https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Oreochocolate&feedformat=atom AoPS Wiki - User contributions [en] 2021-12-02T20:28:02Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_15&diff=154600 2021 AIME I Problems/Problem 15 2021-05-31T22:09:08Z <p>Oreochocolate: /* Solution 1 */ Tweaked the LaTeX a little.</p> <hr /> <div>__TOC__<br /> <br /> ==Problem==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;k&lt;/math&gt; such that the two parabolas&lt;cmath&gt;y=x^2-k~~\text{and}~~x=2(y-20)^2-k&lt;/cmath&gt;intersect in four distinct points, and these four points lie on a circle with radius at most &lt;math&gt;21&lt;/math&gt;. Find the sum of the least element of &lt;math&gt;S&lt;/math&gt; and the greatest element of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Make the translation &lt;math&gt;y \rightarrow y+20&lt;/math&gt; to obtain &lt;math&gt;20+y=x^2-k , x=2y^2-k&lt;/math&gt;. Multiply the first equation by 2 and sum, we see that &lt;math&gt;2(x^2+y^2)=3k+40+2y+x&lt;/math&gt;. Completing the square gives us &lt;math&gt;\left(y- \frac{1}{2}\right)^2+\left(x - \frac{1}{4}\right)^2 = \frac{325+24k}{16}&lt;/math&gt;; this explains why the two parabolas intersect at four points that lie on a circle*. For the upper bound, observe that &lt;math&gt;LHS \leq 21^2=441 \rightarrow 24k \leq 6731&lt;/math&gt;, so &lt;math&gt;k \leq 280&lt;/math&gt;. <br /> <br /> For the lower bound, we need to ensure there are 4 intersections to begin with. (Here I'm using the un-translated coordinates.) Draw up a graph, and realize that two intersections are guaranteed, on the so called &quot;right branch&quot; of &lt;math&gt;y=x^2-k&lt;/math&gt;. As we increase the value of k, two more intersections appear on the &quot;left branch.&quot; <br /> <br /> &lt;math&gt;k=4&lt;/math&gt; does not work because the &quot;leftmost&quot; point of &lt;math&gt;x=2(y-20)^2-4&lt;/math&gt; is &lt;math&gt;(-4,20)&lt;/math&gt; which lies to the right of &lt;math&gt;(-\sqrt{24}, 20)&lt;/math&gt;, which is on the graph &lt;math&gt;y=x^2-4&lt;/math&gt;. While technically speaking this doesn't prove that there are no intersections (why?), drawing the graph should convince you that this is the case. Clearly, no k less than 4 works either.<br /> <br /> &lt;math&gt;k=5&lt;/math&gt; does work because the two graphs intersect at &lt;math&gt;(-5,20)&lt;/math&gt;, and by drawing the graph, you realize this is not a tangent point and there is in fact another intersection nearby, due to slope. Therefore, the answer is &lt;math&gt;5+280=285&lt;/math&gt;.<br /> <br /> <br /> *In general, (Assuming four intersections exist) when two conics intersect, if one conic can be written as &lt;math&gt;ax^2+by^2=f(x,y)&lt;/math&gt; and the other as &lt;math&gt;cx^2+dy^2=g(x,y)&lt;/math&gt; for f,g polynomials of degree at most 1, whenever &lt;math&gt;(a,b),(c,d)&lt;/math&gt; are linearly independent, we can combine the two equations and then complete the square to achieve &lt;math&gt;(x-p)^2+(y-q)^2=r^2&lt;/math&gt;. We can also combine these two equations to form a parabola, or a hyperbola, or an ellipse. When &lt;math&gt;(a,b),(c,d)&lt;/math&gt; are not L.I., the intersection points instead lie on a line, which is a circle of radius infinity. When the two conics only have 3,2 or 1 intersection points, the statement that all these points lie on a circle is trivially true. <br /> <br /> -Ross Gao<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|num-b=14|after=Last problem}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_25&diff=139175 2016 AMC 12A Problems/Problem 25 2020-12-07T04:14:07Z <p>Oreochocolate: Undo revision 129767 by Rainestorm (talk) Non-constructive.</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;k&lt;/math&gt; be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with &lt;math&gt;k+1&lt;/math&gt; digits. Every time Bernardo writes a number, Silvia erases the last &lt;math&gt;k&lt;/math&gt; digits of it. Bernardo then writes the next perfect square, Silvia erases the last &lt;math&gt;k&lt;/math&gt; digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let &lt;math&gt;f(k)&lt;/math&gt; be the smallest positive integer not written on the board. For example, if &lt;math&gt;k = 1&lt;/math&gt;, then the numbers that Bernardo writes are &lt;math&gt;16, 25, 36, 49, 64&lt;/math&gt;, and the numbers showing on the board after Silvia erases are &lt;math&gt;1, 2, 3, 4,&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;, and thus &lt;math&gt;f(1) = 5&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;f(2) + f(4)+ f(6) + \dots + f(2016)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7986\qquad\textbf{(B)}\ 8002\qquad\textbf{(C)}\ 8030\qquad\textbf{(D)}\ 8048\qquad\textbf{(E)}\ 8064&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Consider &lt;math&gt;f(2)&lt;/math&gt;. The numbers left on the blackboard will show the hundreds place at the end. In order for the hundreds place to differ by 2, the difference between two perfect squares needs to be at least &lt;math&gt;100&lt;/math&gt;. Calculus &lt;math&gt;\left(\frac{\text{d}}{\text{d}x} x^2=2x\right)&lt;/math&gt; and a bit of thinking says this first happens at &lt;math&gt;x\ge 100/2 = 50&lt;/math&gt;*. The perfect squares from here go: &lt;math&gt;2500, 2601, 2704, 2809\dots&lt;/math&gt;. Note that the ones and tens also make the perfect squares, &lt;math&gt;1^2,2^2,3^2\dots&lt;/math&gt;. After the ones and tens make &lt;math&gt;100&lt;/math&gt;, the hundreds place will go up by &lt;math&gt;2&lt;/math&gt;, thus reaching our goal. Since &lt;math&gt;10^2=100&lt;/math&gt;, the last perfect square to be written will be &lt;math&gt;\left(50+10\right)^2=60^2=3600&lt;/math&gt;. The missing number is one less than the number of hundreds &lt;math&gt;(k=2)&lt;/math&gt; of &lt;math&gt;3600&lt;/math&gt;, or &lt;math&gt;35&lt;/math&gt;.<br /> <br /> Now consider f(4). Instead of the difference between two squares needing to be &lt;math&gt;100&lt;/math&gt;, the difference must now be &lt;math&gt;10000&lt;/math&gt;. This first happens at &lt;math&gt;x\ge 5000&lt;/math&gt;. After this point, similarly, &lt;math&gt;\sqrt{10000}=100&lt;/math&gt; more numbers are needed to make the &lt;math&gt;10^4&lt;/math&gt; th's place go up by &lt;math&gt;2&lt;/math&gt;. This will take place at &lt;math&gt;\left(5000+100\right)^2=5100^2= 26010000&lt;/math&gt;. Removing the last four digits (the zeros) and subtracting one yields &lt;math&gt;2600&lt;/math&gt; for the skipped value.<br /> <br /> In general, each new value of &lt;math&gt;f(k+2)&lt;/math&gt; will add two digits to the &quot;&lt;math&gt;5&lt;/math&gt;&quot; and one digit to the &quot;&lt;math&gt;1&lt;/math&gt;&quot;. This means that the last number Bernardo writes for &lt;math&gt;k=6&lt;/math&gt; is &lt;math&gt;\left(500000+1000\right)^2&lt;/math&gt;, the last for &lt;math&gt;k = 8&lt;/math&gt; will be &lt;math&gt;\left(50000000+10000\right)^2&lt;/math&gt;, and so on until &lt;math&gt;k=2016&lt;/math&gt;. Removing the last &lt;math&gt;k&lt;/math&gt; digits as Silvia does will be the same as removing &lt;math&gt;k/2&lt;/math&gt; trailing zeroes on the number to be squared. This means that the last number on the board for &lt;math&gt;k=6&lt;/math&gt; is &lt;math&gt;5001^2&lt;/math&gt;, &lt;math&gt;k=8&lt;/math&gt; is &lt;math&gt;50001^2&lt;/math&gt;, and so on. So the first missing number is &lt;math&gt;5001^2-1,50001^2-1\text{ etc.}&lt;/math&gt; The squaring will make a &quot;&lt;math&gt;25&lt;/math&gt;&quot; with two more digits than the last number, a &quot;&lt;math&gt;10&lt;/math&gt;&quot; with one more digit, and a &quot;&lt;math&gt;1&lt;/math&gt;&quot;. The missing number is one less than that, so the &quot;1&quot; will be subtracted from &lt;math&gt;f(k)&lt;/math&gt;. In other words, &lt;math&gt;f(k) = 25\cdot 10^{k-2}+1\cdot 10^{k/2}&lt;/math&gt;.<br /> <br /> Therefore:<br /> <br /> &lt;cmath&gt;f(2) =35 =25 +10&lt;/cmath&gt;<br /> &lt;cmath&gt;f(4) =2600 =2500 +100&lt;/cmath&gt;<br /> &lt;cmath&gt;f(6) =251000 =250000 +1000&lt;/cmath&gt;<br /> &lt;cmath&gt;f(8) = 25010000 = 25000000 + 10000&lt;/cmath&gt;<br /> <br /> And so on. The sum &lt;math&gt;f(2) + f(4) + f(6) +\dots + f(2016)&lt;/math&gt; is:<br /> <br /> &lt;math&gt;2.52525252525\dots 2525\cdot 10^{2015}&lt;/math&gt; + &lt;math&gt;1.11111\dots 110\cdot 10^{1008}&lt;/math&gt;, with &lt;math&gt;2016&lt;/math&gt; repetitions each of &quot;&lt;math&gt;25&lt;/math&gt;&quot; and &quot;&lt;math&gt;1&lt;/math&gt;&quot;. <br /> There is no carrying in this addition. Therefore each &lt;math&gt;f(k)&lt;/math&gt; adds &lt;math&gt;2 + 5 + 1 = 8&lt;/math&gt; to the sum of the digits. <br /> Since &lt;math&gt;2n = 2016&lt;/math&gt;, &lt;math&gt;n = 1008&lt;/math&gt;, and &lt;math&gt;8n = 8064&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(E)}\text{ 8064}}&lt;/math&gt;.<br /> <br /> Addendum: *You could also use the fact that &lt;cmath&gt;(x+1)^2 = x^2 +2x+1&lt;/cmath&gt; <br /> In other words, the difference between &lt;math&gt;x^2&lt;/math&gt; and &lt;math&gt;(x+1)^2&lt;/math&gt; is equal to &lt;math&gt;2x+1&lt;/math&gt;. We can set the inequality &lt;math&gt;2x+1 \geq 100&lt;/math&gt;. Obviously, the first integer &lt;math&gt;x&lt;/math&gt; that satisfies this is 50.<br /> This way, while being longer, is IMO more motivated and doesn't use calculus.<br /> <br /> ==Solution 2(Cheap Realization)==<br /> <br /> If you are one of those people who are willing take educated guesses, then just realize that &lt;math&gt;\textbf{(E)}\text{ 8064}&lt;/math&gt; is the only answer choice that is a multiple of &lt;math&gt;2016&lt;/math&gt;.<br /> <br /> ==See Also==<br /> Related Question: https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_6<br /> {{AMC12 box|year=2016|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=User_talk:Oreochocolate&diff=127573 User talk:Oreochocolate 2020-07-06T02:19:17Z <p>Oreochocolate: Added redirect.</p> <hr /> <div>#REDIRECT [[User:Oreochocolate]]</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_22&diff=127572 2020 AMC 12A Problems/Problem 22 2020-07-06T02:16:09Z <p>Oreochocolate: /* Solution 2 (DeMoivre's Formula) */ Revamped.</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;(a_n)&lt;/math&gt; and &lt;math&gt;(b_n)&lt;/math&gt; be the sequences of real numbers such that<br /> &lt;cmath&gt;$<br /> (2 + i)^n = a_n + b_ni<br />$&lt;/cmath&gt;for all integers &lt;math&gt;n\geq 0&lt;/math&gt;, where &lt;math&gt;i = \sqrt{-1}&lt;/math&gt;. What is&lt;cmath&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Square the given equality to yield<br /> &lt;cmath&gt;<br /> (3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,<br /> &lt;/cmath&gt;<br /> so &lt;math&gt;a_nb_n = \tfrac12\operatorname{Im}((3+4i)^n)&lt;/math&gt; and<br /> &lt;cmath&gt;<br /> \sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.<br /> &lt;/cmath&gt;<br /> <br /> == Solution 2 (DeMoivre's Formula) ==<br /> Note that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left(\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}i \right)&lt;/math&gt;. Let &lt;math&gt;\theta = \arctan (1/2)&lt;/math&gt;, then, we know that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left( \cos \theta + i\sin \theta \right)&lt;/math&gt;, so &lt;math&gt;(2+i)^n = (\cos (n \theta) + i\sin (n\theta))(\sqrt{5})^n&lt;/math&gt;. Therefore, &lt;math&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n} = \sum_{n=0}^\infty\frac{\cos(n\theta)\sin(n\theta) (5)^n}{7^n} =&lt;/math&gt; &lt;math&gt;\frac{1}{2}\sum_{n=0}^\infty \left( \frac{5}{7}\right)^n \sin (2n\theta) = \frac{1}{2} \text{im} \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right)&lt;/math&gt;. Aha &lt;math&gt;\sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} &lt;/math&gt; is a geometric sequence that evaluates to &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}}&lt;/math&gt;. We can quickly see that &lt;math&gt;\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5}&lt;/math&gt;. &lt;math&gt;\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}} = \frac{1}{1 - \frac{5}{7}\left( \frac{3}{5} + \frac{4}{5}i\right)} = \frac{7}{8} + \frac{7}{8}i&lt;/math&gt;. The imaginary part is &lt;math&gt;\frac{7}{8}&lt;/math&gt;, so our answer is &lt;math&gt;\frac{1}{2} \cdot \frac{7}{8} = \boxed{\frac{7}{16}} \Rightarrow \textbf{(B)}&lt;/math&gt;.<br /> <br /> ~AopsUser101, minor edit by vsamc stating that the answer choice is B, revamped by OreoChocolate<br /> <br /> ==Solution 3==<br /> Clearly &lt;math&gt;a_n=\tfrac{(2+i)^n+(2-i)^n}{2}, b_n=\tfrac{(2+i)^n-(2-i)^n}{2i}&lt;/math&gt;. So we have &lt;math&gt;\sum_{n\ge 0}\tfrac{a_nb_n}{7^n}=\sum_{n\ge 0}\tfrac{((2+i)^n+(2-i)^n))((2+i)^n-(2-i)^n))}{4i(7^n)}&lt;/math&gt;. By linearity, we have the latter is equivalent to &lt;math&gt;\tfrac{1}{4i}\sum_{n\ge 0}\tfrac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}&lt;/math&gt;. Expanding the summand yields &lt;math&gt;\tfrac{1}{4i}\sum_{n\ge 0}\tfrac{(3+4i)^n-(3-4i)^n}{7^n}=\tfrac{1}{4}[\tfrac{1}{1-(\tfrac{3+4i}{7})}-\tfrac{1}{1-(\tfrac{3-4i}{7})}]=\tfrac{1}{4i}[\tfrac{7}{7-(3+4i)}-\tfrac{7}{7-(3-4i)}]=\tfrac{1}{4}[\tfrac{7}{4-4i}-\tfrac{7}{4+4i}]=\tfrac{1}{4i}[\tfrac{7(4+4i)}{32}-\tfrac{7(4-4i)}{32}]=\tfrac{1}{4}\cdot \tfrac{56}{32}=\boxed{\tfrac{7}{16}}\textbf{(B)}&lt;/math&gt;<br /> -vsamc<br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_3&diff=124616 2020 AIME II Problems/Problem 3 2020-06-09T18:23:41Z <p>Oreochocolate: /* Solution 2 */ added initial</p> <hr /> <div>==Problem==<br /> <br /> The value of &lt;math&gt;x&lt;/math&gt; that satisfies &lt;math&gt;\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}&lt;/math&gt; can be written as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;\log _{2^x}3^{20}=\log _{2^{x+3}}3^{2020}=n&lt;/math&gt;. Based on the equation, we get &lt;math&gt;(2^x)^n=3^{20}&lt;/math&gt; and &lt;math&gt;(2^{x+3})^n=3^{2020}&lt;/math&gt;. Expanding the second equation, we get &lt;math&gt;8^n\cdot2^{xn}=3^{2020}&lt;/math&gt;. Substituting the first equation in, we get &lt;math&gt;8^n\cdot3^{20}=3^{2020}&lt;/math&gt;, so &lt;math&gt;8^n=3^{2000}&lt;/math&gt;. Taking the 100th root, we get &lt;math&gt;8^{\frac{n}{100}}=3^{20}&lt;/math&gt;. Therefore, &lt;math&gt;(2^{\frac{3}{100}})^n=3^{20}&lt;/math&gt;, so &lt;math&gt;n=\frac{3}{100}&lt;/math&gt; and the answer is &lt;math&gt;\boxed{103}&lt;/math&gt;.<br /> ~rayfish<br /> <br /> ==Easiest Solution==<br /> Recall the identity &lt;math&gt;\log_{a^n} b^{m} = \frac{m}{n}\log_{a} b &lt;/math&gt; (which is easily proven using exponents or change of base)<br /> Then this problem turns into &lt;cmath&gt;\frac{20}{x}\log_{2} 3 = \frac{2020}{x+3}\log_{2} 3&lt;/cmath&gt;<br /> Divide &lt;math&gt;\log_{2} 3&lt;/math&gt; from both sides. And we are left with &lt;math&gt;\frac{20}{x}=\frac{2020}{x+3}&lt;/math&gt;.Solving this simple equation we get &lt;cmath&gt;x = \tfrac{3}{100} \Rightarrow \boxed{103}&lt;/cmath&gt;<br /> ~mlgjeffdoge21<br /> <br /> ==Solution 2==<br /> Because &lt;math&gt;\log_a{b^c}=c\log_a{b},&lt;/math&gt; we have that &lt;math&gt;20\log_{2^x} 3 = 2020\log_{2^{x+3}} 3,&lt;/math&gt; or &lt;math&gt;\log_{2^x} 3 = 101\log_{2^{x+3}} 3.&lt;/math&gt; Since &lt;math&gt;\log_a{b}=\dfrac{1}{\log_b{a}},&lt;/math&gt; &lt;math&gt;\log_{2^x} 3=\dfrac{1}{\log_{3} 2^x},&lt;/math&gt; and &lt;math&gt;101\log_{2^{x+3}} 3=101\dfrac{1}{\log_{3}2^{x+3}},&lt;/math&gt; thus resulting in &lt;math&gt;\log_{3}2^{x+3}=101\log_{3} 2^x,&lt;/math&gt; or &lt;math&gt;\log_{3}2^{x+3}=\log_{3} 2^{101x}.&lt;/math&gt; We remove the base 3 logarithm and the power of 2 to yield &lt;math&gt;x+3=101x,&lt;/math&gt; or &lt;math&gt;x=\dfrac{3}{100}.&lt;/math&gt;<br /> <br /> Our answer is &lt;math&gt;\boxed{3+100=103}.&lt;/math&gt;<br /> ~ OreoChocolate<br /> <br /> ==Solution 3 (Official MAA)==<br /> Using the Change of Base Formula to convert the logarithms in the given equation to base &lt;math&gt;2&lt;/math&gt; yields<br /> &lt;cmath&gt;\frac{\log_2 3^{20}}{\log_2 2^x} = \frac{\log_2 3^{2020}}{\log_2 2^{x+3}}, \text{~ and then ~}<br /> \frac{20\log_2 3}{x\cdot\log_2 2} = \frac{2020\log_2 3}{(x+3)\log_2 2}.&lt;/cmath&gt;Canceling the logarithm factors then yields&lt;cmath&gt;\frac{20}x = \frac{2020}{x+3},&lt;/cmath&gt;which has solution &lt;math&gt;x = \frac3{100}.&lt;/math&gt; The requested sum is &lt;math&gt;3 + 100 = 103&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/lPr4fYEoXi0 ~ CNCM<br /> ==Video Solution 2==<br /> https://www.youtube.com/watch?v=x0QznvXcwHY?t=528<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> {{AIME box|year=2020|n=II|num-b=2|num-a=4}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_3&diff=124615 2020 AIME II Problems/Problem 3 2020-06-09T18:20:52Z <p>Oreochocolate: /* Solution 2 */ oops</p> <hr /> <div>==Problem==<br /> <br /> The value of &lt;math&gt;x&lt;/math&gt; that satisfies &lt;math&gt;\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}&lt;/math&gt; can be written as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;\log _{2^x}3^{20}=\log _{2^{x+3}}3^{2020}=n&lt;/math&gt;. Based on the equation, we get &lt;math&gt;(2^x)^n=3^{20}&lt;/math&gt; and &lt;math&gt;(2^{x+3})^n=3^{2020}&lt;/math&gt;. Expanding the second equation, we get &lt;math&gt;8^n\cdot2^{xn}=3^{2020}&lt;/math&gt;. Substituting the first equation in, we get &lt;math&gt;8^n\cdot3^{20}=3^{2020}&lt;/math&gt;, so &lt;math&gt;8^n=3^{2000}&lt;/math&gt;. Taking the 100th root, we get &lt;math&gt;8^{\frac{n}{100}}=3^{20}&lt;/math&gt;. Therefore, &lt;math&gt;(2^{\frac{3}{100}})^n=3^{20}&lt;/math&gt;, so &lt;math&gt;n=\frac{3}{100}&lt;/math&gt; and the answer is &lt;math&gt;\boxed{103}&lt;/math&gt;.<br /> ~rayfish<br /> <br /> ==Easiest Solution==<br /> Recall the identity &lt;math&gt;\log_{a^n} b^{m} = \frac{m}{n}\log_{a} b &lt;/math&gt; (which is easily proven using exponents or change of base)<br /> Then this problem turns into &lt;cmath&gt;\frac{20}{x}\log_{2} 3 = \frac{2020}{x+3}\log_{2} 3&lt;/cmath&gt;<br /> Divide &lt;math&gt;\log_{2} 3&lt;/math&gt; from both sides. And we are left with &lt;math&gt;\frac{20}{x}=\frac{2020}{x+3}&lt;/math&gt;.Solving this simple equation we get &lt;cmath&gt;x = \tfrac{3}{100} \Rightarrow \boxed{103}&lt;/cmath&gt;<br /> ~mlgjeffdoge21<br /> <br /> ==Solution 2==<br /> Because &lt;math&gt;\log_a{b^c}=c\log_a{b},&lt;/math&gt; we have that &lt;math&gt;20\log_{2^x} 3 = 2020\log_{2^{x+3}} 3,&lt;/math&gt; or &lt;math&gt;\log_{2^x} 3 = 101\log_{2^{x+3}} 3.&lt;/math&gt; Since &lt;math&gt;\log_a{b}=\dfrac{1}{\log_b{a}},&lt;/math&gt; &lt;math&gt;\log_{2^x} 3=\dfrac{1}{\log_{3} 2^x},&lt;/math&gt; and &lt;math&gt;101\log_{2^{x+3}} 3=101\dfrac{1}{\log_{3}2^{x+3}},&lt;/math&gt; thus resulting in &lt;math&gt;\log_{3}2^{x+3}=101\log_{3} 2^x,&lt;/math&gt; or &lt;math&gt;\log_{3}2^{x+3}=\log_{3} 2^{101x}.&lt;/math&gt; We remove the base 3 logarithm and the power of 2 to yield &lt;math&gt;x+3=101x,&lt;/math&gt; or &lt;math&gt;x=\dfrac{3}{100}.&lt;/math&gt;<br /> <br /> Our answer is &lt;math&gt;\boxed{3+100=103}.&lt;/math&gt;<br /> <br /> ==Solution 3 (Official MAA)==<br /> Using the Change of Base Formula to convert the logarithms in the given equation to base &lt;math&gt;2&lt;/math&gt; yields<br /> &lt;cmath&gt;\frac{\log_2 3^{20}}{\log_2 2^x} = \frac{\log_2 3^{2020}}{\log_2 2^{x+3}}, \text{~ and then ~}<br /> \frac{20\log_2 3}{x\cdot\log_2 2} = \frac{2020\log_2 3}{(x+3)\log_2 2}.&lt;/cmath&gt;Canceling the logarithm factors then yields&lt;cmath&gt;\frac{20}x = \frac{2020}{x+3},&lt;/cmath&gt;which has solution &lt;math&gt;x = \frac3{100}.&lt;/math&gt; The requested sum is &lt;math&gt;3 + 100 = 103&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/lPr4fYEoXi0 ~ CNCM<br /> ==Video Solution 2==<br /> https://www.youtube.com/watch?v=x0QznvXcwHY?t=528<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> {{AIME box|year=2020|n=II|num-b=2|num-a=4}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_3&diff=124614 2020 AIME II Problems/Problem 3 2020-06-09T18:18:05Z <p>Oreochocolate: added solution</p> <hr /> <div>==Problem==<br /> <br /> The value of &lt;math&gt;x&lt;/math&gt; that satisfies &lt;math&gt;\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}&lt;/math&gt; can be written as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;\log _{2^x}3^{20}=\log _{2^{x+3}}3^{2020}=n&lt;/math&gt;. Based on the equation, we get &lt;math&gt;(2^x)^n=3^{20}&lt;/math&gt; and &lt;math&gt;(2^{x+3})^n=3^{2020}&lt;/math&gt;. Expanding the second equation, we get &lt;math&gt;8^n\cdot2^{xn}=3^{2020}&lt;/math&gt;. Substituting the first equation in, we get &lt;math&gt;8^n\cdot3^{20}=3^{2020}&lt;/math&gt;, so &lt;math&gt;8^n=3^{2000}&lt;/math&gt;. Taking the 100th root, we get &lt;math&gt;8^{\frac{n}{100}}=3^{20}&lt;/math&gt;. Therefore, &lt;math&gt;(2^{\frac{3}{100}})^n=3^{20}&lt;/math&gt;, so &lt;math&gt;n=\frac{3}{100}&lt;/math&gt; and the answer is &lt;math&gt;\boxed{103}&lt;/math&gt;.<br /> ~rayfish<br /> <br /> ==Easiest Solution==<br /> Recall the identity &lt;math&gt;\log_{a^n} b^{m} = \frac{m}{n}\log_{a} b &lt;/math&gt; (which is easily proven using exponents or change of base)<br /> Then this problem turns into &lt;cmath&gt;\frac{20}{x}\log_{2} 3 = \frac{2020}{x+3}\log_{2} 3&lt;/cmath&gt;<br /> Divide &lt;math&gt;\log_{2} 3&lt;/math&gt; from both sides. And we are left with &lt;math&gt;\frac{20}{x}=\frac{2020}{x+3}&lt;/math&gt;.Solving this simple equation we get &lt;cmath&gt;x = \tfrac{3}{100} \Rightarrow \boxed{103}&lt;/cmath&gt;<br /> ~mlgjeffdoge21<br /> <br /> ==Solution 2==<br /> Because &lt;math&gt;\log_a{b^c}=c\log_a{b},&lt;/math&gt; we have that &lt;math&gt;20\log_{2^x} 3 = 2020\log_{2^{x+3}} 3,&lt;/math&gt; or &lt;math&gt;\log_{2^x} 3 = 101\log_{2^{x+3}} 3.&lt;/math&gt; Since &lt;math&gt;\log_a{b}=\dfrac{1}{\log_b{a},&lt;/math&gt; &lt;math&gt;\log_{2^x} 3=\dfrac{1}{\log_{3} 2^x},&lt;/math&gt; and &lt;math&gt;101\log_{2^{x+3}} 3=101\dfrac{1}{\log_{3}2^{x+3}},&lt;/math&gt; thus resulting in &lt;math&gt;\log_{3}2^{x+3}=101\log_{3} 2^x,&lt;/math&gt; or &lt;math&gt;\log_{3}2^{x+3}=\log_{3} 2^{101x}.&lt;/math&gt; We remove the base 3 logarithm and the power of 2 to yield &lt;math&gt;x+3=101x,&lt;/math&gt; or &lt;math&gt;x=\dfrac{3}{100}.&lt;/math&gt;<br /> <br /> Our answer is &lt;math&gt;\boxed{3+100=103}.&lt;/math&gt;<br /> ==Solution 3 (Official MAA)==<br /> Using the Change of Base Formula to convert the logarithms in the given equation to base &lt;math&gt;2&lt;/math&gt; yields<br /> &lt;cmath&gt;\frac{\log_2 3^{20}}{\log_2 2^x} = \frac{\log_2 3^{2020}}{\log_2 2^{x+3}}, \text{~ and then ~}<br /> \frac{20\log_2 3}{x\cdot\log_2 2} = \frac{2020\log_2 3}{(x+3)\log_2 2}.&lt;/cmath&gt;Canceling the logarithm factors then yields&lt;cmath&gt;\frac{20}x = \frac{2020}{x+3},&lt;/cmath&gt;which has solution &lt;math&gt;x = \frac3{100}.&lt;/math&gt; The requested sum is &lt;math&gt;3 + 100 = 103&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/lPr4fYEoXi0 ~ CNCM<br /> ==Video Solution 2==<br /> https://www.youtube.com/watch?v=x0QznvXcwHY?t=528<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> {{AIME box|year=2020|n=II|num-b=2|num-a=4}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems&diff=124474 2020 AIME II Problems 2020-06-08T18:40:51Z <p>Oreochocolate: Page does not exist. Yet.</p> <hr /> <div>{{AIME Problems|year=2020|n=II}}<br /> <br /> ==Problem 1==<br /> <br /> Find the number of ordered pairs of positive integers &lt;math&gt;(m,n)&lt;/math&gt; such that &lt;math&gt;{m^2n = 20 ^{20}}&lt;/math&gt;.<br /> <br /> [[2020 AIME II Problems/Problem 1 | Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Let &lt;math&gt;P&lt;/math&gt; be a point chosen uniformly at random in the interior of the unit square with vertices at &lt;math&gt;(0,0), (1,0), (1,1)&lt;/math&gt;, and &lt;math&gt;(0,1)&lt;/math&gt;. The probability that the slope of the line determined by &lt;math&gt;P&lt;/math&gt; and the point &lt;math&gt;\left(\frac58, \frac38 \right)&lt;/math&gt; is greater than &lt;math&gt;\frac12&lt;/math&gt; can be written as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2020 AIME II Problems/Problem 2 | Solution]]<br /> <br /> ==Problem 3==<br /> <br /> The value of &lt;math&gt;x&lt;/math&gt; that satisfies &lt;math&gt;\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}&lt;/math&gt; can be written as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2020 AIME II Problems/Problem 3 | Solution]]<br /> <br /> ==Problem 4==<br /> Triangles &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle A'B'C'&lt;/math&gt; lie in the coordinate plane with vertices &lt;math&gt;A(0,0)&lt;/math&gt;, &lt;math&gt;B(0,12)&lt;/math&gt;, &lt;math&gt;C(16,0)&lt;/math&gt;, &lt;math&gt;A'(24,18)&lt;/math&gt;, &lt;math&gt;B'(36,18)&lt;/math&gt;, &lt;math&gt;C'(24,2)&lt;/math&gt;. A rotation of &lt;math&gt;m&lt;/math&gt; degrees clockwise around the point &lt;math&gt;(x,y)&lt;/math&gt; where &lt;math&gt;0&lt;m&lt;180&lt;/math&gt;, will transform &lt;math&gt;\triangle ABC&lt;/math&gt; to &lt;math&gt;\triangle A'B'C'&lt;/math&gt;. Find &lt;math&gt;m+x+y&lt;/math&gt;.<br /> <br /> [[2020 AIME II Problems/Problem 4 | Solution]]<br /> <br /> ==Problem 5==<br /> <br /> For each positive integer &lt;math&gt;n&lt;/math&gt;, left &lt;math&gt;f(n)&lt;/math&gt; be the sum of the digits in the base-four representation of &lt;math&gt;n&lt;/math&gt; and let &lt;math&gt;g(n)&lt;/math&gt; be the sum of the digits in the base-eight representation of &lt;math&gt;f(n)&lt;/math&gt;. For example, &lt;math&gt;f(2020) = f(133210_{\text{four}}) = 10 = 12_{\text{eight}}&lt;/math&gt;, and &lt;math&gt;g(2020) = \text{the digit sum of }12_{\text{eight}} = 3&lt;/math&gt;. Let &lt;math&gt;N&lt;/math&gt; be the least value of &lt;math&gt;n&lt;/math&gt; such that the base-sixteen representation of &lt;math&gt;g(n)&lt;/math&gt; cannot be expressed using only the digits &lt;math&gt;0&lt;/math&gt; through &lt;math&gt;9&lt;/math&gt;. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> [[2020 AIME II Problems/Problem 5 | Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Define a sequence recursively by &lt;math&gt;t_1 = 20&lt;/math&gt;, &lt;math&gt;t_2 = 21&lt;/math&gt;, and&lt;cmath&gt;t_n = \frac{5t_{n-1}+1}{25t_{n-2}}&lt;/cmath&gt;for all &lt;math&gt;n \ge 3&lt;/math&gt;. Then &lt;math&gt;t_{2020}&lt;/math&gt; can be written as &lt;math&gt;\frac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> [[2020 AIME II Problems/Problem 6 | Solution]]<br /> <br /> ==Problem 7==<br /> Two congruent right circular cones each with base radius &lt;math&gt;3&lt;/math&gt; and height &lt;math&gt;8&lt;/math&gt; have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance &lt;math&gt;3&lt;/math&gt; from the base of each cone. A sphere with radius &lt;math&gt;r&lt;/math&gt; lies withing both cones. The maximum possible value of &lt;math&gt;r^2&lt;/math&gt; is &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2020 AIME II Problems/Problem 7 | Solution]]<br /> <br /> ==Problem 8==<br /> Define a sequence recursively by &lt;math&gt;f_1(x)=|x-1|&lt;/math&gt; and &lt;math&gt;f_n(x)=f_{n-1}(|x-n|)&lt;/math&gt; for integers &lt;math&gt;n&gt;1&lt;/math&gt;. Find the least value of &lt;math&gt;n&lt;/math&gt; such that the sum of the zeros of &lt;math&gt;f_n&lt;/math&gt; exceeds &lt;math&gt;500,000&lt;/math&gt;.<br /> <br /> [[2020 AIME II Problems/Problem 8 | Solution]]<br /> <br /> ==Problem 9==<br /> While watching a show, Ayako, Billy, Carlos, Dahlia, Ehuang, and Frank sat in that order in a row of six chairs. During the break, they went to the kitchen for a snack. When they came back, they sat on those six chairs in such a way that if two of them sat next to each other before the break, then they did not sit next to each other after the break. Find the number of possible seating orders they could have chosen after the break.<br /> <br /> [[2020 AIME II Problems/Problem 9 | Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that when &lt;math&gt;1^3+2^3+3^3+\cdots +n^3&lt;/math&gt; is divided by &lt;math&gt;n+5&lt;/math&gt;, the remainder is &lt;math&gt;17&lt;/math&gt;.<br /> <br /> [[2020 AIME II Problems/Problem 10 | Solution]]<br /> <br /> ==Problem 11==<br /> <br /> Let &lt;math&gt;P(x) = x^2 - 3x - 7&lt;/math&gt;, and let &lt;math&gt;Q(x)&lt;/math&gt; and &lt;math&gt;R(x)&lt;/math&gt; be two quadratic polynomials also with the coefficient of &lt;math&gt;x^2&lt;/math&gt; equal to &lt;math&gt;1&lt;/math&gt;. David computes each of the three sums &lt;math&gt;P + Q&lt;/math&gt;, &lt;math&gt;P + R&lt;/math&gt;, and &lt;math&gt;Q + R&lt;/math&gt; and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If &lt;math&gt;Q(0) = 2&lt;/math&gt;, then &lt;math&gt;R(0) = \frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> [[2020 AIME II Problems/Problem 11 | Solution]]<br /> <br /> ==Problem 12==<br /> <br /> Let &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; be odd integers greater than &lt;math&gt;1.&lt;/math&gt; An &lt;math&gt;m\times n&lt;/math&gt; rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;n&lt;/math&gt;, those in the second row are numbered left to right with the integers &lt;math&gt;n + 1&lt;/math&gt; through &lt;math&gt;2n&lt;/math&gt;, and so on. Square &lt;math&gt;200&lt;/math&gt; is in the top row, and square &lt;math&gt;2000&lt;/math&gt; is in the bottom row. Find the number of ordered pairs &lt;math&gt;(m,n)&lt;/math&gt; of odd integers greater than &lt;math&gt;1&lt;/math&gt; with the property that, in the &lt;math&gt;m\times n&lt;/math&gt; rectangle, the line through the centers of squares &lt;math&gt;200&lt;/math&gt; and &lt;math&gt;2000&lt;/math&gt; intersects the interior of square &lt;math&gt;1099&lt;/math&gt;.<br /> <br /> [[2020 AIME II Problems/Problem 12 | Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Convex pentagon &lt;math&gt;ABCDE&lt;/math&gt; has side lengths &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=CD=DE=6&lt;/math&gt;, and &lt;math&gt;EA=7&lt;/math&gt;. Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of &lt;math&gt;ABCDE&lt;/math&gt;.<br /> <br /> [[2020 AIME II Problems/Problem 13 | Solution]]<br /> <br /> ==Problem 14==<br /> <br /> For real number &lt;math&gt;x&lt;/math&gt; let &lt;math&gt;\lfloor x\rfloor&lt;/math&gt; be the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;, and define &lt;math&gt;\{x\} = x - \lfloor x \rfloor&lt;/math&gt; to be the fractional part of &lt;math&gt;x&lt;/math&gt;. For example, &lt;math&gt;\{3\} = 0&lt;/math&gt; and &lt;math&gt;\{4.56\} = 0.56&lt;/math&gt;. Define &lt;math&gt;f(x)=x\{x\}&lt;/math&gt;, and let &lt;math&gt;N&lt;/math&gt; be the number of real-valued solutions to the equation &lt;math&gt;f(f(f(x)))=17&lt;/math&gt; for &lt;math&gt;0\leq x\leq 2020&lt;/math&gt;. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> [[2020 AIME II Problems/Problem 14 | Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an acute scalene triangle with circumcircle &lt;math&gt;\omega&lt;/math&gt;. The tangents to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; intersect at &lt;math&gt;T&lt;/math&gt;. Let &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt; be the projections of &lt;math&gt;T&lt;/math&gt; onto lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;, respectively. Suppose &lt;math&gt;BT = CT = 16&lt;/math&gt;, &lt;math&gt;BC = 22&lt;/math&gt;, and &lt;math&gt;TX^2 + TY^2 + XY^2 = 1143&lt;/math&gt;. Find &lt;math&gt;XY^2&lt;/math&gt;.<br /> <br /> [[2020 AIME II Problems/Problem 15 | Solution]]<br /> <br /> {{AIME box|year=2020|n=II|before=[[2020 AIME I Problems]]|after=2021 AIME I Problems}}<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_11&diff=122837 2020 AIME I Problems/Problem 11 2020-05-22T19:43:28Z <p>Oreochocolate: /* Notes For * */ For bold, italic, underlining, or Asymptote, instead of using brackets, like usual, use &lt;&gt; instead, but only works here.</p> <hr /> <div><br /> == Problem ==<br /> For integers &lt;math&gt;a,b,c&lt;/math&gt; and &lt;math&gt;d,&lt;/math&gt; let &lt;math&gt;f(x)=x^2+ax+b&lt;/math&gt; and &lt;math&gt;g(x)=x^2+cx+d.&lt;/math&gt; Find the number of ordered triples &lt;math&gt;(a,b,c)&lt;/math&gt; of integers with absolute values not exceeding &lt;math&gt;10&lt;/math&gt; for which there is an integer &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;g(f(2))=g(f(4))=0.&lt;/math&gt;<br /> <br /> == Solution 1 (Strategic Casework)==<br /> Either &lt;math&gt;f(2)=f(4)&lt;/math&gt; or not. If it is, note that Vieta's forces &lt;math&gt;a = -6&lt;/math&gt;. Then, &lt;math&gt;b&lt;/math&gt; can be anything. However, &lt;math&gt;c&lt;/math&gt; can also be anything, as we can set the root of &lt;math&gt;g&lt;/math&gt; (not equal to &lt;math&gt;f(2) = f(4)&lt;/math&gt;) to any integer, producing a possible integer value of &lt;math&gt;d&lt;/math&gt;. Therefore there are &lt;math&gt;21^2 = 441&lt;/math&gt; in this case*. If it isn't, then &lt;math&gt;f(2),f(4)&lt;/math&gt; are the roots of &lt;math&gt;g&lt;/math&gt;. This means by Vieta's, that:<br /> <br /> &lt;cmath&gt;f(2)+f(4) = -c \in [-10,10]&lt;/cmath&gt;<br /> &lt;cmath&gt;20 + 6a + 2b \in [-10,10]&lt;/cmath&gt;<br /> &lt;cmath&gt;3a + b \in [-15,-5].&lt;/cmath&gt;<br /> <br /> Solving these inequalities while considering that &lt;math&gt;a \neq -6&lt;/math&gt; to prevent &lt;math&gt;f(2) = f(4)&lt;/math&gt;, we obtain &lt;math&gt;69&lt;/math&gt; possible tuples and adding gives &lt;math&gt;441+69=\boxed{510}&lt;/math&gt;. <br /> ~awang11<br /> <br /> == Solution 2 (Bash) ==<br /> Define &lt;math&gt;h(x)=x^2+cx&lt;/math&gt;. Since &lt;math&gt;g(f(2))=g(f(4))=0&lt;/math&gt;, we know &lt;math&gt;h(f(2))=h(f(4))=-d&lt;/math&gt;. Plugging in &lt;math&gt;f(x)&lt;/math&gt; into &lt;math&gt;h(x)&lt;/math&gt;, we get &lt;math&gt;h(f(x))=x^4+2ax^3+(2b+a^2+c)x^2+(2ab+ac)x+(b^2+bc)&lt;/math&gt;. Setting &lt;math&gt;h(f(2))=h(f(4))&lt;/math&gt;, &lt;cmath&gt;16+16a+8b+4a^2+4ab+b^2+4c+2ac+bc=256+128a+32b+16a^2+8ab+b^2+16c+4ac+bc&lt;/cmath&gt;. Simplifying and cancelling terms, &lt;cmath&gt;240+112a+24b+12a^2+4ab+12c+2ac=0&lt;/cmath&gt; &lt;cmath&gt;120+56a+12b+6a^2+2ab+6c+ac=0&lt;/cmath&gt; &lt;cmath&gt;6a^2+2ab+ac+56a+12b+6c+120=0&lt;/cmath&gt; &lt;cmath&gt;6a^2+2ab+ac+20a+36a+12b+6c+120=0&lt;/cmath&gt; &lt;cmath&gt;a(6a+2b+c+20)+6(6a+2b+c+20)=0&lt;/cmath&gt; &lt;cmath&gt;(a+6)(6a+2b+c+20)=0&lt;/cmath&gt;<br /> <br /> Therefore, either &lt;math&gt;a+6=0&lt;/math&gt; or &lt;math&gt;6a+2b+c=-20&lt;/math&gt;. The first case is easy: &lt;math&gt;a=-6&lt;/math&gt; and there are &lt;math&gt;441&lt;/math&gt; tuples in that case. In the second case, we simply perform casework on even values of &lt;math&gt;c&lt;/math&gt;, to get &lt;math&gt;77&lt;/math&gt; tuples, subtracting the &lt;math&gt;8&lt;/math&gt; tuples in both cases we get &lt;math&gt;441+77-8=\boxed{510}&lt;/math&gt;.<br /> <br /> -EZmath2006<br /> <br /> == Notes For * ==<br /> In case anyone is confused by this (as I initially was). In the case where &lt;math&gt;f(2)=f(4)&lt;/math&gt;, this does not mean that g has a double root of &lt;math&gt;f(2)=f(4)=c&lt;/math&gt;, ONLY that &lt;math&gt;c&lt;/math&gt; is one of the roots of g. So basically since &lt;math&gt;a=-6&lt;/math&gt; in this case, &lt;math&gt;f(2)=f(4)=b-8&lt;/math&gt;, and we have &lt;math&gt;21&lt;/math&gt; choices for b and we &lt;i&gt;still can&lt;/i&gt; ensure c is an integer with absolute value less than or equal to 10 simply by having another integer root of g that when added to &lt;math&gt;b-8&lt;/math&gt; ensures this, and of course an integer multiplied by an integer is an integer so &lt;math&gt;d&lt;/math&gt; will still be an integer. In other words, you have can have &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; be any integer with absolute value less than or equal to 10 with &lt;math&gt;d&lt;/math&gt; still being an integer. Now refer back to the 1st solution.<br /> ~First<br /> <br /> ==See Also==<br /> <br /> {{AIME box|year=2020|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_35&diff=121385 1964 AHSME Problems/Problem 35 2020-04-21T17:20:58Z <p>Oreochocolate: /* Solution 2 (coordinates) */ oops. i forgot some stuff.</p> <hr /> <div>==Problem==<br /> <br /> The sides of a triangle are of lengths &lt;math&gt;13&lt;/math&gt;, &lt;math&gt;14&lt;/math&gt;, and &lt;math&gt;15&lt;/math&gt;. The altitudes of the triangle meet at point &lt;math&gt;H&lt;/math&gt;. if &lt;math&gt;AD&lt;/math&gt; is the altitude to the side of length &lt;math&gt;14&lt;/math&gt;, the ratio &lt;math&gt;HD:HA&lt;/math&gt; is:<br /> <br /> &lt;math&gt;\textbf{(A) }3:11\qquad\textbf{(B) }5:11\qquad\textbf{(C) }1:2\qquad\textbf{(D) }2:3\qquad \textbf{(E) }25:33&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using Law of Cosines<br /> and the fact that the ratio equals cos(a)/[cos(b)cos(c)]<br /> B 5:11<br /> <br /> ==Solution 2 (coordinates)==<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(15,0)--(6.6,11.2)--(0,0));<br /> draw((0,0)--(9.6,7.2));<br /> draw((6.6,0)--(6.6,11.2));<br /> draw((15,0)--(3267/845,5544/845));<br /> label(&quot;$B$&quot;,(15,0),SE);<br /> label(&quot;$C$&quot;,(6.6,11.2),N);<br /> label(&quot;$E$&quot;,(6.6,0),S);<br /> label(&quot;$15$&quot;,(7.5,-0.75),S);<br /> label(&quot;$14$&quot;,(11,5.75),ENE);<br /> label(&quot;$13$&quot;,(3,6),WNW);<br /> label(&quot;$A$&quot;,(0,0),SW);<br /> label(&quot;$D$&quot;,(9.6,7.2),NE);<br /> label(&quot;$H$&quot;,(6.6,3.5),E);<br /> &lt;/asy&gt;<br /> The reason why we have &lt;math&gt;HD&lt;/math&gt; shorter than &lt;math&gt;HA&lt;/math&gt; is that all the ratios' left hand side (&lt;math&gt;HD&lt;/math&gt;) is less than the ratios' right hand side (&lt;math&gt;HA&lt;/math&gt;).<br /> <br /> We label point &lt;math&gt;A&lt;/math&gt; as the origin and point &lt;math&gt;B&lt;/math&gt;, logically, as &lt;math&gt;(15,0)&lt;/math&gt;. By Heron's Formula, the area of this triangle is &lt;math&gt;84.&lt;/math&gt; Thus the height perpendicular to &lt;math&gt;AB&lt;/math&gt; has a length of &lt;math&gt;11.2,&lt;/math&gt; and by the Pythagorean Theorem, &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;EB&lt;/math&gt; have lengths &lt;math&gt;6.6&lt;/math&gt; and &lt;math&gt;8.4,&lt;/math&gt; respectively. These lengths tell us that &lt;math&gt;C&lt;/math&gt; is at &lt;math&gt;(6.6,11.2)&lt;/math&gt;.<br /> <br /> The slope of &lt;math&gt;BC&lt;/math&gt; is &lt;math&gt;\dfrac{0-11.2}{15-6.6}=-\dfrac{4}{3},&lt;/math&gt; and the slope of &lt;math&gt;AD&lt;/math&gt; is &lt;math&gt;\dfrac{3}{4}&lt;/math&gt; by taking the negative reciprocal of &lt;math&gt;-\dfrac{4}{3}.&lt;/math&gt; Therefore, the equation of line &lt;math&gt;AD&lt;/math&gt; can best be represented by &lt;math&gt;y=\dfrac{3}{4}x.&lt;/math&gt;<br /> <br /> We next find the intersection of &lt;math&gt;CE&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;. We automatically know the &lt;math&gt;x&lt;/math&gt;-value; it is just &lt;math&gt;6.6&lt;/math&gt; because &lt;math&gt;CE&lt;/math&gt; is a straight line hitting &lt;math&gt;(6.6,0).&lt;/math&gt; Therefore, the &lt;math&gt;y&lt;/math&gt;-value is at &lt;math&gt;\dfrac{3}{4}\times 6.6=4.95.&lt;/math&gt; Therefore, the intersection between &lt;math&gt;CE&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; is at &lt;math&gt;(6.6,4.95)&lt;/math&gt;.<br /> <br /> We also need to find the intersection between &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;. To do that, we know that the line of &lt;math&gt;AD&lt;/math&gt; is represented as &lt;math&gt;y=\dfrac{3}{4}x,&lt;/math&gt; and the slope of line &lt;math&gt;BC&lt;/math&gt; is &lt;math&gt;-\dfrac{4}{3}.&lt;/math&gt; We just need to find line &lt;math&gt;BC&lt;/math&gt;'s y-intercept. So far, we have &lt;math&gt;y=-\dfrac{4}{3}x+b,&lt;/math&gt; where &lt;math&gt;b&lt;/math&gt; is a real y-intercept. We know that &lt;math&gt;B&lt;/math&gt; is located at &lt;math&gt;(15,0),&lt;/math&gt; so we plug that into the equation and yield &lt;math&gt;b=20.&lt;/math&gt; Therefore, the intersection between the two lines is<br /> &lt;cmath&gt;<br /> \dfrac{3}{4}x=-\dfrac{4}{3}x+20,<br /> 9x=-16x+240,<br /> 25x=240,<br /> x=9.6,<br /> y=\dfrac{3}{4}\times 9.6,<br /> y=7.2.<br /> &lt;/cmath&gt;<br /> After that, we use the distance formula: &lt;math&gt;HA&lt;/math&gt; has a length of &lt;cmath&gt;\sqrt{(6.6-0)^2+(4.95-0)^2}=\sqrt{\dfrac{1089}{25}+\dfrac{9801}{400}}=\sqrt{\dfrac{1089*16+9801}{400}}=\sqrt{\dfrac{27225}{400}}=\sqrt{\dfrac{1089}{16}}=\dfrac{33}{4}=8.25,&lt;/cmath&gt; and &lt;math&gt;HD&lt;/math&gt; has a length of &lt;cmath&gt;\sqrt{(9.6-6.6)^2+(7.2-4.95)^2}=\sqrt{3^2+(\dfrac{36}{5}-\dfrac{99}{20})^2}=\sqrt{9+\dfrac{81}{16}}=\sqrt{\dfrac{225}{16}}=3.75.&lt;/cmath&gt;<br /> Thus, we have that &lt;math&gt;\dfrac{3.75}{8.25}=\dfrac{\frac{15}{4}}{\frac{33}{4}}=\dfrac{15}{33}=\dfrac{5}{11}=\boxed{\bold{B}}.&lt;/math&gt;-OreoChocolate<br /> <br /> ==See Also==<br /> {{AHSME 40p box|year=1964|num-b=34|num-a=36}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> <br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_35&diff=121384 1964 AHSME Problems/Problem 35 2020-04-21T17:19:41Z <p>Oreochocolate: added another possibly time-consuming but good solution, i guess</p> <hr /> <div>==Problem==<br /> <br /> The sides of a triangle are of lengths &lt;math&gt;13&lt;/math&gt;, &lt;math&gt;14&lt;/math&gt;, and &lt;math&gt;15&lt;/math&gt;. The altitudes of the triangle meet at point &lt;math&gt;H&lt;/math&gt;. if &lt;math&gt;AD&lt;/math&gt; is the altitude to the side of length &lt;math&gt;14&lt;/math&gt;, the ratio &lt;math&gt;HD:HA&lt;/math&gt; is:<br /> <br /> &lt;math&gt;\textbf{(A) }3:11\qquad\textbf{(B) }5:11\qquad\textbf{(C) }1:2\qquad\textbf{(D) }2:3\qquad \textbf{(E) }25:33&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using Law of Cosines<br /> and the fact that the ratio equals cos(a)/[cos(b)cos(c)]<br /> B 5:11<br /> <br /> ==Solution 2 (coordinates)==<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(15,0)--(6.6,11.2)--(0,0));<br /> draw((0,0)--(9.6,7.2));<br /> draw((6.6,0)--(6.6,11.2));<br /> draw((15,0)--(3267/845,5544/845));<br /> label(&quot;$B$&quot;,(15,0),SE);<br /> label(&quot;$C$&quot;,(6.6,11.2),N);<br /> label(&quot;$E$&quot;,(6.6,0),S);<br /> label(&quot;$15$&quot;,(7.5,-0.75),S);<br /> label(&quot;$14$&quot;,(11,5.75),ENE);<br /> label(&quot;$13$&quot;,(3,6),WNW);<br /> label(&quot;$A$&quot;,(0,0),SW);<br /> label(&quot;$D$&quot;,(9.6,7.2),NE);<br /> label(&quot;$H$&quot;,(6.6,3.5),E);<br /> &lt;/asy&gt;<br /> The reason why we have &lt;math&gt;HD&lt;/math&gt; shorter than &lt;math&gt;HA&lt;/math&gt; is that all the ratios' left hand side (&lt;math&gt;HD&lt;/math&gt;) is less than the ratios' right hand side (&lt;math&gt;HA&lt;/math&gt;).<br /> <br /> We label point &lt;math&gt;A&lt;/math&gt; as the origin and point &lt;math&gt;B&lt;/math&gt;, logically, as &lt;math&gt;(15,0)&lt;/math&gt;. By Heron's Formula, the area of this triangle is &lt;math&gt;84.&lt;/math&gt; Thus the height perpendicular to &lt;math&gt;AB&lt;/math&gt; has a length of &lt;math&gt;11.2,&lt;/math&gt; and by the Pythagorean Theorem, &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;EB&lt;/math&gt; have lengths &lt;math&gt;6.6&lt;/math&gt; and &lt;math&gt;8.4,&lt;/math&gt; respectively. These lengths tell us that &lt;math&gt;C&lt;/math&gt; is at &lt;math&gt;(6.6,11.2)&lt;/math&gt;.<br /> <br /> The slope of &lt;math&gt;BC&lt;/math&gt; is &lt;math&gt;\dfrac{0-11.2}{15-6.6}=-\dfrac{4}{3},&lt;/math&gt; and the slope of &lt;math&gt;AD&lt;/math&gt; is &lt;math&gt;\dfrac{3}{4}&lt;/math&gt; by taking the negative reciprocal of &lt;math&gt;-\dfrac{4}{3}.&lt;/math&gt; Therefore, the equation of line &lt;math&gt;AD&lt;/math&gt; can best be represented by &lt;math&gt;y=\dfrac{3}{4}x.&lt;/math&gt;<br /> <br /> We next find the intersection of &lt;math&gt;CE&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;. We automatically know the &lt;math&gt;x&lt;/math&gt;-value; it is just &lt;math&gt;6.6&lt;/math&gt; because &lt;math&gt;CE&lt;/math&gt; is a straight line hitting &lt;math&gt;(6.6,0).&lt;/math&gt; Therefore, the &lt;math&gt;y&lt;/math&gt;-value is at &lt;math&gt;\dfrac{3}{4}\times 6.6=4.95.&lt;/math&gt; Therefore, the intersection between &lt;math&gt;CE&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; is at &lt;math&gt;(6.6,4.95)&lt;/math&gt;.<br /> <br /> We also need to find the intersection between &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;. To do that, we know that the line of &lt;math&gt;AD&lt;/math&gt; is represented as &lt;math&gt;y=\dfrac{3}{4}x,&lt;/math&gt; and the slope of line &lt;math&gt;BC&lt;/math&gt; is &lt;math&gt;-\dfrac{4}{3}.&lt;/math&gt; We just need to find line &lt;math&gt;BC&lt;/math&gt;'s y-intercept. So far, we have &lt;math&gt;y=-\dfrac{4}{3}x+b,&lt;/math&gt; where &lt;math&gt;b&lt;/math&gt; is a real y-intercept. We know that &lt;math&gt;B&lt;/math&gt; is located at &lt;math&gt;(15,0),&lt;/math&gt; so we plug that into the equation and yield &lt;math&gt;b=20.&lt;/math&gt; Therefore, the intersection between the two lines is<br /> &lt;cmath&gt;<br /> \dfrac{3}{4}x=-\dfrac{4}{3}x+20,<br /> 9x=-16x+240,<br /> 25x=240,<br /> x=9.6,<br /> y=\dfrac{3}{4}\times 9.6,<br /> y=7.2.<br /> &lt;/cmath&gt;<br /> After that, we use the distance formula: &lt;math&gt;HA&lt;/math&gt; has a length of &lt;cmath&gt;\sqrt{(6.6-0)^2+(4.95-0)^2}=\sqrt{\dfrac{1089}{25}+\dfrac{9801}{400}}=\sqrt{\dfrac{1089*16+9801}{400}}=\sqrt{\dfrac{27225}{400}}=\sqrt{\dfrac{1089}{16}}=\dfrac{33}{4}=8.25,&lt;/cmath&gt; and &lt;math&gt;HD&lt;/math&gt; has a length of &lt;cmath&gt;\sqrt{(9.6-6.6)^2+(7.2-4.95)^2}=\sqrt{3^2+(\dfrac{36}{5}-\dfrac{99}{20})^2}=\sqrt{9+\dfrac{81}{16}}=\sqrt{\dfrac{225}{16}}=3.75.&lt;/cmath&gt;<br /> Thus, we have that &lt;math&gt;\dfrac{3.75}{8.25}=\dfrac{\frac{15}{4}}{\frac{33}{4}}=\dfrac{15}{33}=\dfrac{5}{11}=\bold{\boxed{B}}.&lt;/math&gt;<br /> ==See Also==<br /> {{AHSME 40p box|year=1964|num-b=34|num-a=36}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> <br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_19&diff=121334 2020 AMC 10A Problems/Problem 19 2020-04-21T14:30:38Z <p>Oreochocolate: Undo revision 121153 by Einsteinstudent (talk) Um...hmmmm, seems unnecessary to do that...</p> <hr /> <div>== Problem ==<br /> As shown in the figure below, a regular dodecahedron (the polyhedron consisting of &lt;math&gt;12&lt;/math&gt; congruent regular pentagonal faces) floats in space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?<br /> <br /> &lt;math&gt;\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810&lt;/math&gt;<br /> <br /> == Diagram ==<br /> &lt;asy&gt;<br /> import graph;<br /> unitsize(5cm);<br /> pair A = (0.082, 0.378);<br /> pair B = (0.091, 0.649);<br /> pair C = (0.249, 0.899);<br /> pair D = (0.479, 0.939);<br /> pair E = (0.758, 0.893);<br /> pair F = (0.862, 0.658);<br /> pair G = (0.924, 0.403);<br /> pair H = (0.747, 0.194);<br /> pair I = (0.526, 0.075);<br /> pair J = (0.251, 0.170);<br /> pair K = (0.568, 0.234);<br /> pair L = (0.262, 0.449);<br /> pair M = (0.373, 0.813);<br /> pair N = (0.731, 0.813);<br /> pair O = (0.851, 0.461);<br /> path[] f;<br /> f = A--B--C--M--L--cycle;<br /> f = C--D--E--N--M--cycle;<br /> f = E--F--G--O--N--cycle;<br /> f = G--H--I--K--O--cycle;<br /> f = I--J--A--L--K--cycle;<br /> f = K--L--M--N--O--cycle;<br /> draw(f);<br /> axialshade(f, white, M, gray(0.5), (C+2*D)/3);<br /> draw(f);<br /> filldraw(f, gray);<br /> filldraw(f, gray);<br /> axialshade(f, white, L, gray(0.7), J);<br /> draw(f);<br /> draw(f);<br /> &lt;/asy&gt;<br /> <br /> == Solution 1 ==<br /> Since we start at the top face and end at the bottom face without moving from the lower ring to the upper ring or revisiting a face, our journey must consist of the top face, a series of faces in the upper ring, a series of faces in the lower ring, and the bottom face, in that order.<br /> <br /> We have &lt;math&gt;5&lt;/math&gt; choices for which face we visit first on the top ring. From there, we have &lt;math&gt;9&lt;/math&gt; choices for how far around the top ring we go before moving down: &lt;math&gt;1,2,3,&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; faces around clockwise, &lt;math&gt;1,2,3,&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; faces around counterclockwise, or immediately going down to the lower ring without visiting any other faces in the upper ring.<br /> <br /> We then have &lt;math&gt;2&lt;/math&gt; choices for which lower ring face to visit first (since every upper-ring face is adjacent to exactly &lt;math&gt;2&lt;/math&gt; lower-ring faces) and then once again &lt;math&gt;9&lt;/math&gt; choices for how to travel around the lower ring. We then proceed to the bottom face, completing the trip.<br /> <br /> Multiplying together all the numbers of choices we have, we get &lt;math&gt;5 \cdot 9 \cdot 2 \cdot 9 = \boxed{\textbf{(E) } 810}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Swap the faces as vertices and the vertices as faces. Then, this problem is the same as<br /> [https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3 2016 AIME I #3]<br /> which had an answer of &lt;math&gt;\boxed{\textbf{(E) } 810}&lt;/math&gt;.<br /> &lt;math&gt;\textbf{\textbf{- Emathmaster}}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3<br /> {{AMC10 box|year=2020|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_19&diff=121151 2020 AMC 10A Problems/Problem 19 2020-04-19T18:08:05Z <p>Oreochocolate: extra gap there</p> <hr /> <div>== Problem ==<br /> As shown in the figure below, a regular dodecahedron (the polyhedron consisting of &lt;math&gt;12&lt;/math&gt; congruent regular pentagonal faces) floats in space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?<br /> <br /> &lt;math&gt;\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810&lt;/math&gt;<br /> <br /> == Diagram ==<br /> &lt;asy&gt;<br /> import graph;<br /> unitsize(5cm);<br /> pair A = (0.082, 0.378);<br /> pair B = (0.091, 0.649);<br /> pair C = (0.249, 0.899);<br /> pair D = (0.479, 0.939);<br /> pair E = (0.758, 0.893);<br /> pair F = (0.862, 0.658);<br /> pair G = (0.924, 0.403);<br /> pair H = (0.747, 0.194);<br /> pair I = (0.526, 0.075);<br /> pair J = (0.251, 0.170);<br /> pair K = (0.568, 0.234);<br /> pair L = (0.262, 0.449);<br /> pair M = (0.373, 0.813);<br /> pair N = (0.731, 0.813);<br /> pair O = (0.851, 0.461);<br /> path[] f;<br /> f = A--B--C--M--L--cycle;<br /> f = C--D--E--N--M--cycle;<br /> f = E--F--G--O--N--cycle;<br /> f = G--H--I--K--O--cycle;<br /> f = I--J--A--L--K--cycle;<br /> f = K--L--M--N--O--cycle;<br /> draw(f);<br /> axialshade(f, white, M, gray(0.5), (C+2*D)/3);<br /> draw(f);<br /> filldraw(f, gray);<br /> filldraw(f, gray);<br /> axialshade(f, white, L, gray(0.7), J);<br /> draw(f);<br /> draw(f);<br /> &lt;/asy&gt;<br /> <br /> == Solution 1 ==<br /> Since we start at the top face and end at the bottom face without moving from the lower ring to the upper ring or revisiting a face, our journey must consist of the top face, a series of faces in the upper ring, a series of faces in the lower ring, and the bottom face, in that order.<br /> <br /> We have &lt;math&gt;5&lt;/math&gt; choices for which face we visit first on the top ring. From there, we have &lt;math&gt;9&lt;/math&gt; choices for how far around the top ring we go before moving down: &lt;math&gt;1,2,3,&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; faces around clockwise, &lt;math&gt;1,2,3,&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; faces around counterclockwise, or immediately going down to the lower ring without visiting any other faces in the upper ring.<br /> <br /> We then have &lt;math&gt;2&lt;/math&gt; choices for which lower ring face to visit first (since every upper-ring face is adjacent to exactly &lt;math&gt;2&lt;/math&gt; lower-ring faces) and then once again &lt;math&gt;9&lt;/math&gt; choices for how to travel around the lower ring. We then proceed to the bottom face, completing the trip.<br /> <br /> Multiplying together all the numbers of choices we have, we get &lt;math&gt;5 \cdot 9 \cdot 2 \cdot 9 = \boxed{\textbf{(E) } 810}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Swap the faces as vertices and the vertices as faces. Then, this problem is the same as<br /> [https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3 2016 AIME I #3]<br /> which had an answer of &lt;math&gt;\boxed{\textbf{(E) } 810}&lt;/math&gt;.<br /> &lt;math&gt;\textbf{\textbf{- Emathmaster}}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3<br /> {{AMC10 box|year=2020|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_9&diff=120556 2013 AIME I Problems/Problem 9 2020-04-04T22:03:50Z <p>Oreochocolate: left out a &quot;W&quot; in &quot;We&quot;</p> <hr /> <div>==Problem 9==<br /> A paper equilateral triangle &lt;math&gt;ABC&lt;/math&gt; has side length &lt;math&gt;12&lt;/math&gt;. The paper triangle is folded so that vertex &lt;math&gt;A&lt;/math&gt; touches a point on side &lt;math&gt;\overline{BC}&lt;/math&gt; a distance &lt;math&gt;9&lt;/math&gt; from point &lt;math&gt;B&lt;/math&gt;. The length of the line segment along which the triangle is folded can be written as &lt;math&gt;\frac{m\sqrt{p}}{n}&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;n&lt;/math&gt; are relatively prime, 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 /> &lt;asy&gt;<br /> import cse5;<br /> size(12cm);<br /> pen tpen = defaultpen + 1.337;<br /> real a = 39/5.0;<br /> real b = 39/7.0;<br /> pair B = MP(&quot;B&quot;, (0,0), dir(200));<br /> pair A = MP(&quot;A&quot;, (9,0), dir(-80));<br /> pair C = MP(&quot;C&quot;, (12,0), dir(-20));<br /> pair K = (6,10.392);<br /> pair M = (a*B+(12-a)*K) / 12;<br /> pair N = (b*C+(12-b)*K) / 12;<br /> draw(B--M--N--C--cycle, tpen);<br /> draw(M--A--N--cycle);<br /> fill(M--A--N--cycle, mediumgrey);<br /> pair shift = (-20.13, 0);<br /> pair B1 = MP(&quot;B&quot;, B+shift, dir(200));<br /> pair A1 = MP(&quot;A&quot;, K+shift, dir(90));<br /> pair C1 = MP(&quot;C&quot;, C+shift, dir(-20));<br /> draw(A1--B1--C1--cycle, tpen);&lt;/asy&gt;<br /> <br /> == Solution 1 ==<br /> Let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be the points on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, where the paper is folded.<br /> <br /> Let &lt;math&gt;D&lt;/math&gt; be the point on &lt;math&gt;\overline{BC}&lt;/math&gt; where the folded &lt;math&gt;A&lt;/math&gt; touches it.<br /> <br /> Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;x&lt;/math&gt; be the lengths &lt;math&gt;AP&lt;/math&gt;, &lt;math&gt;AQ&lt;/math&gt;, and &lt;math&gt;PQ&lt;/math&gt;, respectively.<br /> <br /> We have &lt;math&gt;PD = a&lt;/math&gt;, &lt;math&gt;QD = b&lt;/math&gt;, &lt;math&gt;BP = 12 - a&lt;/math&gt;, &lt;math&gt;CQ = 12 - b&lt;/math&gt;, &lt;math&gt;BD = 9&lt;/math&gt;, and &lt;math&gt;CD = 3&lt;/math&gt;.<br /> <br /> Using the Law of Cosines on &lt;math&gt;BPD&lt;/math&gt;:<br /> <br /> &lt;math&gt;a^{2} = (12 - a)^{2} + 9^{2} - 2 \times (12 - a) \times 9 \times \cos{60}&lt;/math&gt;<br /> <br /> &lt;math&gt;a^{2} = 144 - 24a + a^{2} + 81 - 108 + 9a&lt;/math&gt;<br /> <br /> &lt;math&gt;a = \frac{39}{5}&lt;/math&gt;<br /> <br /> Using the Law of Cosines on &lt;math&gt;CQD&lt;/math&gt;:<br /> <br /> &lt;math&gt;b^{2} = (12 - b)^{2} +3^{2} - 2 \times (12 - b) \times 3 \times \cos{60}&lt;/math&gt;<br /> <br /> &lt;math&gt;b^{2} = 144 - 24b + b^{2} + 9 - 36 + 3b&lt;/math&gt;<br /> <br /> &lt;math&gt;b = \frac{39}{7}&lt;/math&gt;<br /> <br /> Using the Law of Cosines on &lt;math&gt;DPQ&lt;/math&gt;:<br /> <br /> &lt;math&gt;x^{2} = a^{2} + b^{2} - 2ab \cos{60}&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{39}{7})&lt;/math&gt;<br /> <br /> &lt;math&gt;x = \frac{39 \sqrt{39}}{35}&lt;/math&gt;<br /> <br /> The solution is &lt;math&gt;39 + 39 + 35 = \boxed{113}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Proceed with the same labeling as in Solution 1. <br /> <br /> &lt;math&gt;\angle B = \angle C = \angle A = \angle PDQ = 60^\circ&lt;/math&gt;<br /> <br /> &lt;math&gt;\angle PDB + \angle PDQ + \angle QDC = \angle QDC + \angle CQD + \angle C = 180^\circ&lt;/math&gt;<br /> <br /> Therefore, &lt;math&gt;\angle PDB = \angle CQD&lt;/math&gt;.<br /> <br /> Similarly, &lt;math&gt;\angle BPD = \angle QDC&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;\bigtriangleup BPD&lt;/math&gt; and &lt;math&gt;\bigtriangleup CDQ&lt;/math&gt; are similar triangles, so<br /> <br /> &lt;math&gt;\frac{3}{12-a} = \frac{12-b}{9} = \frac{b}{a}&lt;/math&gt;.<br /> <br /> Solving this system of equations yields &lt;math&gt;a = \frac{39}{5}&lt;/math&gt; and &lt;math&gt;b = \frac{39}{7}&lt;/math&gt;.<br /> <br /> Using the Law of Cosines on &lt;math&gt;APQ&lt;/math&gt;:<br /> <br /> &lt;math&gt;x^{2} = a^{2} + b^{2} - 2ab \cos{60}&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{39}{7})&lt;/math&gt;<br /> <br /> &lt;math&gt;x = \frac{39 \sqrt{39}}{35}&lt;/math&gt;<br /> <br /> The solution is &lt;math&gt;39 + 39 + 35 = \boxed{113}&lt;/math&gt;.<br /> <br /> == Solution 3 (Coordinate Bash) ==<br /> <br /> We let the original position of &lt;math&gt;A&lt;/math&gt; be &lt;math&gt;A&lt;/math&gt;, and the position of &lt;math&gt;A&lt;/math&gt; after folding be &lt;math&gt;D&lt;/math&gt;. Also, we put the triangle on the coordinate plane such that &lt;math&gt;A=(0,0)&lt;/math&gt;, &lt;math&gt;B=(-6,-6\sqrt3)&lt;/math&gt;, &lt;math&gt;C=(6,-6\sqrt3)&lt;/math&gt;, and &lt;math&gt;D=(3,-6\sqrt3)&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> pen tpen = defaultpen + 1.337;<br /> real a = 39/5.0;<br /> real b = 39/7.0;<br /> pair B = MP(&quot;B&quot;, (0,0), dir(200));<br /> pair A = (9,0);<br /> pair C = MP(&quot;C&quot;, (12,0), dir(-20));<br /> pair K = (6,10.392);<br /> pair M = (a*B+(12-a)*K) / 12;<br /> pair N = (b*C+(12-b)*K) / 12;<br /> draw(B--M--N--C--cycle);<br /> draw(M--A--N--cycle);<br /> label(&quot;$D$&quot;, A, S);<br /> pair X = (6,6*sqrt(3));<br /> draw(B--X--C);<br /> label(&quot;$A$&quot;,X,dir(90));<br /> draw(A--X);<br /> &lt;/asy&gt;<br /> <br /> Note that since &lt;math&gt;A&lt;/math&gt; is reflected over the fold line to &lt;math&gt;D&lt;/math&gt;, the fold line is the perpendicular bisector of &lt;math&gt;AD&lt;/math&gt;. We know &lt;math&gt;A=(0,0)&lt;/math&gt; and &lt;math&gt;D=(3,-6\sqrt3)&lt;/math&gt;. The midpoint of &lt;math&gt;AD&lt;/math&gt; (which is a point on the fold line) is &lt;math&gt;(\tfrac32, -3\sqrt3)&lt;/math&gt;. Also, the slope of &lt;math&gt;AD&lt;/math&gt; is &lt;math&gt;\frac{-6\sqrt3}{3}=-2\sqrt3&lt;/math&gt;, so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of &lt;math&gt;AD&lt;/math&gt;, or &lt;math&gt;\frac{1}{2\sqrt3}=\frac{\sqrt3}{6}&lt;/math&gt;. Then, using point slope form, the equation of the fold line is<br /> &lt;cmath&gt;y+3\sqrt3=\frac{\sqrt3}{6}\left(x-\frac32\right)&lt;/cmath&gt;&lt;cmath&gt;y=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}&lt;/cmath&gt;<br /> Note that the equations of lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; are &lt;math&gt;y=\sqrt3x&lt;/math&gt; and &lt;math&gt;y=-\sqrt3x&lt;/math&gt;, respectively. We will first find the intersection of &lt;math&gt;AB&lt;/math&gt; and the fold line by substituting for &lt;math&gt;y&lt;/math&gt;:<br /> &lt;cmath&gt;\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}&lt;/cmath&gt;&lt;cmath&gt;\frac{5\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=-\frac{39}{10}&lt;/cmath&gt;<br /> Therefore, the point of intersection is &lt;math&gt;\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)&lt;/math&gt;. Now, lets find the intersection with &lt;math&gt;AC&lt;/math&gt;. Substituting for &lt;math&gt;y&lt;/math&gt; yields<br /> &lt;cmath&gt;-\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}&lt;/cmath&gt;&lt;cmath&gt;\frac{-7\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=\frac{39}{14}&lt;/cmath&gt;<br /> Therefore, the point of intersection is &lt;math&gt;\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)&lt;/math&gt;. Now, we just need to use the distance formula to find the distance between &lt;math&gt;\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)&lt;/math&gt; and &lt;math&gt;\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)&lt;/math&gt;.<br /> &lt;cmath&gt;\sqrt{\left(\frac{39}{14}+\frac{39}{10}\right)^2+\left(-\frac{39\sqrt3}{14}+\frac{39\sqrt3}{10}\right)^2}&lt;/cmath&gt;<br /> The number 39 is in all of the terms, so let's factor it out:<br /> &lt;cmath&gt;39\sqrt{\left(\frac{1}{14}+\frac{1}{10}\right)^2+\left(-\frac{\sqrt3}{14}+\frac{\sqrt3}{10}\right)^2}=39\sqrt{\left(\frac{6}{35}\right)^2+\left(\frac{\sqrt3}{35}\right)^2}&lt;/cmath&gt;&lt;cmath&gt;\frac{39}{35}\sqrt{6^2+\sqrt3^2}=\frac{39\sqrt{39}}{35}&lt;/cmath&gt;<br /> Therefore, our answer is &lt;math&gt;39+39+35=\boxed{113}&lt;/math&gt;, and we are done.<br /> <br /> Solution by nosaj.<br /> <br /> == See also ==<br /> {{AIME box|year=2013|n=I|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=User:Oreochocolate&diff=120288 User:Oreochocolate 2020-03-30T20:28:24Z <p>Oreochocolate: stuff on silly proofs</p> <hr /> <div>Math stuff into reality:<br /> <br /> AM-GM inequality:<br /> <br /> American Motors-General Motors Inequality.<br /> <br /> Proof that this is true:<br /> <br /> American Motors encompasses Ford, General Motors, Tesla, and Fiat Chrysler. Therefore, American Motors is greater than General Motors. Equality only holds whenever one is describing is Buick, Cadillac, Chevrolet, GMC, and other American Motors all the same.<br /> <br /> Cube:<br /> <br /> Nissan Cube.<br /> <br /> Proof that this is true:<br /> <br /> A Nissan Cube is classified as a &quot;box car,&quot; the others being the Scion xB, the Honda Element, and the Kia Soul. When looking at them, one can see a similar characteristic: they are all box-like, therefore being a cube.<br /> <br /> AIME:<br /> <br /> Artificial Intelligence of Windows ME.<br /> <br /> Proof that this is true:<br /> <br /> Sometimes, when doing AIME too excessfully, one's brain may crash and freeze, thus causing one unable to do more problems. Windows Millenium Edition, also known as Windows ME, is one of the worst operating systems that freeze, crash, etc. The congruencies are present.<br /> <br /> AMC (8):<br /> <br /> American Motors Corporation.<br /> <br /> Proof that this is true:<br /> <br /> Now that I am in 8th grade and above, I cannot do the AMC (8) anymore. Similarly, the AMC has been discontinued; therefore, we cannot buy any AMCs. Therefore, AMC (8) and the American Motors Corporation have congruencies.<br /> <br /> And more to come!<br /> <br /> Contradictions:<br /> <br /> 9:<br /> <br /> Windows.<br /> <br /> Proof that this is false:<br /> <br /> The number 9 does exist. But Windows 9 does not; Windows 8.1 does. Therefore, we have a contradiction; 9 does not exist in the Windows operating system but exists in the real world (the main reason why I say &quot;eight point one&quot; rather than &quot;nine&quot; and why I partially have enneaphobia).<br /> <br /> And more to come!</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_19&diff=120285 2020 AMC 10A Problems/Problem 19 2020-03-30T19:04:40Z <p>Oreochocolate: Please do not change the diagram. Leave it as it is, and the dodecahedron will stay the same.</p> <hr /> <div>== Problem ==<br /> As shown in the figure below, a regular dodecahedron (the polyhedron consisting of &lt;math&gt;12&lt;/math&gt; congruent regular pentagonal faces) floats in space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?<br /> <br /> &lt;math&gt;\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810&lt;/math&gt;<br /> <br /> == Diagram ==<br /> &lt;asy&gt;<br /> import graph;<br /> unitsize(5cm);<br /> pair A = (0.082, 0.378);<br /> pair B = (0.091, 0.649);<br /> pair C = (0.249, 0.899);<br /> pair D = (0.479, 0.939);<br /> pair E = (0.758, 0.893);<br /> pair F = (0.862, 0.658);<br /> pair G = (0.924, 0.403);<br /> pair H = (0.747, 0.194);<br /> pair I = (0.526, 0.075);<br /> pair J = (0.251, 0.170);<br /> pair K = (0.568, 0.234);<br /> pair L = (0.262, 0.449);<br /> pair M = (0.373, 0.813);<br /> pair N = (0.731, 0.813);<br /> pair O = (0.851, 0.461);<br /> path[] f;<br /> f = A--B--C--M--L--cycle;<br /> f = C--D--E--N--M--cycle;<br /> f = E--F--G--O--N--cycle;<br /> f = G--H--I--K--O--cycle;<br /> f = I--J--A--L--K--cycle;<br /> f = K--L--M--N--O--cycle;<br /> draw(f);<br /> axialshade(f, white, M, gray(0.5), (C+2*D)/3);<br /> draw(f);<br /> filldraw(f, gray);<br /> filldraw(f, gray);<br /> axialshade(f, white, L, gray(0.7), J);<br /> draw(f);<br /> draw(f);<br /> &lt;/asy&gt;<br /> <br /> == Solution 1 ==<br /> Since we start at the top face and end at the bottom face without moving from the lower ring to the upper ring or revisiting a face, our journey must consist of the top face, a series of faces in the upper ring, a series of faces in the lower ring, and the bottom face, in that order.<br /> <br /> We have &lt;math&gt;5&lt;/math&gt; choices for which face we visit first on the top ring. From there, we have &lt;math&gt;9&lt;/math&gt; choices for how far around the top ring we go before moving down: &lt;math&gt;1,2,3,&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; faces around clockwise, &lt;math&gt;1,2,3,&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; faces around counterclockwise, or immediately going down to the lower ring without visiting any other faces in the upper ring.<br /> <br /> We then have &lt;math&gt;2&lt;/math&gt; choices for which lower ring face to visit first (since every upper-ring face is adjacent to exactly &lt;math&gt;2&lt;/math&gt; lower-ring faces) and then once again &lt;math&gt;9&lt;/math&gt; choices for how to travel around the lower ring. We then proceed to the bottom face, completing the trip.<br /> <br /> Multiplying together all the numbers of choices we have, we get &lt;math&gt;5 \cdot 9 \cdot 2 \cdot 9 = \boxed{\textbf{(E) } 810}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Swap the faces as vertices and the vertices as faces. Then, this problem is the same as<br /> [https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3 2016 AIME I #3]<br /> which had an answer of &lt;math&gt;\boxed{\textbf{(E) } 810}&lt;/math&gt;.<br /> &lt;math&gt;\textbf{\textbf{- Emathmaster}}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3<br /> {{AMC10 box|year=2020|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_12&diff=120157 2020 AMC 10B Problems/Problem 12 2020-03-27T03:47:04Z <p>Oreochocolate: NOT a contributor to Solution 3...sigh</p> <hr /> <div>==Problem==<br /> <br /> The decimal representation of&lt;cmath&gt;\dfrac{1}{20^{20}}&lt;/cmath&gt;consists of a string of zeros after the decimal point, followed by a &lt;math&gt;9&lt;/math&gt; and then several more digits. How many zeros are in that initial string of zeros after the decimal point?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;cmath&gt;\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}&lt;/cmath&gt;<br /> <br /> Now we do some estimation. Notice that &lt;math&gt;2^{20} = 1024^2&lt;/math&gt;, which means that &lt;math&gt;2^{20}&lt;/math&gt; is a little more than &lt;math&gt;1000^2=1,000,000&lt;/math&gt;. Multiplying it with &lt;math&gt;10^{20}&lt;/math&gt;, we get that the denominator is about &lt;math&gt;1\underbrace{00\dots0}_{26 \text{ zeros}}&lt;/math&gt;. Notice that when we divide &lt;math&gt;1&lt;/math&gt; by an &lt;math&gt;n&lt;/math&gt; digit number, there are &lt;math&gt;n-1&lt;/math&gt; zeros before the first nonzero digit. This means that when we divide &lt;math&gt;1&lt;/math&gt; by the &lt;math&gt;27&lt;/math&gt; digit integer &lt;math&gt;1\underbrace{00\dots0}_{26 \text{ zeros}}&lt;/math&gt;, there are &lt;math&gt;\boxed{\textbf{(D) } \text{26}}&lt;/math&gt; zeros in the initial string after the decimal point. -PCChess<br /> <br /> ==Solution 2==<br /> First rewrite &lt;math&gt;\frac{1}{20^{20}}&lt;/math&gt; as &lt;math&gt;\frac{5^{20}}{10^{40}}&lt;/math&gt;. Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in &lt;math&gt;{5^{20}}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\log{5^{20}} = 20\log{5}&lt;/math&gt; and memming &lt;math&gt;\log{5}\approx0.69&lt;/math&gt; (alternatively use the fact that &lt;math&gt;\log{5} = 1 - \log{2}&lt;/math&gt;), &lt;math&gt;\lfloor{20\log{5}}\rfloor+1=\lfloor{20\cdot0.69}\rfloor+1=13+1=14&lt;/math&gt; digits. <br /> <br /> Our answer is &lt;math&gt;\boxed{\textbf{(D) } \text{26}}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Brute Force)==<br /> Just as in Solution &lt;math&gt;2,&lt;/math&gt; we rewrite &lt;math&gt;\dfrac{1}{20^{20}}&lt;/math&gt; as &lt;math&gt;\dfrac{5^{20}}{10^{40}}.&lt;/math&gt; We then calculate &lt;math&gt;5^{20}&lt;/math&gt; entirely by hand, first doing &lt;math&gt;5^5 \cdot 5^5,&lt;/math&gt; then multiplying that product by itself, resulting in &lt;math&gt;95,367,431,640,625.&lt;/math&gt; Because this is &lt;math&gt;14&lt;/math&gt; digits, after dividing this number by &lt;math&gt;10&lt;/math&gt; fourteen times, the decimal point is before the &lt;math&gt;9.&lt;/math&gt; Dividing the number again by &lt;math&gt;10&lt;/math&gt; twenty-six more times allows a string of&lt;math&gt;\boxed{\textbf{(D) } \text{26}}&lt;/math&gt; zeroes to be formed. -OreoChocolate<br /> <br /> ==Solution 4 (Smarter Brute Force)==<br /> Just as in Solutions &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3,&lt;/math&gt; we rewrite &lt;math&gt;\dfrac{1}{20^{20}}&lt;/math&gt; as &lt;math&gt;\dfrac{5^{20}}{10^{40}}.&lt;/math&gt; We can then look at the number of digits in powers of &lt;math&gt;5&lt;/math&gt;. &lt;math&gt;5^1=5&lt;/math&gt;, &lt;math&gt;5^2=25&lt;/math&gt;, &lt;math&gt;5^3=125&lt;/math&gt;, &lt;math&gt;5^4=625&lt;/math&gt;, &lt;math&gt;5^5=3125&lt;/math&gt;, &lt;math&gt;5^6=15625&lt;/math&gt;, &lt;math&gt;5^7=78125&lt;/math&gt; and so on. We notice after a few iterations that every power of five with an exponent of &lt;math&gt;1 \mod 3&lt;/math&gt;, the number of digits doesn't increase. This means &lt;math&gt;5^{20}&lt;/math&gt; should have &lt;math&gt;20 - 6&lt;/math&gt; digits since there are &lt;math&gt;6&lt;/math&gt; numbers which are &lt;math&gt;1 \mod 3&lt;/math&gt; from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;20&lt;/math&gt;, or &lt;math&gt;14&lt;/math&gt; digits total. This means our expression can be written as &lt;math&gt;\dfrac{k\cdot10^{14}}{10^{40}}&lt;/math&gt;, where &lt;math&gt;k&lt;/math&gt; is in the range &lt;math&gt;[1,10)&lt;/math&gt;. Canceling gives &lt;math&gt;\dfrac{k}{10^{26}}&lt;/math&gt;, or &lt;math&gt;26&lt;/math&gt; zeroes before the &lt;math&gt;k&lt;/math&gt; since the number &lt;math&gt;k&lt;/math&gt; should start on where the one would be in &lt;math&gt;10^{26}&lt;/math&gt;. ~aop2014<br /> <br /> ==Solution 5 (Logarithms)==<br /> <br /> &lt;cmath&gt;|\lceil \log \dfrac{1}{20^{20}} \rceil|<br /> = |\lceil \log 20^{-20} \rceil|<br /> = |\lceil -20 \log(20) \rceil|<br /> = |\lceil -20(\log 10 + \log 2) \rceil|<br /> = |\lceil -20(1 + 0.301) \rceil|<br /> = |\lceil -26.02 \rceil|<br /> = |-26|<br /> = \boxed{\textbf{(D) } \text{26}}&lt;/cmath&gt;<br /> <br /> ~phoenixfire<br /> <br /> ==Video Solution==<br /> https://youtu.be/t6yjfKXpwDs<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_15&diff=120114 2014 AIME II Problems/Problem 15 2020-03-25T22:44:00Z <p>Oreochocolate: always that i before e except after c rule...</p> <hr /> <div>==Problem==<br /> For any integer &lt;math&gt;k\geq 1&lt;/math&gt;, let &lt;math&gt;p(k)&lt;/math&gt; be the smallest prime which does not divide &lt;math&gt;k.&lt;/math&gt; Define the integer function &lt;math&gt;X(k)&lt;/math&gt; to be the product of all primes less than &lt;math&gt;p(k)&lt;/math&gt; if &lt;math&gt;p(k)&gt;2&lt;/math&gt;, and &lt;math&gt;X(k)=1&lt;/math&gt; if &lt;math&gt;p(k)=2.&lt;/math&gt; Let &lt;math&gt;\{x_n\}&lt;/math&gt; be the sequence defined by &lt;math&gt;x_0=1&lt;/math&gt;, and &lt;math&gt;x_{n+1}X(x_n)=x_np(x_n)&lt;/math&gt; for &lt;math&gt;n\geq 0.&lt;/math&gt; Find the smallest positive integer &lt;math&gt;t&lt;/math&gt; such that &lt;math&gt;x_t=2090.&lt;/math&gt;<br /> <br /> ==Solution==<br /> Note that for any &lt;math&gt;x_i&lt;/math&gt;, for any prime &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;p^2 \nmid x_i&lt;/math&gt;. This provides motivation to translate &lt;math&gt;x_i&lt;/math&gt; into a binary sequence &lt;math&gt;y_i&lt;/math&gt;. <br /> <br /> Let the prime factorization of &lt;math&gt;x_i&lt;/math&gt; be written as &lt;math&gt;p_{a_1} \cdot p_{a_2} \cdot p_{a_3} \cdots&lt;/math&gt;, where &lt;math&gt;p_i&lt;/math&gt; is the &lt;math&gt;i&lt;/math&gt;th prime number. Then, for every &lt;math&gt;p_{a_k}&lt;/math&gt; in the prime factorization of &lt;math&gt;x_i&lt;/math&gt;, place a &lt;math&gt;1&lt;/math&gt; in the &lt;math&gt;a_k&lt;/math&gt;th digit of &lt;math&gt;y_i&lt;/math&gt;. This will result in the conversion &lt;math&gt;x_1 = 2, x_{2} = 3, x_{3} = 2 * 3 = 6, \cdots&lt;/math&gt;.<br /> <br /> Multiplication for the sequence &lt;math&gt;x_i&lt;/math&gt; will translate to addition for the sequence &lt;math&gt;y_i&lt;/math&gt;. Thus, we see that &lt;math&gt;x_{n+1}X(x_n) = x_np(x_n)&lt;/math&gt; translates into &lt;math&gt;y_{n+1} = y_n+1&lt;/math&gt;. Since &lt;math&gt;x_0=1&lt;/math&gt;, and &lt;math&gt;y_0=0&lt;/math&gt;, &lt;math&gt;x_i&lt;/math&gt; corresponds to &lt;math&gt;y_i&lt;/math&gt;, which is &lt;math&gt;i&lt;/math&gt; in binary. Since &lt;math&gt;x_{10010101_2} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090&lt;/math&gt;, &lt;math&gt;t = 10010101_2&lt;/math&gt; = &lt;math&gt;\boxed{149}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Painful Bash)==<br /> We go through the terms and look for a pattern. We find that<br /> <br /> &lt;math&gt;x_0 = 1&lt;/math&gt; &lt;math&gt;x_8 = 7&lt;/math&gt;<br /> <br /> &lt;math&gt;x_1 = 2&lt;/math&gt; &lt;math&gt;x_9 = 14&lt;/math&gt;<br /> <br /> &lt;math&gt;x_2 = 3&lt;/math&gt; &lt;math&gt;x_{10} = 21&lt;/math&gt;<br /> <br /> &lt;math&gt;x_3 = 6&lt;/math&gt; &lt;math&gt;x_{11} = 42&lt;/math&gt;<br /> <br /> &lt;math&gt;x_4 = 5&lt;/math&gt; &lt;math&gt;x_{12} = 35&lt;/math&gt;<br /> <br /> &lt;math&gt;x_5 = 10&lt;/math&gt; &lt;math&gt;x_{13} = 70&lt;/math&gt;<br /> <br /> &lt;math&gt;x_6 = 15&lt;/math&gt; &lt;math&gt;x_{14} = 105&lt;/math&gt;<br /> <br /> &lt;math&gt;x_7 = 30&lt;/math&gt; &lt;math&gt;x_{15} = 210&lt;/math&gt;<br /> <br /> Commit to the bash. Eventually, you will receive that &lt;math&gt;x_{149} = 2090&lt;/math&gt;, so &lt;math&gt;\boxed{149}&lt;/math&gt; is the answer. Trust me, this is worth the 10 index points.<br /> <br /> &lt;math&gt;\textbf{-RootThreeOverTwo}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_12&diff=118300 2020 AMC 10B Problems/Problem 12 2020-02-22T04:40:13Z <p>Oreochocolate: making the logarithm symbols clearer.</p> <hr /> <div>==Problem==<br /> <br /> The decimal representation of&lt;cmath&gt;\dfrac{1}{20^{20}}&lt;/cmath&gt;consists of a string of zeros after the decimal point, followed by a &lt;math&gt;9&lt;/math&gt; and then several more digits. How many zeros are in that initial string of zeros after the decimal point?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;cmath&gt;\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}&lt;/cmath&gt;<br /> <br /> Now we do some estimation. Notice that &lt;math&gt;2^{20} = 1024^2&lt;/math&gt;, which means that &lt;math&gt;2^{20}&lt;/math&gt; is a little more than &lt;math&gt;1000^2=1,000,000&lt;/math&gt;. Multiplying it with &lt;math&gt;10^{20}&lt;/math&gt;, we get that the denominator is about &lt;math&gt;1\underbrace{00\dots0}_{26 \text{ zeros}}&lt;/math&gt;. Notice that when we divide &lt;math&gt;1&lt;/math&gt; by an &lt;math&gt;n&lt;/math&gt; digit number, there are &lt;math&gt;n-1&lt;/math&gt; zeros before the first nonzero digit. This means that when we divide &lt;math&gt;1&lt;/math&gt; by the &lt;math&gt;27&lt;/math&gt; digit integer &lt;math&gt;1\underbrace{00\dots0}_{26 \text{ zeros}}&lt;/math&gt;, there are &lt;math&gt;\boxed{\textbf{(D) } \text{26}}&lt;/math&gt; zeros in the initial string after the decimal point. -PCChess<br /> <br /> ==Solution 2==<br /> First rewrite &lt;math&gt;\frac{1}{20^{20}}&lt;/math&gt; as &lt;math&gt;\frac{5^{20}}{10^{40}}&lt;/math&gt;. Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in &lt;math&gt;{5^{20}}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\log{5^{20}} = 20\log{5}&lt;/math&gt; and memming &lt;math&gt;\log{5}\approx0.69&lt;/math&gt; (alternatively use the fact that &lt;math&gt;\log{5} = 1 - \log{2}&lt;/math&gt;), &lt;math&gt;\lfloor{20\log{5}}\rfloor+1=\lfloor{20\cdot0.69}\rfloor+1=13+1=14&lt;/math&gt; digits. <br /> <br /> Our answer is &lt;math&gt;\boxed{\textbf{(D) } \text{26}}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Brute Force)==<br /> Just as in Solution &lt;math&gt;2,&lt;/math&gt; we rewrite &lt;math&gt;\dfrac{1}{20^{20}}&lt;/math&gt; as &lt;math&gt;\dfrac{5^{20}}{10^{40}}.&lt;/math&gt; We then calculate &lt;math&gt;5^{20}&lt;/math&gt; entirely by hand, first doing &lt;math&gt;5^5 \cdot 5^5,&lt;/math&gt; then multiplying that product by itself, resulting in &lt;math&gt;95,367,431,640,625.&lt;/math&gt; Because this is &lt;math&gt;14&lt;/math&gt; digits, after dividing this number by &lt;math&gt;10&lt;/math&gt; fourteen times, the decimal point is before the &lt;math&gt;9.&lt;/math&gt; Dividing the number again by &lt;math&gt;10&lt;/math&gt; twenty-six more times allows a string of&lt;math&gt;\boxed{\textbf{(D) } \text{26}}&lt;/math&gt; zeroes to be formed. -OreoChocolate<br /> <br /> ==Solution 4 (Smarter Brute Force)==<br /> Just as in Solutions &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3,&lt;/math&gt; we rewrite &lt;math&gt;\dfrac{1}{20^{20}}&lt;/math&gt; as &lt;math&gt;\dfrac{5^{20}}{10^{40}}.&lt;/math&gt; We can then look at the number of digits in powers of &lt;math&gt;5&lt;/math&gt;. &lt;math&gt;5^1=5&lt;/math&gt;, &lt;math&gt;5^2=25&lt;/math&gt;, &lt;math&gt;5^3=125&lt;/math&gt;, &lt;math&gt;5^4=625&lt;/math&gt;, &lt;math&gt;5^5=3125&lt;/math&gt;, &lt;math&gt;5^6=15625&lt;/math&gt;, &lt;math&gt;5^7=78125&lt;/math&gt; and so on. We notice after a few iterations that every power of five with an exponent of &lt;math&gt;1 \mod 3&lt;/math&gt;, the number of digits doesn't increase. This means &lt;math&gt;5^{20}&lt;/math&gt; should have &lt;math&gt;20 - 6&lt;/math&gt; digits since there are &lt;math&gt;6&lt;/math&gt; numbers which are &lt;math&gt;1 \mod 3&lt;/math&gt; from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;20&lt;/math&gt;, or &lt;math&gt;14&lt;/math&gt; digits total. This means our expression can be written as &lt;math&gt;\dfrac{k\cdot10^{14}}{10^{40}}&lt;/math&gt;, where &lt;math&gt;k&lt;/math&gt; is in the range &lt;math&gt;[1,10)&lt;/math&gt;. Canceling gives &lt;math&gt;\dfrac{k}{10^{26}}&lt;/math&gt;, or &lt;math&gt;26&lt;/math&gt; zeroes before the &lt;math&gt;k&lt;/math&gt; since the number &lt;math&gt;k&lt;/math&gt; should start on where the one would be in &lt;math&gt;10^{26}&lt;/math&gt;. ~aop2014<br /> <br /> ==Solution 5 (Logarithms)==<br /> <br /> &lt;cmath&gt;|\lceil \log \dfrac{1}{20^{20}} \rceil|<br /> = |\lceil \log 20^{-20} \rceil|<br /> = |\lceil -20 \log(20) \rceil|<br /> = |\lceil -20(\log 10 + \log 2) \rceil|<br /> = |\lceil -20(1 + 0.301) \rceil|<br /> = |\lceil -26.02 \rceil|<br /> = |-26|<br /> = \boxed{\textbf{(D) } \text{26}}&lt;/cmath&gt;<br /> <br /> ~phoenixfire<br /> <br /> ==Video Solution==<br /> https://youtu.be/t6yjfKXpwDs<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_12&diff=117764 2020 AMC 10B Problems/Problem 12 2020-02-13T16:10:31Z <p>Oreochocolate: typo...</p> <hr /> <div>==Problem==<br /> <br /> The decimal representation of&lt;cmath&gt;\dfrac{1}{20^{20}}&lt;/cmath&gt;consists of a string of zeros after the decimal point, followed by a &lt;math&gt;9&lt;/math&gt; and then several more digits. How many zeros are in that initial string of zeros after the decimal point?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;cmath&gt;\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}&lt;/cmath&gt;<br /> <br /> Now we do some estimation. Notice that &lt;math&gt;2^{20} = 1024^2&lt;/math&gt;, which means that &lt;math&gt;2^{20}&lt;/math&gt; is a little more than &lt;math&gt;1000^2=1,000,000&lt;/math&gt;. Multiplying it with &lt;math&gt;10^{20}&lt;/math&gt;, we get that the denominator is about &lt;math&gt;1\underbrace{00\dots0}_{26 \text{ zeros}}&lt;/math&gt;. Notice that when we divide &lt;math&gt;1&lt;/math&gt; by an &lt;math&gt;n&lt;/math&gt; digit number, there are &lt;math&gt;n-1&lt;/math&gt; zeros before the first nonzero digit. This means that when we divide &lt;math&gt;1&lt;/math&gt; by the &lt;math&gt;27&lt;/math&gt; digit integer &lt;math&gt;1\underbrace{00\dots0}_{26 \text{ zeros}}&lt;/math&gt;, there are &lt;math&gt;\boxed{\textbf{(D) } \text{26}}&lt;/math&gt; zeros in the initial string after the decimal point. -PCChess<br /> <br /> ==Solution 2==<br /> First rewrite &lt;math&gt;\frac{1}{20^{20}}&lt;/math&gt; as &lt;math&gt;\frac{5^{20}}{10^{40}}&lt;/math&gt;. Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in &lt;math&gt;{5^{20}}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\log{5^{20}} = 20\log{5}&lt;/math&gt; and memming &lt;math&gt;\log{5}\approx0.69&lt;/math&gt; (alternatively use the fact that &lt;math&gt;\log{5} = 1 - \log{2}&lt;/math&gt;), &lt;math&gt;\lfloor{20\log{5}}\rfloor+1=\lfloor{20\cdot0.69}\rfloor+1=13+1=14&lt;/math&gt; digits. <br /> <br /> Our answer is &lt;math&gt;\boxed{\textbf{(D) } \text{26}}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Brute Force)==<br /> Just as in Solution &lt;math&gt;2,&lt;/math&gt; we rewrite &lt;math&gt;\dfrac{1}{20^{20}}&lt;/math&gt; as &lt;math&gt;\dfrac{5^{20}}{10^{40}}.&lt;/math&gt; We then calculate &lt;math&gt;5^{20}&lt;/math&gt; entirely by hand, first doing &lt;math&gt;5^5 \cdot 5^5,&lt;/math&gt; then multiplying that product by itself, resulting in &lt;math&gt;95,367,431,640,625.&lt;/math&gt; Because this is &lt;math&gt;14&lt;/math&gt; digits, after dividing this number by &lt;math&gt;10&lt;/math&gt; fourteen times, the decimal point is before the &lt;math&gt;9.&lt;/math&gt; Dividing the number again by &lt;math&gt;10&lt;/math&gt; twenty-six more times allows a string of&lt;math&gt;\boxed{\textbf{(D) } \text{26}}&lt;/math&gt; zeroes to be formed. -OreoChocolate<br /> <br /> ==Solution 4 (Smarter Brute Force)==<br /> Just as in Solutions &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3,&lt;/math&gt; we rewrite &lt;math&gt;\dfrac{1}{20^{20}}&lt;/math&gt; as &lt;math&gt;\dfrac{5^{20}}{10^{40}}.&lt;/math&gt; We can then look at the number of digits in powers of &lt;math&gt;5&lt;/math&gt;. &lt;math&gt;5^1=5&lt;/math&gt;, &lt;math&gt;5^2=25&lt;/math&gt;, &lt;math&gt;5^3=125&lt;/math&gt;, &lt;math&gt;5^4=625&lt;/math&gt;, &lt;math&gt;5^5=3125&lt;/math&gt;, &lt;math&gt;5^6=15625&lt;/math&gt;, &lt;math&gt;5^7=78125&lt;/math&gt; and so on. We notice after a few iterations that every power of five with an exponent of &lt;math&gt;1 \mod 3&lt;/math&gt;, the number of digits doesn't increase. This means &lt;math&gt;5^{20}&lt;/math&gt; should have &lt;math&gt;20 - 6&lt;/math&gt; digits since there are &lt;math&gt;6&lt;/math&gt; numbers which are &lt;math&gt;1 \mod 3&lt;/math&gt; from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;20&lt;/math&gt;, or &lt;math&gt;14&lt;/math&gt; digits total. This means our expression can be written as &lt;math&gt;\dfrac{k\cdot10^{14}}{10^{40}}&lt;/math&gt;, where &lt;math&gt;k&lt;/math&gt; is in the range &lt;math&gt;[1,10)&lt;/math&gt;. Canceling gives &lt;math&gt;\dfrac{k}{10^{26}}&lt;/math&gt;, or &lt;math&gt;26&lt;/math&gt; zeroes before the &lt;math&gt;k&lt;/math&gt; since the number &lt;math&gt;k&lt;/math&gt; should start on where the one would be in &lt;math&gt;10^{26}&lt;/math&gt;. ~aop2014<br /> <br /> ==Video Solution==<br /> https://youtu.be/t6yjfKXpwDs<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_12&diff=117559 2020 AMC 10B Problems/Problem 12 2020-02-09T18:49:41Z <p>Oreochocolate: typo</p> <hr /> <div>==Problem==<br /> <br /> The decimal representation of&lt;cmath&gt;\dfrac{1}{20^{20}}&lt;/cmath&gt;consists of a string of zeros after the decimal point, followed by a &lt;math&gt;9&lt;/math&gt; and then several more digits. How many zeros are in that initial string of zeros after the decimal point?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;cmath&gt;\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}&lt;/cmath&gt;<br /> <br /> Now we do some estimation. Notice that &lt;math&gt;2^{20} = 1024^2&lt;/math&gt;, which means that &lt;math&gt;2^{20}&lt;/math&gt; is a little more than &lt;math&gt;1000^2=1,000,000&lt;/math&gt;. Multiplying it with &lt;math&gt;10^{20}&lt;/math&gt;, we get that the denominator is about &lt;math&gt;1\underbrace{00\dots0}_{26 \text{ zeros}}&lt;/math&gt;. Notice that when we divide &lt;math&gt;1&lt;/math&gt; by an &lt;math&gt;n&lt;/math&gt; digit number, there are &lt;math&gt;n-1&lt;/math&gt; zeros before the first nonzero digit. This means that when we divide &lt;math&gt;1&lt;/math&gt; by the &lt;math&gt;27&lt;/math&gt; digit integer &lt;math&gt;1\underbrace{00\dots0}_{26 \text{ zeros}}&lt;/math&gt;, there are &lt;math&gt;\boxed{\textbf{(D) } \text{26}}&lt;/math&gt; zeros in the initial string after the decimal point. -PCChess<br /> <br /> ==Solution 2==<br /> First rewrite &lt;math&gt;\frac{1}{20^{20}}&lt;/math&gt; as &lt;math&gt;\frac{5^{20}}{10^{40}}&lt;/math&gt;. Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in &lt;math&gt;{5^{20}}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\log{5^{20}} = 20\log{5}&lt;/math&gt; and memming &lt;math&gt;\log{5}\approx0.69&lt;/math&gt; (alternatively use the fact that &lt;math&gt;\log{5} = 1 - \log{2}&lt;/math&gt;), &lt;math&gt;\lfloor{20\log{5}}\rfloor+1=\lfloor{20\cdot0.69}\rfloor+1=13+1=14&lt;/math&gt; digits. <br /> <br /> Our answer is &lt;math&gt;40-14=\boxed{26}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Brute Force)==<br /> Just as in Solution &lt;math&gt;2,&lt;/math&gt; we rewrite &lt;math&gt;\dfrac{1}{20^{20}}&lt;/math&gt; as &lt;math&gt;\dfrac{5^{20}}{10^{40}}.&lt;/math&gt; We then calculate &lt;math&gt;5^{20}&lt;/math&gt; entirely by hand, first doing &lt;math&gt;5^5 \cdot 5^5,&lt;/math&gt; then multiplying that product by itself, resulting in &lt;math&gt;95,367,431,640,625.&lt;/math&gt; Because this is &lt;math&gt;14&lt;/math&gt; digits, after dividing this number by &lt;math&gt;10&lt;/math&gt; fourteen times, the decimal point is before the &lt;math&gt;9.&lt;/math&gt; Dividing the number again by &lt;math&gt;10&lt;/math&gt; twenty-six more times allows a string of &lt;math&gt;\boxed{26~(D)}&lt;/math&gt; zeroes to be formed. -OreoChocolate<br /> <br /> ==Video Solution==<br /> https://youtu.be/t6yjfKXpwDs<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_12&diff=117268 2020 AMC 10B Problems/Problem 12 2020-02-08T03:48:48Z <p>Oreochocolate: Adding an okay but time-consuming solution...</p> <hr /> <div>==Problem==<br /> <br /> The decimal representation of&lt;cmath&gt;\dfrac{1}{20^{20}}&lt;/cmath&gt;consists of a string of zeros after the decimal point, followed by a &lt;math&gt;9&lt;/math&gt; and then several more digits. How many zeros are in that initial string of zeros after the decimal point?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;cmath&gt;\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}&lt;/cmath&gt;<br /> <br /> Now we do some estimation. Notice that &lt;math&gt;2^{20} = 1024^2&lt;/math&gt;, which means that &lt;math&gt;2^{20}&lt;/math&gt; is a little more than &lt;math&gt;1000^2=1,000,000&lt;/math&gt;. Multiplying it with &lt;math&gt;10^{20}&lt;/math&gt;, we get that the denominator is about &lt;math&gt;1\underbrace{00\dots0}_{26 \text{ zeros}}&lt;/math&gt;. Notice that when we divide &lt;math&gt;1&lt;/math&gt; by an &lt;math&gt;n&lt;/math&gt; digit number, there are &lt;math&gt;n-1&lt;/math&gt; zeros before the first nonzero digit. This means that when we divide &lt;math&gt;1&lt;/math&gt; by the &lt;math&gt;27&lt;/math&gt; digit integer &lt;math&gt;1\underbrace{00\dots0}_{26 \text{ zeros}}&lt;/math&gt;, there are &lt;math&gt;\boxed{\textbf{(D) } \text{26}}&lt;/math&gt; zeros in the initial string after the decimal point. -PCChess<br /> <br /> ==Solution 2==<br /> First rewrite &lt;math&gt;\frac{1}{20^{20}}&lt;/math&gt; as &lt;math&gt;\frac{5^{20}}{10^{40}}&lt;/math&gt;. Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in &lt;math&gt;{5^{20}}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\log{5^{20}} = 20\log{5}&lt;/math&gt; and memming &lt;math&gt;\log{5}\approx0.69&lt;/math&gt; (alternatively use the fact that &lt;math&gt;\log{5} = 1 - \log{2}&lt;/math&gt;), &lt;math&gt;\lfloor{20\log{5}}\rfloor+1=\lfloor{20\cdot0.69}\rfloor+1=13+1=14&lt;/math&gt; digits. <br /> <br /> Our answer is &lt;math&gt;40-14=\boxed{26}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Brute Force)==<br /> Just as in Solution &lt;math&gt;2,&lt;/math&gt; we rewrite &lt;math&gt;\dfrac{1}{20^{20}}&lt;/math&gt; as &lt;math&gt;\dfrac{5^{20}}{10^{40}}.&lt;/math&gt; We then calculate &lt;math&gt;5^{20}&lt;/math&gt; entirely by hand, first doing &lt;math&gt;5^5 \cdot 5^5,&lt;/math&gt; then multiplying that product by itself, resulting in &lt;math&gt;95,367,431,640,625.&lt;/math&gt; Because this is &lt;math&gt;14&lt;/math&gt; digits, after dividing this number by &lt;math&gt;10&lt;/math&gt; fourteen times, the decimal point is before the &lt;math&gt;9.&lt;/math&gt; Dividing the number again by &lt;math&gt;10&lt;/math&gt; twenty-six more times allows a string of &lt;math&gt;\boxed{26~(D)}&lt;/math&gt; numbers to be formed. -OreoChocolate<br /> <br /> ==Video Solution==<br /> https://youtu.be/t6yjfKXpwDs<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_19&diff=117244 2020 AMC 10A Problems/Problem 19 2020-02-08T03:23:52Z <p>Oreochocolate: Listing out solutions number by number</p> <hr /> <div>== Problem ==<br /> As shown in the figure below, a regular dodecahedron (the polyhedron consisting of &lt;math&gt;12&lt;/math&gt; congruent regular pentagonal faces) floats in space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?<br /> <br /> &lt;math&gt;\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810&lt;/math&gt;<br /> <br /> == Diagram ==<br /> &lt;asy&gt;<br /> import graph;<br /> unitsize(5cm);<br /> pair A = (0.082, 0.378);<br /> pair B = (0.091, 0.649);<br /> pair C = (0.249, 0.899);<br /> pair D = (0.479, 0.939);<br /> pair E = (0.758, 0.893);<br /> pair F = (0.862, 0.658);<br /> pair G = (0.924, 0.403);<br /> pair H = (0.747, 0.194);<br /> pair I = (0.526, 0.075);<br /> pair J = (0.251, 0.170);<br /> pair K = (0.568, 0.234);<br /> pair L = (0.262, 0.449);<br /> pair M = (0.373, 0.813);<br /> pair N = (0.731, 0.813);<br /> pair O = (0.851, 0.461);<br /> path[] f;<br /> f = A--B--C--M--L--cycle;<br /> f = C--D--E--N--M--cycle;<br /> f = E--F--G--O--N--cycle;<br /> f = G--H--I--K--O--cycle;<br /> f = I--J--A--L--K--cycle;<br /> f = K--L--M--N--O--cycle;<br /> draw(f);<br /> axialshade(f, white, M, gray(0.5), (C+2*D)/3);<br /> draw(f);<br /> filldraw(f, gray);<br /> filldraw(f, gray);<br /> axialshade(f, white, L, gray(0.7), J);<br /> draw(f);<br /> draw(f);<br /> &lt;/asy&gt;<br /> <br /> == Solution 1 ==<br /> Since we start at the top face and end at the bottom face without moving from the lower ring to the upper ring or revisiting a face, our journey must consist of the top face, a series of faces in the upper ring, a series of faces in the lower ring, and the bottom face, in that order.<br /> <br /> We have &lt;math&gt;5&lt;/math&gt; choices for which face we visit first on the top ring. From there, we have &lt;math&gt;9&lt;/math&gt; choices for how far around the top ring we go before moving down: &lt;math&gt;1,2,3,&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; faces around clockwise, &lt;math&gt;1,2,3,&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; faces around counterclockwise, or immediately going down to the lower ring without visiting any other faces in the upper ring.<br /> <br /> We then have &lt;math&gt;2&lt;/math&gt; choices for which lower ring face to visit first (since every upper-ring face is adjacent to exactly &lt;math&gt;2&lt;/math&gt; lower-ring faces) and then once again &lt;math&gt;9&lt;/math&gt; choices for how to travel around the lower ring. We then proceed to the bottom face, completing the trip.<br /> <br /> Multiplying together all the numbers of choices we have, we get &lt;math&gt;5 \cdot 9 \cdot 2 \cdot 9 = \boxed{\textbf{(E) } 810}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Swap the faces as vertices and the vertices as faces. Then, this problem is the same as<br /> [https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3 2016 AIME I #3]<br /> which had an answer of &lt;math&gt;\boxed{\textbf{(E) } 810}&lt;/math&gt;.<br /> &lt;math&gt;\textbf{\textbf{- Emathmaster}}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3<br /> {{AMC10 box|year=2020|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Oreochocolate https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_21&diff=105396 2019 AMC 10A Problems/Problem 21 2019-04-20T02:33:51Z <p>Oreochocolate: /* Solution 3 (similar triangles) */</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #21]] and [[2019 AMC 12A Problems|2019 AMC 12A #18]]}}<br /> <br /> ==Problem==<br /> <br /> A sphere with center &lt;math&gt;O&lt;/math&gt; has radius &lt;math&gt;6&lt;/math&gt;. A triangle with sides of length &lt;math&gt;15, 15,&lt;/math&gt; and &lt;math&gt;24&lt;/math&gt; is situated in space so that each of its sides is tangent to the sphere. What is the distance between &lt;math&gt;O&lt;/math&gt; and the plane determined by the triangle?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }2\sqrt{3}\qquad<br /> \textbf{(B) }4\qquad<br /> \textbf{(C) }3\sqrt{2}\qquad<br /> \textbf{(D) }2\sqrt{5}\qquad<br /> \textbf{(E) }5\qquad<br /> &lt;/math&gt;<br /> <br /> ==Diagram==<br /> 3D:<br /> &lt;asy&gt;<br /> import graph3;<br /> import palette;<br /> size(200);<br /> currentprojection=orthographic(0,4,2);<br /> <br /> triple f(pair z) {return expi(z.x,z.y);}<br /> <br /> surface s=surface(f,(0,0),(pi,2pi),70,Spline);<br /> draw((0,-5/6,sqrt(5)/3)--(2,2/3,sqrt(5)/3)--(-2,2/3,sqrt(5)/3)--cycle);<br /> draw(s,mean(palette(s.map(zpart),Grayscale())),nolight);<br /> draw((2,2/3,sqrt(5)/3)--(-2,2/3,sqrt(5)/3));<br /> &lt;/asy&gt;<br /> Plane through triangle:<br /> &lt;asy&gt;<br /> draw((0,0)--(12,9)--(24,0)--cycle);<br /> draw((12,9)--(12,0), dashed);<br /> draw((11.5,0)--(11.5,0.5)--(12,0.5));<br /> draw(circle((12,4),4));<br /> draw((12,4)--(48/5, 36/5));<br /> dot((12,4));<br /> label(&quot;$15$&quot;, (6,9/2),NW);<br /> label(&quot;$15$&quot;, (18,9/2),NE);<br /> label(&quot;$24$&quot;, (12,-1),S);<br /> label(&quot;$r$&quot;,(54/5, 28/5), SW);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> The triangle is placed on the sphere so that its three sides are tangent to the sphere. The cross-section of the sphere created by the plane of the triangle is also the incircle of the triangle. To find the inradius, use &lt;math&gt;\text{area} = \text{inradius} \cdot \text{semiperimeter}&lt;/math&gt;. The area of the triangle can be found by drawing an altitude from the vertex between sides with length &lt;math&gt;15&lt;/math&gt; to the midpoint of the side with length &lt;math&gt;24&lt;/math&gt;. The Pythagorean triple &lt;math&gt;9&lt;/math&gt; - &lt;math&gt;12&lt;/math&gt; - &lt;math&gt;15&lt;/math&gt; allows us easily to determine that the base is &lt;math&gt;24&lt;/math&gt; and the height is &lt;math&gt;9&lt;/math&gt;. The formula &lt;math&gt;\frac {\text{base} \cdot \text{height}} {2}&lt;/math&gt; can also be used to find the area of the triangle as &lt;math&gt;108&lt;/math&gt;, while the semiperimeter is simply &lt;math&gt;\frac {15 + 15 + 24} {2} = 27&lt;/math&gt;. After plugging into the equation, we thus get &lt;math&gt;108 = \text{inradius} \cdot 27&lt;/math&gt;, so the inradius is &lt;math&gt;4&lt;/math&gt;. Now, let the distance between &lt;math&gt;O&lt;/math&gt; and the triangle be &lt;math&gt;x&lt;/math&gt;. Choose a point on the incircle and denote it by &lt;math&gt;A&lt;/math&gt;. The distance &lt;math&gt;OA&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt;, because it is just the radius of the sphere. The distance from point &lt;math&gt;A&lt;/math&gt; to the center of the incircle is &lt;math&gt;4&lt;/math&gt;, because it is the radius of the incircle. By using the Pythagorean Theorem, we thus find &lt;math&gt;x = \sqrt{6^2-4^2}=\sqrt{20} = \boxed{\textbf {(D) } 2 \sqrt {5}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> As in Solution 1, we note that by the Pythagorean Theorem, the height of the triangle is &lt;math&gt;9&lt;/math&gt;, and that the three sides of the triangle are tangent to the sphere, so the circle in the cross-section of the sphere is the incenter of the triangle. <br /> <br /> Recall that the inradius is the intersection of the angle bisectors. To find the inradius of the incircle, we use the Angle Bisector Theorem.<br /> &lt;asy&gt;<br /> draw((0,0)--(12,9)--(24,0)--cycle);<br /> dot((0,0));<br /> dot((12,9));<br /> dot((24,0));<br /> dot((12,0));<br /> label(&quot;$A$&quot;,(0,0),SW);<br /> label(&quot;$B$&quot;,(12,9),N);<br /> label(&quot;$C$&quot;,(24,0),SE);<br /> label(&quot;$D$&quot;,(12,-1/2),S);<br /> label(&quot;$I$&quot;,(12,4),SE);<br /> <br /> draw((12,9)--(12,0), dashed);<br /> draw(circle((12,4),4));<br /> draw((0,0)--(216/13,216/39));<br /> dot((12,4));<br /> label(&quot;$15$&quot;, (6,9/2),NW);<br /> label(&quot;$12$&quot;, (6,-1),S);<br /> <br /> &lt;/asy&gt;<br /> &lt;cmath&gt;\begin{split}&amp;\frac{AB}{BI}=\frac{AD}{DI} \\<br /> \Rightarrow \ &amp;\frac{15}{BI}=\frac{12}{DI} \\<br /> \Rightarrow \ &amp;\frac{BI}{5}=\frac{DI}{4}\end{split}&lt;/cmath&gt;<br /> Since we know that &lt;math&gt;BI+DI&lt;/math&gt; (the height) is equal to &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;DI&lt;/math&gt; (the inradius) is &lt;math&gt;4&lt;/math&gt;.<br /> From here, the problem can be solved in the same way as in Solution 1. The answer is &lt;math&gt;\boxed{\textbf {(D) } 2 \sqrt{5}}&lt;/math&gt;.<br /> <br /> ==Solution 3 (similar triangles)==<br /> First, we label a few points:<br /> &lt;asy&gt;<br /> draw((0,0)--(12,9)--(24,0)--cycle);<br /> draw((12,9)--(12,0), dashed);<br /> draw((11.5,0)--(11.5,0.5)--(12,0.5));<br /> draw(circle((12,4),4));<br /> draw((12,4)--(48/5, 36/5));<br /> dot((12,4));<br /> label(&quot;$15$&quot;, (6,9/2),NW);<br /> label(&quot;$15$&quot;, (18,9/2),NE);<br /> <br /> label(&quot;$r$&quot;,(54/5, 28/5), SW);<br /> label(&quot;$12$&quot;, (6,-1),S);<br /> label(&quot;$I$&quot;,(12,4),SE);<br /> label(&quot;$A$&quot;,(0,0),SW);<br /> label(&quot;$B$&quot;,(12,9),N);<br /> label(&quot;$C$&quot;,(24,0),SE);<br /> label(&quot;$D$&quot;,(12,-1/2),S);<br /> label(&quot;$E$&quot;,(48/5, 36/5),NW);<br /> &lt;/asy&gt;<br /> <br /> We have that &lt;math&gt;\triangle{BDC}&lt;/math&gt; is a &lt;math&gt;3-4-5&lt;/math&gt; triangle, so, as in Solution 1, &lt;math&gt;BD = 9&lt;/math&gt;.<br /> From this, we know that &lt;math&gt;\overline{BI}=9-r&lt;/math&gt;.<br /> Since &lt;math&gt;AB&lt;/math&gt; is tangent to circle &lt;math&gt;I&lt;/math&gt;, we also know &lt;math&gt;IEB&lt;/math&gt; is a right triangle. &lt;math&gt;\triangle{BIE}&lt;/math&gt; and &lt;math&gt;\triangle{BDA}&lt;/math&gt; share angle &lt;math&gt;DBA&lt;/math&gt;, so &lt;math&gt;\triangle{BIE} \sim \triangle{BDA}&lt;/math&gt; since they have two equal angles.<br /> Hence, by this similarity, &lt;math&gt;\dfrac{9-r}{5}=\dfrac{r}{4}&lt;/math&gt;. Cross-multiplying, we get &lt;math&gt;36-4r =5r&lt;/math&gt;, which gives &lt;math&gt;r=4&lt;/math&gt;.<br /> We now take another cross section of the sphere, perpendicular to the plane of the triangle.<br /> <br /> &lt;asy&gt;<br /> draw(circle((6,6),6));<br /> draw((6,6)--(1.75735931,1.75735931)--(6,1.75735931)--cycle);<br /> dot((6,6));<br /> dot((1.75735931,1.75735931));<br /> dot((6,1.75735931));<br /> <br /> label(&quot;$O$&quot;, (6,6),N);<br /> label(&quot;$6$&quot;, (3.87867965,3.87867965),NW);<br /> label(&quot;$4$&quot;, (3.87867965,1.75735931),SE);<br /> &lt;/asy&gt;<br /> <br /> Using the Pythagorean Theorem, we find that the distance from the center to the plane is &lt;math&gt;\boxed{\textbf {(D) } 2 \sqrt{5}}&lt;/math&gt;.<br /> <br /> solution by woofle628 and GeniusKid1221<br /> <br /> ==Solution 4 (educated guess)==<br /> Test all the answer choices by plugging them into the expression &lt;math&gt;\sqrt{6^2 - x^2}&lt;/math&gt; to find the inradius of the triangle. Seeing that only &lt;math&gt;\sqrt{20} = 2\sqrt{5}&lt;/math&gt; gives an integer inradius, we pick &lt;math&gt;\boxed{\textbf {(D) } 2 \sqrt{5}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2019|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Oreochocolate