https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Firebolt360&feedformat=atom AoPS Wiki - User contributions [en] 2020-12-02T19:06:41Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_15&diff=138796 2013 AIME II Problems/Problem 15 2020-11-29T22:45:14Z <p>Firebolt360: /* Problem 15 */</p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;A,B,C&lt;/math&gt; be angles of an acute triangle with<br /> &lt;cmath&gt; \begin{align*}<br /> \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &amp;= \frac{15}{8} \text{ and} \\<br /> \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &amp;= \frac{14}{9}<br /> \end{align*} &lt;/cmath&gt;<br /> There are positive integers &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt;, &lt;math&gt;r&lt;/math&gt;, and &lt;math&gt;s&lt;/math&gt; for which &lt;cmath&gt; \cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s}, &lt;/cmath&gt; where &lt;math&gt;p+q&lt;/math&gt; and &lt;math&gt;s&lt;/math&gt; are relatively prime and &lt;math&gt;r&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;p+q+r+s&lt;/math&gt;.<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> Let's draw the triangle. Since the problem only deals with angles, we can go ahead and set one of the sides to a convenient value. Let &lt;math&gt;BC = \sin{A}&lt;/math&gt;.<br /> <br /> By the Law of Sines, we must have &lt;math&gt;CA = \sin{B}&lt;/math&gt; and &lt;math&gt;AB = \sin{C}&lt;/math&gt;.<br /> <br /> Now let us analyze the given:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> \cos^2A + \cos^2B + 2\sin A\sin B\cos C &amp;= 1-\sin^2A + 1-\sin^2B + 2\sin A\sin B\cos C \\<br /> &amp;= 2-(\sin^2A + \sin^2B - 2\sin A\sin B\cos C)<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Now we can use the Law of Cosines to simplify this:<br /> <br /> &lt;cmath&gt;= 2-\sin^2C&lt;/cmath&gt;<br /> <br /> <br /> Therefore: &lt;cmath&gt;\sin C = \sqrt{\dfrac{1}{8}},\cos C = \sqrt{\dfrac{7}{8}}.&lt;/cmath&gt; Similarly, &lt;cmath&gt;\sin A = \sqrt{\dfrac{4}{9}},\cos A = \sqrt{\dfrac{5}{9}}.&lt;/cmath&gt; Note that the desired value is equivalent to &lt;math&gt;2-\sin^2B&lt;/math&gt;, which is &lt;math&gt;2-\sin^2(A+C)&lt;/math&gt;. All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of &lt;math&gt;\dfrac{111-4\sqrt{35}}{72}&lt;/math&gt;. Thus, the answer is &lt;math&gt;111+4+35+72 = \boxed{222}&lt;/math&gt;.<br /> <br /> Note that the problem has a flaw because &lt;math&gt;\cos B &lt; 0&lt;/math&gt; which contradicts with the statement that it's an acute triangle. Would be more accurate to state that &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are smaller than 90. -Mathdummy<br /> <br /> ===Solution 2===<br /> Let us use the identity &lt;math&gt;\cos^2A+\cos^2B+\cos^2C+2\cos A \cos B \cos C=1&lt;/math&gt; .<br /> <br /> Add &lt;cmath&gt;\begin{align*} \cos^2 C+2(\cos A\cos B-\sin A \sin B)\cos C \end{align*}&lt;/cmath&gt; to both sides of the first given equation. <br /> <br /> <br /> <br /> <br /> Thus, as &lt;cmath&gt;\begin{align*} \cos A\cos B-\sin A\sin B=\cos (A+B)=-\cos C ,\end{align*}&lt;/cmath&gt; <br /> we have &lt;cmath&gt;\begin{align*} \dfrac{15}{8}-2\cos^2 C +\cos^2 C=1, <br /> \end{align*}&lt;/cmath&gt; so &lt;math&gt;\cos C&lt;/math&gt; is &lt;math&gt;\sqrt{\dfrac{7}{8}}&lt;/math&gt; and therefore &lt;math&gt; \sin C&lt;/math&gt; is &lt;math&gt;\sqrt{\dfrac{1}{8}}&lt;/math&gt;. <br /> <br /> Similarily, we have &lt;math&gt;\sin A =\dfrac{2}{3}&lt;/math&gt; and &lt;math&gt;\cos A=\sqrt{\dfrac{14}{9}-1}=\sqrt{\dfrac{5}{9}}&lt;/math&gt; and the rest of the solution proceeds as above.<br /> <br /> ===Solution 3===<br /> <br /> Let <br /> &lt;cmath&gt; \begin{align*}<br /> \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &amp;= \frac{15}{8} \text{ ------ (1)}\\<br /> \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &amp;= \frac{14}{9} \text{ ------ (2)}\\<br /> \cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B &amp;= x \text{ ------ (3)}\\<br /> \end{align*} &lt;/cmath&gt;<br /> <br /> Adding (1) and (3) we get:<br /> &lt;cmath&gt; 2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A( \sin B \cos C + \sin C \cos B) = \frac{15}{8} + x&lt;/cmath&gt; or<br /> &lt;cmath&gt; 2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A \sin (B+C) = \frac{15}{8} + x&lt;/cmath&gt; or<br /> &lt;cmath&gt; 2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin ^2 A = \frac{15}{8} + x&lt;/cmath&gt; or<br /> &lt;cmath&gt; \cos^2 B + \cos^2 C = x - \frac{1}{8} \text{ ------ (4)}&lt;/cmath&gt;<br /> <br /> Similarly adding (2) and (3) we get:<br /> &lt;cmath&gt; \cos^2 A + \cos^2 B = x - \frac{4}{9} \text{ ------ (5)} &lt;/cmath&gt;<br /> Similarly adding (1) and (2) we get:<br /> &lt;cmath&gt; \cos^2 A + \cos^2 C = \frac{14}{9} - \frac{1}{8} \text{ ------ (6)} &lt;/cmath&gt;<br /> <br /> And (4) - (5) gives:<br /> &lt;cmath&gt; \cos^2 C - \cos^2 A = \frac{4}{9} - \frac{1}{8} \text{ ------ (7)} &lt;/cmath&gt;<br /> <br /> Now (6) - (7) gives:<br /> &lt;math&gt; \cos^2 A = \frac{5}{9} &lt;/math&gt; or<br /> &lt;math&gt;\cos A = \sqrt{\dfrac{5}{9}}&lt;/math&gt; and &lt;math&gt;\sin A = \frac{2}{3} &lt;/math&gt;<br /> so &lt;math&gt;\cos C&lt;/math&gt; is &lt;math&gt;\sqrt{\dfrac{7}{8}}&lt;/math&gt; and therefore &lt;math&gt; \sin C&lt;/math&gt; is &lt;math&gt;\sqrt{\dfrac{1}{8}}&lt;/math&gt;<br /> <br /> Now &lt;math&gt;\sin B = \sin(A+C)&lt;/math&gt; can be computed first and then &lt;math&gt;\cos^2 B&lt;/math&gt; is easily found.<br /> <br /> Thus &lt;math&gt;\cos^2 B&lt;/math&gt; and &lt;math&gt;\cos^2 C&lt;/math&gt; can be plugged into (4) above to give x = &lt;math&gt;\dfrac{111-4\sqrt{35}}{72}&lt;/math&gt;.<br /> <br /> Hence the answer is = &lt;math&gt;111+4+35+72 = \boxed{222}&lt;/math&gt;.<br /> <br /> Kris17<br /> <br /> === Solution 4 ===<br /> <br /> Let's take the first equation &lt;math&gt;\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C = \frac{15}{8}&lt;/math&gt;. Substituting &lt;math&gt;180 - A - B&lt;/math&gt; for C, given A, B, and C form a triangle, and that &lt;math&gt;\cos C = \cos(A + B)&lt;/math&gt;, gives us:<br /> <br /> &lt;math&gt;\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos (A+B) = \frac{15}{8}&lt;/math&gt;<br /> <br /> Expanding out gives us &lt;math&gt;\cos^2 A + \cos^2 B + 2 \sin^2 A \sin^2 B - 2 \sin A \sin B \cos A \cos B = \frac{15}{8}&lt;/math&gt;. <br /> <br /> Using the double angle formula &lt;math&gt;\cos^2 k = \frac{\cos (2k) + 1}{2}&lt;/math&gt;, we can substitute for each of the squares &lt;math&gt;\cos^2 A&lt;/math&gt; and &lt;math&gt;\cos^2 B&lt;/math&gt;. Next we can use the Pythagorean identity on the &lt;math&gt;\sin^2 A&lt;/math&gt; and &lt;math&gt;\sin^2 B&lt;/math&gt; terms. Lastly we can use the sine double angle to simplify.<br /> <br /> &lt;math&gt;\cos^2 A + \cos^2 B + 2(1 - \cos^2 A)(1 - \cos^2 B) - \frac{1}{2} \cdot \sin 2A \sin 2B = \frac{15}{8}&lt;/math&gt;. <br /> <br /> Expanding and canceling yields, and again using double angle substitution, <br /> <br /> &lt;math&gt;1 + 2 \cdot \frac{\cos (2A) + 1}{2} \cdot \frac{\cos (2B) + 1}{2} - \frac{1}{2} \cdot \sin 2A \sin 2B = \frac{15}{8}&lt;/math&gt;. <br /> <br /> Further simplifying yields:<br /> <br /> &lt;math&gt;\frac{3}{2} + \frac{\cos 2A \cos 2B - \sin 2A \sin 2B}{2} = \frac{15}{8}&lt;/math&gt;. <br /> <br /> Using cosine angle addition formula and simplifying further yields, and applying the same logic to Equation &lt;math&gt;2&lt;/math&gt; yields:<br /> <br /> &lt;math&gt;\cos (2A + 2B) = \frac{3}{4}&lt;/math&gt; and &lt;math&gt;\cos (2B + 2C) = \frac{1}{9}&lt;/math&gt;.<br /> <br /> Substituting the identity &lt;math&gt;\cos (2A + 2B) = \cos(2C)&lt;/math&gt;, we get:<br /> <br /> &lt;math&gt;\cos (2C) = \frac{3}{4}&lt;/math&gt; and &lt;math&gt;\cos (2A) = \frac{1}{9}&lt;/math&gt;.<br /> <br /> Since the third expression simplifies to the expression &lt;math&gt;\frac{3}{2} + \frac{\cos (2A + 2C)}{2}&lt;/math&gt;, taking inverse cosine and using the angles in angle addition formula yields the answer, &lt;math&gt;\frac{111 - 4\sqrt{35}}{72}&lt;/math&gt;, giving us the answer &lt;math&gt;\boxed{222}&lt;/math&gt;.<br /> <br /> ===Solution 5===<br /> We will use the sum to product formula to simply these equations. Recall &lt;cmath&gt;2\sin{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}} = \cos{\alpha}+\cos{\beta}.&lt;/cmath&gt; Using this, let's rewrite the first equation: &lt;cmath&gt;\cos^2(A) + \cos^2(B) + 2 \sin(A) \sin(B) \cos(C) = \frac{15}{8}&lt;/cmath&gt; &lt;cmath&gt;\cos^2(A) + \cos^2(B) + (\cos(A+B)+\cos(A-B))\cos(C).&lt;/cmath&gt; Now, note that &lt;math&gt;\cos(C)=-\cos(A+B)&lt;/math&gt;. &lt;cmath&gt;\cos^2(A) + \cos^2(B) + (\cos(A+B)+\cos(A-B))(-\cos(A+B))&lt;/cmath&gt; &lt;cmath&gt;\cos^2(A) + \cos^2(B) - \cos(A+B)\cos(A-B)+cos^2(A+B)=\frac{15}{8}.&lt;/cmath&gt; We apply the sum to product formula again. &lt;cmath&gt;\cos^2(A) + \cos^2(B) - \frac{\cos(2A)+\cos(2B)}{2}+cos^2(A+B)=\frac{15}{8}.&lt;/cmath&gt; Now, recall that &lt;math&gt;\cos(2\alpha)=2\cos^2(\alpha)-1&lt;/math&gt;. We apply this and simplify our expression to get: &lt;cmath&gt;\cos^2(A+B)=\frac{7}{8}&lt;/cmath&gt; &lt;cmath&gt;\cos^2(C)=\frac{7}{8}.&lt;/cmath&gt; Analogously, &lt;cmath&gt;\cos^2(A)=\frac{5}{9}.&lt;/cmath&gt; &lt;cmath&gt;\cos^2(A+C)=\frac{p-q\sqrt{r}}{s}-1.&lt;/cmath&gt; We can find this value easily by angle sum formula. After a few calculations, we get &lt;math&gt;\frac{111 - 4\sqrt{35}}{72}&lt;/math&gt;, giving us the answer &lt;math&gt;\boxed{222}&lt;/math&gt;.<br /> <br /> <br /> ~superagh<br /> <br /> ==See Also==<br /> {{AIME box|year=2013|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_21&diff=138685 2009 AMC 12B Problems/Problem 21 2020-11-29T00:58:33Z <p>Firebolt360: /* Clarification of Solution 2 */</p> <hr /> <div>== Problem ==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;Ten women sit in &lt;math&gt;10&lt;/math&gt; seats in a line. All of the &lt;math&gt;10&lt;/math&gt; get up and then reseat themselves using all &lt;math&gt;10&lt;/math&gt; seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 2^{10}\qquad \textbf{(E)}\ 2^2 3^8&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Notice that either a woman stays in her own seat after the rearrangement, or two adjacent women swap places. Thus, our answer is counting the number of ways to arrange 1x1 and 2x1 blocks to form a 1x10 rectangle. This can be done via casework depending on the number of 2x1 blocks. The cases of 0, 1, 2, 3, 4, 5 2x1 blocks correspond to 10, 8, 6, 4, 2, 0 1x1 blocks, and so the sum of the cases is<br /> &lt;cmath&gt;\binom{10}{0} + \binom{9}{1} + \binom{8}{2} + \binom{7}{3} + \binom{6}{4} + \binom{5}{5} = 1 + 9 + 28 + 35 + 15 + 1 = \boxed{89}.&lt;/cmath&gt;<br /> <br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;S_n&lt;/math&gt; be the number of possible seating arrangements with &lt;math&gt;n&lt;/math&gt; women. Consider &lt;math&gt;n \ge 3,&lt;/math&gt; and focus on the rightmost woman. If she returns back to her seat, then there are &lt;math&gt;S_{n-1}&lt;/math&gt; ways to seat the remaining &lt;math&gt;n-1&lt;/math&gt; women. If she sits in the second to last seat, then the woman who previously sat there must now sit at the rightmost seat. This gives us &lt;math&gt;S_{n-2}&lt;/math&gt; ways to seat the other &lt;math&gt;n-2&lt;/math&gt; women, so we obtain the recursion<br /> &lt;cmath&gt;S_n = S_{n-1}+S_{n-2}.&lt;/cmath&gt;<br /> <br /> Starting with &lt;math&gt;S_1=1&lt;/math&gt; and &lt;math&gt;S_2=2,&lt;/math&gt; we can calculate &lt;math&gt;S_{10}=\boxed{89}.&lt;/math&gt;<br /> <br /> ==Clarification of Solution 2==<br /> The seating possibilities of woman #&lt;math&gt;10&lt;/math&gt; become the two cases which we work out. &lt;math&gt;S_n&lt;/math&gt; was defined to be the number of different seating arrangements with &lt;math&gt;n&lt;/math&gt; women.<br /> <br /> In the first &quot;case&quot; if woman #&lt;math&gt;10&lt;/math&gt; sits in the seat #&lt;math&gt;10&lt;/math&gt;, this leads to a similar scenario, but with &lt;math&gt;9&lt;/math&gt; women instead. That means that for this case, there are a total of &lt;math&gt;S_{9}&lt;/math&gt; possible arrangements. We don't know how many exactly, but we are able to define it in terms of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> During the second &quot;case&quot;, woman #&lt;math&gt;10&lt;/math&gt; sits in seat #&lt;math&gt;9&lt;/math&gt;. This time, woman #9 must go to seat #&lt;math&gt;10&lt;/math&gt;, as she is the only other person who can go there. This leaves us with &lt;math&gt;8&lt;/math&gt; women, and we again represent this in terms of &lt;math&gt;S \Rightarrow S_8&lt;/math&gt;.<br /> <br /> Therefore, we can write &lt;math&gt;S_{10}&lt;/math&gt; in terms of &lt;math&gt;S_8&lt;/math&gt; and &lt;math&gt;S_9&lt;/math&gt;, like so:<br /> <br /> &lt;cmath&gt;S_{10} = S_8 + S_9.&lt;/cmath&gt;<br /> <br /> We can then generalize this to say<br /> <br /> &lt;cmath&gt;S_n = S_{n-1}+S_{n-2}.&lt;/cmath&gt;<br /> <br /> Calculating &lt;math&gt;S_1 = 1&lt;/math&gt; and &lt;math&gt;S_2 =2,&lt;/math&gt; then following the recursive rule from above, we get &lt;math&gt;S_{10} = 89 \Rightarrow \boxed{\text{A}}&lt;/math&gt;.<br /> ==Solution 3==<br /> let &lt;math&gt;a_n=&lt;/math&gt; the number of ways to order n women as per the constraints of the problem. We wish to find &lt;math&gt;a_{10}&lt;/math&gt;. Lets take a look at 3 cases regarding the 2 women on the outside:<br /> <br /> Case 1: Both women stay in the same spot<br /> Then, we wish to order the inner people. This gives us &lt;math&gt;a_8&lt;/math&gt; ways to do so<br /> <br /> Case 2: One woman stays in her spot, and the other moves<br /> In this case, the woman that moves has to swap spots with the person right next to her, because that is the only way to use all 10 seats. If this doesn't make sense, look at the diagram below (each star is a woman, and the a represents the woman who will move:<br /> <br /> a * * * * * * * * b<br /> <br /> After &quot;b&quot; moves:<br /> <br /> a * * * * * * * b ?<br /> <br /> Now, the * has to go somewhere. If the star doesn't fill the spot that &quot;b&quot; was in, then only 9 seats would be used. Thus, the star has to go to the spot where &quot;b&quot; was, giving:<br /> <br /> a * * * * * * * b *<br /> <br /> This leaves us to order the 7 people who have not yet decided where they want to go. Since either of the women can move, this gives us &lt;math&gt;2a_7&lt;/math&gt;.<br /> <br /> Finally, Case 3: Both women switch<br /> In this case, we only have to order the inner 6 people, because the two women will swap spots with the people right next to them (see logic above). Thus, this yeilds &lt;math&gt;a_6&lt;/math&gt;<br /> <br /> So: &lt;math&gt;a_10=a_8+a_6+2a_7&lt;/math&gt; We can generalize this to &lt;math&gt;a_{n+2}=a_n+a_{n-2}+2a_{n-1}&lt;/math&gt;<br /> <br /> Now, we calculate by hand that:<br /> &lt;cmath&gt;a_1=1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_2=2&lt;/cmath&gt;<br /> &lt;cmath&gt;a_3=3&lt;/cmath&gt;<br /> &lt;cmath&gt;a_4=5&lt;/cmath&gt;<br /> Finally, we use the recurrence relation we developed to build our way to &lt;math&gt;a_{10}&lt;/math&gt;<br /> &lt;cmath&gt;a_5=8&lt;/cmath&gt;<br /> &lt;cmath&gt;a_6=13&lt;/cmath&gt;<br /> &lt;cmath&gt;a_7=21&lt;/cmath&gt;<br /> &lt;cmath&gt;a_8=34&lt;/cmath&gt;<br /> &lt;cmath&gt;a_9=55&lt;/cmath&gt;<br /> &lt;cmath&gt;a_{10}=89&lt;/cmath&gt;<br /> So, we get \boxed{10 \left(\text{A}\right)}<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=B|num-b=20|num-a=22}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_25&diff=138650 2009 AMC 12A Problems/Problem 25 2020-11-28T18:34:54Z <p>Firebolt360: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> The first two terms of a sequence are &lt;math&gt;a_1 = 1&lt;/math&gt; and &lt;math&gt;a_2 = \frac {1}{\sqrt3}&lt;/math&gt;. For &lt;math&gt;n\ge1&lt;/math&gt;,<br /> <br /> &lt;center&gt;&lt;cmath&gt;a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> What is &lt;math&gt;|a_{2009}|&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2 - \sqrt3\qquad \textbf{(C)}\ \frac {1}{\sqrt3}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2 + \sqrt3&lt;/math&gt;<br /> <br /> == Solution ==<br /> Consider another sequence &lt;math&gt;\{\theta_1, \theta_2, \theta_3...\}&lt;/math&gt; such that &lt;math&gt;a_n = \tan{\theta_n}&lt;/math&gt;, and &lt;math&gt;0 \leq \theta_n &lt; 180&lt;/math&gt;.<br /> <br /> The given recurrence becomes<br /> <br /> &lt;center&gt;&lt;cmath&gt;\begin{align*} a_{n + 2} &amp; = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}} \\<br /> \tan{\theta_{n + 2}} &amp; = \frac {\tan{\theta_n} + \tan{\theta_{n + 1}}}{1 - \tan{\theta_n}\tan{\theta_{n + 1}}} \\<br /> \tan{\theta_{n + 2}} &amp; = \tan(\theta_{n + 1} + \theta_n) \end{align*}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> It follows that &lt;math&gt;\theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}&lt;/math&gt;. Since &lt;math&gt;\theta_1 = 45, \theta_2 = 30&lt;/math&gt;, all terms in the sequence &lt;math&gt;\{\theta_1, \theta_2, \theta_3...\}&lt;/math&gt; will be a multiple of &lt;math&gt;15&lt;/math&gt;.<br /> <br /> Now consider another sequence &lt;math&gt;\{b_1, b_2, b_3...\}&lt;/math&gt; such that &lt;math&gt;b_n = \theta_n/15&lt;/math&gt;, and &lt;math&gt;0 \leq b_n &lt; 12&lt;/math&gt;. The sequence &lt;math&gt;b_n&lt;/math&gt; satisfies &lt;math&gt;b_1 = 3, b_2 = 2, b_{n + 2} \equiv b_{n + 1} + b_n \pmod{12}&lt;/math&gt;.<br /> <br /> As the number of possible consecutive two terms is finite, we know that the sequence &lt;math&gt;b_n&lt;/math&gt; is periodic. Write out the first few terms of the sequence until it starts to repeat.<br /> <br /> &lt;math&gt;\{b_n\} = \{3,2,5,7,0,7,7,2,9,11,8,7,3,10,1,11,0,11,11,10,9,7,4,11,3,2,5,7,...\}&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;b_{25} = b_1 = 3&lt;/math&gt; and &lt;math&gt;b_{26} = b_2 = 2&lt;/math&gt;. Thus &lt;math&gt;\{b_n\}&lt;/math&gt; has a period of &lt;math&gt;24&lt;/math&gt;: &lt;math&gt;b_{n + 24} = b_n&lt;/math&gt;.<br /> <br /> It follows that &lt;math&gt;b_{2009} = b_{17} = 0&lt;/math&gt; and &lt;math&gt;\theta_{2009} = 15 b_{2009} = 0&lt;/math&gt;. Thus &lt;math&gt;a_{2009} = \tan{\theta_{2009}} = \tan{0} = 0.&lt;/math&gt;<br /> <br /> Our answer is &lt;math&gt;|a_{2009}| = \boxed{\textbf{(A)}\ 0}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> First, some intuition. The given recurrence relation: &lt;math&gt;a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}&lt;/math&gt; looks much like the tangent addition formula. So, we let &lt;math&gt;a_n=\tan{\theta_n}&lt;/math&gt;. Then, we get:<br /> &lt;cmath&gt;\tan{\theta_n}=\tan{(\theta_{n-1}+\theta_{n-2})}&lt;/cmath&gt;<br /> This gives us:<br /> &lt;cmath&gt;\theta_n \equiv \theta_{n-1}+\theta_{n-2} \pmod{180}&lt;/cmath&gt;<br /> Now, observe that <br /> &lt;cmath&gt;\theta_1=45&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_2=30&lt;/cmath&gt;<br /> We know that this sequence of values for &lt;math&gt;\theta_n&lt;/math&gt; will repeat eventually because it is mod &lt;math&gt;180&lt;/math&gt;.<br /> It is just a matter of when, so we start bashing:<br /> <br /> &lt;cmath&gt;\theta_3=75&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_4=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_6=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_7=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_8=30&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_9=135&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{10}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{11}=120&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{12}=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{13}=45&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{14}=150&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{15}=15&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{16}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{17}=0&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{18}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{19}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{20}=150&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{21}=135&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{22}=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{23}=60&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{24}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{25}=45&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{26}=30&lt;/cmath&gt;<br /> And there is the repetition. So, this series has a period of 24. &lt;math&gt;2009 \equiv 17 \pmod{24}&lt;/math&gt;, so &lt;math&gt;|a_{2009}|=|\tan{\theta_{17}}|=|\tan{0}|=|0|= \boxed{\textbf{(A)}\ 0}&lt;/math&gt; ~Firebolt360<br /> <br /> ==Clarification==<br /> (While the above solution is quite intuitive, it's kinda hard for people who don't know many proofing symbols to understand.)<br /> <br /> <br /> Notice that the formula given in the problem looks incredibly similar to the [[Trigonometric_identities#Angle_Addition/Subtraction_Identities|Tangent Addition Formula]].<br /> Since &lt;math&gt;a_1 = 1&lt;/math&gt;, let &lt;math&gt;a_1&lt;/math&gt; be &lt;math&gt;\tan{45}&lt;/math&gt;. Similarly, let &lt;math&gt;a_2&lt;/math&gt; be &lt;math&gt;\tan{30}&lt;/math&gt;. Then the formula for &lt;math&gt;a_3&lt;/math&gt; reads<br /> <br /> <br /> &lt;center&gt;&lt;cmath&gt;a_3 = \frac {\tan{45} + \tan{30} }{1 - \tan{45}*\tan{30}}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> <br /> But from the [[Trigonometric_identities#Angle_Addition/Subtraction_Identities|Tangent Addition Formula]] we know that this is just the formula for &lt;math&gt;\tan{(45+30)}&lt;/math&gt; or &lt;math&gt;\tan{75}&lt;/math&gt;, meaning &lt;math&gt;a_3 = \tan{75}&lt;/math&gt;. So, the sequence becomes a sort of [[Fibonacci sequence]] with angle measures. We continue to sum angle measures, like so:<br /> <br /> <br /> &lt;math&gt;a_4 = \tan{(75+30)} = \tan{105}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_5 = \tan{(105+75)} = \tan{180}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_6 = \tan{(180+105)} =&lt;/math&gt; ... wait a minute! &lt;math&gt;\tan{180} = \tan{0}&lt;/math&gt;!<br /> <br /> <br /> So now we have<br /> <br /> <br /> &lt;math&gt;a_5 = \tan{0}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_6 = \tan{(105+0)} = \tan{105}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_7 = \tan{(105+105)} = \tan{210}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_8 = \tan{(210+105)} = &lt;/math&gt; .... Ok this is getting tiring. Let's stop and think about whether we can simplify this. Notice that all of the angle measures are a multiple of &lt;math&gt;15&lt;/math&gt;. So, let's express the angle measures as multiples of &lt;math&gt;15&lt;/math&gt;.<br /> <br /> <br /> Let &lt;math&gt;b_n = \frac{\arctan{(a_n)}}{15}&lt;/math&gt;.<br /> <br /> <br /> (Basically, &lt;math&gt;b_n&lt;/math&gt; is the angle measure of the corresponding &lt;math&gt;a_n,&lt;/math&gt; divided by &lt;math&gt;15&lt;/math&gt;)<br /> <br /> Now we have<br /> <br /> <br /> &lt;math&gt;b_1 = 3&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_2 = 2&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_3 = 5&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_4 = 7&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_5 = 12&lt;/math&gt;<br /> <br /> <br /> But wait... we're dealing with the &lt;math&gt;\tan&lt;/math&gt; function, which has a period (recurrence rate) of &lt;math&gt;2\pi&lt;/math&gt; or &lt;math&gt;180^{\circ}&lt;/math&gt;. Since we divided the angle measures by &lt;math&gt;15&lt;/math&gt;, the period is now &lt;math&gt;12&lt;/math&gt; (which aligns with what we got earlier: &lt;math&gt;a_5 = 0&lt;/math&gt;). This means that we can reduce the terms of the sequence based on the &lt;math&gt;\bmod&lt;/math&gt; function, which returns the remainder after dividing by a certain amount. So now we can say &lt;math&gt;b_n \equiv b_n \bmod{12}&lt;/math&gt; (&lt;math&gt;\equiv&lt;/math&gt; denotes modular equivalence in this context). Now we continue our sequence:<br /> <br /> &lt;math&gt;3, 2, 5, 7, 0, 7, 7, 2, 9, 11, 8, 7, 3, 10, 1, 11, 0, 11, 11, 10, 9, 7, 4, 11, 3, 2, 5&lt;/math&gt; (It's starting to repeat now)<br /> <br /> So &lt;math&gt;a_1 = 3, a_{25} = 3, a_{49} = 3&lt;/math&gt;, and so on. The sequence repeats every &lt;math&gt;24&lt;/math&gt; terms. The problem asks us for the value of &lt;math&gt;a_{2009}&lt;/math&gt;. Let's whip out the &lt;math&gt;\bmod&lt;/math&gt; function again.<br /> <br /> <br /> &lt;math&gt;2009 \bmod 24 = 17&lt;/math&gt;<br /> <br /> <br /> So we want the value of the &lt;math&gt;17&lt;/math&gt;th number in the sequence, and looking back at it (hopefully you wrote it down somewhere) we find that &lt;math&gt;b_{17} = 0 \Rightarrow \boxed{\text{A}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2009|ab=A|num-b=24|after=Last question}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_25&diff=138649 2009 AMC 12A Problems/Problem 25 2020-11-28T18:32:53Z <p>Firebolt360: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> The first two terms of a sequence are &lt;math&gt;a_1 = 1&lt;/math&gt; and &lt;math&gt;a_2 = \frac {1}{\sqrt3}&lt;/math&gt;. For &lt;math&gt;n\ge1&lt;/math&gt;,<br /> <br /> &lt;center&gt;&lt;cmath&gt;a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> What is &lt;math&gt;|a_{2009}|&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2 - \sqrt3\qquad \textbf{(C)}\ \frac {1}{\sqrt3}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2 + \sqrt3&lt;/math&gt;<br /> <br /> == Solution ==<br /> Consider another sequence &lt;math&gt;\{\theta_1, \theta_2, \theta_3...\}&lt;/math&gt; such that &lt;math&gt;a_n = \tan{\theta_n}&lt;/math&gt;, and &lt;math&gt;0 \leq \theta_n &lt; 180&lt;/math&gt;.<br /> <br /> The given recurrence becomes<br /> <br /> &lt;center&gt;&lt;cmath&gt;\begin{align*} a_{n + 2} &amp; = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}} \\<br /> \tan{\theta_{n + 2}} &amp; = \frac {\tan{\theta_n} + \tan{\theta_{n + 1}}}{1 - \tan{\theta_n}\tan{\theta_{n + 1}}} \\<br /> \tan{\theta_{n + 2}} &amp; = \tan(\theta_{n + 1} + \theta_n) \end{align*}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> It follows that &lt;math&gt;\theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}&lt;/math&gt;. Since &lt;math&gt;\theta_1 = 45, \theta_2 = 30&lt;/math&gt;, all terms in the sequence &lt;math&gt;\{\theta_1, \theta_2, \theta_3...\}&lt;/math&gt; will be a multiple of &lt;math&gt;15&lt;/math&gt;.<br /> <br /> Now consider another sequence &lt;math&gt;\{b_1, b_2, b_3...\}&lt;/math&gt; such that &lt;math&gt;b_n = \theta_n/15&lt;/math&gt;, and &lt;math&gt;0 \leq b_n &lt; 12&lt;/math&gt;. The sequence &lt;math&gt;b_n&lt;/math&gt; satisfies &lt;math&gt;b_1 = 3, b_2 = 2, b_{n + 2} \equiv b_{n + 1} + b_n \pmod{12}&lt;/math&gt;.<br /> <br /> As the number of possible consecutive two terms is finite, we know that the sequence &lt;math&gt;b_n&lt;/math&gt; is periodic. Write out the first few terms of the sequence until it starts to repeat.<br /> <br /> &lt;math&gt;\{b_n\} = \{3,2,5,7,0,7,7,2,9,11,8,7,3,10,1,11,0,11,11,10,9,7,4,11,3,2,5,7,...\}&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;b_{25} = b_1 = 3&lt;/math&gt; and &lt;math&gt;b_{26} = b_2 = 2&lt;/math&gt;. Thus &lt;math&gt;\{b_n\}&lt;/math&gt; has a period of &lt;math&gt;24&lt;/math&gt;: &lt;math&gt;b_{n + 24} = b_n&lt;/math&gt;.<br /> <br /> It follows that &lt;math&gt;b_{2009} = b_{17} = 0&lt;/math&gt; and &lt;math&gt;\theta_{2009} = 15 b_{2009} = 0&lt;/math&gt;. Thus &lt;math&gt;a_{2009} = \tan{\theta_{2009}} = \tan{0} = 0.&lt;/math&gt;<br /> <br /> Our answer is &lt;math&gt;|a_{2009}| = \boxed{\textbf{(A)}\ 0}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> First, some intuition. The given recurrence relation: &lt;math&gt;a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}&lt;/math&gt; looks much like the tangent addition formula. So, we let &lt;math&gt;a_n=\tan{\theta_n}&lt;/math&gt;. Then, we get:<br /> &lt;cmath&gt;\tan{\theta_n}=\tan{(\theta_{n-1}+\theta_{n-2})}&lt;/cmath&gt;<br /> This gives us:<br /> &lt;cmath&gt;\theta_n \equiv \theta_{n-1}+\theta_{n-2} \pmod{180}&lt;/cmath&gt;<br /> Now, observe that <br /> &lt;cmath&gt;\theta_1=45&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_2=30&lt;/cmath&gt;<br /> We know that this sequence of values for &lt;math&gt;\theta_n&lt;/math&gt; will repeat eventually because it is mod &lt;math&gt;180&lt;/math&gt;.<br /> It is just a matter of when, so we start bashing:<br /> <br /> &lt;cmath&gt;\theta_3=75&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_4=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_6=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_7=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_8=30&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_9=135&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{10}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{11}=120&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{12}=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{13}=45&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{14}=150&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{15}=15&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{16}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{17}=0&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{18}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{19}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{20}=150&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{21}=135&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{22}=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{23}=60&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{24}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{25}=45&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{26}=30&lt;/cmath&gt;<br /> And there is the repetition. So, this series has a period of 24. &lt;math&gt;2009 \equiv 17 \pmod{24}&lt;/math&gt;, so &lt;math&gt;|a_{2009}|=|\tan{\theta_{17}}|=|\tan{0}|=|0|= \boxed{\textbf{(A)}\ 0}&lt;/math&gt;<br /> <br /> ==Clarification==<br /> (While the above solution is quite intuitive, it's kinda hard for people who don't know many proofing symbols to understand.)<br /> <br /> <br /> Notice that the formula given in the problem looks incredibly similar to the [[Trigonometric_identities#Angle_Addition/Subtraction_Identities|Tangent Addition Formula]].<br /> Since &lt;math&gt;a_1 = 1&lt;/math&gt;, let &lt;math&gt;a_1&lt;/math&gt; be &lt;math&gt;\tan{45}&lt;/math&gt;. Similarly, let &lt;math&gt;a_2&lt;/math&gt; be &lt;math&gt;\tan{30}&lt;/math&gt;. Then the formula for &lt;math&gt;a_3&lt;/math&gt; reads<br /> <br /> <br /> &lt;center&gt;&lt;cmath&gt;a_3 = \frac {\tan{45} + \tan{30} }{1 - \tan{45}*\tan{30}}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> <br /> But from the [[Trigonometric_identities#Angle_Addition/Subtraction_Identities|Tangent Addition Formula]] we know that this is just the formula for &lt;math&gt;\tan{(45+30)}&lt;/math&gt; or &lt;math&gt;\tan{75}&lt;/math&gt;, meaning &lt;math&gt;a_3 = \tan{75}&lt;/math&gt;. So, the sequence becomes a sort of [[Fibonacci sequence]] with angle measures. We continue to sum angle measures, like so:<br /> <br /> <br /> &lt;math&gt;a_4 = \tan{(75+30)} = \tan{105}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_5 = \tan{(105+75)} = \tan{180}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_6 = \tan{(180+105)} =&lt;/math&gt; ... wait a minute! &lt;math&gt;\tan{180} = \tan{0}&lt;/math&gt;!<br /> <br /> <br /> So now we have<br /> <br /> <br /> &lt;math&gt;a_5 = \tan{0}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_6 = \tan{(105+0)} = \tan{105}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_7 = \tan{(105+105)} = \tan{210}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_8 = \tan{(210+105)} = &lt;/math&gt; .... Ok this is getting tiring. Let's stop and think about whether we can simplify this. Notice that all of the angle measures are a multiple of &lt;math&gt;15&lt;/math&gt;. So, let's express the angle measures as multiples of &lt;math&gt;15&lt;/math&gt;.<br /> <br /> <br /> Let &lt;math&gt;b_n = \frac{\arctan{(a_n)}}{15}&lt;/math&gt;.<br /> <br /> <br /> (Basically, &lt;math&gt;b_n&lt;/math&gt; is the angle measure of the corresponding &lt;math&gt;a_n,&lt;/math&gt; divided by &lt;math&gt;15&lt;/math&gt;)<br /> <br /> Now we have<br /> <br /> <br /> &lt;math&gt;b_1 = 3&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_2 = 2&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_3 = 5&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_4 = 7&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_5 = 12&lt;/math&gt;<br /> <br /> <br /> But wait... we're dealing with the &lt;math&gt;\tan&lt;/math&gt; function, which has a period (recurrence rate) of &lt;math&gt;2\pi&lt;/math&gt; or &lt;math&gt;180^{\circ}&lt;/math&gt;. Since we divided the angle measures by &lt;math&gt;15&lt;/math&gt;, the period is now &lt;math&gt;12&lt;/math&gt; (which aligns with what we got earlier: &lt;math&gt;a_5 = 0&lt;/math&gt;). This means that we can reduce the terms of the sequence based on the &lt;math&gt;\bmod&lt;/math&gt; function, which returns the remainder after dividing by a certain amount. So now we can say &lt;math&gt;b_n \equiv b_n \bmod{12}&lt;/math&gt; (&lt;math&gt;\equiv&lt;/math&gt; denotes modular equivalence in this context). Now we continue our sequence:<br /> <br /> &lt;math&gt;3, 2, 5, 7, 0, 7, 7, 2, 9, 11, 8, 7, 3, 10, 1, 11, 0, 11, 11, 10, 9, 7, 4, 11, 3, 2, 5&lt;/math&gt; (It's starting to repeat now)<br /> <br /> So &lt;math&gt;a_1 = 3, a_{25} = 3, a_{49} = 3&lt;/math&gt;, and so on. The sequence repeats every &lt;math&gt;24&lt;/math&gt; terms. The problem asks us for the value of &lt;math&gt;a_{2009}&lt;/math&gt;. Let's whip out the &lt;math&gt;\bmod&lt;/math&gt; function again.<br /> <br /> <br /> &lt;math&gt;2009 \bmod 24 = 17&lt;/math&gt;<br /> <br /> <br /> So we want the value of the &lt;math&gt;17&lt;/math&gt;th number in the sequence, and looking back at it (hopefully you wrote it down somewhere) we find that &lt;math&gt;b_{17} = 0 \Rightarrow \boxed{\text{A}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2009|ab=A|num-b=24|after=Last question}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_25&diff=138648 2009 AMC 12A Problems/Problem 25 2020-11-28T18:32:14Z <p>Firebolt360: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> The first two terms of a sequence are &lt;math&gt;a_1 = 1&lt;/math&gt; and &lt;math&gt;a_2 = \frac {1}{\sqrt3}&lt;/math&gt;. For &lt;math&gt;n\ge1&lt;/math&gt;,<br /> <br /> &lt;center&gt;&lt;cmath&gt;a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> What is &lt;math&gt;|a_{2009}|&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2 - \sqrt3\qquad \textbf{(C)}\ \frac {1}{\sqrt3}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2 + \sqrt3&lt;/math&gt;<br /> <br /> == Solution ==<br /> Consider another sequence &lt;math&gt;\{\theta_1, \theta_2, \theta_3...\}&lt;/math&gt; such that &lt;math&gt;a_n = \tan{\theta_n}&lt;/math&gt;, and &lt;math&gt;0 \leq \theta_n &lt; 180&lt;/math&gt;.<br /> <br /> The given recurrence becomes<br /> <br /> &lt;center&gt;&lt;cmath&gt;\begin{align*} a_{n + 2} &amp; = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}} \\<br /> \tan{\theta_{n + 2}} &amp; = \frac {\tan{\theta_n} + \tan{\theta_{n + 1}}}{1 - \tan{\theta_n}\tan{\theta_{n + 1}}} \\<br /> \tan{\theta_{n + 2}} &amp; = \tan(\theta_{n + 1} + \theta_n) \end{align*}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> It follows that &lt;math&gt;\theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}&lt;/math&gt;. Since &lt;math&gt;\theta_1 = 45, \theta_2 = 30&lt;/math&gt;, all terms in the sequence &lt;math&gt;\{\theta_1, \theta_2, \theta_3...\}&lt;/math&gt; will be a multiple of &lt;math&gt;15&lt;/math&gt;.<br /> <br /> Now consider another sequence &lt;math&gt;\{b_1, b_2, b_3...\}&lt;/math&gt; such that &lt;math&gt;b_n = \theta_n/15&lt;/math&gt;, and &lt;math&gt;0 \leq b_n &lt; 12&lt;/math&gt;. The sequence &lt;math&gt;b_n&lt;/math&gt; satisfies &lt;math&gt;b_1 = 3, b_2 = 2, b_{n + 2} \equiv b_{n + 1} + b_n \pmod{12}&lt;/math&gt;.<br /> <br /> As the number of possible consecutive two terms is finite, we know that the sequence &lt;math&gt;b_n&lt;/math&gt; is periodic. Write out the first few terms of the sequence until it starts to repeat.<br /> <br /> &lt;math&gt;\{b_n\} = \{3,2,5,7,0,7,7,2,9,11,8,7,3,10,1,11,0,11,11,10,9,7,4,11,3,2,5,7,...\}&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;b_{25} = b_1 = 3&lt;/math&gt; and &lt;math&gt;b_{26} = b_2 = 2&lt;/math&gt;. Thus &lt;math&gt;\{b_n\}&lt;/math&gt; has a period of &lt;math&gt;24&lt;/math&gt;: &lt;math&gt;b_{n + 24} = b_n&lt;/math&gt;.<br /> <br /> It follows that &lt;math&gt;b_{2009} = b_{17} = 0&lt;/math&gt; and &lt;math&gt;\theta_{2009} = 15 b_{2009} = 0&lt;/math&gt;. Thus &lt;math&gt;a_{2009} = \tan{\theta_{2009}} = \tan{0} = 0.&lt;/math&gt;<br /> <br /> Our answer is &lt;math&gt;|a_{2009}| = \boxed{\textbf{(A)}\ 0}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> First, some intuition. The given recurrence relation: &lt;math&gt;a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}&lt;/math&gt; looks much like the tangent addition formula. So, we let &lt;math&gt;a_n=\tan{\theta_n}&lt;/math&gt;. Then, we get:<br /> &lt;cmath&gt;\tan{\theta_n}=\tan{(\theta_{n-1}+\theta_{n-2})}&lt;/cmath&gt;<br /> This gives us:<br /> &lt;cmath&gt;\theta_n \equiv \theta_{n-1}+\theta_{n-2} \pmod{180}&lt;/cmath&gt;<br /> Now, observe that <br /> &lt;cmath&gt;\theta_1=45&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_2=30&lt;/cmath&gt;<br /> We know that this sequence of values for &lt;math&gt;\theta_n&lt;/math&gt; will repeat eventually because it is mod &lt;math&gt;180&lt;/math&gt;.<br /> It is just a matter of when, so we start bashing:<br /> <br /> &lt;cmath&gt;\theta_3=75&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_4=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_6=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_7=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_8=30&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_9=135&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{10}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{11}=120&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{12}=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{13}=45&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{14}=150&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{15}=15&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{16}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{17}=0&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{18}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{19}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{20}=150&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{21}=135&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{22}=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{23}=60&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{24}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{25}=45&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{26}=30&lt;/cmath&gt;<br /> And there is the repetition. So, this series has a period of 24. &lt;math&gt;2009 \equiv 17 \pmod{24}&lt;/math&gt;, so &lt;math&gt;|a_{2009}|=|\tan{\theta_{17}|=|\tan{0}|=|0|= \boxed{\textbf{(A)}\ 0}&lt;/math&gt;<br /> <br /> ==Clarification==<br /> (While the above solution is quite intuitive, it's kinda hard for people who don't know many proofing symbols to understand.)<br /> <br /> <br /> Notice that the formula given in the problem looks incredibly similar to the [[Trigonometric_identities#Angle_Addition/Subtraction_Identities|Tangent Addition Formula]].<br /> Since &lt;math&gt;a_1 = 1&lt;/math&gt;, let &lt;math&gt;a_1&lt;/math&gt; be &lt;math&gt;\tan{45}&lt;/math&gt;. Similarly, let &lt;math&gt;a_2&lt;/math&gt; be &lt;math&gt;\tan{30}&lt;/math&gt;. Then the formula for &lt;math&gt;a_3&lt;/math&gt; reads<br /> <br /> <br /> &lt;center&gt;&lt;cmath&gt;a_3 = \frac {\tan{45} + \tan{30} }{1 - \tan{45}*\tan{30}}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> <br /> But from the [[Trigonometric_identities#Angle_Addition/Subtraction_Identities|Tangent Addition Formula]] we know that this is just the formula for &lt;math&gt;\tan{(45+30)}&lt;/math&gt; or &lt;math&gt;\tan{75}&lt;/math&gt;, meaning &lt;math&gt;a_3 = \tan{75}&lt;/math&gt;. So, the sequence becomes a sort of [[Fibonacci sequence]] with angle measures. We continue to sum angle measures, like so:<br /> <br /> <br /> &lt;math&gt;a_4 = \tan{(75+30)} = \tan{105}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_5 = \tan{(105+75)} = \tan{180}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_6 = \tan{(180+105)} =&lt;/math&gt; ... wait a minute! &lt;math&gt;\tan{180} = \tan{0}&lt;/math&gt;!<br /> <br /> <br /> So now we have<br /> <br /> <br /> &lt;math&gt;a_5 = \tan{0}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_6 = \tan{(105+0)} = \tan{105}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_7 = \tan{(105+105)} = \tan{210}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_8 = \tan{(210+105)} = &lt;/math&gt; .... Ok this is getting tiring. Let's stop and think about whether we can simplify this. Notice that all of the angle measures are a multiple of &lt;math&gt;15&lt;/math&gt;. So, let's express the angle measures as multiples of &lt;math&gt;15&lt;/math&gt;.<br /> <br /> <br /> Let &lt;math&gt;b_n = \frac{\arctan{(a_n)}}{15}&lt;/math&gt;.<br /> <br /> <br /> (Basically, &lt;math&gt;b_n&lt;/math&gt; is the angle measure of the corresponding &lt;math&gt;a_n,&lt;/math&gt; divided by &lt;math&gt;15&lt;/math&gt;)<br /> <br /> Now we have<br /> <br /> <br /> &lt;math&gt;b_1 = 3&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_2 = 2&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_3 = 5&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_4 = 7&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_5 = 12&lt;/math&gt;<br /> <br /> <br /> But wait... we're dealing with the &lt;math&gt;\tan&lt;/math&gt; function, which has a period (recurrence rate) of &lt;math&gt;2\pi&lt;/math&gt; or &lt;math&gt;180^{\circ}&lt;/math&gt;. Since we divided the angle measures by &lt;math&gt;15&lt;/math&gt;, the period is now &lt;math&gt;12&lt;/math&gt; (which aligns with what we got earlier: &lt;math&gt;a_5 = 0&lt;/math&gt;). This means that we can reduce the terms of the sequence based on the &lt;math&gt;\bmod&lt;/math&gt; function, which returns the remainder after dividing by a certain amount. So now we can say &lt;math&gt;b_n \equiv b_n \bmod{12}&lt;/math&gt; (&lt;math&gt;\equiv&lt;/math&gt; denotes modular equivalence in this context). Now we continue our sequence:<br /> <br /> &lt;math&gt;3, 2, 5, 7, 0, 7, 7, 2, 9, 11, 8, 7, 3, 10, 1, 11, 0, 11, 11, 10, 9, 7, 4, 11, 3, 2, 5&lt;/math&gt; (It's starting to repeat now)<br /> <br /> So &lt;math&gt;a_1 = 3, a_{25} = 3, a_{49} = 3&lt;/math&gt;, and so on. The sequence repeats every &lt;math&gt;24&lt;/math&gt; terms. The problem asks us for the value of &lt;math&gt;a_{2009}&lt;/math&gt;. Let's whip out the &lt;math&gt;\bmod&lt;/math&gt; function again.<br /> <br /> <br /> &lt;math&gt;2009 \bmod 24 = 17&lt;/math&gt;<br /> <br /> <br /> So we want the value of the &lt;math&gt;17&lt;/math&gt;th number in the sequence, and looking back at it (hopefully you wrote it down somewhere) we find that &lt;math&gt;b_{17} = 0 \Rightarrow \boxed{\text{A}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2009|ab=A|num-b=24|after=Last question}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_25&diff=138647 2009 AMC 12A Problems/Problem 25 2020-11-28T18:31:47Z <p>Firebolt360: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> The first two terms of a sequence are &lt;math&gt;a_1 = 1&lt;/math&gt; and &lt;math&gt;a_2 = \frac {1}{\sqrt3}&lt;/math&gt;. For &lt;math&gt;n\ge1&lt;/math&gt;,<br /> <br /> &lt;center&gt;&lt;cmath&gt;a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> What is &lt;math&gt;|a_{2009}|&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2 - \sqrt3\qquad \textbf{(C)}\ \frac {1}{\sqrt3}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2 + \sqrt3&lt;/math&gt;<br /> <br /> == Solution ==<br /> Consider another sequence &lt;math&gt;\{\theta_1, \theta_2, \theta_3...\}&lt;/math&gt; such that &lt;math&gt;a_n = \tan{\theta_n}&lt;/math&gt;, and &lt;math&gt;0 \leq \theta_n &lt; 180&lt;/math&gt;.<br /> <br /> The given recurrence becomes<br /> <br /> &lt;center&gt;&lt;cmath&gt;\begin{align*} a_{n + 2} &amp; = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}} \\<br /> \tan{\theta_{n + 2}} &amp; = \frac {\tan{\theta_n} + \tan{\theta_{n + 1}}}{1 - \tan{\theta_n}\tan{\theta_{n + 1}}} \\<br /> \tan{\theta_{n + 2}} &amp; = \tan(\theta_{n + 1} + \theta_n) \end{align*}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> It follows that &lt;math&gt;\theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}&lt;/math&gt;. Since &lt;math&gt;\theta_1 = 45, \theta_2 = 30&lt;/math&gt;, all terms in the sequence &lt;math&gt;\{\theta_1, \theta_2, \theta_3...\}&lt;/math&gt; will be a multiple of &lt;math&gt;15&lt;/math&gt;.<br /> <br /> Now consider another sequence &lt;math&gt;\{b_1, b_2, b_3...\}&lt;/math&gt; such that &lt;math&gt;b_n = \theta_n/15&lt;/math&gt;, and &lt;math&gt;0 \leq b_n &lt; 12&lt;/math&gt;. The sequence &lt;math&gt;b_n&lt;/math&gt; satisfies &lt;math&gt;b_1 = 3, b_2 = 2, b_{n + 2} \equiv b_{n + 1} + b_n \pmod{12}&lt;/math&gt;.<br /> <br /> As the number of possible consecutive two terms is finite, we know that the sequence &lt;math&gt;b_n&lt;/math&gt; is periodic. Write out the first few terms of the sequence until it starts to repeat.<br /> <br /> &lt;math&gt;\{b_n\} = \{3,2,5,7,0,7,7,2,9,11,8,7,3,10,1,11,0,11,11,10,9,7,4,11,3,2,5,7,...\}&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;b_{25} = b_1 = 3&lt;/math&gt; and &lt;math&gt;b_{26} = b_2 = 2&lt;/math&gt;. Thus &lt;math&gt;\{b_n\}&lt;/math&gt; has a period of &lt;math&gt;24&lt;/math&gt;: &lt;math&gt;b_{n + 24} = b_n&lt;/math&gt;.<br /> <br /> It follows that &lt;math&gt;b_{2009} = b_{17} = 0&lt;/math&gt; and &lt;math&gt;\theta_{2009} = 15 b_{2009} = 0&lt;/math&gt;. Thus &lt;math&gt;a_{2009} = \tan{\theta_{2009}} = \tan{0} = 0.&lt;/math&gt;<br /> <br /> Our answer is &lt;math&gt;|a_{2009}| = \boxed{\textbf{(A)}\ 0}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> First, some intuition. The given recurrence relation: &lt;math&gt;a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}&lt;/math&gt; looks much like the tangent addition formula. So, we let &lt;math&gt;a_n=\tan{\theta_n}&lt;/math&gt;. Then, we get:<br /> &lt;cmath&gt;\tan{\theta_n}=\tan{\theta_{n-1}+\theta_{n-2}}&lt;/cmath&gt;<br /> This gives us:<br /> &lt;cmath&gt;\theta_n \equiv \theta_{n-1}+\theta_{n-2} \pmod{180}&lt;/cmath&gt;<br /> Now, observe that <br /> &lt;cmath&gt;\theta_1=45&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_2=30&lt;/cmath&gt;<br /> We know that this sequence of values for &lt;math&gt;\theta_n&lt;/math&gt; will repeat eventually because it is mod &lt;math&gt;180&lt;/math&gt;.<br /> It is just a matter of when, so we start bashing:<br /> <br /> &lt;cmath&gt;\theta_3=75&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_4=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_6=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_7=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_8=30&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_9=135&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{10}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{11}=120&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{12}=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{13}=45&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{14}=150&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{15}=15&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{16}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{17}=0&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{18}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{19}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{20}=150&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{21}=135&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{22}=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{23}=60&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{24}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{25}=45&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{26}=30&lt;/cmath&gt;<br /> And there is the repetition. So, this series has a period of 24. &lt;math&gt;2009 \equiv 17 \pmod{24}&lt;/math&gt;, so &lt;math&gt;|a_{2009}|=|\tan{\theta_{17}|=|\tan{0}|=|0|= \boxed{\textbf{(A)}\ 0}&lt;/math&gt;<br /> <br /> ==Clarification==<br /> (While the above solution is quite intuitive, it's kinda hard for people who don't know many proofing symbols to understand.)<br /> <br /> <br /> Notice that the formula given in the problem looks incredibly similar to the [[Trigonometric_identities#Angle_Addition/Subtraction_Identities|Tangent Addition Formula]].<br /> Since &lt;math&gt;a_1 = 1&lt;/math&gt;, let &lt;math&gt;a_1&lt;/math&gt; be &lt;math&gt;\tan{45}&lt;/math&gt;. Similarly, let &lt;math&gt;a_2&lt;/math&gt; be &lt;math&gt;\tan{30}&lt;/math&gt;. Then the formula for &lt;math&gt;a_3&lt;/math&gt; reads<br /> <br /> <br /> &lt;center&gt;&lt;cmath&gt;a_3 = \frac {\tan{45} + \tan{30} }{1 - \tan{45}*\tan{30}}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> <br /> But from the [[Trigonometric_identities#Angle_Addition/Subtraction_Identities|Tangent Addition Formula]] we know that this is just the formula for &lt;math&gt;\tan{(45+30)}&lt;/math&gt; or &lt;math&gt;\tan{75}&lt;/math&gt;, meaning &lt;math&gt;a_3 = \tan{75}&lt;/math&gt;. So, the sequence becomes a sort of [[Fibonacci sequence]] with angle measures. We continue to sum angle measures, like so:<br /> <br /> <br /> &lt;math&gt;a_4 = \tan{(75+30)} = \tan{105}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_5 = \tan{(105+75)} = \tan{180}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_6 = \tan{(180+105)} =&lt;/math&gt; ... wait a minute! &lt;math&gt;\tan{180} = \tan{0}&lt;/math&gt;!<br /> <br /> <br /> So now we have<br /> <br /> <br /> &lt;math&gt;a_5 = \tan{0}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_6 = \tan{(105+0)} = \tan{105}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_7 = \tan{(105+105)} = \tan{210}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_8 = \tan{(210+105)} = &lt;/math&gt; .... Ok this is getting tiring. Let's stop and think about whether we can simplify this. Notice that all of the angle measures are a multiple of &lt;math&gt;15&lt;/math&gt;. So, let's express the angle measures as multiples of &lt;math&gt;15&lt;/math&gt;.<br /> <br /> <br /> Let &lt;math&gt;b_n = \frac{\arctan{(a_n)}}{15}&lt;/math&gt;.<br /> <br /> <br /> (Basically, &lt;math&gt;b_n&lt;/math&gt; is the angle measure of the corresponding &lt;math&gt;a_n,&lt;/math&gt; divided by &lt;math&gt;15&lt;/math&gt;)<br /> <br /> Now we have<br /> <br /> <br /> &lt;math&gt;b_1 = 3&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_2 = 2&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_3 = 5&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_4 = 7&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_5 = 12&lt;/math&gt;<br /> <br /> <br /> But wait... we're dealing with the &lt;math&gt;\tan&lt;/math&gt; function, which has a period (recurrence rate) of &lt;math&gt;2\pi&lt;/math&gt; or &lt;math&gt;180^{\circ}&lt;/math&gt;. Since we divided the angle measures by &lt;math&gt;15&lt;/math&gt;, the period is now &lt;math&gt;12&lt;/math&gt; (which aligns with what we got earlier: &lt;math&gt;a_5 = 0&lt;/math&gt;). This means that we can reduce the terms of the sequence based on the &lt;math&gt;\bmod&lt;/math&gt; function, which returns the remainder after dividing by a certain amount. So now we can say &lt;math&gt;b_n \equiv b_n \bmod{12}&lt;/math&gt; (&lt;math&gt;\equiv&lt;/math&gt; denotes modular equivalence in this context). Now we continue our sequence:<br /> <br /> &lt;math&gt;3, 2, 5, 7, 0, 7, 7, 2, 9, 11, 8, 7, 3, 10, 1, 11, 0, 11, 11, 10, 9, 7, 4, 11, 3, 2, 5&lt;/math&gt; (It's starting to repeat now)<br /> <br /> So &lt;math&gt;a_1 = 3, a_{25} = 3, a_{49} = 3&lt;/math&gt;, and so on. The sequence repeats every &lt;math&gt;24&lt;/math&gt; terms. The problem asks us for the value of &lt;math&gt;a_{2009}&lt;/math&gt;. Let's whip out the &lt;math&gt;\bmod&lt;/math&gt; function again.<br /> <br /> <br /> &lt;math&gt;2009 \bmod 24 = 17&lt;/math&gt;<br /> <br /> <br /> So we want the value of the &lt;math&gt;17&lt;/math&gt;th number in the sequence, and looking back at it (hopefully you wrote it down somewhere) we find that &lt;math&gt;b_{17} = 0 \Rightarrow \boxed{\text{A}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2009|ab=A|num-b=24|after=Last question}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_25&diff=138646 2009 AMC 12A Problems/Problem 25 2020-11-28T18:31:21Z <p>Firebolt360: /* Solution */</p> <hr /> <div>== Problem ==<br /> The first two terms of a sequence are &lt;math&gt;a_1 = 1&lt;/math&gt; and &lt;math&gt;a_2 = \frac {1}{\sqrt3}&lt;/math&gt;. For &lt;math&gt;n\ge1&lt;/math&gt;,<br /> <br /> &lt;center&gt;&lt;cmath&gt;a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> What is &lt;math&gt;|a_{2009}|&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2 - \sqrt3\qquad \textbf{(C)}\ \frac {1}{\sqrt3}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2 + \sqrt3&lt;/math&gt;<br /> <br /> == Solution ==<br /> Consider another sequence &lt;math&gt;\{\theta_1, \theta_2, \theta_3...\}&lt;/math&gt; such that &lt;math&gt;a_n = \tan{\theta_n}&lt;/math&gt;, and &lt;math&gt;0 \leq \theta_n &lt; 180&lt;/math&gt;.<br /> <br /> The given recurrence becomes<br /> <br /> &lt;center&gt;&lt;cmath&gt;\begin{align*} a_{n + 2} &amp; = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}} \\<br /> \tan{\theta_{n + 2}} &amp; = \frac {\tan{\theta_n} + \tan{\theta_{n + 1}}}{1 - \tan{\theta_n}\tan{\theta_{n + 1}}} \\<br /> \tan{\theta_{n + 2}} &amp; = \tan(\theta_{n + 1} + \theta_n) \end{align*}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> It follows that &lt;math&gt;\theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}&lt;/math&gt;. Since &lt;math&gt;\theta_1 = 45, \theta_2 = 30&lt;/math&gt;, all terms in the sequence &lt;math&gt;\{\theta_1, \theta_2, \theta_3...\}&lt;/math&gt; will be a multiple of &lt;math&gt;15&lt;/math&gt;.<br /> <br /> Now consider another sequence &lt;math&gt;\{b_1, b_2, b_3...\}&lt;/math&gt; such that &lt;math&gt;b_n = \theta_n/15&lt;/math&gt;, and &lt;math&gt;0 \leq b_n &lt; 12&lt;/math&gt;. The sequence &lt;math&gt;b_n&lt;/math&gt; satisfies &lt;math&gt;b_1 = 3, b_2 = 2, b_{n + 2} \equiv b_{n + 1} + b_n \pmod{12}&lt;/math&gt;.<br /> <br /> As the number of possible consecutive two terms is finite, we know that the sequence &lt;math&gt;b_n&lt;/math&gt; is periodic. Write out the first few terms of the sequence until it starts to repeat.<br /> <br /> &lt;math&gt;\{b_n\} = \{3,2,5,7,0,7,7,2,9,11,8,7,3,10,1,11,0,11,11,10,9,7,4,11,3,2,5,7,...\}&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;b_{25} = b_1 = 3&lt;/math&gt; and &lt;math&gt;b_{26} = b_2 = 2&lt;/math&gt;. Thus &lt;math&gt;\{b_n\}&lt;/math&gt; has a period of &lt;math&gt;24&lt;/math&gt;: &lt;math&gt;b_{n + 24} = b_n&lt;/math&gt;.<br /> <br /> It follows that &lt;math&gt;b_{2009} = b_{17} = 0&lt;/math&gt; and &lt;math&gt;\theta_{2009} = 15 b_{2009} = 0&lt;/math&gt;. Thus &lt;math&gt;a_{2009} = \tan{\theta_{2009}} = \tan{0} = 0.&lt;/math&gt;<br /> <br /> Our answer is &lt;math&gt;|a_{2009}| = \boxed{\textbf{(A)}\ 0}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> First, some intuition. The given recurrence relation: &lt;math&gt;a_{n + 2} &amp; = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}&lt;/math&gt; looks much like the tangent addition formula. So, we let &lt;math&gt;a_n=\tan{\theta_n}&lt;/math&gt;. Then, we get:<br /> &lt;cmath&gt;\tan{\theta_n}=\tan{\theta_{n-1}+\theta_{n-2}}&lt;/cmath&gt;<br /> This gives us:<br /> &lt;cmath&gt;\theta_n \equiv \theta_{n-1}+\theta_{n-2} \pmod{180}&lt;/cmath&gt;<br /> Now, observe that <br /> &lt;cmath&gt;\theta_1=45&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_2=30&lt;/cmath&gt;<br /> We know that this sequence of values for &lt;math&gt;\theta_n&lt;/math&gt; will repeat eventually because it is mod &lt;math&gt;180&lt;/math&gt;.<br /> It is just a matter of when, so we start bashing:<br /> <br /> &lt;cmath&gt;\theta_3=75&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_4=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_6=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_7=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_8=30&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_9=135&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{10}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{11}=120&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{12}=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{13}=45&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{14}=150&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{15}=15&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{16}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{17}=0&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{18}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{19}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{20}=150&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{21}=135&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{22}=105&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{23}=60&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{24}=165&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{25}=45&lt;/cmath&gt;<br /> &lt;cmath&gt;\theta_{26}=30&lt;/cmath&gt;<br /> And there is the repetition. So, this series has a period of 24. &lt;math&gt;2009 \equiv 17 \pmod{24}&lt;/math&gt;, so &lt;math&gt;|a_{2009}|=|\tan{\theta_{17}|=|\tan{0}|=|0|= \boxed{\textbf{(A)}\ 0}&lt;/math&gt;<br /> <br /> ==Clarification==<br /> (While the above solution is quite intuitive, it's kinda hard for people who don't know many proofing symbols to understand.)<br /> <br /> <br /> Notice that the formula given in the problem looks incredibly similar to the [[Trigonometric_identities#Angle_Addition/Subtraction_Identities|Tangent Addition Formula]].<br /> Since &lt;math&gt;a_1 = 1&lt;/math&gt;, let &lt;math&gt;a_1&lt;/math&gt; be &lt;math&gt;\tan{45}&lt;/math&gt;. Similarly, let &lt;math&gt;a_2&lt;/math&gt; be &lt;math&gt;\tan{30}&lt;/math&gt;. Then the formula for &lt;math&gt;a_3&lt;/math&gt; reads<br /> <br /> <br /> &lt;center&gt;&lt;cmath&gt;a_3 = \frac {\tan{45} + \tan{30} }{1 - \tan{45}*\tan{30}}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> <br /> But from the [[Trigonometric_identities#Angle_Addition/Subtraction_Identities|Tangent Addition Formula]] we know that this is just the formula for &lt;math&gt;\tan{(45+30)}&lt;/math&gt; or &lt;math&gt;\tan{75}&lt;/math&gt;, meaning &lt;math&gt;a_3 = \tan{75}&lt;/math&gt;. So, the sequence becomes a sort of [[Fibonacci sequence]] with angle measures. We continue to sum angle measures, like so:<br /> <br /> <br /> &lt;math&gt;a_4 = \tan{(75+30)} = \tan{105}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_5 = \tan{(105+75)} = \tan{180}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_6 = \tan{(180+105)} =&lt;/math&gt; ... wait a minute! &lt;math&gt;\tan{180} = \tan{0}&lt;/math&gt;!<br /> <br /> <br /> So now we have<br /> <br /> <br /> &lt;math&gt;a_5 = \tan{0}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_6 = \tan{(105+0)} = \tan{105}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_7 = \tan{(105+105)} = \tan{210}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;a_8 = \tan{(210+105)} = &lt;/math&gt; .... Ok this is getting tiring. Let's stop and think about whether we can simplify this. Notice that all of the angle measures are a multiple of &lt;math&gt;15&lt;/math&gt;. So, let's express the angle measures as multiples of &lt;math&gt;15&lt;/math&gt;.<br /> <br /> <br /> Let &lt;math&gt;b_n = \frac{\arctan{(a_n)}}{15}&lt;/math&gt;.<br /> <br /> <br /> (Basically, &lt;math&gt;b_n&lt;/math&gt; is the angle measure of the corresponding &lt;math&gt;a_n,&lt;/math&gt; divided by &lt;math&gt;15&lt;/math&gt;)<br /> <br /> Now we have<br /> <br /> <br /> &lt;math&gt;b_1 = 3&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_2 = 2&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_3 = 5&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_4 = 7&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;b_5 = 12&lt;/math&gt;<br /> <br /> <br /> But wait... we're dealing with the &lt;math&gt;\tan&lt;/math&gt; function, which has a period (recurrence rate) of &lt;math&gt;2\pi&lt;/math&gt; or &lt;math&gt;180^{\circ}&lt;/math&gt;. Since we divided the angle measures by &lt;math&gt;15&lt;/math&gt;, the period is now &lt;math&gt;12&lt;/math&gt; (which aligns with what we got earlier: &lt;math&gt;a_5 = 0&lt;/math&gt;). This means that we can reduce the terms of the sequence based on the &lt;math&gt;\bmod&lt;/math&gt; function, which returns the remainder after dividing by a certain amount. So now we can say &lt;math&gt;b_n \equiv b_n \bmod{12}&lt;/math&gt; (&lt;math&gt;\equiv&lt;/math&gt; denotes modular equivalence in this context). Now we continue our sequence:<br /> <br /> &lt;math&gt;3, 2, 5, 7, 0, 7, 7, 2, 9, 11, 8, 7, 3, 10, 1, 11, 0, 11, 11, 10, 9, 7, 4, 11, 3, 2, 5&lt;/math&gt; (It's starting to repeat now)<br /> <br /> So &lt;math&gt;a_1 = 3, a_{25} = 3, a_{49} = 3&lt;/math&gt;, and so on. The sequence repeats every &lt;math&gt;24&lt;/math&gt; terms. The problem asks us for the value of &lt;math&gt;a_{2009}&lt;/math&gt;. Let's whip out the &lt;math&gt;\bmod&lt;/math&gt; function again.<br /> <br /> <br /> &lt;math&gt;2009 \bmod 24 = 17&lt;/math&gt;<br /> <br /> <br /> So we want the value of the &lt;math&gt;17&lt;/math&gt;th number in the sequence, and looking back at it (hopefully you wrote it down somewhere) we find that &lt;math&gt;b_{17} = 0 \Rightarrow \boxed{\text{A}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2009|ab=A|num-b=24|after=Last question}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_11&diff=137781 2007 AIME I Problems/Problem 11 2020-11-19T16:32:48Z <p>Firebolt360: /* Solution */</p> <hr /> <div>== Problem ==<br /> For each [[positive]] [[integer]] &lt;math&gt;p&lt;/math&gt;, let &lt;math&gt;b(p)&lt;/math&gt; denote the unique positive integer &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;|k-\sqrt{p}| &lt; \frac{1}{2}&lt;/math&gt;. For example, &lt;math&gt;b(6) = 2&lt;/math&gt; and &lt;math&gt;b(23) = 5&lt;/math&gt;. If &lt;math&gt;S = \sum_{p=1}^{2007} b(p),&lt;/math&gt; find the [[remainder]] when &lt;math&gt;S&lt;/math&gt; is divided by 1000.<br /> <br /> == Solution 1 ==<br /> &lt;math&gt;\left(k- \frac 12\right)^2=k^2-k+\frac 14&lt;/math&gt; and &lt;math&gt;\left(k+ \frac 12\right)^2=k^2+k+ \frac 14&lt;/math&gt;. Therefore &lt;math&gt;b(p)=k&lt;/math&gt; if and only if &lt;math&gt;p&lt;/math&gt; is in this range, or &lt;math&gt;k^2-k&lt;p\leq k^2+k&lt;/math&gt;. There are &lt;math&gt;2k&lt;/math&gt; numbers in this range, so the sum of &lt;math&gt;b(p)&lt;/math&gt; over this range is &lt;math&gt;(2k)k=2k^2&lt;/math&gt;. &lt;math&gt;44&lt;\sqrt{2007}&lt;45&lt;/math&gt;, so all numbers &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;44&lt;/math&gt; have their full range. Summing this up with the formula for the sum of the first &lt;math&gt;n&lt;/math&gt; squares (&lt;math&gt;\frac{n(n+1)(2n+1)}{6}&lt;/math&gt;), we get &lt;math&gt;\sum_{k=1}^{44}2k^2=2\frac{44(44+1)(2*44+1)}{6}=58740&lt;/math&gt;. We need only consider the &lt;math&gt;740&lt;/math&gt; because we are working with modulo &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> Now consider the range of numbers such that &lt;math&gt;b(p)=45&lt;/math&gt;. These numbers are &lt;math&gt;\left\lceil\frac{44^2 + 45^2}{2}\right\rceil = 1981&lt;/math&gt; to &lt;math&gt;2007&lt;/math&gt;. There are &lt;math&gt;2007 - 1981 + 1 = 27&lt;/math&gt; (1 to be inclusive) of them. &lt;math&gt;27*45=1215&lt;/math&gt;, and &lt;math&gt;215+740=<br /> \boxed{955}&lt;/math&gt;, the answer.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;p&lt;/math&gt; be in the range of &lt;math&gt;a^2 \le p &lt; (a+1)^2&lt;/math&gt;. Then, we need to find the point where the value of &lt;math&gt;b(p)&lt;/math&gt; flips from &lt;math&gt;k&lt;/math&gt; to &lt;math&gt;k+1&lt;/math&gt;. This will happen when &lt;math&gt;p&lt;/math&gt; exceeds &lt;math&gt;(a+\frac{1}{2})^2&lt;/math&gt; or &lt;math&gt;a(a+1)+\frac{1}{4}&lt;/math&gt;. Thus, if &lt;math&gt;a^2 \le p \le a(a+1)&lt;/math&gt; then &lt;math&gt;b(p)=a&lt;/math&gt;. For &lt;math&gt;a(a+1) &lt; p &lt; (a+1)^2&lt;/math&gt;, then &lt;math&gt;b(p)=a+1&lt;/math&gt;. There are &lt;math&gt;a+1&lt;/math&gt; terms in the first set of &lt;math&gt;p&lt;/math&gt;, and &lt;math&gt;a&lt;/math&gt; terms in the second set. Thus, the sum of &lt;math&gt;b(p)&lt;/math&gt; from &lt;math&gt;a^2 \le p &lt;(a+1)^2&lt;/math&gt; is &lt;math&gt;2a(a+1)&lt;/math&gt; or &lt;math&gt;4\cdot\binom{a+1}{2}&lt;/math&gt;. For the time being, consider that &lt;math&gt;S = \sum_{p=1}^{44^2-1} b(p)&lt;/math&gt;. Then, the sum of the values of &lt;math&gt;b(p)&lt;/math&gt; is &lt;math&gt;4\binom{2}{2}+4\binom{3}{2}+\cdots +4\binom{44}{2}=4\left(\binom{2}{2}+\binom{3}{2}+\cdots +\binom{44}{2}\right)&lt;/math&gt;. We can collapse this to &lt;math&gt;4\binom{45}{3}=56760&lt;/math&gt;. Now, we have to consider &lt;math&gt;p&lt;/math&gt; from &lt;math&gt;44^2 \le p &lt; 2007&lt;/math&gt;. Considering &lt;math&gt;p&lt;/math&gt; from just &lt;math&gt;44^2 \le p \le 1980&lt;/math&gt;, we see that all of these values have &lt;math&gt;b(p)=44&lt;/math&gt;. Because there are &lt;math&gt;45&lt;/math&gt; values of &lt;math&gt;p&lt;/math&gt; in that range, the sum of &lt;math&gt;b(p)&lt;/math&gt; in that range is &lt;math&gt;45\cdot44=1980&lt;/math&gt;. Adding this to &lt;math&gt;56760&lt;/math&gt; we get &lt;math&gt;58740&lt;/math&gt; or &lt;math&gt;740&lt;/math&gt; mod &lt;math&gt;1000&lt;/math&gt;. Now, take the range &lt;math&gt;1980 &lt; p \le 2007&lt;/math&gt;. There are &lt;math&gt;27&lt;/math&gt; values of &lt;math&gt;p&lt;/math&gt; in this range, and each has &lt;math&gt;b(p)=45&lt;/math&gt;. Thus, that contributes &lt;math&gt;27*45=1215&lt;/math&gt; or &lt;math&gt;215&lt;/math&gt; to the sum. Finally, adding &lt;math&gt;740&lt;/math&gt; and &lt;math&gt;215&lt;/math&gt; we get &lt;math&gt;740+215=\boxed{955}&lt;/math&gt;.<br /> <br /> ~firebolt360<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_11&diff=137780 2007 AIME I Problems/Problem 11 2020-11-19T16:32:06Z <p>Firebolt360: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> For each [[positive]] [[integer]] &lt;math&gt;p&lt;/math&gt;, let &lt;math&gt;b(p)&lt;/math&gt; denote the unique positive integer &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;|k-\sqrt{p}| &lt; \frac{1}{2}&lt;/math&gt;. For example, &lt;math&gt;b(6) = 2&lt;/math&gt; and &lt;math&gt;b(23) = 5&lt;/math&gt;. If &lt;math&gt;S = \sum_{p=1}^{2007} b(p),&lt;/math&gt; find the [[remainder]] when &lt;math&gt;S&lt;/math&gt; is divided by 1000.<br /> <br /> == Solution ==<br /> &lt;math&gt;\left(k- \frac 12\right)^2=k^2-k+\frac 14&lt;/math&gt; and &lt;math&gt;\left(k+ \frac 12\right)^2=k^2+k+ \frac 14&lt;/math&gt;. Therefore &lt;math&gt;b(p)=k&lt;/math&gt; if and only if &lt;math&gt;p&lt;/math&gt; is in this range, or &lt;math&gt;k^2-k&lt;p\leq k^2+k&lt;/math&gt;. There are &lt;math&gt;2k&lt;/math&gt; numbers in this range, so the sum of &lt;math&gt;b(p)&lt;/math&gt; over this range is &lt;math&gt;(2k)k=2k^2&lt;/math&gt;. &lt;math&gt;44&lt;\sqrt{2007}&lt;45&lt;/math&gt;, so all numbers &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;44&lt;/math&gt; have their full range. Summing this up with the formula for the sum of the first &lt;math&gt;n&lt;/math&gt; squares (&lt;math&gt;\frac{n(n+1)(2n+1)}{6}&lt;/math&gt;), we get &lt;math&gt;\sum_{k=1}^{44}2k^2=2\frac{44(44+1)(2*44+1)}{6}=58740&lt;/math&gt;. We need only consider the &lt;math&gt;740&lt;/math&gt; because we are working with modulo &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> Now consider the range of numbers such that &lt;math&gt;b(p)=45&lt;/math&gt;. These numbers are &lt;math&gt;\left\lceil\frac{44^2 + 45^2}{2}\right\rceil = 1981&lt;/math&gt; to &lt;math&gt;2007&lt;/math&gt;. There are &lt;math&gt;2007 - 1981 + 1 = 27&lt;/math&gt; (1 to be inclusive) of them. &lt;math&gt;27*45=1215&lt;/math&gt;, and &lt;math&gt;215+740=<br /> \boxed{955}&lt;/math&gt;, the answer.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;p&lt;/math&gt; be in the range of &lt;math&gt;a^2 \le p &lt; (a+1)^2&lt;/math&gt;. Then, we need to find the point where the value of &lt;math&gt;b(p)&lt;/math&gt; flips from &lt;math&gt;k&lt;/math&gt; to &lt;math&gt;k+1&lt;/math&gt;. This will happen when &lt;math&gt;p&lt;/math&gt; exceeds &lt;math&gt;(a+\frac{1}{2})^2&lt;/math&gt; or &lt;math&gt;a(a+1)+\frac{1}{4}&lt;/math&gt;. Thus, if &lt;math&gt;a^2 \le p \le a(a+1)&lt;/math&gt; then &lt;math&gt;b(p)=a&lt;/math&gt;. For &lt;math&gt;a(a+1) &lt; p &lt; (a+1)^2&lt;/math&gt;, then &lt;math&gt;b(p)=a+1&lt;/math&gt;. There are &lt;math&gt;a+1&lt;/math&gt; terms in the first set of &lt;math&gt;p&lt;/math&gt;, and &lt;math&gt;a&lt;/math&gt; terms in the second set. Thus, the sum of &lt;math&gt;b(p)&lt;/math&gt; from &lt;math&gt;a^2 \le p &lt;(a+1)^2&lt;/math&gt; is &lt;math&gt;2a(a+1)&lt;/math&gt; or &lt;math&gt;4\cdot\binom{a+1}{2}&lt;/math&gt;. For the time being, consider that &lt;math&gt;S = \sum_{p=1}^{44^2-1} b(p)&lt;/math&gt;. Then, the sum of the values of &lt;math&gt;b(p)&lt;/math&gt; is &lt;math&gt;4\binom{2}{2}+4\binom{3}{2}+\cdots +4\binom{44}{2}=4\left(\binom{2}{2}+\binom{3}{2}+\cdots +\binom{44}{2}\right)&lt;/math&gt;. We can collapse this to &lt;math&gt;4\binom{45}{3}=56760&lt;/math&gt;. Now, we have to consider &lt;math&gt;p&lt;/math&gt; from &lt;math&gt;44^2 \le p &lt; 2007&lt;/math&gt;. Considering &lt;math&gt;p&lt;/math&gt; from just &lt;math&gt;44^2 \le p \le 1980&lt;/math&gt;, we see that all of these values have &lt;math&gt;b(p)=44&lt;/math&gt;. Because there are &lt;math&gt;45&lt;/math&gt; values of &lt;math&gt;p&lt;/math&gt; in that range, the sum of &lt;math&gt;b(p)&lt;/math&gt; in that range is &lt;math&gt;45\cdot44=1980&lt;/math&gt;. Adding this to &lt;math&gt;56760&lt;/math&gt; we get &lt;math&gt;58740&lt;/math&gt; or &lt;math&gt;740&lt;/math&gt; mod &lt;math&gt;1000&lt;/math&gt;. Now, take the range &lt;math&gt;1980 &lt; p \le 2007&lt;/math&gt;. There are &lt;math&gt;27&lt;/math&gt; values of &lt;math&gt;p&lt;/math&gt; in this range, and each has &lt;math&gt;b(p)=45&lt;/math&gt;. Thus, that contributes &lt;math&gt;27*45=1215&lt;/math&gt; or &lt;math&gt;215&lt;/math&gt; to the sum. Finally, adding &lt;math&gt;740&lt;/math&gt; and &lt;math&gt;215&lt;/math&gt; we get &lt;math&gt;740+215=\boxed{955}&lt;/math&gt;.<br /> <br /> ~firebolt360<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_11&diff=137779 2007 AIME I Problems/Problem 11 2020-11-19T16:31:18Z <p>Firebolt360: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> For each [[positive]] [[integer]] &lt;math&gt;p&lt;/math&gt;, let &lt;math&gt;b(p)&lt;/math&gt; denote the unique positive integer &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;|k-\sqrt{p}| &lt; \frac{1}{2}&lt;/math&gt;. For example, &lt;math&gt;b(6) = 2&lt;/math&gt; and &lt;math&gt;b(23) = 5&lt;/math&gt;. If &lt;math&gt;S = \sum_{p=1}^{2007} b(p),&lt;/math&gt; find the [[remainder]] when &lt;math&gt;S&lt;/math&gt; is divided by 1000.<br /> <br /> == Solution ==<br /> &lt;math&gt;\left(k- \frac 12\right)^2=k^2-k+\frac 14&lt;/math&gt; and &lt;math&gt;\left(k+ \frac 12\right)^2=k^2+k+ \frac 14&lt;/math&gt;. Therefore &lt;math&gt;b(p)=k&lt;/math&gt; if and only if &lt;math&gt;p&lt;/math&gt; is in this range, or &lt;math&gt;k^2-k&lt;p\leq k^2+k&lt;/math&gt;. There are &lt;math&gt;2k&lt;/math&gt; numbers in this range, so the sum of &lt;math&gt;b(p)&lt;/math&gt; over this range is &lt;math&gt;(2k)k=2k^2&lt;/math&gt;. &lt;math&gt;44&lt;\sqrt{2007}&lt;45&lt;/math&gt;, so all numbers &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;44&lt;/math&gt; have their full range. Summing this up with the formula for the sum of the first &lt;math&gt;n&lt;/math&gt; squares (&lt;math&gt;\frac{n(n+1)(2n+1)}{6}&lt;/math&gt;), we get &lt;math&gt;\sum_{k=1}^{44}2k^2=2\frac{44(44+1)(2*44+1)}{6}=58740&lt;/math&gt;. We need only consider the &lt;math&gt;740&lt;/math&gt; because we are working with modulo &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> Now consider the range of numbers such that &lt;math&gt;b(p)=45&lt;/math&gt;. These numbers are &lt;math&gt;\left\lceil\frac{44^2 + 45^2}{2}\right\rceil = 1981&lt;/math&gt; to &lt;math&gt;2007&lt;/math&gt;. There are &lt;math&gt;2007 - 1981 + 1 = 27&lt;/math&gt; (1 to be inclusive) of them. &lt;math&gt;27*45=1215&lt;/math&gt;, and &lt;math&gt;215+740=<br /> \boxed{955}&lt;/math&gt;, the answer.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;p&lt;/math&gt; be in the range of &lt;math&gt;a^2 \le p &lt; (a+1)^2&lt;/math&gt;. Then, we need to find the point where the value of &lt;math&gt;b(p)&lt;/math&gt; flips from &lt;math&gt;k&lt;/math&gt; to &lt;math&gt;k+1&lt;/math&gt;. This will happen when &lt;math&gt;p&lt;/math&gt; exceeds &lt;math&gt;(a+\frac{1}{2})^2&lt;/math&gt; or &lt;math&gt;a(a+1)+\frac{1}{4}&lt;/math&gt;. Thus, if &lt;math&gt;a^2 \le p \le a(a+1)&lt;/math&gt; then &lt;math&gt;b(p)=a&lt;/math&gt;. For &lt;math&gt;a(a+1) &lt; p &lt; (a+1)^2&lt;/math&gt;, then &lt;math&gt;b(p)=a+1&lt;/math&gt;. There are &lt;math&gt;a+1&lt;/math&gt; terms in the first set of &lt;math&gt;p&lt;/math&gt;, and &lt;math&gt;a&lt;/math&gt; terms in the second set. Thus, the sum of &lt;math&gt;b(p)&lt;/math&gt; from &lt;math&gt;a^2 \le p &lt;(a+1)^2&lt;/math&gt; is &lt;math&gt;2a(a+1)&lt;/math&gt; or &lt;math&gt;4\cdot\binom{a+1}{2}&lt;/math&gt;. For the time being, consider that &lt;math&gt;S = \sum_{p=1}^{44^2-1} b(p)&lt;/math&gt;. Then, the sum of the values of &lt;math&gt;b(p)&lt;/math&gt; is &lt;math&gt;4\binom{2}{2}+4\binom{3}{2}+\cdots +4\binom{44}{2}=4\left(\binom{2}{2}+\binom{3}{2}+\cdots +\binom{44}{2}\right)&lt;/math&gt;. We can collapse this to &lt;math&gt;4\binom{45}{3}=56760&lt;/math&gt;. Now, we have to consider &lt;math&gt;p&lt;/math&gt; from &lt;math&gt;44^2 \le p &lt; 2007&lt;/math&gt;. Considering &lt;math&gt;p&lt;/math&gt; from just &lt;math&gt;44^2 \le p \le 1980&lt;/math&gt;, we see that all of these values have &lt;math&gt;b(p)=44&lt;/math&gt;. Because there are &lt;math&gt;45&lt;/math&gt; values of &lt;math&gt;p&lt;/math&gt; in that range, the sum of &lt;math&gt;b(p)&lt;/math&gt; in that range is &lt;math&gt;45\cdot44=1980&lt;/math&gt;. Adding this to &lt;math&gt;56760&lt;/math&gt; we get &lt;math&gt;58740&lt;/math&gt; or &lt;math&gt;740&lt;/math&gt; mod &lt;math&gt;1000&lt;/math&gt;. Now, take the range &lt;math&gt;1980 &lt; p \le 2007&lt;/math&gt;. There are &lt;math&gt;27&lt;/math&gt; values of &lt;math&gt;p&lt;/math&gt; in this range, and each has &lt;math&gt;b(p)=45&lt;/math&gt;. Thus, that contributes &lt;math&gt;27*45=1215&lt;/math&gt; or &lt;math&gt;215&lt;/math&gt; to the sum. Finally, adding &lt;math&gt;740&lt;/math&gt; and &lt;math&gt;215&lt;/math&gt; we get &lt;math&gt;740+215=\boxed{955}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_11&diff=137777 2007 AIME I Problems/Problem 11 2020-11-19T16:14:05Z <p>Firebolt360: /* Solution */</p> <hr /> <div>== Problem ==<br /> For each [[positive]] [[integer]] &lt;math&gt;p&lt;/math&gt;, let &lt;math&gt;b(p)&lt;/math&gt; denote the unique positive integer &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;|k-\sqrt{p}| &lt; \frac{1}{2}&lt;/math&gt;. For example, &lt;math&gt;b(6) = 2&lt;/math&gt; and &lt;math&gt;b(23) = 5&lt;/math&gt;. If &lt;math&gt;S = \sum_{p=1}^{2007} b(p),&lt;/math&gt; find the [[remainder]] when &lt;math&gt;S&lt;/math&gt; is divided by 1000.<br /> <br /> == Solution ==<br /> &lt;math&gt;\left(k- \frac 12\right)^2=k^2-k+\frac 14&lt;/math&gt; and &lt;math&gt;\left(k+ \frac 12\right)^2=k^2+k+ \frac 14&lt;/math&gt;. Therefore &lt;math&gt;b(p)=k&lt;/math&gt; if and only if &lt;math&gt;p&lt;/math&gt; is in this range, or &lt;math&gt;k^2-k&lt;p\leq k^2+k&lt;/math&gt;. There are &lt;math&gt;2k&lt;/math&gt; numbers in this range, so the sum of &lt;math&gt;b(p)&lt;/math&gt; over this range is &lt;math&gt;(2k)k=2k^2&lt;/math&gt;. &lt;math&gt;44&lt;\sqrt{2007}&lt;45&lt;/math&gt;, so all numbers &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;44&lt;/math&gt; have their full range. Summing this up with the formula for the sum of the first &lt;math&gt;n&lt;/math&gt; squares (&lt;math&gt;\frac{n(n+1)(2n+1)}{6}&lt;/math&gt;), we get &lt;math&gt;\sum_{k=1}^{44}2k^2=2\frac{44(44+1)(2*44+1)}{6}=58740&lt;/math&gt;. We need only consider the &lt;math&gt;740&lt;/math&gt; because we are working with modulo &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> Now consider the range of numbers such that &lt;math&gt;b(p)=45&lt;/math&gt;. These numbers are &lt;math&gt;\left\lceil\frac{44^2 + 45^2}{2}\right\rceil = 1981&lt;/math&gt; to &lt;math&gt;2007&lt;/math&gt;. There are &lt;math&gt;2007 - 1981 + 1 = 27&lt;/math&gt; (1 to be inclusive) of them. &lt;math&gt;27*45=1215&lt;/math&gt;, and &lt;math&gt;215+740=<br /> \boxed{955}&lt;/math&gt;, the answer.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;p&lt;/math&gt; be in the range of &lt;math&gt;a^2 \le p &lt; (a+1)^2&lt;/math&gt; TBC<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_17&diff=137737 2020 AMC 8 Problems/Problem 17 2020-11-19T02:09:59Z <p>Firebolt360: </p> <hr /> <div>How many positive integer factors of &lt;math&gt;2020&lt;/math&gt; have more than &lt;math&gt;3&lt;/math&gt; factors?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We list out the factors of &lt;math&gt;2020&lt;/math&gt;: &lt;cmath&gt;1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.&lt;/cmath&gt; Of these, only &lt;math&gt;1, 2, 4, 5, 101&lt;/math&gt; (&lt;math&gt;5&lt;/math&gt; of them) do not have more than &lt;math&gt;3&lt;/math&gt; factors. Therefore the answer is &lt;math&gt;\tau\left({2020}\right)-5=\boxed{\textbf{(B) }7}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The prime factorization of &lt;math&gt;2020&lt;/math&gt; is &lt;math&gt;2^2\cdot5\cdot101&lt;/math&gt;. The total number of factors of &lt;math&gt;2020&lt;/math&gt; is given by the product of one more than each of the prime powers which comes out to &lt;math&gt;3\cdot2\cdot2=12&lt;/math&gt;. Instead of finding how many factors of &lt;math&gt;2020&lt;/math&gt; have more than three factors, we will instead find how many have one, two, or three factors and subtract this number from &lt;math&gt;12&lt;/math&gt; to find the answer. The only number which has one factor is &lt;math&gt;1&lt;/math&gt;. For a number to have exactly two factors, it must be prime. From the prime factorization of &lt;math&gt;2020&lt;/math&gt;, we know that these can only be &lt;math&gt;2,5,&lt;/math&gt; and &lt;math&gt;101&lt;/math&gt;. For a number to have three factors, it must be a square of a prime. The only square of a prime that is a factor of &lt;math&gt;2020&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;. Our list of factors is &lt;math&gt;1,2,4,5,&lt;/math&gt; and &lt;math&gt;101&lt;/math&gt; which is a total five factors. Thus, the number of factors of &lt;math&gt;2020&lt;/math&gt; that have more than three factors is &lt;math&gt;12-5=7 \implies\boxed{\textbf{(B) }7}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 3==<br /> <br /> The prime factorization of &lt;math&gt;2020&lt;/math&gt; is &lt;math&gt;2^2\cdot5\cdot101&lt;/math&gt; so it has &lt;math&gt;(2+1)(1+1)(1+1)=12&lt;/math&gt; factors. Then we can count that &lt;math&gt;1,2,4,5,101&lt;/math&gt; all have &lt;math&gt;3&lt;/math&gt; or fewer divisors so by complementary counting our answer is &lt;math&gt;12-5=\textbf{(B)}\ 7&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|num-b=16|num-a=18}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_18&diff=137736 2020 AMC 8 Problems/Problem 18 2020-11-19T02:07:52Z <p>Firebolt360: /* Solution 5 -SweetMango77 */</p> <hr /> <div>Rectangle &lt;math&gt;ABCD&lt;/math&gt; is inscribed in a semicircle with diameter &lt;math&gt;\overline{FE},&lt;/math&gt; as shown in the figure. Let &lt;math&gt;DA=16,&lt;/math&gt; and let &lt;math&gt;FD=AE=9.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt; <br /> draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(&quot;A$&quot;,(8,0), 1.25*S); dot(&quot;$B$&quot;,(8,15), 1.25*N); dot(&quot;$C$&quot;,(-8,15), 1.25*N); dot(&quot;$D$&quot;,(-8,0), 1.25*S); dot(&quot;$E$&quot;,(17,0), 1.25*S); dot(&quot;$F$&quot;,(-17,0), 1.25*S); label(&quot;$16$&quot;,(0,0),N); label(&quot;$9$&quot;,(12.5,0),N); label(&quot;$9$&quot;,(-12.5,0),N); <br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> First, realize &lt;math&gt;ABCD&lt;/math&gt; is not a square. It can easily be seen that the diameter of the semicircle is &lt;math&gt;9+16+9=34&lt;/math&gt;, so the radius is &lt;math&gt;\frac{34}{2}=17&lt;/math&gt;. Express the area of Rectangle &lt;math&gt;ABCD&lt;/math&gt; as &lt;math&gt;16h&lt;/math&gt;, where &lt;math&gt;h=AB&lt;/math&gt;. Notice that by the Pythagorean theorem &lt;math&gt;8^2+h^{2}=17^{2}\implies h=15&lt;/math&gt;. Then, the area of Rectangle &lt;math&gt;ABCD&lt;/math&gt; is equal to &lt;math&gt;16\cdot 15=\boxed{\textbf{(A) }240}&lt;/math&gt;. ~icematrix<br /> <br /> ==Solution 2==<br /> <br /> &lt;asy&gt; draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(&quot;$A$&quot;,(8,0), 1.25*S); dot(&quot;$B$&quot;,(8,15), 1.25*N); dot(&quot;$C$&quot;,(-8,15), 1.25*N); dot(&quot;$D$&quot;,(-8,0), 1.25*S); dot(&quot;$E$&quot;,(17,0), 1.25*S); dot(&quot;$F$&quot;,(-17,0), 1.25*S); label(&quot;$16$&quot;,(0,0),N); label(&quot;$9$&quot;,(12.5,0),N); label(&quot;$9$&quot;,(-12.5,0),N); dot(&quot;$O$&quot;, (0,0), 1.25*S); draw((0,0)--(-8,15));&lt;/asy&gt;<br /> <br /> We have &lt;math&gt;OC=17&lt;/math&gt;, as it is a radius, and &lt;math&gt;OD=8&lt;/math&gt; since it is half of &lt;math&gt;AD&lt;/math&gt;. This means that &lt;math&gt;CD=\sqrt{17^2-8^2}=15&lt;/math&gt;. So &lt;math&gt;16*15=\boxed{\textbf{(A)}240}&lt;/math&gt;<br /> <br /> ~yofro<br /> <br /> ==Solution 3 (coordinate bashing)==<br /> <br /> Let the midpoint of segment &lt;math&gt;FE&lt;/math&gt; be the origin. Evidently, point &lt;math&gt;D&lt;/math&gt; is at &lt;math&gt;(-8, 0)&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; is at &lt;math&gt;(8, 0)&lt;/math&gt;. Since points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; share x-coordinates with &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt;, respectively, we can just find the y-coordinate of &lt;math&gt;B&lt;/math&gt; (which is just the width of the rectangle) and multiply this by &lt;math&gt;DA&lt;/math&gt;, or &lt;math&gt;16&lt;/math&gt;. Since the radius of the semicircle is &lt;math&gt;\frac{9+16+9}{2}&lt;/math&gt;, or &lt;math&gt;17&lt;/math&gt;, the equation of the circle that our semicircle is a part of is &lt;math&gt;x^2+y^2=289&lt;/math&gt;. Since we know that the x-coordinate of &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;8&lt;/math&gt;, we can plug this into our equation to obtain that &lt;math&gt;y=\pm15&lt;/math&gt;. Since &lt;math&gt;y&gt;0&lt;/math&gt;, as the diagram suggests, we know that the y-coordinate of &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;15&lt;/math&gt;. Therefore, our answer is &lt;math&gt;16\cdot 15&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> <br /> NOTE: The synthetic solution is definitely faster and more elegant. However, this is the solution that you should use if you can't see any other easier solution.<br /> <br /> - StarryNight7210<br /> <br /> ==Solution 4==<br /> <br /> First, realize that &lt;math&gt;ABCD&lt;/math&gt; is not a square. Let &lt;math&gt;O&lt;/math&gt; be the midpoint of &lt;math&gt;FE&lt;/math&gt;. Since &lt;math&gt;FE=9+9+16=34&lt;/math&gt;, we have &lt;math&gt;OF=OE=\frac{34}{2}=17=OB&lt;/math&gt; because they are all radii. Since &lt;math&gt;O&lt;/math&gt; is also the midpoint of &lt;math&gt;AD&lt;/math&gt;, we have &lt;math&gt;OA=\frac{16}2=8&lt;/math&gt;. By the Pythagorean Theorem on &lt;math&gt;\triangle BAO&lt;/math&gt;, we find that &lt;math&gt;AB=15&lt;/math&gt;. The answer is then &lt;math&gt;16\cdot 15=\textbf{(A) }240&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==Solution 5 -SweetMango77==<br /> <br /> This is an example of a formula in the Introduction to Algebra book (a sidenote): with a semicircle: if the diameter is &lt;math&gt;1+n&lt;/math&gt; with the &lt;math&gt;1&lt;/math&gt; part at one side, and the &lt;math&gt;n&lt;/math&gt; part at the other side, then the height from the end of the &lt;math&gt;1&lt;/math&gt; side and the start of the &lt;math&gt;n&lt;/math&gt; side is &lt;math&gt;\sqrt{n}&lt;/math&gt;. <br /> <br /> Using this, we can scale the image down by &lt;math&gt;9&lt;/math&gt; to get what we note: The other side will be &lt;math&gt;\frac{16+9}{9}=\frac{25}{9}=\left(\frac{5}{3}\right)^2&lt;/math&gt;. Then, the height of that part will be &lt;math&gt;\frac{5}{3}&lt;/math&gt;. But, we have to scale it back up by &lt;math&gt;9&lt;/math&gt; to get a height of &lt;math&gt;15&lt;/math&gt;. Multiplying by &lt;math&gt;16&lt;/math&gt; gives our desired answer: &lt;math&gt;\boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> <br /> ==Solution 6==<br /> A good diagram goes a long way. With a proper compass and ruler, this problem can be easily solved by a quick sketch. However, use this as a last resort.<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|num-b=17|num-a=19}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_18&diff=137735 2020 AMC 8 Problems/Problem 18 2020-11-19T02:07:05Z <p>Firebolt360: /* Solution */</p> <hr /> <div>Rectangle &lt;math&gt;ABCD&lt;/math&gt; is inscribed in a semicircle with diameter &lt;math&gt;\overline{FE},&lt;/math&gt; as shown in the figure. Let &lt;math&gt;DA=16,&lt;/math&gt; and let &lt;math&gt;FD=AE=9.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt; <br /> draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(&quot;$A$&quot;,(8,0), 1.25*S); dot(&quot;$B$&quot;,(8,15), 1.25*N); dot(&quot;$C$&quot;,(-8,15), 1.25*N); dot(&quot;$D$&quot;,(-8,0), 1.25*S); dot(&quot;$E$&quot;,(17,0), 1.25*S); dot(&quot;$F$&quot;,(-17,0), 1.25*S); label(&quot;$16$&quot;,(0,0),N); label(&quot;$9$&quot;,(12.5,0),N); label(&quot;$9$&quot;,(-12.5,0),N); <br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> First, realize &lt;math&gt;ABCD&lt;/math&gt; is not a square. It can easily be seen that the diameter of the semicircle is &lt;math&gt;9+16+9=34&lt;/math&gt;, so the radius is &lt;math&gt;\frac{34}{2}=17&lt;/math&gt;. Express the area of Rectangle &lt;math&gt;ABCD&lt;/math&gt; as &lt;math&gt;16h&lt;/math&gt;, where &lt;math&gt;h=AB&lt;/math&gt;. Notice that by the Pythagorean theorem &lt;math&gt;8^2+h^{2}=17^{2}\implies h=15&lt;/math&gt;. Then, the area of Rectangle &lt;math&gt;ABCD&lt;/math&gt; is equal to &lt;math&gt;16\cdot 15=\boxed{\textbf{(A) }240}&lt;/math&gt;. ~icematrix<br /> <br /> ==Solution 2==<br /> <br /> &lt;asy&gt; draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(&quot;$A$&quot;,(8,0), 1.25*S); dot(&quot;$B$&quot;,(8,15), 1.25*N); dot(&quot;$C$&quot;,(-8,15), 1.25*N); dot(&quot;$D$&quot;,(-8,0), 1.25*S); dot(&quot;$E$&quot;,(17,0), 1.25*S); dot(&quot;$F$&quot;,(-17,0), 1.25*S); label(&quot;$16$&quot;,(0,0),N); label(&quot;$9$&quot;,(12.5,0),N); label(&quot;$9$&quot;,(-12.5,0),N); dot(&quot;$O$&quot;, (0,0), 1.25*S); draw((0,0)--(-8,15));&lt;/asy&gt;<br /> <br /> We have &lt;math&gt;OC=17&lt;/math&gt;, as it is a radius, and &lt;math&gt;OD=8&lt;/math&gt; since it is half of &lt;math&gt;AD&lt;/math&gt;. This means that &lt;math&gt;CD=\sqrt{17^2-8^2}=15&lt;/math&gt;. So &lt;math&gt;16*15=\boxed{\textbf{(A)}240}&lt;/math&gt;<br /> <br /> ~yofro<br /> <br /> ==Solution 3 (coordinate bashing)==<br /> <br /> Let the midpoint of segment &lt;math&gt;FE&lt;/math&gt; be the origin. Evidently, point &lt;math&gt;D&lt;/math&gt; is at &lt;math&gt;(-8, 0)&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; is at &lt;math&gt;(8, 0)&lt;/math&gt;. Since points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; share x-coordinates with &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt;, respectively, we can just find the y-coordinate of &lt;math&gt;B&lt;/math&gt; (which is just the width of the rectangle) and multiply this by &lt;math&gt;DA&lt;/math&gt;, or &lt;math&gt;16&lt;/math&gt;. Since the radius of the semicircle is &lt;math&gt;\frac{9+16+9}{2}&lt;/math&gt;, or &lt;math&gt;17&lt;/math&gt;, the equation of the circle that our semicircle is a part of is &lt;math&gt;x^2+y^2=289&lt;/math&gt;. Since we know that the x-coordinate of &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;8&lt;/math&gt;, we can plug this into our equation to obtain that &lt;math&gt;y=\pm15&lt;/math&gt;. Since &lt;math&gt;y&gt;0&lt;/math&gt;, as the diagram suggests, we know that the y-coordinate of &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;15&lt;/math&gt;. Therefore, our answer is &lt;math&gt;16\cdot 15&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> <br /> NOTE: The synthetic solution is definitely faster and more elegant. However, this is the solution that you should use if you can't see any other easier solution.<br /> <br /> - StarryNight7210<br /> <br /> ==Solution 4==<br /> <br /> First, realize that &lt;math&gt;ABCD&lt;/math&gt; is not a square. Let &lt;math&gt;O&lt;/math&gt; be the midpoint of &lt;math&gt;FE&lt;/math&gt;. Since &lt;math&gt;FE=9+9+16=34&lt;/math&gt;, we have &lt;math&gt;OF=OE=\frac{34}{2}=17=OB&lt;/math&gt; because they are all radii. Since &lt;math&gt;O&lt;/math&gt; is also the midpoint of &lt;math&gt;AD&lt;/math&gt;, we have &lt;math&gt;OA=\frac{16}2=8&lt;/math&gt;. By the Pythagorean Theorem on &lt;math&gt;\triangle BAO&lt;/math&gt;, we find that &lt;math&gt;AB=15&lt;/math&gt;. The answer is then &lt;math&gt;16\cdot 15=\textbf{(A) }240&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==Solution 5 -SweetMango77==<br /> <br /> This is an example of a formula in the Introduction to Algebra book (a sidenote): with a semicircle: if the diameter is &lt;math&gt;1+n&lt;/math&gt; with the &lt;math&gt;1&lt;/math&gt; part at one side, and the &lt;math&gt;n&lt;/math&gt; part at the other side, then the height from the end of the &lt;math&gt;1&lt;/math&gt; side and the start of the &lt;math&gt;n&lt;/math&gt; side is &lt;math&gt;\sqrt{n}&lt;/math&gt;. <br /> <br /> Using this, we can scale the image down by &lt;math&gt;9&lt;/math&gt; to get what we note: The other side will be &lt;math&gt;\frac{16+9}{9}=\frac{25}{9}=\left(\frac{5}{3}\right)^2&lt;/math&gt;. Then, the height of that part will be &lt;math&gt;\frac{5}{3}&lt;/math&gt;. But, we have to scale it back up by &lt;math&gt;9&lt;/math&gt; to get a height of &lt;math&gt;15&lt;/math&gt;. Multiplying by &lt;math&gt;16&lt;/math&gt; gives our desired answer: &lt;math&gt;\boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|num-b=17|num-a=19}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_15&diff=136405 2019 AIME I Problems/Problem 15 2020-11-02T16:01:07Z <p>Firebolt360: /* Problem 15 */</p> <hr /> <div>==Problem 15==<br /> <br /> Let &lt;math&gt;\overline{AB}&lt;/math&gt; be a chord of a circle &lt;math&gt;\omega&lt;/math&gt;, and let &lt;math&gt;P&lt;/math&gt; be a point on the chord &lt;math&gt;\overline{AB}&lt;/math&gt;. Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; intersect at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Line &lt;math&gt;PQ&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;. Assume that &lt;math&gt;AP=5&lt;/math&gt;, &lt;math&gt;PB=3&lt;/math&gt;, &lt;math&gt;XY=11&lt;/math&gt;, and &lt;math&gt;PQ^2 = \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 1==<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair O, A, B, P, O1, O2, Q, X, Y;<br /> O=(0, 0);<br /> A=dir(140); B=dir(40);<br /> P=(3A+5B)/8;<br /> O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);<br /> O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);<br /> Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)));<br /> X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));<br /> Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));<br /> <br /> draw(circle(O, 1));<br /> draw(circle(O1, length(A-O1)));<br /> draw(circle(O2, length(B-O2)));<br /> draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);<br /> <br /> dot(&quot;$O$&quot;, O, S);<br /> dot(&quot;$A$&quot;, A, A);<br /> dot(&quot;$B$&quot;, B, B);<br /> dot(&quot;$P$&quot;, P, dir(70));<br /> dot(&quot;$Q$&quot;, Q, dir(200));<br /> dot(&quot;$O_1$&quot;, O1, SW);<br /> dot(&quot;$O_2$&quot;, O2, SE);<br /> dot(&quot;$X$&quot;, X, X);<br /> dot(&quot;$Y$&quot;, Y, Y);<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; be the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, respectively. There is a homothety at &lt;math&gt;A&lt;/math&gt; sending &lt;math&gt;\omega&lt;/math&gt; to &lt;math&gt;\omega_1&lt;/math&gt; that sends &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;O_1&lt;/math&gt;, so &lt;math&gt;\overline{OO_2}\parallel\overline{O_1P}&lt;/math&gt;. Similarly, &lt;math&gt;\overline{OO_1}\parallel\overline{O_2P}&lt;/math&gt;, so &lt;math&gt;OO_1PO_2&lt;/math&gt; is a parallelogram. Moreover, &lt;cmath&gt;\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,&lt;/cmath&gt;whence &lt;math&gt;OO_1O_2Q&lt;/math&gt; is cyclic. However, &lt;cmath&gt;OO_1=O_2P=O_2Q,&lt;/cmath&gt;so &lt;math&gt;OO_1O_2Q&lt;/math&gt; is an isosceles trapezoid. Since &lt;math&gt;\overline{O_1O_2}\perp\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{OQ}\perp\overline{XY}&lt;/math&gt;, so &lt;math&gt;Q&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;.<br /> <br /> By Power of a Point, &lt;math&gt;PX\cdot PY=PA\cdot PB=15&lt;/math&gt;. Since &lt;math&gt;PX+PY=XY=11&lt;/math&gt; and &lt;math&gt;XQ=11/2&lt;/math&gt;, &lt;cmath&gt;XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,&lt;/cmath&gt;<br /> and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ===Note===<br /> One may solve for &lt;math&gt;PX&lt;/math&gt; first using PoAP, &lt;math&gt;PX = \frac{11}{2} - \frac{\sqrt{61}}{2}&lt;/math&gt;. Then, notice that &lt;math&gt;PQ^2&lt;/math&gt; is rational but &lt;math&gt;PX^2&lt;/math&gt; is not, also &lt;math&gt;PX = \frac{XY}{2} - \frac{\sqrt{61}}{2}&lt;/math&gt;. The most likely explanation for this is that &lt;math&gt;Q&lt;/math&gt; is the midpoint of &lt;math&gt;XY&lt;/math&gt;, so that &lt;math&gt;XQ = \frac{11}{2}&lt;/math&gt; and &lt;math&gt;PQ=\frac{\sqrt{61}}{2}&lt;/math&gt;. Then our answer is &lt;math&gt;m+n=61+4=\boxed{065}&lt;/math&gt;. One can rigorously prove this using the methods above<br /> <br /> ==Solution 2==<br /> <br /> Let the tangents to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at &lt;math&gt;R&lt;/math&gt;. Then, since &lt;math&gt;RA^2=RB^2&lt;/math&gt;, &lt;math&gt;R&lt;/math&gt; lies on the radical axis of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, which is &lt;math&gt;\overline{PQ}&lt;/math&gt;. It follows that &lt;cmath&gt;-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).&lt;/cmath&gt;<br /> Let &lt;math&gt;Q'&lt;/math&gt; denote the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;. By the Midpoint of Harmonic Bundles Lemma, &lt;cmath&gt;RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,&lt;/cmath&gt;<br /> whence &lt;math&gt;Q=Q'&lt;/math&gt;. Like above, &lt;math&gt;XP=\tfrac{11-\sqrt{61}}2&lt;/math&gt;. Since &lt;math&gt;XQ=\tfrac{11}2&lt;/math&gt;, we establish that &lt;math&gt;PQ=\tfrac{\sqrt{61}}2&lt;/math&gt;, from which &lt;math&gt;PQ^2=\tfrac{61}4&lt;/math&gt;, and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 3==<br /> <br /> Firstly we need to notice that &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;. Assume the center of circle &lt;math&gt;w, w_1, w_2&lt;/math&gt; are &lt;math&gt;O, O_1, O_2&lt;/math&gt;, respectively. Then &lt;math&gt;A, O_2, O&lt;/math&gt; are collinear and &lt;math&gt;O, O_1, B&lt;/math&gt; are collinear. Link &lt;math&gt;O_1P, O_2P, O_1Q, O_2Q&lt;/math&gt;. Notice that, &lt;math&gt;\angle B=\angle A=\angle APO_2=\angle BPO_1&lt;/math&gt;. As a result, &lt;math&gt;PO_1\parallel O_2O&lt;/math&gt; and &lt;math&gt;QO_1\parallel O_2P&lt;/math&gt;. So we have parallelogram &lt;math&gt;PO_2O_1O&lt;/math&gt;. So &lt;math&gt;\angle O_2PO_1=\angle O&lt;/math&gt; Notice that, &lt;math&gt;O_1O_2\bot PQ&lt;/math&gt; and &lt;math&gt;O_1O_2&lt;/math&gt; divide &lt;math&gt;PQ&lt;/math&gt; into two equal length pieces, So we have &lt;math&gt;\angle O_2PO_1=\angle O_2QO_1=\angle O&lt;/math&gt;. As a result, &lt;math&gt;O_2, Q, O, O_1,&lt;/math&gt; lie on one circle. So &lt;math&gt;\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P&lt;/math&gt;. Notice that &lt;math&gt;\angle O_1PQ+\angle O_2O_1P=90^{\circ}&lt;/math&gt;, we have &lt;math&gt;\angle OQP=90^{\circ}&lt;/math&gt;. As a result, &lt;math&gt;OQ\bot PQ&lt;/math&gt;. So &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;.<br /> <br /> Back to our problem. Assume &lt;math&gt;XP=x&lt;/math&gt;, &lt;math&gt;PY=y&lt;/math&gt; and &lt;math&gt;x&lt;y&lt;/math&gt;. Then we have &lt;math&gt;AP\cdot PB=XP\cdot PY&lt;/math&gt;, that is, &lt;math&gt;xy=15&lt;/math&gt;. Also, &lt;math&gt;XP+PY=x+y=XY=11&lt;/math&gt;. Solve these above, we have &lt;math&gt;x=\frac{11-\sqrt{61}}{2}=XP&lt;/math&gt;. As a result, we have &lt;math&gt;PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}&lt;/math&gt;. So, we have &lt;math&gt;PQ^2=\frac{61}{4}&lt;/math&gt;. As a result, our answer is &lt;math&gt;m+n=61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> <br /> <br /> Solution By BladeRunnerAUG (Fanyuchen20020715).<br /> <br /> ==Solution 4==<br /> Note that the tangents to the circles at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at a point &lt;math&gt;Z&lt;/math&gt; on &lt;math&gt;XY&lt;/math&gt; by radical center. Then, since &lt;math&gt;\angle ZAB = \angle ZQA&lt;/math&gt; and &lt;math&gt;\angle ZBA = \angle ZQB&lt;/math&gt;, we have <br /> &lt;cmath&gt;\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},&lt;/cmath&gt;<br /> so &lt;math&gt;ZAQB&lt;/math&gt; is cyclic. But if &lt;math&gt;O&lt;/math&gt; is the center of &lt;math&gt;\omega&lt;/math&gt;, clearly &lt;math&gt;ZAOB&lt;/math&gt; is cyclic with diameter &lt;math&gt;ZO&lt;/math&gt;, so &lt;math&gt;\angle ZQO = 90^{\circ} \implies Q&lt;/math&gt; is the midpoint of &lt;math&gt;XY&lt;/math&gt;. Then, by Power of a Point, &lt;math&gt;PY \cdot PX = PA \cdot PB = 15&lt;/math&gt; and it is given that &lt;math&gt;PY+PX = 11&lt;/math&gt;. Thus &lt;math&gt;PY, PX = \frac{11 \pm \sqrt{61}}{2}&lt;/math&gt; so &lt;math&gt;PQ = \frac{\sqrt{61}}{2} \implies PQ^2 = \frac{61}{4}&lt;/math&gt; and the answer is &lt;math&gt;61+4 = \boxed{065}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_15&diff=136404 2019 AIME I Problems/Problem 15 2020-11-02T15:57:33Z <p>Firebolt360: /* Note */</p> <hr /> <div>==Problem 15==<br /> <br /> Let &lt;math&gt;\overline{AB}&lt;/math&gt; be a chord of a circle &lt;math&gt;\omega&lt;/math&gt;, and let &lt;math&gt;P&lt;/math&gt; be a point on the chord &lt;math&gt;\overline{AB}&lt;/math&gt;. Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; intersect at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Line &lt;math&gt;PQ&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;. Assume that &lt;math&gt;AP=5&lt;/math&gt;, &lt;math&gt;PB=3&lt;/math&gt;, &lt;math&gt;XY=11&lt;/math&gt;, and &lt;math&gt;PQ = \sqrt{\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 1==<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair O, A, B, P, O1, O2, Q, X, Y;<br /> O=(0, 0);<br /> A=dir(140); B=dir(40);<br /> P=(3A+5B)/8;<br /> O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);<br /> O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);<br /> Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)));<br /> X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));<br /> Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));<br /> <br /> draw(circle(O, 1));<br /> draw(circle(O1, length(A-O1)));<br /> draw(circle(O2, length(B-O2)));<br /> draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);<br /> <br /> dot(&quot;$O$&quot;, O, S);<br /> dot(&quot;$A$&quot;, A, A);<br /> dot(&quot;$B$&quot;, B, B);<br /> dot(&quot;$P$&quot;, P, dir(70));<br /> dot(&quot;$Q$&quot;, Q, dir(200));<br /> dot(&quot;$O_1$&quot;, O1, SW);<br /> dot(&quot;$O_2$&quot;, O2, SE);<br /> dot(&quot;$X$&quot;, X, X);<br /> dot(&quot;$Y$&quot;, Y, Y);<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; be the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, respectively. There is a homothety at &lt;math&gt;A&lt;/math&gt; sending &lt;math&gt;\omega&lt;/math&gt; to &lt;math&gt;\omega_1&lt;/math&gt; that sends &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;O_1&lt;/math&gt;, so &lt;math&gt;\overline{OO_2}\parallel\overline{O_1P}&lt;/math&gt;. Similarly, &lt;math&gt;\overline{OO_1}\parallel\overline{O_2P}&lt;/math&gt;, so &lt;math&gt;OO_1PO_2&lt;/math&gt; is a parallelogram. Moreover, &lt;cmath&gt;\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,&lt;/cmath&gt;whence &lt;math&gt;OO_1O_2Q&lt;/math&gt; is cyclic. However, &lt;cmath&gt;OO_1=O_2P=O_2Q,&lt;/cmath&gt;so &lt;math&gt;OO_1O_2Q&lt;/math&gt; is an isosceles trapezoid. Since &lt;math&gt;\overline{O_1O_2}\perp\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{OQ}\perp\overline{XY}&lt;/math&gt;, so &lt;math&gt;Q&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;.<br /> <br /> By Power of a Point, &lt;math&gt;PX\cdot PY=PA\cdot PB=15&lt;/math&gt;. Since &lt;math&gt;PX+PY=XY=11&lt;/math&gt; and &lt;math&gt;XQ=11/2&lt;/math&gt;, &lt;cmath&gt;XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,&lt;/cmath&gt;<br /> and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ===Note===<br /> One may solve for &lt;math&gt;PX&lt;/math&gt; first using PoAP, &lt;math&gt;PX = \frac{11}{2} - \frac{\sqrt{61}}{2}&lt;/math&gt;. Then, notice that &lt;math&gt;PQ^2&lt;/math&gt; is rational but &lt;math&gt;PX^2&lt;/math&gt; is not, also &lt;math&gt;PX = \frac{XY}{2} - \frac{\sqrt{61}}{2}&lt;/math&gt;. The most likely explanation for this is that &lt;math&gt;Q&lt;/math&gt; is the midpoint of &lt;math&gt;XY&lt;/math&gt;, so that &lt;math&gt;XQ = \frac{11}{2}&lt;/math&gt; and &lt;math&gt;PQ=\frac{\sqrt{61}}{2}&lt;/math&gt;. Then our answer is &lt;math&gt;m+n=61+4=\boxed{065}&lt;/math&gt;. One can rigorously prove this using the methods above<br /> <br /> ==Solution 2==<br /> <br /> Let the tangents to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at &lt;math&gt;R&lt;/math&gt;. Then, since &lt;math&gt;RA^2=RB^2&lt;/math&gt;, &lt;math&gt;R&lt;/math&gt; lies on the radical axis of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, which is &lt;math&gt;\overline{PQ}&lt;/math&gt;. It follows that &lt;cmath&gt;-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).&lt;/cmath&gt;<br /> Let &lt;math&gt;Q'&lt;/math&gt; denote the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;. By the Midpoint of Harmonic Bundles Lemma, &lt;cmath&gt;RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,&lt;/cmath&gt;<br /> whence &lt;math&gt;Q=Q'&lt;/math&gt;. Like above, &lt;math&gt;XP=\tfrac{11-\sqrt{61}}2&lt;/math&gt;. Since &lt;math&gt;XQ=\tfrac{11}2&lt;/math&gt;, we establish that &lt;math&gt;PQ=\tfrac{\sqrt{61}}2&lt;/math&gt;, from which &lt;math&gt;PQ^2=\tfrac{61}4&lt;/math&gt;, and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 3==<br /> <br /> Firstly we need to notice that &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;. Assume the center of circle &lt;math&gt;w, w_1, w_2&lt;/math&gt; are &lt;math&gt;O, O_1, O_2&lt;/math&gt;, respectively. Then &lt;math&gt;A, O_2, O&lt;/math&gt; are collinear and &lt;math&gt;O, O_1, B&lt;/math&gt; are collinear. Link &lt;math&gt;O_1P, O_2P, O_1Q, O_2Q&lt;/math&gt;. Notice that, &lt;math&gt;\angle B=\angle A=\angle APO_2=\angle BPO_1&lt;/math&gt;. As a result, &lt;math&gt;PO_1\parallel O_2O&lt;/math&gt; and &lt;math&gt;QO_1\parallel O_2P&lt;/math&gt;. So we have parallelogram &lt;math&gt;PO_2O_1O&lt;/math&gt;. So &lt;math&gt;\angle O_2PO_1=\angle O&lt;/math&gt; Notice that, &lt;math&gt;O_1O_2\bot PQ&lt;/math&gt; and &lt;math&gt;O_1O_2&lt;/math&gt; divide &lt;math&gt;PQ&lt;/math&gt; into two equal length pieces, So we have &lt;math&gt;\angle O_2PO_1=\angle O_2QO_1=\angle O&lt;/math&gt;. As a result, &lt;math&gt;O_2, Q, O, O_1,&lt;/math&gt; lie on one circle. So &lt;math&gt;\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P&lt;/math&gt;. Notice that &lt;math&gt;\angle O_1PQ+\angle O_2O_1P=90^{\circ}&lt;/math&gt;, we have &lt;math&gt;\angle OQP=90^{\circ}&lt;/math&gt;. As a result, &lt;math&gt;OQ\bot PQ&lt;/math&gt;. So &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;.<br /> <br /> Back to our problem. Assume &lt;math&gt;XP=x&lt;/math&gt;, &lt;math&gt;PY=y&lt;/math&gt; and &lt;math&gt;x&lt;y&lt;/math&gt;. Then we have &lt;math&gt;AP\cdot PB=XP\cdot PY&lt;/math&gt;, that is, &lt;math&gt;xy=15&lt;/math&gt;. Also, &lt;math&gt;XP+PY=x+y=XY=11&lt;/math&gt;. Solve these above, we have &lt;math&gt;x=\frac{11-\sqrt{61}}{2}=XP&lt;/math&gt;. As a result, we have &lt;math&gt;PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}&lt;/math&gt;. So, we have &lt;math&gt;PQ^2=\frac{61}{4}&lt;/math&gt;. As a result, our answer is &lt;math&gt;m+n=61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> <br /> <br /> Solution By BladeRunnerAUG (Fanyuchen20020715).<br /> <br /> ==Solution 4==<br /> Note that the tangents to the circles at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at a point &lt;math&gt;Z&lt;/math&gt; on &lt;math&gt;XY&lt;/math&gt; by radical center. Then, since &lt;math&gt;\angle ZAB = \angle ZQA&lt;/math&gt; and &lt;math&gt;\angle ZBA = \angle ZQB&lt;/math&gt;, we have <br /> &lt;cmath&gt;\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},&lt;/cmath&gt;<br /> so &lt;math&gt;ZAQB&lt;/math&gt; is cyclic. But if &lt;math&gt;O&lt;/math&gt; is the center of &lt;math&gt;\omega&lt;/math&gt;, clearly &lt;math&gt;ZAOB&lt;/math&gt; is cyclic with diameter &lt;math&gt;ZO&lt;/math&gt;, so &lt;math&gt;\angle ZQO = 90^{\circ} \implies Q&lt;/math&gt; is the midpoint of &lt;math&gt;XY&lt;/math&gt;. Then, by Power of a Point, &lt;math&gt;PY \cdot PX = PA \cdot PB = 15&lt;/math&gt; and it is given that &lt;math&gt;PY+PX = 11&lt;/math&gt;. Thus &lt;math&gt;PY, PX = \frac{11 \pm \sqrt{61}}{2}&lt;/math&gt; so &lt;math&gt;PQ = \frac{\sqrt{61}}{2} \implies PQ^2 = \frac{61}{4}&lt;/math&gt; and the answer is &lt;math&gt;61+4 = \boxed{065}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_15&diff=136403 2019 AIME I Problems/Problem 15 2020-11-02T15:57:19Z <p>Firebolt360: /* Solution 1 */</p> <hr /> <div>==Problem 15==<br /> <br /> Let &lt;math&gt;\overline{AB}&lt;/math&gt; be a chord of a circle &lt;math&gt;\omega&lt;/math&gt;, and let &lt;math&gt;P&lt;/math&gt; be a point on the chord &lt;math&gt;\overline{AB}&lt;/math&gt;. Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; intersect at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Line &lt;math&gt;PQ&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;. Assume that &lt;math&gt;AP=5&lt;/math&gt;, &lt;math&gt;PB=3&lt;/math&gt;, &lt;math&gt;XY=11&lt;/math&gt;, and &lt;math&gt;PQ = \sqrt{\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 1==<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair O, A, B, P, O1, O2, Q, X, Y;<br /> O=(0, 0);<br /> A=dir(140); B=dir(40);<br /> P=(3A+5B)/8;<br /> O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);<br /> O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);<br /> Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)));<br /> X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));<br /> Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));<br /> <br /> draw(circle(O, 1));<br /> draw(circle(O1, length(A-O1)));<br /> draw(circle(O2, length(B-O2)));<br /> draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);<br /> <br /> dot(&quot;$O$&quot;, O, S);<br /> dot(&quot;$A$&quot;, A, A);<br /> dot(&quot;$B$&quot;, B, B);<br /> dot(&quot;$P$&quot;, P, dir(70));<br /> dot(&quot;$Q$&quot;, Q, dir(200));<br /> dot(&quot;$O_1$&quot;, O1, SW);<br /> dot(&quot;$O_2$&quot;, O2, SE);<br /> dot(&quot;$X$&quot;, X, X);<br /> dot(&quot;$Y$&quot;, Y, Y);<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; be the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, respectively. There is a homothety at &lt;math&gt;A&lt;/math&gt; sending &lt;math&gt;\omega&lt;/math&gt; to &lt;math&gt;\omega_1&lt;/math&gt; that sends &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;O_1&lt;/math&gt;, so &lt;math&gt;\overline{OO_2}\parallel\overline{O_1P}&lt;/math&gt;. Similarly, &lt;math&gt;\overline{OO_1}\parallel\overline{O_2P}&lt;/math&gt;, so &lt;math&gt;OO_1PO_2&lt;/math&gt; is a parallelogram. Moreover, &lt;cmath&gt;\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,&lt;/cmath&gt;whence &lt;math&gt;OO_1O_2Q&lt;/math&gt; is cyclic. However, &lt;cmath&gt;OO_1=O_2P=O_2Q,&lt;/cmath&gt;so &lt;math&gt;OO_1O_2Q&lt;/math&gt; is an isosceles trapezoid. Since &lt;math&gt;\overline{O_1O_2}\perp\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{OQ}\perp\overline{XY}&lt;/math&gt;, so &lt;math&gt;Q&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;.<br /> <br /> By Power of a Point, &lt;math&gt;PX\cdot PY=PA\cdot PB=15&lt;/math&gt;. Since &lt;math&gt;PX+PY=XY=11&lt;/math&gt; and &lt;math&gt;XQ=11/2&lt;/math&gt;, &lt;cmath&gt;XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,&lt;/cmath&gt;<br /> and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ===Note===<br /> <br /> One may solve for &lt;math&gt;PX&lt;/math&gt; first using PoAP, &lt;math&gt;PX = \frac{11}{2} - \frac{\sqrt{61}}{2}&lt;/math&gt;. Then, notice that &lt;math&gt;PQ^2&lt;/math&gt; is rational but &lt;math&gt;PX^2&lt;/math&gt; is not, also &lt;math&gt;PX = \frac{XY}{2} - \frac{\sqrt{61}}{2}&lt;/math&gt;. The most likely explanation for this is that &lt;math&gt;Q&lt;/math&gt; is the midpoint of &lt;math&gt;XY&lt;/math&gt;, so that &lt;math&gt;XQ = \frac{11}{2}&lt;/math&gt; and &lt;math&gt;PQ=\frac{\sqrt{61}}{2}&lt;/math&gt;. Then our answer is &lt;math&gt;m+n=61+4=\boxed{065}&lt;/math&gt;. One can rigorously prove this using the methods above<br /> <br /> ==Solution 2==<br /> <br /> Let the tangents to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at &lt;math&gt;R&lt;/math&gt;. Then, since &lt;math&gt;RA^2=RB^2&lt;/math&gt;, &lt;math&gt;R&lt;/math&gt; lies on the radical axis of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, which is &lt;math&gt;\overline{PQ}&lt;/math&gt;. It follows that &lt;cmath&gt;-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).&lt;/cmath&gt;<br /> Let &lt;math&gt;Q'&lt;/math&gt; denote the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;. By the Midpoint of Harmonic Bundles Lemma, &lt;cmath&gt;RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,&lt;/cmath&gt;<br /> whence &lt;math&gt;Q=Q'&lt;/math&gt;. Like above, &lt;math&gt;XP=\tfrac{11-\sqrt{61}}2&lt;/math&gt;. Since &lt;math&gt;XQ=\tfrac{11}2&lt;/math&gt;, we establish that &lt;math&gt;PQ=\tfrac{\sqrt{61}}2&lt;/math&gt;, from which &lt;math&gt;PQ^2=\tfrac{61}4&lt;/math&gt;, and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 3==<br /> <br /> Firstly we need to notice that &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;. Assume the center of circle &lt;math&gt;w, w_1, w_2&lt;/math&gt; are &lt;math&gt;O, O_1, O_2&lt;/math&gt;, respectively. Then &lt;math&gt;A, O_2, O&lt;/math&gt; are collinear and &lt;math&gt;O, O_1, B&lt;/math&gt; are collinear. Link &lt;math&gt;O_1P, O_2P, O_1Q, O_2Q&lt;/math&gt;. Notice that, &lt;math&gt;\angle B=\angle A=\angle APO_2=\angle BPO_1&lt;/math&gt;. As a result, &lt;math&gt;PO_1\parallel O_2O&lt;/math&gt; and &lt;math&gt;QO_1\parallel O_2P&lt;/math&gt;. So we have parallelogram &lt;math&gt;PO_2O_1O&lt;/math&gt;. So &lt;math&gt;\angle O_2PO_1=\angle O&lt;/math&gt; Notice that, &lt;math&gt;O_1O_2\bot PQ&lt;/math&gt; and &lt;math&gt;O_1O_2&lt;/math&gt; divide &lt;math&gt;PQ&lt;/math&gt; into two equal length pieces, So we have &lt;math&gt;\angle O_2PO_1=\angle O_2QO_1=\angle O&lt;/math&gt;. As a result, &lt;math&gt;O_2, Q, O, O_1,&lt;/math&gt; lie on one circle. So &lt;math&gt;\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P&lt;/math&gt;. Notice that &lt;math&gt;\angle O_1PQ+\angle O_2O_1P=90^{\circ}&lt;/math&gt;, we have &lt;math&gt;\angle OQP=90^{\circ}&lt;/math&gt;. As a result, &lt;math&gt;OQ\bot PQ&lt;/math&gt;. So &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;.<br /> <br /> Back to our problem. Assume &lt;math&gt;XP=x&lt;/math&gt;, &lt;math&gt;PY=y&lt;/math&gt; and &lt;math&gt;x&lt;y&lt;/math&gt;. Then we have &lt;math&gt;AP\cdot PB=XP\cdot PY&lt;/math&gt;, that is, &lt;math&gt;xy=15&lt;/math&gt;. Also, &lt;math&gt;XP+PY=x+y=XY=11&lt;/math&gt;. Solve these above, we have &lt;math&gt;x=\frac{11-\sqrt{61}}{2}=XP&lt;/math&gt;. As a result, we have &lt;math&gt;PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}&lt;/math&gt;. So, we have &lt;math&gt;PQ^2=\frac{61}{4}&lt;/math&gt;. As a result, our answer is &lt;math&gt;m+n=61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> <br /> <br /> Solution By BladeRunnerAUG (Fanyuchen20020715).<br /> <br /> ==Solution 4==<br /> Note that the tangents to the circles at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at a point &lt;math&gt;Z&lt;/math&gt; on &lt;math&gt;XY&lt;/math&gt; by radical center. Then, since &lt;math&gt;\angle ZAB = \angle ZQA&lt;/math&gt; and &lt;math&gt;\angle ZBA = \angle ZQB&lt;/math&gt;, we have <br /> &lt;cmath&gt;\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},&lt;/cmath&gt;<br /> so &lt;math&gt;ZAQB&lt;/math&gt; is cyclic. But if &lt;math&gt;O&lt;/math&gt; is the center of &lt;math&gt;\omega&lt;/math&gt;, clearly &lt;math&gt;ZAOB&lt;/math&gt; is cyclic with diameter &lt;math&gt;ZO&lt;/math&gt;, so &lt;math&gt;\angle ZQO = 90^{\circ} \implies Q&lt;/math&gt; is the midpoint of &lt;math&gt;XY&lt;/math&gt;. Then, by Power of a Point, &lt;math&gt;PY \cdot PX = PA \cdot PB = 15&lt;/math&gt; and it is given that &lt;math&gt;PY+PX = 11&lt;/math&gt;. Thus &lt;math&gt;PY, PX = \frac{11 \pm \sqrt{61}}{2}&lt;/math&gt; so &lt;math&gt;PQ = \frac{\sqrt{61}}{2} \implies PQ^2 = \frac{61}{4}&lt;/math&gt; and the answer is &lt;math&gt;61+4 = \boxed{065}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_15&diff=136402 2019 AIME I Problems/Problem 15 2020-11-02T15:55:19Z <p>Firebolt360: /* Solution 5 (Easy) */</p> <hr /> <div>==Problem 15==<br /> <br /> Let &lt;math&gt;\overline{AB}&lt;/math&gt; be a chord of a circle &lt;math&gt;\omega&lt;/math&gt;, and let &lt;math&gt;P&lt;/math&gt; be a point on the chord &lt;math&gt;\overline{AB}&lt;/math&gt;. Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; intersect at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Line &lt;math&gt;PQ&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;. Assume that &lt;math&gt;AP=5&lt;/math&gt;, &lt;math&gt;PB=3&lt;/math&gt;, &lt;math&gt;XY=11&lt;/math&gt;, and &lt;math&gt;PQ = \sqrt{\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 1==<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair O, A, B, P, O1, O2, Q, X, Y;<br /> O=(0, 0);<br /> A=dir(140); B=dir(40);<br /> P=(3A+5B)/8;<br /> O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);<br /> O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);<br /> Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)));<br /> X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));<br /> Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));<br /> <br /> draw(circle(O, 1));<br /> draw(circle(O1, length(A-O1)));<br /> draw(circle(O2, length(B-O2)));<br /> draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);<br /> <br /> dot(&quot;$O$&quot;, O, S);<br /> dot(&quot;$A$&quot;, A, A);<br /> dot(&quot;$B$&quot;, B, B);<br /> dot(&quot;$P$&quot;, P, dir(70));<br /> dot(&quot;$Q$&quot;, Q, dir(200));<br /> dot(&quot;$O_1$&quot;, O1, SW);<br /> dot(&quot;$O_2$&quot;, O2, SE);<br /> dot(&quot;$X$&quot;, X, X);<br /> dot(&quot;$Y&quot;, Y, Y);<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; be the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, respectively. There is a homothety at &lt;math&gt;A&lt;/math&gt; sending &lt;math&gt;\omega&lt;/math&gt; to &lt;math&gt;\omega_1&lt;/math&gt; that sends &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;O_1&lt;/math&gt;, so &lt;math&gt;\overline{OO_2}\parallel\overline{O_1P}&lt;/math&gt;. Similarly, &lt;math&gt;\overline{OO_1}\parallel\overline{O_2P}&lt;/math&gt;, so &lt;math&gt;OO_1PO_2&lt;/math&gt; is a parallelogram. Moreover, &lt;cmath&gt;\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,&lt;/cmath&gt;whence &lt;math&gt;OO_1O_2Q&lt;/math&gt; is cyclic. However, &lt;cmath&gt;OO_1=O_2P=O_2Q,&lt;/cmath&gt;so &lt;math&gt;OO_1O_2Q&lt;/math&gt; is an isosceles trapezoid. Since &lt;math&gt;\overline{O_1O_2}\perp\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{OQ}\perp\overline{XY}&lt;/math&gt;, so &lt;math&gt;Q&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;.<br /> <br /> By Power of a Point, &lt;math&gt;PX\cdot PY=PA\cdot PB=15&lt;/math&gt;. Since &lt;math&gt;PX+PY=XY=11&lt;/math&gt; and &lt;math&gt;XQ=11/2&lt;/math&gt;, &lt;cmath&gt;XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,&lt;/cmath&gt;<br /> and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 2==<br /> <br /> Let the tangents to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at &lt;math&gt;R&lt;/math&gt;. Then, since &lt;math&gt;RA^2=RB^2&lt;/math&gt;, &lt;math&gt;R&lt;/math&gt; lies on the radical axis of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, which is &lt;math&gt;\overline{PQ}&lt;/math&gt;. It follows that &lt;cmath&gt;-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).&lt;/cmath&gt;<br /> Let &lt;math&gt;Q'&lt;/math&gt; denote the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;. By the Midpoint of Harmonic Bundles Lemma, &lt;cmath&gt;RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,&lt;/cmath&gt;<br /> whence &lt;math&gt;Q=Q'&lt;/math&gt;. Like above, &lt;math&gt;XP=\tfrac{11-\sqrt{61}}2&lt;/math&gt;. Since &lt;math&gt;XQ=\tfrac{11}2&lt;/math&gt;, we establish that &lt;math&gt;PQ=\tfrac{\sqrt{61}}2&lt;/math&gt;, from which &lt;math&gt;PQ^2=\tfrac{61}4&lt;/math&gt;, and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 3==<br /> <br /> Firstly we need to notice that &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;. Assume the center of circle &lt;math&gt;w, w_1, w_2&lt;/math&gt; are &lt;math&gt;O, O_1, O_2&lt;/math&gt;, respectively. Then &lt;math&gt;A, O_2, O&lt;/math&gt; are collinear and &lt;math&gt;O, O_1, B&lt;/math&gt; are collinear. Link &lt;math&gt;O_1P, O_2P, O_1Q, O_2Q&lt;/math&gt;. Notice that, &lt;math&gt;\angle B=\angle A=\angle APO_2=\angle BPO_1&lt;/math&gt;. As a result, &lt;math&gt;PO_1\parallel O_2O&lt;/math&gt; and &lt;math&gt;QO_1\parallel O_2P&lt;/math&gt;. So we have parallelogram &lt;math&gt;PO_2O_1O&lt;/math&gt;. So &lt;math&gt;\angle O_2PO_1=\angle O&lt;/math&gt; Notice that, &lt;math&gt;O_1O_2\bot PQ&lt;/math&gt; and &lt;math&gt;O_1O_2&lt;/math&gt; divide &lt;math&gt;PQ&lt;/math&gt; into two equal length pieces, So we have &lt;math&gt;\angle O_2PO_1=\angle O_2QO_1=\angle O&lt;/math&gt;. As a result, &lt;math&gt;O_2, Q, O, O_1,&lt;/math&gt; lie on one circle. So &lt;math&gt;\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P&lt;/math&gt;. Notice that &lt;math&gt;\angle O_1PQ+\angle O_2O_1P=90^{\circ}&lt;/math&gt;, we have &lt;math&gt;\angle OQP=90^{\circ}&lt;/math&gt;. As a result, &lt;math&gt;OQ\bot PQ&lt;/math&gt;. So &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;.<br /> <br /> Back to our problem. Assume &lt;math&gt;XP=x&lt;/math&gt;, &lt;math&gt;PY=y&lt;/math&gt; and &lt;math&gt;x&lt;y&lt;/math&gt;. Then we have &lt;math&gt;AP\cdot PB=XP\cdot PY&lt;/math&gt;, that is, &lt;math&gt;xy=15&lt;/math&gt;. Also, &lt;math&gt;XP+PY=x+y=XY=11&lt;/math&gt;. Solve these above, we have &lt;math&gt;x=\frac{11-\sqrt{61}}{2}=XP&lt;/math&gt;. As a result, we have &lt;math&gt;PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}&lt;/math&gt;. So, we have &lt;math&gt;PQ^2=\frac{61}{4}&lt;/math&gt;. As a result, our answer is &lt;math&gt;m+n=61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> <br /> <br /> Solution By BladeRunnerAUG (Fanyuchen20020715).<br /> <br /> ==Solution 4==<br /> Note that the tangents to the circles at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at a point &lt;math&gt;Z&lt;/math&gt; on &lt;math&gt;XY&lt;/math&gt; by radical center. Then, since &lt;math&gt;\angle ZAB = \angle ZQA&lt;/math&gt; and &lt;math&gt;\angle ZBA = \angle ZQB&lt;/math&gt;, we have <br /> &lt;cmath&gt;\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},&lt;/cmath&gt;<br /> so &lt;math&gt;ZAQB&lt;/math&gt; is cyclic. But if &lt;math&gt;O&lt;/math&gt; is the center of &lt;math&gt;\omega&lt;/math&gt;, clearly &lt;math&gt;ZAOB&lt;/math&gt; is cyclic with diameter &lt;math&gt;ZO&lt;/math&gt;, so &lt;math&gt;\angle ZQO = 90^{\circ} \implies Q&lt;/math&gt; is the midpoint of &lt;math&gt;XY&lt;/math&gt;. Then, by Power of a Point, &lt;math&gt;PY \cdot PX = PA \cdot PB = 15&lt;/math&gt; and it is given that &lt;math&gt;PY+PX = 11&lt;/math&gt;. Thus &lt;math&gt;PY, PX = \frac{11 \pm \sqrt{61}}{2}&lt;/math&gt; so &lt;math&gt;PQ = \frac{\sqrt{61}}{2} \implies PQ^2 = \frac{61}{4}&lt;/math&gt; and the answer is &lt;math&gt;61+4 = \boxed{065}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=135725 User:Piphi 2020-10-24T16:44:18Z <p>Firebolt360: /* User Count */</p> <hr /> <div>{{User:Piphi/Template:Header}}<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#dddddd&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;100px&quot;&gt;360&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;Piphi is legendary and made the USA IMO team in 2019.&lt;br&gt;<br /> <br /> Piphi is the creator of the [[User:Piphi/Games|AoPS Wiki Games by Piphi]], the future of games on AoPS.&lt;br&gt;<br /> <br /> Piphi started the signature trend at around May 2020.&lt;br&gt;<br /> <br /> Piphi is an extremely OP person - LJCoder619. &lt;br&gt;<br /> <br /> Piphi is OP --[[User:Aray10|Aray10]] ([[User talk:Aray10|talk]]) 23:22, 17 June 2020 (EDT) &lt;br&gt;<br /> <br /> Piphi is 100 percent OP compared to the rest of us --[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 09:24, 4 September 2020 (EDT) &lt;br&gt;<br /> <br /> According to my studies Piphi is 50 percent asymptote, 50 percent AoPS Wiki, and 100 percent fun. --[[User:CreativeHedgehog|Creativehedgehog]] ([[User talk:CreativeHedgehog|talk]]) 16:14, 19 September 2020 (EDT) &lt;br&gt;<br /> <br /> Piphi has been very close to winning multiple [[Greed Control]] games, piphi placed 5th in game #18 and 2nd in game #19. Thanks to piphi, Greed Control games have started to be kept track of. Piphi made a spreadsheet that has all of Greed Control history [https://artofproblemsolving.com/community/c19451h2126208p15569802 here].&lt;br&gt;<br /> <br /> Piphi also found out who won [[Reaper]] games #1 and #2 as seen [https://artofproblemsolving.com/community/c19451h1826745p15526330 here].&lt;br&gt;<br /> <br /> Piphi has been called op by many AoPSers, including the legendary [[User:Radio2|Radio2]] himself [https://artofproblemsolving.com/community/c19451h1826745p15526800 here]. (note: Radio2 calls many users op.)&lt;br&gt;<br /> <br /> Piphi created the [[AoPS Administrators]] page, added most of the AoPS Admins to it, and created the scrollable table.&lt;br&gt;<br /> <br /> Piphi has also added a lot of the info that is in the [[Reaper Archives]].&lt;br&gt;<br /> <br /> Piphi has a side-project that is making the Wiki's [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]].&lt;br&gt;<br /> <br /> Piphi published Greed Control Game 19 statistics [https://artofproblemsolving.com/community/c19451h2126212 here].<br /> <br /> Piphi has a post that was made an announcement on a official AoPS Forum [https://artofproblemsolving.com/community/c68h2175116 here].<br /> <br /> Piphi is a proud member of [https://artofproblemsolving.com/community/c562043 The Interuniversal GMAAS Society].&lt;br&gt;<br /> <br /> Piphi is amazing at asymptote. [[User:Sonone]]<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;<br /> You can check out more goals/statistics [[User:Piphi/Statistics|here]].<br /> <br /> A User Count of 500<br /> {{User:Piphi/Template:Progress_Bar|71.4|width=100%}}<br /> <br /> 200 subpages of [[User:Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|63.5|width=100%}}<br /> <br /> 200 signups for [[User:Piphi/Games|AoPS Wiki Games by Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|46.5|width=100%}}<br /> <br /> Make 10,000 edits<br /> {{User:Piphi/Template:Progress_Bar|19.72|width=100%}}&lt;/div&gt;<br /> &lt;/div&gt;</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=Mock_AMC&diff=135244 Mock AMC 2020-10-18T20:52:47Z <p>Firebolt360: /* Mock AMC 12 */</p> <hr /> <div>A '''Mock AMC''' is a contest intended to mimic an actual [[AMC]] (American Mathematics Competitions 8, 10, or 12) exam. A number of '''Mock AMC''' competitions have been hosted on the [[Art of Problem Solving]] message boards. They are generally made by one community member and then administered for any of the other community members to take. Sometimes, the administrator may ask other people to sign up to write problems for the contest.<br /> <br /> Mock AMCs are usually very popular in the months leading up to the actual [[AMC]] competition. There is no guarantee that community members will make Mock AMCs in any given year, but there probably will be one.<br /> <br /> Feel free to remove mock tests that are not high quality or to recommend others.<br /> <br /> Ongoing mock AMCs, as well as other mock contests, can be found in the [https://artofproblemsolving.com/community/c594864_aops_mock_contests AoPS Mock Contests forum].<br /> <br /> == Tips for Writing a Mock AMC ==<br /> Anyone can write a Mock AMC and administer it. If you are interested in writing one, here are some tips:<br /> <br /> * Look at past [[AMC]]/[[AHSME]] tests to get a feel for what kind of problems you should write and what difficulty level they should be.<br /> * Look at famous theorems and formulas and see if there's any way you can make a good problem out of them.<br /> * If you are running out of creative juice and decide to pull problems from contests, try using problems from obscure contests first, if possible. This way, even the more experienced test takers will hopefully find problems that they do not already know how to do.<br /> * Pair up with another user on AoPS and write it together. Two minds are much better than one. With just one person, the problems may be biased toward one subject, but with two people, the chances of this happening are smaller.<br /> <br /> == Past Mock AMCs ==<br /> <br /> Listed below are the [higher-quality] Mock AMCs that have been hosted over AoPS in the past. Feel free to add, remove, or recommend.<br /> <br /> Note that the &quot;level&quot; column represents the originally intended difficulty. In other words, if a person makes a mock [[AMC 12]], the level would be &quot;12&quot;, even if the problems themselves are much easier. Recommended mock tests are starred.<br /> <br /> === Mock AMC 12 ===<br /> <br /> {| class=&quot;wikitable&quot; style=&quot;text-align:center;width:100%&quot;<br /> |-<br /> |<br /> ! scope=&quot;col&quot; | '''Author'''<br /> ! scope=&quot;col&quot; | '''Year'''<br /> ! scope=&quot;col&quot; | '''Initial Discussion'''<br /> ! scope=&quot;col&quot; | '''Problems'''<br /> ! scope=&quot;col&quot; width=80 | '''Answers'''<br /> ! scope=&quot;col&quot; | '''Results/Discussion'''<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC #1'''<br /> | mathfanatic<br /> | 2003<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9321 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9353 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9572 1-5]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9573 6-10]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9574 11-15]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9575 16-20]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9576 21-25]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9365 Results]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC #2'''<br /> | mathfanatic<br /> | 2004<br /> | [http://www.artofproblemsolving.com/community/c5h10497p67837 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10497 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10497 Solutions]<br /> | n/a<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC A'''<br /> | JSRosen3<br /> | 2004<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14138 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14361 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=102483#p102483 Answers] [http://www.artofproblemsolving.com/community/c5h14516 Solutions]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14489 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC B'''<br /> | beta<br /> | 2004<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14735 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14764 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14894 Answers] [http://www.artofproblemsolving.com/community/c5h14884 Solutions]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=105741#p105741 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC C'''<br /> | JGeneson<br /> | 2004<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15001 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15134 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15251 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15251 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC D'''<br /> | joml88<br /> | 2004<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16886 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17888 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17891 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17891 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC E'''<br /> | Silverfalcon<br /> | 2004<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21997 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22141 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22344 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22344 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC F'''<br /> | joml88<br /> | 2005<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22049 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23163 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23177 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23177 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC G'''<br /> | Lucky707<br /> | 2005<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24355 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24974 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25087 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h25087p157983 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC H'''<br /> | Silverfalcon<br /> | 2005<br /> | [http://www.artofproblemsolving.com/community/c5h24437p154514 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h24437p154514 Problems]<br /> | throughout thread<br /> | n/a<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC I'''<br /> | white_horse_king88<br /> | 2005<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21280 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25181 Problems]<br /> | n/a<br /> | n/a<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC J'''<br /> | Silverfalcon<br /> | 2005<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47625 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48129 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=304246#p304246 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48132 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC K'''<br /> | amirhtlusa<br /> | 2005<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49958 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50515 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=321102#p321102 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=321102#p321102 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC L'''<br /> | amirhtlusa<br /> | 2005<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61330 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63041 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63258 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63258 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC M'''<br /> | Silverfalcon<br /> | 2006<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63542 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78982 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79749 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79749 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC N'''<br /> | chess64<br /> | 2006<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98894 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99307 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99344 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99344 Results] / [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99566 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC O'''<br /> | mustafa<br /> | 2006<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121312 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122126 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126071 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126071 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC P'''<br /> | Anirudh<br /> | 2006<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125029 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=709655#p709655 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=716240#p716240 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=716262#p716262 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125029 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC Q'''<br /> | calc rulz<br /> | 2006<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125194 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125886 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128102 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128102 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC R'''<br /> | rnwang2<br /> | 2006<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126107 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=715597#p715597 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127421 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127421 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC S'''<br /> | mysmartmouth<br /> | 2007<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127221 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128689 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131403 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131403 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC T'''<br /> | paladin8<br /> | 2007<br /> | [http://www.artofproblemsolving.com/community/c5h127979p726003 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127979 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129159 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129159 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC U'''<br /> | Silverfalcon<br /> | 2008<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184067 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185233 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185238 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185236 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185238 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC V'''<br /> | gfour84<br /> | 2009<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=298452 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1634175#p1634175 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1637429#p1637429 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1637006#p1637006 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=302030 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 12'''<br /> | MathTwo<br /> | 2010<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=319184 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1759276#p1759276 Problems]<br /> | n/a<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1778080#p1778080 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=328510 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 2/11'''<br /> | Caelestor<br /> | 2011<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=390907 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2183293#p2183293 Problems]<br /> | n/a<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=390907 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 1/12'''<br /> | Lord.of.AMC<br /> | 2012<br /> | [http://www.artofproblemsolving.com/community/c5h456321p2563731 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2605537#p2605537 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2610048#p2610048 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2610048#p2610048 Results]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 12'''<br /> | Diehard<br /> | 2012<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=459149 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=459710 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2580327#p2580327 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=459710 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 12 2012'''<br /> | python123<br /> | 2012<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=456256 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2576205#p2576205 Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h456256p2580610 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h456256p2580610 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC A'''<br /> | Binomial-theorem<br /> | 2012<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=473867 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2692082#p2692082 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2717578#p2717578 Solutions]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2707954#p2707954 Results]<br /> |-<br /> ! scope = &quot;row&quot; | '''Almost 2016 Mock AMC 11.5'''<br /> | whatshisbucket<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h1175087p5665743 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h1175087p5665743 Problems]<br /> | [http://artofproblemsolving.com/community/c209194_almost_2016_mock_amc_11.5 Solutions]<br /> | [http://www.artofproblemsolving.com/community/c5h1175087p5665743 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''hnkevin42 Mock AMC 12'''<br /> | hnkevin42<br /> | 2016<br /> | [http://www.artofproblemsolving.com/community/c5h1187484p5778754 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h1187484p5778754 Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h1187484p5913294 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h1187484p5778754 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Last-Minute Mock AMC 12'''<br /> | CountofMC<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c594864h1582044_lastminute_mock_amc_12 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1582044_lastminute_mock_amc_12 Problems]<br /> | [https://artofproblemsolving.com/community/c594864h1582044_lastminute_mock_amc_12 Answers]<br /> | [https://artofproblemsolving.com/community/c4t334444f4_lastminute_mock_amc_12 Results/Discussion]<br /> |-<br /> !scope=&quot;row&quot; | Christmas Math Competitions Year 2*<br /> | CMC Committee<br /> | 2018-2019<br /> | [https://artofproblemsolving.com/community/c594864h1747367_aime_ii_released_christmas_mathematics_competition_cmc_year_2 Initial Discussion]<br /> | [https://drive.google.com/file/d/1uS3YqAK10jB1RkCQCLQgeBHUACydJph-/view CMC 12A] [https://drive.google.com/file/d/1txKy-MfZCPPnUNTLCIad8dbatV-EGu71/view CMC 12B]<br /> | [https://artofproblemsolving.com/community/c798404h1760024_cmc_10a12a_amp_10b12b_year_2__problems_and_answer_keys Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1747367p11379904 Results] / [https://artofproblemsolving.com/community/c798404_christmas_mathematics_competitions_year_2 Discussion]<br /> |-<br /> !scope=&quot;row&quot; | 2019 Mock AMC 12B*<br /> | fidgetboss_4000<br /> | 2019<br /> | [https://artofproblemsolving.com/community/c594864h1897633p13160252 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1897633p13160252 Problems] (or click here: [[Mock AMC 12B Problems]]<br /> | [https://artofproblemsolving.com/community/c963448h1919463_answer_keys_d Answers]<br /> | N/A / N/A<br /> |-<br /> !scope=&quot;row&quot; | Christmas Math Competitions Year 3*<br /> | CMC Committee<br /> | 2019-2020<br /> | [https://artofproblemsolving.com/community/c594864h1967029_cime_ii_released_christmas_math_competition_cmc_year_3 Initial Discussion]<br /> | [http://cmc.ericshen.net/CMC-2020/CMC-2020-12A.pdf CMC 12A] [http://cmc.ericshen.net/CMC-2020/CMC-2020-12B-booklet.pdf CMC 12B]<br /> | [https://artofproblemsolving.com/community/c1035147h1980361p13760203 Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1967029p13623910 Results] / [https://artofproblemsolving.com/community/c1035147_christmas_mathematics_competitions_year_3 Discussion]<br /> |-<br /> !scope=&quot;row&quot; | January 2020 Mock AMC 10/12<br /> | P_Groudon<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h1984167_concluded_january_2020_mock_amc_1012 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1984167_concluded_january_2020_mock_amc_1012 Problems]<br /> | [https://artofproblemsolving.com/community/c1035224h1984207_january_2020_answer_keys Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1984167p13963794 Results]<br /> |-<br /> !scope=&quot;row&quot; | OTSS 2020<br /> | kevinmathz<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h2066002_otss_olympiad_test_spring_series Initial Discussion]<br /> | [https://drive.google.com/file/d/1JfuEvDGw9i99XeQYACPUDzHebxyrcmtE/edit TMC 10A] [https://drive.google.com/file/d/1FkKjAgehDwpmmNOj_mjvYuCCV1r7p08p/edit TMC 12A]<br /> | [https://artofproblemsolving.com/community/c1130807h2094651_tmc_1012_a_solutions Answers]<br /> | [https://artofproblemsolving.com/community/c1130807h2081571_links_to_mocks_solutions_and_leaderboards Results]<br /> |-<br /> !scope=&quot;row&quot; | FMC 2020<br /> | fidgetboss_4000<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h2069998 Initial Discussion]<br /> | [https://drive.google.com/file/d/17jscyJzVCFDlV6YL0OpPKczyk04Tltsx/view FMC 10A] [https://drive.google.com/file/d/1pMdXlGy9F5hbWSP8CGANAhAA9_ct7pdS/view FMC 12A]<br /> | [https://artofproblemsolving.com/community/c1177836_2020_fmc_public_discussion_forum_uwu Answers]<br /> | [https://artofproblemsolving.com/community/q2h2069998p15525666 Results]<br /> |-<br /> !scope=&quot;row&quot; | 2020 Mock AMC 12<br /> | scrabbler94<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h2159345_contest_over_2020_mock_amc_1012_scrabbler94 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h2159345_contest_over_2020_mock_amc_1012_scrabbler94 Problems]<br /> | [https://artofproblemsolving.com/community/c594864h2159345p16460394 Answers]<br /> | [https://artofproblemsolving.com/community/c594864h2159345p16460394 Results]<br /> |-<br /> |}<br /> <br /> === Mock AMC 10 ===<br /> <br /> {| class=&quot;wikitable&quot; style=&quot;text-align:center;width:100%&quot;<br /> |-<br /> |<br /> ! scope=&quot;col&quot; | '''Author'''<br /> ! scope=&quot;col&quot; | '''Year'''<br /> ! scope=&quot;col&quot; | '''Initial Discussion'''<br /> ! scope=&quot;col&quot; | '''Problems'''<br /> ! scope=&quot;col&quot; | '''Answers'''<br /> ! scope=&quot;col&quot; | '''Results/Discussion'''<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 10'''<br /> | #H34N1<br /> | 2008<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212730 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214081 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213727 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213732 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''agent's Mock AMC 10'''<br /> | agentcx<br /> | 2009<br /> | [http://www.artofproblemsolving.com/community/c5h331823p1775678 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1781208#p1781208 Problems]<br /> | n/a<br /> | [http://www.artofproblemsolving.com/community/c5h331823p1782769 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 10 Set'''<br /> | AwesomeToad<br /> | 2010<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311120 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1762829#p1762829 Problems]<br /> | n/a<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1778740#p1778740 Results]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 10/12'''<br /> | djmathman<br /> | 2013<br /> | [http://www.artofproblemsolving.com/community/c5h556673s1_first_mock_amcs_of_the_20132014_season Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h556673p3294854 Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h556673p3324794 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h556673p3324794 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 10 2014-2015*'''<br /> | AlcumusGuy<br /> | 2014<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=618080&amp;hilit=AlcumusGuy%27s+mock+amc+10 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=618080&amp;hilit=AlcumusGuy%27s+mock+amc+10 Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h618080p3710795 Answers/Results]<br /> | [http://artofproblemsolving.com/community/c56018 Problem Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Kelvin the Frog v2015'''<br /> | BOGTRO<br /> | 2015<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=623373&amp;hilit=Kelvin+the+frog Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=623373&amp;hilit=Kelvin+the+frog Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3731642#p3731642 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3731642#p3731642 Results]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 10/12'''<br /> | joey8189681<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h624714p3741743 Initial Discussion]<br /> | [https://www.dropbox.com/s/1c8hhl6yt2awpix/Mock%20AMC%2010.pdf?dl=0 Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h624714p3754593 Answers]<br /> | [http://www.artofproblemsolving.com/community/q2h626092p3756574 Results]<br /> |-<br /> ! scope=&quot;row&quot; | '''May Mock AMC 10 Contest'''<br /> | azmath333<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h1082684_mock_amc_1012_contest Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h1082684_mock_amc_1012_contest Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h1082684p4812575 Answers] [http://www.artofproblemsolving.com/community/c5h1082684p4825662 Solutions]<br /> | [http://www.artofproblemsolving.com/community/c5h1082684p4812575 Results / Discussion] <br /> |-<br /> ! scope=&quot;row&quot; | '''2015 Mock AMC 10*'''<br /> | droid347<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h1094505p4893513 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h1094505p4893513 Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h1094505p5051006 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h1094505p5045295 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''July Mock AMC 10 Contest'''<br /> | akshaygowrish<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c5h1104125_july_mock_amc_10_contest Initial Discussion]<br /> | [https://docs.google.com/document/d/12OXd-mmS_T7SyMcqw1hfvykNlcj-4TR4mMXlRJ09uBg/edit?usp=sharing Problems]<br /> | [https://docs.google.com/document/d/1-Q_w5XNKcRaffvpCLSdoJ51TpRpcDAgYLmV9OlczBCs/edit?usp=sharing Answer Key]<br /> | [http://artofproblemsolving.com/community/c5h1104125p5154283 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; |'''August Mock AMC 10'''<br /> | azmath333<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h1115462_august_mock_amc_10 Initial Discussion]<br /> | [http://latex.artofproblemsolving.com/miscpdf/augustmockamc10.pdf Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h1115462p5270430 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h1115462p5270430 Results] / [http://www.artofproblemsolving.com/community/c121880_august_mock_amc_10_discussion_forum Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''September Mock AMC 10'''<br /> | cpma213<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c5h1137447p5320386 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c5h1137447p5320386 Problems]<br /> | n/a<br /> | [http://artofproblemsolving.com/community/c5h1137447p5479954 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''December 2015 Mock AMC 1^3*(3+7)'''<br /> | mathisawesome2169<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h1173465p5647504 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h1173465p5647504 Problems]<br /> | n/a<br /> | [http://www.artofproblemsolving.com/community/c5h1173465p5707101 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''New Year's Mock AMC10*'''<br /> | checkmatetang<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c5h1171366p5627439 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c5h1171366p5627439 Problems]<br /> | [http://artofproblemsolving.com/community/c5h1171366p5711384 Answers]<br /> | [http://artofproblemsolving.com/community/c5h1171366p5711384 Results] / [http://artofproblemsolving.com/community/c202656_new_years_mock_amc10_discussion Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''2015-2016 Mock AMC 10*'''<br /> | PersonPsychopath<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c5h1168398p5595105 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c5h1168398p5595105 Problems]<br /> | [http://artofproblemsolving.com/community/c147536h1182005_mock_amc10_wrapup Answers]<br /> | [http://artofproblemsolving.com/community/c5h1168398p5595105 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''Mock AMC 10 2015-2016*'''<br /> | AlcumusGuy<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h1177413p5687990 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c5h1177413_mock_amc_10_20152016_released Problems]<br /> | [http://artofproblemsolving.com/community/c212844_mock_amc_10_20152016 Solutions]<br /> | [http://artofproblemsolving.com/community/c5h1177413p5769839 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''January 2016 Mock AMC 10*'''<br /> | atmchallenge<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h1178971p5700377 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h1178971p5700377 Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h1178971p5838468 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h1178971p5838468 Results/Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''2015-2016 Mock AMC 10* #2'''<br /> | PersonPsychopath<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c5h1188467 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c5h1188467 Problems]<br /> | [http://artofproblemsolving.com/community/c147536h1175930 Solutions]<br /> | [http://artofproblemsolving.com/community/c5h1188467 Results / Discussion]<br /> |-<br /> <br /> ! scope=&quot;row&quot; | '''MeepyMeepMeep Mock AMC 10'''<br /> | MeepyMeepMeep, speck<br /> | 2016<br /> | [http://artofproblemsolving.com/community/c5h1195432p5852002 Initial Discussion]<br /> | [https://www.dropbox.com/s/uw0herhuqk8zbp5/Mock%20AMC%2010.pdf?dl=0 Problems]<br /> | n/a<br /> | n/a<br /> |- <br /> <br /> !scope=&quot;row&quot; | eisirrational's Mock AMC 10<br /> | eisirrational, illogical_21, whatshisbucket<br /> | 2017<br /> | [https://artofproblemsolving.com/community/c5h1372991_eisirrationals_mock_amc_10 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c5h1372991_eisirrationals_mock_amc_10 Problems]<br /> | [http://artofproblemsolving.com/community/c418591h1392426p7811150 Answer Key]<br /> | [https://artofproblemsolving.com/community/c418591_mock_amc_10_discussion Discussion]<br /> |-<br /> <br /> !scope=&quot;row&quot; | Summer Mock AMC 10<br /> | Rowechen, OmicronGamma, FedeX333X, KenV, kvedula2004, ItzVineeth<br /> | 2017<br /> | [https://artofproblemsolving.com/community/c5h1471834_summer_mock_amc_series Initial Discussion]<br /> | [https://drive.google.com/file/d/0B3M-fxa6QG0_ODNpSnJZcEt3djQ/view Problems]<br /> | [http://latex.artofproblemsolving.com/miscpdf/hehdrwpi.pdf?t=1500270651030 Answers]<br /> | [https://artofproblemsolving.com/community/c481237_2017_summer_mock_amc_10_discussion_forum Statistics / Discussion]<br /> |-<br /> <br /> !scope=&quot;row&quot; | scrabbler94's Mock AMC 10<br /> | scrabbler94<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c5h1569848_mock_2018_amc_10_test_scrabbler94 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c5h1569848_mock_2018_amc_10_test_scrabbler94 Problems]<br /> | [https://artofproblemsolving.com/community/c5h1569848_mock_2018_amc_10_test_scrabbler94 Answers]<br /> | [http://artofproblemsolving.com/community/c5h1569848p9704902 Results] / [https://artofproblemsolving.com/community/c593716_scrabbler94s_mock_amc_10_discussion Discussion]<br /> |-<br /> <br /> ! scope=&quot;row&quot; | '''2018 Mock AMC 10'''<br /> | blue8931<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c594864h1606066 Initial Discussion]<br /> | [https://artofproblemsolving.com/downloads/printable_post_collections/628064 Problems]<br /> | [https://artofproblemsolving.com/community/c628116h1632436_official_answer_key Answer Key]<br /> | [https://artofproblemsolving.com/community/c628116_2018_mock_amc_10_discussion_forum Results / Discussion]<br /> |-<br /> <br /> !scope=&quot;row&quot; | 2018 Memorial Day Mock AMC 10<br /> | QIDb602<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c594864h1645806_released_2018_memorial_day_mock_amc_10 Initial Discussion]<br /> | [https://drive.google.com/file/d/1paAHspy5PH1J9fMoNYTb7cIEH3Di72K5/view Problems]<br /> | [https://drive.google.com/file/d/1tipZuHE11jmmOfnSDnhkwKnvIOnbsbaL/view?usp=sharing Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1645806_released_2018_memorial_day_mock_amc_10 Results] / [https://artofproblemsolving.com/community/c671484 Discussion]<br /> |-<br /> <br /> !scope=&quot;row&quot; | Autumn Mock AMC 10<br /> | Krypton36, AlastorMoody, alphaone001, dchenmathcounts, InternetPerson10, kootrapali, mathdragon2000, MathGeek2018, Stormersyle<br /> | 2018<br /> | [http://artofproblemsolving.com/community/c594864h1710983p11032970 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c594864h1710983p11032970 Problems]<br /> | [http://artofproblemsolving.com/community/c594864h1710983p11331788 Answers]<br /> | [http://artofproblemsolving.com/community/c594864h1710983p11332566 Results] / [https://artofproblemsolving.com/community/c761744_autumn_mock_amc_10_discussion_forum_d Discussion]<br /> |-<br /> <br /> !scope=&quot;row&quot; | 2018 December Mock AMC 10<br /> | mathchampion1, kcbhatraju, kootrapali, chocolatelover111<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c594864h1743064_2018_december_mock_amc_10_released Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1743064_2018_december_mock_amc_10_released Problems]<br /> | [https://artofproblemsolving.com/community/c784638_december_mock_amc_10_2018_discussion_forum Answers]<br /> | [https://artofproblemsolving.com/community/c784638_december_mock_amc_10_2018_discussion_forum Results] / [https://artofproblemsolving.com/community/c784638_december_mock_amc_10_2018_discussion_forum Discussion]<br /> |-<br /> <br /> !scope=&quot;row&quot; | Christmas Math Competitions Year 2*<br /> | CMC Committee<br /> | 2018-2019<br /> | [https://artofproblemsolving.com/community/c594864h1747367_aime_ii_released_christmas_mathematics_competition_cmc_year_2 Initial Discussion]<br /> | [https://drive.google.com/file/d/15heOR_6rnH70joxJRfZBhQXa8Wezb2AW/view CMC 10A] [https://drive.google.com/file/d/1ZNayjDykKYj4889nXtOb7k9KBu_bqXki/view CMC 10B]<br /> | [https://artofproblemsolving.com/community/c798404h1760024_cmc_10a12a_amp_10b12b_year_2__problems_and_answer_keys Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1747367p11379904 Results] / [https://artofproblemsolving.com/community/c798404_christmas_mathematics_competitions_year_2 Discussion]<br /> |-<br /> <br /> !scope=&quot;row&quot; | 2019 Mock AMC 10C*<br /> | fidgetboss_4000<br /> | 2019<br /> | [https://artofproblemsolving.com/community/c594864h1786263_released_2019_amc_10c Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1786263_released_2019_amc_10c Problems] (or click here: [[2019 AMC 10C Problems]]<br /> | [https://artofproblemsolving.com/community/c594864h1786263_released_2019_amc_10c Answers - In Thread]<br /> | N/A / N/A<br /> |-<br /> !scope=&quot;row&quot; | 2019 Mock AMC 10B*<br /> | fidgetboss_4000<br /> | 2019<br /> | [https://artofproblemsolving.com/community/c594864h1897633p13160252 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1897633p13160252 Problems] (or click here: [[Mock AMC 10B Problems]]<br /> | [https://artofproblemsolving.com/community/c963448h1919463_answer_keys_d Answers]<br /> | N/A / N/A<br /> |-<br /> !scope=&quot;row&quot; | Christmas Math Competitions Year 3*<br /> | CMC Committee<br /> | 2019-2020<br /> | [https://artofproblemsolving.com/community/c594864h1967029_cime_ii_released_christmas_math_competition_cmc_year_3 Initial Discussion]<br /> | [http://cmc.ericshen.net/CMC-2020/CMC-2020-10A.pdf CMC 10A] [http://cmc.ericshen.net/CMC-2020/CMC-2020-10B-booklet.pdf CMC 10B]<br /> | [https://artofproblemsolving.com/community/c1035147h1980361p13760203 Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1967029p13623910 Results] / [https://artofproblemsolving.com/community/c1035147_christmas_mathematics_competitions_year_3 Discussion]<br /> |-<br /> !scope=&quot;row&quot; | Stormersyle's Mock AMC 10*<br /> | Stormersyle<br /> | 2019-2020<br /> | [https://artofproblemsolving.com/community/c594864h1974214p13694448 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1974214p13694448 Problems]<br /> | [https://artofproblemsolving.com/community/c594864h1974214p13694448 Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1974214p13694448 Results] / [https://artofproblemsolving.com/community/c594864h1974214p13694448 Discussion]<br /> |-<br /> !scope=&quot;row&quot; | January 2020 Mock AMC 10/12<br /> | P_Groudon<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h1984167_concluded_january_2020_mock_amc_1012 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1984167_concluded_january_2020_mock_amc_1012 Problems]<br /> | [https://artofproblemsolving.com/community/c1035224h1984207_january_2020_answer_keys Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1984167p13963794 Results]<br /> |-<br /> !scope=&quot;row&quot; | 2020 Mock Combo AMC 10<br /> | fidgetboss_4000<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h2005582_amc_1012_released_mock_combo_amc_1012_ii Initial Discussion]<br /> | [https://artofproblemsolving.com/wiki/index.php?title=2020_Mock_Combo_AMC_10&amp;action=view Problems]<br /> | [https://artofproblemsolving.com/community/c1121049h2053291_answer_key Answers]<br /> | [https://artofproblemsolving.com/community/c594864h2005582_amc_1012_released_mock_combo_amc_1012_ii Results]<br /> |-<br /> !scope=&quot;row&quot; | OTSS 2020<br /> | kevinmathz<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h2066002_otss_olympiad_test_spring_series Initial Discussion]<br /> | [https://drive.google.com/file/d/1JfuEvDGw9i99XeQYACPUDzHebxyrcmtE/edit TMC 10A] [https://drive.google.com/file/d/1FkKjAgehDwpmmNOj_mjvYuCCV1r7p08p/edit TMC 12A]<br /> | [https://artofproblemsolving.com/community/c1130807h2094651_tmc_1012_a_solutions Answers]<br /> | [https://artofproblemsolving.com/community/c1130807h2081571_links_to_mocks_solutions_and_leaderboards Results]<br /> |-<br /> !scope=&quot;row&quot; | FMC 2020<br /> | fidgetboss_4000<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h2069998 Initial Discussion]<br /> | [https://drive.google.com/file/d/17jscyJzVCFDlV6YL0OpPKczyk04Tltsx/view FMC 10A] [https://drive.google.com/file/d/1pMdXlGy9F5hbWSP8CGANAhAA9_ct7pdS/view FMC 12A]<br /> | [https://artofproblemsolving.com/community/c1177836_2020_fmc_public_discussion_forum_uwu Answers]<br /> | [https://artofproblemsolving.com/community/q2h2069998p15525666 Results]<br /> |-<br /> !scope=&quot;row&quot; | 2020 Mock AMC 10<br /> | scrabbler94<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h2159345_contest_over_2020_mock_amc_1012_scrabbler94 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h2159345_contest_over_2020_mock_amc_1012_scrabbler94 Problems]<br /> | [https://artofproblemsolving.com/community/c594864h2159345p16460394 Answers]<br /> | [https://artofproblemsolving.com/community/c594864h2159345p16460394 Results]<br /> |-<br /> !scope=&quot;row&quot; | JMC 2020*<br /> | skyscraper<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h2164953_jmc_10_concluded_july_mock_amc Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h2164953_jmc_10_concluded_july_mock_amc Problems]<br /> | [https://artofproblemsolving.com/community/c594864h2164953p16895447 Answers]<br /> | [https://artofproblemsolving.com/community/c594864h2164953p16895447 Results]<br /> |-<br /> |}<br /> <br /> === Mock AMC 8 ===<br /> <br /> {| class=&quot;wikitable&quot; style=&quot;text-align:center;width:100%&quot;<br /> |-<br /> |<br /> ! scope=&quot;col&quot; | '''Author'''<br /> ! scope=&quot;col&quot; | '''Year'''<br /> ! scope=&quot;col&quot; | '''Initial Discussion'''<br /> ! scope=&quot;col&quot; | '''Problems'''<br /> ! scope=&quot;col&quot; width=80 | '''Answers'''<br /> ! scope=&quot;col&quot; | '''Results/Discussion'''<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 8'''<br /> | mathfanatic<br /> | 2004<br /> | [http://www.artofproblemsolving.com/community/c5h15878p111575 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15878 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=114070#p114070 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h15878p114404 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 8'''<br /> | 13375P34K43V312<br /> | 2006<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107507 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=108455 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=630802#p630802 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h108455p623522 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 8'''<br /> | ahaanomegas<br /> | 2012<br /> | [http://www.artofproblemsolving.com/community/c5h459346p2577870 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2598111#p2598111 Problems]<br /> | n/a<br /> | n/a<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 8'''<br /> | utahjazz<br /> | 2012<br /> | [http://www.artofproblemsolving.com/community/c5h485041p2717878 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2728186#p2728186 Problems]<br /> | n/a<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2731391#p2731391 Results]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 8'''<br /> | iNomOnCountdown<br /> | 2014<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&amp;t=590451&amp;hilit=iNomonCountdown%27s+mock+amc+8 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&amp;t=590451&amp;hilit=iNomonCountdown%27s+mock+amc+8 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&amp;t=590451&amp;start=0 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&amp;t=590451&amp;start=0 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''Mock AMC 8'''<br /> | Tan<br /> | 2014<br /> | [http://www.artofproblemsolving.com/community/c5h613297p3648226 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h613297p3648227 Problems]<br /> | [https://docs.google.com/spreadsheets/d/1U1QJ9r4r2xFbRByk9gXOhEVV48ScjoLWZp4IFanbKF0/edit#gid=1475864782 Answers]<br /> | [https://docs.google.com/spreadsheets/d/1U1QJ9r4r2xFbRByk9gXOhEVV48ScjoLWZp4IFanbKF0/edit#gid=1475864782 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''2015 Hard Mock AMC 8*'''<br /> | Not_a_Username, 8invalid8<br /> | 2015<br /> | [http://artofproblemsolving.com/community/q1h1128391p5227860 Initial Discussion]<br /> | [http://latex.artofproblemsolving.com/miscpdf/kashimyo.pdf Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h1128391p5423312 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h1128391p5423312 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''Mock AMC 8'''<br /> | PersonPsychopath<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c3h1152332p5457736 Initial Discussion]<br /> | [http://artofproblemsolving.com/downloads/printable_post_collections/163667 Problems]<br /> | [http://artofproblemsolving.com/community/category-admin/165328 Forum]<br /> | [http://artofproblemsolving.com/community/c165328h1155882p5484131 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''Mock AMC 8 #2'''<br /> | PersonPsychopath<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c3h1167425p5586244 Initial Discussion]<br /> | [http://artofproblemsolving.com/downloads/printable_post_collections/165838 Problems]<br /> | [http://artofproblemsolving.com/community/c147536_the_mock_amc_8_forum Forum]<br /> | [http://artofproblemsolving.com/community/c3h1167425p5586244 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''Mock AMC 8 2015'''<br /> | Alberty44<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c194276_discusssion Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c5h1168822p5674003 Problems]<br /> | n/a<br /> | [http://artofproblemsolving.com/community/c194276h1175928p5673996 Results/Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''Christmas AMC 8'''<br /> | Mudkipswims42<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c5h1177489p5688588 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c5h1177489p5688588 Problems]<br /> | [http://www.artofproblemsolving.com/community/c229455_christmas_amc8_discussion_forum Forum]<br /> | [http://artofproblemsolving.com/community/c5h1177489p5880200 Results/Discussion]<br /> |- <br /> <br /> ! scope = &quot;row&quot; | '''Mock AMC 8!'''<br /> | eisirrational, pretzel, AOPS12142015, Th3Numb3rThr33, e_power_pi_times_i<br /> | 2017<br /> | [https://artofproblemsolving.com/community/c5h1526187_mock_amc_8 Initial Discussion]<br /> | [http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi84LzFkNzUyMWYwZTViOTQ1MzRmZmU3ODE0NmI2MzIzMGUxNjA2MTcwLnBkZg==&amp;rn=TW9jayBBTUMgOCB2Ny5wZGY= Problems]<br /> | [http://artofproblemsolving.com/community/c3h1545004p9368333 Answer Key/Solutions]<br /> | [https://artofproblemsolving.com/community/c561318_mock_amc_8_2017_discussion Discussion]<br /> |-<br /> <br /> ! scope = &quot;row&quot; | '''Mock Summ(er)ation AMC 8!'''<br /> | mathchamp1, kevinmathz, Gali, mathdragon2000, reddragon644, ShreyJ, Quadrastic, Mathnerd1223334444<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c594864h1672811_2018_summeration_mock_amc_8 Initial Discussion]<br /> | [https://math-adventures.weebly.com/uploads/1/1/8/8/118819793/amc_8_problems.pdf Problems]<br /> | n/a<br /> | n/a<br /> |-<br /> <br /> ! scope = &quot;row&quot; | '''popcorn1's AMC 8 2018'''<br /> | popcorn1<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c594864h1715072_popcorn1s_amc_8_2018 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c594864h1715072p11074872 Problems]<br /> | Answer Key / Solutions - not released<br /> | Discussion - not released<br /> |-<br /> <br /> ! scope = &quot;row&quot; | ''' Stormersyle's Mock AMC 8<br /> | Stormersyle <br /> | 2018<br /> | [http://artofproblemsolving.com/community/c594864h1722790p11146826 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c594864h1722790p11146826 Problems]<br /> | [https://latex.artofproblemsolving.com/miscpdf/qgdafjlp.pdf?t=1544494957792 Answer Key/Solutions]<br /> | Discussion - not released<br /> |-<br /> <br /> ! scope = &quot;row&quot; | '''mathchampion1's Christmas Mock AMC 8'''<br /> | mathchampion1<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c594864h1752494_released_mock_amc_8_christmas_edition Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1752494_released_mock_amc_8_christmas_edition Problems]<br /> | Answer Key / Solutions - not released<br /> | Discussion - not released<br /> |-<br /> <br /> ! scope = &quot;row&quot; | '''June 2019 Mock AMC 8'''<br /> | fidgetboss_4000<br /> | 2019<br /> | [https://artofproblemsolving.com/community/c594864h1853912_june_2019_mock_amc_8 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1853912_june_2019_mock_amc_8 Problems]<br /> | Answer Key / Solutions - not released<br /> | Discussion - not released<br /> |-<br /> <br /> ! scope = &quot;row&quot; | '''popcorn1's AMC 8 2019'''<br /> | popcorn1<br /> | 2019<br /> | [https://artofproblemsolving.com/community/c594864h1896786 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1896786 Problems]<br /> | Answer Key / Solutions - https://artofproblemsolving.com/community/c947561h1950746_answer_keys<br /> | Discussion - https://artofproblemsolving.com/community/c947561_popcorn1s_amc_8_2019_discussion_forum<br /> |-<br /> <br /> ! scope = &quot;row&quot; | '''lethan3's Mock AMC 8'''<br /> | lethan3<br /> | 2020<br /> | [https://artofproblemsolving.com/community/c594864h2216496_lethan3s_mock_amc_8_ended Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h2216496_lethan3s_mock_amc_8_ended Problems]<br /> | Answer Key / Solutions - https://artofproblemsolving.com/community/c594864h2216496_lethan3s_mock_amc_8_ended<br /> | Discussion - https://artofproblemsolving.com/community/c1266375_lethan3s_mock_amc_8_discussion_forum<br /> |-<br /> <br /> }<br /> <br /> == See also ==<br /> * [[American Mathematics Competitions]]<br /> * [[Math books]]<br /> * [[Mathematics competitions]]<br /> * [[Mock AIME]]<br /> * [[Mock MathCounts]]<br /> * [[Mock USAMO]]<br /> * [[Mock USAJMO]]<br /> * [[Resources for mathematics competitions]]<br /> * [[AoPS Past Contests]]</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems&diff=135242 2021 AMC 12B Problems 2020-10-18T19:41:19Z <p>Firebolt360: Created page with &quot;{{AMC12 Problems|year=2021|ab=B}} ==Problem 1== These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021. 2021 AMC 12B Problems/P...&quot;</p> <hr /> <div>{{AMC12 Problems|year=2021|ab=B}}<br /> ==Problem 1==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> These problems will not be posted until the 2021 AMC 12B is released on Thursday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=B|before=[[2021 AMC 12A Problems]]|after=[[2022 AMC 12A Problems]]}}<br /> <br /> [[Category:AMC 12 Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_6&diff=134438 1991 AIME Problems/Problem 6 2020-10-01T15:23:45Z <p>Firebolt360: /* Solution 2 (Faster) */</p> <hr /> <div>== Problem ==<br /> Suppose &lt;math&gt;r^{}_{}&lt;/math&gt; is a [[real number]] for which<br /> &lt;div style=&quot;text-align:center&quot;&gt;&lt;math&gt;<br /> \left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546.<br /> &lt;/math&gt;&lt;/div&gt;<br /> Find &lt;math&gt;\lfloor 100r \rfloor&lt;/math&gt;. (For real &lt;math&gt;x^{}_{}&lt;/math&gt;, &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is the [[floor function|greatest integer]] less than or equal to &lt;math&gt;x^{}_{}&lt;/math&gt;.)<br /> <br /> == Solution ==<br /> There are &lt;math&gt;91 - 19 + 1 = 73&lt;/math&gt; numbers in the [[sequence]]. Since the terms of the sequence can be at most &lt;math&gt;1&lt;/math&gt; apart, all of the numbers in the sequence can take one of two possible values. Since &lt;math&gt;\frac{546}{73} = 7 R 35&lt;/math&gt;, the values of each of the terms of the sequence must be either &lt;math&gt;7&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt;. As the remainder is &lt;math&gt;35&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt; must take on &lt;math&gt;35&lt;/math&gt; of the values, with &lt;math&gt;7&lt;/math&gt; being the value of the remaining &lt;math&gt;73 - 35 = 38&lt;/math&gt; numbers. The 39th number is &lt;math&gt;\lfloor r+\frac{19 + 39 - 1}{100}\rfloor= \lfloor r+\frac{57}{100}\rfloor&lt;/math&gt;, which is also the first term of this sequence with a value of &lt;math&gt;8&lt;/math&gt;, so &lt;math&gt;8 \le r + \frac{57}{100} &lt; 8.01&lt;/math&gt;. Solving shows that &lt;math&gt;\frac{743}{100} \le r &lt; \frac{744}{100}&lt;/math&gt;, so &lt;math&gt;743\le 100r &lt; 744&lt;/math&gt;, and &lt;math&gt;\lfloor 100r \rfloor = \boxed{743}&lt;/math&gt;.<br /> <br /> == Solution 2 (Faster) ==<br /> Recall by Hermite's Identity that &lt;math&gt;\lfloor x\rfloor +\lfloor x+\frac{1}{n}\rfloor +...+\lfloor x+\frac{n-1}{n}\rfloor = \lfloor nx\rfloor&lt;/math&gt; for positive integers &lt;math&gt;n&lt;/math&gt;, and real &lt;math&gt;x&lt;/math&gt;. Similar to above, we quickly observe that the last 35 take the value of 8, and remaining first ones take a value of 7. So, &lt;math&gt;\lfloor r\rfloor \le ...\le \lfloor r+\frac{18}{100}\rfloor \le 7&lt;/math&gt; and &lt;math&gt;\lfloor r+\frac{92}{100}\rfloor \ge ...\ge \lfloor r+1\rfloor \ge 8&lt;/math&gt;. We can see that &lt;math&gt;\lfloor r\rfloor +1=\lfloor r+1\rfloor&lt;/math&gt;. Because &lt;math&gt;\lfloor r\rfloor&lt;/math&gt; is at most 7, and &lt;math&gt;\lfloor r+1\rfloor&lt;/math&gt; is at least 8, we can clearly see their values are &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; respectively. <br /> So, &lt;math&gt;\lfloor r\rfloor = ... = \lfloor r+\frac{18}{100}\rfloor = 7&lt;/math&gt;, and &lt;math&gt;\lfloor r+\frac{92}{100}\rfloor = ...= \lfloor r+1\rfloor = 8&lt;/math&gt;. Since there are 19 terms in the former equation and 8 terms in the latter, our anser is &lt;math&gt;\lfloor nx\rfloor = 546+19\cdot 7+8\cdot 8=\boxed{743}&lt;/math&gt;<br /> <br /> === Note ===<br /> In the contest, you would just observe this mentally, and then calculate &lt;math&gt;546+19\cdot 7+8\cdot 8= 743&lt;/math&gt;, hence the speed at which one can carry out this solution.<br /> <br /> == See also ==<br /> {{AIME box|year=1991|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_6&diff=134437 1991 AIME Problems/Problem 6 2020-10-01T15:22:17Z <p>Firebolt360: /* Solution 2 (Faster) */</p> <hr /> <div>== Problem ==<br /> Suppose &lt;math&gt;r^{}_{}&lt;/math&gt; is a [[real number]] for which<br /> &lt;div style=&quot;text-align:center&quot;&gt;&lt;math&gt;<br /> \left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546.<br /> &lt;/math&gt;&lt;/div&gt;<br /> Find &lt;math&gt;\lfloor 100r \rfloor&lt;/math&gt;. (For real &lt;math&gt;x^{}_{}&lt;/math&gt;, &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is the [[floor function|greatest integer]] less than or equal to &lt;math&gt;x^{}_{}&lt;/math&gt;.)<br /> <br /> == Solution ==<br /> There are &lt;math&gt;91 - 19 + 1 = 73&lt;/math&gt; numbers in the [[sequence]]. Since the terms of the sequence can be at most &lt;math&gt;1&lt;/math&gt; apart, all of the numbers in the sequence can take one of two possible values. Since &lt;math&gt;\frac{546}{73} = 7 R 35&lt;/math&gt;, the values of each of the terms of the sequence must be either &lt;math&gt;7&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt;. As the remainder is &lt;math&gt;35&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt; must take on &lt;math&gt;35&lt;/math&gt; of the values, with &lt;math&gt;7&lt;/math&gt; being the value of the remaining &lt;math&gt;73 - 35 = 38&lt;/math&gt; numbers. The 39th number is &lt;math&gt;\lfloor r+\frac{19 + 39 - 1}{100}\rfloor= \lfloor r+\frac{57}{100}\rfloor&lt;/math&gt;, which is also the first term of this sequence with a value of &lt;math&gt;8&lt;/math&gt;, so &lt;math&gt;8 \le r + \frac{57}{100} &lt; 8.01&lt;/math&gt;. Solving shows that &lt;math&gt;\frac{743}{100} \le r &lt; \frac{744}{100}&lt;/math&gt;, so &lt;math&gt;743\le 100r &lt; 744&lt;/math&gt;, and &lt;math&gt;\lfloor 100r \rfloor = \boxed{743}&lt;/math&gt;.<br /> <br /> == Solution 2 (Faster) ==<br /> Recall by Hermite's Identity that &lt;math&gt;\lfloor x\rfloor +\lfloor x+\frac{1}{n}\rfloor +...+\lfloor x+\frac{n-1}{n}\rfloor = \lfloor nx\rfloor&lt;/math&gt; for positive integers &lt;math&gt;n&lt;/math&gt;, and real &lt;math&gt;x&lt;/math&gt;. Similar to above, we quickly observe that the last 35 take the value of 8, and remaining first ones take a value of 7. So, &lt;math&gt;\lfloor r\rfloor \le ...\le \lfloor r+\frac{18}{100}\rfloor \le 7&lt;/math&gt; and &lt;math&gt;\lfloor r+\frac{92}{100}\rfloor \ge ...\ge \lfloor r+1\rfloor \ge 8&lt;/math&gt;. We can see that &lt;math&gt;\lfloor r\rfloor +1=\lfloor r+1\rfloor&lt;/math&gt;. Because &lt;math&gt;\lfloor r\rfloor&lt;/math&gt; is at most 7, and &lt;math&gt;\lfloor r+1\rfloor&lt;/math&gt; is at least 8, we can clearly see their values are &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; respectively. <br /> So, &lt;math&gt;\lfloor r\rfloor = ... = \lfloor r+\frac{18}{100}\rfloor = 7&lt;/math&gt;, and &lt;math&gt;\lfloor r+\frac{92}{100}\rfloor = ...= \lfloor r+1\rfloor = 8&lt;/math&gt;. Since there are 19 terms in the former equation and 8 terms in the latter, our anser is &lt;math&gt;\lfloor nx\rfloor = 546+19\cdot 7+8\cdot 8=\boxed{743}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1991|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_6&diff=134436 1991 AIME Problems/Problem 6 2020-10-01T15:20:15Z <p>Firebolt360: /* Solution 2 (Faster) */</p> <hr /> <div>== Problem ==<br /> Suppose &lt;math&gt;r^{}_{}&lt;/math&gt; is a [[real number]] for which<br /> &lt;div style=&quot;text-align:center&quot;&gt;&lt;math&gt;<br /> \left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546.<br /> &lt;/math&gt;&lt;/div&gt;<br /> Find &lt;math&gt;\lfloor 100r \rfloor&lt;/math&gt;. (For real &lt;math&gt;x^{}_{}&lt;/math&gt;, &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is the [[floor function|greatest integer]] less than or equal to &lt;math&gt;x^{}_{}&lt;/math&gt;.)<br /> <br /> == Solution ==<br /> There are &lt;math&gt;91 - 19 + 1 = 73&lt;/math&gt; numbers in the [[sequence]]. Since the terms of the sequence can be at most &lt;math&gt;1&lt;/math&gt; apart, all of the numbers in the sequence can take one of two possible values. Since &lt;math&gt;\frac{546}{73} = 7 R 35&lt;/math&gt;, the values of each of the terms of the sequence must be either &lt;math&gt;7&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt;. As the remainder is &lt;math&gt;35&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt; must take on &lt;math&gt;35&lt;/math&gt; of the values, with &lt;math&gt;7&lt;/math&gt; being the value of the remaining &lt;math&gt;73 - 35 = 38&lt;/math&gt; numbers. The 39th number is &lt;math&gt;\lfloor r+\frac{19 + 39 - 1}{100}\rfloor= \lfloor r+\frac{57}{100}\rfloor&lt;/math&gt;, which is also the first term of this sequence with a value of &lt;math&gt;8&lt;/math&gt;, so &lt;math&gt;8 \le r + \frac{57}{100} &lt; 8.01&lt;/math&gt;. Solving shows that &lt;math&gt;\frac{743}{100} \le r &lt; \frac{744}{100}&lt;/math&gt;, so &lt;math&gt;743\le 100r &lt; 744&lt;/math&gt;, and &lt;math&gt;\lfloor 100r \rfloor = \boxed{743}&lt;/math&gt;.<br /> <br /> == Solution 2 (Faster) ==<br /> Recall by Hermite's Identity that &lt;math&gt;\lfloor x\rfloor +\lfloor x+\frac{1}{n}\rfloor +...+\lfloor x+\frac{n-1}{n}\rfloor = \lfloor nx\rfloor&lt;/math&gt; for positive integers &lt;math&gt;n&lt;/math&gt;, and real &lt;math&gt;x&lt;/math&gt;. Similar to above, we quickly observe that the last 35 take the value of 8, and remaining first ones take a value of 7. So, &lt;math&gt;\lfloor r\rfloor \le ...\le \lfloor r+\frac{18}{100}\rfloor \le 7&lt;/math&gt; and &lt;math&gt;\lfloor r+\frac{92}{100}\rfloor \ge ...\ge \lfloor r+1\rfloor \ge 8&lt;/math&gt;. We can see that &lt;math&gt;\lfloor r\rfloor +1=\lfloor r+1\rfloor&lt;/math&gt;. Because &lt;math&gt;\lfloor r\rfloor&lt;/math&gt; is at most 7, and &lt;math&gt;\lfloor r+1\rfloor&lt;/math&gt; is at least 8, we can clearly see their values are &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; respectively. So, &lt;math&gt;\lfloor nx\rfloor&lt;/math&gt; is &lt;math&gt;546+19\cdot 7+8\cdot 8=\boxed{743}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1991|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_6&diff=134435 1991 AIME Problems/Problem 6 2020-10-01T15:18:47Z <p>Firebolt360: /* Solution 2 (Faster) */</p> <hr /> <div>== Problem ==<br /> Suppose &lt;math&gt;r^{}_{}&lt;/math&gt; is a [[real number]] for which<br /> &lt;div style=&quot;text-align:center&quot;&gt;&lt;math&gt;<br /> \left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546.<br /> &lt;/math&gt;&lt;/div&gt;<br /> Find &lt;math&gt;\lfloor 100r \rfloor&lt;/math&gt;. (For real &lt;math&gt;x^{}_{}&lt;/math&gt;, &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is the [[floor function|greatest integer]] less than or equal to &lt;math&gt;x^{}_{}&lt;/math&gt;.)<br /> <br /> == Solution ==<br /> There are &lt;math&gt;91 - 19 + 1 = 73&lt;/math&gt; numbers in the [[sequence]]. Since the terms of the sequence can be at most &lt;math&gt;1&lt;/math&gt; apart, all of the numbers in the sequence can take one of two possible values. Since &lt;math&gt;\frac{546}{73} = 7 R 35&lt;/math&gt;, the values of each of the terms of the sequence must be either &lt;math&gt;7&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt;. As the remainder is &lt;math&gt;35&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt; must take on &lt;math&gt;35&lt;/math&gt; of the values, with &lt;math&gt;7&lt;/math&gt; being the value of the remaining &lt;math&gt;73 - 35 = 38&lt;/math&gt; numbers. The 39th number is &lt;math&gt;\lfloor r+\frac{19 + 39 - 1}{100}\rfloor= \lfloor r+\frac{57}{100}\rfloor&lt;/math&gt;, which is also the first term of this sequence with a value of &lt;math&gt;8&lt;/math&gt;, so &lt;math&gt;8 \le r + \frac{57}{100} &lt; 8.01&lt;/math&gt;. Solving shows that &lt;math&gt;\frac{743}{100} \le r &lt; \frac{744}{100}&lt;/math&gt;, so &lt;math&gt;743\le 100r &lt; 744&lt;/math&gt;, and &lt;math&gt;\lfloor 100r \rfloor = \boxed{743}&lt;/math&gt;.<br /> <br /> == Solution 2 (Faster) ==<br /> Recall by Hermite's Identity that &lt;math&gt;\lfloor x\rfloor +\lfloor x+\frac{1}{n}\rfloor +...+\lfloor x+\frac{n-1}{n}\rfloor = \lfloor nx\rfloor&lt;/math&gt; for positive integers &lt;math&gt;n&lt;/math&gt;, and real &lt;math&gt;x&lt;/math&gt;. Similar to above, we quickly observe that the last 35 take the value of 8, and remaining first ones take a value of 7. So, &lt;math&gt;\lfloor r\rfloor,...,\lfloor r+\frac{18}{100}\rfloor \le 7&lt;/math&gt; and &lt;math&gt;\lfloor r+\frac{92}{100}\rfloor,...,\lfloor r+1\rfloor \ge 8&lt;/math&gt;. We can see that &lt;math&gt;\lfloor r\rfloor +1=\lfloor r+1\rfloor&lt;/math&gt;. Because &lt;math&gt;\lfloor r\rfloor&lt;/math&gt; is at most 7, and &lt;math&gt;\lfloor r+1\rfloor&lt;/math&gt; is at least 8, we can clearly see their values are &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; respectively. So, &lt;math&gt;\lfloor nx\rfloor&lt;/math&gt; is &lt;math&gt;546+19\cdot 7+8\cdot 8=\boxed{743}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1991|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_6&diff=134434 1991 AIME Problems/Problem 6 2020-10-01T15:18:31Z <p>Firebolt360: /* Solution 2 (Faster) */</p> <hr /> <div>== Problem ==<br /> Suppose &lt;math&gt;r^{}_{}&lt;/math&gt; is a [[real number]] for which<br /> &lt;div style=&quot;text-align:center&quot;&gt;&lt;math&gt;<br /> \left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546.<br /> &lt;/math&gt;&lt;/div&gt;<br /> Find &lt;math&gt;\lfloor 100r \rfloor&lt;/math&gt;. (For real &lt;math&gt;x^{}_{}&lt;/math&gt;, &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is the [[floor function|greatest integer]] less than or equal to &lt;math&gt;x^{}_{}&lt;/math&gt;.)<br /> <br /> == Solution ==<br /> There are &lt;math&gt;91 - 19 + 1 = 73&lt;/math&gt; numbers in the [[sequence]]. Since the terms of the sequence can be at most &lt;math&gt;1&lt;/math&gt; apart, all of the numbers in the sequence can take one of two possible values. Since &lt;math&gt;\frac{546}{73} = 7 R 35&lt;/math&gt;, the values of each of the terms of the sequence must be either &lt;math&gt;7&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt;. As the remainder is &lt;math&gt;35&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt; must take on &lt;math&gt;35&lt;/math&gt; of the values, with &lt;math&gt;7&lt;/math&gt; being the value of the remaining &lt;math&gt;73 - 35 = 38&lt;/math&gt; numbers. The 39th number is &lt;math&gt;\lfloor r+\frac{19 + 39 - 1}{100}\rfloor= \lfloor r+\frac{57}{100}\rfloor&lt;/math&gt;, which is also the first term of this sequence with a value of &lt;math&gt;8&lt;/math&gt;, so &lt;math&gt;8 \le r + \frac{57}{100} &lt; 8.01&lt;/math&gt;. Solving shows that &lt;math&gt;\frac{743}{100} \le r &lt; \frac{744}{100}&lt;/math&gt;, so &lt;math&gt;743\le 100r &lt; 744&lt;/math&gt;, and &lt;math&gt;\lfloor 100r \rfloor = \boxed{743}&lt;/math&gt;.<br /> <br /> == Solution 2 (Faster) ==<br /> Recall by Hermite's Identity that &lt;math&gt;\lfloor x\rfloor +\lfloor x+\frac{1}{n}\rfloor +...+\lfloor x+\frac{n-1}{n}\rfloor = \lfloor nx\rfloor&lt;/math&gt; for positive integers &lt;math&gt;n&lt;/math&gt;, and real &lt;math&gt;x&lt;/math&gt;. Similar to above, we quickly observe that the last 35 take the value of 8, and remaining first ones take a value of 7. So, &lt;math&gt;\lfloor r\rfloor,...,\lfloor r+\frac{18}{100}\rfloor \le 7&lt;/math&gt; and &lt;math&gt;\lfloor r+\frac{92}{100}\rfloor,...,\lfloor r+1\rfloor \ge 8&lt;/math&gt;. We can see that &lt;math&gt;\lfloor r\rfloor +1=\lfloor r+1\rfloor&lt;/math&gt;. Because &lt;math&gt;\lfloor r\rfloor&lt;/math&gt; is at most 7, and &lt;math&gt;\lfloor r+1\rfloor&lt;/math&gt; is at least 8, we can clearly see their values are &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; respectively. So, &lt;math&gt;lfloor nx\rfloor&lt;/math&gt; is &lt;math&gt;546+19\cdot 7+8\cdot 8=\boxed{743}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1991|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_6&diff=134433 1991 AIME Problems/Problem 6 2020-10-01T15:17:56Z <p>Firebolt360: /* Solution 2 (Faster) */</p> <hr /> <div>== Problem ==<br /> Suppose &lt;math&gt;r^{}_{}&lt;/math&gt; is a [[real number]] for which<br /> &lt;div style=&quot;text-align:center&quot;&gt;&lt;math&gt;<br /> \left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546.<br /> &lt;/math&gt;&lt;/div&gt;<br /> Find &lt;math&gt;\lfloor 100r \rfloor&lt;/math&gt;. (For real &lt;math&gt;x^{}_{}&lt;/math&gt;, &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is the [[floor function|greatest integer]] less than or equal to &lt;math&gt;x^{}_{}&lt;/math&gt;.)<br /> <br /> == Solution ==<br /> There are &lt;math&gt;91 - 19 + 1 = 73&lt;/math&gt; numbers in the [[sequence]]. Since the terms of the sequence can be at most &lt;math&gt;1&lt;/math&gt; apart, all of the numbers in the sequence can take one of two possible values. Since &lt;math&gt;\frac{546}{73} = 7 R 35&lt;/math&gt;, the values of each of the terms of the sequence must be either &lt;math&gt;7&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt;. As the remainder is &lt;math&gt;35&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt; must take on &lt;math&gt;35&lt;/math&gt; of the values, with &lt;math&gt;7&lt;/math&gt; being the value of the remaining &lt;math&gt;73 - 35 = 38&lt;/math&gt; numbers. The 39th number is &lt;math&gt;\lfloor r+\frac{19 + 39 - 1}{100}\rfloor= \lfloor r+\frac{57}{100}\rfloor&lt;/math&gt;, which is also the first term of this sequence with a value of &lt;math&gt;8&lt;/math&gt;, so &lt;math&gt;8 \le r + \frac{57}{100} &lt; 8.01&lt;/math&gt;. Solving shows that &lt;math&gt;\frac{743}{100} \le r &lt; \frac{744}{100}&lt;/math&gt;, so &lt;math&gt;743\le 100r &lt; 744&lt;/math&gt;, and &lt;math&gt;\lfloor 100r \rfloor = \boxed{743}&lt;/math&gt;.<br /> <br /> == Solution 2 (Faster) ==<br /> Recall by Hermite's Identity that &lt;math&gt;\lfloor x\rfloor +\lfloor x+\frac{1}{n}\rfloor +...+\lfloor x+\frac{n-1}{n}\rfloor = \lfloor nx\rfloor&lt;/math&gt; for positive integers &lt;math&gt;n&lt;/math&gt;, and real &lt;math&gt;x&lt;/math&gt;. Similar to above, we quickly observe that the last 35 take the value of 8, and remaining first ones take a value of 7. So, &lt;math&gt;\lfloor r\rfloor,...,\lfloor r+\frac{18}{100}\rfloor \le 7&lt;/math&gt; and &lt;math&gt;\lfloor r+\frac{92}{100},...,\lfloor r+1\rfloor \ge 8&lt;/math&gt;. We can see that &lt;math&gt;\lfloor r\rfloor +1=\lfloor r+1\rfloor&lt;/math&gt;. Because &lt;math&gt;\lfloor r\rfloor&lt;/math&gt; is at most 7, and &lt;math&gt;\lfloor r+1\rfloor&lt;/math&gt; is at least 8, we can clearly see their values are &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; respectively. So, &lt;math&gt;lfloor nx\rfloor&lt;/math&gt; is &lt;math&gt;546+19\cdot 7+8\cdot 8=\boxed{743}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1991|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_6&diff=134432 1991 AIME Problems/Problem 6 2020-10-01T15:17:21Z <p>Firebolt360: /* Solution */</p> <hr /> <div>== Problem ==<br /> Suppose &lt;math&gt;r^{}_{}&lt;/math&gt; is a [[real number]] for which<br /> &lt;div style=&quot;text-align:center&quot;&gt;&lt;math&gt;<br /> \left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546.<br /> &lt;/math&gt;&lt;/div&gt;<br /> Find &lt;math&gt;\lfloor 100r \rfloor&lt;/math&gt;. (For real &lt;math&gt;x^{}_{}&lt;/math&gt;, &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is the [[floor function|greatest integer]] less than or equal to &lt;math&gt;x^{}_{}&lt;/math&gt;.)<br /> <br /> == Solution ==<br /> There are &lt;math&gt;91 - 19 + 1 = 73&lt;/math&gt; numbers in the [[sequence]]. Since the terms of the sequence can be at most &lt;math&gt;1&lt;/math&gt; apart, all of the numbers in the sequence can take one of two possible values. Since &lt;math&gt;\frac{546}{73} = 7 R 35&lt;/math&gt;, the values of each of the terms of the sequence must be either &lt;math&gt;7&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt;. As the remainder is &lt;math&gt;35&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt; must take on &lt;math&gt;35&lt;/math&gt; of the values, with &lt;math&gt;7&lt;/math&gt; being the value of the remaining &lt;math&gt;73 - 35 = 38&lt;/math&gt; numbers. The 39th number is &lt;math&gt;\lfloor r+\frac{19 + 39 - 1}{100}\rfloor= \lfloor r+\frac{57}{100}\rfloor&lt;/math&gt;, which is also the first term of this sequence with a value of &lt;math&gt;8&lt;/math&gt;, so &lt;math&gt;8 \le r + \frac{57}{100} &lt; 8.01&lt;/math&gt;. Solving shows that &lt;math&gt;\frac{743}{100} \le r &lt; \frac{744}{100}&lt;/math&gt;, so &lt;math&gt;743\le 100r &lt; 744&lt;/math&gt;, and &lt;math&gt;\lfloor 100r \rfloor = \boxed{743}&lt;/math&gt;.<br /> <br /> == Solution 2 (Faster) ==<br /> Recall by Hermite's Identity that &lt;math&gt;\lfloor x\rfloor +\lfloor x+\frac{1}{n}\rfloor +...+\lfloor x+\frac{n-1}{n}\rfloor = \lfloor nx\rfloor&lt;/math&gt; for positive integers &lt;math&gt;n&lt;/math&gt;, and real &lt;math&gt;x&lt;/math&gt;. Similar to above, we quickly observe that the last 35 take the value of 8, and remaining first ones take a value of 7. So, &lt;math&gt;\lfloor r\rfloor,...,\lfloor r+\frac{18}{100} = 7&lt;/math&gt; and &lt;math&gt;\lfloor r+\frac{92}{100},...,\lfloor r+1\rfloor = 8&lt;/math&gt;. We can see that &lt;math&gt;\lfloor r\rfloor +1=\lfloor r+1\rfloor&lt;/math&gt;. Because &lt;math&gt;\lfloor r\rfloor&lt;/math&gt; is at most 7, and &lt;math&gt;\lfloor r+1\rfloor&lt;/math&gt; is at least 8, we can clearly see their values are &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; respectively. So, &lt;math&gt;lfloor nx\rfloor&lt;/math&gt; is &lt;math&gt;546+19\cdot 7+8\cdot 8=\boxed{743}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1991|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems_Raw&diff=132393 2010 AMC 12A Problems Raw 2020-08-23T16:24:23Z <p>Firebolt360: Blanked the page</p> <hr /> <div></div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems_Raw&diff=132392 2010 AMC 12A Problems Raw 2020-08-23T16:22:49Z <p>Firebolt360: Created page with &quot;== Problem 1 == What is &lt;math&gt;\left(20-\left(2010-201\right)\right)+\left(2010-\left(201-20\right)\right)&lt;/math&gt;? &lt;math&gt;\textbf{(A)}\ -4020 \qquad \textbf{(B)}\ 0 \qquad \tex...&quot;</p> <hr /> <div>== Problem 1 ==<br /> What is &lt;math&gt;\left(20-\left(2010-201\right)\right)+\left(2010-\left(201-20\right)\right)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -4020 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 401 \qquad \textbf{(E)}\ 4020&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> A ferry boat shuttles tourists to an island every hour starting at 10 AM until its last trip, which starts at 3 PM. One day the boat captain notes that on the 10 AM trip there were 100 tourists on the ferry boat, and that on each successive trip, the number of tourists was 1 fewer than on the previous trip. How many tourists did the ferry take to the island that day?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 585 \qquad \textbf{(B)}\ 594 \qquad \textbf{(C)}\ 672 \qquad \textbf{(D)}\ 679 \qquad \textbf{(E)}\ 694&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt;, pictured below, shares &lt;math&gt;50\%&lt;/math&gt; of its area with square &lt;math&gt;EFGH&lt;/math&gt;. Square &lt;math&gt;EFGH&lt;/math&gt; shares &lt;math&gt;20\%&lt;/math&gt; of its area with rectangle &lt;math&gt;ABCD&lt;/math&gt;. What is &lt;math&gt;\frac{AB}{AD}&lt;/math&gt;?<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(1mm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> <br /> draw((0,0)--(0,25)--(25,25)--(25,0)--cycle);<br /> fill((0,20)--(0,15)--(25,15)--(25,20)--cycle,gray);<br /> draw((0,15)--(0,20)--(25,20)--(25,15)--cycle);<br /> draw((25,15)--(25,20)--(50,20)--(50,15)--cycle);<br /> <br /> label(&quot;A$&quot;,(0,20),W);<br /> label(&quot;$B$&quot;,(50,20),E);<br /> label(&quot;$C$&quot;,(50,15),E);<br /> label(&quot;$D$&quot;,(0,15),W);<br /> label(&quot;$E$&quot;,(0,25),NW);<br /> label(&quot;$F$&quot;,(25,25),NE);<br /> label(&quot;$G$&quot;,(25,0),SE);<br /> label(&quot;$H&quot;,(0,0),SW);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> If &lt;math&gt;x&lt;0&lt;/math&gt;, then which of the following must be positive?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{x}{\left|x\right|} \qquad \textbf{(B)}\ -x^2 \qquad \textbf{(C)}\ -2^x \qquad \textbf{(D)}\ -x^{-1} \qquad \textbf{(E)}\ \sqrt{x}&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> Halfway through a 100-shot archery tournament, Chelsea leads by 50 points. For each shot a bullseye scores 10 points, with other possible scores being 8, 4, 2, and 0 points. Chelsea always scores at least 4 points on each shot. If Chelsea's next &lt;math&gt;n&lt;/math&gt; shots are bullseyes she will be guaranteed victory. What is the minimum value for &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 38 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 42 \qquad \textbf{(D)}\ 44 \qquad \textbf{(E)}\ 46&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> A &lt;math&gt;\textit{palindrome}&lt;/math&gt;, such as 83438, is a number that remains the same when its digits are reversed. The numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;x+32&lt;/math&gt; are three-digit and four-digit palindromes, respectively. What is the sum of the digits of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0.04 \qquad \textbf{(B)}\ \frac{0.4}{\pi} \qquad \textbf{(C)}\ 0.4 \qquad \textbf{(D)}\ \frac{4}{\pi} \qquad \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has &lt;math&gt;AB=2 \cdot AC&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;, respectively, such that &lt;math&gt;\angle BAE = \angle ACD&lt;/math&gt;. Let &lt;math&gt;F&lt;/math&gt; be the intersection of segments &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, and suppose that &lt;math&gt;\triangle CFE&lt;/math&gt; is equilateral. What is &lt;math&gt;\angle ACB&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> A solid cube has side length &lt;math&gt;3&lt;/math&gt; inches. A &lt;math&gt;2&lt;/math&gt;-inch by &lt;math&gt;2&lt;/math&gt;-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The first four terms of an arithmetic sequence are &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;3p-q&lt;/math&gt;, and &lt;math&gt;3p+q&lt;/math&gt;. What is the &lt;math&gt;2010^\text{th}&lt;/math&gt; term of this sequence?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 8041 \qquad \textbf{(B)}\ 8043 \qquad \textbf{(C)}\ 8045 \qquad \textbf{(D)}\ 8047 \qquad \textbf{(E)}\ 8049&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> The solution of the equation &lt;math&gt;7^{x+7} = 8^x&lt;/math&gt; can be expressed in the form &lt;math&gt;x = \log_b 7^7&lt;/math&gt;. What is &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{7}{15} \qquad \textbf{(B)}\ \frac{7}{8} \qquad \textbf{(C)}\ \frac{8}{7} \qquad \textbf{(D)}\ \frac{15}{8} \qquad \textbf{(E)}\ \frac{15}{7}&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.<br /> <br /> Brian: &quot;Mike and I are different species.&quot;<br /> <br /> Chris: &quot;LeRoy is a frog.&quot;<br /> <br /> LeRoy: &quot;Chris is a frog.&quot;<br /> <br /> Mike: &quot;Of the four of us, at least two are toads.&quot;<br /> <br /> How many of these amphibians are frogs?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> For how many integer values of &lt;math&gt;k&lt;/math&gt; do the graphs of &lt;math&gt;x^2+y^2=k^2&lt;/math&gt; and &lt;math&gt;xy = k&lt;/math&gt; not intersect?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Nondegenerate &lt;math&gt;\triangle ABC&lt;/math&gt; has integer side lengths, &lt;math&gt;\overline{BD}&lt;/math&gt; is an angle bisector, &lt;math&gt;AD = 3&lt;/math&gt;, and &lt;math&gt;DC=8&lt;/math&gt;. What is the smallest possible value of the perimeter?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> A coin is altered so that the probability that it lands on heads is less than &lt;math&gt;\frac{1}{2}&lt;/math&gt; and when the coin is flipped four times, the probability of an equal number of heads and tails is &lt;math&gt;\frac{1}{6}&lt;/math&gt;. What is the probability that the coin lands on heads?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{\sqrt{15}-3}{6} \qquad \textbf{(B)}\ \frac{6-\sqrt{6\sqrt{6}+2}}{12} \qquad \textbf{(C)}\ \frac{\sqrt{2}-1}{2} \qquad \textbf{(D)}\ \frac{3-\sqrt{3}}{6} \qquad \textbf{(E)}\ \frac{\sqrt{3}-1}{2}&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> Bernardo randomly picks 3 distinct numbers from the set &lt;math&gt;\{1,2,3,...,7,8,9\}&lt;/math&gt; and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set &lt;math&gt;\{1,2,3,...,6,7,8\}&lt;/math&gt; and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{47}{72} \qquad \textbf{(B)}\ \frac{37}{56} \qquad \textbf{(C)}\ \frac{2}{3} \qquad \textbf{(D)}\ \frac{49}{72} \qquad \textbf{(E)}\ \frac{39}{56}&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> Equiangular hexagon &lt;math&gt;ABCDEF&lt;/math&gt; has side lengths &lt;math&gt;AB=CD=EF=1&lt;/math&gt; and &lt;math&gt;BC=DE=FA=r&lt;/math&gt;. The area of &lt;math&gt;\triangle ACE&lt;/math&gt; is &lt;math&gt;70\%&lt;/math&gt; of the area of the hexagon. What is the sum of all possible values of &lt;math&gt;r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> A 16-step path is to go from &lt;math&gt;(-4,-4)&lt;/math&gt; to &lt;math&gt;(4,4)&lt;/math&gt; with each step increasing either the &lt;math&gt;x&lt;/math&gt;-coordinate or the &lt;math&gt;y&lt;/math&gt;-coordinate by 1. How many such paths stay outside or on the boundary of the square &lt;math&gt;-2 \le x \le 2&lt;/math&gt;, &lt;math&gt;-2 \le y \le 2&lt;/math&gt; at each step?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> Each of 2010 boxes in a line contains a single red marble, and for &lt;math&gt;1 \le k \le 2010&lt;/math&gt;, the box in the &lt;math&gt;k\text{th}&lt;/math&gt; position also contains &lt;math&gt;k&lt;/math&gt; white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let &lt;math&gt;P(n)&lt;/math&gt; be the probability that Isabella stops after drawing exactly &lt;math&gt;n&lt;/math&gt; marbles. What is the smallest value of &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;P(n) &lt; \frac{1}{2010}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> Arithmetic sequences &lt;math&gt;\left(a_n\right)&lt;/math&gt; and &lt;math&gt;\left(b_n\right)&lt;/math&gt; have integer terms with &lt;math&gt;a_1=b_1=1&lt;a_2 \le b_2&lt;/math&gt; and &lt;math&gt;a_n b_n = 2010&lt;/math&gt; for some &lt;math&gt;n&lt;/math&gt;. What is the largest possible value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 288 \qquad \textbf{(E)}\ 2009&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> The graph of &lt;math&gt;y=x^6-10x^5+29x^4-4x^3+ax^2&lt;/math&gt; lies above the line &lt;math&gt;y=bx+c&lt;/math&gt; except at three values of &lt;math&gt;x&lt;/math&gt;, where the graph and the line intersect. What is the largest of these values?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> What is the minimum value of &lt;math&gt;\left|x-1\right| + \left|2x-1\right| + \left|3x-1\right| + \cdots + \left|119x - 1 \right|&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 53&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> The number obtained from the last two nonzero digits of &lt;math&gt;90!&lt;/math&gt; is equal to &lt;math&gt;n&lt;/math&gt;. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> Let &lt;math&gt;f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)&lt;/math&gt;. The intersection of the domain of &lt;math&gt;f(x)&lt;/math&gt; with the interval &lt;math&gt;[0,1]&lt;/math&gt; is a union of &lt;math&gt;n&lt;/math&gt; disjoint open intervals. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad \textbf{(E)}\ 2255&lt;/math&gt;<br /> <br /> [[2010 AMC 12A Problems/Problem 25|Solution]]</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2008_JBMO_Problems&diff=130716 2008 JBMO Problems 2020-08-06T00:58:37Z <p>Firebolt360: /* Problem 3 */</p> <hr /> <div>==Problem 1==<br /> ==Problem 2==<br /> ==Problem 3==<br /> <br /> Find all prime numbers &lt;math&gt; p,q,r&lt;/math&gt;, such that &lt;math&gt; \frac{p}{q}-\frac{4}{r+1}=1&lt;/math&gt;<br /> ==Problem 4==</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2008_JBMO_Problems&diff=130715 2008 JBMO Problems 2020-08-06T00:57:33Z <p>Firebolt360: Created page with &quot;==Problem 3== Find all prime numbers &lt;math&gt; p,q,r&lt;/math&gt;, such that &lt;math&gt; \frac{p}{q}-\frac{4}{r+1}=1&lt;/math&gt;&quot;</p> <hr /> <div>==Problem 3==<br /> <br /> Find all prime numbers &lt;math&gt; p,q,r&lt;/math&gt;, such that &lt;math&gt; \frac{p}{q}-\frac{4}{r+1}=1&lt;/math&gt;</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=Competition_Ratings&diff=130714 Competition Ratings 2020-08-06T00:55:04Z <p>Firebolt360: Created page with &quot;This page contains an approximate estimation of the difficulty level of various competitions. It is designed with the intention of introdu...&quot;</p> <hr /> <div>This page contains an approximate estimation of the difficulty level of various [[List of mathematics competitions|competitions]]. It is designed with the intention of introducing contests of similar difficulty levels (but possibly different styles of problems) that readers may like to try to gain more experience.<br /> <br /> Each entry groups the problems into sets of similar difficulty levels and suggests an approximate difficulty rating, on a scale from 1 to 10 (from easiest to hardest). Note that many of these ratings are not directly comparable, because the actual competitions have many different rules; the ratings are generally synchronized with the amount of available time, etc. Also, due to variances within a contest, ranges shown may overlap. A sample problem is provided with each entry, with a link to a solution. <br /> <br /> As you may have guessed with time many competitions got more challenging because many countries got more access to books targeted at olympiad preparation. But especially web site where one can discuss Olympiads such as our very own AoPS!<br /> <br /> If you have some experience with mathematical competitions, we hope that you can help us make the difficulty rankings more accurate. Currently, the system is on a scale from 1 to 10 where 1 is the easiest level, e.g. [http://www.mathlinks.ro/resources.php?c=182&amp;cid=44 early AMC problems] and 10 is hardest level, e.g. [http://www.mathlinks.ro/resources.php?c=37&amp;cid=47 China IMO Team Selection Test.] When considering problem difficulty '''put more emphasis on problem-solving aspects and less so on technical skill requirements'''.<br /> <br /> = Scale =<br /> All levels are estimated and refer to ''averages''. The following is a rough standard based on the USA tier system AMC 8 – AMC 10 – AMC 12 – AIME – USAMO/USAJMO, representing Middle School – Junior High – High School – Challenging High School – Olympiad levels. Other contests can be interpolated against this. <br /> # Problems strictly for beginner, on the easiest elementary school or middle school levels (MOEMS, easy Mathcounts questions, #1-20 on AMC 8s, #1-10 AMC 10s, and others that involve standard techniques introduced up to the middle school level), most traditional middle/high school word problems<br /> # For motivated beginners, harder questions from the previous categories (#21-25 on AMC 8, Challenging Mathcounts questions, #11-20 on AMC 10, #5-10 on AMC 12, the easiest AIME questions, etc), traditional middle/high school word problems with extremely complex problem solving<br /> # Beginner/novice problems that require more creative thinking (MathCounts National, #21-25 on AMC 10, #11-20ish on AMC 12, easier #1-5 on AIMEs, etc.)<br /> # Intermediate-leveled problems, the most difficult questions on AMC 12s (#21-25s), more difficult AIME-styled questions such as #6-9.<br /> # More difficult AIME problems (#10-12), simple proof-based problems (JBMO), etc<br /> # High-leveled AIME-styled questions (#13-15). Introductory-leveled Olympiad-level questions (#1,4s).<br /> # Tougher Olympiad-level questions, #1,4s that require more technical knowledge than new students to Olympiad-type questions have, easier #2,5s, etc.<br /> # High-level Olympiad-level questions, eg #2,5s on difficult Olympiad contest and easier #3,6s, etc.<br /> # Expert Olympiad-level questions, eg #3,6s on difficult Olympiad contests.<br /> # Super Expert problems, problems occasionally even unsuitable for very hard competitions (like the IMO) due to being exceedingly tedious/long/difficult (e.g. very few students are capable of solving, even on a worldwide basis).<br /> <br /> = Competitions =<br /> <br /> ==Introductory Competitions==<br /> Most middle school and first-stage high school competitions would fall under this category. Problems in these competitions are usually ranked from 1 to 3. A full list is available [https://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AIntroductory+mathematics+competitions here].<br /> <br /> === [[MOEMS]] ===<br /> *Division E: '''1'''<br /> *: ''The whole number &lt;math&gt;N&lt;/math&gt; is divisible by &lt;math&gt;7&lt;/math&gt;. &lt;math&gt;N&lt;/math&gt; leaves a remainder of &lt;math&gt;1&lt;/math&gt; when divided by &lt;math&gt;2,3,4,&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt;. What is the smallest value that &lt;math&gt;N&lt;/math&gt; can be?'' ([http://www.moems.org/sample_files/SampleE.pdf Solution])<br /> *Division M: '''1'''<br /> *: ''The value of a two-digit number is &lt;math&gt;10&lt;/math&gt; times more than the sum of its digits. The units digit is 1 more than twice the tens digit. Find the two-digit number.'' ([http://www.moems.org/sample_files/SampleM.pdf Solution])<br /> <br /> === [[AMC 8]] ===<br /> <br /> * Problem 1 - Problem 12: '''1''' <br /> *: ''The &lt;math&gt;\emph{harmonic mean}&lt;/math&gt; of a set of non-zero numbers is the reciprocal of the average of the reciprocals of the numbers. What is the harmonic mean of 1, 2, and 4?'' ([[2018 AMC 8 Problems/Problem 10|Solution]])<br /> * Problem 13 - Problem 25: '''1.5-2'''<br /> *: ''How many positive factors does &lt;math&gt;23,232&lt;/math&gt; have?'' ([[2018 AMC 8 Problems/Problem 18|Solution]])<br /> <br /> === [[Mathcounts]] ===<br /> <br /> * Countdown: '''1-2.'''<br /> * Sprint: '''1-1.5''' (school/chapter), '''1.5-2''' (State), '''2-2.5''' (National)<br /> * Target: '''1-2''' (school/chapter), '''1.5-2.5''' (State), '''2-3''' (National)<br /> <br /> === [[AMC 10]] ===<br /> <br /> * Problem 1 - 10: '''1-2'''<br /> *: ''A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?'' ([[2016 AMC 10A Problems/Problem 5|Solution]])<br /> * Problem 11 - 20: '''2-3'''<br /> *: ''For some positive integer &lt;math&gt;k&lt;/math&gt;, the repeating base-&lt;math&gt;k&lt;/math&gt; representation of the (base-ten) fraction &lt;math&gt;\frac{7}{51}&lt;/math&gt; is &lt;math&gt;0.\overline{23}_k = 0.232323..._k&lt;/math&gt;. What is &lt;math&gt;k&lt;/math&gt;?'' ([[2019 AMC 10A Problems/Problem 18|Solution]])<br /> * Problem 21 - 25: '''3.5-4.5'''<br /> *: ''The vertices of an equilateral triangle lie on the hyperbola &lt;math&gt;xy=1&lt;/math&gt;, and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?'' ([[2017 AMC 10B Problems/Problem 24|Solution]])<br /> <br /> ===[[CEMC|CEMC Multiple Choice Tests]]===<br /> This covers the CEMC Gauss, Pascal, Cayley, and Fermat tests.<br /> <br /> * Part A: '''0.5-1.5'''<br /> *: ''How many different 3-digit whole numbers can be formed using the digits 4, 7, and 9, assuming that no digit can be repeated in a number?'' (2015 Gauss 7 Problem 10)<br /> * Part B: '''1-2'''<br /> *: ''Two lines with slopes &lt;math&gt;\tfrac14&lt;/math&gt; and &lt;math&gt;\tfrac54&lt;/math&gt; intersect at &lt;math&gt;(1,1)&lt;/math&gt;. What is the area of the triangle formed by these two lines and the vertical line &lt;math&gt;x = 5&lt;/math&gt;?'' (2017 Cayley Problem 19)<br /> * Part C (Gauss/Pascal): '''2-2.5'''<br /> *: ''Suppose that &lt;math&gt;\tfrac{2009}{2014} + \tfrac{2019}{n} = \tfrac{a}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;n&lt;/math&gt; are positive integers with &lt;math&gt;\tfrac{a}{b}&lt;/math&gt; in lowest terms. What is the sum of the digits of the smallest positive integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;a&lt;/math&gt; is a multiple of 1004?'' (2014 Pascal Problem 25)<br /> * Part C (Cayley/Fermat): '''2.5-3'''<br /> *: ''Wayne has 3 green buckets, 3 red buckets, 3 blue buckets, and 3 yellow buckets. He randomly distributes 4 hockey pucks among the green buckets, with each puck equally likely to be put in each bucket. Similarly, he distributes 3 pucks among the red buckets, 2 pucks among the blue buckets, and 1 puck among the yellow buckets. Once he is ﬁnished, what is the probability that a green bucket contains more pucks than each of the other 11 buckets?'' (2018 Fermat Problem 24)<br /> <br /> ===[[CEMC|CEMC Fryer/Galois/Hypatia]]===<br /> <br /> * Problem 1-2: '''1-2'''<br /> * Problem 3-4 (early parts): '''2-3'''<br /> * Problem 3-4 (later parts): '''3-5'''<br /> <br /> ===Problem Solving Books for Introductory Students===<br /> <br /> Remark: There are many other problem books for Introductory Students that are not published by AoPS. Typically the rating on the left side is equivalent to the difficulty of the easiest review problems and the difficulty on the right side is the difficulty of the hardest challenge problems. The difficulty may vary greatly between sections of a book.<br /> <br /> ===[[Prealgebra by AoPS]]===<br /> 1-2<br /> ===[[Introduction to Algebra by AoPS]]===<br /> 1-3.5<br /> ===[[Introduction to Counting and Probability by AoPS]]===<br /> 1-3.5<br /> ===[[Introduction to Number Theory by AoPS]]===<br /> 1-3<br /> ===[[Introduction to Geometry by AoPS]]===<br /> 1-4<br /> <br /> ==Intermediate Competitions==<br /> This category consists of all the non-proof math competitions for the middle stages of high school. The difficulty range would normally be from 3 to 6. A full list is available [https://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AIntermediate+mathematics+competitions here].<br /> <br /> === [[AMC 12]] ===<br /> <br /> * Problem 1-10: '''1.5-2'''<br /> *: ''What is the value of &lt;cmath&gt;\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27?&lt;/cmath&gt;'' ([[2018 AMC 12B Problems/Problem 7|Solution]])<br /> * Problem 11-20: '''2.5-3.5'''<br /> *: ''Points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; lie on a circle centered at &lt;math&gt;O&lt;/math&gt;, and &lt;math&gt;\angle AOB = 60^\circ&lt;/math&gt;. A second circle is internally tangent to the first and tangent to both &lt;math&gt;\overline{OA}&lt;/math&gt; and &lt;math&gt;\overline{OB}&lt;/math&gt;. What is the ratio of the area of the smaller circle to that of the larger circle?'' ([[2008 AMC 12A Problems/Problem 13|Solution]])<br /> * Problem 21-25: '''4-5.5'''<br /> *: ''Functions &lt;math&gt;f&lt;/math&gt; and &lt;math&gt;g&lt;/math&gt; are quadratic, &lt;math&gt;g(x) = - f(100 - x)&lt;/math&gt;, and the graph of &lt;math&gt;g&lt;/math&gt; contains the vertex of the graph of &lt;math&gt;f&lt;/math&gt;. The four &lt;math&gt;x&lt;/math&gt;-intercepts on the two graphs have &lt;math&gt;x&lt;/math&gt;-coordinates &lt;math&gt;x_1&lt;/math&gt;, &lt;math&gt;x_2&lt;/math&gt;, &lt;math&gt;x_3&lt;/math&gt;, and &lt;math&gt;x_4&lt;/math&gt;, in increasing order, and &lt;math&gt;x_3 - x_2 = 150&lt;/math&gt;. The value of &lt;math&gt;x_4 - x_1&lt;/math&gt; is &lt;math&gt;m + n\sqrt p&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;p&lt;/math&gt; are positive integers, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. What is &lt;math&gt;m + n + p&lt;/math&gt;?'' ([[2009 AMC 12A Problems/Problem 23|Solution]])<br /> <br /> === [[AIME]] ===<br /> <br /> * Problem 1 - 5: '''3-3.5'''<br /> *: ''Consider the integer &lt;cmath&gt;N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.&lt;/cmath&gt;Find the sum of the digits of &lt;math&gt;N&lt;/math&gt;.'' ([[2019 AIME I Problems/Problem 1|Solution]])<br /> * Problem 6 - 9: '''4-4.5''' <br /> *: ''Define &lt;math&gt;n!!&lt;/math&gt; to be &lt;math&gt;n(n-2)(n-4)\cdots 3\cdot 1&lt;/math&gt; for &lt;math&gt;n&lt;/math&gt; odd and &lt;math&gt;n(n-2)(n-4)\cdots 4\cdot 2&lt;/math&gt; for &lt;math&gt;n&lt;/math&gt; even. When &lt;math&gt;\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}&lt;/math&gt; is expressed as a fraction in lowest terms, its denominator is &lt;math&gt;2^ab&lt;/math&gt; with &lt;math&gt;b&lt;/math&gt; odd. Find &lt;math&gt;\dfrac{ab}{10}&lt;/math&gt;.'' ([[2009 AIME II Problems/Problem 7|Solution]])<br /> * Problem 10 - 12: '''5-5.5'''<br /> *: Let &lt;math&gt;R&lt;/math&gt; be the set of all possible remainders when a number &lt;math&gt;2^n&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; a nonnegative integer, is divided by &lt;math&gt;1000&lt;/math&gt;.Let &lt;math&gt;S&lt;/math&gt; be the sum of all elements in &lt;math&gt;R&lt;/math&gt;. Find the remainder when &lt;math&gt;S&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt; ([[2011 AIME I Problems/Problem 11|Solution]])<br /> * Problem 13 - 15: '''6-6.5'''<br /> *: ''Let<br /> <br /> &lt;cmath&gt;P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).&lt;/cmath&gt;<br /> Let &lt;math&gt;z_{1},z_{2},\ldots,z_{r}&lt;/math&gt; be the distinct zeros of &lt;math&gt;P(x),&lt;/math&gt; and let &lt;math&gt;z_{k}^{2} = a_{k} + b_{k}i&lt;/math&gt; for &lt;math&gt;k = 1,2,\ldots,r,&lt;/math&gt; where &lt;math&gt;i = \sqrt { - 1},&lt;/math&gt; and &lt;math&gt;a_{k}&lt;/math&gt; and &lt;math&gt;b_{k}&lt;/math&gt; are real numbers. Let<br /> <br /> &lt;cmath&gt;\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},&lt;/cmath&gt;<br /> where &lt;math&gt;m,&lt;/math&gt; &lt;math&gt;n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m + n + p.&lt;/math&gt;.'' ([[2003 AIME II Problems/Problem 15|Solution]])<br /> <br /> === [[ARML]] ===<br /> <br /> * Individuals, Problem 1: '''2'''<br /> <br /> * Individuals, Problems 2, 3, 4, 5, 7, and 9: '''3'''<br /> <br /> * Individuals, Problems 6 and 8: '''4''' <br /> <br /> * Individuals, Problem 10: '''5.5'''<br /> <br /> * Team/power, Problem 1-5: '''3.5''' <br /> <br /> * Team/power, Problem 6-10: '''5'''<br /> <br /> ===[[HMMT|HMMT (November)]]===<br /> * Individual Round, Problem 6-8: '''4'''<br /> * Individual Round, Problem 10: '''4.5'''<br /> * Team Round: '''4-5'''<br /> * Guts: '''3.5-5.25'''<br /> <br /> ===[[CEMC|CEMC Euclid]]===<br /> <br /> * Problem 1-6: '''1-3'''<br /> * Problem 7-10: '''3-5'''<br /> <br /> ===[[Purple Comet! Math Meet|Purple Comet]]===<br /> <br /> * Problems 1-10 (MS): '''1.5-2.5'''<br /> * Problems 11-20 (MS): '''2.5-4'''<br /> * Problems 1-10 (HS): '''1.5-2.5'''<br /> * Problems 11-20 (HS): '''2-3.5'''<br /> * Problems 21-30 (HS): '''3.5-4.5'''<br /> <br /> === [[Philippine Mathematical Olympiad Qualifying Round]] ===<br /> <br /> * Problem 1-15: '''2'''<br /> * Problem 16-25: '''3'''<br /> * Problem 26-30: '''4'''<br /> <br /> ==Beginner Olympiad Competitions==<br /> This category consists of beginning Olympiad math competitions. Most junior and first stage Olympiads fall under this category. The range from the difficulty scale would be around 4 to 6. A full list is available [http://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3ABeginner+Olympiad+mathematics+competitions here].<br /> <br /> === [[USAMTS]] ===<br /> USAMTS generally has a different feel to it than olympiads, and is mainly for proofwriting practice instead of olympiad practice depending on how one takes the test. USAMTS allows an entire month to solve problems, with internet resources and books being allowed. However, the ultimate gap is that it permits computer programs to be used, and that Problem 1 is not a proof problem. However, it can still be roughly put to this rating scale:<br /> * Problem 1-2: '''3-4'''<br /> *: ''Find three isosceles triangles, no two of which are congruent, with integer sides, such that each triangle’s area is numerically equal to 6 times its perimeter.'' ([http://usamts.org/Solutions/Solution2_3_16.pdf Solution])<br /> * Problem 3-5: '''4-6'''<br /> *: ''Call a positive real number groovy if it can be written in the form &lt;math&gt;\sqrt{n} + \sqrt{n + 1}&lt;/math&gt; for some positive integer &lt;math&gt;n&lt;/math&gt;. Show that if &lt;math&gt;x&lt;/math&gt; is groovy, then for any positive integer &lt;math&gt;r&lt;/math&gt;, the number &lt;math&gt;x^r&lt;/math&gt; is groovy as well.'' ([http://usamts.org/Solutions/Solutions_20_1.pdf Solution])<br /> <br /> === [[Indonesia Mathematical Olympiad|Indonesia MO]] ===<br /> * Problem 1/5: '''3.5'''<br /> *: '' In a drawer, there are at most &lt;math&gt;2009&lt;/math&gt; balls, some of them are white, the rest are blue, which are randomly distributed. If two balls were taken at the same time, then the probability that the balls are both blue or both white is &lt;math&gt;\frac12&lt;/math&gt;. Determine the maximum amount of white balls in the drawer, such that the probability statement is true?'' &lt;url&gt;viewtopic.php?t=294065 (Solution)&lt;/url&gt;<br /> * Problem 2/6: '''4.5'''<br /> *: ''Find the lowest possible values from the function<br /> &lt;math&gt;f(x) = x^{2008} - 2x^{2007} + 3x^{2006} - 4x^{2005} + 5x^{2004} - \cdots - 2006x^3 + 2007x^2 - 2008x + 2009&lt;/math&gt;<br /> <br /> for any real numbers &lt;math&gt;x&lt;/math&gt;.''&lt;url&gt;viewtopic.php?t=294067 (Solution)&lt;/url&gt;<br /> * Problem 3/7: '''5'''<br /> *: ''A pair of integers &lt;math&gt;(m,n)&lt;/math&gt; is called ''good'' if<br /> &lt;math&gt;m\mid n^2 + n \ \text{and} \ n\mid m^2 + m&lt;/math&gt;<br /> <br /> Given 2 positive integers &lt;math&gt;a,b &gt; 1&lt;/math&gt; which are relatively prime, prove that there exists a ''good'' pair &lt;math&gt;(m,n)&lt;/math&gt; with &lt;math&gt;a\mid m&lt;/math&gt; and &lt;math&gt;b\mid n&lt;/math&gt;, but &lt;math&gt;a\nmid n&lt;/math&gt; and &lt;math&gt;b\nmid m&lt;/math&gt;.'' &lt;url&gt;viewtopic.php?t=294068 (Solution)&lt;/url&gt;<br /> * Problem 4/8: '''6'''<br /> *: ''Given an acute triangle &lt;math&gt;ABC&lt;/math&gt;. The incircle of triangle &lt;math&gt;ABC&lt;/math&gt; touches &lt;math&gt;BC,CA,AB&lt;/math&gt; respectively at &lt;math&gt;D,E,F&lt;/math&gt;. The angle bisector of &lt;math&gt;\angle A&lt;/math&gt; cuts &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;DF&lt;/math&gt; respectively at &lt;math&gt;K&lt;/math&gt; and &lt;math&gt;L&lt;/math&gt;. Suppose &lt;math&gt;AA_1&lt;/math&gt; is one of the altitudes of triangle &lt;math&gt;ABC&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; be the midpoint of &lt;math&gt;BC&lt;/math&gt;.<br /> <br /> (a) Prove that &lt;math&gt;BK&lt;/math&gt; and &lt;math&gt;CL&lt;/math&gt; are perpendicular with the angle bisector of &lt;math&gt;\angle BAC&lt;/math&gt;.<br /> <br /> (b) Show that &lt;math&gt;A_1KML&lt;/math&gt; is a cyclic quadrilateral.'' &lt;url&gt;viewtopic.php?t=294069 (Solution)&lt;/url&gt;<br /> <br /> === [[Central American Olympiad]] ===<br /> * Problem 1: '''4'''<br /> *: ''Find all three-digit numbers &lt;math&gt;abc&lt;/math&gt; (with &lt;math&gt;a \neq 0&lt;/math&gt;) such that &lt;math&gt;a^{2} + b^{2} + c^{2}&lt;/math&gt; is a divisor of 26.'' (&lt;url&gt;viewtopic.php?p=903856#903856 Solution&lt;/url&gt;)<br /> * Problem 2,4,5: '''5-6'''<br /> *: ''Show that the equation &lt;math&gt;a^{2}b^{2} + b^{2}c^{2} + 3b^{2} - c^{2} - a^{2} = 2005&lt;/math&gt; has no integer solutions.'' (&lt;url&gt;viewtopic.php?p=291301#291301 Solution&lt;/url&gt;)<br /> * Problem 3/6: '''6.5''' <br /> *: ''Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral. &lt;math&gt;I = AC\cap BD&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; are points on &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;CD&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt; respectively, such that &lt;math&gt;EF \cap GH = I&lt;/math&gt;. If &lt;math&gt;M = EG \cap AC&lt;/math&gt;, &lt;math&gt;N = HF \cap AC&lt;/math&gt;, show that &lt;math&gt;\frac {AM}{IM}\cdot \frac {IN}{CN} = \frac {IA}{IC}&lt;/math&gt;.'' (&lt;url&gt;viewtopic.php?p=828841#p828841 Solution&lt;/url&gt;<br /> <br /> === [[JBMO]] ===<br /> <br /> * Problem 1: '''4'''<br /> *: ''Find all real numbers &lt;math&gt;a,b,c,d&lt;/math&gt; such that <br /> &lt;cmath&gt; \left\{\begin{array}{cc}a+b+c+d = 20,\\ ab+ac+ad+bc+bd+cd = 150.\end{array}\right. &lt;/cmath&gt;''<br /> * Problem 2: '''4.5-5'''<br /> *: ''Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral with &lt;math&gt;\angle DAC=\angle BDC=36^\circ&lt;/math&gt;, &lt;math&gt;\angle CBD=18^\circ&lt;/math&gt; and &lt;math&gt;\angle BAC=72^\circ&lt;/math&gt;. The diagonals intersect at point &lt;math&gt;P&lt;/math&gt;. Determine the measure of &lt;math&gt;\angle APD&lt;/math&gt;.''<br /> * Problem 3: '''5'''<br /> *: ''Find all prime numbers &lt;math&gt;p,q,r&lt;/math&gt;, such that &lt;math&gt;\frac pq-\frac4{r+1}=1&lt;/math&gt;.''<br /> * Problem 4: '''6'''<br /> *: ''A &lt;math&gt;4\times4&lt;/math&gt; table is divided into &lt;math&gt;16&lt;/math&gt; white unit square cells. Two cells are called neighbors if they share a common side. A '''move''' consists in choosing a cell and changing the colors of neighbors from white to black or from black to white. After exactly &lt;math&gt;n&lt;/math&gt; moves all the &lt;math&gt;16&lt;/math&gt; cells were black. Find all possible values of &lt;math&gt;n&lt;/math&gt;.''<br /> <br /> ==Olympiad Competitions==<br /> This category consists of standard Olympiad competitions, usually ones from national Olympiads. Average difficulty is from 5 to 8. A full list is available [http://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AOlympiad+mathematics+competitions here].<br /> <br /> === [[USAJMO]] ===<br /> * Problem 1/4: '''5'''<br /> *: ''There are &lt;math&gt;a+b&lt;/math&gt; bowls arranged in a row, numbered &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;a+b&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are given positive integers. Initially, each of the first &lt;math&gt;a&lt;/math&gt; bowls contains an apple, and each of the last &lt;math&gt;b&lt;/math&gt; bowls contains a pear.''<br /> <br /> ''A legal move consists of moving an apple from bowl &lt;math&gt;i&lt;/math&gt; to bowl &lt;math&gt;i+1&lt;/math&gt; and a pear from bowl &lt;math&gt;j&lt;/math&gt; to bowl &lt;math&gt;j-1&lt;/math&gt;, provided that the difference &lt;math&gt;i-j&lt;/math&gt; is even. We permit multiple fruits in the same bowl at the same time. The goal is to end up with the first &lt;math&gt;b&lt;/math&gt; bowls each containing a pear and the last &lt;math&gt;a&lt;/math&gt; bowls each containing an apple. Show that this is possible if and only if the product &lt;math&gt;ab&lt;/math&gt; is even.'' ([[2019 USAJMO Problems/Problem 1|Solution]])<br /> <br /> * Problem 2/5: '''6'''<br /> *: ''Let &lt;math&gt;a,b,c&lt;/math&gt; be positive real numbers such that &lt;math&gt;a+b+c=4\sqrt{abc}&lt;/math&gt;. Prove that &lt;cmath&gt;2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.&lt;/cmath&gt;'' ([[2018 USAJMO Problems/Problem 2|Solution]])<br /> <br /> * Problem 3/6: '''7'''<br /> *: ''Two rational numbers &lt;math&gt;\tfrac{m}{n}&lt;/math&gt; and &lt;math&gt;\tfrac{n}{m}&lt;/math&gt; are written on a blackboard, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. At any point, Evan may pick two of the numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; written on the board and write either their arithmetic mean &lt;math&gt;\tfrac{x+y}{2}&lt;/math&gt; or their harmonic mean &lt;math&gt;\tfrac{2xy}{x+y}&lt;/math&gt; on the board as well. Find all pairs &lt;math&gt;(m,n)&lt;/math&gt; such that Evan can write &lt;math&gt;1&lt;/math&gt; on the board in finitely many steps.'' ([[2019 USAJMO Problems/Problem 6|Solution]])<br /> <br /> ===[[HMMT|HMMT (February)]]===<br /> * Individual Round, Problem 1-5: '''5'''<br /> * Individual Round, Problem 6-10: '''5.5-6'''<br /> * Team Round: '''7.5'''<br /> * HMIC: '''8'''<br /> <br /> === [[Canadian MO]] ===<br /> <br /> * Problem 1: '''5.5'''<br /> * Problem 2: '''6'''<br /> * Problem 3: '''6.5''' <br /> * Problem 4: '''7-7.5'''<br /> * Problem 5: '''7.5-8'''<br /> <br /> === Austrian MO ===<br /> <br /> * Regional Competition for Advanced Students, Problems 1-4: '''5''' <br /> * Federal Competition for Advanced Students, Part 1. Problems 1-4: '''6''' <br /> * Federal Competition for Advanced Students, Part 2, Problems 1-6: '''7'''<br /> <br /> === [[Ibero American Olympiad]] ===<br /> <br /> * Problem 1/4: '''5.5'''<br /> * Problem 2/5: '''6.5'''<br /> * Problem 3/6: '''7.5'''<br /> <br /> === [[APMO]] ===<br /> *Problem 1: '''6'''<br /> *Problem 2: '''7'''<br /> *Problem 3: '''7'''<br /> *Problem 4: '''7.5'''<br /> *Problem 5: '''8.5'''<br /> <br /> === Balkan MO ===<br /> <br /> * Problem 1: '''6'''<br /> *: '' Solve the equation &lt;math&gt;3^x - 5^y = z^2&lt;/math&gt; in positive integers. '' <br /> * Problem 2: '''6.5'''<br /> *: '' Let &lt;math&gt;MN&lt;/math&gt; be a line parallel to the side &lt;math&gt;BC&lt;/math&gt; of a triangle &lt;math&gt;ABC&lt;/math&gt;, with &lt;math&gt;M&lt;/math&gt; on the side &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; on the side &lt;math&gt;AC&lt;/math&gt;. The lines &lt;math&gt;BN&lt;/math&gt; and &lt;math&gt;CM&lt;/math&gt; meet at point &lt;math&gt;P&lt;/math&gt;. The circumcircles of triangles &lt;math&gt;BMP&lt;/math&gt; and &lt;math&gt;CNP&lt;/math&gt; meet at two distinct points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Prove that &lt;math&gt;\angle BAQ = \angle CAP&lt;/math&gt;. ''<br /> * Problem 3: '''7.5'''<br /> *: '' A &lt;math&gt;9 \times 12&lt;/math&gt; rectangle is partitioned into unit squares. The centers of all the unit squares, except for the four corner squares and eight squares sharing a common side with one of them, are coloured red. Is it possible to label these red centres &lt;math&gt;C_1,C_2...,C_{96}&lt;/math&gt; in such way that the following to conditions are both fulfilled<br /> <br /> &lt;math&gt;(i)&lt;/math&gt; the distances &lt;math&gt;C_1C_2,...C_{95}C_{96}, C_{96}C_{1}&lt;/math&gt; are all equal to &lt;math&gt;\sqrt {13}&lt;/math&gt;<br /> <br /> &lt;math&gt;(ii)&lt;/math&gt; the closed broken line &lt;math&gt;C_1C_2...C_{96}C_1&lt;/math&gt; has a centre of symmetry? ''<br /> * Problem 4: '''8'''<br /> *: '' Denote by &lt;math&gt;S&lt;/math&gt; the set of all positive integers. Find all functions &lt;math&gt;f: S \rightarrow S&lt;/math&gt; such that<br /> <br /> &lt;math&gt;f \bigg(f^2(m) + 2f^2(n)\bigg) = m^2 + 2 n^2&lt;/math&gt; for all &lt;math&gt;m,n \in S&lt;/math&gt;. '<br /> <br /> ==Hard Olympiad Competitions==<br /> This category consists of harder Olympiad contests. Difficulty is usually from 7 to 10. A full list is available [https://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AHard+Olympiad+mathematics+competitions here].<br /> <br /> === [[USAMO]] ===<br /> * Problem 1/4: '''6-7'''<br /> *: ''Let &lt;math&gt;\mathcal{P}&lt;/math&gt; be a convex polygon with &lt;math&gt;n&lt;/math&gt; sides, &lt;math&gt;n\ge3&lt;/math&gt;. Any set of &lt;math&gt;n - 3&lt;/math&gt; diagonals of &lt;math&gt;\mathcal{P}&lt;/math&gt; that do not intersect in the interior of the polygon determine a ''triangulation'' of &lt;math&gt;\mathcal{P}&lt;/math&gt; into &lt;math&gt;n - 2&lt;/math&gt; triangles. If &lt;math&gt;\mathcal{P}&lt;/math&gt; is regular and there is a triangulation of &lt;math&gt;\mathcal{P}&lt;/math&gt; consisting of only isosceles triangles, find all the possible values of &lt;math&gt;n&lt;/math&gt;.'' ([[2008 USAMO Problems/Problem 4|Solution]]) <br /> * Problem 2/5: '''7-8'''<br /> *: ''Three nonnegative real numbers &lt;math&gt;r_1&lt;/math&gt;, &lt;math&gt;r_2&lt;/math&gt;, &lt;math&gt;r_3&lt;/math&gt; are written on a blackboard. These numbers have the property that there exist integers &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;a_3&lt;/math&gt;, not all zero, satisfying &lt;math&gt;a_1r_1 + a_2r_2 + a_3r_3 = 0&lt;/math&gt;. We are permitted to perform the following operation: find two numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; on the blackboard with &lt;math&gt;x \le y&lt;/math&gt;, then erase &lt;math&gt;y&lt;/math&gt; and write &lt;math&gt;y - x&lt;/math&gt; in its place. Prove that after a finite number of such operations, we can end up with at least one &lt;math&gt;0&lt;/math&gt; on the blackboard.'' ([[2008 USAMO Problems/Problem 5|Solution]])<br /> * Problem 3/6: '''8-9'''<br /> *: ''Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree &lt;math&gt;n &lt;/math&gt; with real coefficients is the average of two monic polynomials of degree &lt;math&gt;n &lt;/math&gt; with &lt;math&gt;n &lt;/math&gt; real roots.'' ([[2002 USAMO Problems/Problem 3|Solution]])<br /> <br /> === [[USA TST]] ===<br /> <br /> <br /> <br /> * Problem 1/4/7: '''6.5-7'''<br /> * Problem 2/5/8: '''7.5-8'''<br /> * Problem 3/6/9: '''8.5-9'''<br /> <br /> === [[Putnam]] ===<br /> <br /> * Problem A/B,1-2: '''7'''<br /> *: ''Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola &lt;math&gt;xy = 1&lt;/math&gt; and both branches of the hyperbola &lt;math&gt;xy = - 1.&lt;/math&gt; (A set &lt;math&gt;S&lt;/math&gt; in the plane is called ''convex'' if for any two points in &lt;math&gt;S&lt;/math&gt; the line segment connecting them is contained in &lt;math&gt;S.&lt;/math&gt;)'' ([https://artofproblemsolving.com/community/c7h177227p978383 Solution])<br /> * Problem A/B,3-4: '''8'''<br /> *: ''Let &lt;math&gt;H&lt;/math&gt; be an &lt;math&gt;n\times n&lt;/math&gt; matrix all of whose entries are &lt;math&gt;\pm1&lt;/math&gt; and whose rows are mutually orthogonal. Suppose &lt;math&gt;H&lt;/math&gt; has an &lt;math&gt;a\times b&lt;/math&gt; submatrix whose entries are all &lt;math&gt;1.&lt;/math&gt; Show that &lt;math&gt;ab\le n&lt;/math&gt;.'' ([https://artofproblemsolving.com/community/c7h64435p383280 Solution])<br /> * Problem A/B,5-6: '''9'''<br /> *: ''For any &lt;math&gt;a &gt; 0&lt;/math&gt;, define the set &lt;math&gt;S(a) = \{[an]|n = 1,2,3,...\}&lt;/math&gt;. Show that there are no three positive reals &lt;math&gt;a,b,c&lt;/math&gt; such that &lt;math&gt;S(a)\cap S(b) = S(b)\cap S(c) = S(c)\cap S(a) = \emptyset,S(a)\cup S(b)\cup S(c) = \{1,2,3,...\}&lt;/math&gt;.'' (&lt;url&gt;viewtopic.php?t=127810 Solution&lt;/url&gt;)<br /> <br /> === [[China TST]] ===<br /> <br /> * Problem 1/4: '''8-8.5''' <br /> *: ''Given an integer &lt;math&gt;m,&lt;/math&gt; prove that there exist odd integers &lt;math&gt;a,b&lt;/math&gt; and a positive integer &lt;math&gt;k&lt;/math&gt; such that &lt;cmath&gt;2m=a^{19}+b^{99}+k*2^{1000}.&lt;/cmath&gt;''<br /> * Problem 2/5: '''9''' <br /> *: ''Given a positive integer &lt;math&gt;n&gt;1&lt;/math&gt; and real numbers &lt;math&gt;a_1 &lt; a_2 &lt; \ldots &lt; a_n,&lt;/math&gt; such that &lt;math&gt;\dfrac{1}{a_1} + \dfrac{1}{a_2} + \ldots + \dfrac{1}{a_n} \le 1,&lt;/math&gt; prove that for any positive real number &lt;math&gt;x,&lt;/math&gt; &lt;cmath&gt;\left(\dfrac{1}{a_1^2+x} + \dfrac{1}{a_2^2+x} + \ldots + \dfrac{1}{a_n^2+x}\right)^2 \ge \dfrac{1}{2a_1(a_1-1)+2x}.&lt;/cmath&gt;''<br /> * Problem 3/6: '''9.5-10'''<br /> *: ''Let &lt;math&gt;n&gt;1&lt;/math&gt; be an integer and let &lt;math&gt;a_0,a_1,\ldots,a_n&lt;/math&gt; be non-negative real numbers. Define &lt;math&gt;S_k=\sum_{i=0}^k \binom{k}{i}a_i&lt;/math&gt; for &lt;math&gt;k=0,1,\ldots,n&lt;/math&gt;. Prove that&lt;cmath&gt;\frac{1}{n} \sum_{k=0}^{n-1} S_k^2-\frac{1}{n^2}\left(\sum_{k=0}^{n} S_k\right)^2\le \frac{4}{45} (S_n-S_0)^2.&lt;/cmath&gt;''<br /> <br /> === [[IMO]] ===<br /> <br /> * Problem 1/4: '''5.5-7'''<br /> *: ''Let &lt;math&gt;\Gamma&lt;/math&gt; be the circumcircle of acute triangle &lt;math&gt;ABC&lt;/math&gt;. Points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; are on segments &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; respectively such that &lt;math&gt;AD = AE&lt;/math&gt;. The perpendicular bisectors of &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;CE&lt;/math&gt; intersect minor arcs &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; of &lt;math&gt;\Gamma&lt;/math&gt; at points &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; respectively. Prove that lines &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;FG&lt;/math&gt; are either parallel or they are the same line.'' ([[2018 IMO Problems/Problem 1|Solution]])<br /> <br /> * Problem 2/5: '''7-8'''<br /> *: ''Let &lt;math&gt;P(x)&lt;/math&gt; be a polynomial of degree &lt;math&gt;n&gt;1&lt;/math&gt; with integer coefficients, and let &lt;math&gt;k&lt;/math&gt; be a positive integer. Consider the polynomial &lt;math&gt;Q(x) = P( P ( \ldots P(P(x)) \ldots ))&lt;/math&gt;, where &lt;math&gt;P&lt;/math&gt; occurs &lt;math&gt;k&lt;/math&gt; times. Prove that there are at most &lt;math&gt;n&lt;/math&gt; integers &lt;math&gt;t&lt;/math&gt; such that &lt;math&gt;Q(t)=t&lt;/math&gt;.'' ([[2006 IMO Problems/Problem 5|Solution]])<br /> <br /> * Problem 3/6: '''9-10'''<br /> *: ''Assign to each side &lt;math&gt;b&lt;/math&gt; of a convex polygon &lt;math&gt;P&lt;/math&gt; the maximum area of a triangle that has &lt;math&gt;b&lt;/math&gt; as a side and is contained in &lt;math&gt;P&lt;/math&gt;. Show that the sum of the areas assigned to the sides of &lt;math&gt;P&lt;/math&gt; is at least twice the area of &lt;math&gt;P&lt;/math&gt;.'' (&lt;url&gt;viewtopic.php?p=572824#572824 Solution&lt;/url&gt;)<br /> <br /> === [[IMO Shortlist]] ===<br /> <br /> * Problem 1-2: '''5.5-7'''<br /> * Problem 3-4: '''7-8'''<br /> * Problem 5+: '''8-10'''<br /> <br /> [[Category:Mathematics competitions]]</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=123628 User:Piphi 2020-06-04T15:18:11Z <p>Firebolt360: /* User Count */</p> <hr /> <div>{{User:Piphi/Template:Header}}<br /> &lt;br&gt;<br /> &lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#dddddd&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;100px&quot;&gt;113&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;A legendary AoPSer.&lt;br&gt;<br /> <br /> My main project on the AoPS wiki is [[AoPS_Administrators#Current_Admins | a list of all the AoPS admins]], everyone is welcome to add more admins to the list by clicking [https://artofproblemsolving.com/wiki/index.php?title=AoPS_Administrators&amp;action=edit&amp;section=1 here]. I've also added a lot of the info in the [[Reaper Archives]]. Another one of my projects on the Wiki is trying to make the [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]]<br /> <br /> You can also check out Greed Control Game 19 statistics that I found, [https://artofproblemsolving.com/community/c19451h2126212 here].&lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;[[User:Piphi/Asymptote|Asymptote]]&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;For a complete list of my Asymptote drawings, go [[User:Piphi/Asymptote|here]].&lt;/div&gt;<br /> &lt;/div&gt;</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_15&diff=123501 2019 AIME I Problems/Problem 15 2020-06-02T23:17:25Z <p>Firebolt360: /* Solution 5 (Easy) */</p> <hr /> <div>==Problem 15==<br /> <br /> Let &lt;math&gt;\overline{AB}&lt;/math&gt; be a chord of a circle &lt;math&gt;\omega&lt;/math&gt;, and let &lt;math&gt;P&lt;/math&gt; be a point on the chord &lt;math&gt;\overline{AB}&lt;/math&gt;. Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; intersect at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Line &lt;math&gt;PQ&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;. Assume that &lt;math&gt;AP=5&lt;/math&gt;, &lt;math&gt;PB=3&lt;/math&gt;, &lt;math&gt;XY=11&lt;/math&gt;, and &lt;math&gt;PQ^2 = \tfrac{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 1==<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair O, A, B, P, O1, O2, Q, X, Y;<br /> O=(0, 0);<br /> A=dir(140); B=dir(40);<br /> P=(3A+5B)/8;<br /> O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);<br /> O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);<br /> Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)));<br /> X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));<br /> Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));<br /> <br /> draw(circle(O, 1));<br /> draw(circle(O1, length(A-O1)));<br /> draw(circle(O2, length(B-O2)));<br /> draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);<br /> <br /> dot(&quot;O$&quot;, O, S);<br /> dot(&quot;$A$&quot;, A, A);<br /> dot(&quot;$B$&quot;, B, B);<br /> dot(&quot;$P$&quot;, P, dir(70));<br /> dot(&quot;$Q$&quot;, Q, dir(200));<br /> dot(&quot;$O_1$&quot;, O1, SW);<br /> dot(&quot;$O_2$&quot;, O2, SE);<br /> dot(&quot;$X$&quot;, X, X);<br /> dot(&quot;$Y$&quot;, Y, Y);<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; be the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, respectively. There is a homothety at &lt;math&gt;A&lt;/math&gt; sending &lt;math&gt;\omega&lt;/math&gt; to &lt;math&gt;\omega_1&lt;/math&gt; that sends &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;O_1&lt;/math&gt;, so &lt;math&gt;\overline{OO_2}\parallel\overline{O_1P}&lt;/math&gt;. Similarly, &lt;math&gt;\overline{OO_1}\parallel\overline{O_2P}&lt;/math&gt;, so &lt;math&gt;OO_1PO_2&lt;/math&gt; is a parallelogram. Moreover, &lt;cmath&gt;\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,&lt;/cmath&gt;whence &lt;math&gt;OO_1O_2Q&lt;/math&gt; is cyclic. However, &lt;cmath&gt;OO_1=O_2P=O_2Q,&lt;/cmath&gt;so &lt;math&gt;OO_1O_2Q&lt;/math&gt; is an isosceles trapezoid. Since &lt;math&gt;\overline{O_1O_2}\perp\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{OQ}\perp\overline{XY}&lt;/math&gt;, so &lt;math&gt;Q&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;.<br /> <br /> By Power of a Point, &lt;math&gt;PX\cdot PY=PA\cdot PB=15&lt;/math&gt;. Since &lt;math&gt;PX+PY=XY=11&lt;/math&gt; and &lt;math&gt;XQ=11/2&lt;/math&gt;, &lt;cmath&gt;XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,&lt;/cmath&gt;<br /> and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 2==<br /> <br /> Let the tangents to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at &lt;math&gt;R&lt;/math&gt;. Then, since &lt;math&gt;RA^2=RB^2&lt;/math&gt;, &lt;math&gt;R&lt;/math&gt; lies on the radical axis of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, which is &lt;math&gt;\overline{PQ}&lt;/math&gt;. It follows that &lt;cmath&gt;-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).&lt;/cmath&gt;<br /> Let &lt;math&gt;Q'&lt;/math&gt; denote the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;. By the Midpoint of Harmonic Bundles Lemma, &lt;cmath&gt;RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,&lt;/cmath&gt;<br /> whence &lt;math&gt;Q=Q'&lt;/math&gt;. Like above, &lt;math&gt;XP=\tfrac{11-\sqrt{61}}2&lt;/math&gt;. Since &lt;math&gt;XQ=\tfrac{11}2&lt;/math&gt;, we establish that &lt;math&gt;PQ=\tfrac{\sqrt{61}}2&lt;/math&gt;, from which &lt;math&gt;PQ^2=\tfrac{61}4&lt;/math&gt;, and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 3==<br /> <br /> Firstly we need to notice that &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;. Assume the center of circle &lt;math&gt;w, w_1, w_2&lt;/math&gt; are &lt;math&gt;O, O_1, O_2&lt;/math&gt;, respectively. Then &lt;math&gt;A, O_2, O&lt;/math&gt; are collinear and &lt;math&gt;O, O_1, B&lt;/math&gt; are collinear. Link &lt;math&gt;O_1P, O_2P, O_1Q, O_2Q&lt;/math&gt;. Notice that, &lt;math&gt;\angle B=\angle A=\angle APO_2=\angle BPO_1&lt;/math&gt;. As a result, &lt;math&gt;PO_1\parallel O_2O&lt;/math&gt; and &lt;math&gt;QO_1\parallel O_2P&lt;/math&gt;. So we have parallelogram &lt;math&gt;PO_2O_1O&lt;/math&gt;. So &lt;math&gt;\angle O_2PO_1=\angle O&lt;/math&gt; Notice that, &lt;math&gt;O_1O_2\bot PQ&lt;/math&gt; and &lt;math&gt;O_1O_2&lt;/math&gt; divide &lt;math&gt;PQ&lt;/math&gt; into two equal length pieces, So we have &lt;math&gt;\angle O_2PO_1=\angle O_2QO_1=\angle O&lt;/math&gt;. As a result, &lt;math&gt;O_2, Q, O, O_1,&lt;/math&gt; lie on one circle. So &lt;math&gt;\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P&lt;/math&gt;. Notice that &lt;math&gt;\angle O_1PQ+\angle O_2O_1P=90^{\circ}&lt;/math&gt;, we have &lt;math&gt;\angle OQP=90^{\circ}&lt;/math&gt;. As a result, &lt;math&gt;OQ\bot PQ&lt;/math&gt;. So &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;.<br /> <br /> Back to our problem. Assume &lt;math&gt;XP=x&lt;/math&gt;, &lt;math&gt;PY=y&lt;/math&gt; and &lt;math&gt;x&lt;y&lt;/math&gt;. Then we have &lt;math&gt;AP\cdot PB=XP\cdot PY&lt;/math&gt;, that is, &lt;math&gt;xy=15&lt;/math&gt;. Also, &lt;math&gt;XP+PY=x+y=XY=11&lt;/math&gt;. Solve these above, we have &lt;math&gt;x=\frac{11-\sqrt{61}}{2}=XP&lt;/math&gt;. As a result, we have &lt;math&gt;PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}&lt;/math&gt;. So, we have &lt;math&gt;PQ^2=\frac{61}{4}&lt;/math&gt;. As a result, our answer is &lt;math&gt;m+n=61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> <br /> <br /> Solution By BladeRunnerAUG (Fanyuchen20020715).<br /> <br /> ==Solution 4==<br /> Note that the tangents to the circles at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at a point &lt;math&gt;Z&lt;/math&gt; on &lt;math&gt;XY&lt;/math&gt; by radical center. Then, since &lt;math&gt;\angle ZAB = \angle ZQA&lt;/math&gt; and &lt;math&gt;\angle ZBA = \angle ZQB&lt;/math&gt;, we have <br /> &lt;cmath&gt;\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},&lt;/cmath&gt;<br /> so &lt;math&gt;ZAQB&lt;/math&gt; is cyclic. But if &lt;math&gt;O&lt;/math&gt; is the center of &lt;math&gt;\omega&lt;/math&gt;, clearly &lt;math&gt;ZAOB&lt;/math&gt; is cyclic with diameter &lt;math&gt;ZO&lt;/math&gt;, so &lt;math&gt;\angle ZQO = 90^{\circ} \implies Q&lt;/math&gt; is the midpoint of &lt;math&gt;XY&lt;/math&gt;. Then, by Power of a Point, &lt;math&gt;PY \cdot PX = PA \cdot PB = 15&lt;/math&gt; and it is given that &lt;math&gt;PY+PX = 11&lt;/math&gt;. Thus &lt;math&gt;PY, PX = \frac{11 \pm \sqrt{61}}{2}&lt;/math&gt; so &lt;math&gt;PQ = \frac{\sqrt{61}}{2} \implies PQ^2 = \frac{61}{4}&lt;/math&gt; and the answer is &lt;math&gt;61+4 = \boxed{065}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_11&diff=123498 2019 AIME I Problems/Problem 11 2020-06-02T22:43:26Z <p>Firebolt360: /* Solution 3 */</p> <hr /> <div>==Problem 11==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, the sides have integers lengths and &lt;math&gt;AB=AC&lt;/math&gt;. Circle &lt;math&gt;\omega&lt;/math&gt; has its center at the incenter of &lt;math&gt;\triangle ABC&lt;/math&gt;. An ''excircle'' of &lt;math&gt;\triangle ABC&lt;/math&gt; is a circle in the exterior of &lt;math&gt;\triangle ABC&lt;/math&gt; that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to &lt;math&gt;\overline{BC}&lt;/math&gt; is internally tangent to &lt;math&gt;\omega&lt;/math&gt;, and the other two excircles are both externally tangent to &lt;math&gt;\omega&lt;/math&gt;. Find the minimum possible value of the perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let the tangent circle be &lt;math&gt;\omega&lt;/math&gt;. Some notation first: let &lt;math&gt;BC=a&lt;/math&gt;, &lt;math&gt;AB=b&lt;/math&gt;, &lt;math&gt;s&lt;/math&gt; be the semiperimeter, &lt;math&gt;\theta=\angle ABC&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; be the inradius. Intuition tells us that the radius of &lt;math&gt;\omega&lt;/math&gt; is &lt;math&gt;r+\frac{2rs}{s-a}&lt;/math&gt; (using the exradius formula). However, the sum of the radius of &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\frac{rs}{s-b}&lt;/math&gt; is equivalent to the distance between the incenter and the the &lt;math&gt;B/C&lt;/math&gt; excenter. Denote the B excenter as &lt;math&gt;I_B&lt;/math&gt; and the incenter as &lt;math&gt;I&lt;/math&gt;. <br /> Lemma: &lt;math&gt;I_BI=\frac{2b*IB}{a}&lt;/math&gt;<br /> We draw the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;. Let the angle bisector of &lt;math&gt;\angle ABC&lt;/math&gt; hit the circumcircle at a second point &lt;math&gt;M&lt;/math&gt;. By the incenter-excenter lemma, &lt;math&gt;AM=CM=IM&lt;/math&gt;. Let this distance be &lt;math&gt;\alpha&lt;/math&gt;. Ptolemy's theorem on &lt;math&gt;ABCM&lt;/math&gt; gives us &lt;cmath&gt;a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}&lt;/cmath&gt; Again, by the incenter-excenter lemma, &lt;math&gt;II_B=2IM&lt;/math&gt; so &lt;math&gt;II_b=\frac{2b*IB}{a}&lt;/math&gt; as desired.<br /> Using this gives us the following equation: &lt;cmath&gt;\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}&lt;/cmath&gt; <br /> Motivated by the &lt;math&gt;s-a&lt;/math&gt; and &lt;math&gt;s-b&lt;/math&gt;, we make the following substitution: &lt;math&gt;x=s-a, y=s-b&lt;/math&gt;<br /> This changes things quite a bit. Here's what we can get from it: &lt;cmath&gt;a=2y, b=x+y, s=x+2y&lt;/cmath&gt; It is known (easily proved with Heron's and a=rs) that &lt;cmath&gt;r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}&lt;/cmath&gt; Using this, we can also find &lt;math&gt;IB&lt;/math&gt;: let the midpoint of &lt;math&gt;BC&lt;/math&gt; be &lt;math&gt;N&lt;/math&gt;. Using Pythagorean's Theorem on &lt;math&gt;\triangle INB&lt;/math&gt;, &lt;cmath&gt;IB^2=r^2+(\frac{a}{2})^2=\frac{xy^2}{x+2y}+y^2=\frac{2xy^2+2y^3}{x+2y}=\frac{2y^2(x+y)}{x+2y} &lt;/cmath&gt; We now look at the RHS of the main equation: &lt;cmath&gt;r+\frac{2rs}{s-a}+\frac{rs}{s-b}=r(1+\frac{2(x+2y)}{x}+\frac{x+2y}{y})=r(\frac{x^2+5xy+4y^2}{xy})=\frac{r(x+4y)(x+y)}{xy}=\frac{2(x+y)IB}{2y}&lt;/cmath&gt;<br /> Cancelling some terms, we have &lt;cmath&gt;\frac{r(x+4y)}{x}=IB&lt;/cmath&gt; <br /> Squaring, &lt;cmath&gt;\frac{2y^2(x+y)}{x+2y}=\frac{(x+4y)^2*xy^2}{x^2(x+2y)}\to \frac{(x+4y)^2}{x}=2(x+y)&lt;/cmath&gt; Expanding and moving terms around gives &lt;cmath&gt;(x-8y)(x+2y)=0\to x=8y&lt;/cmath&gt; Reverse substituting, &lt;cmath&gt;s-a=8s-8b\to b=\frac{9}{2}a&lt;/cmath&gt; Clearly the smallest solution is &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;b=9&lt;/math&gt;, so our answer is &lt;math&gt;2+9+9=\boxed{020}&lt;/math&gt;<br /> -franchester<br /> <br /> ==Solution 2==<br /> <br /> Video Link Here: https://www.youtube.com/watch?v=zKHwTJBhKdM<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_11&diff=123497 2019 AIME I Problems/Problem 11 2020-06-02T22:43:00Z <p>Firebolt360: /* Solution 2 */</p> <hr /> <div>==Problem 11==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, the sides have integers lengths and &lt;math&gt;AB=AC&lt;/math&gt;. Circle &lt;math&gt;\omega&lt;/math&gt; has its center at the incenter of &lt;math&gt;\triangle ABC&lt;/math&gt;. An ''excircle'' of &lt;math&gt;\triangle ABC&lt;/math&gt; is a circle in the exterior of &lt;math&gt;\triangle ABC&lt;/math&gt; that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to &lt;math&gt;\overline{BC}&lt;/math&gt; is internally tangent to &lt;math&gt;\omega&lt;/math&gt;, and the other two excircles are both externally tangent to &lt;math&gt;\omega&lt;/math&gt;. Find the minimum possible value of the perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let the tangent circle be &lt;math&gt;\omega&lt;/math&gt;. Some notation first: let &lt;math&gt;BC=a&lt;/math&gt;, &lt;math&gt;AB=b&lt;/math&gt;, &lt;math&gt;s&lt;/math&gt; be the semiperimeter, &lt;math&gt;\theta=\angle ABC&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; be the inradius. Intuition tells us that the radius of &lt;math&gt;\omega&lt;/math&gt; is &lt;math&gt;r+\frac{2rs}{s-a}&lt;/math&gt; (using the exradius formula). However, the sum of the radius of &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\frac{rs}{s-b}&lt;/math&gt; is equivalent to the distance between the incenter and the the &lt;math&gt;B/C&lt;/math&gt; excenter. Denote the B excenter as &lt;math&gt;I_B&lt;/math&gt; and the incenter as &lt;math&gt;I&lt;/math&gt;. <br /> Lemma: &lt;math&gt;I_BI=\frac{2b*IB}{a}&lt;/math&gt;<br /> We draw the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;. Let the angle bisector of &lt;math&gt;\angle ABC&lt;/math&gt; hit the circumcircle at a second point &lt;math&gt;M&lt;/math&gt;. By the incenter-excenter lemma, &lt;math&gt;AM=CM=IM&lt;/math&gt;. Let this distance be &lt;math&gt;\alpha&lt;/math&gt;. Ptolemy's theorem on &lt;math&gt;ABCM&lt;/math&gt; gives us &lt;cmath&gt;a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}&lt;/cmath&gt; Again, by the incenter-excenter lemma, &lt;math&gt;II_B=2IM&lt;/math&gt; so &lt;math&gt;II_b=\frac{2b*IB}{a}&lt;/math&gt; as desired.<br /> Using this gives us the following equation: &lt;cmath&gt;\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}&lt;/cmath&gt; <br /> Motivated by the &lt;math&gt;s-a&lt;/math&gt; and &lt;math&gt;s-b&lt;/math&gt;, we make the following substitution: &lt;math&gt;x=s-a, y=s-b&lt;/math&gt;<br /> This changes things quite a bit. Here's what we can get from it: &lt;cmath&gt;a=2y, b=x+y, s=x+2y&lt;/cmath&gt; It is known (easily proved with Heron's and a=rs) that &lt;cmath&gt;r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}&lt;/cmath&gt; Using this, we can also find &lt;math&gt;IB&lt;/math&gt;: let the midpoint of &lt;math&gt;BC&lt;/math&gt; be &lt;math&gt;N&lt;/math&gt;. Using Pythagorean's Theorem on &lt;math&gt;\triangle INB&lt;/math&gt;, &lt;cmath&gt;IB^2=r^2+(\frac{a}{2})^2=\frac{xy^2}{x+2y}+y^2=\frac{2xy^2+2y^3}{x+2y}=\frac{2y^2(x+y)}{x+2y} &lt;/cmath&gt; We now look at the RHS of the main equation: &lt;cmath&gt;r+\frac{2rs}{s-a}+\frac{rs}{s-b}=r(1+\frac{2(x+2y)}{x}+\frac{x+2y}{y})=r(\frac{x^2+5xy+4y^2}{xy})=\frac{r(x+4y)(x+y)}{xy}=\frac{2(x+y)IB}{2y}&lt;/cmath&gt;<br /> Cancelling some terms, we have &lt;cmath&gt;\frac{r(x+4y)}{x}=IB&lt;/cmath&gt; <br /> Squaring, &lt;cmath&gt;\frac{2y^2(x+y)}{x+2y}=\frac{(x+4y)^2*xy^2}{x^2(x+2y)}\to \frac{(x+4y)^2}{x}=2(x+y)&lt;/cmath&gt; Expanding and moving terms around gives &lt;cmath&gt;(x-8y)(x+2y)=0\to x=8y&lt;/cmath&gt; Reverse substituting, &lt;cmath&gt;s-a=8s-8b\to b=\frac{9}{2}a&lt;/cmath&gt; Clearly the smallest solution is &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;b=9&lt;/math&gt;, so our answer is &lt;math&gt;2+9+9=\boxed{020}&lt;/math&gt;<br /> -franchester<br /> <br /> ==Solution 3==<br /> <br /> On the Spot STEM solves this problem here: https://www.youtube.com/watch?v=zKHwTJBhKdM<br /> <br /> Please like, share, and subscribe!<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_11&diff=123362 2019 AIME II Problems/Problem 11 2020-05-31T22:19:35Z <p>Firebolt360: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has side lengths &lt;math&gt;AB=7, BC=8, &lt;/math&gt; and &lt;math&gt;CA=9.&lt;/math&gt; Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and is tangent to line &lt;math&gt;AC&lt;/math&gt; at &lt;math&gt;A.&lt;/math&gt; Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;C&lt;/math&gt; and is tangent to line &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;A.&lt;/math&gt; Let &lt;math&gt;K&lt;/math&gt; be the intersection of circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; not equal to &lt;math&gt;A.&lt;/math&gt; Then &lt;math&gt;AK=\tfrac mn,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n.&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;asy&gt;<br /> unitsize(20);<br /> pair B = (0,0);<br /> pair A = (2,sqrt(45));<br /> pair C = (8,0);<br /> draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7));<br /> draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7));<br /> draw(B--A--C--cycle);<br /> label(&quot;$A$&quot;,A,dir(105));<br /> label(&quot;$B$&quot;,B,dir(-135));<br /> label(&quot;$C$&quot;,C,dir(-75));<br /> dot((2.68,2.25));<br /> label(&quot;$K$&quot;,(2.68,2.25),dir(-150));<br /> label(&quot;$\omega_1$&quot;,(-6,1));<br /> label(&quot;$\omega_2$&quot;,(14,6));<br /> label(&quot;$7$&quot;,(A+B)/2,dir(140));<br /> label(&quot;$8$&quot;,(B+C)/2,dir(-90));<br /> label(&quot;$9&quot;,(A+C)/2,dir(60));<br /> &lt;/asy&gt;<br /> -Diagram by Brendanb4321<br /> <br /> <br /> Note that from the tangency condition that the supplement of &lt;math&gt;\angle CAB&lt;/math&gt; with respects to lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; are equal to &lt;math&gt;\angle AKB&lt;/math&gt; and &lt;math&gt;\angle AKC&lt;/math&gt;, respectively, so from tangent-chord, &lt;cmath&gt;\angle AKC=\angle AKB=180^{\circ}-\angle BAC&lt;/cmath&gt; Also note that &lt;math&gt;\angle ABK=\angle KAC&lt;/math&gt;, so &lt;math&gt;\triangle AKB\sim \triangle CKA&lt;/math&gt;. Using similarity ratios, we can easily find &lt;cmath&gt;AK^2=BK*KC&lt;/cmath&gt; However, since &lt;math&gt;AB=7&lt;/math&gt; and &lt;math&gt;CA=9&lt;/math&gt;, we can use similarity ratios to get &lt;cmath&gt;BK=\frac{7}{9}AK, CK=\frac{9}{7}AK&lt;/cmath&gt; Now we use Law of Cosines on &lt;math&gt;\triangle AKB&lt;/math&gt;: From reverse Law of Cosines, &lt;math&gt;\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}&lt;/math&gt;. This gives us &lt;cmath&gt;AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49&lt;/cmath&gt; &lt;cmath&gt;\implies \frac{196}{81}AK^2=49&lt;/cmath&gt; &lt;cmath&gt;AK=\frac{9}{2}&lt;/cmath&gt; so our answer is &lt;math&gt;9+2=\boxed{011}&lt;/math&gt;.<br /> -franchester<br /> <br /> ==Solution 2 (Video)==<br /> <br /> Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_II_Problems/Problem_11&diff=123350 2016 AIME II Problems/Problem 11 2020-05-31T18:27:53Z <p>Firebolt360: /* Solution */</p> <hr /> <div>==Problem==<br /> For positive integers &lt;math&gt;N&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt;, define &lt;math&gt;N&lt;/math&gt; to be &lt;math&gt;k&lt;/math&gt;-nice if there exists a positive integer &lt;math&gt;a&lt;/math&gt; such that &lt;math&gt;a^{k}&lt;/math&gt; has exactly &lt;math&gt;N&lt;/math&gt; positive divisors. Find the number of positive integers less than &lt;math&gt;1000&lt;/math&gt; that are neither &lt;math&gt;7&lt;/math&gt;-nice nor &lt;math&gt;8&lt;/math&gt;-nice.<br /> <br /> ==Solution==<br /> We claim that an integer &lt;math&gt;N&lt;/math&gt; is only &lt;math&gt;k&lt;/math&gt;-nice if and only if &lt;math&gt;N \equiv 1 \pmod k&lt;/math&gt;. By the number of divisors formula, the number of divisors of &lt;math&gt;\prod_{i=1}^n p_i^{a_i}&lt;/math&gt; is &lt;math&gt;\prod_{i=1}^n (a_i+1)&lt;/math&gt;. Since all the &lt;math&gt;a_i&lt;/math&gt;s are divisible by &lt;math&gt;k&lt;/math&gt; in a perfect &lt;math&gt;k&lt;/math&gt; power, the only if part of the claim follows. To show that all numbers &lt;math&gt;N \equiv 1 \pmod k&lt;/math&gt; are &lt;math&gt;k&lt;/math&gt;-nice, write &lt;math&gt;N=bk+1&lt;/math&gt;. Note that &lt;math&gt;2^{kb}&lt;/math&gt; has the desired number of factors and is a perfect kth power. By PIE, the number of positive integers less than &lt;math&gt;1000&lt;/math&gt; that are either &lt;math&gt;1 \pmod 7&lt;/math&gt; or &lt;math&gt;1\pmod 8&lt;/math&gt; is &lt;math&gt;143+125-18=250&lt;/math&gt;, so the desired answer is &lt;math&gt;999-250=\boxed{749}&lt;/math&gt;.<br /> <br /> Solution by Shaddoll and firebolt360<br /> <br /> ==Solution 2==<br /> All integers &lt;math&gt;a&lt;/math&gt; will have factorization &lt;math&gt;2^a3^b5^c7^d...&lt;/math&gt;. Therefore, the number of factors in &lt;math&gt;a^7&lt;/math&gt; is &lt;math&gt;(7a+1)(7b+1)...&lt;/math&gt;, and for &lt;math&gt;a^8&lt;/math&gt; is &lt;math&gt;(8a+1)(8b+1)...&lt;/math&gt;. The most salient step afterwards is to realize that all numbers &lt;math&gt;N&lt;/math&gt; not &lt;math&gt;1 \pmod{7}&lt;/math&gt; and also not &lt;math&gt;1 \pmod{8}&lt;/math&gt; satisfy the criterion. The cycle repeats every &lt;math&gt;56&lt;/math&gt; integers, and by PIE, &lt;math&gt;7+8-1=14&lt;/math&gt; of them are either &lt;math&gt;7&lt;/math&gt;-nice or &lt;math&gt;8&lt;/math&gt;-nice or both. Therefore, we can take &lt;math&gt;\frac{42}{56} * 1008 = 756&lt;/math&gt; numbers minus the &lt;math&gt;7&lt;/math&gt; that work between &lt;math&gt;1000-1008&lt;/math&gt; inclusive, to get &lt;math&gt;\boxed{749}&lt;/math&gt; positive integers less than &lt;math&gt;1000&lt;/math&gt; that are not nice for &lt;math&gt;k=7, 8&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2016|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category: Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_3&diff=123326 2015 AIME I Problems/Problem 3 2020-05-30T17:50:09Z <p>Firebolt360: /* Solution 2 (Similar to 1) */</p> <hr /> <div>==Problem==<br /> <br /> There is a prime number &lt;math&gt;p&lt;/math&gt; such that &lt;math&gt;16p+1&lt;/math&gt; is the cube of a positive integer. Find &lt;math&gt;p&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Let the positive integer mentioned be &lt;math&gt;a&lt;/math&gt;, so that &lt;math&gt;a^3 = 16p+1&lt;/math&gt;. Note that &lt;math&gt;a&lt;/math&gt; must be odd, because &lt;math&gt;16p+1&lt;/math&gt; is odd.<br /> <br /> Rearrange this expression and factor the left side (this factoring can be done using &lt;math&gt;(a^3-b^3) = (a-b)(a^2+a b+b^2)&lt;/math&gt; or synthetic divison once it is realized that &lt;math&gt;a = 1&lt;/math&gt; is a root):<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> a^3-1 &amp;= 16p\\<br /> (a-1)(a^2+a+1) &amp;= 16p\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;a&lt;/math&gt; is odd, &lt;math&gt;a-1&lt;/math&gt; is even and &lt;math&gt;a^2+a+1&lt;/math&gt; is odd. If &lt;math&gt;a^2+a+1&lt;/math&gt; is odd, &lt;math&gt;a-1&lt;/math&gt; must be some multiple of &lt;math&gt;16&lt;/math&gt;. However, for &lt;math&gt;a-1&lt;/math&gt; to be any multiple of &lt;math&gt;16&lt;/math&gt; other than &lt;math&gt;16&lt;/math&gt; would mean &lt;math&gt;p&lt;/math&gt; is not a prime. Therefore, &lt;math&gt;a-1 = 16&lt;/math&gt; and &lt;math&gt;a = 17&lt;/math&gt;.<br /> <br /> Then our other factor, &lt;math&gt;a^2+a+1&lt;/math&gt;, is the prime &lt;math&gt;p&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> (a-1)(a^2+a+1) &amp;= 16p\\<br /> (17-1)(17^2+17+1) &amp;=16p\\<br /> p = 289+17+1 &amp;= \boxed{307}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> ==Solution 2 (Similar to 1)==<br /> <br /> Observe that this is the same as &lt;math&gt;16p+1=n^3&lt;/math&gt; for some integer &lt;math&gt;n&lt;/math&gt;.<br /> So:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> 16p &amp;= n^3-1\\<br /> 16p &amp;= n^3-1^3\\<br /> 16p &amp;= (n-1)(n^2+n+1)\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Observe that either &lt;math&gt;p=n-1&lt;/math&gt; or &lt;math&gt;p=n^2+n+1&lt;/math&gt; because &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;16&lt;/math&gt; share no factors (&lt;math&gt;p&lt;/math&gt; can't be &lt;math&gt;2&lt;/math&gt;).<br /> Let &lt;math&gt;p=n-1&lt;/math&gt;.<br /> Then:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> p &amp;= n-1\\<br /> 16 &amp;= n^2+n+1\\<br /> n^2+n &amp;= 15\\<br /> n(n+1) &amp;= 15\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Which is obviously impossible for integer n. So &lt;math&gt;p=n^2+n+1&lt;/math&gt; and<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> 16 &amp;= n-1\\<br /> n &amp;= 17\\<br /> p &amp;= 17^2+17+1\\<br /> p = 289+17+1 &amp;= \boxed{307}\\<br /> \end{align*}&lt;/cmath&gt; - firebolt360<br /> <br /> ==Solution 3==<br /> <br /> Since &lt;math&gt;16p+1&lt;/math&gt; is odd, let &lt;math&gt;16p+1 = (2a+1)^3&lt;/math&gt;. Therefore, &lt;math&gt;16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1&lt;/math&gt;. From this, we get &lt;math&gt;8p=a(4a^2+6a+3)&lt;/math&gt;. We know &lt;math&gt;p&lt;/math&gt; is a prime number and it is not an even number. Since &lt;math&gt;4a^2+6a+3&lt;/math&gt; is an odd number, we know that &lt;math&gt;a=8&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> Let &lt;math&gt;16p+1=a^3&lt;/math&gt;. Realize that &lt;math&gt;a&lt;/math&gt; congruent to &lt;math&gt;1\mod 4&lt;/math&gt;, so let &lt;math&gt;a=4n+1&lt;/math&gt;. Expansion, then division by 4, gets &lt;math&gt;16n^3+12n^2+3n=4p&lt;/math&gt;. Clearly &lt;math&gt;n=4m&lt;/math&gt; for some &lt;math&gt;m&lt;/math&gt;. Substitution and another division by 4 gets &lt;math&gt;256m^3+48m^4+3m=p&lt;/math&gt;. Since &lt;math&gt;p&lt;/math&gt; is prime and there is a factor of &lt;math&gt;m&lt;/math&gt; in the LHS, &lt;math&gt;m=1&lt;/math&gt;. Therefore, &lt;math&gt;p=\boxed{307}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> <br /> Notice that &lt;math&gt;16p+1&lt;/math&gt; must be in the form &lt;math&gt;(a+1)^3 = a^3 + 3a^2 + 3a + 1&lt;/math&gt;. Thus &lt;math&gt;16p = a^3 + 3a^2 + 3a&lt;/math&gt;, or &lt;math&gt;16p = a\cdot (a^2 + 3a + 3)&lt;/math&gt;. Since &lt;math&gt;p&lt;/math&gt; must be prime, we either have &lt;math&gt;p = a&lt;/math&gt; or &lt;math&gt;a = 16&lt;/math&gt;. Upon further inspection and/or using the quadratic formula, we can deduce &lt;math&gt;p \neq a&lt;/math&gt;. Thus we have &lt;math&gt;a = 16&lt;/math&gt;, and &lt;math&gt;p = 16^2 + 3\cdot 16 + 3 = \boxed{307}&lt;/math&gt;.<br /> <br /> ==Solution 6==<br /> Notice that the cube 16p+1 is equal to is congruent to 1 mod 16. The only cubic numbers that leave a residue of 1 mod 16 are 1 and 15. <br /> Case one: The cube is of the form 16k+1--&gt;Plugging this in, and taking note that p is prime and has only 1 factor gives p=307<br /> Case two: The cube is of the form 16k+15--&gt; Plugging this in, we quickly realize that this case is invalid, as that implies p is even, and p=2 doesn't work here<br /> <br /> Hence, &lt;math&gt;p=\boxed{307}&lt;/math&gt; is our only answer<br /> <br /> <br /> pi_is_3.141<br /> <br /> == See also ==<br /> {{AIME box|year=2015|n=I|num-b=2|num-a=4}}<br /> {{MAA Notice}}<br /> [[Category:Introductory Number Theory Problems]]</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_3&diff=123325 2015 AIME I Problems/Problem 3 2020-05-30T17:48:30Z <p>Firebolt360: </p> <hr /> <div>==Problem==<br /> <br /> There is a prime number &lt;math&gt;p&lt;/math&gt; such that &lt;math&gt;16p+1&lt;/math&gt; is the cube of a positive integer. Find &lt;math&gt;p&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Let the positive integer mentioned be &lt;math&gt;a&lt;/math&gt;, so that &lt;math&gt;a^3 = 16p+1&lt;/math&gt;. Note that &lt;math&gt;a&lt;/math&gt; must be odd, because &lt;math&gt;16p+1&lt;/math&gt; is odd.<br /> <br /> Rearrange this expression and factor the left side (this factoring can be done using &lt;math&gt;(a^3-b^3) = (a-b)(a^2+a b+b^2)&lt;/math&gt; or synthetic divison once it is realized that &lt;math&gt;a = 1&lt;/math&gt; is a root):<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> a^3-1 &amp;= 16p\\<br /> (a-1)(a^2+a+1) &amp;= 16p\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;a&lt;/math&gt; is odd, &lt;math&gt;a-1&lt;/math&gt; is even and &lt;math&gt;a^2+a+1&lt;/math&gt; is odd. If &lt;math&gt;a^2+a+1&lt;/math&gt; is odd, &lt;math&gt;a-1&lt;/math&gt; must be some multiple of &lt;math&gt;16&lt;/math&gt;. However, for &lt;math&gt;a-1&lt;/math&gt; to be any multiple of &lt;math&gt;16&lt;/math&gt; other than &lt;math&gt;16&lt;/math&gt; would mean &lt;math&gt;p&lt;/math&gt; is not a prime. Therefore, &lt;math&gt;a-1 = 16&lt;/math&gt; and &lt;math&gt;a = 17&lt;/math&gt;.<br /> <br /> Then our other factor, &lt;math&gt;a^2+a+1&lt;/math&gt;, is the prime &lt;math&gt;p&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> (a-1)(a^2+a+1) &amp;= 16p\\<br /> (17-1)(17^2+17+1) &amp;=16p\\<br /> p = 289+17+1 &amp;= \boxed{307}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> ==Solution 2 (Similar to 1)==<br /> <br /> Observe that this is the same as &lt;math&gt;16p+1=n^3&lt;/math&gt; for some integer &lt;math&gt;n&lt;/math&gt;.<br /> So:<br /> &lt;cmath&gt;\begin{align*}<br /> 16p &amp;= n^3-1\\<br /> 16p &amp;= n^3-1^3\\<br /> 16p &amp;= (n-1)(n^2+n+1)\\<br /> \end{align*}&lt;/cmath&gt;<br /> Observe that either &lt;math&gt;p=n-1&lt;/math&gt; or &lt;math&gt;p=n^2+n+1&lt;/math&gt; because &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;16&lt;/math&gt; share no factors (&lt;math&gt;p&lt;/math&gt; can't be &lt;math&gt;2&lt;/math&gt;).<br /> Let &lt;math&gt;p=n-1&lt;/math&gt;.<br /> Then:<br /> &lt;cmath&gt;\begin{align*}<br /> p &amp;= n-1\\<br /> 16 &amp;= n^2+n+1\\<br /> n^2+n &amp;= 15\\<br /> n(n+1) &amp;= 15\\<br /> \end{align*}&lt;/cmath&gt;<br /> Which is obviously impossible for integer n. So &lt;math&gt;p=n^2+n+1&lt;/math&gt; and<br /> &lt;cmath&gt;\begin{align*}<br /> 16 &amp;= n-1\\<br /> n &amp;= 17\\<br /> p &amp;= 17^2+17+1\\<br /> p = 289+17+1 &amp;= \boxed{307}\\<br /> \end{align*}&lt;/cmath&gt; - firebolt360<br /> <br /> ==Solution 3==<br /> <br /> Since &lt;math&gt;16p+1&lt;/math&gt; is odd, let &lt;math&gt;16p+1 = (2a+1)^3&lt;/math&gt;. Therefore, &lt;math&gt;16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1&lt;/math&gt;. From this, we get &lt;math&gt;8p=a(4a^2+6a+3)&lt;/math&gt;. We know &lt;math&gt;p&lt;/math&gt; is a prime number and it is not an even number. Since &lt;math&gt;4a^2+6a+3&lt;/math&gt; is an odd number, we know that &lt;math&gt;a=8&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> Let &lt;math&gt;16p+1=a^3&lt;/math&gt;. Realize that &lt;math&gt;a&lt;/math&gt; congruent to &lt;math&gt;1\mod 4&lt;/math&gt;, so let &lt;math&gt;a=4n+1&lt;/math&gt;. Expansion, then division by 4, gets &lt;math&gt;16n^3+12n^2+3n=4p&lt;/math&gt;. Clearly &lt;math&gt;n=4m&lt;/math&gt; for some &lt;math&gt;m&lt;/math&gt;. Substitution and another division by 4 gets &lt;math&gt;256m^3+48m^4+3m=p&lt;/math&gt;. Since &lt;math&gt;p&lt;/math&gt; is prime and there is a factor of &lt;math&gt;m&lt;/math&gt; in the LHS, &lt;math&gt;m=1&lt;/math&gt;. Therefore, &lt;math&gt;p=\boxed{307}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> <br /> Notice that &lt;math&gt;16p+1&lt;/math&gt; must be in the form &lt;math&gt;(a+1)^3 = a^3 + 3a^2 + 3a + 1&lt;/math&gt;. Thus &lt;math&gt;16p = a^3 + 3a^2 + 3a&lt;/math&gt;, or &lt;math&gt;16p = a\cdot (a^2 + 3a + 3)&lt;/math&gt;. Since &lt;math&gt;p&lt;/math&gt; must be prime, we either have &lt;math&gt;p = a&lt;/math&gt; or &lt;math&gt;a = 16&lt;/math&gt;. Upon further inspection and/or using the quadratic formula, we can deduce &lt;math&gt;p \neq a&lt;/math&gt;. Thus we have &lt;math&gt;a = 16&lt;/math&gt;, and &lt;math&gt;p = 16^2 + 3\cdot 16 + 3 = \boxed{307}&lt;/math&gt;.<br /> <br /> ==Solution 6==<br /> Notice that the cube 16p+1 is equal to is congruent to 1 mod 16. The only cubic numbers that leave a residue of 1 mod 16 are 1 and 15. <br /> Case one: The cube is of the form 16k+1--&gt;Plugging this in, and taking note that p is prime and has only 1 factor gives p=307<br /> Case two: The cube is of the form 16k+15--&gt; Plugging this in, we quickly realize that this case is invalid, as that implies p is even, and p=2 doesn't work here<br /> <br /> Hence, &lt;math&gt;p=\boxed{307}&lt;/math&gt; is our only answer<br /> <br /> <br /> pi_is_3.141<br /> <br /> == See also ==<br /> {{AIME box|year=2015|n=I|num-b=2|num-a=4}}<br /> {{MAA Notice}}<br /> [[Category:Introductory Number Theory Problems]]</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_12&diff=118914 2012 AIME II Problems/Problem 12 2020-03-08T19:12:45Z <p>Firebolt360: /* Problem 12 */</p> <hr /> <div>== Problem 12 ==<br /> For a positive integer &lt;math&gt;p&lt;/math&gt;, define the positive integer &lt;math&gt;n&lt;/math&gt; to be &lt;math&gt;p&lt;/math&gt;''-safe'' if &lt;math&gt;n&lt;/math&gt; differs in absolute value by more than &lt;math&gt;2&lt;/math&gt; from all multiples of &lt;math&gt;p&lt;/math&gt;. For example, the set of &lt;math&gt;10&lt;/math&gt;-safe numbers is &lt;math&gt;\{ 3, 4, 5, 6, 7, 13, 14, 15, 16, 17, 23, \ldots\}&lt;/math&gt;. Find the number of positive integers less than or equal to &lt;math&gt;10,000&lt;/math&gt; which are simultaneously &lt;math&gt;7&lt;/math&gt;-safe, &lt;math&gt;11&lt;/math&gt;-safe, and &lt;math&gt;13&lt;/math&gt;-safe.<br /> <br /> == Solution ==<br /> <br /> We see that a number &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;p&lt;/math&gt;-safe if and only if the residue of &lt;math&gt;n \mod p&lt;/math&gt; is greater than &lt;math&gt;2&lt;/math&gt; and less than &lt;math&gt;p-2&lt;/math&gt;; thus, there are &lt;math&gt;p-5&lt;/math&gt; residues &lt;math&gt;\mod p&lt;/math&gt; that a &lt;math&gt;p&lt;/math&gt;-safe number can have. Therefore, a number &lt;math&gt;n&lt;/math&gt; satisfying the conditions of the problem can have &lt;math&gt;2&lt;/math&gt; different residues &lt;math&gt;\mod 7&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt; different residues &lt;math&gt;\mod 11&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt; different residues &lt;math&gt;\mod 13&lt;/math&gt;. The Chinese Remainder Theorem states that for a number &lt;math&gt;x&lt;/math&gt; that is<br /> &lt;math&gt;a&lt;/math&gt; (mod b)<br /> &lt;math&gt;c&lt;/math&gt; (mod d)<br /> &lt;math&gt;e&lt;/math&gt; (mod f)<br /> has one solution if &lt;math&gt;gcd(b,d,f)=1&lt;/math&gt;. For example, in our case, the number &lt;math&gt;n&lt;/math&gt; can be:<br /> 3 (mod 7)<br /> 3 (mod 11)<br /> 7 (mod 13)<br /> so since &lt;math&gt;gcd(7,11,13)&lt;/math&gt;=1, there is 1 solution for n for this case of residues of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> This means that by the Chinese Remainder Theorem, &lt;math&gt;n&lt;/math&gt; can have &lt;math&gt;2\cdot 6 \cdot 8 = 96&lt;/math&gt; different residues mod &lt;math&gt;7 \cdot 11 \cdot 13 = 1001&lt;/math&gt;. Thus, there are &lt;math&gt;960&lt;/math&gt; values of &lt;math&gt;n&lt;/math&gt; satisfying the conditions in the range &lt;math&gt;0 \le n &lt; 10010&lt;/math&gt;. However, we must now remove any values greater than &lt;math&gt;10000&lt;/math&gt; that satisfy the conditions. By checking residues, we easily see that the only such values are &lt;math&gt;10006&lt;/math&gt; and &lt;math&gt;10007&lt;/math&gt;, so there remain &lt;math&gt;\fbox{958}&lt;/math&gt; values satisfying the conditions of the problem.<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_14&diff=118913 2012 AIME II Problems/Problem 14 2020-03-08T19:05:28Z <p>Firebolt360: /* Solution */</p> <hr /> <div>== Problem 14 ==<br /> In a group of nine people each person shakes hands with exactly two of the other people from the group. Let &lt;math&gt;N&lt;/math&gt; be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> <br /> == Solution 1==<br /> Given that each person shakes hands with two people, we can view all of these through graph theory as 'rings'. This will split it into four cases: Three rings of three, one ring of three and one ring of six, one ring of four and one ring of five, and one ring of nine. (All other cases that sum to nine won't work, since they have at least one 'ring' of two or fewer points, which doesn't satisfy the handshaking conditions of the problem.)<br /> <br /> '''Case 1:''' To create our groups of three, there are &lt;math&gt;\dfrac{\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{3!}&lt;/math&gt;. In general, the number of ways we can arrange people within the rings to count properly is &lt;math&gt;\dfrac{(n-1)!}{2}&lt;/math&gt;, since there are &lt;math&gt;(n-1)!&lt;/math&gt; ways to arrange items in the circle, and then we don't want to want to consider reflections as separate entities. Thus, each of the three cases has &lt;math&gt;\dfrac{(3-1)!}{2}=1&lt;/math&gt; arrangements. Therefore, for this case, there are &lt;math&gt;\left(\dfrac{\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{3!}\right)(1)^3=280&lt;/math&gt;<br /> <br /> '''Case 2:''' For three and six, there are &lt;math&gt;\dbinom{9}{6}=84&lt;/math&gt; sets for the rings. For organization within the ring, as before, there is only one way to arrange the ring of three. For six, there is &lt;math&gt;\dfrac{(6-1)!}{2}=60&lt;/math&gt;. This means there are &lt;math&gt;(84)(1)(60)=5040&lt;/math&gt; arrangements.<br /> <br /> '''Case 3:''' For four and five, there are &lt;math&gt;\dbinom{9}{5}=126&lt;/math&gt; sets for the rings. Within the five, there are &lt;math&gt;\dfrac{4!}{2}=12&lt;/math&gt;, and within the four there are &lt;math&gt;\dfrac{3!}{2}=3&lt;/math&gt; arrangements. This means the total is &lt;math&gt;(126)(12)(3)=4536&lt;/math&gt;.<br /> <br /> '''Case 4:''' For the nine case, there is &lt;math&gt;\dbinom{9}{9}=1&lt;/math&gt; arrangement for the ring. Within it, there are &lt;math&gt;\dfrac{8!}{2}=20160&lt;/math&gt; arrangements.<br /> <br /> Summing the cases, we have &lt;math&gt;280+5040+4536+20160=30016 \to \boxed{016}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Video Solution)==<br /> Very Neat solution: https://www.youtube.com/watch?v=lG8N9RuI-8o<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_14&diff=118912 2012 AIME II Problems/Problem 14 2020-03-08T19:05:09Z <p>Firebolt360: /* Solution */</p> <hr /> <div>== Problem 14 ==<br /> In a group of nine people each person shakes hands with exactly two of the other people from the group. Let &lt;math&gt;N&lt;/math&gt; be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> <br /> == Solution ==<br /> Given that each person shakes hands with two people, we can view all of these through graph theory as 'rings'. This will split it into four cases: Three rings of three, one ring of three and one ring of six, one ring of four and one ring of five, and one ring of nine. (All other cases that sum to nine won't work, since they have at least one 'ring' of two or fewer points, which doesn't satisfy the handshaking conditions of the problem.)<br /> <br /> '''Case 1:''' To create our groups of three, there are &lt;math&gt;\dfrac{\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{3!}&lt;/math&gt;. In general, the number of ways we can arrange people within the rings to count properly is &lt;math&gt;\dfrac{(n-1)!}{2}&lt;/math&gt;, since there are &lt;math&gt;(n-1)!&lt;/math&gt; ways to arrange items in the circle, and then we don't want to want to consider reflections as separate entities. Thus, each of the three cases has &lt;math&gt;\dfrac{(3-1)!}{2}=1&lt;/math&gt; arrangements. Therefore, for this case, there are &lt;math&gt;\left(\dfrac{\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{3!}\right)(1)^3=280&lt;/math&gt;<br /> <br /> '''Case 2:''' For three and six, there are &lt;math&gt;\dbinom{9}{6}=84&lt;/math&gt; sets for the rings. For organization within the ring, as before, there is only one way to arrange the ring of three. For six, there is &lt;math&gt;\dfrac{(6-1)!}{2}=60&lt;/math&gt;. This means there are &lt;math&gt;(84)(1)(60)=5040&lt;/math&gt; arrangements.<br /> <br /> '''Case 3:''' For four and five, there are &lt;math&gt;\dbinom{9}{5}=126&lt;/math&gt; sets for the rings. Within the five, there are &lt;math&gt;\dfrac{4!}{2}=12&lt;/math&gt;, and within the four there are &lt;math&gt;\dfrac{3!}{2}=3&lt;/math&gt; arrangements. This means the total is &lt;math&gt;(126)(12)(3)=4536&lt;/math&gt;.<br /> <br /> '''Case 4:''' For the nine case, there is &lt;math&gt;\dbinom{9}{9}=1&lt;/math&gt; arrangement for the ring. Within it, there are &lt;math&gt;\dfrac{8!}{2}=20160&lt;/math&gt; arrangements.<br /> <br /> Summing the cases, we have &lt;math&gt;280+5040+4536+20160=30016 \to \boxed{016}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Video Solution)==<br /> Very Neat solution: https://www.youtube.com/watch?v=lG8N9RuI-8o<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_24&diff=112636 2014 AMC 10B Problems/Problem 24 2019-12-07T02:09:09Z <p>Firebolt360: </p> <hr /> <div>{{duplicate|[[2014 AMC 12B Problems|2014 AMC 12B #18]] and [[2014 AMC 10B Problems|2014 AMC 10B #24]]}}<br /> <br /> ==Problem==<br /> The numbers &lt;math&gt;1, 2, 3, 4, 5&lt;/math&gt; are to be arranged in a circle. An arrangement is &lt;math&gt;\textit{bad}&lt;/math&gt; if it is not true that for every &lt;math&gt;n&lt;/math&gt; from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;15&lt;/math&gt; one can find a subset of the numbers that appear consecutively on the circle that sum to &lt;math&gt;n&lt;/math&gt;. Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?<br /> <br /> &lt;math&gt; \textbf {(A) } 1 \qquad \textbf {(B) } 2 \qquad \textbf {(C) } 3 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 5 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We see that there are &lt;math&gt;5!&lt;/math&gt; total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number &lt;math&gt;1&lt;/math&gt; is always at the top of the circle. Thus, there are only &lt;math&gt;4!&lt;/math&gt; ways under rotation. Every case has exactly 1 reflection, so that gives us only &lt;math&gt;4!/2&lt;/math&gt;, or 12 cases, which is not difficult to list out. We systematically list out all &lt;math&gt;12&lt;/math&gt; cases. <br /> <br /> Now, we must examine if they satisfy the conditions. We can see that by choosing one number at a time, we can always obtain subsets with sums &lt;math&gt;1, 2, 3, 4,&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;. By choosing the full circle, we can obtain &lt;math&gt;15&lt;/math&gt;. By choosing everything except for &lt;math&gt;1, 2, 3, 4,&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;, we can obtain subsets with sums of &lt;math&gt;10, 11, 12, 13,&lt;/math&gt; and &lt;math&gt;14&lt;/math&gt;. <br /> <br /> This means that we now only need to check for &lt;math&gt;6, 7, 8,&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt;. However, once we have found a set summing to &lt;math&gt;6&lt;/math&gt;, we can choose everything else and obtain a set summing to &lt;math&gt;9&lt;/math&gt;, and similarly for &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;. Thus, we only need to check each case for whether or not we can obtain &lt;math&gt;6&lt;/math&gt; or &lt;math&gt;7&lt;/math&gt;. <br /> <br /> We can make &lt;math&gt;6&lt;/math&gt; by having &lt;math&gt;4, 2&lt;/math&gt;, or &lt;math&gt;3, 2, 1&lt;/math&gt;, or &lt;math&gt;5, 1&lt;/math&gt;. We can start with the group of three. To separate &lt;math&gt;3, 2, 1&lt;/math&gt; from each other, they must be grouped two together and one separate, like this.<br /> <br /> &lt;asy&gt;<br /> draw(circle((0, 0), 5));<br /> pair O, A, B, C, D, E;<br /> O=origin;<br /> A=(0, 5);<br /> B=rotate(72)*A;<br /> C=rotate(144)*A;<br /> D=rotate(216)*A;<br /> E=rotate(288)*A;<br /> label(&quot;x$&quot;, A, N);<br /> label(&quot;$y$&quot;, C, SW);<br /> label(&quot;$z$&quot;, D, SE);<br /> &lt;/asy&gt;<br /> <br /> Now, we note that &lt;math&gt;x&lt;/math&gt; is next to both blank spots, so we can't have a number from one of the pairs. So since we can't have &lt;math&gt;1&lt;/math&gt;, because it is part of the &lt;math&gt;5, 1&lt;/math&gt; pair, and we can't have &lt;math&gt;2&lt;/math&gt; there, because it's part of the &lt;math&gt;4, 2&lt;/math&gt; pair, we must have &lt;math&gt;3&lt;/math&gt; inserted into the &lt;math&gt;x&lt;/math&gt; spot. We can insert &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; in &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; interchangeably, since reflections are considered the same.<br /> <br /> &lt;asy&gt;<br /> draw(circle((0, 0), 5));<br /> pair O, A, B, C, D, E;<br /> O=origin;<br /> A=(0, 5);<br /> B=rotate(72)*A;<br /> C=rotate(144)*A;<br /> D=rotate(216)*A;<br /> E=rotate(288)*A;<br /> label(&quot;$3$&quot;, A, N);<br /> label(&quot;$2$&quot;, C, SW);<br /> label(&quot;$1$&quot;, D, SE);<br /> &lt;/asy&gt;<br /> <br /> We have &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt; left to insert. We can't place the &lt;math&gt;4&lt;/math&gt; next to the &lt;math&gt;2&lt;/math&gt; or the &lt;math&gt;5&lt;/math&gt; next to the &lt;math&gt;1&lt;/math&gt;, so we must place &lt;math&gt;4&lt;/math&gt; next to the &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt; next to the &lt;math&gt;2&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> draw(circle((0, 0), 5));<br /> pair O, A, B, C, D, E;<br /> O=origin;<br /> A=(0, 5);<br /> B=rotate(72)*A;<br /> C=rotate(144)*A;<br /> D=rotate(216)*A;<br /> E=rotate(288)*A;<br /> label(&quot;$3$&quot;, A, N);<br /> label(&quot;$5$&quot;, B, NW);<br /> label(&quot;$2$&quot;, C, SW);<br /> label(&quot;$1$&quot;, D, SE);<br /> label(&quot;$4$&quot;, E, NE);<br /> &lt;/asy&gt;<br /> <br /> This is the only solution to make &lt;math&gt;6&lt;/math&gt; &quot;bad.&quot;<br /> <br /> Next we move on to &lt;math&gt;7&lt;/math&gt;, which can be made by &lt;math&gt;3, 4&lt;/math&gt;, or &lt;math&gt;5, 2&lt;/math&gt;, or &lt;math&gt;4, 2, 1&lt;/math&gt;. We do this the same way as before. We start with the three group. Since we can't have 4 or 2 in the top slot, we must have one there, and 4 and 2 are next to each other on the bottom. When we have &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt; left to insert, we place them such that we don't have the two pairs adjacent.<br /> <br /> &lt;asy&gt;<br /> draw(circle((0, 0), 5));<br /> pair O, A, B, C, D, E;<br /> O=origin;<br /> A=(0, 5);<br /> B=rotate(72)*A;<br /> C=rotate(144)*A;<br /> D=rotate(216)*A;<br /> E=rotate(288)*A;<br /> label(&quot;$1$&quot;, A, N);<br /> label(&quot;$3$&quot;, B, NW);<br /> label(&quot;$2$&quot;, C, SW);<br /> label(&quot;$4$&quot;, D, SE);<br /> label(&quot;$5\$&quot;, E, NE);<br /> &lt;/asy&gt;<br /> <br /> This is the only solution to make &lt;math&gt;7&lt;/math&gt; &quot;bad.&quot;<br /> <br /> We've covered all needed cases, and the two examples we found are distinct, therefore the answer is &lt;math&gt;\boxed{\textbf {(B) }2}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2014|ab=B|num-b=23|num-a=25}}<br /> <br /> {{AMC12 box|year=2014|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A&diff=111378 2007 AMC 10A 2019-11-13T15:29:48Z <p>Firebolt360: </p> <hr /> <div>'''2007 AMC 10A''' problems and solutions. The first link contains the full set of test problems. The rest contain each individual problem and its solution.<br /> * [[2007 AMC 10A Problems]]<br /> * [[2007 AMC 10A Answer Key|Answer Key]]<br /> ** [[2007 AMC 10A Problems/Problem 1|Problem 1]]<br /> ** [[2007 AMC 10A Problems/Problem 2|Problem 2]]<br /> ** [[2007 AMC 10A Problems/Problem 3|Problem 3]]<br /> ** [[2007 AMC 10A Problems/Problem 4|Problem 4]]<br /> ** [[2007 AMC 10A Problems/Problem 5|Problem 5]]<br /> ** [[2007 AMC 10A Problems/Problem 6|Problem 6]]<br /> ** [[2007 AMC 10A Problems/Problem 7]]<br /> ** [[2007 AMC 10A Problems/Problem 8]]<br /> ** [[2007 AMC 10A Problems/Problem 9]]<br /> ** [[2007 AMC 10A Problems/Problem 10]]<br /> ** [[2007 AMC 10A Problems/Problem 11]]<br /> ** [[2007 AMC 10A Problems/Problem 12]]<br /> ** [[2007 AMC 10A Problems/Problem 13]]<br /> ** [[2007 AMC 10A Problems/Problem 14]]<br /> ** [[2007 AMC 10A Problems/Problem 15]]<br /> ** [[2007 AMC 10A Problems/Problem 16]]<br /> ** [[2007 AMC 10A Problems/Problem 17]]<br /> ** [[2007 AMC 10A Problems/Problem 18]]<br /> ** [[2007 AMC 10A Problems/Problem 19]]<br /> ** [[2007 AMC 10A Problems/Problem 20]]<br /> ** [[2007 AMC 10A Problems/Problem 21]]<br /> ** [[2007 AMC 10A Problems/Problem 22]]<br /> ** [[2007 AMC 10A Problems/Problem 23]]<br /> ** [[2007 AMC 10A Problems/Problem 24]]<br /> ** [[2007 AMC 10A Problems/Problem 25]]<br /> <br /> == See also ==<br /> {{AMC10 box|year=2007|before=[[2006 AMC 10A|2006 AMC 10A]], [[2006 AMC 10B|B]]|after=[[2008 AMC 10A|2008 AMC 10A]], [[2008 AMC 10B|B]]|ab=A}}<br /> <br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=142 2007 AMC 10A Math Jam Transcript]<br /> * [[Mathematics competition resources]]</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A&diff=111377 2007 AMC 10A 2019-11-13T15:28:29Z <p>Firebolt360: </p> <hr /> <div>'''2007 AMC 10A''' problems and solutions. The first link contains the full set of test problems. The rest contain each individual problem and its solution.<br /> * [[2007 AMC 10A Problems]]<br /> * [[2007 AMC 10A Answer Key|Answer Key]]<br /> ** [[2007 AMC 10A Problems/Problem 1]]<br /> ** [[2007 AMC 10A Problems/Problem 2]]<br /> ** [[2007 AMC 10A Problems/Problem 3]]<br /> ** [[2007 AMC 10A Problems/Problem 4]]<br /> ** [[2007 AMC 10A Problems/Problem 5]]<br /> ** [[2007 AMC 10A Problems/Problem 6]]<br /> ** [[2007 AMC 10A Problems/Problem 7]]<br /> ** [[2007 AMC 10A Problems/Problem 8]]<br /> ** [[2007 AMC 10A Problems/Problem 9]]<br /> ** [[2007 AMC 10A Problems/Problem 10]]<br /> ** [[2007 AMC 10A Problems/Problem 11]]<br /> ** [[2007 AMC 10A Problems/Problem 12]]<br /> ** [[2007 AMC 10A Problems/Problem 13]]<br /> ** [[2007 AMC 10A Problems/Problem 14]]<br /> ** [[2007 AMC 10A Problems/Problem 15]]<br /> ** [[2007 AMC 10A Problems/Problem 16]]<br /> ** [[2007 AMC 10A Problems/Problem 17]]<br /> ** [[2007 AMC 10A Problems/Problem 18]]<br /> ** [[2007 AMC 10A Problems/Problem 19]]<br /> ** [[2007 AMC 10A Problems/Problem 20]]<br /> ** [[2007 AMC 10A Problems/Problem 21]]<br /> ** [[2007 AMC 10A Problems/Problem 22]]<br /> ** [[2007 AMC 10A Problems/Problem 23]]<br /> ** [[2007 AMC 10A Problems/Problem 24]]<br /> ** [[2007 AMC 10A Problems/Problem 25]]<br /> <br /> == See also ==<br /> {{AMC10 box|year=2007|before=[[2006 AMC 10A|2006 AMC 10A]], [[2006 AMC 10B|B]]|after=[[2008 AMC 10A|2008 AMC 10A]], [[2008 AMC 10B|B]]|ab=A}}<br /> <br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=142 2007 AMC 10A Math Jam Transcript]<br /> * [[Mathematics competition resources]]</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_22&diff=111295 2011 AMC 8 Problems/Problem 22 2019-11-12T15:50:53Z <p>Firebolt360: </p> <hr /> <div>==Problem==<br /> What is the tens digit of &lt;math&gt;7^{2011}&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We want the tens digit<br /> So, we take &lt;math&gt;7^{2011}\ (\text{mod }100)&lt;/math&gt;. That is congruent to &lt;math&gt;7^{11}\ (\text{mod}100)&lt;/math&gt;. From here, it is an easy bash, 7, 49, 43, 01, 07, 49, 43, 01, 07, 49, 43. So the answer is &lt;math&gt;\boxed{\textbf{(D)}\ 4}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2011|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_22&diff=111294 2011 AMC 8 Problems/Problem 22 2019-11-12T15:50:19Z <p>Firebolt360: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> What is the tens digit of &lt;math&gt;7^{2011}&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The first couple powers of &lt;math&gt;7&lt;/math&gt; are &lt;math&gt;7, 49, 343, 2401, 16807.&lt;/math&gt; As you can see, the last two digits cycle after every 4 powers. &lt;math&gt;7^{1}\ (\text{mod }100) \equiv 7^{5}\ (\text{mod }100) \equiv 7^{2009}\ (\text{mod }100).&lt;/math&gt; From there, we go two more powers. The last two digits are &lt;math&gt;43&lt;/math&gt; so the tens digit is &lt;math&gt;\boxed{\textbf{(D)}\ 4}&lt;/math&gt;<br /> ==Solution 2==<br /> We want the tens digit<br /> So, we take &lt;math&gt;7^{2011}\ (\text{mod }100)&lt;/math&gt;. That is congruent to &lt;math&gt;7^{11}\ (\text{mod}100)&lt;/math&gt;. From here, it is an easy bash, 7, 49, 43, 01, 07, 49, 43, 01, 07, 49, 43. So the answer is 4<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2011|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_22&diff=111293 2011 AMC 8 Problems/Problem 22 2019-11-12T15:49:32Z <p>Firebolt360: </p> <hr /> <div>==Problem==<br /> What is the tens digit of &lt;math&gt;7^{2011}&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The first couple powers of &lt;math&gt;7&lt;/math&gt; are &lt;math&gt;7, 49, 343, 2401, 16807.&lt;/math&gt; As you can see, the last two digits cycle after every 4 powers. &lt;math&gt;7^{1}\ (\text{mod }100) \equiv 7^{5}\ (\text{mod }100) \equiv 7^{2009}\ (\text{mod }100).&lt;/math&gt; From there, we go two more powers. The last two digits are &lt;math&gt;43&lt;/math&gt; so the tens digit is &lt;math&gt;\boxed{\textbf{(D)}\ 4}&lt;/math&gt;<br /> ==Solution 2==<br /> We want the tens digit<br /> So, we take &lt;math&gt;7^{2009}\ (\text{mod }100)&lt;/math&gt;. That is congruent to &lt;math&gt;7^9\ (\text{mod}100)&lt;/math&gt;. From here, it is an easy bash, 7, 49, 43, 01, 07, 49, 43, 01, 07<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2011|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Firebolt360