /> &lt;math&gt;y_0 = 6-\frac{45}{24} = \frac{33}{8}&lt;/math&gt;<br /> <br /> So &lt;math&gt;X = (7, \frac{33}{8})&lt;/math&gt;<br /> <br /> and &lt;math&gt;XA +XB+XC = 3XB = 3\sqrt{7^2 + \left(\frac{33}{8}\right)^2} = 3\times\frac{65}{8}=\frac{195}{8}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Consider an additional circumcircle on &lt;math&gt;\triangle ADF&lt;/math&gt;. After drawing the diagram, it is noticed that each triangle has side values: &lt;math&gt;7&lt;/math&gt;, &lt;math&gt;\frac{15}{2}&lt;/math&gt;, &lt;math&gt;\frac{13}{2}&lt;/math&gt;. Thus they are congruent, and their respective circumcircles are. By inspection, we see that &lt;math&gt;XA&lt;/math&gt;, &lt;math&gt;XB&lt;/math&gt;, and &lt;math&gt;XC&lt;/math&gt; are the circumdiameters, and so they are congruent. Therefore, the solution can be found by calculating one of these circumdiameters and multiplying it by a factor of &lt;math&gt;3&lt;/math&gt;. We can find the circumradius quite easily with the formula &lt;math&gt;\sqrt{(s)(s-a)(s-b)(s-c)} = \frac{abc}{4R}&lt;/math&gt;, s.t. &lt;math&gt;s=\frac{a+b+c}{2}&lt;/math&gt; and R is the circumradius. Since &lt;math&gt;s = \frac{21}{2}&lt;/math&gt;:<br /> <br /> &lt;cmath&gt; \sqrt{(\frac{21}{2})(4)(3)(\frac{7}{2})} = \frac{\frac{15}{2}\cdot\frac{13}{2}\cdot 7}{4R} &lt;/cmath&gt;<br /> <br /> After a few algebraic manipulations:<br /> <br /> &lt;math&gt;\Rightarrow R = \frac{65}{16} \Rightarrow D=2R=\frac{65}{8} \Rightarrow 3D = \boxed{\frac{195}{8}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=19|num-a=21|ab=B}}<br /> {{MAA Notice}}</div> ZekromReshiram https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_20&diff=67624 2011 AMC 12B Problems/Problem 20 2015-02-05T19:35:47Z <p>ZekromReshiram: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has &lt;math&gt;AB = 13, BC = 14&lt;/math&gt;, and &lt;math&gt;AC = 15&lt;/math&gt;. The points &lt;math&gt;D, E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; are the midpoints of &lt;math&gt;\overline{AB}, \overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt; respectively. Let &lt;math&gt;X \neq E&lt;/math&gt; be the intersection of the circumcircles of &lt;math&gt;\triangle BDE&lt;/math&gt; and &lt;math&gt;\triangle CEF&lt;/math&gt;. What is &lt;math&gt;XA + XB + XC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 14\sqrt{3} \qquad \textbf{(C)}\ \frac{195}{8} \qquad \textbf{(D)}\ \frac{129\sqrt{7}}{14} \qquad \textbf{(E)}\ \frac{69\sqrt{2}}{4}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Answer: (C)<br /> <br /> Let us also consider the circumcircle of &lt;math&gt;\triangle ADF&lt;/math&gt;.<br /> <br /> Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt; which is &lt;math&gt;P&lt;/math&gt;, Also, since &lt;math&gt;m\angle ADP = m\angle AFP = 90^\circ&lt;/math&gt;. &lt;math&gt;ADPF&lt;/math&gt; is cyclic, similarly, &lt;math&gt;BDPE&lt;/math&gt; and &lt;math&gt;CEPF&lt;/math&gt; are also cyclic. With this, we know that the circumcircles of &lt;math&gt;\triangle ADF&lt;/math&gt;, &lt;math&gt;\triangle BDE&lt;/math&gt; and &lt;math&gt;\triangle CEF&lt;/math&gt; all intercept at &lt;math&gt;P&lt;/math&gt;, so &lt;math&gt;P&lt;/math&gt; is &lt;math&gt;X&lt;/math&gt;.<br /> <br /> The question now becomes calculate the sum of distance from each vertices to the circumcenter. <br /> <br /> We can calculate the distances with coordinate geometry. (Note that &lt;math&gt;XA = XB = XC&lt;/math&gt; because &lt;math&gt;X&lt;/math&gt; is the circumcenter.<br /> <br /> Let &lt;math&gt;A = (5,12)&lt;/math&gt;, &lt;math&gt;B = (0,0)&lt;/math&gt;, &lt;math&gt;C = (14, 0)&lt;/math&gt;, &lt;math&gt;X= (x_0, y_0)&lt;/math&gt;<br /> <br /> Then &lt;math&gt;X&lt;/math&gt; is on the line &lt;math&gt;x = 7&lt;/math&gt; and also the line with slope &lt;math&gt;-\frac{5}{12}&lt;/math&gt; and passes through &lt;math&gt;(2.5, 6)&lt;/math&gt;. <br /> <br /> &lt;math&gt;y_0 = 6-\frac{45}{24} = \frac{33}{8}&lt;/math&gt;<br /> <br /> So &lt;math&gt;X = (7, \frac{33}{8})&lt;/math&gt;<br /> <br /> and &lt;math&gt;XA +XB+XC = 3XB = 3\sqrt{7^2 + \left(\frac{33}{8}\right)^2} = 3\times\frac{65}{8}=\frac{195}{8}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Consider an additional circumcircle on &lt;math&gt;\triangle ADF&lt;/math&gt;. After drawing the diagram, it is noticed that each triangle has side values: &lt;math&gt;7&lt;/math&gt;, &lt;math&gt;\frac{15}{2}&lt;/math&gt;, &lt;math&gt;\frac{13}{2}&lt;/math&gt;. Thus they are congruent, and their respective circumcircles are. By inspection, we see that &lt;math&gt;XA&lt;/math&gt;, &lt;math&gt;XB&lt;/math&gt;, and &lt;math&gt;XC&lt;/math&gt; are the circumdiameters, and so they are congruent. Therefore, the solution can be found by calculating one of these circumdiameters and multiplying it by a factor of &lt;math&gt;3&lt;/math&gt;. We can find the circumradius quite easily with the formula &lt;math&gt;\sqrt{(s)(s-a)(s-b)(s-c)} = \frac{abc}{4R}&lt;/math&gt;, s.t. &lt;math&gt;s=\frac{a+b+c}{2}&lt;/math&gt; and R is the circumradius. Since &lt;math&gt;s = \frac{21}{2}&lt;/math&gt;:<br /> <br /> &lt;cmath&gt; \sqrt{(\frac{21}{2})(4)(3)(\frac{7}{2})} = \frac{\frac{15}{2}\cdot\frac{13}{2}\cdot 7}{4R} &lt;/cmath&gt;<br /> <br /> After a few algebraic manipulations:<br /> <br /> &lt;math&gt;\Rightarrow R = \frac{65}{16} \Rightarrow D=2R=\frac{65}{8} \Rightarrow 3D = \boxed{\frac{195}{8}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=19|num-a=21|ab=B}}<br /> {{MAA Notice}}</div> ZekromReshiram https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_19&diff=67201 2010 AMC 12B Problems/Problem 19 2015-02-02T20:58:41Z <p>ZekromReshiram: /* Solution */</p> <hr /> <div>== Problem 19 ==<br /> A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than &lt;math&gt;100&lt;/math&gt; points. What was the total number of points scored by the two teams in the first half?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let &lt;math&gt;a,ar,ar^{2},ar^{3}&lt;/math&gt; be the quarterly scores for the Raiders. We know that the Raiders and Wildcats both scored the same number of points in the first quarter so let &lt;math&gt;a,a+d,a+2d,a+3d&lt;/math&gt; be the quarterly scores for the Wildcats. The sum of the Raiders scores is &lt;math&gt;a(1+r+r^{2}+r^{3})&lt;/math&gt; and the sum of the Wildcats scores is &lt;math&gt;4a+6d&lt;/math&gt;. Now we can narrow our search for the values of &lt;math&gt;a,d&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt;. Because points are always measured in positive integers, we can conclude that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive integers. We can also conclude that &lt;math&gt;r&lt;/math&gt; is a positive integer by writing down the equation:<br /> <br /> &lt;cmath&gt;a(1+r+r^{2}+r^{3})=4a+6d+1&lt;/cmath&gt;<br /> <br /> Now we can start trying out some values of &lt;math&gt;r&lt;/math&gt;. We start at &lt;math&gt;r=1&lt;/math&gt; which gives that &lt;math&gt;6d=-1&lt;/math&gt; which is a contradiction. Next we try &lt;math&gt;r=2&lt;/math&gt;, which gives<br /> <br /> &lt;cmath&gt;15a=4a+6d+1&lt;/cmath&gt;<br /> &lt;cmath&gt;11a=6d+1&lt;/cmath&gt;<br /> <br /> We need the smallest multiple of &lt;math&gt;11&lt;/math&gt; (to satisfy the &lt;100 condition) that is &lt;math&gt;\equiv 1 \pmod{6}&lt;/math&gt;. We see that this is &lt;math&gt;55&lt;/math&gt;, and therefore &lt;math&gt;a=5&lt;/math&gt; and &lt;math&gt;d=9&lt;/math&gt;.<br /> <br /> So the Raiders first two scores were &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;10&lt;/math&gt; and the Wildcats first two scores were &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;14&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;5+10+5+14=34 \longrightarrow \boxed{\textbf{(E)}}&lt;/cmath&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=18|num-a=20|ab=B}}<br /> {{MAA Notice}}</div> ZekromReshiram https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_19&diff=67200 2010 AMC 12B Problems/Problem 19 2015-02-02T20:57:45Z <p>ZekromReshiram: /* Solution */</p> <hr /> <div>== Problem 19 ==<br /> A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than &lt;math&gt;100&lt;/math&gt; points. What was the total number of points scored by the two teams in the first half?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let &lt;math&gt;a,ar,ar^{2},ar^{3}&lt;/math&gt; be the quarterly scores for the Raiders. We know that the Raiders and Wildcats both scored the same number of points in the first quarter so let &lt;math&gt;a,a+d,a+2d,a+3d&lt;/math&gt; be quarterly scores for the Raiders. The sum of the Raiders scores is &lt;math&gt;a(1+r+r^{2}+r^{3})&lt;/math&gt; and the sum of the Wildcats scores is &lt;math&gt;4a+6d&lt;/math&gt;. Now we can narrow our search for the values of &lt;math&gt;a,d&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt;. Because points are always measured in positive integers, we can conclude that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive integers. We can also conclude that &lt;math&gt;r&lt;/math&gt; is a positive integer by writing down the equation:<br /> <br /> &lt;cmath&gt;a(1+r+r^{2}+r^{3})=4a+6d+1&lt;/cmath&gt;<br /> <br /> Now we can start trying out some values of &lt;math&gt;r&lt;/math&gt;. We start at &lt;math&gt;r=1&lt;/math&gt; which gives that &lt;math&gt;6d=-1&lt;/math&gt; which is a contradiction. Next we try &lt;math&gt;r=2&lt;/math&gt;, which gives<br /> <br /> &lt;cmath&gt;15a=4a+6d+1&lt;/cmath&gt;<br /> &lt;cmath&gt;11a=6d+1&lt;/cmath&gt;<br /> <br /> We need the smallest multiple of &lt;math&gt;11&lt;/math&gt; (to satisfy the &lt;100 condition) that is &lt;math&gt;\equiv 1 \pmod{6}&lt;/math&gt;. We see that this is &lt;math&gt;55&lt;/math&gt;, and therefore &lt;math&gt;a=5&lt;/math&gt; and &lt;math&gt;d=9&lt;/math&gt;.<br /> <br /> So the Raiders first two scores were &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;10&lt;/math&gt; and the Wildcats first two scores were &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;14&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;5+10+5+14=34 \longrightarrow \boxed{\textbf{(E)}}&lt;/cmath&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=18|num-a=20|ab=B}}<br /> {{MAA Notice}}</div> ZekromReshiram https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems&diff=67197 2010 AMC 12B Problems 2015-02-02T18:30:02Z <p>ZekromReshiram: /* Problem 14 */</p> <hr /> <div>== Problem 1 ==<br /> Makarla attended two meetings during her &lt;math&gt;9&lt;/math&gt;-hour work day. The first meeting took &lt;math&gt;45&lt;/math&gt; minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 35&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> A big &lt;math&gt;L&lt;/math&gt; is formed as shown. What is its area?<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(4mm);<br /> defaultpen(linewidth(.8pt));<br /> <br /> draw((0,0)--(5,0)--(5,2)--(2,2)--(2,8)--(0,8)--cycle);<br /> label(&quot;8&quot;,(0,4),W);<br /> label(&quot;5&quot;,(5/2,0),S);<br /> label(&quot;2&quot;,(5,1),E);<br /> label(&quot;2&quot;,(1,8),N);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 26 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 30&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> A ticket to a school play cost &lt;math&gt;x&lt;/math&gt; dollars, where &lt;math&gt;x&lt;/math&gt; is a whole number. A group of 9&lt;sub&gt;th&lt;/sub&gt; graders buys tickets costing a total of &amp;#36;&lt;math&gt;48&lt;/math&gt;, and a group of 10&lt;sub&gt;th&lt;/sub&gt; graders buys tickets costing a total of &amp;#36;&lt;math&gt;64&lt;/math&gt;. How many values for &lt;math&gt;x&lt;/math&gt; are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> A month with &lt;math&gt;31&lt;/math&gt; days has the same number of Mondays and Wednesdays.How many of the seven days of the week could be the first day of this month?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> Lucky Larry's teacher asked him to substitute numbers for &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt; in the expression &lt;math&gt;a-(b-(c-(d+e)))&lt;/math&gt; and evaluate the result. Larry ignored the parenthese but added and subtracted correctly and obtained the correct result by coincidence. The numbers Larry substituted for &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; were &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt;, respectively. What number did Larry substitute for &lt;math&gt;e&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -5 \qquad \textbf{(B)}\ -3 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> At the beginning of the school year, &lt;math&gt;50\%&lt;/math&gt; of all students in Mr. Wells' math class answered &quot;Yes&quot; to the question &quot;Do you love math&quot;, and &lt;math&gt;50\%&lt;/math&gt; answered &quot;No.&quot; At the end of the school year, &lt;math&gt;70\%&lt;/math&gt; answered &quot;Yes&quot; and &lt;math&gt;30\%&lt;/math&gt; answerws &quot;No.&quot; Altogether, &lt;math&gt;x\%&lt;/math&gt; of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 6|Solution]]<br /> <br /> <br /> == Problem 7 ==<br /> Shelby drives her scooter at a speed of &lt;math&gt;30&lt;/math&gt; miles per hour if it is not raining, and &lt;math&gt;20&lt;/math&gt; miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of &lt;math&gt;16&lt;/math&gt; miles in &lt;math&gt;40&lt;/math&gt; minutes. How many minutes did she drive in the rain?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 7|Solution]]<br /> <br /> <br /> <br /> == Problem 8 ==<br /> Every high school in the city of Euclid sent a team of &lt;math&gt;3&lt;/math&gt; students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed &lt;math&gt;37&lt;/math&gt;&lt;sup&gt;th&lt;/sup&gt; and &lt;math&gt;64&lt;/math&gt;&lt;sup&gt;th&lt;/sup&gt;, respectively. How many schools are in the city?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Let &lt;math&gt;n&lt;/math&gt; be the smallest positive integer such that &lt;math&gt;n&lt;/math&gt; is divisible by &lt;math&gt;20&lt;/math&gt;, &lt;math&gt;n^2&lt;/math&gt; is a perfect cube, and &lt;math&gt;n^3&lt;/math&gt; is a perfect square. What is the number of digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The average of the numbers &lt;math&gt;1, 2, 3,\cdots, 98, 99,&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;100x&lt;/math&gt;. What is &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{49}{101} \qquad \textbf{(B)}\ \dfrac{50}{101} \qquad \textbf{(C)}\ \dfrac{1}{2} \qquad \textbf{(D)}\ \dfrac{51}{101} \qquad \textbf{(E)}\ \dfrac{50}{99}&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 10|Solution]]<br /> <br /> <br /> == Problem 11 ==<br /> A palindrome between &lt;math&gt;1000&lt;/math&gt; and &lt;math&gt;10,000&lt;/math&gt; is chosen at random. What is the probability that it is divisible by &lt;math&gt;7&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1}{10} \qquad \textbf{(B)}\ \dfrac{1}{9} \qquad \textbf{(C)}\ \dfrac{1}{7} \qquad \textbf{(D)}\ \dfrac{1}{6} \qquad \textbf{(E)}\ \dfrac{1}{5}&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> For what value of &lt;math&gt;x&lt;/math&gt; does<br /> <br /> &lt;cmath&gt;\log_{\sqrt{2}}\sqrt{x}+\log_{2}{x}+\log_{4}{x^2}+\log_{8}{x^3}+\log_{16}{x^4}=40?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 256 \qquad \textbf{(E)}\ 1024&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;\cos(2A-B)+\sin(A+B)=2&lt;/math&gt; and &lt;math&gt;AB=4&lt;/math&gt;. What is &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \sqrt{2} \qquad \textbf{(B)}\ \sqrt{3} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 2\sqrt{2} \qquad \textbf{(E)}\ 2\sqrt{3}&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt; be positive integers with &lt;math&gt;a+b+c+d+e=2010&lt;/math&gt; and let &lt;math&gt;M&lt;/math&gt; be the largest of the sum &lt;math&gt;a+b&lt;/math&gt;, &lt;math&gt;b+c&lt;/math&gt;, &lt;math&gt;c+d&lt;/math&gt; and &lt;math&gt;d+e&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;M&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> For how many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of nonnegative integers less than &lt;math&gt;20&lt;/math&gt; are there exactly two distinct elements in the set &lt;math&gt;\{i^x, (1+i)^y, z\}&lt;/math&gt;, where &lt;math&gt;i=\sqrt{-1}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 149 \qquad \textbf{(B)}\ 205 \qquad \textbf{(C)}\ 215 \qquad \textbf{(D)}\ 225 \qquad \textbf{(E)}\ 235&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> Positive integers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are randomly and independently selected with replacement from the set &lt;math&gt;\{1, 2, 3,\dots, 2010\}&lt;/math&gt;. What is the probability that &lt;math&gt;abc + ab + a&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> The entries in a &lt;math&gt;3 \times 3&lt;/math&gt; array include all the digits from &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;9&lt;/math&gt;, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> A frog makes &lt;math&gt;3&lt;/math&gt; jumps, each exactly &lt;math&gt;1&lt;/math&gt; meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than &lt;math&gt;1&lt;/math&gt; meter from its starting position?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than &lt;math&gt;100&lt;/math&gt; points. What was the total number of points scored by the two teams in the first half?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> A geometric sequence &lt;math&gt;(a_n)&lt;/math&gt; has &lt;math&gt;a_1=\sin x&lt;/math&gt;, &lt;math&gt;a_2=\cos x&lt;/math&gt;, and &lt;math&gt;a_3= \tan x&lt;/math&gt; for some real number &lt;math&gt;x&lt;/math&gt;. For what value of &lt;math&gt;n&lt;/math&gt; does &lt;math&gt;a_n=1+\cos x&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> Let &lt;math&gt;a &gt; 0&lt;/math&gt;, and let &lt;math&gt;P(x)&lt;/math&gt; be a polynomial with integer coefficients such that<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;P(1) = P(3) = P(5) = P(7) = a&lt;/math&gt;, and&lt;br/&gt;<br /> &lt;math&gt;P(2) = P(4) = P(6) = P(8) = -a&lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> What is the smallest possible value of &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a cyclic quadrilateral. The side lengths of &lt;math&gt;ABCD&lt;/math&gt; are distinct integers less than &lt;math&gt;15&lt;/math&gt; such that &lt;math&gt;BC\cdot CD=AB\cdot DA&lt;/math&gt;. What is the largest possible value of &lt;math&gt;BD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \sqrt{\dfrac{325}{2}} \qquad \textbf{(B)}\ \sqrt{185} \qquad \textbf{(C)}\ \sqrt{\dfrac{389}{2}} \qquad \textbf{(D)}\ \sqrt{\dfrac{425}{2}} \qquad \textbf{(E)}\ \sqrt{\dfrac{533}{2}}&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> Monic quadratic polynomials &lt;math&gt;P(x)&lt;/math&gt; and &lt;math&gt;Q(x)&lt;/math&gt; have the property that &lt;math&gt;P(Q(x))&lt;/math&gt; has zeros at &lt;math&gt;x=-23, -21, -17,&lt;/math&gt; and &lt;math&gt;-15&lt;/math&gt;, and &lt;math&gt;Q(P(x))&lt;/math&gt; has zeros at &lt;math&gt;x=-59,-57,-51&lt;/math&gt; and &lt;math&gt;-49&lt;/math&gt;. What is the sum of the minimum values of &lt;math&gt;P(x)&lt;/math&gt; and &lt;math&gt;Q(x)&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> The set of real numbers &lt;math&gt;x&lt;/math&gt; for which <br /> <br /> &lt;cmath&gt;\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge1&lt;/cmath&gt;<br /> <br /> is the union of intervals of the form &lt;math&gt;a&lt;x\le b&lt;/math&gt;. What is the sum of the lengths of these intervals?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1003}{335} \qquad \textbf{(B)}\ \dfrac{1004}{335} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \dfrac{403}{134} \qquad \textbf{(E)}\ \dfrac{202}{67}&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> For every integer &lt;math&gt;n\ge2&lt;/math&gt;, let &lt;math&gt;\text{pow}(n)&lt;/math&gt; be the largest power of the largest prime that divides &lt;math&gt;n&lt;/math&gt;. For example &lt;math&gt;\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2&lt;/math&gt;. What is the largest integer &lt;math&gt;m&lt;/math&gt; such that &lt;math&gt;2010^m&lt;/math&gt; divides<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;\prod_{n=2}^{5300}\text{pow}(n)&lt;/math&gt;?<br /> &lt;/center&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ 74 \qquad \textbf{(B)}\ 75 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 77 \qquad \textbf{(E)}\ 78&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 25|Solution]]<br /> {{MAA Notice}}</div> ZekromReshiram https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems&diff=67196 2010 AMC 12B Problems 2015-02-02T18:14:56Z <p>ZekromReshiram: /* Problem 5 */</p> <hr /> <div>== Problem 1 ==<br /> Makarla attended two meetings during her &lt;math&gt;9&lt;/math&gt;-hour work day. The first meeting took &lt;math&gt;45&lt;/math&gt; minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 35&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> A big &lt;math&gt;L&lt;/math&gt; is formed as shown. What is its area?<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(4mm);<br /> defaultpen(linewidth(.8pt));<br /> <br /> draw((0,0)--(5,0)--(5,2)--(2,2)--(2,8)--(0,8)--cycle);<br /> label(&quot;8&quot;,(0,4),W);<br /> label(&quot;5&quot;,(5/2,0),S);<br /> label(&quot;2&quot;,(5,1),E);<br /> label(&quot;2&quot;,(1,8),N);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 26 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 30&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> A ticket to a school play cost &lt;math&gt;x&lt;/math&gt; dollars, where &lt;math&gt;x&lt;/math&gt; is a whole number. A group of 9&lt;sub&gt;th&lt;/sub&gt; graders buys tickets costing a total of &amp;#36;&lt;math&gt;48&lt;/math&gt;, and a group of 10&lt;sub&gt;th&lt;/sub&gt; graders buys tickets costing a total of &amp;#36;&lt;math&gt;64&lt;/math&gt;. How many values for &lt;math&gt;x&lt;/math&gt; are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> A month with &lt;math&gt;31&lt;/math&gt; days has the same number of Mondays and Wednesdays.How many of the seven days of the week could be the first day of this month?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> Lucky Larry's teacher asked him to substitute numbers for &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt; in the expression &lt;math&gt;a-(b-(c-(d+e)))&lt;/math&gt; and evaluate the result. Larry ignored the parenthese but added and subtracted correctly and obtained the correct result by coincidence. The numbers Larry substituted for &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; were &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt;, respectively. What number did Larry substitute for &lt;math&gt;e&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -5 \qquad \textbf{(B)}\ -3 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> At the beginning of the school year, &lt;math&gt;50\%&lt;/math&gt; of all students in Mr. Wells' math class answered &quot;Yes&quot; to the question &quot;Do you love math&quot;, and &lt;math&gt;50\%&lt;/math&gt; answered &quot;No.&quot; At the end of the school year, &lt;math&gt;70\%&lt;/math&gt; answered &quot;Yes&quot; and &lt;math&gt;30\%&lt;/math&gt; answerws &quot;No.&quot; Altogether, &lt;math&gt;x\%&lt;/math&gt; of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 6|Solution]]<br /> <br /> <br /> == Problem 7 ==<br /> Shelby drives her scooter at a speed of &lt;math&gt;30&lt;/math&gt; miles per hour if it is not raining, and &lt;math&gt;20&lt;/math&gt; miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of &lt;math&gt;16&lt;/math&gt; miles in &lt;math&gt;40&lt;/math&gt; minutes. How many minutes did she drive in the rain?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 7|Solution]]<br /> <br /> <br /> <br /> == Problem 8 ==<br /> Every high school in the city of Euclid sent a team of &lt;math&gt;3&lt;/math&gt; students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed &lt;math&gt;37&lt;/math&gt;&lt;sup&gt;th&lt;/sup&gt; and &lt;math&gt;64&lt;/math&gt;&lt;sup&gt;th&lt;/sup&gt;, respectively. How many schools are in the city?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Let &lt;math&gt;n&lt;/math&gt; be the smallest positive integer such that &lt;math&gt;n&lt;/math&gt; is divisible by &lt;math&gt;20&lt;/math&gt;, &lt;math&gt;n^2&lt;/math&gt; is a perfect cube, and &lt;math&gt;n^3&lt;/math&gt; is a perfect square. What is the number of digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The average of the numbers &lt;math&gt;1, 2, 3,\cdots, 98, 99,&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;100x&lt;/math&gt;. What is &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{49}{101} \qquad \textbf{(B)}\ \dfrac{50}{101} \qquad \textbf{(C)}\ \dfrac{1}{2} \qquad \textbf{(D)}\ \dfrac{51}{101} \qquad \textbf{(E)}\ \dfrac{50}{99}&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 10|Solution]]<br /> <br /> <br /> == Problem 11 ==<br /> A palindrome between &lt;math&gt;1000&lt;/math&gt; and &lt;math&gt;10,000&lt;/math&gt; is chosen at random. What is the probability that it is divisible by &lt;math&gt;7&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1}{10} \qquad \textbf{(B)}\ \dfrac{1}{9} \qquad \textbf{(C)}\ \dfrac{1}{7} \qquad \textbf{(D)}\ \dfrac{1}{6} \qquad \textbf{(E)}\ \dfrac{1}{5}&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> For what value of &lt;math&gt;x&lt;/math&gt; does<br /> <br /> &lt;cmath&gt;\log_{\sqrt{2}}\sqrt{x}+\log_{2}{x}+\log_{4}{x^2}+\log_{8}{x^3}+\log_{16}{x^4}=40?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 256 \qquad \textbf{(E)}\ 1024&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;\cos(2A-B)+\sin(A+B)=2&lt;/math&gt; and &lt;math&gt;AB=4&lt;/math&gt;. What is &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \sqrt{2} \qquad \textbf{(B)}\ \sqrt{3} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 2\sqrt{2} \qquad \textbf{(E)}\ 2\sqrt{3}&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt; be postive integers with &lt;math&gt;a+b+c+d+e=2010&lt;/math&gt; and let &lt;math&gt;M&lt;/math&gt; be the largest of the sum &lt;math&gt;a+b&lt;/math&gt;, &lt;math&gt;b+c&lt;/math&gt;, &lt;math&gt;c+d&lt;/math&gt; and &lt;math&gt;d+e&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;M&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> For how many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of nonnegative integers less than &lt;math&gt;20&lt;/math&gt; are there exactly two distinct elements in the set &lt;math&gt;\{i^x, (1+i)^y, z\}&lt;/math&gt;, where &lt;math&gt;i=\sqrt{-1}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 149 \qquad \textbf{(B)}\ 205 \qquad \textbf{(C)}\ 215 \qquad \textbf{(D)}\ 225 \qquad \textbf{(E)}\ 235&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> Positive integers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are randomly and independently selected with replacement from the set &lt;math&gt;\{1, 2, 3,\dots, 2010\}&lt;/math&gt;. What is the probability that &lt;math&gt;abc + ab + a&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> The entries in a &lt;math&gt;3 \times 3&lt;/math&gt; array include all the digits from &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;9&lt;/math&gt;, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> A frog makes &lt;math&gt;3&lt;/math&gt; jumps, each exactly &lt;math&gt;1&lt;/math&gt; meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than &lt;math&gt;1&lt;/math&gt; meter from its starting position?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than &lt;math&gt;100&lt;/math&gt; points. What was the total number of points scored by the two teams in the first half?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> A geometric sequence &lt;math&gt;(a_n)&lt;/math&gt; has &lt;math&gt;a_1=\sin x&lt;/math&gt;, &lt;math&gt;a_2=\cos x&lt;/math&gt;, and &lt;math&gt;a_3= \tan x&lt;/math&gt; for some real number &lt;math&gt;x&lt;/math&gt;. For what value of &lt;math&gt;n&lt;/math&gt; does &lt;math&gt;a_n=1+\cos x&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> Let &lt;math&gt;a &gt; 0&lt;/math&gt;, and let &lt;math&gt;P(x)&lt;/math&gt; be a polynomial with integer coefficients such that<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;P(1) = P(3) = P(5) = P(7) = a&lt;/math&gt;, and&lt;br/&gt;<br /> &lt;math&gt;P(2) = P(4) = P(6) = P(8) = -a&lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> What is the smallest possible value of &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a cyclic quadrilateral. The side lengths of &lt;math&gt;ABCD&lt;/math&gt; are distinct integers less than &lt;math&gt;15&lt;/math&gt; such that &lt;math&gt;BC\cdot CD=AB\cdot DA&lt;/math&gt;. What is the largest possible value of &lt;math&gt;BD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \sqrt{\dfrac{325}{2}} \qquad \textbf{(B)}\ \sqrt{185} \qquad \textbf{(C)}\ \sqrt{\dfrac{389}{2}} \qquad \textbf{(D)}\ \sqrt{\dfrac{425}{2}} \qquad \textbf{(E)}\ \sqrt{\dfrac{533}{2}}&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> Monic quadratic polynomials &lt;math&gt;P(x)&lt;/math&gt; and &lt;math&gt;Q(x)&lt;/math&gt; have the property that &lt;math&gt;P(Q(x))&lt;/math&gt; has zeros at &lt;math&gt;x=-23, -21, -17,&lt;/math&gt; and &lt;math&gt;-15&lt;/math&gt;, and &lt;math&gt;Q(P(x))&lt;/math&gt; has zeros at &lt;math&gt;x=-59,-57,-51&lt;/math&gt; and &lt;math&gt;-49&lt;/math&gt;. What is the sum of the minimum values of &lt;math&gt;P(x)&lt;/math&gt; and &lt;math&gt;Q(x)&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> The set of real numbers &lt;math&gt;x&lt;/math&gt; for which <br /> <br /> &lt;cmath&gt;\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge1&lt;/cmath&gt;<br /> <br /> is the union of intervals of the form &lt;math&gt;a&lt;x\le b&lt;/math&gt;. What is the sum of the lengths of these intervals?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1003}{335} \qquad \textbf{(B)}\ \dfrac{1004}{335} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \dfrac{403}{134} \qquad \textbf{(E)}\ \dfrac{202}{67}&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> For every integer &lt;math&gt;n\ge2&lt;/math&gt;, let &lt;math&gt;\text{pow}(n)&lt;/math&gt; be the largest power of the largest prime that divides &lt;math&gt;n&lt;/math&gt;. For example &lt;math&gt;\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2&lt;/math&gt;. What is the largest integer &lt;math&gt;m&lt;/math&gt; such that &lt;math&gt;2010^m&lt;/math&gt; divides<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;\prod_{n=2}^{5300}\text{pow}(n)&lt;/math&gt;?<br /> &lt;/center&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ 74 \qquad \textbf{(B)}\ 75 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 77 \qquad \textbf{(E)}\ 78&lt;/math&gt;<br /> <br /> [[2010 AMC 12B Problems/Problem 25|Solution]]<br /> {{MAA Notice}}</div> ZekromReshiram https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems&diff=67174 2010 AMC 12A Problems 2015-02-02T04:45:09Z <p>ZekromReshiram: </p> <hr /> <div>== Problem 1 ==<br /> What is &lt;math&gt;\left(20-\left(2010-201\right)\right)+\left(2010-\left(201-20\right)\right)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -4020 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 401 \qquad \textbf{(E)}\ 4020&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> A ferry boat shuttles tourists to an island every hour starting at 10 AM until its last trip, which starts at 3 PM. One day the boat captain notes that on the 10 AM trip there were 100 tourists on the ferry boat, and that on each successive trip, the number of tourists was 1 fewer than on the previous trip. How many tourists did the ferry take to the island that day?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 585 \qquad \textbf{(B)}\ 594 \qquad \textbf{(C)}\ 672 \qquad \textbf{(D)}\ 679 \qquad \textbf{(E)}\ 694&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt;, pictured below, shares &lt;math&gt;50\%&lt;/math&gt; of its area with square &lt;math&gt;EFGH&lt;/math&gt;. Square &lt;math&gt;EFGH&lt;/math&gt; shares &lt;math&gt;20\%&lt;/math&gt; of its area with rectangle &lt;math&gt;ABCD&lt;/math&gt;. What is &lt;math&gt;\frac{AB}{AD}&lt;/math&gt;?<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(1mm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> <br /> draw((0,0)--(0,25)--(25,25)--(25,0)--cycle);<br /> fill((0,20)--(0,15)--(25,15)--(25,20)--cycle,gray);<br /> draw((0,15)--(0,20)--(25,20)--(25,15)--cycle);<br /> draw((25,15)--(25,20)--(50,20)--(50,15)--cycle);<br /> <br /> label(&quot;$A$&quot;,(0,20),W);<br /> label(&quot;$B$&quot;,(50,20),E);<br /> label(&quot;$C$&quot;,(50,15),E);<br /> label(&quot;$D$&quot;,(0,15),W);<br /> label(&quot;$E$&quot;,(0,25),NW);<br /> label(&quot;$F$&quot;,(25,25),NE);<br /> label(&quot;$G$&quot;,(25,0),SE);<br /> label(&quot;$H$&quot;,(0,0),SW);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> If &lt;math&gt;x&lt;0&lt;/math&gt;, then which of the following must be positive?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{x}{\left|x\right|} \qquad \textbf{(B)}\ -x^2 \qquad \textbf{(C)}\ -2^x \qquad \textbf{(D)}\ -x^{-1} \qquad \textbf{(E)}\ \sqrt{x}&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> Halfway through a 100-shot archery tournament, Chelsea leads by 50 points. For each shot a bullseye scores 10 points, with other possible scores being 8, 4, 2, and 0 points. Chelsea always scores at least 4 points on each shot. If Chelsea's next &lt;math&gt;n&lt;/math&gt; shots are bullseyes she will be guaranteed victory. What is the minimum value for &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 38 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 42 \qquad \textbf{(D)}\ 44 \qquad \textbf{(E)}\ 46&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> A &lt;math&gt;\texti{palindrome}&lt;/math&gt;, such as 83438, is a number that remains the same when its digits are reversed. The numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;x+32&lt;/math&gt; are three-digit and four-digit palindromes, respectively. What is the sum of the digits of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0.04 \qquad \textbf{(B)}\ \frac{0.4}{\pi} \qquad \textbf{(C)}\ 0.4 \qquad \textbf{(D)}\ \frac{4}{\pi} \qquad \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has &lt;math&gt;AB=2 \cdot AC&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;, respectively, such that &lt;math&gt;\angle BAE = \angle ACD&lt;/math&gt;. Let &lt;math&gt;F&lt;/math&gt; be the intersection of segments &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, and suppose that &lt;math&gt;\triangle CFE&lt;/math&gt; is equilateral. What is &lt;math&gt;\angle ACB&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> A solid cube has side length &lt;math&gt;3&lt;/math&gt; inches. A &lt;math&gt;2&lt;/math&gt;-inch by &lt;math&gt;2&lt;/math&gt;-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The first four terms of an arithmetic sequence are &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;3p-q&lt;/math&gt;, and &lt;math&gt;3p+q&lt;/math&gt;. What is the &lt;math&gt;2010^\text{th}&lt;/math&gt; term of this sequence?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 8041 \qquad \textbf{(B)}\ 8043 \qquad \textbf{(C)}\ 8045 \qquad \textbf{(D)}\ 8047 \qquad \textbf{(E)}\ 8049&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> The solution of the equation &lt;math&gt;7^{x+7} = 8^x&lt;/math&gt; can be expressed in the form &lt;math&gt;x = \log_b 7^7&lt;/math&gt;. What is &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{7}{15} \qquad \textbf{(B)}\ \frac{7}{8} \qquad \textbf{(C)}\ \frac{8}{7} \qquad \textbf{(D)}\ \frac{15}{8} \qquad \textbf{(E)}\ \frac{15}{7}&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.<br /> <br /> Brian: &quot;Mike and I are different species.&quot;<br /> <br /> Chris: &quot;LeRoy is a frog.&quot;<br /> <br /> LeRoy: &quot;Chris is a frog.&quot;<br /> <br /> Mike: &quot;Of the four of us, at least two are toads.&quot;<br /> <br /> How many of these amphibians are frogs?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> For how many integer values of &lt;math&gt;k&lt;/math&gt; do the graphs of &lt;math&gt;x^2+y^2=k^2&lt;/math&gt; and &lt;math&gt;xy = k&lt;/math&gt; not intersect?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Nondegenerate &lt;math&gt;\triangle ABC&lt;/math&gt; has integer side lengths, &lt;math&gt;\overline{BD}&lt;/math&gt; is an angle bisector, &lt;math&gt;AD = 3&lt;/math&gt;, and &lt;math&gt;DC=8&lt;/math&gt;. What is the smallest possible value of the perimeter?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> A coin is altered so that the probability that it lands on heads is less than &lt;math&gt;\frac{1}{2}&lt;/math&gt; and when the coin is flipped four times, the probability of an equal number of heads and tails is &lt;math&gt;\frac{1}{6}&lt;/math&gt;. What is the probability that the coin lands on heads?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{\sqrt{15}-3}{6} \qquad \textbf{(B)}\ \frac{6-\sqrt{6\sqrt{6}+2}}{12} \qquad \textbf{(C)}\ \frac{\sqrt{2}-1}{2} \qquad \textbf{(D)}\ \frac{3-\sqrt{3}}{6} \qquad \textbf{(E)}\ \frac{\sqrt{3}-1}{2}&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> Bernardo randomly picks 3 distinct numbers from the set &lt;math&gt;\{1,2,3,4,5,6,7,8,9\}&lt;/math&gt; and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set &lt;math&gt;\{1,2,3,4,5,6,7,8\}&lt;/math&gt; and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{47}{72} \qquad \textbf{(B)}\ \frac{37}{56} \qquad \textbf{(C)}\ \frac{2}{3} \qquad \textbf{(D)}\ \frac{49}{72} \qquad \textbf{(E)}\ \frac{39}{56}&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> Equiangular hexagon &lt;math&gt;ABCDEF&lt;/math&gt; has side lengths &lt;math&gt;AB=CD=EF=1&lt;/math&gt; and &lt;math&gt;BC=DE=FA=r&lt;/math&gt;. The area of &lt;math&gt;\triangle ACE&lt;/math&gt; is &lt;math&gt;70\%&lt;/math&gt; of the area of the hexagon. What is the sum of all possible values of &lt;math&gt;r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> A 16-step path is to go from &lt;math&gt;(-4,-4)&lt;/math&gt; to &lt;math&gt;(4,4)&lt;/math&gt; with each step increasing either the &lt;math&gt;x&lt;/math&gt;-coordinate or the &lt;math&gt;y&lt;/math&gt;-coordinate by 1. How many such paths stay outside or on the boundary of the square &lt;math&gt;-2 \le x \le 2&lt;/math&gt;, &lt;math&gt;-2 \le y \le 2&lt;/math&gt; at each step?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> Each of 2010 boxes in a line contains a single red marble, and for &lt;math&gt;1 \le k \le 2010&lt;/math&gt;, the box in the &lt;math&gt;k\text{th}&lt;/math&gt; position also contains &lt;math&gt;k&lt;/math&gt; white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let &lt;math&gt;P(n)&lt;/math&gt; be the probability that Isabella stops after drawing exactly &lt;math&gt;n&lt;/math&gt; marbles. What is the smallest value of &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;P(n) &lt; \frac{1}{2010}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> Arithmetic sequences &lt;math&gt;\left(a_n\right)&lt;/math&gt; and &lt;math&gt;\left(b_n\right)&lt;/math&gt; have integer terms with &lt;math&gt;a_1=b_1=1&lt;a_2 \le b_2&lt;/math&gt; and &lt;math&gt;a_n b_n = 2010&lt;/math&gt; for some &lt;math&gt;n&lt;/math&gt;. What is the largest possible value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 288 \qquad \textbf{(E)}\ 2009&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> The graph of &lt;math&gt;y=x^6-10x^5+29x^4-4x^3+ax^2&lt;/math&gt; lies above the line &lt;math&gt;y=bx+c&lt;/math&gt; except at three values of &lt;math&gt;x&lt;/math&gt;, where the graph and the line intersect. What is the largest of these values?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> What is the minimum value of &lt;math&gt;\left|x-1\right| + \left|2x-1\right| + \left|3x-1\right| + \cdots + \left|119x - 1 \right|&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 53&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> The number obtained from the last two nonzero digits of &lt;math&gt;90!&lt;/math&gt; is equal to &lt;math&gt;n&lt;/math&gt;. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> Let &lt;math&gt;f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)&lt;/math&gt;. The intersection of the domain of &lt;math&gt;f(x)&lt;/math&gt; with the interval &lt;math&gt;[0,1]&lt;/math&gt; is a union of &lt;math&gt;n&lt;/math&gt; disjoint open intervals. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad \textbf{(E)}\ 2255&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 25|Solution]]<br /> {{MAA Notice}}</div> ZekromReshiram https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_20&diff=62259 2002 AMC 12B Problems/Problem 20 2014-06-11T23:57:31Z <p>ZekromReshiram: </p> <hr /> <div>{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #20]] and [[2002 AMC 10B Problems|2002 AMC 10B #22]]}}<br /> == Problem ==<br /> Let &lt;math&gt;\triangle XOY&lt;/math&gt; be a [[right triangle|right-angled triangle]] with &lt;math&gt;m\angle XOY = 90^{\circ}&lt;/math&gt;. Let &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; be the [[midpoint]]s of legs &lt;math&gt;OX&lt;/math&gt; and &lt;math&gt;OY&lt;/math&gt;, respectively. Given that &lt;math&gt;XN = 19&lt;/math&gt; and &lt;math&gt;YM = 22&lt;/math&gt;, find &lt;math&gt;XY&lt;/math&gt;.<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 24<br /> \qquad\mathrm{(B)}\ 26<br /> \qquad\mathrm{(C)}\ 28<br /> \qquad\mathrm{(D)}\ 30<br /> \qquad\mathrm{(E)}\ 32&lt;/math&gt;<br /> == Solution ==<br /> [[Image:2002_12B_AMC-20.png]]<br /> <br /> Let &lt;math&gt;OM = x&lt;/math&gt;, &lt;math&gt;ON = y&lt;/math&gt;. By the [[Pythagorean Theorem]] on &lt;math&gt;\triangle XON, MOY&lt;/math&gt; respectively,<br /> &lt;cmath&gt;\begin{align*}<br /> (2x)^2 + y^2 &amp;= 19^2\\<br /> x^2 + (2y)^2 &amp;= 22^2\end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Summing these gives &lt;math&gt;5x^2 + 5y^2 = 845 \Longrightarrow x^2 + y^2 = 169&lt;/math&gt;. <br /> <br /> By the Pythagorean Theorem again, we have <br /> <br /> &lt;cmath&gt;(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = \sqrt{4(169)} = \sqrt{676} = \boxed{\mathrm{(B)}\ 26}&lt;/cmath&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2002|ab=B|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2002|ab=B|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> ZekromReshiram https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_25&diff=61686 2005 AMC 10B Problems/Problem 25 2014-04-20T23:45:17Z <p>ZekromReshiram: /* Solution */</p> <hr /> <div>== Problem ==<br /> A subset &lt;math&gt;B&lt;/math&gt; of the set of integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;100&lt;/math&gt;, inclusive, has the property that no two elements of &lt;math&gt;B&lt;/math&gt; sum to &lt;math&gt;125&lt;/math&gt;. What is the maximum possible number of elements in &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)} 50 \qquad \mathrm{(B)} 51 \qquad \mathrm{(C)} 62 \qquad \mathrm{(D)} 65 \qquad \mathrm{(E)} 68 &lt;/math&gt;<br /> <br /> == Solution ==<br /> The question asks for the maximum possible number of elements. The integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;24&lt;/math&gt; can be included because you cannot make &lt;math&gt;125&lt;/math&gt; with integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;24&lt;/math&gt; without the other number being greater than &lt;math&gt;100&lt;/math&gt;. The integers from &lt;math&gt;25&lt;/math&gt; to &lt;math&gt;100&lt;/math&gt; are left. They can be paired so the sum is &lt;math&gt;125&lt;/math&gt;: &lt;math&gt;25+100&lt;/math&gt;, &lt;math&gt;26+99&lt;/math&gt;, &lt;math&gt;27+98&lt;/math&gt;, &lt;math&gt;\ldots&lt;/math&gt;, &lt;math&gt;62+63&lt;/math&gt;. That is &lt;math&gt;38&lt;/math&gt; pairs, and at most one number from each pair can be included in the set. The total is &lt;math&gt;24 + 38 = \boxed{\mathrm{(C)}\ 62}&lt;/math&gt;.<br /> Also, it is possible to see that since the numbers &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;24&lt;/math&gt; are in the set there are only the numbers &lt;math&gt;25&lt;/math&gt; to &lt;math&gt;100&lt;/math&gt; to consider. As &lt;math&gt;62+63&lt;/math&gt; gives &lt;math&gt;125&lt;/math&gt;, the numbers &lt;math&gt;25&lt;/math&gt; to &lt;math&gt;62&lt;/math&gt; can be put in subset &lt;math&gt;B&lt;/math&gt; without having two numbers add up to &lt;math&gt;125&lt;/math&gt;. In this way, subset &lt;math&gt;B&lt;/math&gt; will have the numbers &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;62&lt;/math&gt;, and so &lt;math&gt;\boxed{\mathrm{(C)}\ 62}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2005|ab=B|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> ZekromReshiram https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_24&diff=61682 2005 AMC 10B Problems/Problem 24 2014-04-20T23:32:57Z <p>ZekromReshiram: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be two-digit integers such that &lt;math&gt;y&lt;/math&gt; is obtained by reversing the digits<br /> of &lt;math&gt;x&lt;/math&gt;. The integers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; satisfy &lt;math&gt;x^2 - y^2 = m^2&lt;/math&gt; for some positive integer &lt;math&gt;m&lt;/math&gt;.<br /> What is &lt;math&gt;x + y + m&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)} 88 \qquad \mathrm{(B)} 112 \qquad \mathrm{(C)} 116 \qquad \mathrm{(D)} 144 \qquad \mathrm{(E)} 154 &lt;/math&gt;<br /> <br /> == Solution ==<br /> Let &lt;math&gt;x = 10a+b, y = 10b+a&lt;/math&gt;, [[without loss of generality]] with &lt;math&gt;a&gt;b&lt;/math&gt;. Then &lt;math&gt;x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2&lt;/math&gt;. It follows that &lt;math&gt;11|(a-b)(a+b)&lt;/math&gt;, but &lt;math&gt;a-b &lt; 10&lt;/math&gt; so &lt;math&gt;11|a+b \Longrightarrow a+b=11&lt;/math&gt;. Then we have &lt;math&gt;33^2(a-b) = m^2&lt;/math&gt;. Thus &lt;math&gt;a-b&lt;/math&gt; is a perfect square. Also, because &lt;math&gt;a-b&lt;/math&gt; and &lt;math&gt;a+b&lt;/math&gt; have the same parity, &lt;math&gt;a-b&lt;/math&gt; is a one-digit odd perfect square, namely &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;. The latter case gives &lt;math&gt;(a,b) = (10,1)&lt;/math&gt;, which does not work. The former case gives &lt;math&gt;(a,b) = (6,5)&lt;/math&gt;, which works, and we have &lt;math&gt;x+y+m = 65 + 56 + 33 = 154\ \mathbf{(E)}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> The first steps are the same as above. Let &lt;math&gt;x = 10a+b, y = 10b+a&lt;/math&gt;, where we know that a and b are digits (whole numbers less than 10). Like above, we end up getting &lt;math&gt;(9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2&lt;/math&gt;. This is where the solution diverges. <br /> <br /> We know that the left side of the equation is a perfect square because m is an integer. If we factor 99 into its prime factors, we get &lt;math&gt;3^2 * 11&lt;/math&gt;. In order to get a perfect square on the left side, &lt;math&gt;(a-b)(a+b)&lt;/math&gt; must make both prime exponents even. Because the a and b are digits, a simple guess would be that &lt;math&gt;(a+b)&lt;/math&gt; (the bigger number) equals 11 while &lt;math&gt;(a-b)&lt;/math&gt; is a factor of nine (1 or 9). The correct guesses are &lt;math&gt;a = 6, b = 5&lt;/math&gt; causing &lt;math&gt;x = 65, y = 56,&lt;/math&gt; and &lt;math&gt;m = 33&lt;/math&gt;. The sum of the numbers is &lt;math&gt;\boxed{\textbf{(E) }154}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2005|ab=B|num-b=23|num-a=25}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> ZekromReshiram https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_21&diff=61681 2005 AMC 10B Problems/Problem 21 2014-04-20T23:15:22Z <p>ZekromReshiram: /* Solution (where the order of drawing slips matters) */</p> <hr /> <div>== Problem ==<br /> Forty slips are placed into a hat, each bearing a number &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;9&lt;/math&gt;, or &lt;math&gt;10&lt;/math&gt;, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let &lt;math&gt;p&lt;/math&gt; be the probability that all four slips bear the same number. Let &lt;math&gt;q&lt;/math&gt; be the probability that two of the slips bear a number &lt;math&gt;a&lt;/math&gt; and the other two bear a number &lt;math&gt;b \neq a&lt;/math&gt;. What is the value of &lt;math&gt;q/p&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)} 162 \qquad \mathrm{(B)} 180 \qquad \mathrm{(C)} 324 \qquad \mathrm{(D)} 360 \qquad \mathrm{(E)} 720 &lt;/math&gt;<br /> <br /> == Solution (where the order of drawing slips matters) ==<br /> There are &lt;math&gt;10&lt;/math&gt; ways to determine which number to pick. There are &lt;math&gt;4!&lt;/math&gt; way to then draw those four slips with that number, and &lt;math&gt;40 \cdot 39 \cdot 38 \cdot 37&lt;/math&gt; total ways to draw four slips. Thus &lt;math&gt;p = \frac{10\cdot 4!}{40 \cdot 39 \cdot 38 \cdot 37}&lt;/math&gt;. <br /> <br /> There are &lt;math&gt;{10 \choose 2} = 45&lt;/math&gt; ways to determine which two numbers to pick for the second probability. There are &lt;math&gt;{4 \choose 2}&lt;/math&gt; ways to arrange the order which we draw the non-equal slips, and in each order there are &lt;math&gt;4 \times 3 \times 4 \times 3&lt;/math&gt; ways to pick the slips, so &lt;math&gt;q = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{40 \times 39 \times 38 \times 37}&lt;/math&gt;. <br /> <br /> Hence, the answer is &lt;math&gt;\frac{q}{p} = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{10\cdot 4!} = \boxed{\ \mathbf{(A)}162}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2005|ab=B|num-b=20|num-a=22}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> ZekromReshiram