https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Zeroman&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-26T22:33:12Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_12&diff=94908 2012 AIME I Problems/Problem 12 2018-06-03T21:10:10Z <p>Zeroman: /* Solutions */</p> <hr /> <div>==Problem 12==<br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be a right triangle with right angle at &lt;math&gt;C.&lt;/math&gt; Let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be points on &lt;math&gt;\overline{AB}&lt;/math&gt; with &lt;math&gt;D&lt;/math&gt; between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; such that &lt;math&gt;\overline{CD}&lt;/math&gt; and &lt;math&gt;\overline{CE}&lt;/math&gt; trisect &lt;math&gt;\angle C.&lt;/math&gt; If &lt;math&gt;\frac{DE}{BE} = \frac{8}{15},&lt;/math&gt; then &lt;math&gt;\tan B&lt;/math&gt; can be written as &lt;math&gt;\frac{m \sqrt{p}}{n},&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, and &lt;math&gt;p&lt;/math&gt; is a positive integer not divisible by the square of any prime. Find &lt;math&gt;m+n+p.&lt;/math&gt;<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> Let &lt;math&gt;CD = 2a&lt;/math&gt;. Using angle bisector theorem on &lt;math&gt;\triangle CDB&lt;/math&gt;, &lt;math&gt;\frac{2a}{8}=\frac{CB}{15}&lt;/math&gt;, so &lt;math&gt;CB = \frac{15a}{4}&lt;/math&gt;. Then, drop the altitude from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;CB&lt;/math&gt; and call the foot &lt;math&gt;F&lt;/math&gt;. Thus, &lt;math&gt;CF = a&lt;/math&gt;, &lt;math&gt;FD = a\sqrt{3}&lt;/math&gt;, and &lt;math&gt;FB = \frac{11a}{4}&lt;/math&gt;. Finally, &lt;math&gt;\tan{B} = \frac{a\sqrt{3}}{\frac{11a}{4}} = \frac{4\sqrt{3}}{11}&lt;/math&gt;. Our answer is &lt;math&gt;\boxed{018}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Without loss of generality, set &lt;math&gt;CB = 1&lt;/math&gt;. Then, by the Angle Bisector Theorem on triangle &lt;math&gt;DCB&lt;/math&gt;, we have &lt;math&gt;CD = \frac{8}{15}&lt;/math&gt;. We apply the Law of Cosines to triangle &lt;math&gt;DCB&lt;/math&gt; to get &lt;math&gt;1 + \frac{64}{225} - \frac{8}{15} = BD^{2}&lt;/math&gt;, which we can simplify to get &lt;math&gt;BD = \frac{13}{15}&lt;/math&gt;. <br /> <br /> Now, we have &lt;math&gt;\cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}&lt;/math&gt; by another application of the Law of Cosines to triangle &lt;math&gt;DCB&lt;/math&gt;, so &lt;math&gt;\cos \angle B = \frac{11}{13}&lt;/math&gt;. In addition, &lt;math&gt;\sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}&lt;/math&gt;, so &lt;math&gt;\tan \angle B = \frac{4\sqrt{3}}{11}&lt;/math&gt;. <br /> <br /> Our final answer is &lt;math&gt;4+3+11 = \boxed{018}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> (This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig)<br /> <br /> Find values for all angles in terms of &lt;math&gt;\angle B&lt;/math&gt;. &lt;math&gt;\angle CEB = 150-B&lt;/math&gt;, &lt;math&gt;\angle CED = 30+B&lt;/math&gt;, &lt;math&gt;\angle CDE = 120-B&lt;/math&gt;, &lt;math&gt;\angle CDA = 60+B&lt;/math&gt;, and &lt;math&gt;\angle A = 90-B&lt;/math&gt;.<br /> <br /> Use the law of sines on &lt;math&gt;\triangle CED&lt;/math&gt; and &lt;math&gt;\triangle CEB&lt;/math&gt;:<br /> <br /> In &lt;math&gt;\triangle CED&lt;/math&gt;, &lt;math&gt;\frac{8}{\sin 30} = \frac{CE}{\sin (120-B)}&lt;/math&gt;. This simplifies to &lt;math&gt;16 = \frac{CE}{\sin (120-B)}&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\triangle CEB&lt;/math&gt;, &lt;math&gt;\frac{15}{\sin 30} = \frac{CE}{\sin B}&lt;/math&gt;. This simplifies to &lt;math&gt;30 = \frac{CE}{\sin B}&lt;/math&gt;.<br /> <br /> Solve for &lt;math&gt;CE&lt;/math&gt; and equate them so that you get &lt;math&gt;16\sin (120-B) = 30\sin B&lt;/math&gt;. <br /> <br /> From this, &lt;math&gt;\frac{8}{15} = \frac{\sin B}{\sin (120-B)}&lt;/math&gt;. <br /> <br /> Use a trig identity on the denominator on the right to obtain: &lt;math&gt;\frac{8}{15} = \frac{\sin B}{\sin 120 \cos B - \cos 120 \sin B}&lt;/math&gt;<br /> <br /> This simplifies to &lt;math&gt;\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \cos B + \sin B}{2}} = \frac{2\sin B}{\sqrt{3}\cos B + \sin B}&lt;/math&gt;<br /> <br /> This gives &lt;math&gt;8\sqrt{3}\cos B+8\sin B=30\sin B&lt;/math&gt;<br /> Dividing by &lt;math&gt;\cos B&lt;/math&gt;, we have &lt;math&gt;{8\sqrt{3}}= 22\tan B&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan{B} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}&lt;/math&gt;. Our final answer is &lt;math&gt;4 + 3 + 11 = \boxed{018}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> (This solution avoids advanced trigonometry)<br /> <br /> Let &lt;math&gt;X&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;Y&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;E&lt;/math&gt; to &lt;math&gt;\overline{BC}&lt;/math&gt;.<br /> <br /> Now let &lt;math&gt;EY=x&lt;/math&gt;. Clearly, triangles &lt;math&gt;EYB&lt;/math&gt; and &lt;math&gt;DXB&lt;/math&gt; are similar with &lt;math&gt;\frac{BE}{BD}=\frac{15}{15+8}=\frac{15}{23}=\frac{EY}{DX}&lt;/math&gt;, so &lt;math&gt;DX=\frac{23}{15}x&lt;/math&gt;.<br /> <br /> Since triangles &lt;math&gt;CDX&lt;/math&gt; and &lt;math&gt;CEY&lt;/math&gt; are 30-60-90 right triangles, we can easily find other lengths in terms of &lt;math&gt;x&lt;/math&gt;. For example, we see that &lt;math&gt;CY=x\sqrt{3}&lt;/math&gt; and &lt;math&gt;CX=\frac{\frac{23}{15}x}{\sqrt{3}}=\frac{23\sqrt{3}}{45}x&lt;/math&gt;. Therefore &lt;math&gt;XY=CY-CX=x\sqrt{3}-\frac{23\sqrt{3}}{45}x=\frac{22\sqrt{3}}{45}x&lt;/math&gt;.<br /> <br /> Again using the fact that triangles &lt;math&gt;EYB&lt;/math&gt; and &lt;math&gt;DXB&lt;/math&gt; are similar, we see that &lt;math&gt;\frac{BX}{BY}=\frac{XY+BY}{BY}=\frac{XY}{BY}+1=\frac{23}{15}&lt;/math&gt;, so &lt;math&gt;BY=\frac{15}{8}XY=\frac{15}{8}*\frac{22\sqrt{3}}{45}=\frac{11\sqrt{3}}{2}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\tan \angle B = \frac{x}{\frac{11\sqrt{3}}{12}x}=\frac{4\sqrt{3}}{11}&lt;/math&gt;, and our answer is &lt;math&gt;4+3+11=\boxed{018}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> (Another solution without trigonometry)<br /> <br /> Extend &lt;math&gt;CD&lt;/math&gt; to point &lt;math&gt;F&lt;/math&gt; such that &lt;math&gt;\overline{AF} \parallel \overline{CB}&lt;/math&gt;. It is then clear that &lt;math&gt;\triangle AFD&lt;/math&gt; is similar to &lt;math&gt;\triangle BCD&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;AC=p&lt;/math&gt;, &lt;math&gt;BC=q&lt;/math&gt;. Then &lt;math&gt;\tan \angle B = p/q&lt;/math&gt;.<br /> <br /> With the Angle Bisector Theorem, we get that &lt;math&gt;CD=\frac{8}{15}q&lt;/math&gt;. From 30-60-90 &lt;math&gt;\triangle CAF&lt;/math&gt;, we get that &lt;math&gt;AF=\frac{1}{\sqrt{3}}p&lt;/math&gt; and &lt;math&gt;FD=FC-CD=\frac{2}{\sqrt{3}}p-\frac{8}{15}q&lt;/math&gt;.<br /> <br /> From &lt;math&gt;\triangle AFD \sim \triangle BCD&lt;/math&gt;, we have that &lt;math&gt;\frac{FD}{CD}=\frac{FA}{CB}=\frac{\frac{2}{\sqrt{3}}p-\frac{8}{15}q}{\frac{8}{15}q}=\frac{\frac{1}{\sqrt{3}}p}{q}&lt;/math&gt;. Simplifying yields &lt;math&gt;\left(\frac{p}{q}\right)\left(\frac{2\sqrt{3}}{3}*\frac{15}{8}-\frac{\sqrt{3}}{3}\right)=1&lt;/math&gt;, and &lt;math&gt;\tan \angle B=\frac{p}{q}=\frac{4\sqrt{3}}{11}&lt;/math&gt;, so our answer is &lt;math&gt;4+3+11=\boxed{018}&lt;/math&gt;.<br /> <br /> === Solution 6 ===<br /> Let &lt;math&gt;CB = 1&lt;/math&gt;, and let the feet of the altitudes from &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; to &lt;math&gt;\overline{CB}&lt;/math&gt; be &lt;math&gt;D'&lt;/math&gt; and &lt;math&gt;E'&lt;/math&gt;, respectively. Also, let &lt;math&gt;DE = 8k&lt;/math&gt; and &lt;math&gt;EB = 15k&lt;/math&gt;. We see that &lt;math&gt;BD' = 15k\cos B&lt;/math&gt; and &lt;math&gt;BE' = 23k\cos B&lt;/math&gt; by right triangles &lt;math&gt;\triangle{BDD'}&lt;/math&gt; and &lt;math&gt;\triangle{BEE'}&lt;/math&gt;. From this we have that &lt;math&gt;D'E' = 8k\cos B&lt;/math&gt;. With the same triangles we have &lt;math&gt;DD' = 23k\sin B&lt;/math&gt; and &lt;math&gt;EE' = 15k\sin B&lt;/math&gt;. From 30-60-90 triangles &lt;math&gt;\triangle{CDD'}&lt;/math&gt; and &lt;math&gt;\triangle{CEE'}&lt;/math&gt;, we see that &lt;math&gt;CD' = \frac{23k\sqrt{3}\sin B}{3}&lt;/math&gt; and &lt;math&gt;CE' = 15k\sqrt{3}\sin B&lt;/math&gt;, so &lt;math&gt;D'E' = \frac{22k\sqrt{3}\sin B}{3}&lt;/math&gt;. From our two values of &lt;math&gt;D'E'&lt;/math&gt; we get:<br /> &lt;math&gt;8k\cos B = \frac{22k\sqrt{3}\sin B}{3}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{\sin B}{\cos B} = \frac{8k}{\frac{22k\sqrt{3}}{3}} = \tan B&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan B = \frac{8}{\frac{22\sqrt{3}}{3}} = \frac{24}{22\sqrt{3}} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}&lt;/math&gt;<br /> Our answer is then &lt;math&gt;4+3+11 = \boxed{018}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_7&diff=93471 2018 AIME II Problems/Problem 7 2018-03-24T18:41:57Z <p>Zeroman: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has side lengths &lt;math&gt;AB = 9&lt;/math&gt;, &lt;math&gt;BC =&lt;/math&gt; &lt;math&gt;5\sqrt{3}&lt;/math&gt;, and &lt;math&gt;AC = 12&lt;/math&gt;. Points &lt;math&gt;A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B&lt;/math&gt; are on segment &lt;math&gt;\overline{AB}&lt;/math&gt; with &lt;math&gt;P_{k}&lt;/math&gt; between &lt;math&gt;P_{k-1}&lt;/math&gt; and &lt;math&gt;P_{k+1}&lt;/math&gt; for &lt;math&gt;k = 1, 2, ..., 2449&lt;/math&gt;, and points &lt;math&gt;A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C&lt;/math&gt; are on segment &lt;math&gt;\overline{AC}&lt;/math&gt; with &lt;math&gt;Q_{k}&lt;/math&gt; between &lt;math&gt;Q_{k-1}&lt;/math&gt; and &lt;math&gt;Q_{k+1}&lt;/math&gt; for &lt;math&gt;k = 1, 2, ..., 2449&lt;/math&gt;. Furthermore, each segment &lt;math&gt;\overline{P_{k}Q_{k}}&lt;/math&gt;, &lt;math&gt;k = 1, 2, ..., 2449&lt;/math&gt;, is parallel to &lt;math&gt;\overline{BC}&lt;/math&gt;. The segments cut the triangle into &lt;math&gt;2450&lt;/math&gt; regions, consisting of &lt;math&gt;2449&lt;/math&gt; trapezoids and &lt;math&gt;1&lt;/math&gt; triangle. Each of the &lt;math&gt;2450&lt;/math&gt; regions has the same area. Find the number of segments &lt;math&gt;\overline{P_{k}Q_{k}}&lt;/math&gt;, &lt;math&gt;k = 1, 2, ..., 2450&lt;/math&gt;, that have rational length.<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> For each &lt;math&gt;k&lt;/math&gt; between &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;2450&lt;/math&gt;, the area of the trapezoid with &lt;math&gt;\overline{P_kQ_k}&lt;/math&gt; as its bottom base is the difference between the areas of two triangles, both similar to &lt;math&gt;\triangle{ABC}&lt;/math&gt;. Let &lt;math&gt;d_k&lt;/math&gt; be the length of segment &lt;math&gt;\overline{P_kQ_k}&lt;/math&gt;. The area of the trapezoid with bases &lt;math&gt;\overline{P_{k-1}Q_{k-1}}&lt;/math&gt; and &lt;math&gt;P_kQ_k&lt;/math&gt; is &lt;math&gt;(\frac{d_k}{5\sqrt{3}})^2 - (\frac{d_{k-1}}{5\sqrt{3}})^2 = \frac{d_k^2-d_{k-1}^2}{75}&lt;/math&gt; times the area of &lt;math&gt;\triangle{ABC}&lt;/math&gt;. (This logic also applies to the topmost triangle if we notice that &lt;math&gt;d_0 = 0&lt;/math&gt;.) However, we also know that the area of each shape is &lt;math&gt;\frac{1}{2450}&lt;/math&gt; times the area of &lt;math&gt;\triangle{ABC}&lt;/math&gt;. We then have &lt;math&gt;\frac{d_k^2-d_{k-1}^2}{75} = \frac{1}{2450}&lt;/math&gt;. Simplifying, &lt;math&gt;d_k^2-d_{k-1}^2 = \frac{3}{98}&lt;/math&gt;. However, we know that &lt;math&gt;d_0^2 = 0&lt;/math&gt;, so &lt;math&gt;d_1^2 = \frac{3}{98}&lt;/math&gt;, and in general, &lt;math&gt;d_k^2 = \frac{3k}{98}&lt;/math&gt; and &lt;math&gt;d_k = \frac{\sqrt{\frac{3k}{2}}}{7}&lt;/math&gt;. The smallest &lt;math&gt;k&lt;/math&gt; that gives a rational &lt;math&gt;d_k&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt;, so &lt;math&gt;d_k&lt;/math&gt; is rational if and only if &lt;math&gt;k = 6n^2&lt;/math&gt; for some integer &lt;math&gt;n&lt;/math&gt;.The largest &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;6n^2&lt;/math&gt; is less than &lt;math&gt;2450&lt;/math&gt; is &lt;math&gt;20&lt;/math&gt;, so &lt;math&gt;k&lt;/math&gt; has &lt;math&gt;\boxed{020}&lt;/math&gt; possible values.<br /> <br /> Solution by zeroman<br /> {{AIME box|year=2018|n=II|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_7&diff=93470 2018 AIME II Problems/Problem 7 2018-03-24T18:41:35Z <p>Zeroman: </p> <hr /> <div>== Problem ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has side lengths &lt;math&gt;AB = 9&lt;/math&gt;, &lt;math&gt;BC =&lt;/math&gt; &lt;math&gt;5\sqrt{3}&lt;/math&gt;, and &lt;math&gt;AC = 12&lt;/math&gt;. Points &lt;math&gt;A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B&lt;/math&gt; are on segment &lt;math&gt;\overline{AB}&lt;/math&gt; with &lt;math&gt;P_{k}&lt;/math&gt; between &lt;math&gt;P_{k-1}&lt;/math&gt; and &lt;math&gt;P_{k+1}&lt;/math&gt; for &lt;math&gt;k = 1, 2, ..., 2449&lt;/math&gt;, and points &lt;math&gt;A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C&lt;/math&gt; are on segment &lt;math&gt;\overline{AC}&lt;/math&gt; with &lt;math&gt;Q_{k}&lt;/math&gt; between &lt;math&gt;Q_{k-1}&lt;/math&gt; and &lt;math&gt;Q_{k+1}&lt;/math&gt; for &lt;math&gt;k = 1, 2, ..., 2449&lt;/math&gt;. Furthermore, each segment &lt;math&gt;\overline{P_{k}Q_{k}}&lt;/math&gt;, &lt;math&gt;k = 1, 2, ..., 2449&lt;/math&gt;, is parallel to &lt;math&gt;\overline{BC}&lt;/math&gt;. The segments cut the triangle into &lt;math&gt;2450&lt;/math&gt; regions, consisting of &lt;math&gt;2449&lt;/math&gt; trapezoids and &lt;math&gt;1&lt;/math&gt; triangle. Each of the &lt;math&gt;2450&lt;/math&gt; regions has the same area. Find the number of segments &lt;math&gt;\overline{P_{k}Q_{k}}&lt;/math&gt;, &lt;math&gt;k = 1, 2, ..., 2450&lt;/math&gt;, that have rational length.<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> For each &lt;math&gt;k&lt;/math&gt; between &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;2450&lt;/math&gt;, the area of the trapezoid with &lt;math&gt;\overline{P_kQ_k}&lt;/math&gt; as its bottom base is the difference between the areas of two triangles, both similar to &lt;math&gt;\triangle{ABC}&lt;/math&gt;. Let &lt;math&gt;d_k&lt;/math&gt; be the length of segment &lt;math&gt;\overline{P_kQ_k}&lt;/math&gt;. The area of the trapezoid with bases &lt;math&gt;\overline{P_{k-1}Q_{k-1}}&lt;/math&gt; and &lt;math&gt;P_kQ_k&lt;/math&gt; is &lt;math&gt;(\frac{d_k}{5\sqrt{3}})^2 - (\frac{d_{k-1}}{5\sqrt{3}})^2 = \frac{d_k^2-d_{k-1}^2}{75}&lt;/math&gt; times the area of &lt;math&gt;\triangle{ABC}&lt;/math&gt;. (This logic also applies to the topmost triangle if we notice that &lt;math&gt;d_0 = 0&lt;/math&gt;.) However, we also know that the area of each shape is &lt;math&gt;\frac{1}{2450}&lt;/math&gt; times the area of &lt;math&gt;\triangle{ABC}&lt;/math&gt;. We then have &lt;math&gt;\frac{d_k^2-d_{k-1}^2}{75} = \frac{1}{2450}&lt;/math&gt;. Simplifying, &lt;math&gt;d_k^2-d_{k-1}^2 = \frac{3}{98}&lt;/math&gt;. However, we know that &lt;math&gt;d_0^2 = 0&lt;/math&gt;, so &lt;math&gt;d_1^2 = \frac{3}{98}&lt;/math&gt;, and in general, &lt;math&gt;d_k^2 = \frac{3k}{98}&lt;/math&gt; and &lt;math&gt;d_k = \frac{\sqrt{\frac{3k}{2}}}{7}&lt;/math&gt;. The smallest &lt;math&gt;k&lt;/math&gt; that gives a rational &lt;math&gt;d_k&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt;, so &lt;math&gt;d_k&lt;/math&gt; is rational if and only if &lt;math&gt;k = 6n^2&lt;/math&gt; for some integer &lt;math&gt;n&lt;/math&gt;.The largest &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;6n^2&lt;/math&gt; is less than &lt;math&gt;2450&lt;/math&gt; is &lt;math&gt;20&lt;/math&gt;, so &lt;math&gt;k&lt;/math&gt; has &lt;math&gt;\boxed{020}&lt;/math&gt; possible values.<br /> {{AIME box|year=2018|n=II|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_10&diff=93396 2006 AIME II Problems/Problem 10 2018-03-22T19:47:33Z <p>Zeroman: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a &lt;math&gt; 50\% &lt;/math&gt; chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumilated to decide the ranks of the teams. In the first game of the tournament, team &lt;math&gt; A &lt;/math&gt; beats team &lt;math&gt; B. &lt;/math&gt; The [[probability]] that team &lt;math&gt; A &lt;/math&gt; finishes with more points than team &lt;math&gt; B &lt;/math&gt; is &lt;math&gt; m/n, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt; m+n. &lt;/math&gt;<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> The results of the five remaining games are independent of the first game, so by symmetry, the probability that &lt;math&gt;A&lt;/math&gt; scores higher than &lt;math&gt;B&lt;/math&gt; in these five games is equal to the probability that &lt;math&gt;B&lt;/math&gt; scores higher than &lt;math&gt;A&lt;/math&gt;. We let this probability be &lt;math&gt;p&lt;/math&gt;; then the probability that &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; end with the same score in these five games is &lt;math&gt;1-2p&lt;/math&gt;.<br /> <br /> Of these three cases (&lt;math&gt;|A| &gt; |B|, |A| &lt; |B|, |A|=|B|&lt;/math&gt;), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases). <br /> <br /> There are &lt;math&gt;{5\choose k}&lt;/math&gt; ways to &lt;math&gt;A&lt;/math&gt; to have &lt;math&gt;k&lt;/math&gt; victories, and &lt;math&gt;{5\choose k}&lt;/math&gt; ways for &lt;math&gt;B&lt;/math&gt; to have &lt;math&gt;k&lt;/math&gt; victories. Summing for all values of &lt;math&gt;k&lt;/math&gt;,<br /> &lt;center&gt;&lt;math&gt;1-2p = \frac{1}{2^{5} \times 2^{5}}\left(\sum_{k=0}^{5} {5\choose k}^2\right) = \frac{1^2+5^2+10^2+10^2+5^2+1^2}{1024} = \frac{126}{512}.&lt;/math&gt;&lt;/center&gt;<br /> Thus &lt;math&gt;p = \frac 12 \left(1-\frac{126}{512}\right) = \frac{193}{512}&lt;/math&gt;. The desired probability is the sum of the cases when &lt;math&gt;|A| \ge |B|&lt;/math&gt;, so the answer is &lt;math&gt;\frac{126}{512} + \frac{193}{512} = \frac{319}{512}&lt;/math&gt;, and &lt;math&gt;m+n = \boxed{831}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> You can break this into cases based on how many rounds &lt;math&gt;A&lt;/math&gt; wins out of the remaining &lt;math&gt;5&lt;/math&gt; games.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 0 games, then &lt;math&gt;B&lt;/math&gt; must win 0 games and the probability of this is &lt;math&gt; \frac{{5 \choose 0}}{2^5} \frac{{5 \choose 0}}{2^5} = \frac{1}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 1 games, then &lt;math&gt;B&lt;/math&gt; must win 1 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 1}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}}{2^5} = \frac{30}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 2 games, then &lt;math&gt;B&lt;/math&gt; must win 2 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 2}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}}{2^5} = \frac{160}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 3 games, then &lt;math&gt;B&lt;/math&gt; must win 3 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 3}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}}{2^5} = \frac{260}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 4 games, then &lt;math&gt;B&lt;/math&gt; must win 4 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 4}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}}{2^5} = \frac{155}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 5 games, then &lt;math&gt;B&lt;/math&gt; must win 5 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 5}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}+{5 \choose 5}}{2^5} = \frac{32}{1024} &lt;/math&gt;.<br /> <br /> Summing these 6 cases, we get &lt;math&gt; \frac{638}{1024} &lt;/math&gt;, which simplifies to &lt;math&gt; \frac{319}{512} &lt;/math&gt;, so our answer is &lt;math&gt;319 + 512 = \boxed{831}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> We can apply the concept of generating functions here. <br /> <br /> The generating function for &lt;math&gt;B&lt;/math&gt; is &lt;math&gt; (1 + 0x^{1}) &lt;/math&gt; for the first game where &lt;math&gt;x^{n}&lt;/math&gt; is winning n games. Since &lt;math&gt;B&lt;/math&gt; lost the first game, the coefficient for &lt;math&gt;x^{1}&lt;/math&gt; is 0. The generating function for the next 5 games is &lt;math&gt;(1 + x)^{5}&lt;/math&gt;. Thus, the total generating function for number of games he wins is<br /> <br /> &lt;math&gt;{(1 + 0x)(1 + x)^{5}} = (1 + 5x^{1} + 10x^{2} + 10x^{3} + 5x^{4} + x^{5})&lt;/math&gt;.<br /> <br /> The generating function for &lt;math&gt;A&lt;/math&gt; is the same except that it is multiplied by &lt;math&gt;x&lt;/math&gt; instead of &lt;math&gt;(1+0x)&lt;/math&gt;. <br /> Thus, the generating function for &lt;math&gt;A&lt;/math&gt; is <br /> <br /> &lt;math&gt;1x + 5x^{2} + 10x^{3} + 10x^{4} + 5x^{5} + x^{6}&lt;/math&gt;. <br /> <br /> The probability that &lt;math&gt;B&lt;/math&gt; wins 0 games is &lt;math&gt;\frac{1}{32}&lt;/math&gt;. Since the coefficients for all &lt;math&gt;x^{n}&lt;/math&gt; where <br /> <br /> &lt;math&gt;n \geq 1&lt;/math&gt; sums to 32, the probability that &lt;math&gt;A&lt;/math&gt; wins more games is &lt;math&gt;\frac{32}{32}&lt;/math&gt;. <br /> <br /> Thus, the probability that &lt;math&gt;A&lt;/math&gt; has more wins than &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;\frac{1}{32} \times \frac{32}{32} + \frac{5}{32} \times \frac{31}{32} + \frac{10}{32} \times \frac{26}{32} + \frac{10}{32} \times \frac{16}{32} + \frac{5}{32} \times \frac{6}{32} +\frac{1}{32} \times \frac{1}{32} = \frac{638}{1024} = \frac{319}{512}&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;319 + 512 = \boxed{831} &lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> After the first game, there are &lt;math&gt;10&lt;/math&gt; games we care about-- those involving &lt;math&gt;A&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt;. There are &lt;math&gt;3&lt;/math&gt; cases of these &lt;math&gt;10&lt;/math&gt; games: &lt;math&gt;A&lt;/math&gt; wins more than &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; wins more than &lt;math&gt;A&lt;/math&gt;, or &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; win the same number of games. Also, there are &lt;math&gt;2^{10} = 1024&lt;/math&gt; total outcomes. By symmetry, the first and second cases are equally likely, and the third case occurs &lt;math&gt;\binom{5}{0}^2+\binom{5}{1}^2+\binom{5}{2}^2+\binom{5}{3}^2+\binom{5}{4}^2+\binom{5}{5}^2 = \binom{10}{5} = 252&lt;/math&gt; times, by [[Combinatorial identity#Another Identity|a special case of Vandermonde's Identity]]. There are therefore &lt;math&gt;\frac{1024-252}{2} = 386&lt;/math&gt; possibilities for each of the other two cases.<br /> <br /> If &lt;math&gt;B&lt;/math&gt; has more wins than &lt;math&gt;A&lt;/math&gt; in its &lt;math&gt;5&lt;/math&gt; remaining games, then &lt;math&gt;A&lt;/math&gt; cannot beat &lt;math&gt;B&lt;/math&gt; overall. However, if &lt;math&gt;A&lt;/math&gt; has more wins or if &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are tied, &lt;math&gt;A&lt;/math&gt; will beat &lt;math&gt;B&lt;/math&gt; overall. Therefore, out of the &lt;math&gt;1024&lt;/math&gt; possibilites, &lt;math&gt;386+252 = 638&lt;/math&gt; ways where &lt;math&gt;A&lt;/math&gt; wins, so the desired probability is &lt;math&gt;\frac{638}{1024} = \frac{319}{512}&lt;/math&gt;, and &lt;math&gt;m+n=\boxed{831}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_10&diff=93395 2006 AIME II Problems/Problem 10 2018-03-22T19:46:29Z <p>Zeroman: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a &lt;math&gt; 50\% &lt;/math&gt; chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumilated to decide the ranks of the teams. In the first game of the tournament, team &lt;math&gt; A &lt;/math&gt; beats team &lt;math&gt; B. &lt;/math&gt; The [[probability]] that team &lt;math&gt; A &lt;/math&gt; finishes with more points than team &lt;math&gt; B &lt;/math&gt; is &lt;math&gt; m/n, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt; m+n. &lt;/math&gt;<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> The results of the five remaining games are independent of the first game, so by symmetry, the probability that &lt;math&gt;A&lt;/math&gt; scores higher than &lt;math&gt;B&lt;/math&gt; in these five games is equal to the probability that &lt;math&gt;B&lt;/math&gt; scores higher than &lt;math&gt;A&lt;/math&gt;. We let this probability be &lt;math&gt;p&lt;/math&gt;; then the probability that &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; end with the same score in these five games is &lt;math&gt;1-2p&lt;/math&gt;.<br /> <br /> Of these three cases (&lt;math&gt;|A| &gt; |B|, |A| &lt; |B|, |A|=|B|&lt;/math&gt;), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases). <br /> <br /> There are &lt;math&gt;{5\choose k}&lt;/math&gt; ways to &lt;math&gt;A&lt;/math&gt; to have &lt;math&gt;k&lt;/math&gt; victories, and &lt;math&gt;{5\choose k}&lt;/math&gt; ways for &lt;math&gt;B&lt;/math&gt; to have &lt;math&gt;k&lt;/math&gt; victories. Summing for all values of &lt;math&gt;k&lt;/math&gt;,<br /> &lt;center&gt;&lt;math&gt;1-2p = \frac{1}{2^{5} \times 2^{5}}\left(\sum_{k=0}^{5} {5\choose k}^2\right) = \frac{1^2+5^2+10^2+10^2+5^2+1^2}{1024} = \frac{126}{512}.&lt;/math&gt;&lt;/center&gt;<br /> Thus &lt;math&gt;p = \frac 12 \left(1-\frac{126}{512}\right) = \frac{193}{512}&lt;/math&gt;. The desired probability is the sum of the cases when &lt;math&gt;|A| \ge |B|&lt;/math&gt;, so the answer is &lt;math&gt;\frac{126}{512} + \frac{193}{512} = \frac{319}{512}&lt;/math&gt;, and &lt;math&gt;m+n = \boxed{831}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> You can break this into cases based on how many rounds &lt;math&gt;A&lt;/math&gt; wins out of the remaining &lt;math&gt;5&lt;/math&gt; games.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 0 games, then &lt;math&gt;B&lt;/math&gt; must win 0 games and the probability of this is &lt;math&gt; \frac{{5 \choose 0}}{2^5} \frac{{5 \choose 0}}{2^5} = \frac{1}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 1 games, then &lt;math&gt;B&lt;/math&gt; must win 1 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 1}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}}{2^5} = \frac{30}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 2 games, then &lt;math&gt;B&lt;/math&gt; must win 2 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 2}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}}{2^5} = \frac{160}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 3 games, then &lt;math&gt;B&lt;/math&gt; must win 3 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 3}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}}{2^5} = \frac{260}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 4 games, then &lt;math&gt;B&lt;/math&gt; must win 4 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 4}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}}{2^5} = \frac{155}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 5 games, then &lt;math&gt;B&lt;/math&gt; must win 5 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 5}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}+{5 \choose 5}}{2^5} = \frac{32}{1024} &lt;/math&gt;.<br /> <br /> Summing these 6 cases, we get &lt;math&gt; \frac{638}{1024} &lt;/math&gt;, which simplifies to &lt;math&gt; \frac{319}{512} &lt;/math&gt;, so our answer is &lt;math&gt;319 + 512 = \boxed{831}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> We can apply the concept of generating functions here. <br /> <br /> The generating function for &lt;math&gt;B&lt;/math&gt; is &lt;math&gt; (1 + 0x^{1}) &lt;/math&gt; for the first game where &lt;math&gt;x^{n}&lt;/math&gt; is winning n games. Since &lt;math&gt;B&lt;/math&gt; lost the first game, the coefficient for &lt;math&gt;x^{1}&lt;/math&gt; is 0. The generating function for the next 5 games is &lt;math&gt;(1 + x)^{5}&lt;/math&gt;. Thus, the total generating function for number of games he wins is<br /> <br /> &lt;math&gt;{(1 + 0x)(1 + x)^{5}} = (1 + 5x^{1} + 10x^{2} + 10x^{3} + 5x^{4} + x^{5})&lt;/math&gt;.<br /> <br /> The generating function for &lt;math&gt;A&lt;/math&gt; is the same except that it is multiplied by &lt;math&gt;x&lt;/math&gt; instead of &lt;math&gt;(1+0x)&lt;/math&gt;. <br /> Thus, the generating function for &lt;math&gt;A&lt;/math&gt; is <br /> <br /> &lt;math&gt;1x + 5x^{2} + 10x^{3} + 10x^{4} + 5x^{5} + x^{6}&lt;/math&gt;. <br /> <br /> The probability that &lt;math&gt;B&lt;/math&gt; wins 0 games is &lt;math&gt;\frac{1}{32}&lt;/math&gt;. Since the coefficients for all &lt;math&gt;x^{n}&lt;/math&gt; where <br /> <br /> &lt;math&gt;n \geq 1&lt;/math&gt; sums to 32, the probability that &lt;math&gt;A&lt;/math&gt; wins more games is &lt;math&gt;\frac{32}{32}&lt;/math&gt;. <br /> <br /> Thus, the probability that &lt;math&gt;A&lt;/math&gt; has more wins than &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;\frac{1}{32} \times \frac{32}{32} + \frac{5}{32} \times \frac{31}{32} + \frac{10}{32} \times \frac{26}{32} + \frac{10}{32} \times \frac{16}{32} + \frac{5}{32} \times \frac{6}{32} +\frac{1}{32} \times \frac{1}{32} = \frac{638}{1024} = \frac{319}{512}&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;319 + 512 = \boxed{831} &lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> After the first game, there are &lt;math&gt;10&lt;/math&gt; games we care about-- those involving &lt;math&gt;A&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt;. There are &lt;math&gt;3&lt;/math&gt; cases of these &lt;math&gt;10&lt;/math&gt; games: &lt;math&gt;A&lt;/math&gt; wins more than &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; wins more than &lt;math&gt;A&lt;/math&gt;, or &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; win the same number of games. Also, there are &lt;math&gt;2^{10} = 1024&lt;/math&gt; total outcomes. By symmetry, the first and second cases are equally likely, and the third case occurs &lt;math&gt;\binom{5}{0}^2+\binom{5}{1}^2+\binom{5}{2}^2+\binom{5}{3}^2+\binom{5}{4}^2+\binom{5}{5}^2 = \binom{10}{5} = 252&lt;/math&gt; times, by a special case of [[Combinatorial identity|Vandermonde's Identity]]. There are therefore &lt;math&gt;\frac{1024-252}{2} = 386&lt;/math&gt; possibilities for each of the other two cases.<br /> <br /> If &lt;math&gt;B&lt;/math&gt; has more wins than &lt;math&gt;A&lt;/math&gt; in its &lt;math&gt;5&lt;/math&gt; remaining games, then &lt;math&gt;A&lt;/math&gt; cannot beat &lt;math&gt;B&lt;/math&gt; overall. However, if &lt;math&gt;A&lt;/math&gt; has more wins or if &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are tied, &lt;math&gt;A&lt;/math&gt; will beat &lt;math&gt;B&lt;/math&gt; overall. Therefore, out of the &lt;math&gt;1024&lt;/math&gt; possibilites, &lt;math&gt;386+252 = 638&lt;/math&gt; ways where &lt;math&gt;A&lt;/math&gt; wins, so the desired probability is &lt;math&gt;\frac{638}{1024} = \frac{319}{512}&lt;/math&gt;, and &lt;math&gt;m+n=\boxed{831}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Answer_Key&diff=93394 2018 AIME II Answer Key 2018-03-22T19:44:25Z <p>Zeroman: /* 2018 AIME II Answer Key */</p> <hr /> <div></div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Answer_Key&diff=93393 2018 AIME II Answer Key 2018-03-22T19:44:14Z <p>Zeroman: Created page with &quot;== 2018 AIME II Answer Key ==&quot;</p> <hr /> <div>== 2018 AIME II Answer Key ==</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_11&diff=93385 2006 AIME II Problems/Problem 11 2018-03-22T02:33:24Z <p>Zeroman: </p> <hr /> <div>== Problem ==<br /> A [[sequence]] is defined as follows &lt;math&gt; a_1=a_2=a_3=1, &lt;/math&gt; and, for all positive integers &lt;math&gt; n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. &lt;/math&gt; Given that &lt;math&gt; a_{28}=6090307, a_{29}=11201821, &lt;/math&gt; and &lt;math&gt; a_{30}=20603361, &lt;/math&gt; find the [[remainder]] when &lt;math&gt;\sum^{28}_{k=1} a_k &lt;/math&gt; is divided by 1000.<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Define the sum as &lt;math&gt;s&lt;/math&gt;. Since &lt;math&gt;a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} &lt;/math&gt;, the sum will be:<br /> &lt;center&gt;&lt;math&gt;s = a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\<br /> s = a_{28} + \left(\sum^{30}_{k=4} a_{k} - \sum^{29}_{k=3} a_{k}\right) - \left(\sum^{28}_{k=2} a_{k}\right)\\<br /> s = a_{28} + (a_{30} - a_{3}) - \left(\sum^{28}_{k=2} a_{k}\right) = a_{28} + a_{30} - a_{3} - (s - a_{1})\\<br /> s = -s + a_{28} + a_{30}<br /> &lt;/math&gt;&lt;/center&gt;<br /> <br /> Thus &lt;math&gt;s = \frac{a_{28} + a_{30}}{2}&lt;/math&gt;, and &lt;math&gt;a_{28},\,a_{30}&lt;/math&gt; are both given; the last four digits of their sum is &lt;math&gt;3668&lt;/math&gt;, and half of that is &lt;math&gt;1834&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{834}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Very bad solution: Brute Force. Since the problem asks for the answer of the end value when divided by 1000, it wouldn't be that difficult because you only need to keep track of the last 3 digits.<br /> <br /> === Solution 3 (some guessing involved) ===<br /> All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given &lt;math&gt;a_{28}, a_{29}, &lt;/math&gt; and &lt;math&gt;a_{30}&lt;/math&gt;, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some &lt;math&gt;p, q, r&lt;/math&gt; such that &lt;math&gt;\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}&lt;/math&gt;. From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that &lt;math&gt;(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})&lt;/math&gt;, at least for the first few terms. From this, we have that &lt;math&gt;\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}}(\mod 1000)&lt;/math&gt;.<br /> <br /> Solution by zeroman<br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_10&diff=93382 2006 AIME II Problems/Problem 10 2018-03-22T02:09:25Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a &lt;math&gt; 50\% &lt;/math&gt; chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumilated to decide the ranks of the teams. In the first game of the tournament, team &lt;math&gt; A &lt;/math&gt; beats team &lt;math&gt; B. &lt;/math&gt; The [[probability]] that team &lt;math&gt; A &lt;/math&gt; finishes with more points than team &lt;math&gt; B &lt;/math&gt; is &lt;math&gt; m/n, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt; m+n. &lt;/math&gt;<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> The results of the five remaining games are independent of the first game, so by symmetry, the probability that &lt;math&gt;A&lt;/math&gt; scores higher than &lt;math&gt;B&lt;/math&gt; in these five games is equal to the probability that &lt;math&gt;B&lt;/math&gt; scores higher than &lt;math&gt;A&lt;/math&gt;. We let this probability be &lt;math&gt;p&lt;/math&gt;; then the probability that &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; end with the same score in these five games is &lt;math&gt;1-2p&lt;/math&gt;.<br /> <br /> Of these three cases (&lt;math&gt;|A| &gt; |B|, |A| &lt; |B|, |A|=|B|&lt;/math&gt;), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases). <br /> <br /> There are &lt;math&gt;{5\choose k}&lt;/math&gt; ways to &lt;math&gt;A&lt;/math&gt; to have &lt;math&gt;k&lt;/math&gt; victories, and &lt;math&gt;{5\choose k}&lt;/math&gt; ways for &lt;math&gt;B&lt;/math&gt; to have &lt;math&gt;k&lt;/math&gt; victories. Summing for all values of &lt;math&gt;k&lt;/math&gt;,<br /> &lt;center&gt;&lt;math&gt;1-2p = \frac{1}{2^{5} \times 2^{5}}\left(\sum_{k=0}^{5} {5\choose k}^2\right) = \frac{1^2+5^2+10^2+10^2+5^2+1^2}{1024} = \frac{126}{512}.&lt;/math&gt;&lt;/center&gt;<br /> Thus &lt;math&gt;p = \frac 12 \left(1-\frac{126}{512}\right) = \frac{193}{512}&lt;/math&gt;. The desired probability is the sum of the cases when &lt;math&gt;|A| \ge |B|&lt;/math&gt;, so the answer is &lt;math&gt;\frac{126}{512} + \frac{193}{512} = \frac{319}{512}&lt;/math&gt;, and &lt;math&gt;m+n = \boxed{831}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> You can break this into cases based on how many rounds &lt;math&gt;A&lt;/math&gt; wins out of the remaining &lt;math&gt;5&lt;/math&gt; games.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 0 games, then &lt;math&gt;B&lt;/math&gt; must win 0 games and the probability of this is &lt;math&gt; \frac{{5 \choose 0}}{2^5} \frac{{5 \choose 0}}{2^5} = \frac{1}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 1 games, then &lt;math&gt;B&lt;/math&gt; must win 1 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 1}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}}{2^5} = \frac{30}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 2 games, then &lt;math&gt;B&lt;/math&gt; must win 2 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 2}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}}{2^5} = \frac{160}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 3 games, then &lt;math&gt;B&lt;/math&gt; must win 3 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 3}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}}{2^5} = \frac{260}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 4 games, then &lt;math&gt;B&lt;/math&gt; must win 4 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 4}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}}{2^5} = \frac{155}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 5 games, then &lt;math&gt;B&lt;/math&gt; must win 5 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 5}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}+{5 \choose 5}}{2^5} = \frac{32}{1024} &lt;/math&gt;.<br /> <br /> Summing these 6 cases, we get &lt;math&gt; \frac{638}{1024} &lt;/math&gt;, which simplifies to &lt;math&gt; \frac{319}{512} &lt;/math&gt;, so our answer is &lt;math&gt;319 + 512 = \boxed{831}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> We can apply the concept of generating functions here. <br /> <br /> The generating function for &lt;math&gt;B&lt;/math&gt; is &lt;math&gt; (1 + 0x^{1}) &lt;/math&gt; for the first game where &lt;math&gt;x^{n}&lt;/math&gt; is winning n games. Since &lt;math&gt;B&lt;/math&gt; lost the first game, the coefficient for &lt;math&gt;x^{1}&lt;/math&gt; is 0. The generating function for the next 5 games is &lt;math&gt;(1 + x)^{5}&lt;/math&gt;. Thus, the total generating function for number of games he wins is<br /> <br /> &lt;math&gt;{(1 + 0x)(1 + x)^{5}} = (1 + 5x^{1} + 10x^{2} + 10x^{3} + 5x^{4} + x^{5})&lt;/math&gt;.<br /> <br /> The generating function for &lt;math&gt;A&lt;/math&gt; is the same except that it is multiplied by &lt;math&gt;x&lt;/math&gt; instead of &lt;math&gt;(1+0x)&lt;/math&gt;. <br /> Thus, the generating function for &lt;math&gt;A&lt;/math&gt; is <br /> <br /> &lt;math&gt;1x + 5x^{2} + 10x^{3} + 10x^{4} + 5x^{5} + x^{6}&lt;/math&gt;. <br /> <br /> The probability that &lt;math&gt;B&lt;/math&gt; wins 0 games is &lt;math&gt;\frac{1}{32}&lt;/math&gt;. Since the coefficients for all &lt;math&gt;x^{n}&lt;/math&gt; where <br /> <br /> &lt;math&gt;n \geq 1&lt;/math&gt; sums to 32, the probability that &lt;math&gt;A&lt;/math&gt; wins more games is &lt;math&gt;\frac{32}{32}&lt;/math&gt;. <br /> <br /> Thus, the probability that &lt;math&gt;A&lt;/math&gt; has more wins than &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;\frac{1}{32} \times \frac{32}{32} + \frac{5}{32} \times \frac{31}{32} + \frac{10}{32} \times \frac{26}{32} + \frac{10}{32} \times \frac{16}{32} + \frac{5}{32} \times \frac{6}{32} +\frac{1}{32} \times \frac{1}{32} = \frac{638}{1024} = \frac{319}{512}&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;319 + 512 = \boxed{831} &lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> After the first game, there are &lt;math&gt;10&lt;/math&gt; games we care about-- those involving &lt;math&gt;A&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt;. There are &lt;math&gt;3&lt;/math&gt; cases of these &lt;math&gt;10&lt;/math&gt; games: &lt;math&gt;A&lt;/math&gt; wins more than &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; wins more than &lt;math&gt;A&lt;/math&gt;, or &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; win the same number of games. Also, there are &lt;math&gt;2^{10} = 1024&lt;/math&gt; total outcomes. By symmetry, the first and second cases are equally likely, and the third case occurs &lt;math&gt;\binom{5}{0}^2+\binom{5}{1}^2+\binom{5}{2}^2+\binom{5}{3}^2+\binom{5}{4}^2+\binom{5}{5}^2 = \binom{10}{5} = 252&lt;/math&gt; times, by a special case of Vandermonde's Identity (see [[Combinatorial identity]]). There are therefore &lt;math&gt;\frac{1024-252}{2} = 386&lt;/math&gt; possibilities for each of the other two cases.<br /> <br /> If &lt;math&gt;B&lt;/math&gt; has more wins than &lt;math&gt;A&lt;/math&gt; in its &lt;math&gt;5&lt;/math&gt; remaining games, then &lt;math&gt;A&lt;/math&gt; cannot beat &lt;math&gt;B&lt;/math&gt; overall. However, if &lt;math&gt;A&lt;/math&gt; has more wins or if &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are tied, &lt;math&gt;A&lt;/math&gt; will beat &lt;math&gt;B&lt;/math&gt; overall. Therefore, out of the &lt;math&gt;1024&lt;/math&gt; possibilites, &lt;math&gt;386+252 = 638&lt;/math&gt; ways where &lt;math&gt;A&lt;/math&gt; wins, so the desired probability is &lt;math&gt;\frac{638}{1024} = \frac{319}{512}&lt;/math&gt;, and &lt;math&gt;m+n=\boxed{831}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=91276 AMC historical results 2018-02-16T02:13:15Z <p>Zeroman: </p> <hr /> <div>&lt;!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --&gt;<br /> This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br /> <br /> ==2018==<br /> ===AMC 10A===<br /> *Average score: TBD<br /> *AIME floor: TBD<br /> *Distinguished Honor Roll: TBD<br /> <br /> ===AMC 10B===<br /> *Average score: TBD<br /> *AIME floor: TBD<br /> *Distinguished Honor Roll: TBD<br /> <br /> ===AMC 12A===<br /> *Average score: TBD<br /> *AIME floor: TBD<br /> *Distinguished Honor Roll floor: TBD<br /> <br /> ===AMC 12B===<br /> *Average score: TBD<br /> *AIME floor: TBD<br /> *Distinguished Honor Roll: TBD<br /> <br /> ==2017==<br /> ===AMC 10A===<br /> *Average score: 59.33<br /> *AIME floor: 112.5<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 66.55<br /> *AIME floor: 120<br /> *DHR: 136.5<br /> <br /> ===AMC 12A===<br /> *Average score: 60.32<br /> *AIME floor: 96<br /> *DHR: 115.5<br /> <br /> ===AMC 12B===<br /> *Average score: 58.35<br /> *AIME floor: 100.5<br /> *DHR: 129<br /> <br /> ===AIME I===<br /> *Average score: 5.69<br /> *Median score: 6<br /> *USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br /> *USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.64<br /> *Median score: 6<br /> *USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br /> *USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br /> <br /> ==2016==<br /> ===AMC 10A===<br /> *Average score: 65.3<br /> *AIME floor: 110<br /> *DHR: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 65.4<br /> *AIME floor: 110<br /> *DHR: 124.5<br /> <br /> ===AMC 12A===<br /> *Average score: 59.06<br /> *AIME floor: 92<br /> *DHR: 110<br /> <br /> ===AMC 12B===<br /> *Average score: 67.96<br /> *AIME floor: 100<br /> *DHR: 127.5<br /> <br /> ===AIME I===<br /> *Average score: 5.83<br /> *Median score: 6<br /> *USAMO cutoff: 220<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 4.33<br /> *Median score: 4<br /> *USAMO cutoff: 205<br /> *USAJMO cutoff: 200<br /> <br /> ==2015==<br /> ===AMC 10A===<br /> *Average score: 73.39<br /> *AIME floor: 106.5<br /> *DHR: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.10<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 69.90<br /> *AIME floor: 99<br /> *DHR: 117<br /> <br /> ===AMC 12B===<br /> *Average score: 66.92<br /> *AIME floor: 100<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.29<br /> *Median score: 5<br /> *USAMO cutoff: 219.0<br /> *USAJMO cutoff: 213.0<br /> <br /> ===AIME II===<br /> *Average score: 6.63<br /> *Median score: 6<br /> *USAMO cutoff: 229.0<br /> *USAJMO cutoff: 223.5<br /> <br /> ==2014==<br /> ===AMC 10A===<br /> *Average score: 63.83<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 10B===<br /> *Average score: 71.44<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 64.01<br /> *AIME floor: 93<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.11<br /> *AIME floor: 100<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 4.88<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AIME II===<br /> *Average score: 5.49<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ==2013==<br /> ===AMC 10A===<br /> *Average score: 72.50<br /> *AIME floor: 108<br /> *DHR: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 72.62<br /> *AIME floor: 120<br /> *DHR: 129<br /> <br /> ===AMC 12A===<br /> *Average score: 65.06<br /> *AIME floor: 88.5<br /> *DHR: 106.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.21<br /> *AIME floor: 93<br /> *DHR: 108<br /> <br /> ===AIME I===<br /> *Average score: 4.69<br /> *Median score: 4<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 6.56<br /> *Median score: 6<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ==2012==<br /> ===AMC 10A===<br /> *Average score: 72.51<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.59<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 64.62<br /> *AIME floor: 94.5<br /> <br /> ===AMC 12B===<br /> *Average score: 70.08<br /> *AIME floor: 99<br /> <br /> ===AIME I===<br /> *Average score: 5.13<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AIME II===<br /> *Average score: 4.94<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ==2011==<br /> ===AMC 10A===<br /> *Average score: 67.61<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 71.78<br /> *AIME floor: 117<br /> <br /> ===AMC 12A===<br /> *Average score: 66.77<br /> *AIME floor: 93<br /> <br /> ===AMC 12B===<br /> *Average score: 64.71<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: 5.47<br /> *Median score: <br /> *USAMO cutoff: 215.5<br /> *USAJMO cutoff: 196.5<br /> <br /> ===AIME II===<br /> *Average score: 2.23<br /> *Median score: <br /> *USAMO cutoff: 188<br /> *USAJMO cutoff: 179<br /> <br /> ==2010==<br /> ===AMC 10A===<br /> *Average score: 68.11<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 68.57<br /> *AIME floor: 118.5<br /> <br /> ===AMC 12A===<br /> *Average score: 61.02<br /> *AIME floor: 88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 59.58<br /> *AIME floor: 88.5<br /> <br /> ===AIME I===<br /> *Average score: 5.90<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AIME II===<br /> *Average score: 3.39<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ==2009==<br /> ===AMC 10A===<br /> *Average score: 67.41<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 74.73<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 66.37<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 71.88<br /> *AIME floor: 100 (Top 5% (1.00))<br /> <br /> ===AIME I===<br /> *Average score: 4.17<br /> *Median score: 4<br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score: 3.27<br /> *Median score: 3<br /> *USAMO floor:<br /> <br /> ==2008==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 12A===<br /> *Average score: 65.6<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.9<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2007==<br /> <br /> ===AMC 10A===<br /> *Average score: 67.9<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 61.5<br /> *AIME floor: 115.5<br /> <br /> ===AMC 12A===<br /> *Average score: 66.8<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 73.1<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5<br /> *Median score: 3<br /> *USAMO floor: 6<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2006==<br /> ===AMC 10A===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 68.5<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 85.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 85.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2005==<br /> ===AMC 10A===<br /> *Average score: 74.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 78.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 83.4<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2004==<br /> ===AMC 10A===<br /> *Average score: 69.1<br /> *AIME floor: 110<br /> <br /> ===AMC 10B===<br /> *Average score: 80.4<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 73.9<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 84.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2003==<br /> ===AMC 10A===<br /> *Average score: 74.4<br /> *AIME floor: 119<br /> <br /> ===AMC 10B===<br /> *Average score: 79.6<br /> *AIME floor: 121<br /> <br /> ===AMC 12A===<br /> *Average score: 77.8<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 76.6<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2002==<br /> ===AMC 10A===<br /> *Average score: 68.5<br /> *AIME floor: 115<br /> <br /> ===AMC 10B===<br /> *Average score: 74.9<br /> *AIME floor: 118<br /> <br /> ===AMC 12A===<br /> *Average score: 72.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 80.8<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2001==<br /> ===AMC 10===<br /> *Average score: 67.8<br /> *AIME floor: 116<br /> <br /> ===AMC 12===<br /> *Average score: 56.6<br /> *AIME floor: 84<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2000==<br /> ===AMC 10===<br /> *Average score: 64.2<br /> *AIME floor: <br /> <br /> ===AMC 12===<br /> *Average score: 64.9<br /> *AIME floor: <br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1999==<br /> ===AHSME===<br /> *Average score: 68.8<br /> *AIME floor:<br /> <br /> ===AIME===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1998==<br /> none<br /> <br /> ==1997==<br /> ==1996==<br /> ==1995==<br /> ==1994==<br /> ==1993==<br /> ==1992==<br /> ==1991==<br /> ==1990==<br /> ==1989==<br /> ==1988==<br /> ==1987==<br /> ==1986==<br /> ==1985==<br /> ==1984==<br /> ==1983==<br /> ==1982==<br /> ==1981==<br /> ==1980==<br /> ==1979==<br /> ==1978==<br /> ==1977==<br /> ==1976==<br /> ==1975==<br /> ==1974==<br /> ==1973==<br /> ==1972==<br /> ==1971==<br /> ==1970==<br /> ==1969==<br /> ==1968==<br /> ==1967==<br /> ==1966==<br /> ==1965==<br /> ==1964==<br /> ==1963==<br /> ==1962==<br /> ==1961==<br /> ==1960==<br /> <br /> ==1959==</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_8&diff=90865 2015 AIME I Problems/Problem 8 2018-02-10T01:13:40Z <p>Zeroman: </p> <hr /> <div>==Problem==<br /> <br /> For positive integer &lt;math&gt;n&lt;/math&gt;, let &lt;math&gt;s(n)&lt;/math&gt; denote the sum of the digits of &lt;math&gt;n&lt;/math&gt;. Find the smallest positive integer satisfying &lt;math&gt;s(n) = s(n+864) = 20&lt;/math&gt;.<br /> <br /> ==Solution==<br /> You know whatever &lt;math&gt;n&lt;/math&gt; is, it has to have 3 digits, because if it had only two, the maximum of &lt;math&gt;s(n)&lt;/math&gt; is 18. <br /> <br /> Now let &lt;math&gt;n=100a_2+10a_1+a_0&lt;/math&gt;<br /> <br /> So first we know, &lt;math&gt;a_2+a_1+a_0=20&lt;/math&gt;. Okay now we have to split into cases based on which digit gets carried. This meaning, when you add a 3 digit number to 864, we have to know when to carry the digits. Note that if you don't understand any of the steps I take, just try adding any 3-digit number to 864 regularly (using the old-fashioned &quot;put one number over the other&quot; method, not mental calculation), and observe what you do at each step.<br /> <br /> (1) None of the digit gets carried over to the next space:<br /> So this means &lt;math&gt;a_2&lt;2, a_1&lt;4&lt;/math&gt; and &lt;math&gt;a_0&lt;6&lt;/math&gt;. So <br /> <br /> &lt;math&gt;s(864+n)=(8+a_2)+(6+a_1)+(4+a_0)=38&lt;/math&gt;<br /> So it doesn't work. Now:<br /> <br /> (2) &lt;math&gt;a_2+8&lt;/math&gt; is the only one that carries over<br /> So this means &lt;math&gt;a_2&gt;1, a_1&lt;4&lt;/math&gt; and &lt;math&gt;a_0&lt;6&lt;/math&gt;. So<br /> <br /> &lt;math&gt;s(864+n)=1+(8+a_2-10)+(6+a_1)+(a_0+4)=29&lt;/math&gt; <br /> <br /> (3)&lt;math&gt;a_0+4&lt;/math&gt; is the only one that carries over.<br /> So<br /> <br /> &lt;math&gt;s(864+n)=(8+a_2)+(6+a_1+1)+(4+a_0-10)=29&lt;/math&gt;<br /> <br /> (4)The first and second digit carry over (but not the third)<br /> <br /> &lt;math&gt;s(864+n)=1+(8+a_2-10+1)+(6+a_1-10)+(4+a_0)=20&lt;/math&gt;<br /> <br /> Aha! This case works but we still have to make sure it's possible for &lt;math&gt;a_2+a_1+a_0=20&lt;/math&gt; (We assumed this is true, so we have to find a number that works.) Since only the second and first digit carry over, &lt;math&gt;a_2&gt;0, a_1&gt;3&lt;/math&gt; and &lt;math&gt;a_0&lt;6&lt;/math&gt;. The smallest value we can get with this is 695. Let's see if we can find a smaller one:<br /> <br /> (5)The first and third digit carry over (but not the second)<br /> <br /> &lt;math&gt;s(864+n)=1+(8+a_2-10)+(7+a_1)+(4+a_0-10)=20&lt;/math&gt;<br /> <br /> The largest value for the middle digit is 2, so the other digits have to be both 9's. So the smallest possible value is 929<br /> <br /> (6) All the digits carry over<br /> <br /> &lt;math&gt;s(864+n)=1+(9+a_2-10)+(7+a_1-10)+(4+a_0-10)=\text{Way less than 20}&lt;/math&gt;<br /> <br /> <br /> So the answer is &lt;math&gt;\boxed{695}&lt;/math&gt; which after a quick test, does indeed work.<br /> <br /> ==Solution 2==<br /> First, it is easy to verify that &lt;math&gt;695&lt;/math&gt; works and that no other numbers beginning with the digit 6 work (i.e. &lt;math&gt;686, 677, 668, 659&lt;/math&gt; do not work).<br /> <br /> Suppose by contradiction that there is a smaller valid &lt;math&gt;n&lt;/math&gt;, where the leading digit of the three-digit number &lt;math&gt;n&lt;/math&gt; is 5 or less. (Two-digit &lt;math&gt;n&lt;/math&gt; obviously do not work because 9 + 9 &lt; 20.) Clearly &lt;math&gt;n &gt; 200&lt;/math&gt; because the smallest three-digit number whose digits sum to 20 is &lt;math&gt;299&lt;/math&gt;. Also, because the second digit is at most 9, the units digit is at least 6, which means that the addition &lt;math&gt;N = n + 864&lt;/math&gt; regroups in the ones place. Then the units digit of &lt;math&gt;N&lt;/math&gt; is clearly less than 4. But as &lt;math&gt;1000 &lt; 200 + 864 &lt; N &lt; 600 + 864 = 1464&lt;/math&gt;, the sum of the thousands digit and the hundredth digit is at most 5. Because the second digit is at most 9, the sum of the digits of &lt;math&gt;N&lt;/math&gt; is at most &lt;math&gt;5 + 9 + 4 &lt; 20&lt;/math&gt;, contradiction. Hence &lt;math&gt;\boxed{695}&lt;/math&gt; is the answer.<br /> <br /> ==Motivation for Solution 2==<br /> During the real test, I immediately noticed that &lt;math&gt;n&lt;/math&gt; must be less than 1000 (AIME problem) and that &lt;math&gt;n&lt;/math&gt; must be a three-digit number. Therefore, I began casework on the leading digit of &lt;math&gt;n&lt;/math&gt;. The casework was not intensive (how many ways are there to have digits sum to 20?) and I eventually got 695 as my answer. The rigorous proof that 695 was the smallest came afterwards.<br /> <br /> <br /> ==Solution 3==<br /> First of all, notice that the smallest &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;s(n) = 20&lt;/math&gt; is &lt;math&gt;299&lt;/math&gt;. Also, if &lt;math&gt;s(n + 864) = 20&lt;/math&gt;, &lt;math&gt;s(n - 136) = 19&lt;/math&gt; (because subtracting &lt;math&gt;1000&lt;/math&gt; from the number removes the &lt;math&gt;1&lt;/math&gt; in the thousands place). After checking &lt;math&gt;s(n - 136)&lt;/math&gt; for various &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;s(n) = 20&lt;/math&gt;, we see that we need to have a carry when subtracting &lt;math&gt;136&lt;/math&gt;. To have this, we must either have a &lt;math&gt;2&lt;/math&gt; in the tens place or a &lt;math&gt;5&lt;/math&gt; in the units place. The minimum &lt;math&gt;n&lt;/math&gt; for the former is &lt;math&gt;929&lt;/math&gt;, and for the latter it is &lt;math&gt;695&lt;/math&gt;. We check and see that &lt;math&gt;s(695-136) = s(559) = 19&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{695}&lt;/math&gt;.<br /> == See also ==<br /> {{AIME box|year=2015|n=I|num-b=7|num-a=9}}<br /> {{MAA Notice}}<br /> [[Category:Introductory Number Theory Problems]]</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1994_AIME_Problems/Problem_10&diff=87286 1994 AIME Problems/Problem 10 2017-08-30T20:01:17Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> In triangle &lt;math&gt;ABC,\,&lt;/math&gt; angle &lt;math&gt;C&lt;/math&gt; is a right angle and the altitude from &lt;math&gt;C\,&lt;/math&gt; meets &lt;math&gt;\overline{AB}\,&lt;/math&gt; at &lt;math&gt;D.\,&lt;/math&gt; The lengths of the sides of &lt;math&gt;\triangle ABC\,&lt;/math&gt; are integers, &lt;math&gt;BD=29^3,\,&lt;/math&gt; and &lt;math&gt;\cos B=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 /> Since &lt;math&gt;\triangle ABC \sim \triangle CBD&lt;/math&gt;, we have &lt;math&gt;\frac{BC}{AB} = \frac{29^3}{BC} \Longrightarrow BC^2 = 29^3 AB&lt;/math&gt;. It follows that &lt;math&gt;29^2 | BC&lt;/math&gt; and &lt;math&gt;29 | AB&lt;/math&gt;, so &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; are in the form &lt;math&gt;29^2 a&lt;/math&gt; and &lt;math&gt;29 a^2&lt;/math&gt;, respectively. <br /> <br /> By the [[Pythagorean Theorem]], we find that &lt;math&gt;AC^2 + BC^2 = AB^2 \Longrightarrow (29^2a)^2 + AC^2 = (29 a^2)^2&lt;/math&gt;, so &lt;math&gt;29a | AC&lt;/math&gt;. Letting &lt;math&gt;b = AC / 29a&lt;/math&gt;, we obtain after dividing through by &lt;math&gt;(29a)^2&lt;/math&gt;, &lt;math&gt;29^2 = a^2 - b^2 = (a-b)(a+b)&lt;/math&gt;. As &lt;math&gt;a,b \in \mathbb{Z}&lt;/math&gt;, the pairs of factors of &lt;math&gt;29^2&lt;/math&gt; are &lt;math&gt;(1,29^2)(29,29)&lt;/math&gt;; clearly &lt;math&gt;b = \frac{AC}{29a} \neq 0&lt;/math&gt;, so &lt;math&gt;a-b = 1, a+b= 29^2&lt;/math&gt;. Then, &lt;math&gt;a = \frac{1+29^2}{2} = 421&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;\cos B = \frac{BC}{AB} = \frac{29^2 a}{29a^2} = \frac{29}{421}&lt;/math&gt;, and &lt;math&gt;m+n = \boxed{450}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> We will solve for &lt;math&gt;\cos B&lt;/math&gt; using &lt;math&gt;\triangle CBD&lt;/math&gt;, which gives us &lt;math&gt;\cos B = \frac{29^3}{BC}&lt;/math&gt;. By the Pythagorean Theorem on &lt;math&gt;\triangle CBD&lt;/math&gt;, we have &lt;math&gt;BC^2 - DC^2 = (BC + DC)(BC - DC) = 29^6&lt;/math&gt;. Trying out factors of &lt;math&gt;29^6&lt;/math&gt;, we can either guess and check or just guess to find that &lt;math&gt;BC + DC = 29^4&lt;/math&gt; and &lt;math&gt;BC - DC = 29^2&lt;/math&gt; (The other pairs give answers over 999). Adding these, we have &lt;math&gt;2BC = 29^4 + 29^2&lt;/math&gt; and &lt;math&gt;\frac{29^3}{BC} = \frac{2*29^3}{29^2 (29^2 +1)} = \frac{58}{842} = \frac{29}{421}&lt;/math&gt;, and our answer is &lt;math&gt;\boxed{450}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1994|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_9&diff=86364 1986 AIME Problems/Problem 9 2017-07-13T19:16:13Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB= 425&lt;/math&gt;, &lt;math&gt;BC=450&lt;/math&gt;, and &lt;math&gt;AC=510&lt;/math&gt;. An interior [[point]] &lt;math&gt;P&lt;/math&gt; is then drawn, and [[segment]]s are drawn through &lt;math&gt;P&lt;/math&gt; [[parallel]] to the sides of the [[triangle]]. If these three segments are of an equal length &lt;math&gt;d&lt;/math&gt;, find &lt;math&gt;d&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solution ==<br /> ===Solution 1 ===<br /> &lt;asy&gt;<br /> size(200);<br /> pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10);<br /> pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425));<br /> /* construct remaining points */<br /> pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C);<br /> pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); <br /> /* Construct P */<br /> pair P = IP(D--Ea,E--Fa); dot(MP(&quot;P&quot;,P,NE)); <br /> pair X = IP(L(A,P,4), B--C); dot(MP(&quot;X&quot;,X,NW));<br /> pair Y = IP(L(B,P,4), C--A); dot(MP(&quot;Y&quot;,Y,NE));<br /> pair Z = IP(L(C,P,4), A--B); dot(MP(&quot;Z&quot;,Z,N));<br /> <br /> D(A--X); D(B--Y); D(C--Z);<br /> D(MP(&quot;A&quot;,A,s)--MP(&quot;B&quot;,B,N,s)--MP(&quot;C&quot;,C,s)--cycle);<br /> MP(&quot;450&quot;,(B+C)/2,NW);MP(&quot;425&quot;,(A+B)/2,NE);MP(&quot;510&quot;,(A+C)/2);<br /> &lt;/asy&gt;<br /> <br /> Construct cevians &lt;math&gt;AX&lt;/math&gt;, &lt;math&gt;BY&lt;/math&gt; and &lt;math&gt;CZ&lt;/math&gt; through &lt;math&gt;P&lt;/math&gt;. Place masses of &lt;math&gt;x,y,z&lt;/math&gt; on &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; respectively; then &lt;math&gt;P&lt;/math&gt; has mass &lt;math&gt;x+y+z&lt;/math&gt;.<br /> <br /> Notice that &lt;math&gt;Z&lt;/math&gt; has mass &lt;math&gt;x+y&lt;/math&gt;. On the other hand, by similar triangles, &lt;math&gt;\frac{CP}{CZ} = \frac{d}{AB}&lt;/math&gt;. Hence by mass points we find that &lt;cmath&gt; \frac{x+y}{x+y+z} = \frac{d}{AB} &lt;/cmath&gt; Similarly, we obtain &lt;cmath&gt; \frac{y+z}{x+y+z} = \frac{d}{BC} \qquad \text{and} \qquad \frac{z+x}{x+y+z} = \frac{d}{CA} &lt;/cmath&gt; Summing these three equations yields &lt;cmath&gt; \frac{d}{AB} + \frac{d}{BC} + \frac{d}{CA} = \frac{x+y}{x+y+z} + \frac{y+z}{x+y+z} + \frac{z+x}{x+y+z} = \frac{2x+2y+2z}{x+y+z} = 2 &lt;/cmath&gt;<br /> <br /> Hence, &lt;center&gt;&lt;math&gt; d = \frac{2}{\frac{1}{AB} + \frac{1}{BC} + \frac{1}{CA}} = \frac{2}{\frac{1}{510} + \frac{1}{450} + \frac{1}{425}} = \frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}&lt;/math&gt;&lt;math&gt;= \frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}=\frac{10}{\frac{10}{306}} = \boxed{306}&lt;/math&gt;&lt;/center&gt;<br /> <br /> === Solution 2 ===<br /> &lt;center&gt;&lt;asy&gt;<br /> size(200);<br /> pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10);<br /> pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425));<br /> /* construct remaining points */<br /> pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C);<br /> pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); <br /> D(MP(&quot;A&quot;,A,s)--MP(&quot;B&quot;,B,N,s)--MP(&quot;C&quot;,C,s)--cycle);<br /> dot(MP(&quot;D&quot;,D,NE,s));dot(MP(&quot;E&quot;,E,NW,s));dot(MP(&quot;F&quot;,F,s));dot(MP(&quot;D'&quot;,Da,NE,s));dot(MP(&quot;E'&quot;,Ea,NW,s));dot(MP(&quot;F'&quot;,Fa,s));<br /> D(D--Ea);D(Da--F);D(Fa--E);<br /> MP(&quot;450&quot;,(B+C)/2,NW);MP(&quot;425&quot;,(A+B)/2,NE);MP(&quot;510&quot;,(A+C)/2);<br /> /*P copied from above solution*/<br /> pair P = IP(D--Ea,E--Fa); dot(MP(&quot;P&quot;,P,N)); <br /> &lt;/asy&gt;&lt;/center&gt; &lt;!-- Asymptote replacement for Image:1986_AIME-9.png by azjps --&gt;<br /> <br /> Let the points at which the segments hit the triangle be called &lt;math&gt;D, D', E, E', F, F'&lt;/math&gt; as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are [[similar triangles|similar]] (&lt;math&gt;\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF&lt;/math&gt;). The remaining three sections are [[parallelogram]]s.<br /> <br /> Since &lt;math&gt;PDAF'&lt;/math&gt; is a parallelogram, we find &lt;math&gt;PF' = AD&lt;/math&gt;, and similarly &lt;math&gt;PE = BD'&lt;/math&gt;. So &lt;math&gt;d = PF' + PE = AD + BD' = 425 - DD'&lt;/math&gt;. Thus &lt;math&gt;DD' = 425 - d&lt;/math&gt;. By the same logic, &lt;math&gt;EE' = 450 - d&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;\triangle DPD' \sim \triangle ABC&lt;/math&gt;, we have the [[proportion]]:<br /> <br /> &lt;center&gt;&lt;math&gt;\frac{425-d}{425} = \frac{PD}{510} \Longrightarrow PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d&lt;/math&gt;&lt;/center&gt;<br /> <br /> Doing the same with &lt;math&gt;\triangle PEE'&lt;/math&gt;, we find that &lt;math&gt;PE' =510 - \frac{17}{15}d&lt;/math&gt;. Now, &lt;math&gt;d = PD + PE' = 510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = \boxed{306}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Define the points the same as above.<br /> <br /> Let &lt;math&gt;[CE'PF] = a&lt;/math&gt;, &lt;math&gt;[E'EP] = b&lt;/math&gt;, &lt;math&gt;[BEPD'] = c&lt;/math&gt;, &lt;math&gt;[D'PD] = d&lt;/math&gt;, &lt;math&gt;[DAF'P] = e&lt;/math&gt; and &lt;math&gt;[F'D'P] = f&lt;/math&gt;<br /> <br /> The key theorem we apply here is that the ratio of the areas of 2 similar triangles is the ratio of a pair of corresponding sides squared.<br /> <br /> Let the length of the segment be &lt;math&gt;x&lt;/math&gt; and the area of the triangle be &lt;math&gt;A&lt;/math&gt;, using the theorem, we get:<br /> <br /> &lt;math&gt;\frac {c + e + d}{A} = \left(\frac {x}{BC}\right)^2&lt;/math&gt;, &lt;math&gt;\frac {b + c + d}{A}= \left(\frac {x}{AC}\right)^2&lt;/math&gt;, &lt;math&gt;\frac {a + b + f}{A} = \left(\frac {x}{AB}\right)^2&lt;/math&gt;<br /> adding all these together and using &lt;math&gt;a + b + c + d + e + f = A&lt;/math&gt; we get<br /> &lt;math&gt;\frac {f + d + b}{A} + 1 = x^2*\left(\frac {1}{BC^2} + \frac {1}{AC^2} + \frac {1}{AB^2}\right)&lt;/math&gt;<br /> <br /> Using [[corresponding angles]] from parallel lines, it is easy to show that &lt;math&gt;\triangle ABC \sim \triangle F'PF&lt;/math&gt;, since &lt;math&gt;ADPF'&lt;/math&gt; and &lt;math&gt;CFPE'&lt;/math&gt; are parallelograms, it is easy to show that &lt;math&gt;FF' = AC - x&lt;/math&gt;<br /> <br /> Now we have the side length [[ratio]], so we have the area ratio<br /> &lt;math&gt;\frac {f}{A} = \left(\frac {AC - x}{AC}\right)^2 = \left(1 - \frac {x}{AC}\right)^2&lt;/math&gt;, by symmetry, we have<br /> &lt;math&gt;\frac {d}{A} = \left(1 - \frac {x}{AB}\right)^2&lt;/math&gt; and &lt;math&gt;\frac {b}{A} = \left(1 - \frac {x}{BC}\right)^2&lt;/math&gt;<br /> <br /> Substituting these into our initial equation, we have<br /> &lt;math&gt;1 + \sum_{cyc}\left(1 - \frac {x}{AB}\right) - \frac {x^2}{AB^2} = 0&lt;/math&gt;<br /> &lt;math&gt;1 + \sum_{cyc}1 - 2*\frac {x}{AB} = 0&lt;/math&gt;<br /> &lt;math&gt;\frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x&lt;/math&gt;<br /> answer follows after some hideous computation.<br /> <br /> ===Solution 4===<br /> Refer to the diagram in solution 2; let &lt;math&gt;a^2=[E'EP]&lt;/math&gt;, &lt;math&gt;b^2=[D'DP]&lt;/math&gt;, and &lt;math&gt;c^2=[F'FP]&lt;/math&gt;. Now, note that &lt;math&gt;[E'BD]&lt;/math&gt;, &lt;math&gt;[D'DP]&lt;/math&gt;, and &lt;math&gt;[E'EP]&lt;/math&gt; are similar, so through some similarities we find that &lt;math&gt;\frac{E'P}{PD}=\frac{a}{b}\implies\frac{E'D}{PD}=\frac{a+b}{b}\implies[E'BD]=b^2\left(\frac{a+b}{b}\right)^2=(a+b)^2&lt;/math&gt;. Similarly, we find that &lt;math&gt;[D'AF]=(b+c)^2&lt;/math&gt; and &lt;math&gt;[F'CE]=(c+a)^2&lt;/math&gt;, so &lt;math&gt;[ABC]=(a+b+c)^2&lt;/math&gt;. Now, again from similarity, it follows that &lt;math&gt;\frac{d}{510}=\frac{a+b}{a+b+c}&lt;/math&gt;, &lt;math&gt;\frac{d}{450}=\frac{b+c}{a+b+c}&lt;/math&gt;, and &lt;math&gt;\frac{d}{425}=\frac{c+a}{a+b+c}&lt;/math&gt;, so adding these together, simplifying, and solving gives &lt;math&gt;d=\frac{2}{\frac{1}{425}+\frac{1}{450}+\frac{1}{510}}=\frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}=\frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}&lt;/math&gt;<br /> &lt;math&gt;=\frac{10}{\frac{10}{306}}=\boxed{306}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> Refer to the diagram from Solution 2. Notice that because &lt;math&gt;CE'PF&lt;/math&gt;, &lt;math&gt;AF'PD&lt;/math&gt;, and &lt;math&gt;BD'PE&lt;/math&gt; are parallelograms, &lt;math&gt;\overline{DD'} = 425-d&lt;/math&gt;, &lt;math&gt;\overline{EE'} = 450-d&lt;/math&gt;, and &lt;math&gt;\overline{FF'} = 510-d&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;F'P = x&lt;/math&gt;. Then, because &lt;math&gt;\triangle ABC \sim \triangle F'PF&lt;/math&gt;, &lt;math&gt;\frac{AB}{AC}=\frac{F'P}{F'F}&lt;/math&gt;, so &lt;math&gt;\frac{425}{510}=\frac{x}{510-d}&lt;/math&gt;. Simplifying the LHS and cross-multiplying, we have &lt;math&gt;6x=2550-5d&lt;/math&gt;. From the same triangles, we can find that &lt;math&gt;FP=\frac{18}{17}x&lt;/math&gt;.<br /> <br /> &lt;math&gt;\triangle PEE'&lt;/math&gt; is also similar to &lt;math&gt;\triangle F'PF&lt;/math&gt;. Since &lt;math&gt;EF'=d&lt;/math&gt;, &lt;math&gt;EP=d-x&lt;/math&gt;. We now have &lt;math&gt;\frac{PE}{EE'}=\frac{F'P}{FP}&lt;/math&gt;, and &lt;math&gt;\frac{d-x}{450-d}=\frac{17}{18}&lt;/math&gt;. Cross multiplying, we have &lt;math&gt;18d-18x=450*17-17d&lt;/math&gt;. Using the previous equation to substitute for &lt;math&gt;x&lt;/math&gt;, we have: &lt;cmath&gt;18d-3*2550+15d=450*17-17d&lt;/cmath&gt; This is a linear equation in one variable, and we can solve to get &lt;math&gt;d=\boxed{306}&lt;/math&gt;<br /> <br /> *I did not show the multiplication in the last equation because most of it cancels out when solving.<br /> <br /> (Note: I chose &lt;math&gt;F'P&lt;/math&gt; to be &lt;math&gt;x&lt;/math&gt; only because that is what I had written when originally solving. The solution would work with other choices for &lt;math&gt;x&lt;/math&gt;.)<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_6&diff=86351 1984 AIME Problems/Problem 6 2017-07-12T17:55:46Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> Three circles, each of [[radius]] &lt;math&gt;\displaystyle 3&lt;/math&gt;, are drawn with centers at &lt;math&gt;\displaystyle (14, 92)&lt;/math&gt;, &lt;math&gt;\displaystyle (17, 76)&lt;/math&gt;, and &lt;math&gt;\displaystyle (19, 84)&lt;/math&gt;. A [[line]] passing through &lt;math&gt;\displaystyle (17,76)&lt;/math&gt; is such that the total area of the parts of the three circles to one side of the line is equal to the total [[area]] of the parts of the three circles to the other side of it. What is the [[absolute value]] of the [[slope]] of this line?<br /> <br /> __TOC__<br /> == Solution ==<br /> [[Image:1984_AIME-6.png]]<br /> <br /> The line passes through the center of the second circle; hence it is the circle's [[diameter]] and splits the circle into two equal areas. For the rest of the problem, we do not have to worry about that circle. <br /> <br /> === Solution 1 ===<br /> Draw the [[midpoint]] of &lt;math&gt;\overline{AC}&lt;/math&gt; (the centers of the other two circles), and call it &lt;math&gt;M&lt;/math&gt;. If we draw the feet of the [[perpendicular]]s from &lt;math&gt;A,C&lt;/math&gt; to the line (call &lt;math&gt;E,F&lt;/math&gt;), we see that &lt;math&gt;\triangle AEM\cong \triangle CFM&lt;/math&gt; by [[HA congruency]]; hence &lt;math&gt;M&lt;/math&gt; lies on the line. The coordinates of &lt;math&gt;M&lt;/math&gt; are &lt;math&gt;\left(\frac{19+14}{2},\frac{84+92}{2}\right) = \left(\frac{33}{2},88\right)&lt;/math&gt;. <br /> <br /> Thus, the slope of the line is &lt;math&gt;\frac{88 - 76}{\frac{33}{2} - 17} = -24&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{024}&lt;/math&gt;.<br /> <br /> ''Remark'': Notice the fact that the radius is 3 is not used in this problem; in fact changing the radius does not affect the answer.<br /> <br /> === Solution 2 ===<br /> Define &lt;math&gt;E,F&lt;/math&gt; to be the feet of the perpendiculars from &lt;math&gt;A,C&lt;/math&gt; to the line (same as above). The equation of the line is &lt;math&gt;y = mx + b&lt;/math&gt;; substituting &lt;math&gt;y=76,x=17&lt;/math&gt; gives us that &lt;math&gt;b = 76 - 17m&lt;/math&gt;, so the line is &lt;math&gt;y = mx + (76 - 17m)&lt;/math&gt;. &lt;math&gt;AE = CF&lt;/math&gt; by the HA argument above and [[CPCTC]], so we can use the distance of a point to a line formula and equate.<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;\left|\frac{m(14) - 92 + (76 - 17m)}{\sqrt{m^2 + 1}}\right| = \left|\frac{m(19) - 84 + (76 - 17m)}{\sqrt{m^2 + 1}}\right|&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;-3m - 16 = -2m + 8&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;m = -24&lt;/math&gt;&lt;/div&gt;<br /> <br /> And &lt;math&gt;|-24| = 24&lt;/math&gt;.<br /> <br /> === Solution 3 (non-rigorous)===<br /> Consider the region of the plane between &lt;math&gt;x=16&lt;/math&gt; and &lt;math&gt;x=17&lt;/math&gt;. The parts of the circles centered at &lt;math&gt;(14,92)&lt;/math&gt; and &lt;math&gt;(19,84)&lt;/math&gt; in this region have equal area. This is by symmetry- the lines defining the region are 2 units away from the centers of each circle and therefore cut off congruent segments. We will draw the line in a way that uses this symmetry and makes identical cuts on the circles. Since &lt;math&gt;(17,76)&lt;/math&gt; is &lt;math&gt;8&lt;/math&gt; units below the center of the lower circle, we will have the line exit the region &lt;math&gt;8&lt;/math&gt; units above the center of the upper circle, at &lt;math&gt;(16,100)&lt;/math&gt;. We then find that the slope of the line is &lt;math&gt;-24&lt;/math&gt; and our answer is &lt;math&gt;\boxed{024}&lt;/math&gt;.<br /> <br /> (Note: this solution does not feel rigorous when working through it, but it can be checked easily. In the above diagram, the point &lt;math&gt;M&lt;/math&gt; is marked. Rotate the aforementioned region of the plane &lt;math&gt;180^\circ&lt;/math&gt; about point &lt;math&gt;M&lt;/math&gt;, and the fact that certain areas are equal is evident.)<br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=5|num-a=7}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=Product_Rule&diff=86303 Product Rule 2017-07-09T21:53:00Z <p>Zeroman: Created page with &quot;The Product Rule is a method of taking the derivative of the product of two functions. It states that the derivative of &lt;math&gt;f(x)*g(x) = f'(x)g(x)+g'(x)f(x)&lt;/math&gt;.&quot;</p> <hr /> <div>The Product Rule is a method of taking the [[derivative]] of the product of two functions. It states that the derivative of &lt;math&gt;f(x)*g(x) = f'(x)g(x)+g'(x)f(x)&lt;/math&gt;.</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=Quotient_Rule&diff=86302 Quotient Rule 2017-07-09T21:51:23Z <p>Zeroman: Created page with &quot;The Quotient Rule is a rule for taking the derivative of the quotient of two functions. It states that the derivative of &lt;math&gt;\frac{f(x)}{g(x)}&lt;/math&gt; is: &lt;cmath&gt;\frac{f'...&quot;</p> <hr /> <div>The Quotient Rule is a rule for taking the [[derivative]] of the quotient of two functions. It states that the derivative of &lt;math&gt;\frac{f(x)}{g(x)}&lt;/math&gt; is: &lt;cmath&gt;\frac{f'(x)g(x)-g'(x)f(x)}{(g'(x)^2)}&lt;/cmath&gt;.<br /> <br /> This result can be derived from using the [[Product Rule]] on the functions &lt;math&gt;f(x)&lt;/math&gt; and &lt;math&gt;\frac{1}{g(x)}&lt;/math&gt;.</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_9&diff=86301 1983 AIME Problems/Problem 9 2017-07-09T21:46:55Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> Find the minimum value of &lt;math&gt;\frac{9x^2\sin^2 x + 4}{x\sin x}&lt;/math&gt; for &lt;math&gt;0 &lt; x &lt; \pi&lt;/math&gt;.<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Let &lt;math&gt;y=x\sin{x}&lt;/math&gt;. We can rewrite the expression as &lt;math&gt;\frac{9y^2+4}{y}=9y+\frac{4}{y}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;x&gt;0&lt;/math&gt; and &lt;math&gt;\sin{x}&gt;0&lt;/math&gt; because &lt;math&gt;0&lt; x&lt;\pi&lt;/math&gt;, we have &lt;math&gt;y&gt;0&lt;/math&gt;. So we can apply [[AM-GM]]:<br /> <br /> &lt;cmath&gt;9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12&lt;/cmath&gt;<br /> <br /> The equality holds when &lt;math&gt;9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23&lt;/math&gt;.<br /> <br /> Therefore, the minimum value is &lt;math&gt;\boxed{012}&lt;/math&gt;. This is reached when plugging in &lt;math&gt;2/3&lt;/math&gt; for &lt;math&gt;x\sin{x}&lt;/math&gt; in the original equation (when &lt;math&gt;x\sin{x}=\frac23&lt;/math&gt;; since &lt;math&gt;x\sin x&lt;/math&gt; is continuous and increasing on the interval &lt;math&gt;0 \le x \le \frac{\pi}{2}&lt;/math&gt; and its range on that interval is from &lt;math&gt;0 \le x\sin x \le \frac{\pi}{2}&lt;/math&gt;, by the [[Intermediate Value Theorem]] this value is attainable).<br /> <br /> === Solution 2 ===<br /> We can rewrite the numerator to be a perfect square by adding &lt;math&gt;-\dfrac{12x \sin x}{x \sin x}&lt;/math&gt;. Thus, we must also add back &lt;math&gt;12&lt;/math&gt;.<br /> <br /> This results in &lt;math&gt;\dfrac{(3x \sin x-2)^2}{x \sin x}+12&lt;/math&gt;.<br /> <br /> Thus, if &lt;math&gt;3x \sin x-2=0&lt;/math&gt;, then the minimum is obviously &lt;math&gt;12&lt;/math&gt;. We show this possible with the same methods in Solution 1; thus the answer is &lt;math&gt;\boxed{012}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> Let &lt;math&gt;y = x\sin{x}&lt;/math&gt; and rewrite the expression as &lt;math&gt;f(y) = 9y + \frac{4}{y}&lt;/math&gt;, similar to the previous solution. To minimize &lt;math&gt;f(y)&lt;/math&gt;, take the [[derivative]] of &lt;math&gt;f(y)&lt;/math&gt; and set it equal to zero. <br /> <br /> The derivative of &lt;math&gt;f(y)&lt;/math&gt;, using the Power Rule, is<br /> <br /> &lt;math&gt;f'(y)&lt;/math&gt; = &lt;math&gt;9 - 4y^{-2}&lt;/math&gt;<br /> <br /> &lt;math&gt;f'(y)&lt;/math&gt; is zero only when &lt;math&gt;y = \frac{2}{3}&lt;/math&gt; or &lt;math&gt;y = -\frac{2}{3}&lt;/math&gt;. It can further be verified that &lt;math&gt;\frac{2}{3}&lt;/math&gt; and &lt;math&gt;-\frac{2}{3}&lt;/math&gt; are relative minima by finding the derivatives of other points near the critical points. However, since &lt;math&gt;x \sin{x}&lt;/math&gt; is always positive in the given domain, &lt;math&gt;y = \frac{2}{3}&lt;/math&gt;. Therefore, &lt;math&gt;x\sin{x}&lt;/math&gt; = &lt;math&gt;\frac{2}{3}&lt;/math&gt;, and the answer is &lt;math&gt;\frac{(9)(\frac{2}{3})^2 + 4}{\frac{2}{3}} = \boxed{012}&lt;/math&gt;.<br /> <br /> === Solution 4 (also uses calculus) ===<br /> <br /> As above, let &lt;math&gt;y = x\sin{x}&lt;/math&gt;. Add &lt;math&gt;\frac{12y}{y}&lt;/math&gt; to the expression and subtract &lt;math&gt;12&lt;/math&gt;, giving &lt;math&gt;f(x) = \frac{(3y+2)^2}{y} - 12&lt;/math&gt;. Taking the [[derivative]] of &lt;math&gt;f(x)&lt;/math&gt; using the [[Chain Rule]] and [[Quotient Rule]], we have &lt;math&gt;\frac{df}{dx} = \frac{6y(3y+2)-(3y+2)^2}{y^2}&lt;/math&gt;. We find the minimum value by setting this to 0. Simplifying, we have &lt;math&gt;6y(3y+2) = (3y+2)^2&lt;/math&gt; and &lt;math&gt;y = \pm{\frac{2}{3}} = x\sin{x}&lt;/math&gt;. Since both &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;\sin{x}&lt;/math&gt; are positive on the given interval, we can ignore the negative result. Plugging &lt;math&gt;y = \frac{2}{3}&lt;/math&gt; into our expression for &lt;math&gt;f(x)&lt;/math&gt;, we have &lt;math&gt;\frac{(3(\frac{2}{3})+2)^2}{y}-12 = \frac{16}{\frac{2}{3}}-12 = \boxed{012}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> [[Category:Intermediate Trigonometry Problems]]</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=Mock_Geometry_AIME_2011_Problems/Problem_4&diff=86076 Mock Geometry AIME 2011 Problems/Problem 4 2017-06-15T22:33:57Z <p>Zeroman: </p> <hr /> <div>==Problem== <br /> In triangle &lt;math&gt;ABC,&lt;/math&gt; &lt;math&gt;AB=6, BC=9, \angle ABC=120^{\circ}&lt;/math&gt; Let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be points on &lt;math&gt;AC&lt;/math&gt; such that &lt;math&gt;BPQ&lt;/math&gt; is equilateral. The perimeter of &lt;math&gt;BPQ&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac{m} {\sqrt{n}},&lt;/math&gt; where &lt;math&gt;m,n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n.&lt;/math&gt; <br /> <br /> ==Solution==<br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> draw((0,3sqrt(3))--(3,0)--(12,0)--cycle);<br /> draw((3,0)--(84/19,36sqrt(3)/19));<br /> draw((3,0)--(48/19, 4.10223));<br /> draw((3,0)--(120/19,2.46134));<br /> label(&quot;$A$&quot;,(0,3sqrt(3)),NNW);<br /> label(&quot;$B$&quot;,(3,0),SW);<br /> label(&quot;$C$&quot;,(12,0),ESE);<br /> label(&quot;$P$&quot;,(48/19,4.10223),NNE);<br /> label(&quot;$Q$&quot;,(120/19,2.46134),NE);<br /> label(&quot;$H$&quot;,(84/19,36sqrt(3)/19),NNE);<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;H&lt;/math&gt; be the midpoint of &lt;math&gt;PQ&lt;/math&gt;. It follows that &lt;math&gt;BH&lt;/math&gt; is perpendicular to &lt;math&gt;PQ&lt;/math&gt; and to &lt;math&gt;AC&lt;/math&gt;. The area of &lt;math&gt;\Delta ABC&lt;/math&gt; can then be calculated two different ways: &lt;math&gt;\frac{1}{2}*AB*BC*\sin{B}&lt;/math&gt;, and &lt;math&gt;\frac{BH*AC}{2}&lt;/math&gt;.<br /> <br /> <br /> By the Law of Cosines, &lt;math&gt;AC^2=9^2+6^2-2*9*6\cos{120}=171&lt;/math&gt; and so &lt;math&gt;AC=3\sqrt{19}&lt;/math&gt;. Therefore, &lt;math&gt;[ABC]=\frac{1}{2}*6*9\sin{120}=\frac{3\sqrt{19}BH}{2}&lt;/math&gt;. Solving for &lt;math&gt;BH&lt;/math&gt; yields &lt;math&gt;BH=\frac{9\sqrt{3}}{\sqrt{19}}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;s&lt;/math&gt; be the side length of &lt;math&gt;BPQ&lt;/math&gt;. The height of an equilateral triangle is given by the formula &lt;math&gt;\frac{s\sqrt3}{2}&lt;/math&gt;. Then &lt;math&gt;BH=\frac{s\sqrt{3}}{2}=\frac{9\sqrt{3}}{\sqrt{19}}&lt;/math&gt;. Solving for &lt;math&gt;s&lt;/math&gt; yields &lt;math&gt;s=\frac{18}{\sqrt{19}}&lt;/math&gt;. Then the perimeter of the triangle is &lt;math&gt;3s=\frac{54}{\sqrt{19}}&lt;/math&gt; and &lt;math&gt;m+n=54+19=\boxed{073}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Let &lt;math&gt;\angle A = \alpha&lt;/math&gt; and &lt;math&gt;BP = PQ = QB = x&lt;/math&gt;. By the Law of Cosines, &lt;math&gt;AC = 3\sqrt{19}&lt;/math&gt;. It is easy to see that &lt;math&gt;\angle APB = 120^\circ&lt;/math&gt;. Since &lt;math&gt;\angle ABC = 120^\circ&lt;/math&gt;, by AA similarity&lt;math&gt;\triangle ABC \sim \triangle APB&lt;/math&gt;. From this, we have: &lt;cmath&gt;\frac{AB}{PB} = \frac{AC}{BC}&lt;/cmath&gt; &lt;cmath&gt;\frac{6}{x}=\frac{3\sqrt{19}}{9}&lt;/cmath&gt; Solving, we find that &lt;math&gt;x = \frac{18}{\sqrt{19}}&lt;/math&gt;, so the perimeter is &lt;math&gt;3x = \frac{54}{\sqrt{19}}&lt;/math&gt;, and our answer is &lt;math&gt;m+n=\boxed{73}&lt;/math&gt;</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_16&diff=86000 2000 AMC 12 Problems/Problem 16 2017-06-11T00:46:58Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> A checkerboard of &lt;math&gt;13&lt;/math&gt; rows and &lt;math&gt;17&lt;/math&gt; columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered &lt;math&gt;1,2,\ldots,17&lt;/math&gt;, the second row &lt;math&gt;18,19,\ldots,34&lt;/math&gt;, and so on down the board. If the board is renumbered so that the left column, top to bottom, is &lt;math&gt;1,2,\ldots,13,&lt;/math&gt;, the second column &lt;math&gt;14,15,\ldots,26&lt;/math&gt; and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system). <br /> <br /> &lt;math&gt;\text {(A)}\ 222 \qquad \text {(B)}\ 333\qquad \text {(C)}\ 444 \qquad \text {(D)}\ 555 \qquad \text {(E)}\ 666&lt;/math&gt;<br /> <br /> == Solution ==<br /> Index the rows with &lt;math&gt;i = 1, 2, 3, ..., 13&lt;/math&gt;<br /> Index the columns with &lt;math&gt;j = 1, 2, 3, ..., 17&lt;/math&gt;<br /> <br /> For the first row number the cells &lt;math&gt;1, 2, 3, ..., 17&lt;/math&gt;<br /> For the second, &lt;math&gt;18, 19, ..., 34&lt;/math&gt;<br /> and so on<br /> <br /> So the number in row = &lt;math&gt;i&lt;/math&gt; and column = &lt;math&gt;j&lt;/math&gt; is<br /> &lt;math&gt;f(i, j) = 17(i-1) + j = 17i + j - 17&lt;/math&gt;<br /> <br /> Similarly, numbering the same cells columnwise we<br /> find the number in row = &lt;math&gt;i&lt;/math&gt; and column = &lt;math&gt;j&lt;/math&gt; is<br /> &lt;math&gt;g(i, j) = i + 13j - 13&lt;/math&gt;<br /> <br /> So we need to solve<br /> <br /> &lt;math&gt;f(i, j) = g(i, j)&lt;/math&gt;<br /> <br /> &lt;math&gt;17i + j - 17 = i + 13j - 13&lt;/math&gt;<br /> <br /> &lt;math&gt;16i = 4 + 12j&lt;/math&gt; <br /> <br /> &lt;math&gt;4i = 1 + 3j&lt;/math&gt;<br /> <br /> &lt;math&gt;i = (1 + 3j)/4&lt;/math&gt;<br /> <br /> We get<br /> &lt;math&gt;(i, j) = (1, 1), f(i, j) = g(i, j) = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;(i, j) = (4, 5), f(i, j) = g(i, j) = 56&lt;/math&gt;<br /> <br /> &lt;math&gt;(i, j) = (7, 9), f(i, j) = g(i, j) = 111&lt;/math&gt;<br /> <br /> &lt;math&gt;(i, j) = (10, 13), f(i, j) = g(i, j) = 166&lt;/math&gt;<br /> <br /> &lt;math&gt;(i, j) = (13, 17), f(i, j) = g(i, j) = 221&lt;/math&gt;<br /> <br /> &lt;math&gt;\boxed{D}&lt;/math&gt; &lt;math&gt;555&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2000|num-b=15|num-a=17}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_16&diff=85999 2000 AMC 12 Problems/Problem 16 2017-06-11T00:45:57Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> A checkerboard of &lt;math&gt;13&lt;/math&gt; rows and &lt;math&gt;17&lt;/math&gt; columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered &lt;math&gt;1,2,\ldots,17&lt;/math&gt;, the second row &lt;math&gt;18,19,\ldots,34&lt;/math&gt;, and so on down the board. If the board is renumbered so that the left column, top to bottom, is &lt;math&gt;1,2,\ldots,13,&lt;/math&gt;, the second column &lt;math&gt;14,15,\ldots,26&lt;/math&gt; and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system). <br /> <br /> &lt;math&gt;\text {(A)}\ 222 \qquad \text {(B)}\ 333\qquad \text {(C)}\ 444 \qquad \text {(D)}\ 555 \qquad \text {(E)}\ 666&lt;/math&gt;<br /> <br /> == Solution ==<br /> Index the rows with &lt;math&gt;i = 1, 2, 3, ..., 13&lt;/math&gt;<br /> Index the columns with &lt;math&gt;j = 1, 2, 3, ..., 17&lt;/math&gt;<br /> <br /> For the first row number the cells &lt;math&gt;1, 2, 3, ..., 17&lt;/math&gt;<br /> For the second, &lt;math&gt;18, 19, ..., 34&lt;/math&gt;<br /> and so on<br /> <br /> So the number in row = &lt;math&gt;i&lt;/math&gt; and column = &lt;math&gt;j&lt;/math&gt; is<br /> &lt;math&gt;f(i, j) = 17(i-1) + j = 17i + j - 17&lt;/math&gt;<br /> <br /> Similarly, numbering the same cells columnwise we<br /> find the number in row = &lt;math&gt;i&lt;/math&gt; and column = &lt;math&gt;j&lt;/math&gt; is<br /> &lt;math&gt;g(i, j) = i + 13j - 13&lt;/math&gt;<br /> <br /> So we need to solve<br /> <br /> &lt;math&gt;f(i, j) = g(i, j)<br /> <br /> 17i + j - 17 = i + 13j - 13<br /> <br /> 16i = 4 + 12j <br /> <br /> 4i = 1 + 3j<br /> <br /> i = (1 + 3j)/4&lt;/math&gt;<br /> <br /> We get<br /> &lt;math&gt;(i, j) = (1, 1), f(i, j) = g(i, j) = 1<br /> <br /> (i, j) = (4, 5), f(i, j) = g(i, j) = 56<br /> <br /> (i, j) = (7, 9), f(i, j) = g(i, j) = 111<br /> <br /> (i, j) = (10, 13), f(i, j) = g(i, j) = 166<br /> <br /> (i, j) = (13, 17), f(i, j) = g(i, j) = 221&lt;/math&gt;<br /> <br /> &lt;math&gt;\boxed{D}&lt;/math&gt; &lt;math&gt;555&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2000|num-b=15|num-a=17}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_9&diff=85673 2005 AIME II Problems/Problem 9 2017-05-15T00:08:27Z <p>Zeroman: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> For how many positive integers &lt;math&gt; n &lt;/math&gt; less than or equal to &lt;math&gt;1000&lt;/math&gt; is &lt;math&gt; (\sin t + i \cos t)^n = \sin nt + i \cos nt &lt;/math&gt; true for all real &lt;math&gt; t &lt;/math&gt;?<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> We know by [[De Moivre's Theorem]] that &lt;math&gt;(\cos t + i \sin t)^n = \cos nt + i \sin nt&lt;/math&gt; for all [[real number]]s &lt;math&gt;t&lt;/math&gt; and all [[integer]]s &lt;math&gt;n&lt;/math&gt;. So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem. <br /> <br /> Recall the [[trigonometric identities]] &lt;math&gt;\cos (\frac{\pi}2 - u) = \sin u&lt;/math&gt; and &lt;math&gt;\sin (\frac{\pi}2 - u) = \cos u&lt;/math&gt; hold for all real &lt;math&gt;u&lt;/math&gt;. If our original equation holds for all &lt;math&gt;t&lt;/math&gt;, it must certainly hold for &lt;math&gt;t = \frac{\pi}2 - u&lt;/math&gt;. Thus, the question is equivalent to asking for how many [[positive integer]]s &lt;math&gt;n \leq 1000&lt;/math&gt; we have that &lt;math&gt;\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \sin n \left(\frac\pi2 -u \right) + i\cos n \left(\frac\pi2 - u\right)&lt;/math&gt; holds for all real &lt;math&gt;u&lt;/math&gt;.<br /> <br /> &lt;math&gt;\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \left(\cos u + i \sin u\right)^n = \cos nu + i\sin nu&lt;/math&gt;. We know that two [[complex number]]s are equal if and only if both their [[real part]] and [[imaginary part]] are equal. Thus, we need to find all &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;\cos n u = \sin n\left(\frac\pi2 - u\right)&lt;/math&gt; and &lt;math&gt;\sin nu = \cos n\left(\frac\pi2 - u\right)&lt;/math&gt; hold for all real &lt;math&gt;u&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sin x = \cos y&lt;/math&gt; if and only if either &lt;math&gt;x + y = \frac \pi 2 + 2\pi \cdot k&lt;/math&gt; or &lt;math&gt;x - y = \frac\pi2 + 2\pi\cdot k&lt;/math&gt; for some integer &lt;math&gt;k&lt;/math&gt;. So from the equality of the real parts we need either &lt;math&gt;nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k&lt;/math&gt;, in which case &lt;math&gt;n = 1 + 4k&lt;/math&gt;, or we need &lt;math&gt;-nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k&lt;/math&gt;, in which case &lt;math&gt;n&lt;/math&gt; will depend on &lt;math&gt;u&lt;/math&gt; and so the equation will not hold for all real values of &lt;math&gt;u&lt;/math&gt;. Checking &lt;math&gt;n = 1 + 4k&lt;/math&gt; in the equation for the imaginary parts, we see that it works there as well, so exactly those values of &lt;math&gt;n&lt;/math&gt; congruent to &lt;math&gt;1 \pmod 4&lt;/math&gt; work. There are &lt;math&gt;\boxed{250}&lt;/math&gt; of them in the given range.<br /> <br /> === Solution 2 ===<br /> This problem begs us to use the familiar identity &lt;math&gt;e^{it} = \cos(t) + i \sin(t)&lt;/math&gt;. Notice, &lt;math&gt;\sin(t) + i \cos(t) = i(\cos(t) - i \sin(t)) = i e^{-it}&lt;/math&gt; since &lt;math&gt;\sin(-t) = -\sin(t)&lt;/math&gt;. Using this, &lt;math&gt;(\sin(t) + i \cos(t))^n = \sin(nt) + i \cos(nt)&lt;/math&gt; is recast as &lt;math&gt;(i e^{-it})^n = i e^{-itn}&lt;/math&gt;. Hence we must have &lt;math&gt;i^n = i \Rightarrow i^{n-1} = 1 \Rightarrow n \equiv 1 \bmod{4}&lt;/math&gt;. Thus since &lt;math&gt;1000&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt; exactly one quarter of the residues are congruent to &lt;math&gt;1&lt;/math&gt; hence we have &lt;math&gt;\boxed{250}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> We can rewrite &lt;math&gt;\sin(t)&lt;/math&gt; as &lt;math&gt;\cos\left(\frac{\pi}{2}-t\right)&lt;/math&gt; and &lt;math&gt;\cos(t)&lt;/math&gt; as &lt;math&gt;\sin\left(\frac{\pi}{2}-t\right)&lt;/math&gt;. This means that &lt;math&gt;\sin t + i\cos t = e^{i\left(\frac{\pi}{2}-t\right)}=\frac{e^{\frac{\pi i}{2}}}{e^{it}}&lt;/math&gt;. This theorem also tells us that &lt;math&gt;e^{\frac{\pi i}{2}}=i&lt;/math&gt;, so &lt;math&gt;\sin t + i\cos t = \frac{i}{e^{it}}&lt;/math&gt;. By the same line of reasoning, we have &lt;math&gt;\sin nt + i\cos nt = \frac{i}{e^{int}}&lt;/math&gt;.<br /> <br /> For the statement in the question to be true, we must have &lt;math&gt;\left(\frac{i}{e^{it}}\right)^n=\frac{i}{e^{int}}&lt;/math&gt;. The left hand side simplifies to &lt;math&gt;\frac{i^n}{e^{int}}&lt;/math&gt;. We cancel the denominators and find that the only thing that needs to be true is that &lt;math&gt;i^n=i&lt;/math&gt;. This is true if &lt;math&gt;n\equiv1\pmod{4}&lt;/math&gt;, and there are &lt;math&gt;\boxed{250}&lt;/math&gt; such numbers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;1000&lt;/math&gt;. Solution by Zeroman<br /> [[Category:Intermediate Algebra Problems]]<br /> <br /> {{MAA Notice}}<br /> <br /> ==See Also==<br /> {{AIME box|year=2005|n=II|num-b=8|num-a=10}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_9&diff=85672 2005 AIME II Problems/Problem 9 2017-05-15T00:07:09Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> For how many positive integers &lt;math&gt; n &lt;/math&gt; less than or equal to &lt;math&gt;1000&lt;/math&gt; is &lt;math&gt; (\sin t + i \cos t)^n = \sin nt + i \cos nt &lt;/math&gt; true for all real &lt;math&gt; t &lt;/math&gt;?<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> We know by [[De Moivre's Theorem]] that &lt;math&gt;(\cos t + i \sin t)^n = \cos nt + i \sin nt&lt;/math&gt; for all [[real number]]s &lt;math&gt;t&lt;/math&gt; and all [[integer]]s &lt;math&gt;n&lt;/math&gt;. So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem. <br /> <br /> Recall the [[trigonometric identities]] &lt;math&gt;\cos (\frac{\pi}2 - u) = \sin u&lt;/math&gt; and &lt;math&gt;\sin (\frac{\pi}2 - u) = \cos u&lt;/math&gt; hold for all real &lt;math&gt;u&lt;/math&gt;. If our original equation holds for all &lt;math&gt;t&lt;/math&gt;, it must certainly hold for &lt;math&gt;t = \frac{\pi}2 - u&lt;/math&gt;. Thus, the question is equivalent to asking for how many [[positive integer]]s &lt;math&gt;n \leq 1000&lt;/math&gt; we have that &lt;math&gt;\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \sin n \left(\frac\pi2 -u \right) + i\cos n \left(\frac\pi2 - u\right)&lt;/math&gt; holds for all real &lt;math&gt;u&lt;/math&gt;.<br /> <br /> &lt;math&gt;\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \left(\cos u + i \sin u\right)^n = \cos nu + i\sin nu&lt;/math&gt;. We know that two [[complex number]]s are equal if and only if both their [[real part]] and [[imaginary part]] are equal. Thus, we need to find all &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;\cos n u = \sin n\left(\frac\pi2 - u\right)&lt;/math&gt; and &lt;math&gt;\sin nu = \cos n\left(\frac\pi2 - u\right)&lt;/math&gt; hold for all real &lt;math&gt;u&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sin x = \cos y&lt;/math&gt; if and only if either &lt;math&gt;x + y = \frac \pi 2 + 2\pi \cdot k&lt;/math&gt; or &lt;math&gt;x - y = \frac\pi2 + 2\pi\cdot k&lt;/math&gt; for some integer &lt;math&gt;k&lt;/math&gt;. So from the equality of the real parts we need either &lt;math&gt;nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k&lt;/math&gt;, in which case &lt;math&gt;n = 1 + 4k&lt;/math&gt;, or we need &lt;math&gt;-nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k&lt;/math&gt;, in which case &lt;math&gt;n&lt;/math&gt; will depend on &lt;math&gt;u&lt;/math&gt; and so the equation will not hold for all real values of &lt;math&gt;u&lt;/math&gt;. Checking &lt;math&gt;n = 1 + 4k&lt;/math&gt; in the equation for the imaginary parts, we see that it works there as well, so exactly those values of &lt;math&gt;n&lt;/math&gt; congruent to &lt;math&gt;1 \pmod 4&lt;/math&gt; work. There are &lt;math&gt;\boxed{250}&lt;/math&gt; of them in the given range.<br /> <br /> === Solution 2 ===<br /> This problem begs us to use the familiar identity &lt;math&gt;e^{it} = \cos(t) + i \sin(t)&lt;/math&gt;. Notice, &lt;math&gt;\sin(t) + i \cos(t) = i(\cos(t) - i \sin(t)) = i e^{-it}&lt;/math&gt; since &lt;math&gt;\sin(-t) = -\sin(t)&lt;/math&gt;. Using this, &lt;math&gt;(\sin(t) + i \cos(t))^n = \sin(nt) + i \cos(nt)&lt;/math&gt; is recast as &lt;math&gt;(i e^{-it})^n = i e^{-itn}&lt;/math&gt;. Hence we must have &lt;math&gt;i^n = i \Rightarrow i^{n-1} = 1 \Rightarrow n \equiv 1 \bmod{4}&lt;/math&gt;. Thus since &lt;math&gt;1000&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt; exactly one quarter of the residues are congruent to &lt;math&gt;1&lt;/math&gt; hence we have &lt;math&gt;\boxed{250}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> We can rewrite &lt;math&gt;\sin(t)&lt;/math&gt; as &lt;math&gt;\cos\left(\frac{\pi}{2}-t\right)&lt;/math&gt; and &lt;math&gt;\cos(t)&lt;/math&gt; as &lt;math&gt;\sin\left(\frac{\pi}{2}-t\right)&lt;/math&gt;. This means that &lt;math&gt;\sin t + i\cos t = e^{i\left(\frac{\pi}{2}-t\right)}=\frac{e^{\frac{\pi i}{2}}}{e^{it}}&lt;/math&gt;, by [[De Moivre's Theorem]]. This theorem also tells us that &lt;math&gt;e^{\frac{\pi i}{2}}=i&lt;/math&gt;, so &lt;math&gt;\sin t + i\cos t = \frac{i}{e^{it}}&lt;/math&gt;. By the same line of reasoning, we have &lt;math&gt;\sin nt + i\cos nt = \frac{i}{e^{int}}&lt;/math&gt;.<br /> <br /> For the statement in the question to be true, we must have &lt;math&gt;\left(\frac{i}{e^{it}}\right)^n=\frac{i}{e^{int}}&lt;/math&gt;. The left hand side simplifies to &lt;math&gt;\frac{i^n}{e^{int}}&lt;/math&gt;. We cancel the denominators and find that the only thing that needs to be true is that &lt;math&gt;i^n=i&lt;/math&gt;. This is true if &lt;math&gt;n\equiv1\pmod{4}&lt;/math&gt;, and there are &lt;math&gt;\boxed{250}&lt;/math&gt; such numbers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;1000&lt;/math&gt;. Solution by Zeroman<br /> [[Category:Intermediate Algebra Problems]]<br /> <br /> {{MAA Notice}}<br /> <br /> ==See Also==<br /> {{AIME box|year=2005|n=II|num-b=8|num-a=10}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_8&diff=85547 2000 AIME I Problems/Problem 8 2017-05-04T01:10:54Z <p>Zeroman: </p> <hr /> <div>== Problem ==<br /> A container in the shape of a right circular [[cone]] is &lt;math&gt;12&lt;/math&gt; inches tall and its base has a &lt;math&gt;5&lt;/math&gt;-inch [[radius]]. The liquid that is sealed inside is &lt;math&gt;9&lt;/math&gt; inches deep when the cone is held with its [[point]] down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is &lt;math&gt;m - n\sqrt {p},&lt;/math&gt; from the base 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 cube of any prime number. Find &lt;math&gt;m + n + p&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solution ==<br /> [[Image:2000_I_AIME-8.png]]<br /> === Solution 1 ===<br /> The scale factor is uniform in all dimensions, so the volume of the liquid is &lt;math&gt;\left(\frac{3}{4}\right)^{3}&lt;/math&gt; of the container. The remaining section of the volume is &lt;math&gt;\frac{1-\left(\frac{3}{4}\right)^{3}}{1}&lt;/math&gt; of the volume, and therefore &lt;math&gt;\frac{\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}&lt;/math&gt; of the height when the vertex is at the top.<br /> <br /> So, the liquid occupies &lt;math&gt;\frac{1-\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}&lt;/math&gt; of the height, or &lt;math&gt;12-12\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}=12-3\left(37^{1/3}\right)&lt;/math&gt;. Thus &lt;math&gt;m+n+p=\boxed{052}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> (Computational) The volume of a cone can be found by &lt;math&gt;V = \frac{\pi}{3}r^2h&lt;/math&gt;. In the second container, if we let &lt;math&gt;h',r'&lt;/math&gt; represent the height, radius (respectively) of the air (so &lt;math&gt;12 -h'&lt;/math&gt; is the height of the liquid), then the volume of the liquid can be found by &lt;math&gt;\frac{\pi}{3}r^2h - \frac{\pi}{3}(r')^2h'&lt;/math&gt;. <br /> <br /> By [[similar triangle]]s, we find that the dimensions of the liquid in the first cone to the entire cone is &lt;math&gt;\frac{3}{4}&lt;/math&gt;, and that &lt;math&gt;r' = \frac{rh'}{h}&lt;/math&gt;; equating,<br /> <br /> &lt;cmath&gt;\begin{align*}\frac{\pi}{3}\left(\frac{3}{4}r\right)^2 \left(\frac{3}{4}h\right) &amp;= \frac{\pi}{3}\left(r^2h - \left(\frac{rh'}{h}\right)^2h'\right)\\<br /> \frac{37}{64}r^2h &amp;= \frac{r^2}{h^2}(h')^3 \\<br /> h' &amp;= \sqrt{\frac{37}{64} \cdot 12^3} = 3\sqrt{37}\end{align*}&lt;/cmath&gt;<br /> <br /> Thus the answer is &lt;math&gt;12 - h' = 12-3\sqrt{37}&lt;/math&gt;, and &lt;math&gt;m+n+p=\boxed{052}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> From the formula &lt;math&gt;V=\frac{\pi r^2h}{3}&lt;/math&gt;, we can find that the volume of the container is &lt;math&gt;100\pi&lt;/math&gt;. The cone formed by the liquid is similar to the original, but scaled down by &lt;math&gt;\frac{3}{4}&lt;/math&gt; in all directions, so its volume is &lt;math&gt;100\pi*\frac{27}{64}=\frac{675\pi}{16}&lt;/math&gt;. The volume of the air in the container is the volume of the container minus the volume of the liquid, which is &lt;math&gt;\frac{925\pi}{64}&lt;/math&gt;, which is &lt;math&gt;\frac{37}{64}&lt;/math&gt; of the volume of the container. When the point faces upwards, the air forms a cone at the top of the container. This cone must have &lt;math&gt;\sqrt{\frac{37}{64}}=\frac{\sqrt{37}}{4}&lt;/math&gt; of the height of the container. This means that the height of the liquid is &lt;math&gt;12\left(1-\frac{\sqrt{37}}{4}\right)=12-3\sqrt{37}&lt;/math&gt; inches, so our answer is &lt;math&gt;\boxed{052}&lt;/math&gt;. Solution by Zeroman<br /> <br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_8&diff=85546 2000 AIME I Problems/Problem 8 2017-05-04T01:10:37Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> A container in the shape of a right circular [[cone]] is &lt;math&gt;12&lt;/math&gt; inches tall and its base has a &lt;math&gt;5&lt;/math&gt;-inch [[radius]]. The liquid that is sealed inside is &lt;math&gt;9&lt;/math&gt; inches deep when the cone is held with its [[point]] down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is &lt;math&gt;m - n\sqrt {p},&lt;/math&gt; from the base 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 cube of any prime number. Find &lt;math&gt;m + n + p&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solution ==<br /> [[Image:2000_I_AIME-8.png]]<br /> === Solution 1 ===<br /> The scale factor is uniform in all dimensions, so the volume of the liquid is &lt;math&gt;\left(\frac{3}{4}\right)^{3}&lt;/math&gt; of the container. The remaining section of the volume is &lt;math&gt;\frac{1-\left(\frac{3}{4}\right)^{3}}{1}&lt;/math&gt; of the volume, and therefore &lt;math&gt;\frac{\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}&lt;/math&gt; of the height when the vertex is at the top.<br /> <br /> So, the liquid occupies &lt;math&gt;\frac{1-\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}&lt;/math&gt; of the height, or &lt;math&gt;12-12\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}=12-3\left(37^{1/3}\right)&lt;/math&gt;. Thus &lt;math&gt;m+n+p=\boxed{052}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> (Computational) The volume of a cone can be found by &lt;math&gt;V = \frac{\pi}{3}r^2h&lt;/math&gt;. In the second container, if we let &lt;math&gt;h',r'&lt;/math&gt; represent the height, radius (respectively) of the air (so &lt;math&gt;12 -h'&lt;/math&gt; is the height of the liquid), then the volume of the liquid can be found by &lt;math&gt;\frac{\pi}{3}r^2h - \frac{\pi}{3}(r')^2h'&lt;/math&gt;. <br /> <br /> By [[similar triangle]]s, we find that the dimensions of the liquid in the first cone to the entire cone is &lt;math&gt;\frac{3}{4}&lt;/math&gt;, and that &lt;math&gt;r' = \frac{rh'}{h}&lt;/math&gt;; equating,<br /> <br /> &lt;cmath&gt;\begin{align*}\frac{\pi}{3}\left(\frac{3}{4}r\right)^2 \left(\frac{3}{4}h\right) &amp;= \frac{\pi}{3}\left(r^2h - \left(\frac{rh'}{h}\right)^2h'\right)\\<br /> \frac{37}{64}r^2h &amp;= \frac{r^2}{h^2}(h')^3 \\<br /> h' &amp;= \sqrt{\frac{37}{64} \cdot 12^3} = 3\sqrt{37}\end{align*}&lt;/cmath&gt;<br /> <br /> Thus the answer is &lt;math&gt;12 - h' = 12-3\sqrt{37}&lt;/math&gt;, and &lt;math&gt;m+n+p=\boxed{052}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> From the formula &lt;math&gt;V=\frac{\pi r^2h}{3}&lt;/math&gt;, we can find that the volume of the container is &lt;math&gt;100\pi&lt;/math&gt;. The cone formed by the liquid is similar to the original, but scaled down by &lt;math&gt;\frac{3}{4}&lt;/math&gt; in all directions, so its volume is &lt;math&gt;100\pi*\frac{27}{64}=\frac{675\pi}{16}&lt;/math&gt;. The volume of the air in the container is the volume of the container minus the volume of the liquid, which is &lt;math&gt;\frac{925\pi}{64}&lt;/math&gt;, which is &lt;math&gt;\frac{37}{64}&lt;/math&gt; of the volume of the container. When the point faces upwards, the air forms a cone at the top of the container. This cone must have &lt;math&gt;\sqrt{\frac{37}{64}}=\frac{\sqrt{37}}{4}&lt;/math&gt; of the height of the container. This means that the height of the liquid is &lt;math&gt;12\left(1-\frac{\sqrt{37}}{4}\right)=12-3\sqrt{37}&lt;/math&gt; inches, so our answer is &lt;math&gt;\boxed{052}&lt;/math&gt;. Solution by Zeroman<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_6&diff=85449 1993 AIME Problems/Problem 6 2017-04-28T20:16:04Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> What is the smallest [[positive]] [[integer]] that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?<br /> <br /> ==Solution==<br /> === Solution 1 ===<br /> Denote the first of each of the series of consecutive integers as &lt;math&gt;a,\ b,\ c&lt;/math&gt;. Therefore, &lt;math&gt;n = a + (a + 1) \ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55&lt;/math&gt;. Simplifying, &lt;math&gt;9a = 10b + 9 = 11c + 19&lt;/math&gt;. The relationship between &lt;math&gt;a,\ b&lt;/math&gt; suggests that &lt;math&gt;b&lt;/math&gt; is divisible by &lt;math&gt;9&lt;/math&gt;. Also, &lt;math&gt;10b -10 = 10(b-1) = 11c&lt;/math&gt;, so &lt;math&gt;b-1&lt;/math&gt; is divisible by &lt;math&gt;11&lt;/math&gt;. We find that the least possible value of &lt;math&gt;b = 45&lt;/math&gt;, so the answer is &lt;math&gt;10(45) + 45 = 495&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Let the desired integer be &lt;math&gt;n&lt;/math&gt;. From the information given, it can be determined that, for positive integers &lt;math&gt;a, \ b, \ c&lt;/math&gt;:<br /> <br /> &lt;math&gt;n = 9a + 36 = 10b + 45 = 11c + 55&lt;/math&gt;<br /> <br /> This can be rewritten as the following congruences:<br /> <br /> &lt;math&gt;n \equiv 0 \pmod{9}&lt;/math&gt; <br /> <br /> &lt;math&gt;n \equiv 5 \pmod{10}&lt;/math&gt;<br /> <br /> &lt;math&gt;n \equiv 0 \pmod{11}&lt;/math&gt;<br /> <br /> Since 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is &lt;math&gt;\boxed{495}&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be the desired integer. From the given information, we have<br /> &lt;cmath&gt; \begin{align*}9x &amp;= a \\ 11y &amp;= a \\ 10z + 5 &amp;= a, \end{align*}&lt;/cmath&gt; here, &lt;math&gt;x,&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are the middle terms of the sequence of 9 and 11 numbers, respectively. Similarly, we have &lt;math&gt;z&lt;/math&gt; as the 4th term of the sequence. Since, &lt;math&gt;a&lt;/math&gt; is a multiple of &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;11,&lt;/math&gt; it is also a multiple of &lt;math&gt;\text{lcm}[9,11]=99.&lt;/math&gt; Hence, &lt;math&gt;a=99m,&lt;/math&gt; for some &lt;math&gt;m.&lt;/math&gt; So, we have &lt;math&gt;10z + 5 = 99m.&lt;/math&gt; It follows that &lt;math&gt;99(5) = \boxed{495}&lt;/math&gt; is the smallest integer that can be represented in such a way.<br /> <br /> === Solution 4 ===<br /> By the method in Solution 1, we find that the number &lt;math&gt;n&lt;/math&gt; can be written as &lt;math&gt;9a+36=10b+45=11c+55&lt;/math&gt; for some integers &lt;math&gt;a,b,c&lt;/math&gt;. From this, we can see that &lt;math&gt;n&lt;/math&gt; must be divisible by 9, 5, and 11. This means &lt;math&gt;n&lt;/math&gt; must be divisible by 495. The only multiples of 495 that are small enough to be AIME answers are 495 and 990. From the second of the three expressions above, we can see that &lt;math&gt;n&lt;/math&gt; cannot be divisible by 10, so &lt;math&gt;n&lt;/math&gt; must equal &lt;math&gt;\boxed{495}&lt;/math&gt;. Solution by Zeroman.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_13&diff=85437 1991 AIME Problems/Problem 13 2017-04-26T20:11:33Z <p>Zeroman: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> A drawer contains a mixture of red socks and blue socks, at most &lt;math&gt;1991&lt;/math&gt; in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly &lt;math&gt;\frac{1}{2}&lt;/math&gt; that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data? <br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; denote the number of red and blue socks, respectively. Also, let &lt;math&gt;t=r+b&lt;/math&gt;. The probability &lt;math&gt;P&lt;/math&gt; that when two socks are drawn randomly, without replacement, both are red or both are blue is given by<br /> <br /> &lt;cmath&gt;<br /> \frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}.<br /> &lt;/cmath&gt;<br /> <br /> Solving the resulting quadratic equation &lt;math&gt;r^{2}-rt+t(t-1)/4=0&lt;/math&gt;, for &lt;math&gt;r&lt;/math&gt; in terms of &lt;math&gt;t&lt;/math&gt;, one obtains that <br /> <br /> &lt;cmath&gt;<br /> r=\frac{t\pm\sqrt{t}}{2}\, .<br /> &lt;/cmath&gt;<br /> <br /> Now, since &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;t&lt;/math&gt; are positive integers, it must be the case that &lt;math&gt;t=n^{2}&lt;/math&gt;, with &lt;math&gt;n\in\mathbb{N}&lt;/math&gt;. Hence, &lt;math&gt;r=n(n\pm 1)/2&lt;/math&gt; would correspond to the general solution. For the present case &lt;math&gt;t\leq 1991&lt;/math&gt;, and so one easily finds that &lt;math&gt;n=44&lt;/math&gt; is the largest possible integer satisfying the problem conditions. <br /> <br /> In summary, the solution is that the maximum number of red socks is &lt;math&gt;r=\boxed{990}&lt;/math&gt;.<br /> === Solution 2 ===<br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; denote the number of red and blue socks such that &lt;math&gt;r+b\le1991&lt;/math&gt;. Then by complementary counting, the number of ways to get a red and a blue sock must be equal to &lt;math&gt;1-\frac12=\frac12=\frac{2rb}{(r+b)(r+b-1)}\implies4rb=(r+b)(r+b-1)&lt;/math&gt;<br /> &lt;math&gt;=(r+b)^2-(r+b)\implies r^2+2rb+b^2-r-b=4rb\implies r^2-2rb+b^2&lt;/math&gt;<br /> &lt;math&gt;=(r-b)^2=r+b&lt;/math&gt;, so &lt;math&gt;r+b&lt;/math&gt; must be a perfect square &lt;math&gt;k^2&lt;/math&gt;. Clearly, &lt;math&gt;r=\frac{k^2+k}2&lt;/math&gt;, so the larger &lt;math&gt;k&lt;/math&gt;, the larger &lt;math&gt;r&lt;/math&gt;: &lt;math&gt;k^2=44^2&lt;/math&gt; is the largest perfect square below &lt;math&gt;1991&lt;/math&gt;, and our answer is &lt;math&gt;\frac{44^2+44}2=\frac12\cdot44(44+1)=22\cdot45=11\cdot90=\boxed{990}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 3===<br /> <br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; denote the number of red and blue socks, respectively. In addition, let &lt;math&gt;t = r + b&lt;/math&gt;, the total number of socks in the drawer.<br /> <br /> From the problem, it is clear that &lt;math&gt;\frac{r(r-1)}{t(t-1)} + \frac{b(b-a)}{t(t-1)} = \frac{1}{2}&lt;/math&gt;<br /> <br /> Expanding, we get &lt;math&gt;\frac{r^2 + b^2 - r - b}{t^2 - t} = \frac{1}{2}&lt;/math&gt;<br /> <br /> Substituting &lt;math&gt;t&lt;/math&gt; for &lt;math&gt;r + b&lt;/math&gt; and cross multiplying, we get &lt;math&gt;2r^2 + 2b^2 - 2r - 2b = r^2 + 2br + b^2 - r - b&lt;/math&gt;<br /> <br /> Combining terms, we get &lt;math&gt;b^2 - 2br + r^2 - b - r = 0&lt;/math&gt;<br /> <br /> To make this expression factorable, we add &lt;math&gt;2r&lt;/math&gt; to both sides, resulting in &lt;math&gt;(b - r)^2 - 1(b-r) = (b - r - 1)(b - r) = 2r&lt;/math&gt;<br /> <br /> From this equation, we can test values for the expression &lt;math&gt;(b - r - 1)(b - r)&lt;/math&gt;, which is the multiplication of two consecutive integers, until we find the highest value of &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;r&lt;/math&gt; such that &lt;math&gt;b + r \leq 1991&lt;/math&gt;. <br /> <br /> By testing &lt;math&gt;(b - r - 1) = 43&lt;/math&gt; and &lt;math&gt;(b - r) = 44&lt;/math&gt;, we get that &lt;math&gt;r = 43(22) = 946&lt;/math&gt; and &lt;math&gt;b = 990&lt;/math&gt;. Testing values one integer higher, we get that &lt;math&gt;r = 990&lt;/math&gt; and &lt;math&gt;b = 1035&lt;/math&gt;. Since &lt;math&gt;990 + 1035 = 2025&lt;/math&gt; is greater than &lt;math&gt;1991&lt;/math&gt;, we conclude that &lt;math&gt;(946, 990)&lt;/math&gt; is our answer. <br /> <br /> Since it doesn't matter whether the number of blue or red socks is &lt;math&gt;990&lt;/math&gt;, we take the higher value for &lt;math&gt;r&lt;/math&gt;, thus the maximum number of red socks is &lt;math&gt;r=\boxed{990}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> As above, let &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;t&lt;/math&gt; denote the number of red socks, the number of blue socks, and the total number of socks, respectively. We see that &lt;math&gt;\frac{r(r-1)}{t(t-1)}+\frac{b(b-1)}{t(t-1)}=\frac{1}{2}&lt;/math&gt;, so &lt;math&gt;r^2+b^2-r-b=\frac{t(t-1)}{2}=r^2+b^2-t=\frac{t^2}{2}-\frac{t}{2}&lt;/math&gt;. <br /> <br /> Seeing that we can rewrite &lt;math&gt;r^2+b^2&lt;/math&gt; as &lt;math&gt;(r+b)^2-2rb&lt;/math&gt;, and remembering that &lt;math&gt;r+b=t&lt;/math&gt;, we have &lt;math&gt;\frac{t^2}{2}-\frac{t}{2}=t^2-2rb-t&lt;/math&gt;, so &lt;math&gt;2rb=\frac{t^2}{2}-\frac{t}{2}&lt;/math&gt;, which equals &lt;math&gt;r^2+b^2-t&lt;/math&gt;.<br /> <br /> We now have &lt;math&gt;r^2+b^2-t=2rb&lt;/math&gt;, so &lt;math&gt;r^2-2rb+b^2=t&lt;/math&gt; and &lt;math&gt;r-b=\pm\sqrt{t}&lt;/math&gt;. Adding this to &lt;math&gt;r+b=t&lt;/math&gt;, we have &lt;math&gt;2r=t\pm\sqrt{t}&lt;/math&gt;. To maximize &lt;math&gt;r&lt;/math&gt;, we must use the positive square root and maximize &lt;math&gt;t&lt;/math&gt;. The largest possible value of &lt;math&gt;t&lt;/math&gt; is the largest perfect square less than 1991, which is &lt;math&gt;1936=44^2&lt;/math&gt;. Therefore, &lt;math&gt;r=\frac{t+\sqrt{t}}{2}=\frac{1936+44}{2}=\boxed{990}&lt;/math&gt;.<br /> <br /> Solution by Zeroman<br /> <br /> == See also ==<br /> {{AIME box|year=1991|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_13&diff=85436 1991 AIME Problems/Problem 13 2017-04-26T20:11:21Z <p>Zeroman: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> A drawer contains a mixture of red socks and blue socks, at most &lt;math&gt;1991&lt;/math&gt; in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly &lt;math&gt;\frac{1}{2}&lt;/math&gt; that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data? <br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; denote the number of red and blue socks, respectively. Also, let &lt;math&gt;t=r+b&lt;/math&gt;. The probability &lt;math&gt;P&lt;/math&gt; that when two socks are drawn randomly, without replacement, both are red or both are blue is given by<br /> <br /> &lt;cmath&gt;<br /> \frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}.<br /> &lt;/cmath&gt;<br /> <br /> Solving the resulting quadratic equation &lt;math&gt;r^{2}-rt+t(t-1)/4=0&lt;/math&gt;, for &lt;math&gt;r&lt;/math&gt; in terms of &lt;math&gt;t&lt;/math&gt;, one obtains that <br /> <br /> &lt;cmath&gt;<br /> r=\frac{t\pm\sqrt{t}}{2}\, .<br /> &lt;/cmath&gt;<br /> <br /> Now, since &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;t&lt;/math&gt; are positive integers, it must be the case that &lt;math&gt;t=n^{2}&lt;/math&gt;, with &lt;math&gt;n\in\mathbb{N}&lt;/math&gt;. Hence, &lt;math&gt;r=n(n\pm 1)/2&lt;/math&gt; would correspond to the general solution. For the present case &lt;math&gt;t\leq 1991&lt;/math&gt;, and so one easily finds that &lt;math&gt;n=44&lt;/math&gt; is the largest possible integer satisfying the problem conditions. <br /> <br /> In summary, the solution is that the maximum number of red socks is &lt;math&gt;r=\boxed{990}&lt;/math&gt;.<br /> === Solution 2 ===<br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; denote the number of red and blue socks such that &lt;math&gt;r+b\le1991&lt;/math&gt;. Then by complementary counting, the number of ways to get a red and a blue sock must be equal to &lt;math&gt;1-\frac12=\frac12=\frac{2rb}{(r+b)(r+b-1)}\implies4rb=(r+b)(r+b-1)&lt;/math&gt;<br /> &lt;math&gt;=(r+b)^2-(r+b)\implies r^2+2rb+b^2-r-b=4rb\implies r^2-2rb+b^2&lt;/math&gt;<br /> &lt;math&gt;=(r-b)^2=r+b&lt;/math&gt;, so &lt;math&gt;r+b&lt;/math&gt; must be a perfect square &lt;math&gt;k^2&lt;/math&gt;. Clearly, &lt;math&gt;r=\frac{k^2+k}2&lt;/math&gt;, so the larger &lt;math&gt;k&lt;/math&gt;, the larger &lt;math&gt;r&lt;/math&gt;: &lt;math&gt;k^2=44^2&lt;/math&gt; is the largest perfect square below &lt;math&gt;1991&lt;/math&gt;, and our answer is &lt;math&gt;\frac{44^2+44}2=\frac12\cdot44(44+1)=22\cdot45=11\cdot90=\boxed{990}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 3===<br /> <br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; denote the number of red and blue socks, respectively. In addition, let &lt;math&gt;t = r + b&lt;/math&gt;, the total number of socks in the drawer.<br /> <br /> From the problem, it is clear that &lt;math&gt;\frac{r(r-1)}{t(t-1)} + \frac{b(b-a)}{t(t-1)} = \frac{1}{2}&lt;/math&gt;<br /> <br /> Expanding, we get &lt;math&gt;\frac{r^2 + b^2 - r - b}{t^2 - t} = \frac{1}{2}&lt;/math&gt;<br /> <br /> Substituting &lt;math&gt;t&lt;/math&gt; for &lt;math&gt;r + b&lt;/math&gt; and cross multiplying, we get &lt;math&gt;2r^2 + 2b^2 - 2r - 2b = r^2 + 2br + b^2 - r - b&lt;/math&gt;<br /> <br /> Combining terms, we get &lt;math&gt;b^2 - 2br + r^2 - b - r = 0&lt;/math&gt;<br /> <br /> To make this expression factorable, we add &lt;math&gt;2r&lt;/math&gt; to both sides, resulting in &lt;math&gt;(b - r)^2 - 1(b-r) = (b - r - 1)(b - r) = 2r&lt;/math&gt;<br /> <br /> From this equation, we can test values for the expression &lt;math&gt;(b - r - 1)(b - r)&lt;/math&gt;, which is the multiplication of two consecutive integers, until we find the highest value of &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;r&lt;/math&gt; such that &lt;math&gt;b + r \leq 1991&lt;/math&gt;. <br /> <br /> By testing &lt;math&gt;(b - r - 1) = 43&lt;/math&gt; and &lt;math&gt;(b - r) = 44&lt;/math&gt;, we get that &lt;math&gt;r = 43(22) = 946&lt;/math&gt; and &lt;math&gt;b = 990&lt;/math&gt;. Testing values one integer higher, we get that &lt;math&gt;r = 990&lt;/math&gt; and &lt;math&gt;b = 1035&lt;/math&gt;. Since &lt;math&gt;990 + 1035 = 2025&lt;/math&gt; is greater than &lt;math&gt;1991&lt;/math&gt;, we conclude that &lt;math&gt;(946, 990)&lt;/math&gt; is our answer. <br /> <br /> Since it doesn't matter whether the number of blue or red socks is &lt;math&gt;990&lt;/math&gt;, we take the higher value for &lt;math&gt;r&lt;/math&gt;, thus the maximum number of red socks is &lt;math&gt;r=\boxed{990}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> As above, let &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;t&lt;/math&gt; denote the number of red socks, the number of blue socks, and the total number of socks, respectively. We see that &lt;math&gt;\frac{r(r-1)}{t(t-1)}+\frac{b(b-1)}{t(t-1)}=\frac{1}{2}&lt;/math&gt;, so &lt;math&gt;r^2+b^2-r-b=\frac{t(t-1)}{2}=r^2+b^2-t=\frac{t^2}{2}-\frac{t}{2}&lt;/math&gt;. <br /> <br /> Seeing that we can rewrite &lt;math&gt;r^2+b^2&lt;/math&gt; as &lt;math&gt;(r+b)^2-2rb&lt;/math&gt;, and remembering that &lt;math&gt;r+b=t&lt;/math&gt;, we have &lt;math&gt;\frac{t^2}{2}-\frac{t}{2}=t^2-2rb-t&lt;/math&gt;, so &lt;math&gt;2rb=\frac{t^2}{2}-\frac{t}{2}&lt;/math&gt;, which equals &lt;math&gt;r^2+b^2-t&lt;/math&gt;.<br /> <br /> We now have &lt;math&gt;r^2+b^2-t=2rb&lt;/math&gt;, so &lt;math&gt;r^2-2rb+b^2=t&lt;/math&gt; and &lt;math&gt;r-b=\pm\sqrt{t}&lt;/math&gt;. Adding this to &lt;math&gt;r+b=t&lt;/math&gt;, we have &lt;math&gt;2r=t\pm\sqrt{t}&lt;/math&gt;. To maximize &lt;math&gt;r&lt;/math&gt;, we must use the positive square root and maximize &lt;math&gt;t&lt;/math&gt;. The largest possible value of &lt;math&gt;t&lt;/math&gt; is the largest perfect square less than 1991, which is &lt;math&gt;1936=44^2&lt;/math&gt;. Therefore, &lt;math&gt;r=\frac{t+\sqrt{t}}{2}=\frac{1936+44}{2}=\boxed{990}&lt;/math&gt;.<br /> Solution by Zeroman<br /> <br /> == See also ==<br /> {{AIME box|year=1991|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_8&diff=85435 1992 AIME Problems/Problem 8 2017-04-26T20:10:31Z <p>Zeroman: </p> <hr /> <div>== Problem ==<br /> For any sequence of real numbers &lt;math&gt;A=(a_1,a_2,a_3,\ldots)&lt;/math&gt;, define &lt;math&gt;\Delta A^{}_{}&lt;/math&gt; to be the sequence &lt;math&gt;(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)&lt;/math&gt;, whose &lt;math&gt;n^{\mbox{th}}_{}&lt;/math&gt; term is &lt;math&gt;a_{n+1}-a_n^{}&lt;/math&gt;. Suppose that all of the terms of the sequence &lt;math&gt;\Delta(\Delta A^{}_{})&lt;/math&gt; are &lt;math&gt;1^{}_{}&lt;/math&gt;, and that &lt;math&gt;a_{19}=a_{92}^{}=0&lt;/math&gt;. Find &lt;math&gt;a_1^{}&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> Note that the &lt;math&gt;\Delta&lt;/math&gt;s are reminiscent of differentiation; from the condition &lt;math&gt;\Delta(\Delta{A}) = 1&lt;/math&gt;, we are led to consider the differential equation<br /> &lt;cmath&gt; \frac{d^2 A}{dn^2} = 1 &lt;/cmath&gt;<br /> This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; <br /> &lt;cmath&gt; a_{n} = \frac{1}{2}(n-19)(n-92) &lt;/cmath&gt;<br /> as we must have roots at &lt;math&gt;n = 19&lt;/math&gt; and &lt;math&gt;n = 92&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;a_1=\frac{1}{2}(1-19)(1-92)=\boxed{819}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;\Delta^1 A=\Delta A&lt;/math&gt;, and &lt;math&gt;\Delta^n A=\Delta(\Delta^{(n-1)}A)&lt;/math&gt;.<br /> <br /> Note that in every sequence of &lt;math&gt;a_i&lt;/math&gt;, &lt;math&gt;a_n=\binom{n-1}{1}\Delta a_n + \binom{n-1}{2}\Delta^2 a_n +\binom{n-1}{3}\Delta^3 a_n + ...&lt;/math&gt;<br /> <br /> Then &lt;math&gt;a_n=a_1 +\binom{n-1}{1}\Delta a_1 +\binom{n-1}{2}\Delta^2 a_1 +\binom{n-1}{3}\Delta^3 a_1 + ...&lt;/math&gt;<br /> <br /> Since &lt;math&gt;\Delta a_1 =a_2 -a_1&lt;/math&gt;, &lt;math&gt;a_n=a_1 +\binom{n-1}{1}(a_2-a_1) +\binom{n-1}{2}\cdot 1=a_1 +n(a_2-a_1) +\binom{n-1}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153&lt;/math&gt;<br /> <br /> &lt;math&gt;a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095&lt;/math&gt;<br /> <br /> Solving, &lt;math&gt;a_1=\boxed{819}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> The sequence &lt;math&gt;\Delta(\Delta A)&lt;/math&gt; is the second finite difference sequence, and the first &lt;math&gt;k-1&lt;/math&gt; terms of this sequence can be computed in terms of the original sequence as shown below.<br /> <br /> &lt;math&gt;\begin{array}{rcl}<br /> a_3+a_1-2a_2&amp;=&amp;1\\<br /> a_4+a_2-2a_3&amp;=&amp;1\\<br /> &amp;\vdots\\<br /> a_k + a_{k-2} - 2a_{k-1} &amp;= &amp;1\\<br /> a_{k+1} + a_{k-1} - 2a_k &amp;=&amp; 1.\\<br /> \end{array}<br /> &lt;/math&gt;<br /> <br /> Adding the above &lt;math&gt;k-1&lt;/math&gt; equations we find that<br /> <br /> &lt;cmath&gt;(a_{k+1} - a_k) = k-1 + (a_2-a_1).\tag{1}&lt;/cmath&gt;<br /> <br /> We can sum equation &lt;math&gt;(1)&lt;/math&gt; from &lt;math&gt;k=1&lt;/math&gt; to &lt;math&gt;18&lt;/math&gt;, finding<br /> &lt;cmath&gt;18(a_1-a_2) - a_1 = 153.\tag{2} &lt;/cmath&gt;<br /> <br /> We can also sum equation &lt;math&gt;(1)&lt;/math&gt; from &lt;math&gt;k=1&lt;/math&gt; to &lt;math&gt;91&lt;/math&gt;, finding<br /> &lt;cmath&gt;91(a_1-a_2) - a_1 = 4095.\tag{3}&lt;/cmath&gt;<br /> Finally, &lt;math&gt;18\cdot (3) - 91\cdot(2)&lt;/math&gt; gives &lt;math&gt;a_1=\boxed{819}&lt;/math&gt;.<br /> <br /> Kris17<br /> <br /> == Solution 4 ==<br /> Since all terms of &lt;math&gt;\Delta(\Delta A)&lt;/math&gt; are 1, we know that &lt;math&gt;\Delta A&lt;/math&gt; looks like &lt;math&gt;(k,k+1,k+2,...)&lt;/math&gt; for some &lt;math&gt;k&lt;/math&gt;. This means &lt;math&gt;A&lt;/math&gt; looks like &lt;math&gt;(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)&lt;/math&gt;. More specifically, &lt;math&gt;A_n=a_1+k(n-1)+\frac{(a_1-1)(a_1-2)}{2}&lt;/math&gt;. Plugging in &lt;math&gt;a_{19}=a_{92}=0&lt;/math&gt;, we have the following linear system: &lt;cmath&gt;a_1+91k=-4095&lt;/cmath&gt; &lt;cmath&gt;a_1+18k=-153&lt;/cmath&gt; From this, we can easily find that &lt;math&gt;k=-54&lt;/math&gt; and &lt;math&gt;a_1=\boxed{819}&lt;/math&gt;.<br /> Solution by Zeroman<br /> == See also ==<br /> {{AIME box|year=1992|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_8&diff=85434 1992 AIME Problems/Problem 8 2017-04-26T20:00:30Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> For any sequence of real numbers &lt;math&gt;A=(a_1,a_2,a_3,\ldots)&lt;/math&gt;, define &lt;math&gt;\Delta A^{}_{}&lt;/math&gt; to be the sequence &lt;math&gt;(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)&lt;/math&gt;, whose &lt;math&gt;n^{\mbox{th}}_{}&lt;/math&gt; term is &lt;math&gt;a_{n+1}-a_n^{}&lt;/math&gt;. Suppose that all of the terms of the sequence &lt;math&gt;\Delta(\Delta A^{}_{})&lt;/math&gt; are &lt;math&gt;1^{}_{}&lt;/math&gt;, and that &lt;math&gt;a_{19}=a_{92}^{}=0&lt;/math&gt;. Find &lt;math&gt;a_1^{}&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> Note that the &lt;math&gt;\Delta&lt;/math&gt;s are reminiscent of differentiation; from the condition &lt;math&gt;\Delta(\Delta{A}) = 1&lt;/math&gt;, we are led to consider the differential equation<br /> &lt;cmath&gt; \frac{d^2 A}{dn^2} = 1 &lt;/cmath&gt;<br /> This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; <br /> &lt;cmath&gt; a_{n} = \frac{1}{2}(n-19)(n-92) &lt;/cmath&gt;<br /> as we must have roots at &lt;math&gt;n = 19&lt;/math&gt; and &lt;math&gt;n = 92&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;a_1=\frac{1}{2}(1-19)(1-92)=\boxed{819}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;\Delta^1 A=\Delta A&lt;/math&gt;, and &lt;math&gt;\Delta^n A=\Delta(\Delta^{(n-1)}A)&lt;/math&gt;.<br /> <br /> Note that in every sequence of &lt;math&gt;a_i&lt;/math&gt;, &lt;math&gt;a_n=\binom{n-1}{1}\Delta a_n + \binom{n-1}{2}\Delta^2 a_n +\binom{n-1}{3}\Delta^3 a_n + ...&lt;/math&gt;<br /> <br /> Then &lt;math&gt;a_n=a_1 +\binom{n-1}{1}\Delta a_1 +\binom{n-1}{2}\Delta^2 a_1 +\binom{n-1}{3}\Delta^3 a_1 + ...&lt;/math&gt;<br /> <br /> Since &lt;math&gt;\Delta a_1 =a_2 -a_1&lt;/math&gt;, &lt;math&gt;a_n=a_1 +\binom{n-1}{1}(a_2-a_1) +\binom{n-1}{2}\cdot 1=a_1 +n(a_2-a_1) +\binom{n-1}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153&lt;/math&gt;<br /> <br /> &lt;math&gt;a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095&lt;/math&gt;<br /> <br /> Solving, &lt;math&gt;a_1=\boxed{819}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> The sequence &lt;math&gt;\Delta(\Delta A)&lt;/math&gt; is the second finite difference sequence, and the first &lt;math&gt;k-1&lt;/math&gt; terms of this sequence can be computed in terms of the original sequence as shown below.<br /> <br /> &lt;math&gt;\begin{array}{rcl}<br /> a_3+a_1-2a_2&amp;=&amp;1\\<br /> a_4+a_2-2a_3&amp;=&amp;1\\<br /> &amp;\vdots\\<br /> a_k + a_{k-2} - 2a_{k-1} &amp;= &amp;1\\<br /> a_{k+1} + a_{k-1} - 2a_k &amp;=&amp; 1.\\<br /> \end{array}<br /> &lt;/math&gt;<br /> <br /> Adding the above &lt;math&gt;k-1&lt;/math&gt; equations we find that<br /> <br /> &lt;cmath&gt;(a_{k+1} - a_k) = k-1 + (a_2-a_1).\tag{1}&lt;/cmath&gt;<br /> <br /> We can sum equation &lt;math&gt;(1)&lt;/math&gt; from &lt;math&gt;k=1&lt;/math&gt; to &lt;math&gt;18&lt;/math&gt;, finding<br /> &lt;cmath&gt;18(a_1-a_2) - a_1 = 153.\tag{2} &lt;/cmath&gt;<br /> <br /> We can also sum equation &lt;math&gt;(1)&lt;/math&gt; from &lt;math&gt;k=1&lt;/math&gt; to &lt;math&gt;91&lt;/math&gt;, finding<br /> &lt;cmath&gt;91(a_1-a_2) - a_1 = 4095.\tag{3}&lt;/cmath&gt;<br /> Finally, &lt;math&gt;18\cdot (3) - 91\cdot(2)&lt;/math&gt; gives &lt;math&gt;a_1=\boxed{819}&lt;/math&gt;.<br /> <br /> Kris17<br /> <br /> == See also ==<br /> {{AIME box|year=1992|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_13&diff=85416 1991 AIME Problems/Problem 13 2017-04-24T01:55:34Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> A drawer contains a mixture of red socks and blue socks, at most &lt;math&gt;1991&lt;/math&gt; in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly &lt;math&gt;\frac{1}{2}&lt;/math&gt; that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data? <br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; denote the number of red and blue socks, respectively. Also, let &lt;math&gt;t=r+b&lt;/math&gt;. The probability &lt;math&gt;P&lt;/math&gt; that when two socks are drawn randomly, without replacement, both are red or both are blue is given by<br /> <br /> &lt;cmath&gt;<br /> \frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}.<br /> &lt;/cmath&gt;<br /> <br /> Solving the resulting quadratic equation &lt;math&gt;r^{2}-rt+t(t-1)/4=0&lt;/math&gt;, for &lt;math&gt;r&lt;/math&gt; in terms of &lt;math&gt;t&lt;/math&gt;, one obtains that <br /> <br /> &lt;cmath&gt;<br /> r=\frac{t\pm\sqrt{t}}{2}\, .<br /> &lt;/cmath&gt;<br /> <br /> Now, since &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;t&lt;/math&gt; are positive integers, it must be the case that &lt;math&gt;t=n^{2}&lt;/math&gt;, with &lt;math&gt;n\in\mathbb{N}&lt;/math&gt;. Hence, &lt;math&gt;r=n(n\pm 1)/2&lt;/math&gt; would correspond to the general solution. For the present case &lt;math&gt;t\leq 1991&lt;/math&gt;, and so one easily finds that &lt;math&gt;n=44&lt;/math&gt; is the largest possible integer satisfying the problem conditions. <br /> <br /> In summary, the solution is that the maximum number of red socks is &lt;math&gt;r=\boxed{990}&lt;/math&gt;.<br /> === Solution 2 ===<br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; denote the number of red and blue socks such that &lt;math&gt;r+b\le1991&lt;/math&gt;. Then by complementary counting, the number of ways to get a red and a blue sock must be equal to &lt;math&gt;1-\frac12=\frac12=\frac{2rb}{(r+b)(r+b-1)}\implies4rb=(r+b)(r+b-1)&lt;/math&gt;<br /> &lt;math&gt;=(r+b)^2-(r+b)\implies r^2+2rb+b^2-r-b=4rb\implies r^2-2rb+b^2&lt;/math&gt;<br /> &lt;math&gt;=(r-b)^2=r+b&lt;/math&gt;, so &lt;math&gt;r+b&lt;/math&gt; must be a perfect square &lt;math&gt;k^2&lt;/math&gt;. Clearly, &lt;math&gt;r=\frac{k^2+k}2&lt;/math&gt;, so the larger &lt;math&gt;k&lt;/math&gt;, the larger &lt;math&gt;r&lt;/math&gt;: &lt;math&gt;k^2=44^2&lt;/math&gt; is the largest perfect square below &lt;math&gt;1991&lt;/math&gt;, and our answer is &lt;math&gt;\frac{44^2+44}2=\frac12\cdot44(44+1)=22\cdot45=11\cdot90=\boxed{990}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 3===<br /> <br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; denote the number of red and blue socks, respectively. In addition, let &lt;math&gt;t = r + b&lt;/math&gt;, the total number of socks in the drawer.<br /> <br /> From the problem, it is clear that &lt;math&gt;\frac{r(r-1)}{t(t-1)} + \frac{b(b-a)}{t(t-1)} = \frac{1}{2}&lt;/math&gt;<br /> <br /> Expanding, we get &lt;math&gt;\frac{r^2 + b^2 - r - b}{t^2 - t} = \frac{1}{2}&lt;/math&gt;<br /> <br /> Substituting &lt;math&gt;t&lt;/math&gt; for &lt;math&gt;r + b&lt;/math&gt; and cross multiplying, we get &lt;math&gt;2r^2 + 2b^2 - 2r - 2b = r^2 + 2br + b^2 - r - b&lt;/math&gt;<br /> <br /> Combining terms, we get &lt;math&gt;b^2 - 2br + r^2 - b - r = 0&lt;/math&gt;<br /> <br /> To make this expression factorable, we add &lt;math&gt;2r&lt;/math&gt; to both sides, resulting in &lt;math&gt;(b - r)^2 - 1(b-r) = (b - r - 1)(b - r) = 2r&lt;/math&gt;<br /> <br /> From this equation, we can test values for the expression &lt;math&gt;(b - r - 1)(b - r)&lt;/math&gt;, which is the multiplication of two consecutive integers, until we find the highest value of &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;r&lt;/math&gt; such that &lt;math&gt;b + r \leq 1991&lt;/math&gt;. <br /> <br /> By testing &lt;math&gt;(b - r - 1) = 43&lt;/math&gt; and &lt;math&gt;(b - r) = 44&lt;/math&gt;, we get that &lt;math&gt;r = 43(22) = 946&lt;/math&gt; and &lt;math&gt;b = 990&lt;/math&gt;. Testing values one integer higher, we get that &lt;math&gt;r = 990&lt;/math&gt; and &lt;math&gt;b = 1035&lt;/math&gt;. Since &lt;math&gt;990 + 1035 = 2025&lt;/math&gt; is greater than &lt;math&gt;1991&lt;/math&gt;, we conclude that &lt;math&gt;(946, 990)&lt;/math&gt; is our answer. <br /> <br /> Since it doesn't matter whether the number of blue or red socks is &lt;math&gt;990&lt;/math&gt;, we take the higher value for &lt;math&gt;r&lt;/math&gt;, thus the maximum number of red socks is &lt;math&gt;r=\boxed{990}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> As above, let &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;t&lt;/math&gt; denote the number of red socks, the number of blue socks, and the total number of socks, respectively. We see that &lt;math&gt;\frac{r(r-1)}{t(t-1)}+\frac{b(b-1)}{t(t-1)}=\frac{1}{2}&lt;/math&gt;, so &lt;math&gt;r^2+b^2-r-b=\frac{t(t-1)}{2}=r^2+b^2-t=\frac{t^2}{2}-\frac{t}{2}&lt;/math&gt;. <br /> <br /> Seeing that we can rewrite &lt;math&gt;r^2+b^2&lt;/math&gt; as &lt;math&gt;(r+b)^2-2rb&lt;/math&gt;, and remembering that &lt;math&gt;r+b=t&lt;/math&gt;, we have &lt;math&gt;\frac{t^2}{2}-\frac{t}{2}=t^2-2rb-t&lt;/math&gt;, so &lt;math&gt;2rb=\frac{t^2}{2}-\frac{t}{2}&lt;/math&gt;, which equals &lt;math&gt;r^2+b^2-t&lt;/math&gt;.<br /> <br /> We now have &lt;math&gt;r^2+b^2-t=2rb&lt;/math&gt;, so &lt;math&gt;r^2-2rb+b^2=t&lt;/math&gt; and &lt;math&gt;r-b=\pm\sqrt{t}&lt;/math&gt;. Adding this to &lt;math&gt;r+b=t&lt;/math&gt;, we have &lt;math&gt;2r=t\pm\sqrt{t}&lt;/math&gt;. To maximize &lt;math&gt;r&lt;/math&gt;, we must use the positive square root and maximize &lt;math&gt;t&lt;/math&gt;. The largest possible value of &lt;math&gt;t&lt;/math&gt; is the largest perfect square less than 1991, which is &lt;math&gt;1936=44^2&lt;/math&gt;. Therefore, &lt;math&gt;r=\frac{t+\sqrt{t}}{2}=\frac{1936+44}{2}=\boxed{990}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1991|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_13&diff=85415 1991 AIME Problems/Problem 13 2017-04-24T01:55:16Z <p>Zeroman: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> A drawer contains a mixture of red socks and blue socks, at most &lt;math&gt;1991&lt;/math&gt; in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly &lt;math&gt;\frac{1}{2}&lt;/math&gt; that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data? <br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; denote the number of red and blue socks, respectively. Also, let &lt;math&gt;t=r+b&lt;/math&gt;. The probability &lt;math&gt;P&lt;/math&gt; that when two socks are drawn randomly, without replacement, both are red or both are blue is given by<br /> <br /> &lt;cmath&gt;<br /> \frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}.<br /> &lt;/cmath&gt;<br /> <br /> Solving the resulting quadratic equation &lt;math&gt;r^{2}-rt+t(t-1)/4=0&lt;/math&gt;, for &lt;math&gt;r&lt;/math&gt; in terms of &lt;math&gt;t&lt;/math&gt;, one obtains that <br /> <br /> &lt;cmath&gt;<br /> r=\frac{t\pm\sqrt{t}}{2}\, .<br /> &lt;/cmath&gt;<br /> <br /> Now, since &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;t&lt;/math&gt; are positive integers, it must be the case that &lt;math&gt;t=n^{2}&lt;/math&gt;, with &lt;math&gt;n\in\mathbb{N}&lt;/math&gt;. Hence, &lt;math&gt;r=n(n\pm 1)/2&lt;/math&gt; would correspond to the general solution. For the present case &lt;math&gt;t\leq 1991&lt;/math&gt;, and so one easily finds that &lt;math&gt;n=44&lt;/math&gt; is the largest possible integer satisfying the problem conditions. <br /> <br /> In summary, the solution is that the maximum number of red socks is &lt;math&gt;r=\boxed{990}&lt;/math&gt;.<br /> === Solution 2 ===<br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; denote the number of red and blue socks such that &lt;math&gt;r+b\le1991&lt;/math&gt;. Then by complementary counting, the number of ways to get a red and a blue sock must be equal to &lt;math&gt;1-\frac12=\frac12=\frac{2rb}{(r+b)(r+b-1)}\implies4rb=(r+b)(r+b-1)&lt;/math&gt;<br /> &lt;math&gt;=(r+b)^2-(r+b)\implies r^2+2rb+b^2-r-b=4rb\implies r^2-2rb+b^2&lt;/math&gt;<br /> &lt;math&gt;=(r-b)^2=r+b&lt;/math&gt;, so &lt;math&gt;r+b&lt;/math&gt; must be a perfect square &lt;math&gt;k^2&lt;/math&gt;. Clearly, &lt;math&gt;r=\frac{k^2+k}2&lt;/math&gt;, so the larger &lt;math&gt;k&lt;/math&gt;, the larger &lt;math&gt;r&lt;/math&gt;: &lt;math&gt;k^2=44^2&lt;/math&gt; is the largest perfect square below &lt;math&gt;1991&lt;/math&gt;, and our answer is &lt;math&gt;\frac{44^2+44}2=\frac12\cdot44(44+1)=22\cdot45=11\cdot90=\boxed{990}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 3===<br /> <br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; denote the number of red and blue socks, respectively. In addition, let &lt;math&gt;t = r + b&lt;/math&gt;, the total number of socks in the drawer.<br /> <br /> From the problem, it is clear that &lt;math&gt;\frac{r(r-1)}{t(t-1)} + \frac{b(b-a)}{t(t-1)} = \frac{1}{2}&lt;/math&gt;<br /> <br /> Expanding, we get &lt;math&gt;\frac{r^2 + b^2 - r - b}{t^2 - t} = \frac{1}{2}&lt;/math&gt;<br /> <br /> Substituting &lt;math&gt;t&lt;/math&gt; for &lt;math&gt;r + b&lt;/math&gt; and cross multiplying, we get &lt;math&gt;2r^2 + 2b^2 - 2r - 2b = r^2 + 2br + b^2 - r - b&lt;/math&gt;<br /> <br /> Combining terms, we get &lt;math&gt;b^2 - 2br + r^2 - b - r = 0&lt;/math&gt;<br /> <br /> To make this expression factorable, we add &lt;math&gt;2r&lt;/math&gt; to both sides, resulting in &lt;math&gt;(b - r)^2 - 1(b-r) = (b - r - 1)(b - r) = 2r&lt;/math&gt;<br /> <br /> From this equation, we can test values for the expression &lt;math&gt;(b - r - 1)(b - r)&lt;/math&gt;, which is the multiplication of two consecutive integers, until we find the highest value of &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;r&lt;/math&gt; such that &lt;math&gt;b + r \leq 1991&lt;/math&gt;. <br /> <br /> By testing &lt;math&gt;(b - r - 1) = 43&lt;/math&gt; and &lt;math&gt;(b - r) = 44&lt;/math&gt;, we get that &lt;math&gt;r = 43(22) = 946&lt;/math&gt; and &lt;math&gt;b = 990&lt;/math&gt;. Testing values one integer higher, we get that &lt;math&gt;r = 990&lt;/math&gt; and &lt;math&gt;b = 1035&lt;/math&gt;. Since &lt;math&gt;990 + 1035 = 2025&lt;/math&gt; is greater than &lt;math&gt;1991&lt;/math&gt;, we conclude that &lt;math&gt;(946, 990)&lt;/math&gt; is our answer. <br /> <br /> Since it doesn't matter whether the number of blue or red socks is &lt;math&gt;990&lt;/math&gt;, we take the higher value for &lt;math&gt;r&lt;/math&gt;, thus the maximum number of red socks is &lt;math&gt;r=\boxed{990}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1991|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_13&diff=85414 1991 AIME Problems/Problem 13 2017-04-24T01:55:02Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> A drawer contains a mixture of red socks and blue socks, at most &lt;math&gt;1991&lt;/math&gt; in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly &lt;math&gt;\frac{1}{2}&lt;/math&gt; that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data? <br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; denote the number of red and blue socks, respectively. Also, let &lt;math&gt;t=r+b&lt;/math&gt;. The probability &lt;math&gt;P&lt;/math&gt; that when two socks are drawn randomly, without replacement, both are red or both are blue is given by<br /> <br /> &lt;cmath&gt;<br /> \frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}.<br /> &lt;/cmath&gt;<br /> <br /> Solving the resulting quadratic equation &lt;math&gt;r^{2}-rt+t(t-1)/4=0&lt;/math&gt;, for &lt;math&gt;r&lt;/math&gt; in terms of &lt;math&gt;t&lt;/math&gt;, one obtains that <br /> <br /> &lt;cmath&gt;<br /> r=\frac{t\pm\sqrt{t}}{2}\, .<br /> &lt;/cmath&gt;<br /> <br /> Now, since &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;t&lt;/math&gt; are positive integers, it must be the case that &lt;math&gt;t=n^{2}&lt;/math&gt;, with &lt;math&gt;n\in\mathbb{N}&lt;/math&gt;. Hence, &lt;math&gt;r=n(n\pm 1)/2&lt;/math&gt; would correspond to the general solution. For the present case &lt;math&gt;t\leq 1991&lt;/math&gt;, and so one easily finds that &lt;math&gt;n=44&lt;/math&gt; is the largest possible integer satisfying the problem conditions. <br /> <br /> In summary, the solution is that the maximum number of red socks is &lt;math&gt;r=\boxed{990}&lt;/math&gt;.<br /> === Solution 2 ===<br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; denote the number of red and blue socks such that &lt;math&gt;r+b\le1991&lt;/math&gt;. Then by complementary counting, the number of ways to get a red and a blue sock must be equal to &lt;math&gt;1-\frac12=\frac12=\frac{2rb}{(r+b)(r+b-1)}\implies4rb=(r+b)(r+b-1)&lt;/math&gt;<br /> &lt;math&gt;=(r+b)^2-(r+b)\implies r^2+2rb+b^2-r-b=4rb\implies r^2-2rb+b^2&lt;/math&gt;<br /> &lt;math&gt;=(r-b)^2=r+b&lt;/math&gt;, so &lt;math&gt;r+b&lt;/math&gt; must be a perfect square &lt;math&gt;k^2&lt;/math&gt;. Clearly, &lt;math&gt;r=\frac{k^2+k}2&lt;/math&gt;, so the larger &lt;math&gt;k&lt;/math&gt;, the larger &lt;math&gt;r&lt;/math&gt;: &lt;math&gt;k^2=44^2&lt;/math&gt; is the largest perfect square below &lt;math&gt;1991&lt;/math&gt;, and our answer is &lt;math&gt;\frac{44^2+44}2=\frac12\cdot44(44+1)=22\cdot45=11\cdot90=\boxed{990}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 3===<br /> <br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; denote the number of red and blue socks, respectively. In addition, let &lt;math&gt;t = r + b&lt;/math&gt;, the total number of socks in the drawer.<br /> <br /> From the problem, it is clear that &lt;math&gt;\frac{r(r-1)}{t(t-1)} + \frac{b(b-a)}{t(t-1)} = \frac{1}{2}&lt;/math&gt;<br /> <br /> Expanding, we get &lt;math&gt;\frac{r^2 + b^2 - r - b}{t^2 - t} = \frac{1}{2}&lt;/math&gt;<br /> <br /> Substituting &lt;math&gt;t&lt;/math&gt; for &lt;math&gt;r + b&lt;/math&gt; and cross multiplying, we get &lt;math&gt;2r^2 + 2b^2 - 2r - 2b = r^2 + 2br + b^2 - r - b&lt;/math&gt;<br /> <br /> Combining terms, we get &lt;math&gt;b^2 - 2br + r^2 - b - r = 0&lt;/math&gt;<br /> <br /> To make this expression factorable, we add &lt;math&gt;2r&lt;/math&gt; to both sides, resulting in &lt;math&gt;(b - r)^2 - 1(b-r) = (b - r - 1)(b - r) = 2r&lt;/math&gt;<br /> <br /> From this equation, we can test values for the expression &lt;math&gt;(b - r - 1)(b - r)&lt;/math&gt;, which is the multiplication of two consecutive integers, until we find the highest value of &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;r&lt;/math&gt; such that &lt;math&gt;b + r \leq 1991&lt;/math&gt;. <br /> <br /> By testing &lt;math&gt;(b - r - 1) = 43&lt;/math&gt; and &lt;math&gt;(b - r) = 44&lt;/math&gt;, we get that &lt;math&gt;r = 43(22) = 946&lt;/math&gt; and &lt;math&gt;b = 990&lt;/math&gt;. Testing values one integer higher, we get that &lt;math&gt;r = 990&lt;/math&gt; and &lt;math&gt;b = 1035&lt;/math&gt;. Since &lt;math&gt;990 + 1035 = 2025&lt;/math&gt; is greater than &lt;math&gt;1991&lt;/math&gt;, we conclude that &lt;math&gt;(946, 990)&lt;/math&gt; is our answer. <br /> <br /> Since it doesn't matter whether the number of blue or red socks is &lt;math&gt;990&lt;/math&gt;, we take the higher value for &lt;math&gt;r&lt;/math&gt;, thus the maximum number of red socks is &lt;math&gt;r=\boxed{990}&lt;/math&gt;.<br /> <br /> == Solution 4 ==<br /> As above, let &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;t&lt;/math&gt; denote the number of red socks, the number of blue socks, and the total number of socks, respectively. We see that &lt;math&gt;\frac{r(r-1)}{t(t-1)}+\frac{b(b-1)}{t(t-1)}=\frac{1}{2}&lt;/math&gt;, so &lt;math&gt;r^2+b^2-r-b=\frac{t(t-1)}{2}=r^2+b^2-t=\frac{t^2}{2}-\frac{t}{2}&lt;/math&gt;. <br /> <br /> Seeing that we can rewrite &lt;math&gt;r^2+b^2&lt;/math&gt; as &lt;math&gt;(r+b)^2-2rb&lt;/math&gt;, and remembering that &lt;math&gt;r+b=t&lt;/math&gt;, we have &lt;math&gt;\frac{t^2}{2}-\frac{t}{2}=t^2-2rb-t&lt;/math&gt;, so &lt;math&gt;2rb=\frac{t^2}{2}-\frac{t}{2}&lt;/math&gt;, which equals &lt;math&gt;r^2+b^2-t&lt;/math&gt;.<br /> <br /> We now have &lt;math&gt;r^2+b^2-t=2rb&lt;/math&gt;, so &lt;math&gt;r^2-2rb+b^2=t&lt;/math&gt; and &lt;math&gt;r-b=\pm\sqrt{t}&lt;/math&gt;. Adding this to &lt;math&gt;r+b=t&lt;/math&gt;, we have &lt;math&gt;2r=t\pm\sqrt{t}&lt;/math&gt;. To maximize &lt;math&gt;r&lt;/math&gt;, we must use the positive square root and maximize &lt;math&gt;t&lt;/math&gt;. The largest possible value of &lt;math&gt;t&lt;/math&gt; is the largest perfect square less than 1991, which is &lt;math&gt;1936=44^2&lt;/math&gt;. Therefore, &lt;math&gt;r=\frac{t+\sqrt{t}}{2}=\frac{1936+44}{2}=\boxed{990}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1991|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_9&diff=85391 1989 AIME Problems/Problem 9 2017-04-23T01:12:44Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that &lt;math&gt;133^5+110^5+84^5+27^5=n^{5}&lt;/math&gt;. Find the value of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == Solution 1 ==<br /> Note that &lt;math&gt;n&lt;/math&gt; is even, since the &lt;math&gt;LHS&lt;/math&gt; consists of two odd and two even numbers. By [[Fermat's Little Theorem]], we know &lt;math&gt;{n^{5}}&lt;/math&gt; is congruent to &lt;math&gt;n&lt;/math&gt; [[modulo]] 5. Hence,<br /> &lt;center&gt;&lt;math&gt;3 + 0 + 4 + 7 \equiv n\pmod{5}&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;4 \equiv n\pmod{5}&lt;/math&gt;&lt;/center&gt;<br /> <br /> Continuing, we examine the equation modulo 3,<br /> &lt;center&gt;&lt;math&gt;1 - 1 + 0 + 0 \equiv n\pmod{3}&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;0 \equiv n\pmod{3}&lt;/math&gt;&lt;/center&gt;<br /> <br /> Thus, &lt;math&gt;n&lt;/math&gt; is divisible by three and leaves a remainder of four when divided by 5. It's obvious that &lt;math&gt;n&gt;133&lt;/math&gt;, so the only possibilities are &lt;math&gt;n = 144&lt;/math&gt; or &lt;math&gt;n \geq 174&lt;/math&gt;. It quickly becomes apparent that 174 is much too large, so &lt;math&gt;n&lt;/math&gt; must be &lt;math&gt;\boxed{144}&lt;/math&gt;.<br /> <br /> <br /> == Solution 2 ==<br /> We can cheat a little bit and approximate, since we are dealing with such large numbers. As above, &lt;math&gt;n^5\equiv n\pmod{5}&lt;/math&gt;, and it is easy to see that &lt;math&gt;n^5\equiv n\pmod 2&lt;/math&gt;. Therefore, &lt;math&gt;133^5+110^5+84^5+27^5\equiv 3+0+4+7\equiv 4\pmod{10}&lt;/math&gt;, so the last digit of &lt;math&gt;n&lt;/math&gt; is 4.<br /> <br /> We notice that &lt;math&gt;133,110,84,&lt;/math&gt; and &lt;math&gt;27&lt;/math&gt; are all very close or equal to multiples of 27. We can rewrite &lt;math&gt;n^5&lt;/math&gt; as approximately equal to &lt;math&gt;27^5(5^5+4^5+3^5+1^5) = 27^5(4393)&lt;/math&gt;. This means &lt;math&gt;\frac{n}{27}&lt;/math&gt; must be close to &lt;math&gt;4393&lt;/math&gt;.<br /> <br /> 134 will obviously be too small, so we try 144. &lt;math&gt;\left(\frac{144}{27}\right)^5=\left(\frac{16}{3}\right)^5&lt;/math&gt;. Bashing through the division, we find that &lt;math&gt;\frac{1048576}{243}\approx 4315&lt;/math&gt;, which is very close to &lt;math&gt;4393&lt;/math&gt;. It is clear that 154 will not give any closer of an answer, given the rate that fifth powers grow, so we can safely assume that &lt;math&gt;\boxed{144}&lt;/math&gt; is the answer.<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_3&diff=85389 1989 AIME Problems/Problem 3 2017-04-22T23:13:51Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> Suppose &lt;math&gt;n&lt;/math&gt; is a [[positive integer]] and &lt;math&gt;d&lt;/math&gt; is a single [[digit]] in [[base 10]]. Find &lt;math&gt;n&lt;/math&gt; if<br /> &lt;center&gt;&lt;math&gt;\frac{n}{810}=0.d25d25d25\ldots&lt;/math&gt;&lt;/center&gt;<br /> <br /> == Solution ==<br /> Repeating decimals represent [[rational number]]s. To figure out which rational number, we sum an [[infinite]] [[geometric series]], &lt;math&gt;0.d25d25d25\ldots = \sum_{i = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}&lt;/math&gt;. Thus &lt;math&gt;\frac{n}{810} = \frac{100d + 25}{999}&lt;/math&gt; so &lt;math&gt;n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}&lt;/math&gt;. Since 750 and 37 are [[relatively prime]], &lt;math&gt;4d + 1&lt;/math&gt; must be [[divisible]] by 37, and the only digit for which this is possible is &lt;math&gt;d = 9&lt;/math&gt;. Thus &lt;math&gt;4d + 1 = 37&lt;/math&gt; and &lt;math&gt;n = 750&lt;/math&gt;.<br /> <br /> <br /> (Note: Any repeating sequence of &lt;math&gt;n&lt;/math&gt; digits that looks like &lt;math&gt;0.a_1a_2a_3...a_{n-1}a_na_1a_2...&lt;/math&gt; can be written as &lt;math&gt;\frac{a_1a_2...a_n}{10^n-1}&lt;/math&gt;, where &lt;math&gt;a_1a_2...a_n&lt;/math&gt; represents an &lt;math&gt;n&lt;/math&gt; digit number.)<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_3&diff=85210 1986 AIME Problems/Problem 3 2017-04-15T18:43:17Z <p>Zeroman: /* Solution */</p> <hr /> <div>== Problem ==<br /> If &lt;math&gt;\tan x+\tan y=25&lt;/math&gt; and &lt;math&gt;\cot x + \cot y=30&lt;/math&gt;, what is &lt;math&gt;\tan(x+y)&lt;/math&gt;?<br /> <br /> == Solution 1 ==<br /> Since &lt;math&gt;\cot&lt;/math&gt; is the reciprocal function of &lt;math&gt;\tan&lt;/math&gt;:<br /> <br /> &lt;math&gt;\cot x + \cot y = \frac{1}{\tan x} + \frac{1}{\tan y} = \frac{\tan x + \tan y}{\tan x \cdot \tan y} = 30&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;\tan x \cdot \tan y = \frac{\tan x + \tan y}{30} = \frac{25}{30} = \frac{5}{6}&lt;/math&gt;<br /> <br /> Using the tangent addition formula: <br /> <br /> &lt;math&gt;\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \cdot \tan y} = \frac{25}{1-\frac{5}{6}} = \boxed{150}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Using the formula for tangent of a sum, &lt;math&gt;\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y} = \frac{25}{1-\tan x \tan y}&lt;/math&gt;. We only need to find &lt;math&gt;\tan x \tan y&lt;/math&gt;.<br /> <br /> We know that &lt;math&gt;25 = \tan x + \tan y = \frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}&lt;/math&gt;. Cross multiplying, we have &lt;math&gt;\frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y} = \frac{\sin(x+y)}{\cos x \cos y} = 25&lt;/math&gt;.<br /> <br /> Similarly, we have &lt;math&gt;30 = \cot x + \cot y = \frac{\cos x}{\sin x} + \frac{\cos y}{\sin y} = \frac{\cos x \sin y + \sin x \cos y}{\sin x \sin y} = \frac{\sin(x+y)}{\sin x \sin y}&lt;/math&gt;.<br /> <br /> Dividing:<br /> <br /> &lt;math&gt;\frac{25}{30} = \frac{\frac{\sin(x+y)}{\cos x \cos y}}{\frac{\sin(x+y)}{\sin x \sin y}} = \frac{\sin x \sin y}{\cos x \cos y} = \tan x \tan y = \frac{5}{6}&lt;/math&gt;. Plugging in to the earlier formula, we have &lt;math&gt;\tan(x+y) = \frac{25}{1-\frac{5}{6}} = \frac{25}{\frac{1}{6}} = \boxed{150}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=2|num-a=4}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_1&diff=85204 1983 AIME Problems/Problem 1 2017-04-14T21:34:44Z <p>Zeroman: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; all exceed &lt;math&gt;1&lt;/math&gt;, and let &lt;math&gt;w&lt;/math&gt; be a [[positive number]] such that &lt;math&gt;\log_x w = 24&lt;/math&gt;, &lt;math&gt;\log_y w = 40&lt;/math&gt;, and &lt;math&gt;\log_{xyz} w = 12&lt;/math&gt;. Find &lt;math&gt;\log_z w&lt;/math&gt;.<br /> <br /> == Solutions ==<br /> <br /> === Solution 1 ===<br /> The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential [[expression]]s. <br /> <br /> &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. If we now convert everything to a power of &lt;math&gt;120&lt;/math&gt;, it will be easy to isolate &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt;.<br /> <br /> &lt;math&gt;x^{120}=w^5&lt;/math&gt;, &lt;math&gt;y^{120}=w^3&lt;/math&gt;, and &lt;math&gt;(xyz)^{120}=w^{10}&lt;/math&gt;.<br /> <br /> With some substitution, we get &lt;math&gt;w^5w^3z^{120}=w^{10}&lt;/math&gt; and &lt;math&gt;\log_zw=\boxed{060}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> First we'll convert everything to exponential form.<br /> &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. The only expression with z is &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. It now becomes clear one way to find &lt;math&gt;\log_z w&lt;/math&gt; is to find what &lt;math&gt;x^{12}&lt;/math&gt; and &lt;math&gt;y^{12}&lt;/math&gt; are in terms of &lt;math&gt;w&lt;/math&gt;. <br /> <br /> Taking the square root of the equation &lt;math&gt;x^{24}=w&lt;/math&gt; results in &lt;math&gt;x^{12}=w^{1/2}&lt;/math&gt;. Taking the &lt;math&gt;12/40&lt;/math&gt; root of &lt;math&gt;y^{40}=w&lt;/math&gt; equates to &lt;math&gt;y^{12}=w^{3/10}&lt;/math&gt;.<br /> <br /> Going back to &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;, we can substitute the &lt;math&gt;x^{12}&lt;/math&gt; and &lt;math&gt;y^{12}&lt;/math&gt; with &lt;math&gt;w^{1/2}&lt;/math&gt; and &lt;math&gt;w^{3/10}&lt;/math&gt;, respectively. We now have &lt;math&gt;w^{1/2}&lt;/math&gt;*&lt;math&gt;w^{3/10}&lt;/math&gt;*&lt;math&gt;z^{12}=w&lt;/math&gt;. Simplify we get &lt;math&gt;z^{60}=w&lt;/math&gt;.<br /> So our answer is &lt;math&gt;\boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Applying the change of base formula,<br /> &lt;cmath&gt;\begin{align*} \log_x w = 24 &amp;\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\<br /> \log_y w = 40 &amp;\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \\<br /> \log_{xyz} w = 12 &amp;\implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} \end{align*}&lt;/cmath&gt;<br /> Therefore, &lt;math&gt; \frac {\log z}{\log<br /> w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}&lt;/math&gt;. <br /> <br /> Hence, &lt;math&gt; \log_z w = \boxed{060}&lt;/math&gt;.<br /> <br /> <br /> <br /> {{alternate solutions}}<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|before=First Question|num-a=2}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_5&diff=85045 2017 AIME II Problems/Problem 5 2017-03-27T22:04:31Z <p>Zeroman: /* Solutions */</p> <hr /> <div>==Problem==<br /> A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are &lt;math&gt;189&lt;/math&gt;, &lt;math&gt;320&lt;/math&gt;, &lt;math&gt;287&lt;/math&gt;, &lt;math&gt;234&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt;, and &lt;math&gt;y&lt;/math&gt;. Find the greatest possible value of &lt;math&gt;x+y&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let these four numbers be &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt;, where &lt;math&gt;a&gt;b&gt;c&gt;d&lt;/math&gt;. &lt;math&gt;x+y&lt;/math&gt; needs to be maximized, so let &lt;math&gt;x=a+b&lt;/math&gt; and &lt;math&gt;y=a+c&lt;/math&gt; because these are the two largest pairwise sums. Now &lt;math&gt;x+y=2a+b+c&lt;/math&gt; needs to be maximized. Notice &lt;math&gt;2a+b+c=3(a+b+c+d)-(a+2b+2c+3d)=3((a+c)+(b+d))-((a+d)+(b+c)+(b+d)+(c+d))&lt;/math&gt;. No matter how the numbers &lt;math&gt;189&lt;/math&gt;, &lt;math&gt;320&lt;/math&gt;, &lt;math&gt;287&lt;/math&gt;, and &lt;math&gt;234&lt;/math&gt; are assigned to the values &lt;math&gt;a+d&lt;/math&gt;, &lt;math&gt;b+c&lt;/math&gt;, &lt;math&gt;b+d&lt;/math&gt;, and &lt;math&gt;c+d&lt;/math&gt;, the sum &lt;math&gt;(a+d)+(b+c)+(b+d)+(c+d)&lt;/math&gt; will always be &lt;math&gt;189+320+287+234&lt;/math&gt;. Therefore we need to maximize &lt;math&gt;3((a+c)+(b+d))-(189+320+287+234)&lt;/math&gt;. The maximum value of &lt;math&gt;(a+c)+(b+d)&lt;/math&gt; is achieved when we let &lt;math&gt;a+c&lt;/math&gt; and &lt;math&gt;b+d&lt;/math&gt; be &lt;math&gt;320&lt;/math&gt; and &lt;math&gt;287&lt;/math&gt; because these are the two largest pairwise sums besides &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;. Therefore, the maximum possible value of &lt;math&gt;x+y=3(320+287)-(189+320+287+234)=\boxed{791}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let the four numbers be &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt;, in no particular order. Adding the pairwise sums, we have &lt;math&gt;3a+3b+3c+3d=1030+x+y&lt;/math&gt;, so &lt;math&gt;x+y=3(a+b+c+d)-1030&lt;/math&gt;. Since we want to maximize &lt;math&gt;x+y&lt;/math&gt;, we must maximize &lt;math&gt;a+b+c+d&lt;/math&gt;.<br /> <br /> Of the four sums whose values we know, there must be two sums that add to &lt;math&gt;a+b+c+d&lt;/math&gt;. To maximize this value, we choose the highest pairwise sums, &lt;math&gt;320&lt;/math&gt; and &lt;math&gt;287&lt;/math&gt;. Therefore, &lt;math&gt;a+b+c+d=320+287=607&lt;/math&gt;.<br /> <br /> We can substitute this value into the earlier equation to find that &lt;math&gt;x+y=3(607)-1030=1821-1030=\boxed{791}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_5&diff=85044 2017 AIME II Problems/Problem 5 2017-03-27T21:53:32Z <p>Zeroman: /* Solution */</p> <hr /> <div>==Problem==<br /> A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are &lt;math&gt;189&lt;/math&gt;, &lt;math&gt;320&lt;/math&gt;, &lt;math&gt;287&lt;/math&gt;, &lt;math&gt;234&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt;, and &lt;math&gt;y&lt;/math&gt;. Find the greatest possible value of &lt;math&gt;x+y&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let these four numbers be &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt;, where &lt;math&gt;a&gt;b&gt;c&gt;d&lt;/math&gt;. &lt;math&gt;x+y&lt;/math&gt; needs to be maximized, so let &lt;math&gt;x=a+b&lt;/math&gt; and &lt;math&gt;y=a+c&lt;/math&gt; because these are the two largest pairwise sums. Now &lt;math&gt;x+y=2a+b+c&lt;/math&gt; needs to be maximized. Notice &lt;math&gt;2a+b+c=3(a+b+c+d)-(a+2b+2c+3d)=3((a+c)+(b+d))-((a+d)+(b+c)+(b+d)+(c+d))&lt;/math&gt;. No matter how the numbers &lt;math&gt;189&lt;/math&gt;, &lt;math&gt;320&lt;/math&gt;, &lt;math&gt;287&lt;/math&gt;, and &lt;math&gt;234&lt;/math&gt; are assigned to the values &lt;math&gt;a+d&lt;/math&gt;, &lt;math&gt;b+c&lt;/math&gt;, &lt;math&gt;b+d&lt;/math&gt;, and &lt;math&gt;c+d&lt;/math&gt;, the sum &lt;math&gt;(a+d)+(b+c)+(b+d)+(c+d)&lt;/math&gt; will always be &lt;math&gt;189+320+287+234&lt;/math&gt;. Therefore we need to maximize &lt;math&gt;3((a+c)+(b+d))-(189+320+287+234)&lt;/math&gt;. The maximum value of &lt;math&gt;(a+c)+(b+d)&lt;/math&gt; is achieved when we let &lt;math&gt;a+c&lt;/math&gt; and &lt;math&gt;b+d&lt;/math&gt; be &lt;math&gt;320&lt;/math&gt; and &lt;math&gt;287&lt;/math&gt; because these are the two largest pairwise sums besides &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;. Therefore, the maximum possible value of &lt;math&gt;x+y=3(320+287)-(189+320+287+234)=\boxed{791}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_4&diff=84850 2017 AIME I Problems/Problem 4 2017-03-21T01:12:20Z <p>Zeroman: /* Solution */</p> <hr /> <div>==Problem 4==<br /> A pyramid has a triangular base with side lengths &lt;math&gt;20&lt;/math&gt;, &lt;math&gt;20&lt;/math&gt;, and &lt;math&gt;24&lt;/math&gt;. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length &lt;math&gt;25&lt;/math&gt;. The volume of the pyramid is &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers, and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let the triangular base be &lt;math&gt;\triangle ABC&lt;/math&gt;, with &lt;math&gt;\overline {AB} = 24&lt;/math&gt;. We find that the altitude to side &lt;math&gt;\overline {AB}&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;, so the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;(24*16)/2 = 192&lt;/math&gt;.<br /> <br /> Let the fourth vertex of the tetrahedron be &lt;math&gt;P&lt;/math&gt;, and let the midpoint of &lt;math&gt;\overline {AB}&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Since &lt;math&gt;P&lt;/math&gt; is equidistant from &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, the line through &lt;math&gt;P&lt;/math&gt; perpendicular to the plane of &lt;math&gt;\triangle ABC&lt;/math&gt; will pass through the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;, which we will call &lt;math&gt;O&lt;/math&gt;. Note that &lt;math&gt;O&lt;/math&gt; is equidistant from each of &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;. Then,<br /> <br /> &lt;cmath&gt;\overline {OM} + \overline {OC} = \overline {CM} = 16&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;\overline {OM} = d&lt;/math&gt;.<br /> Equation &lt;math&gt;(1)&lt;/math&gt;:<br /> &lt;cmath&gt;d + \sqrt {d^2 + 144} = 16&lt;/cmath&gt;<br /> <br /> Squaring both sides, we have<br /> <br /> &lt;cmath&gt;d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2d^2 + 2d\sqrt {d^2+144} = 112&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2d(d + \sqrt {d^2+144}) = 112&lt;/cmath&gt;<br /> <br /> Substituting with equation &lt;math&gt;(1)&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;2d(16) = 112&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;d = 7/2&lt;/cmath&gt;<br /> <br /> We now find that &lt;math&gt;\sqrt{d^2 + 144} = 25/2&lt;/math&gt;.<br /> <br /> Let the distance &lt;math&gt;\overline {OP} = h&lt;/math&gt;. Using the Pythagorean Theorem on triangle &lt;math&gt;AOP&lt;/math&gt;, &lt;math&gt;BOP&lt;/math&gt;, or &lt;math&gt;COP&lt;/math&gt; (all three are congruent by SSS):<br /> <br /> &lt;cmath&gt;25^2 = h^2 + (25/2)^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;625 = h^2 + 625/4&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1875/4 = h^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;25\sqrt {3} / 2 = h&lt;/cmath&gt;<br /> <br /> <br /> Finally, by the formula for volume of a pyramid,<br /> <br /> &lt;cmath&gt;V = Bh/3&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;V = (192)(25\sqrt{3}/2)/3&lt;/cmath&gt;<br /> This simplifies to &lt;math&gt;V = 800\sqrt {3}&lt;/math&gt;, so &lt;math&gt;m+n = \boxed {803}&lt;/math&gt;.<br /> <br /> Solution by Zeroman<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_4&diff=84849 2017 AIME I Problems/Problem 4 2017-03-21T01:11:20Z <p>Zeroman: /* Solution */</p> <hr /> <div>==Problem 4==<br /> A pyramid has a triangular base with side lengths &lt;math&gt;20&lt;/math&gt;, &lt;math&gt;20&lt;/math&gt;, and &lt;math&gt;24&lt;/math&gt;. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length &lt;math&gt;25&lt;/math&gt;. The volume of the pyramid is &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers, and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let the triangular base be &lt;math&gt;\triangle ABC&lt;/math&gt;, with &lt;math&gt;\overline {AB} = 24&lt;/math&gt;. We find that the altitude to side &lt;math&gt;\overline {AB}&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;, so the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;(24*16)/2 = 192&lt;/math&gt;.<br /> <br /> Let the fourth vertex of the tetrahedron be &lt;math&gt;P&lt;/math&gt;, and let the midpoint of &lt;math&gt;\overline {AB}&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Since &lt;math&gt;P&lt;/math&gt; is equidistant from &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, the line through &lt;math&gt;P&lt;/math&gt; perpendicular to the plane of &lt;math&gt;\triangle ABC&lt;/math&gt; will pass through the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;, which we will call &lt;math&gt;O&lt;/math&gt;. Note that &lt;math&gt;O&lt;/math&gt; is equidistant from each of &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;. Then,<br /> <br /> &lt;cmath&gt;\overline {OM} + \overline {OC} = \overline {CM} = 16&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;\overline {OM} = d&lt;/math&gt;.<br /> Equation &lt;math&gt;(1)&lt;/math&gt;:<br /> &lt;cmath&gt;d + \sqrt {d^2 + 144} = 16&lt;/cmath&gt;<br /> <br /> Squaring both sides, we have<br /> <br /> &lt;cmath&gt;d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2d^2 + 2d\sqrt {d^2+144} = 112&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2d(d + \sqrt {d^2+144}) = 112&lt;/cmath&gt;<br /> <br /> Substituting with equation &lt;math&gt;(1)&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;2d(16) = 112&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;d = 7/2&lt;/cmath&gt;<br /> <br /> We now find that &lt;math&gt;\sqrt{d^2 + 144} = 25/2&lt;/math&gt;.<br /> <br /> Let the distance &lt;math&gt;\overline {OP} = h&lt;/math&gt;. Using the Pythagorean Theorem on triangle &lt;math&gt;AOP&lt;/math&gt;, &lt;math&gt;BOP&lt;/math&gt;, or &lt;math&gt;COP&lt;/math&gt; (all three are congruent by SSS):<br /> <br /> &lt;cmath&gt;25^2 = h^2 + (\sqrt {d^2 + 144})^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;625 = h^2 + 625/4&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1875/4 = h^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;25\sqrt {3} / 2 = h&lt;/cmath&gt;<br /> <br /> <br /> Finally, by the formula for volume of a pyramid,<br /> <br /> &lt;cmath&gt;V = Bh/3&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;V = (192)(25\sqrt{3}/2)/3&lt;/cmath&gt;<br /> This simplifies to &lt;math&gt;V = 800\sqrt {3}&lt;/math&gt;, so &lt;math&gt;m+n = \boxed {803}&lt;/math&gt;.<br /> <br /> Solution by Zeroman<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_4&diff=84848 2017 AIME I Problems/Problem 4 2017-03-21T01:10:51Z <p>Zeroman: /* Solution */</p> <hr /> <div>==Problem 4==<br /> A pyramid has a triangular base with side lengths &lt;math&gt;20&lt;/math&gt;, &lt;math&gt;20&lt;/math&gt;, and &lt;math&gt;24&lt;/math&gt;. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length &lt;math&gt;25&lt;/math&gt;. The volume of the pyramid is &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers, and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let the triangular base be &lt;math&gt;\triangle ABC&lt;/math&gt;, with &lt;math&gt;\overline {AB} = 24&lt;/math&gt;. We find that the altitude to side &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;, so the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;(24*16)/2 = 192&lt;/math&gt;.<br /> <br /> Let the fourth vertex of the tetrahedron be &lt;math&gt;P&lt;/math&gt;, and let the midpoint of &lt;math&gt;\overline {AB}&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Since &lt;math&gt;P&lt;/math&gt; is equidistant from &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, the line through &lt;math&gt;P&lt;/math&gt; perpendicular to the plane of &lt;math&gt;\triangle ABC&lt;/math&gt; will pass through the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;, which we will call &lt;math&gt;O&lt;/math&gt;. Note that &lt;math&gt;O&lt;/math&gt; is equidistant from each of &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;. Then,<br /> <br /> &lt;cmath&gt;\overline {OM} + \overline {OC} = \overline {CM} = 16&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;\overline {OM} = d&lt;/math&gt;.<br /> Equation &lt;math&gt;(1)&lt;/math&gt;:<br /> &lt;cmath&gt;d + \sqrt {d^2 + 144} = 16&lt;/cmath&gt;<br /> <br /> Squaring both sides, we have<br /> <br /> &lt;cmath&gt;d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2d^2 + 2d\sqrt {d^2+144} = 112&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2d(d + \sqrt {d^2+144}) = 112&lt;/cmath&gt;<br /> <br /> Substituting with equation &lt;math&gt;(1)&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;2d(16) = 112&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;d = 7/2&lt;/cmath&gt;<br /> <br /> We now find that &lt;math&gt;\sqrt{d^2 + 144} = 25/2&lt;/math&gt;.<br /> <br /> Let the distance &lt;math&gt;\overline {OP} = h&lt;/math&gt;. Using the Pythagorean Theorem on triangle &lt;math&gt;AOP&lt;/math&gt;, &lt;math&gt;BOP&lt;/math&gt;, or &lt;math&gt;COP&lt;/math&gt; (all three are congruent by SSS):<br /> <br /> &lt;cmath&gt;25^2 = h^2 + (\sqrt {d^2 + 144})^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;625 = h^2 + 625/4&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1875/4 = h^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;25\sqrt {3} / 2 = h&lt;/cmath&gt;<br /> <br /> <br /> Finally, by the formula for volume of a pyramid,<br /> <br /> &lt;cmath&gt;V = Bh/3&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;V = (192)(25\sqrt{3}/2)/3&lt;/cmath&gt;<br /> This simplifies to &lt;math&gt;V = 800\sqrt {3}&lt;/math&gt;, so &lt;math&gt;m+n = \boxed {803}&lt;/math&gt;.<br /> <br /> Solution by Zeroman<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_4&diff=84565 2017 AIME I Problems/Problem 4 2017-03-09T20:48:40Z <p>Zeroman: /* Solution */</p> <hr /> <div>==Problem 4==<br /> A pyramid has a triangular base with side lengths &lt;math&gt;20&lt;/math&gt;, &lt;math&gt;20&lt;/math&gt;, and &lt;math&gt;24&lt;/math&gt;. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length &lt;math&gt;25&lt;/math&gt;. The volume of the pyramid is &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers, and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let the triangular base be &lt;math&gt;\triangle ABC&lt;/math&gt;, with &lt;math&gt;\overline {AB} = 24&lt;/math&gt;. Using Simplified Heron's formula for the area of an isosceles triangle gives &lt;math&gt;12\sqrt{32(8)}=192&lt;/math&gt;.<br /> <br /> Let the fourth vertex of the tetrahedron be &lt;math&gt;P&lt;/math&gt;, and let the midpoint of &lt;math&gt;\overline {AB}&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Since &lt;math&gt;P&lt;/math&gt; is equidistant from &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, the line through &lt;math&gt;P&lt;/math&gt; perpendicular to the plane of &lt;math&gt;\triangle ABC&lt;/math&gt; will pass through the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;, which we will call &lt;math&gt;O&lt;/math&gt;. Note that &lt;math&gt;O&lt;/math&gt; is equidistant from each of &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;. We find that &lt;math&gt;\overline {CM} = 16&lt;/math&gt;. Then,<br /> <br /> &lt;cmath&gt;\overline {OM} + \overline {OC} = \overline {CM} = 16&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;\overline {OM} = d&lt;/math&gt;.<br /> Equation &lt;math&gt;(1)&lt;/math&gt;:<br /> &lt;cmath&gt;d + \sqrt {d^2 + 144} = 16&lt;/cmath&gt;<br /> <br /> Squaring both sides, we have<br /> <br /> &lt;cmath&gt;d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2d^2 + 2d\sqrt {d^2+144} = 112&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2d(d + \sqrt {d^2+144}) = 112&lt;/cmath&gt;<br /> <br /> Substituting with equation &lt;math&gt;(1)&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;2d(16) = 112&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;d = 7/2&lt;/cmath&gt;<br /> <br /> We now find that &lt;math&gt;\sqrt{d^2 + 144} = 25/2&lt;/math&gt;.<br /> <br /> Let the distance &lt;math&gt;\overline {OP} = h&lt;/math&gt;. Using the Pythagorean Theorem on triangle &lt;math&gt;AOP&lt;/math&gt;, &lt;math&gt;BOP&lt;/math&gt;, or &lt;math&gt;COP&lt;/math&gt; (all three are congruent by SSS):<br /> <br /> &lt;cmath&gt;25^2 = h^2 + (\sqrt {d^2 + 144})^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;625 = h^2 + 625/4&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1875/4 = h^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;25\sqrt {3} / 2 = h&lt;/cmath&gt;<br /> <br /> <br /> Finally, by the formula for volume of a pyramid,<br /> <br /> &lt;cmath&gt;V = Bh/3&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;V = (192)(25\sqrt{3}/2)/3&lt;/cmath&gt;<br /> This simplifies to &lt;math&gt;V = 800\sqrt {3}&lt;/math&gt;, so &lt;math&gt;m+n = \boxed {803}&lt;/math&gt;.<br /> <br /> Solution by Zeroman<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_4&diff=84564 2017 AIME I Problems/Problem 4 2017-03-09T20:48:23Z <p>Zeroman: /* Solution */</p> <hr /> <div>==Problem 4==<br /> A pyramid has a triangular base with side lengths &lt;math&gt;20&lt;/math&gt;, &lt;math&gt;20&lt;/math&gt;, and &lt;math&gt;24&lt;/math&gt;. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length &lt;math&gt;25&lt;/math&gt;. The volume of the pyramid is &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers, and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let the triangular base be &lt;math&gt;\triangle ABC&lt;/math&gt;, with &lt;math&gt;\overline {AB} = 24&lt;/math&gt;. Using Simplified Heron's formula for the area of an isosceles triangle gives &lt;math&gt;12\sqrt{32(8)}=192&lt;/math&gt;.<br /> <br /> Let the fourth vertex of the tetrahedron be &lt;math&gt;P&lt;/math&gt;, and let the midpoint of &lt;math&gt;\overline {AB}&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Since &lt;math&gt;P&lt;/math&gt; is equidistant from &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, the line through &lt;math&gt;P&lt;/math&gt; perpendicular to the plane of &lt;math&gt;\triangle ABC&lt;/math&gt; will pass through the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;, which we will call &lt;math&gt;O&lt;/math&gt;. Note that &lt;math&gt;O&lt;/math&gt; is equidistant from each of &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;. We find that &lt;math&gt;\overline {CM} = 16&lt;/math&gt;. Then,<br /> <br /> &lt;cmath&gt;\overline {OM} + \overline {OC} = \overline {CM} = 16&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;\overline {OM} = d&lt;/math&gt;.<br /> Equation &lt;math&gt;(1)&lt;/math&gt;:<br /> &lt;cmath&gt;d + \sqrt {d^2 + 144} = 16&lt;/cmath&gt;<br /> <br /> Squaring both sides, we have<br /> <br /> &lt;cmath&gt;d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2d^2 + 2d\sqrt {d^2+144} = 112&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2d(d + \sqrt {d^2+144}) = 112&lt;/cmath&gt;<br /> <br /> Substituting with equation &lt;math&gt;(1)&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;2d(16) = 112&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;d = 7/2&lt;/cmath&gt;<br /> <br /> We now find that &lt;math&gt;\sqrt{d^2 + 144} = 25/2&lt;/math&gt;.<br /> <br /> Let the distance &lt;math&gt;\overline {OP} = h&lt;/math&gt;. Using the Pythagorean Theorem on triangle &lt;math&gt;AOP&lt;/math&gt;, &lt;math&gt;BOP&lt;/math&gt;, or &lt;math&gt;COP&lt;/math&gt; (all three are congruent by SSS):<br /> <br /> &lt;cmath&gt;25^2 = h^2 + (\sqrt {d^2 + 144})^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;625 = h^2 + 625/4&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1875/4 = h^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;25\sqrt {3} / 2 = h&lt;/cmath&gt;<br /> <br /> <br /> Finally, by the formula for volume of a pyramid,<br /> <br /> &lt;cmath&gt;V = Bh/3&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;V = (192)(25\sqrt{3}/2)/3&lt;/cmath&gt;<br /> This simplifies to &lt;math&gt;V = 800\sqrt {3}&lt;/math&gt;, so &lt;math&gt;m+n = \boxed {803}&lt;/math&gt;.<br /> Solution by Zeroman<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_4&diff=84535 2017 AIME I Problems/Problem 4 2017-03-09T01:08:19Z <p>Zeroman: /* Solution */</p> <hr /> <div>==Problem 4==<br /> A pyramid has a triangular base with side lengths &lt;math&gt;20&lt;/math&gt;, &lt;math&gt;20&lt;/math&gt;, and &lt;math&gt;24&lt;/math&gt;. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length &lt;math&gt;25&lt;/math&gt;. The volume of the pyramid is &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers, and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let the triangular base be &lt;math&gt;\triangle ABC&lt;/math&gt;, with &lt;math&gt;\overline {AB} = 24&lt;/math&gt;. Using Simplified Heron's formula for the area of an isosceles triangle gives &lt;math&gt;12\sqrt{32(8)}=192&lt;/math&gt;.<br /> <br /> Let the fourth vertex of the tetrahedron be &lt;math&gt;P&lt;/math&gt;, and let the midpoint of &lt;math&gt;\overline {AB}&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Since &lt;math&gt;P&lt;/math&gt; is equidistant from &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, the line through &lt;math&gt;P&lt;/math&gt; perpendicular to the plane of &lt;math&gt;\triangle ABC&lt;/math&gt; will pass through the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;, which we will call &lt;math&gt;O&lt;/math&gt;. Note that &lt;math&gt;O&lt;/math&gt; is equidistant from each of &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;. We find that &lt;math&gt;\overline {CM} = 16&lt;/math&gt;. Then,<br /> <br /> &lt;cmath&gt;\overline {OM} + \overline {OC} = \overline {CM} = 16&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;\overline {OM} = d&lt;/math&gt;.<br /> Equation &lt;math&gt;(1)&lt;/math&gt;:<br /> &lt;cmath&gt;d + \sqrt {d^2 + 144} = 16&lt;/cmath&gt;<br /> <br /> Squaring both sides, we have<br /> <br /> &lt;cmath&gt;d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2d^2 + 2d\sqrt {d^2+144} = 112&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2d(d + \sqrt {d^2+144}) = 112&lt;/cmath&gt;<br /> <br /> Substituting with equation &lt;math&gt;(1)&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;2d(16) = 112&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;d = 7/2&lt;/cmath&gt;<br /> <br /> We now find that &lt;math&gt;\sqrt{d^2 + 144} = 25/2&lt;/math&gt;.<br /> <br /> Let the distance &lt;math&gt;\overline {OP} = h&lt;/math&gt;. Using the Pythagorean Theorem on triangle &lt;math&gt;AOP&lt;/math&gt;, &lt;math&gt;BOP&lt;/math&gt;, or &lt;math&gt;COP&lt;/math&gt; (all three are congruent by SSS):<br /> <br /> &lt;cmath&gt;25^2 = h^2 + (\sqrt {d^2 + 144})^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;625 = h^2 + 625/4&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1875/4 = h^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;25\sqrt {3} / 2 = h&lt;/cmath&gt;<br /> <br /> <br /> Finally, by the formula for volume of a pyramid,<br /> <br /> &lt;cmath&gt;V = Bh/3&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;V = (192)(25\sqrt{3}/2)/3&lt;/cmath&gt;<br /> This simplifies to &lt;math&gt;V = 800\sqrt {3}&lt;/math&gt;, so &lt;math&gt;m+n = \boxed {803}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_4&diff=84495 2017 AIME I Problems/Problem 4 2017-03-08T22:05:03Z <p>Zeroman: /* Solution */</p> <hr /> <div>==Problem 4==<br /> A pyramid has a triangular base with side lengths &lt;math&gt;20&lt;/math&gt;, &lt;math&gt;20&lt;/math&gt;, and &lt;math&gt;24&lt;/math&gt;. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length &lt;math&gt;25&lt;/math&gt;. The volume of the pyramid is &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers, and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> ==Solution==<br /> Let the triangular base be &lt;math&gt;\triangle ABC&lt;/math&gt;, with &lt;math&gt;\overline {AB} = 24&lt;/math&gt;. Using Simplified Heron's formula for the area of an isosceles triangle gives &lt;math&gt;12\sqrt{32(8)}=192&lt;/math&gt;.<br /> <br /> Let the fourth vertex of the tetrahedron be &lt;math&gt;P&lt;/math&gt;, and let the midpoint of &lt;math&gt;\overline {AB}&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Since &lt;math&gt;P&lt;/math&gt; is equidistant from &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, the line through &lt;math&gt;P&lt;/math&gt; perpendicular to the plane of &lt;math&gt;\triangle ABC&lt;/math&gt; will pass through the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;, which we will call &lt;math&gt;O&lt;/math&gt;. Note that &lt;math&gt;O&lt;/math&gt; is equidistant from each of &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;. We find that &lt;math&gt;\overline {CM} = 16&lt;/math&gt;. Then,<br /> <br /> &lt;math&gt;\overline {OM} + \overline {OC} = \overline {CM} = 16&lt;/math&gt;<br /> <br /> &lt;math&gt;d + \sqrt {d^2 + 144} = 16&lt;/math&gt; (1)<br /> <br /> Squaring both sides, we have<br /> <br /> &lt;math&gt;d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256&lt;/math&gt;<br /> <br /> &lt;math&gt;2d^2 + 2d\sqrt {d^2+144} = 112&lt;/math&gt;<br /> <br /> &lt;math&gt;2d(d + \sqrt {d^2+144}) = 112&lt;/math&gt;<br /> <br /> Substituting with equation (1):<br /> <br /> &lt;math&gt;2d(16) = 112&lt;/math&gt;<br /> <br /> &lt;math&gt;d = 7/2&lt;/math&gt;.<br /> <br /> We now find that &lt;math&gt;\sqrt{d^2 + 144} = 25/2&lt;/math&gt;.<br /> <br /> Let the distance &lt;math&gt;\overline {OP} = h&lt;/math&gt;. Using the Pythagorean Theorem on triangle &lt;math&gt;AOP&lt;/math&gt;, &lt;math&gt;BOP&lt;/math&gt;, or &lt;math&gt;COP&lt;/math&gt; (all three are congruent by SSS):<br /> <br /> &lt;math&gt;25^2 = h^2 + (\sqrt {d^2 + 144})^2&lt;/math&gt;<br /> <br /> &lt;math&gt;625 = h^2 + 625/4&lt;/math&gt;<br /> <br /> &lt;math&gt;1875/4 = h^2&lt;/math&gt;<br /> <br /> &lt;math&gt;25\sqrt {3} / 2 = h&lt;/math&gt;.<br /> <br /> <br /> Finally, by the formula for volume of a pyramid,<br /> <br /> &lt;math&gt;V = Bh/3&lt;/math&gt;<br /> <br /> &lt;math&gt;V = (192)(25\sqrt{3}/2)/3&lt;/math&gt;. This simplifies to &lt;math&gt;V = 800\sqrt {3}&lt;/math&gt;, so &lt;math&gt;m+n = \boxed {803}&lt;/math&gt;.</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_4&diff=84481 2017 AIME I Problems/Problem 4 2017-03-08T21:41:34Z <p>Zeroman: /* Solution */</p> <hr /> <div>==Problem 4==<br /> A pyramid has a triangular base with side lengths &lt;math&gt;20&lt;/math&gt;, &lt;math&gt;20&lt;/math&gt;, and &lt;math&gt;24&lt;/math&gt;. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length &lt;math&gt;25&lt;/math&gt;. The volume of the pyramid is &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers, and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> ==Solution==<br /> Let the triangular base be &lt;math&gt;\triangle ABC&lt;/math&gt;, with &lt;math&gt;\overline {AB} = 24&lt;/math&gt;. Using Simplified Heron's formula for the area of an isosceles triangle gives &lt;math&gt;12\sqrt{32(8)}=192&lt;/math&gt;.<br /> <br /> Let the fourth vertex of the tetrahedron be &lt;math&gt;P&lt;/math&gt;, and let the midpoint of &lt;math&gt;\overline {AB}&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Since &lt;math&gt;P&lt;/math&gt; is equidistant from &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, the line through &lt;math&gt;P&lt;/math&gt; perpendicular to the plane of &lt;math&gt;\triangle ABC&lt;/math&gt; will pass through the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;, which we will call &lt;math&gt;O&lt;/math&gt;. Note that &lt;math&gt;O&lt;/math&gt; is equidistant from each of &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;. We find that &lt;math&gt;\overline {CM} = 16&lt;/math&gt;. Then,<br /> <br /> &lt;math&gt;\overline {OM} + \overline {OC} = \overline {CM} = 16&lt;/math&gt;<br /> <br /> &lt;math&gt;d + \sqrt {d^2 + 144} = 16&lt;/math&gt; (1)<br /> <br /> Squaring both sides, we have<br /> <br /> &lt;math&gt;d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256&lt;/math&gt;<br /> <br /> &lt;math&gt;2d^2 + 2d\sqrt {d^2+144} = 112&lt;/math&gt;<br /> <br /> &lt;math&gt;2d(d + \sqrt {d^2+144}) = 112&lt;/math&gt;<br /> <br /> Substituting with equation (1):<br /> <br /> &lt;math&gt;2d(16) = 112&lt;/math&gt;<br /> <br /> &lt;math&gt;d = 7/2&lt;/math&gt;.<br /> <br /> We now find that &lt;math&gt;\sqrt{d^2 + 144} = 25/2&lt;/math&gt;.<br /> <br /> Let the distance &lt;math&gt;\overline {OP} = h&lt;/math&gt;. Using the Pythagorean Theorem on triangle &lt;math&gt;AOP&lt;/math&gt;, &lt;math&gt;BOP&lt;/math&gt;, or &lt;math&gt;COP&lt;/math&gt; (all three are congruent by SSS):<br /> <br /> &lt;math&gt;25^2 = h^2 + (\sqrt {d^2 + 144})^2&lt;/math&gt;<br /> <br /> &lt;math&gt;625 = h^2 + 625/4&lt;/math&gt;<br /> <br /> &lt;math&gt;1875/4 = h^2&lt;/math&gt;<br /> <br /> &lt;math&gt;25\sqrt {3} / 2 = h&lt;/math&gt;.<br /> <br /> <br /> Finally, by the formula for volume of a pyramid,<br /> <br /> &lt;math&gt;V = Bh/3&lt;/math&gt;<br /> <br /> &lt;math&gt;V = (192)(25\sqrt{3}/2)/3&lt;/math&gt;. This simplifies to &lt;math&gt;V = 800\sqrt {3}&lt;/math&gt;, so &lt;math&gt;m+n = \boxed {803}&lt;/math&gt;.<br /> <br /> Solution by Zeroman</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_4&diff=84479 2017 AIME I Problems/Problem 4 2017-03-08T21:40:51Z <p>Zeroman: </p> <hr /> <div>==Problem 4==<br /> A pyramid has a triangular base with side lengths &lt;math&gt;20&lt;/math&gt;, &lt;math&gt;20&lt;/math&gt;, and &lt;math&gt;24&lt;/math&gt;. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length &lt;math&gt;25&lt;/math&gt;. The volume of the pyramid is &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers, and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> ==Solution==<br /> Let the triangular base be &lt;math&gt;\triangle ABC&lt;/math&gt;, with &lt;math&gt;\overline {AB} = 24&lt;/math&gt;. Using Simplified Heron's formula for the area of an isosceles triangle gives &lt;math&gt;12\sqrt{32(8)}=192&lt;/math&gt;.<br /> <br /> Let the fourth vertex of the tetrahedron be &lt;math&gt;P&lt;/math&gt;, and let the midpoint of &lt;math&gt;\overline {AB}&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Since &lt;math&gt;P&lt;/math&gt; is equidistant from &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, the line through &lt;math&gt;P&lt;/math&gt; perpendicular to the plane of &lt;math&gt;\triangle ABC&lt;/math&gt; will pass through the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;, which we will call &lt;math&gt;O&lt;/math&gt;. Note that &lt;math&gt;O&lt;/math&gt; is equidistant from each of &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;. We find that &lt;math&gt;\overline {CM} = 16&lt;/math&gt;. Then,<br /> <br /> &lt;math&gt;\overline {OM} + \overline {OC} = \overline {CM} = 16&lt;/math&gt;<br /> <br /> &lt;math&gt;d + \sqrt {d^2 + 144} = 16&lt;/math&gt; (1)<br /> <br /> Squaring both sides, we have<br /> <br /> &lt;math&gt;d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256&lt;/math&gt;<br /> <br /> &lt;math&gt;2d^2 + 2d\sqrt {d^2+144} = 112&lt;/math&gt;<br /> <br /> &lt;math&gt;2d(d + \sqrt {d^2+144}) = 112&lt;/math&gt;<br /> <br /> Substituting with equation (1):<br /> <br /> &lt;math&gt;2d(16) = 112&lt;/math&gt;<br /> <br /> &lt;math&gt;d = 7/2&lt;/math&gt;.<br /> <br /> We now find that &lt;math&gt;\sqrt{d^2 + 144} = 25/2&lt;/math&gt;.<br /> <br /> Let the distance &lt;math&gt;\overline {OP} = h&lt;/math&gt;. Using the Pythagorean Theorem on triangle &lt;math&gt;AOP&lt;/math&gt;, &lt;math&gt;BOP&lt;/math&gt;, or &lt;math&gt;COP&lt;/math&gt; (all three are congruent by SSS):<br /> <br /> &lt;math&gt;25^2 = h^2 + (\sqrt {d^2 + 144})^2&lt;/math&gt;<br /> <br /> &lt;math&gt;625 = h^2 + 625/4&lt;/math&gt;<br /> <br /> &lt;math&gt;1875/4 = h^2&lt;/math&gt;<br /> <br /> &lt;math&gt;25\sqrt {3} / 2 = h&lt;/math&gt;.<br /> <br /> <br /> Finally, by the formula for volume of a pyramid,<br /> <br /> &lt;math&gt;V = Bh/3&lt;/math&gt;<br /> <br /> &lt;math&gt;V = (192)(25\sqrt{3}/2)/3&lt;/math&gt;. This simplifies to &lt;math&gt;V = 800\sqrt {3}&lt;/math&gt;, so &lt;math&gt;m+n = \boxed {803}&lt;/math&gt;.</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_14&diff=84476 2017 AIME I Problems/Problem 14 2017-03-08T21:32:09Z <p>Zeroman: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a &gt; 1&lt;/math&gt; and &lt;math&gt;x &gt; 1&lt;/math&gt; satisfy &lt;math&gt;\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128&lt;/math&gt; and &lt;math&gt;\log_a(\log_a x) = 256&lt;/math&gt;. Find the remainder when &lt;math&gt;x&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> The first condition implies<br /> <br /> &lt;cmath&gt;a^{128} = \log_a\log_a 2 + \log_a 24 - 128&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;128+a^{128} = \log_a\log_a 2^{24}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^{a^{128}a^{a^{128}}} = 2^{24}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\left(a^{a^{128}}\right)^{\left(a^{a^{128}}\right)} = 2^{24} = 8^8&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;a^{a^{128}} = 8&lt;/math&gt;. <br /> <br /> Putting each side to the power of &lt;math&gt;128&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\left(a^{128}\right)^{\left(a^{128}\right)} = 8^{128} = 64^{64},&lt;/cmath&gt;<br /> <br /> so &lt;math&gt;a^{128} = 64 \implies a = 2^{\frac{3}{64}}&lt;/math&gt;. Specifically,<br /> <br /> &lt;cmath&gt;\log_a(y) = \frac{\log_2(y)}{\log_2(a)} = \frac{64}{3}\log_2(y)&lt;/cmath&gt;<br /> <br /> so we have that<br /> <br /> &lt;cmath&gt;256 = \log_a(\log_a(y)) = \frac{64}{3}\log_2\left(\frac{64}{3}\log_2(x)\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;12 = \log_2\left(\frac{64}{3}\log_2(x)\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2^{12} = \frac{64}{3}\log_2(x)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;192 = \log_2(x)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;x = 2^{192}&lt;/cmath&gt;<br /> <br /> We only wish to find &lt;math&gt;x\bmod 1000&lt;/math&gt;. To do this, we note that &lt;math&gt;x\equiv 0\bmod 8&lt;/math&gt; and now, by the Chinese Remainder Theorem, wish only to find &lt;math&gt;x\bmod 125&lt;/math&gt;. By Euler's Theorem:<br /> <br /> &lt;cmath&gt;2^{\phi(125)} = 2^{100} \equiv 1\bmod 125&lt;/cmath&gt;<br /> <br /> so<br /> <br /> &lt;cmath&gt;2^{192} \equiv \frac{1}{2^8} \equiv \frac{1}{256} \equiv \frac{1}{6} \bmod 125&lt;/cmath&gt;<br /> <br /> so we only need to find the inverse of &lt;math&gt;6 \bmod 125&lt;/math&gt;. It is easy to realize that &lt;math&gt;6\cdot 21 = 126 \equiv 1\bmod 125&lt;/math&gt;, so <br /> <br /> &lt;cmath&gt;x\equiv 21\bmod 125, x\equiv 0\bmod 8.&lt;/cmath&gt;<br /> <br /> Using CRT, we get that &lt;math&gt;x\equiv \boxed{896}\bmod 1000&lt;/math&gt;, finishing the solution.<br /> <br /> == See also ==<br /> {{AIME box|year=2017|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_4&diff=84467 2017 AIME I Problems/Problem 4 2017-03-08T20:57:45Z <p>Zeroman: Created page with &quot;Let the triangular base be &lt;math&gt;\triangle ABC&lt;/math&gt;.&quot;</p> <hr /> <div>Let the triangular base be &lt;math&gt;\triangle ABC&lt;/math&gt;.</div> Zeroman https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_6&diff=84408 2011 AIME II Problems/Problem 6 2017-03-05T04:11:31Z <p>Zeroman: /* Solution 1 */</p> <hr /> <div>==Problem 6==<br /> <br /> Define an ordered quadruple of integers &lt;math&gt;(a, b, c, d)&lt;/math&gt; as interesting if &lt;math&gt;1 \le a&lt;b&lt;c&lt;d \le 10&lt;/math&gt;, and &lt;math&gt;a+d&gt;b+c&lt;/math&gt;. How many interesting ordered quadruples are there?<br /> <br /> ==Solution 1==<br /> Rearranging the [[inequality]] we get &lt;math&gt;d-c &gt; b-a&lt;/math&gt;. Let &lt;math&gt;e = 11&lt;/math&gt;, then &lt;math&gt;(a, b-a, c-b, d-c, e-d)&lt;/math&gt; is a partition of 11 into 5 positive integers or equivalently:<br /> &lt;math&gt;(a-1, b-a-1, c-b-1, d-c-1, e-d-1)&lt;/math&gt; is a [[partition]] of 6 into 5 non-negative integer parts. Via a standard stars and bars argument, the number of ways to partition 6 into 5 non-negative parts is &lt;math&gt;\binom{6+4}4 = \binom{10}4 = 210&lt;/math&gt;. The interesting quadruples correspond to partitions where the second number is less than the fourth. By symmetry, there are as many partitions where the fourth is less than the second. So, if &lt;math&gt;N&lt;/math&gt; is the number of partitions where the second element is equal to the fourth, our answer is &lt;math&gt;(210-N)/2&lt;/math&gt;.<br /> <br /> We find &lt;math&gt;N&lt;/math&gt; as a sum of 4 cases:<br /> * two parts equal to zero, &lt;math&gt;\binom82 = 28&lt;/math&gt; ways,<br /> * two parts equal to one, &lt;math&gt;\binom62 = 15&lt;/math&gt; ways,<br /> * two parts equal to two, &lt;math&gt;\binom42 = 6&lt;/math&gt; ways,<br /> * two parts equal to three, &lt;math&gt;\binom22 = 1&lt;/math&gt; way.<br /> Therefore, &lt;math&gt;N = 28 + 15 + 6 + 1 = 50&lt;/math&gt; and our answer is &lt;math&gt;(210 - 50)/2 = \fbox{80.}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let us consider our quadruple (a,b,c,d) as the following image xaxbcxxdxx. The location of the letter a,b,c,d represents its value and x is a place holder. Clearly the quadruple is interesting if there are more place holders between c and d than there are between a and b. 0 holders between a and b means we consider a and b as one unit ab and c as cx yielding &lt;math&gt;\binom83 = 56&lt;/math&gt; ways, 1 holder between a and b means we consider a and b as one unit axb and c as cxx yielding &lt;math&gt;\binom 63 = 20&lt;/math&gt; ways, 2 holders between a and b means we consider a and b as one unit axxb and c as cxxx yielding &lt;math&gt;\binom43 = 4&lt;/math&gt; ways and there cannot be 3 holders between a and b so our total is 56+20+4=&lt;math&gt;\fbox{80.}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Zeroman