https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Flamewire&feedformat=atom AoPS Wiki - User contributions [en] 2021-12-02T19:58:20Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_12&diff=45630 2012 AIME I Problems/Problem 12 2012-03-18T16:14:41Z <p>Flamewire: /* Solution 2 */</p> <hr /> <div>==Problem 12==<br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be a right triangle with right angle at &lt;math&gt;C.&lt;/math&gt; Let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be points on &lt;math&gt;\overline{AB}&lt;/math&gt; with &lt;math&gt;D&lt;/math&gt; between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; such that &lt;math&gt;\overline{CD}&lt;/math&gt; and &lt;math&gt;\overline{CE}&lt;/math&gt; trisect &lt;math&gt;\angle C.&lt;/math&gt; If &lt;math&gt;\frac{DE}{BE} = \frac{8}{15},&lt;/math&gt; then &lt;math&gt;\tan B&lt;/math&gt; can be written as &lt;math&gt;\frac{m \sqrt{p}}{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, and &lt;math&gt;p&lt;/math&gt; is a positive integer not divisible by the square of any prime. Find &lt;math&gt;m+n+p.&lt;/math&gt;<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Without loss of generality, set &lt;math&gt;CB = 1&lt;/math&gt;. Then, by the Angle Bisector Theorem on triangle &lt;math&gt;DCB&lt;/math&gt;, we have &lt;math&gt;CD = \frac{8}{15}&lt;/math&gt;. We apply the Law of Cosines to triangle &lt;math&gt;DCB&lt;/math&gt; to get &lt;math&gt;1 + \frac{64}{225} - \frac{8}{15} = BD^{2}&lt;/math&gt;, which we can simplify to get &lt;math&gt;BD = \frac{13}{15}&lt;/math&gt;. <br /> <br /> Now, we have &lt;math&gt;\cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}&lt;/math&gt; by another application of the Law of Cosines to triangle &lt;math&gt;DCB&lt;/math&gt;, so &lt;math&gt;\cos \angle B = \frac{11}{13}&lt;/math&gt;. In addition, &lt;math&gt;\sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}&lt;/math&gt;, so &lt;math&gt;\tan \angle B = \frac{4\sqrt{3}}{11}&lt;/math&gt;. <br /> <br /> Our final answer is &lt;math&gt;4+3+11 = \boxed{018}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> (This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig.)<br /> <br /> Find values for all angles in terms of &lt;math&gt;\angle B&lt;/math&gt;. &lt;math&gt;\angle CEB = 150-B&lt;/math&gt;, &lt;math&gt;\angle CED = 30+B&lt;/math&gt;, &lt;math&gt;\angle CDE = 120-B&lt;/math&gt;, &lt;math&gt;\angle CDA = 60+B&lt;/math&gt;, and &lt;math&gt;\angle A = 90-B&lt;/math&gt;.<br /> <br /> Use the law of sines on &lt;math&gt;\triangle CED&lt;/math&gt; and &lt;math&gt;\triangle CEB&lt;/math&gt;:<br /> <br /> In &lt;math&gt;\triangle CED&lt;/math&gt;, &lt;math&gt;\frac{8}{\sin 30} = \frac{CE}{\sin (120-B)}&lt;/math&gt;. This simplifies to &lt;math&gt;16 = \frac{CE}{\sin (120-B)}&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\triangle CEB&lt;/math&gt;, &lt;math&gt;\frac{15}{\sin 30} = \frac{CE}{\sin B}&lt;/math&gt;. This simplifies to &lt;math&gt;30 = \frac{CE}{\sin B}&lt;/math&gt;.<br /> <br /> Solve for &lt;math&gt;CE&lt;/math&gt; and equate them so that you get &lt;math&gt;16\sin (120-B) = 30\sin B&lt;/math&gt;. <br /> <br /> From this, &lt;math&gt;\frac{8}{15} = \frac{\sin B}{\sin (120-B)}&lt;/math&gt;. <br /> <br /> Use a trig identity on the denominator on the right to obtain: &lt;math&gt;\frac{8}{15} = \frac{\sin B}{\sin 120 \cos B - \cos 120 \sin B}&lt;/math&gt;<br /> <br /> This simplifies to &lt;math&gt;\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \cosB + \sin B}{2}} = \frac{2\sin B}{\sqrt{3}\cos B + \sin B}&lt;/math&gt;<br /> <br /> This gives &lt;math&gt;\frac{15}{8} = \frac{\sqrt{3}\cos B + \sin B}{2\sin B} = \frac{\sqrt{3}\cos B}{2\sinB} + \frac{1}{2} = \frac{\sqrt{3}\cot B}{2} + \frac{1}{2}.&lt;/math&gt;<br /> <br /> Thus: &lt;math&gt;\frac{11}{7} = \frac{\sqrt{3}\cot B}{2}&lt;/math&gt; and &lt;math&gt;\cot B = \frac{11*2}{8\sqrt{3}}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cot B = \frac{1}{\tan B}&lt;/math&gt;, &lt;math&gt;tan{B} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}&lt;/math&gt;. Our final answer is &lt;math&gt;4 + 3 + 11 = 18&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=11|num-a=13}}</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_12&diff=45629 2012 AIME I Problems/Problem 12 2012-03-18T16:14:14Z <p>Flamewire: /* Solution */</p> <hr /> <div>==Problem 12==<br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be a right triangle with right angle at &lt;math&gt;C.&lt;/math&gt; Let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be points on &lt;math&gt;\overline{AB}&lt;/math&gt; with &lt;math&gt;D&lt;/math&gt; between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; such that &lt;math&gt;\overline{CD}&lt;/math&gt; and &lt;math&gt;\overline{CE}&lt;/math&gt; trisect &lt;math&gt;\angle C.&lt;/math&gt; If &lt;math&gt;\frac{DE}{BE} = \frac{8}{15},&lt;/math&gt; then &lt;math&gt;\tan B&lt;/math&gt; can be written as &lt;math&gt;\frac{m \sqrt{p}}{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, and &lt;math&gt;p&lt;/math&gt; is a positive integer not divisible by the square of any prime. Find &lt;math&gt;m+n+p.&lt;/math&gt;<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Without loss of generality, set &lt;math&gt;CB = 1&lt;/math&gt;. Then, by the Angle Bisector Theorem on triangle &lt;math&gt;DCB&lt;/math&gt;, we have &lt;math&gt;CD = \frac{8}{15}&lt;/math&gt;. We apply the Law of Cosines to triangle &lt;math&gt;DCB&lt;/math&gt; to get &lt;math&gt;1 + \frac{64}{225} - \frac{8}{15} = BD^{2}&lt;/math&gt;, which we can simplify to get &lt;math&gt;BD = \frac{13}{15}&lt;/math&gt;. <br /> <br /> Now, we have &lt;math&gt;\cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}&lt;/math&gt; by another application of the Law of Cosines to triangle &lt;math&gt;DCB&lt;/math&gt;, so &lt;math&gt;\cos \angle B = \frac{11}{13}&lt;/math&gt;. In addition, &lt;math&gt;\sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}&lt;/math&gt;, so &lt;math&gt;\tan \angle B = \frac{4\sqrt{3}}{11}&lt;/math&gt;. <br /> <br /> Our final answer is &lt;math&gt;4+3+11 = \boxed{018}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> (This solution does not use the Angle Bisector Theorem but uses more trig.)<br /> <br /> Find values for all angles in terms of &lt;math&gt;\angle B&lt;/math&gt;. &lt;math&gt;\angle CEB = 150-B&lt;/math&gt;, &lt;math&gt;\angle CED = 30+B&lt;/math&gt;, &lt;math&gt;\angle CDE = 120-B&lt;/math&gt;, &lt;math&gt;\angle CDA = 60+B&lt;/math&gt;, and &lt;math&gt;\angle A = 90-B&lt;/math&gt;.<br /> <br /> Use the law of sines on &lt;math&gt;\triangle CED&lt;/math&gt; and &lt;math&gt;\triangle CEB&lt;/math&gt;:<br /> <br /> In &lt;math&gt;\triangle CED&lt;/math&gt;, &lt;math&gt;\frac{8}{\sin 30} = \frac{CE}{\sin (120-B)}&lt;/math&gt;. This simplifies to &lt;math&gt;16 = \frac{CE}{\sin (120-B)}&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\triangle CEB&lt;/math&gt;, &lt;math&gt;\frac{15}{\sin 30} = \frac{CE}{\sin B}&lt;/math&gt;. This simplifies to &lt;math&gt;30 = \frac{CE}{\sin B}&lt;/math&gt;.<br /> <br /> Solve for &lt;math&gt;CE&lt;/math&gt; and equate them so that you get &lt;math&gt;16\sin (120-B) = 30\sin B&lt;/math&gt;. <br /> <br /> From this, &lt;math&gt;\frac{8}{15} = \frac{\sin B}{\sin (120-B)}&lt;/math&gt;. <br /> <br /> Use a trig identity on the denominator on the right to obtain: &lt;math&gt;\frac{8}{15} = \frac{\sin B}{\sin 120 \cos B - \cos 120 \sin B}&lt;/math&gt;<br /> <br /> This simplifies to &lt;math&gt;\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \cosB + \sin B}{2}} = \frac{2\sin B}{\sqrt{3}\cos B + \sin B}&lt;/math&gt;<br /> <br /> This gives &lt;math&gt;\frac{15}{8} = \frac{\sqrt{3}\cos B + \sin B}{2\sin B} = \frac{\sqrt{3}\cos B}{2\sinB} + \frac{1}{2} = \frac{\sqrt{3}\cot B}{2} + \frac{1}{2}.&lt;/math&gt;<br /> <br /> Thus: &lt;math&gt;\frac{11}{7} = \frac{\sqrt{3}\cot B}{2}&lt;/math&gt; and &lt;math&gt;\cot B = \frac{11*2}{8\sqrt{3}}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\cot B = \frac{1}{\tan B}&lt;/math&gt;, &lt;math&gt;tan{B} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}&lt;/math&gt;. Our final answer is &lt;math&gt;4 + 3 + 11 = 18&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=11|num-a=13}}</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_24&diff=44640 2012 AMC 10A Problems/Problem 24 2012-02-10T01:36:36Z <p>Flamewire: /* Problem 24 */</p> <hr /> <div>== Problem 24 ==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; be positive integers with &lt;math&gt;a\ge&lt;/math&gt; &lt;math&gt;b\ge&lt;/math&gt; &lt;math&gt;c&lt;/math&gt; such that<br /> &lt;math&gt;a^2-b^2-c^2+ab=2011&lt;/math&gt; and<br /> &lt;math&gt;a^2+3b^2+3c^2-3ab-2ac-2bc=-1997&lt;/math&gt;.<br /> <br /> What is &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Add the two equations.<br /> <br /> &lt;math&gt;2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14&lt;/math&gt;.<br /> <br /> Now, this can be rearranged:<br /> <br /> &lt;math&gt;(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14&lt;/math&gt;<br /> <br /> and factored:<br /> <br /> &lt;math&gt;(a - b)^2 + (a - c)^2 + (b - c)^2 = 14&lt;/math&gt;<br /> <br /> &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are all integers, so the three terms on the left side of the equation must all be perfect squares. Recognize that &lt;math&gt;14 = 9 + 4 + 1&lt;/math&gt;. <br /> <br /> &lt;math&gt;(a-c)^2 = 9 -&gt; a-c = 3&lt;/math&gt;, since &lt;math&gt;a-c&lt;/math&gt; is the biggest difference. It is impossible to determine by inspection whether &lt;math&gt;a-b = 2&lt;/math&gt; or &lt;math&gt;1&lt;/math&gt;, or whether &lt;math&gt;b-c = 1&lt;/math&gt; or &lt;math&gt;2&lt;/math&gt;.<br /> <br /> We want to solve for &lt;math&gt;a&lt;/math&gt;, so take the two cases and solve them each for an expression in terms of &lt;math&gt;a&lt;/math&gt;. Our two cases are &lt;math&gt;(a, b, c) = (a, a-1, a-3)&lt;/math&gt; or &lt;math&gt;(a, a-2, a-3)&lt;/math&gt;. Plug these values into one of the original equations to see if we can get an integer for &lt;math&gt;a&lt;/math&gt;.<br /> <br /> &lt;math&gt;a^2 - (a-1)^2 - (a-3)^2 + a(a-1) = 2011&lt;/math&gt;, after some algebra, simplifies to<br /> &lt;math&gt;7a = 2021&lt;/math&gt;. 2021 is not divisible by 7, so &lt;math&gt;a&lt;/math&gt; is not an integer. <br /> <br /> The other case gives &lt;math&gt;a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011&lt;/math&gt;, which simplifies to &lt;math&gt;8a = 2024&lt;/math&gt;. Thus, &lt;math&gt;a = 253&lt;/math&gt; and the answer is &lt;math&gt;\qquad\textbf{(E)}&lt;/math&gt;.</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_22&diff=44639 2012 AMC 10A Problems/Problem 22 2012-02-10T01:21:55Z <p>Flamewire: /* Problem 22 */</p> <hr /> <div>== Problem 22 ==<br /> <br /> The sum of the first &lt;math&gt;m&lt;/math&gt; positive odd integers is 212 more than the sum of the first &lt;math&gt;n&lt;/math&gt; positive even integers. What is the sum of all possible values of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> The sum of the first &lt;math&gt;m&lt;/math&gt; odd integers is given by &lt;math&gt;m^2&lt;/math&gt;. The sum of the first &lt;math&gt;n&lt;/math&gt; even integers is given by &lt;math&gt;n(n+1)&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;m^2 = n^2 + n + 212&lt;/math&gt;. Since we want to solve for n, rearrange as a quadratic equation: &lt;math&gt;n^2 + n + (212 - m^2) = 0&lt;/math&gt;.<br /> <br /> Use the quadratic formula: &lt;math&gt;n = \frac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}&lt;/math&gt;. &lt;math&gt;n&lt;/math&gt; is clearly an integer, so &lt;math&gt;1 - 4(212 - m^2) = 4m^2 - 847&lt;/math&gt; must be not only a perfect square, but also an odd perfect square. This is because the entire expression must be an integer, and for the numerator to be even (divisible by 2), &lt;math&gt;4m^2 - 847&lt;/math&gt; must be odd.<br /> <br /> Let &lt;math&gt;x&lt;/math&gt; = &lt;math&gt;\sqrt{4m^2 - 847}&lt;/math&gt;. (Note that this means that &lt;math&gt;n = \frac{-1 + x}{2}&lt;/math&gt;.) This can be rewritten as &lt;math&gt;x^2 = 4m^2 - 847&lt;/math&gt;, which can then be rewritten to &lt;math&gt;4m^2 - x^2 = 847&lt;/math&gt;. Factor the left side by using the difference of squares. &lt;math&gt;(2m + x)(2m - x) = 847 = 7*11^2&lt;/math&gt;.<br /> <br /> Our goal is to find possible values for &lt;math&gt;a&lt;/math&gt;, then use the equation above to find &lt;math&gt;n&lt;/math&gt;. The difference between the factors is &lt;math&gt;(2m + a) - (2m - a) = 2m + a - 2m + a = 2a.&lt;/math&gt; We have three pairs of factors, &lt;math&gt;847*1, 7*121, and 11*77&lt;/math&gt;. The differences between these factors are &lt;math&gt;846&lt;/math&gt;, &lt;math&gt;114&lt;/math&gt;, and &lt;math&gt;66&lt;/math&gt; - those are all possible values for &lt;math&gt;2a&lt;/math&gt;. Thus the possibilities for &lt;math&gt;a&lt;/math&gt; are &lt;math&gt;423&lt;/math&gt;, &lt;math&gt;57&lt;/math&gt;, and &lt;math&gt;33&lt;/math&gt;. <br /> <br /> Now plug in these values into the equation &lt;math&gt;n = \frac{-1 + x}{2}&lt;/math&gt;. &lt;math&gt;n&lt;/math&gt; can equal &lt;math&gt;211&lt;/math&gt;, &lt;math&gt;28&lt;/math&gt;, or &lt;math&gt;16&lt;/math&gt;. Add &lt;math&gt;211 + 28 + 16 = 255&lt;/math&gt;. The answer is &lt;math&gt;\qquad\textbf{(B)}&lt;/math&gt;.</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_11&diff=44638 2012 AMC 10A Problems/Problem 11 2012-02-10T00:46:02Z <p>Flamewire: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Externally tangent circles with centers at points A and B have radii of lengths 5 and 3, respectively. A line externally tangent to both circles intersects ray AB at point C. What is BC?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.8\qquad\textbf{(C)}\ 10.2\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 14.4 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(3.5mm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=4;<br /> <br /> pair A=(0,0), B=(8,0); pair C=(20,0); pair D=(1.25,-0.25sqrt(375)); pair E=(8.75,-0.15sqrt(375));<br /> path a=Circle(A,5);<br /> path b=Circle(B,3);<br /> draw(a); draw(b);<br /> draw(C--D);<br /> draw(A--C);<br /> draw(A--D);<br /> draw(B--E);<br /> <br /> pair[] ps={A,B,C,D,E};<br /> dot(ps);<br /> <br /> label(&quot;$A$&quot;,A,N); label(&quot;$B$&quot;,B,N); label(&quot;$C$&quot;,C,N);<br /> label(&quot;$D$&quot;,D,SE); label(&quot;$E$&quot;,E,SE);<br /> label(&quot;$5$&quot;,(A--D),SW);<br /> label(&quot;$3$&quot;,(B--E),SW);<br /> label(&quot;$8$&quot;,(A--B),N);<br /> label(&quot;$x$&quot;,(C--B),N);<br /> <br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the points of tangency on circles &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; with line &lt;math&gt;CD&lt;/math&gt;. &lt;math&gt;AB=8&lt;/math&gt;. Also, let &lt;math&gt;BC=x&lt;/math&gt;. As &lt;math&gt;\angle ADC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share &lt;math&gt;\angle ACD&lt;/math&gt;, &lt;math&gt;\triangle ADC \sim \triangle BCE&lt;/math&gt;. From this we can get a proportion.<br /> <br /> &lt;math&gt;\frac{BC}{AC}=\frac{BE}{AD} \rightarrow \frac{x}{x+8}=\frac{3}{5} \rightarrow 5x=3x+24 \rightarrow x=\boxed{\textbf{(D)}\ 12}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2012|ab=A|num-b=10|num-a=12}}</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_9&diff=44582 2012 AMC 10A Problems/Problem 9 2012-02-09T03:08:23Z <p>Flamewire: /* Solution */</p> <hr /> <div>== Problem 9 ==<br /> <br /> A pair of six-sided dice are labeled so that one die has only even numbers (two each of 2, 4, and 6), and the other die has only odd numbers (two of each 1, 3, and 5). The pair of dice is rolled. What is the probability that the sum of the numbers on the tops of the two dice is 7?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> To solve this, we need to find the number of ways that we can roll a sum of 7 divided by the total possible rolls.<br /> <br /> The total number of combinations when rolling two dice is &lt;math&gt;6*6 = 36&lt;/math&gt;.<br /> <br /> There are three ways that a sum of 7 can be rolled. &lt;math&gt;2+5&lt;/math&gt;, &lt;math&gt;4+3&lt;/math&gt;, and &lt;math&gt;6+1&lt;/math&gt;. There are two 2's on one die and two 5's on the other, so there are a total of 4 ways to roll the combination of 2 and 5. There are two 4's on one die and two 3's on the other, so there are a total of 4 ways to roll the combination of 4 and 3. There are two 6's on one die and two 1's on the other, so there are a total of 4 ways to roll the combination of 6 and 1. Add &lt;math&gt;4 + 4 + 4 = 12&lt;/math&gt;.<br /> <br /> Thus, our probability is &lt;math&gt;\frac{12}{36} = \frac{1}{3}&lt;/math&gt;. The answer is &lt;math&gt;\qquad\textbf{(D)}&lt;/math&gt;.</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_9&diff=44581 2012 AMC 10A Problems/Problem 9 2012-02-09T03:08:07Z <p>Flamewire: /* Problem 9 */</p> <hr /> <div>== Problem 9 ==<br /> <br /> A pair of six-sided dice are labeled so that one die has only even numbers (two each of 2, 4, and 6), and the other die has only odd numbers (two of each 1, 3, and 5). The pair of dice is rolled. What is the probability that the sum of the numbers on the tops of the two dice is 7?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> To solve this, we need to find the number of ways that we can roll a sum of 7 divided by the total possible rolls.<br /> <br /> The total number of combinations when rolling two dice is &lt;math&gt;6*6 = 36&lt;/math&gt;.<br /> <br /> There are three ways that a sum of 7 can be rolled. 2+5, 4+3, and 6+1. There are two 2's on one die and two 5's on the other, so there are a total of 4 ways to roll the combination of 2 and 5. There are two 4's on one die and two 3's on the other, so there are a total of 4 ways to roll the combination of 4 and 3. There are two 6's on one die and two 1's on the other, so there are a total of 4 ways to roll the combination of 6 and 1. Add &lt;math&gt;4 + 4 + 4 = 12&lt;/math&gt;.<br /> <br /> Thus, our probability is &lt;math&gt;\frac{12}{36} = \frac{1}{3}&lt;/math&gt;. The answer is &lt;math&gt;\qquad\textbf{(D)}&lt;/math&gt;.</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_7&diff=44580 2012 AMC 10A Problems/Problem 7 2012-02-09T03:00:01Z <p>Flamewire: /* Problem 7 */</p> <hr /> <div>== Problem 7 ==<br /> <br /> In a bag of marbles, &lt;math&gt;\frac{3}{5}&lt;/math&gt; of the marbles are blue and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{3}{7}\qquad\textbf{(C)}\ \frac{4}{7}\qquad\textbf{(D)}\ \frac{3}{5}\qquad\textbf{(E)}\ \frac{4}{5} &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Assume that there are 5 total marbles in the bag. The actual number does not matter, since all we care about is the ratios, and the only operation performed on the marbles in the bag is doubling.<br /> <br /> There are 3 blue marbles in the bag and 2 red marbles. If you double the amount of red marbles, there will still be 3 blue marbles but now there will be 4 red marbles. Thus, the answer is &lt;math&gt;\qquad\textbf{(C)} \frac{4}{7}&lt;/math&gt;.</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_4&diff=44554 2012 AMC 10A Problems/Problem 4 2012-02-09T00:30:49Z <p>Flamewire: /* Problem 4 */</p> <hr /> <div>== Problem 4 ==<br /> <br /> Let &lt;math&gt;\angle ABC = 24^\circ &lt;/math&gt; and &lt;math&gt;\angle ABD = 20^\circ &lt;/math&gt;. What is the smallest possible degree measure for angle CBD?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 0\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 12 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> &lt;math&gt;\angle ABD&lt;/math&gt; and &lt;math&gt;\angle ABC&lt;/math&gt; share ray &lt;math&gt;AB&lt;/math&gt;. In order to minimize the value of &lt;math&gt;\angle CBD&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt; should be located between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;. <br /> <br /> &lt;math&gt;\angle ABC = \angle ABD + \angle CBD&lt;/math&gt;, so &lt;math&gt;\angle CBD = 4&lt;/math&gt;. The answer is &lt;math&gt; \qquad\textbf{(C)}&lt;/math&gt;</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_3&diff=44540 2012 AMC 10A Problems/Problem 3 2012-02-09T00:17:41Z <p>Flamewire: /* Solution */</p> <hr /> <div>== Problem 3 ==<br /> <br /> A bug crawls along a number line, starting at -2. It crawls to -6, then turns around and crawls to 5. How many units does the bug crawl altogether?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Crawling from -2 to -6 takes it a distance of 4 units. Crawling from -6 to 5 takes it a distance of 11 units. Add 4 and 11 to get &lt;math&gt;\qquad\textbf{(E)}\ 15&lt;/math&gt;</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_3&diff=44539 2012 AMC 10A Problems/Problem 3 2012-02-09T00:17:31Z <p>Flamewire: /* Problem 3 */</p> <hr /> <div>== Problem 3 ==<br /> <br /> A bug crawls along a number line, starting at -2. It crawls to -6, then turns around and crawls to 5. How many units does the bug crawl altogether?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Crawling from -2 to -6 takes it a distance of 4 units. Crawling from -6 to 5 takes it a distance of 11 units. Add the two to get &lt;math&gt;\qquad\textbf{(E)}\ 15&lt;/math&gt;</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_2&diff=44519 2012 AMC 10A Problems/Problem 2 2012-02-08T23:35:06Z <p>Flamewire: /* Solution */</p> <hr /> <div>== Problem 2 ==<br /> <br /> A square with side length 8 is cut in half, creating two congruent rectangles. What are the dimensions of one of these rectangles?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \emph{2\ by\ 4}\qquad\textbf{(B)}\ \emph{\ 2\ by\ 6}\qquad\textbf{(C)}\ \emph{\ 2\ by\ 8}\qquad\textbf{(D)}\ \emph{4\ by\ 4}\qquad\textbf{(E)}\ \emph{4\ by\ 8} &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Cutting the square in half will bisect one pair of sides while the other side will remain unchanged. Thus, the new square is &lt;math&gt;\frac{8}{2}*8&lt;/math&gt;, or &lt;math&gt;\qquad\textbf{(E)}\ 4*8&lt;/math&gt;.</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_2&diff=44518 2012 AMC 10A Problems/Problem 2 2012-02-08T23:33:49Z <p>Flamewire: /* Problem 2 */</p> <hr /> <div>== Problem 2 ==<br /> <br /> A square with side length 8 is cut in half, creating two congruent rectangles. What are the dimensions of one of these rectangles?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \emph{2\ by\ 4}\qquad\textbf{(B)}\ \emph{\ 2\ by\ 6}\qquad\textbf{(C)}\ \emph{\ 2\ by\ 8}\qquad\textbf{(D)}\ \emph{4\ by\ 4}\qquad\textbf{(E)}\ \emph{4\ by\ 8} &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Cutting the square in half will bisect one pair of sides while the other side will remain unchanged. Thus, the new square is &lt;math&gt;\frac{8}{2}*8 = \qquad\textbf{(E)}\ 4*8&lt;/math&gt;.</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_1&diff=44515 2012 AMC 10A Problems/Problem 1 2012-02-08T23:27:15Z <p>Flamewire: </p> <hr /> <div>== Problem 1 ==<br /> <br /> Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Cagney can frost one in 20 seconds, and Lacey can frost one in 30 seconds. Working together, they can frost one in &lt;math&gt;\frac{20*30}{20+30} = \frac{600}{50} = 12&lt;/math&gt; seconds. In 300 seconds (5 minutes), they can frost &lt;math&gt;\qquad\textbf{(D)}\ 25&lt;/math&gt; cupcakes.</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_18&diff=44479 2008 AMC 10A Problems/Problem 18 2012-02-06T02:40:36Z <p>Flamewire: /* Solution */</p> <hr /> <div>==Problem==<br /> A [[right triangle]] has [[perimeter]] &lt;math&gt;32&lt;/math&gt; and area &lt;math&gt;20&lt;/math&gt;. What is the length of its [[hypotenuse]]?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac{57}{4}\qquad\mathrm{(B)}\ \frac{59}{4}\qquad\mathrm{(C)}\ \frac{61}{4}\qquad\mathrm{(D)}\ \frac{63}{4}\qquad\mathrm{(E)}\ \frac{65}{4}&lt;/math&gt;<br /> <br /> __TOC__<br /> ==Solution==<br /> === Solution 1 ===<br /> Let the legs of the triangle have lengths &lt;math&gt;a,b&lt;/math&gt;. Then, by the [[Pythagorean Theorem]], the length of the hypotenuse is &lt;math&gt;\sqrt{a^2+b^2}&lt;/math&gt;, and the area of the triangle is &lt;math&gt;\frac 12 ab&lt;/math&gt;. So we have the two equations<br /> &lt;center&gt;&lt;math&gt;\begin{align}<br /> a+b+\sqrt{a^2+b^2} &amp;= 32 \\<br /> \frac{1}{2}ab &amp;= 20<br /> \end{align}&lt;/math&gt;&lt;/center&gt;<br /> Re-arranging the first equation and squaring, <br /> &lt;center&gt;&lt;math&gt;\begin{align*}<br /> \sqrt{a^2+b^2} &amp;= 32-(a+b)\\<br /> a^2 + b^2 &amp;= 32^2 - 64(a+b) + (a+b)^2\\<br /> a^2 + b^2 + 64(a+b) &amp;= a^2 + b^2 + 2ab + 32^2\\<br /> a+b &amp;= \frac{2ab+32^2}{64}\end{align*}&lt;/math&gt;&lt;/center&gt;<br /> From &lt;math&gt;(2)&lt;/math&gt; we have &lt;math&gt;2ab = 80&lt;/math&gt;, so<br /> &lt;center&gt;&lt;math&gt;a+b &amp;= \frac{80 + 32^2}{64} = \frac{69}{4}.&lt;/math&gt;&lt;/center&gt;<br /> The length of the hypotenuse is &lt;math&gt;p - a - b = 32 - \frac{69}{4} = \frac{59}{4}\ \mathrm{(B)}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> From the formula &lt;math&gt;A = rs&lt;/math&gt;, where &lt;math&gt;A&lt;/math&gt; is the area of a triangle, &lt;math&gt;r&lt;/math&gt; is its [[inradius]], and &lt;math&gt;s&lt;/math&gt; is the [[semiperimeter]], we can find that &lt;math&gt;r = \frac{20}{32/2} = \frac{5}{4}&lt;/math&gt;. It is known that in a right triangle, &lt;math&gt;r = s - h&lt;/math&gt;, where &lt;math&gt;h&lt;/math&gt; is the hypotenuse, so &lt;math&gt;h = 16 - \frac{5}{4} = \frac{59}{4}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> From the problem, we know that <br /> &lt;center&gt;&lt;math&gt;\begin{align*}<br /> a+b+c &amp;= 32 \\<br /> 2ab &amp;= 80. \\<br /> \end{align*}&lt;/math&gt;&lt;/center&gt;<br /> <br /> Subtracting &lt;math&gt;c&lt;/math&gt; from both sides of the first equation and squaring both sides, we get<br /> &lt;center&gt;&lt;math&gt;\begin{align*}<br /> (a+b)^2 &amp;= (32 - c)^2\\<br /> a^2 + b^2 + 2ab &amp;= 32^2 + c^2 - 64c.\\<br /> \end{align*}&lt;/math&gt;&lt;/center&gt;<br /> <br /> Now we substitute in &lt;math&gt;a^2 + b^2 = c^2&lt;/math&gt; as well as &lt;math&gt;2ab = 80&lt;/math&gt; into the equation to get<br /> &lt;center&gt;&lt;math&gt;\begin{align*}<br /> 80 &amp;= 1024 - 64c\\<br /> c &amp;= \frac{944}{64}.<br /> \end{align*}&lt;/math&gt;&lt;/center&gt;<br /> <br /> Further simplification yields the result of &lt;math&gt;\frac{59}{4}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be the legs of the triangle, and &lt;math&gt;c&lt;/math&gt; the hypotenuse.<br /> <br /> Since the area is 20, we have &lt;math&gt;\frac{1}{2}ab = 20 =&gt; ab=40&lt;/math&gt;.<br /> <br /> Since the perimeter is 32, we have &lt;math&gt;a + b + c = 32&lt;/math&gt;.<br /> <br /> The Pythagorean Theorem gives &lt;math&gt;c^2 = a^2 + b^2&lt;/math&gt;. <br /> <br /> This gives us three equations with three variables:<br /> <br /> &lt;center&gt;&lt;math&gt;ab = 40 \\<br /> a + b + c = 32 \\<br /> c^2 = a^2 + b^2&lt;/math&gt;&lt;/center&gt;<br /> <br /> Rewrite equation 3 as &lt;math&gt;c^2 = (a+b)^2 - 2ab&lt;/math&gt;.<br /> Substitute in equations 1 and 2 to get &lt;math&gt;c^2 = (32-c)^2 - 80&lt;/math&gt;.<br /> <br /> &lt;center&gt;&lt;math&gt;c^2 = (32-c)^2 - 80 \\<br /> c^2 = 1024 - 64c + c^2 - 80 \\<br /> 64c = 944 \\<br /> c = \frac{944}{64} = \frac{236}{16} = \frac{59}{4} &lt;/math&gt;. &lt;/center&gt;<br /> The answer is choice (D).<br /> <br /> ==See also==<br /> {{AMC10 box|year=2008|ab=A|num-b=17|num-a=19}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_10&diff=44474 2008 AMC 12A Problems/Problem 10 2012-02-06T01:37:50Z <p>Flamewire: /* Solution */</p> <hr /> <div>{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #10]] and [[2008 AMC 10A Problems/Problem 14|2008 AMC 10A #13]]}}<br /> ==Problem==<br /> Doug can paint a room in &lt;math&gt;5&lt;/math&gt; hours. Dave can paint the same room in &lt;math&gt;7&lt;/math&gt; hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let &lt;math&gt;t&lt;/math&gt; be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by &lt;math&gt;t&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \left(\frac{1}{5}+\frac{1}{7}\right)\left(t+1\right)=1\qquad\mathrm{(B)}\ \left(\frac{1}{5}+\frac{1}{7}\right)t+1=1\qquad\mathrm{(C)}\ \left(\frac{1}{5}+\frac{1}{7}\right)t=1\\\mathrm{(D)}\ \left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1\qquad\mathrm{(E)}\ \left(5+7\right)t=1&lt;/math&gt;<br /> <br /> ==Solution==<br /> === Solution 1 ===<br /> Doug can paint &lt;math&gt;\frac{1}{5}&lt;/math&gt; of a room per hour, Dave can paint &lt;math&gt;\frac{1}{7}&lt;/math&gt; of a room in an hour, and the time they spend working together is &lt;math&gt;t-1&lt;/math&gt;.<br /> <br /> Since rate times time gives output, &lt;math&gt;\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow \mathrm{(D)}&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> <br /> If one person does a job in &lt;math&gt;a&lt;/math&gt; hours and another person does a job in &lt;math&gt;b&lt;/math&gt; hours, the time it takes to do the job together is &lt;math&gt;\frac{ab}{a+b}&lt;/math&gt; hours.<br /> <br /> Since Doug paints a room in 5 hours and Dave paints a room in 7 hours, they both paint in &lt;math&gt;\frac{5*7}{5+7} = \frac{35}{12}&lt;/math&gt; hours. They also take 1 hour for lunch, so the total time &lt;math&gt;t = \frac{35}{12} + 1&lt;/math&gt; hours.<br /> <br /> Looking at the answer choices, &lt;math&gt;(D)&lt;/math&gt; is the only one satisfied by &lt;math&gt;t = \frac{35}{12} + 1&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2008|ab=A|num-b=9|num-a=11}}<br /> {{AMC10 box|year=2008|ab=A|num-b=12|num-a=14}}</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_7&diff=44451 2010 AMC 10B Problems/Problem 7 2012-02-04T03:59:58Z <p>Flamewire: /* Solution */</p> <hr /> <div>==Problem==<br /> A triangle has side lengths &lt;math&gt;10&lt;/math&gt;, &lt;math&gt;10&lt;/math&gt;, and &lt;math&gt;12&lt;/math&gt;. A rectangle has width &lt;math&gt;4&lt;/math&gt; and area equal to the<br /> area of the triangle. What is the perimeter of this rectangle?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 36&lt;/math&gt;<br /> <br /> ==Solution==<br /> The triangle is isosceles. The height of the triangle is therefore given by &lt;math&gt;h = \sqrt{10^2 - ( \dfrac{12}{2})^2} = \sqrt{64} = 8&lt;/math&gt;<br /> <br /> Now, the area of the triangle is &lt;math&gt;\dfrac{bh}{2} = \dfrac{12*8}{2} = \dfrac{96}{2} = 48 &lt;/math&gt;<br /> <br /> We have that the area of the rectangle is the same as the area of the triangle, namely &lt;math&gt;48&lt;/math&gt;. We also have the width of the rectangle: &lt;math&gt;4&lt;/math&gt;.<br /> <br /> The length of the rectangle therefore is:<br /> &lt;math&gt;l = \dfrac{48}{4} = 12&lt;/math&gt;<br /> <br /> The perimeter of the rectangle then becomes:<br /> &lt;math&gt;2l + 2w = 2*12 + 2*4 = 32&lt;/math&gt;<br /> <br /> The answer is:<br /> <br /> &lt;math&gt; \boxed{\textbf{(D)}\ 32} &lt;/math&gt;<br /> <br /> An alternative way to find the area of the triangle is by using Heron's formula, &lt;math&gt;A=\sqrt{(s)(s-a)(s-b)(s-c)}&lt;/math&gt; where &lt;math&gt;s&lt;/math&gt; is the semi-perimeter of the triangle (meaning half the perimeter). Here, the semi-perimeter is &lt;math&gt;(10+10+12)/2 = 16&lt;/math&gt;. Thus the area equals &lt;math&gt;\sqrt{(16)(16-10)(16-10)(16-12)} = \sqrt{16*6*6*4} = 48. &lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2010|ab=B|num-b=6|num-a=8}}</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_23&diff=44441 2010 AMC 10A Problems/Problem 23 2012-02-03T04:13:35Z <p>Flamewire: /* Solution */</p> <hr /> <div>== Problem ==<br /> Each of &lt;math&gt;2010&lt;/math&gt; 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 /> == Solution ==<br /> === Solution 1 ===<br /> The probability of drawing a white marble from box &lt;math&gt;k&lt;/math&gt; is &lt;math&gt;\frac{k}{k+1}&lt;/math&gt;. The probability of drawing a red marble from box &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\frac{1}{n+1}&lt;/math&gt;.<br /> <br /> The probability of drawing a red marble at box &lt;math&gt;n&lt;/math&gt; is therefore<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;\frac{1}{n+1} \left( \prod_{k=1}^{n-1}\frac{k}{k+1} \right) &lt; \frac{1}{2010}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{n+1} \left( \frac{1}{n} \right) &lt; \frac{1}{2010}&lt;/math&gt;<br /> <br /> &lt;math&gt;(n+1)n &gt; 2010&lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> It is then easy to see that the lowest integer value of &lt;math&gt;n&lt;/math&gt; that satisfies the inequality is &lt;math&gt;\boxed{45\ \textbf{(A)}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Using the first few values of &lt;math&gt;n&lt;/math&gt;, it is easy to derive a formula for &lt;math&gt;P(n)&lt;/math&gt;. The chance that she stops on the second box (&lt;math&gt;n=2&lt;/math&gt;) is the chance of drawing a white marble then a red marble: &lt;math&gt;\frac{1}2 * \frac{1}3&lt;/math&gt;. The chance that she stops on the third box (&lt;math&gt;n=3&lt;/math&gt;) is the chance of drawing two white marbles then a red marble:&lt;math&gt;\frac{1}2 * \frac{2}3 * \frac{1}4&lt;/math&gt;. If &lt;math&gt;n=4&lt;/math&gt;, &lt;math&gt;P(n) = \frac{1}2 * \frac{2}3 * \frac{3}4 * \frac{1}5&lt;/math&gt;.<br /> <br /> Cross-cancelling in the fractions gives &lt;math&gt;P(2) = &lt;/math&gt;\frac{1}{2*3}&lt;math&gt;, &lt;/math&gt;P(3) = &lt;math&gt;\frac{1}{3*4}%, and &lt;/math&gt;P(4) = &lt;math&gt;\frac{1}{4*5}&lt;/math&gt;. From this, it is clear that &lt;math&gt;P(n) = \frac{1}{(n)(n+1)}&lt;/math&gt;. (Alternatively, &lt;math&gt;P(n) = \frac{(n-1)!}{(n+1)!}&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\frac{1}{(n+1)(n)} &lt; \frac{1}{2010}&lt;/math&gt;<br /> <br /> The lowest integer that satisfies the above inequality is 45.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|num-b=22|num-a=24|ab=A}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]</div> Flamewire https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_17&diff=44425 2009 AMC 10A Problems/Problem 17 2012-02-01T04:25:32Z <p>Flamewire: /* Solution */</p> <hr /> <div>== Problem ==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB=4&lt;/math&gt; and &lt;math&gt;BC=3&lt;/math&gt;. Segment &lt;math&gt;EF&lt;/math&gt; is constructed through &lt;math&gt;B&lt;/math&gt; so that &lt;math&gt;EF&lt;/math&gt; is perpendicular to &lt;math&gt;DB&lt;/math&gt;, and &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; lie on &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;DF&lt;/math&gt;, respectively. What is &lt;math&gt;EF&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 9<br /> \qquad<br /> \mathrm{(B)}\ 10<br /> \qquad<br /> \mathrm{(C)}\ \frac {125}{12}<br /> \qquad<br /> \mathrm{(D)}\ \frac {103}{9}<br /> \qquad<br /> \mathrm{(E)}\ 12<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> The situation is shown in the picture below.<br /> <br /> &lt;asy&gt;<br /> unitsize(0.6cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=(4,0), C=(4,3), D=(0,3);<br /> pair EF=rotate(90)*(D-B);<br /> pair E=intersectionpoint( (0,-100)--(0,100), (B-100*EF)--(B+100*EF) );<br /> pair F=intersectionpoint( (-100,3)--(100,3), (B-100*EF)--(B+100*EF) );<br /> draw(A--B--C--D--cycle);<br /> draw(B--D, dashed);<br /> draw(E--F);<br /> draw(A--E, dashed);<br /> draw(C--F, dashed);<br /> label(&quot;$A$&quot;,A,W);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,N);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$E$&quot;,E,SW);<br /> label(&quot;$F$&quot;,F,NE);<br /> label(&quot;$3$&quot;,A--D,W);<br /> label(&quot;$4$&quot;,C--D,N);<br /> &lt;/asy&gt;<br /> <br /> Obviously, from the [[Pythagorean theorem]] we have &lt;math&gt;BD=5&lt;/math&gt;.<br /> <br /> Triangle &lt;math&gt;EAB&lt;/math&gt; is similar to &lt;math&gt;ABD&lt;/math&gt;, as they have the same angles. Hence &lt;math&gt;BE/AB = DB/AD&lt;/math&gt;, and therefore &lt;math&gt;BE = AB\cdot DB/AD = 20/3&lt;/math&gt;.<br /> <br /> Also triangle &lt;math&gt;CBF&lt;/math&gt; is similar to &lt;math&gt;ABD&lt;/math&gt;. Hence &lt;math&gt;BF/BC = DB/AB&lt;/math&gt;, and therefore &lt;math&gt;BF=BC\cdot DB / AB = 15/4&lt;/math&gt;.<br /> <br /> We then have &lt;math&gt;EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Since &lt;math&gt;BD&lt;/math&gt; is the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;EF&lt;/math&gt;, we can use the equation &lt;math&gt;BD^2 = EB\cdot BF&lt;/math&gt;. <br /> <br /> Looking at the angles, we see that triangle &lt;math&gt;EAB&lt;/math&gt; is similar to &lt;math&gt;DCB&lt;/math&gt;. Because of this, &lt;math&gt;\frac{AB}{CB} = \frac{EB}{DB}&lt;/math&gt;. From the given information and the [[Pythagorean theorem]], &lt;math&gt;AB=4&lt;/math&gt;, &lt;math&gt;CB=3&lt;/math&gt;, and &lt;math&gt;DB=5&lt;/math&gt;. Solving gives &lt;math&gt;EB=20/3&lt;/math&gt;. <br /> <br /> We can use the above formula to solve for &lt;math&gt;BF&lt;/math&gt;. &lt;math&gt;BD^2 = 20/3\cdot BF&lt;/math&gt;. Solve to obtain &lt;math&gt;BF=15/4&lt;/math&gt;. <br /> <br /> We now know &lt;math&gt;EB&lt;/math&gt; and &lt;math&gt;BF&lt;/math&gt;. &lt;math&gt;EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=A|num-b=16|num-a=18}}</div> Flamewire