https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Williamhu888&feedformat=atom AoPS Wiki - User contributions [en] 2021-07-31T19:11:14Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_21&diff=60316 2014 AMC 10B Problems/Problem 21 2014-02-21T17:04:40Z <p>Williamhu888: /* Solution */</p> <hr /> <div>==Problem==<br /> Trapezoid &lt;math&gt; ABCD &lt;/math&gt; has parallel sides &lt;math&gt; \overline{AB} &lt;/math&gt; of length &lt;math&gt; 33 &lt;/math&gt; and &lt;math&gt; \overline {CD} &lt;/math&gt; of length &lt;math&gt; 21 &lt;/math&gt;. The other two sides are of lengths &lt;math&gt; 10 &lt;/math&gt; and &lt;math&gt; 14 &lt;/math&gt;. The angles &lt;math&gt; A &lt;/math&gt; and &lt;math&gt; B &lt;/math&gt; are acute. What is the length of the shorter diagonal of &lt;math&gt; ABCD &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qquad\textbf{(D) }18\sqrt{2}\qquad\textbf{(E) }26 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> &lt;asy&gt;<br /> size(7cm);<br /> pair A,B,C,D,CC,DD;<br /> A = (-2,7);<br /> B = (14,7);<br /> C = (10,0);<br /> D = (0,0);<br /> CC = (10,7);<br /> DD = (0,7);<br /> draw(A--B--C--D--cycle);<br /> //label(&quot;33&quot;,(A+B)/2,N);<br /> label(&quot;21&quot;,(C+D)/2,S);<br /> label(&quot;10&quot;,(A+D)/2,W);<br /> label(&quot;14&quot;,(B+C)/2,E);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,SW);<br /> draw(C--CC); draw(D--DD);<br /> &lt;/asy&gt;<br /> <br /> In the diagram, &lt;math&gt;\overline{DE} \perp \overline{AB}, \overline{FC} \perp \overline{AB}&lt;/math&gt;. <br /> Denote &lt;math&gt;\overline{AE} = x&lt;/math&gt; and &lt;math&gt;\overline{DE} = h&lt;/math&gt;. In right triangle &lt;math&gt;AED&lt;/math&gt;, we have from the Pythagorean theorem: &lt;math&gt;x^2+h^2=100&lt;/math&gt;. Note that since &lt;math&gt;EF = DC&lt;/math&gt;, we have &lt;math&gt;BF = 33-DC-x = 12-x&lt;/math&gt;. Using the Pythagorean theorem in right triangle &lt;math&gt;BFC&lt;/math&gt;, we have &lt;math&gt;(12-x)^2 + h^2 = 196&lt;/math&gt;. <br /> <br /> <br /> We isolate the &lt;math&gt;h^2&lt;/math&gt; term in both equations, getting &lt;math&gt;\begin{align*}h^2 &amp;= 100-x^2<br /> \\h^2 &amp;= 196-(12-x)^2\end{align}&lt;/math&gt;.<br /> <br /> Setting these equal, we have &lt;math&gt;100-x^2 = 196 - 144 + 24x -x^2 \implies 24x = 48 \implies x = 2&lt;/math&gt;. Now, we can determine that &lt;math&gt;h^2 = 100-4 \implies h = \sqrt{96}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(7cm);<br /> pair A,B,C,D,CC,DD;<br /> A = (-2,7);<br /> B = (14,7);<br /> C = (10,0);<br /> D = (0,0);<br /> CC = (10,7);<br /> DD = (0,7);<br /> draw(A--B--C--D--cycle);<br /> //label(&quot;33&quot;,(A+B)/2,N);<br /> label(&quot;21&quot;,(C+D)/2,S);<br /> label(&quot;10&quot;,(A+D)/2,W);<br /> label(&quot;14&quot;,(B+C)/2,E);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,SW);<br /> draw(C--CC); draw(D--DD);<br /> label(&quot;21&quot;,(CC+DD)/2,N);<br /> label(&quot;$2$&quot;,(A+DD)/2,N);<br /> label(&quot;$10$&quot;,(CC+B)/2,N);<br /> label(&quot;$\sqrt{96}$&quot;,(C+CC)/2,W);<br /> label(&quot;$\sqrt{96}$&quot;,(D+DD)/2,E);<br /> pair X = (-2,0);<br /> //draw(X--C--A--cycle,black+2bp);<br /> &lt;/asy&gt;<br /> <br /> The two diagonals are &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt;. Using the Pythagorean theorem again on &lt;math&gt;\bigtriangleup AFC&lt;/math&gt; and &lt;math&gt;\bigtriangleup BED&lt;/math&gt;, we can find these lengths to be &lt;math&gt;\sqrt{96+529} = 25&lt;/math&gt; and &lt;math&gt;\sqrt{96+961} = \sqrt{1057}&lt;/math&gt;. Obviously, &lt;math&gt;25&lt;/math&gt; is the shorter length, and thus the answer is &lt;math&gt;\boxed{\textbf{(B) }25}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2014|ab=B|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_24&diff=51333 2013 AMC 10B Problems/Problem 24 2013-02-22T15:31:53Z <p>Williamhu888: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A positive integer &lt;math&gt;n&lt;/math&gt; is ''nice'' if there is a positive integer &lt;math&gt;m&lt;/math&gt; with exactly four positive divisors (including &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt;) such that the sum of the four divisors is equal to &lt;math&gt;n&lt;/math&gt;. How many numbers in the set &lt;math&gt;\{ 2010,2011,2012,\dotsc,2019 \}&lt;/math&gt; are nice?<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> ==Solution==<br /> A positive integer with only four positive divisors has its prime factorization in the form of &lt;math&gt;a*b&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are both prime positive integers. The four factors of this number would be &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;ab&lt;/math&gt;. The sum of these would be &lt;math&gt;ab+a+b+1&lt;/math&gt;, which can be factored into the form &lt;math&gt;(a+1)(b+1)&lt;/math&gt;. Since all odd numbers have an odd number in their prime factorization, the product of &lt;math&gt;(a+1)(b+1)&lt;/math&gt; results in an even number. Therefore we can rule all out all of the odd integers in the set. Considering the even cases, we see that the only one that works is &lt;math&gt;2016&lt;/math&gt;, when &lt;math&gt;a=7&lt;/math&gt; and &lt;math&gt;b=251&lt;/math&gt;. Thus the answer is &lt;math&gt;\boxed{\textbf{(A)}\ 1}&lt;/math&gt;.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_2&diff=51332 2013 AMC 10B Problems/Problem 2 2013-02-22T15:25:04Z <p>Williamhu888: </p> <hr /> <div>==Problem==<br /> <br /> Mr. Green measures his rectangular garden by walking two of the sides and finding that it is &lt;math&gt;15&lt;/math&gt; steps by &lt;math&gt;20&lt;/math&gt; steps. Each of Mr. Green's steps is &lt;math&gt;2&lt;/math&gt; feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 600 \qquad\textbf{(B)}\ 800 \qquad\textbf{(C)}\ 1000 \qquad\textbf{(D)}\ 1200 \qquad\textbf{(E)}\ 1400 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Since each step is &lt;math&gt;2&lt;/math&gt; feet, his garden is &lt;math&gt;30&lt;/math&gt; by &lt;math&gt;40&lt;/math&gt; feet. Thus, the area of &lt;math&gt;30(40) = 1200&lt;/math&gt; square feet. Since he is expecting &lt;math&gt;\frac{1}{2}&lt;/math&gt; of a pound per square foot, the total amount of potatoes expected is &lt;math&gt;1200 \times \frac{1}{2} = \boxed{\textbf{(A) }600}&lt;/math&gt;</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Answer_Key&diff=49053 2009 AMC 10A Answer Key 2012-11-03T00:30:44Z <p>Williamhu888: </p> <hr /> <div>1. E<br /> <br /> 2. A<br /> <br /> 3. C<br /> <br /> 4. A<br /> <br /> 5. E<br /> <br /> 6. A<br /> <br /> 7. C<br /> <br /> 8. B<br /> <br /> 9. B<br /> <br /> 10. B<br /> <br /> 11. D<br /> <br /> 12. C<br /> <br /> 13. E<br /> <br /> 14. A<br /> <br /> 15. E<br /> <br /> 16. D<br /> <br /> 17. C<br /> <br /> 18. D<br /> <br /> 19. B<br /> <br /> 20. D<br /> <br /> 21. C<br /> <br /> 22. D<br /> <br /> 23. E<br /> <br /> 24. C<br /> <br /> 25. B</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Answer_Key&diff=49052 2009 AMC 10A Answer Key 2012-11-03T00:30:14Z <p>Williamhu888: Created page with &quot;1. E 2. A 3. C 4. A 5. E 6. A 7. C 8. B 9. B 10. B 11. D 12. C 13. E 14. A 15. E 16. D 17. C 18. D 19. B 20. D 21. C 22. D 23. E 24. C 25. B&quot;</p> <hr /> <div>1. E<br /> 2. A<br /> 3. C<br /> 4. A<br /> 5. E<br /> 6. A<br /> 7. C<br /> 8. B<br /> 9. B<br /> 10. B<br /> 11. D<br /> 12. C<br /> 13. E<br /> 14. A<br /> 15. E<br /> 16. D<br /> 17. C<br /> 18. D<br /> 19. B<br /> 20. D<br /> 21. C<br /> 22. D<br /> 23. E<br /> 24. C<br /> 25. B</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A&diff=49051 2009 AMC 10A 2012-11-03T00:29:14Z <p>Williamhu888: </p> <hr /> <div>'''2009 AMC 10A''' problems and solutions. The test was held on February 10, 2009. The first link contains the full set of test problems. The rest contain each individual problem and its solution.<br /> <br /> *[[2009 AMC 10A Problems]]<br /> *[[2009 AMC 10A Solutions]]<br /> **[[2009 AMC 10A Problems/Problem 1|Problem 1]]<br /> **[[2009 AMC 10A Problems/Problem 2|Problem 2]]<br /> **[[2009 AMC 10A Problems/Problem 3|Problem 3]]<br /> **[[2009 AMC 10A Problems/Problem 4|Problem 4]]<br /> **[[2009 AMC 10A Problems/Problem 5|Problem 5]]<br /> **[[2009 AMC 10A Problems/Problem 6|Problem 6]]<br /> **[[2009 AMC 10A Problems/Problem 7|Problem 7]]<br /> **[[2009 AMC 10A Problems/Problem 8|Problem 8]]<br /> **[[2009 AMC 10A Problems/Problem 9|Problem 9]]<br /> **[[2009 AMC 10A Problems/Problem 10|Problem 10]]<br /> **[[2009 AMC 10A Problems/Problem 11|Problem 11]]<br /> **[[2009 AMC 10A Problems/Problem 12|Problem 12]]<br /> **[[2009 AMC 10A Problems/Problem 13|Problem 13]]<br /> **[[2009 AMC 10A Problems/Problem 14|Problem 14]]<br /> **[[2009 AMC 10A Problems/Problem 15|Problem 15]]<br /> **[[2009 AMC 10A Problems/Problem 16|Problem 16]]<br /> **[[2009 AMC 10A Problems/Problem 17|Problem 17]]<br /> **[[2009 AMC 10A Problems/Problem 18|Problem 18]]<br /> **[[2009 AMC 10A Problems/Problem 19|Problem 19]]<br /> **[[2009 AMC 10A Problems/Problem 20|Problem 20]]<br /> **[[2009 AMC 10A Problems/Problem 21|Problem 21]]<br /> **[[2009 AMC 10A Problems/Problem 22|Problem 22]]<br /> **[[2009 AMC 10A Problems/Problem 23|Problem 23]]<br /> **[[2009 AMC 10A Problems/Problem 24|Problem 24]]<br /> **[[2009 AMC 10A Problems/Problem 25|Problem 25]]</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_18&diff=49050 2009 AMC 10A Problems/Problem 18 2012-11-03T00:27:16Z <p>Williamhu888: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> <br /> At Jefferson Summer Camp, &lt;math&gt;60\%&lt;/math&gt; of the children play soccer, &lt;math&gt;30\%&lt;/math&gt; of the children swim, and &lt;math&gt;40\%&lt;/math&gt; of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer? <br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 30\%<br /> \qquad<br /> \mathrm{(B)}\ 40\%<br /> \qquad<br /> \mathrm{(C)}\ 49\%<br /> \qquad<br /> \mathrm{(D)}\ 51\%<br /> \qquad<br /> \mathrm{(E)}\ 70\%<br /> &lt;/math&gt;<br /> <br /> === Solution 1 ===<br /> <br /> Out of the soccer players, &lt;math&gt;40\%&lt;/math&gt; swim. As the soccer players are &lt;math&gt;60\%&lt;/math&gt; of the whole, the swimming soccer players are &lt;math&gt;0.4 \cdot 0.6 = 0.24 = 24\%&lt;/math&gt; of all children.<br /> <br /> The non-swimming soccer players then form &lt;math&gt;60\% - 24\% = 36\%&lt;/math&gt; of all the children.<br /> <br /> Out of all the children, &lt;math&gt;30\%&lt;/math&gt; swim. We know that &lt;math&gt;24\%&lt;/math&gt; of all the children swim and play soccer, hence &lt;math&gt;30\%-24\% = 6\%&lt;/math&gt; of all the children swim and don't play soccer.<br /> <br /> Finally, we know that &lt;math&gt;70\%&lt;/math&gt; of all the children are non-swimmers. And as &lt;math&gt;36\%&lt;/math&gt; of all the children do not swim but play soccer, &lt;math&gt;70\% - 36\% = 34\%&lt;/math&gt; of all the children do not engage in any activity.<br /> <br /> A quick summary of what we found out:<br /> * &lt;math&gt;24\%&lt;/math&gt;: swimming yes, soccer yes<br /> * &lt;math&gt;36\%&lt;/math&gt;: swimming no, soccer yes<br /> * &lt;math&gt;6\%&lt;/math&gt;: swimming yes, soccer no<br /> * &lt;math&gt;34\%&lt;/math&gt;: swimming no, soccer no<br /> <br /> Now we can compute the answer. Out of all children, &lt;math&gt;70\%&lt;/math&gt; are non-swimmers, and again out of all children &lt;math&gt;36\%&lt;/math&gt; are non-swimmers that play soccer. Hence the part of non-swimmers that plays soccer is &lt;math&gt;\frac{36}{70} \approx \boxed{51\%}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Let us set that the total number of children is &lt;math&gt;100&lt;/math&gt;. So &lt;math&gt;60&lt;/math&gt; children play soccer, &lt;math&gt;30&lt;/math&gt; swim, and &lt;math&gt;0.4\times60=24&lt;/math&gt; play soccer and swim. <br /> <br /> Thus, &lt;math&gt;60-24=36&lt;/math&gt; children only play soccer.<br /> <br /> So our numerator is &lt;math&gt;36&lt;/math&gt;.<br /> <br /> Our denominator is simply &lt;math&gt;100-\text{Swimmers}=100-30=70&lt;/math&gt;<br /> <br /> And so we get &lt;math&gt;\frac{36}{70}&lt;/math&gt; which is roughly &lt;math&gt;51.4=\boxed{\text{D}}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=A|num-b=17|num-a=19}}</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_5&diff=49049 2009 AMC 12A Problems/Problem 5 2012-11-03T00:24:51Z <p>Williamhu888: /* Solution */</p> <hr /> <div>{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #5]] and [[2009 AMC 10A Problems|2009 AMC 10A #11]]}}<br /> <br /> == Problem ==<br /> One dimension of a cube is increased by &lt;math&gt;1&lt;/math&gt;, another is decreased by &lt;math&gt;1&lt;/math&gt;, and the third is left unchanged. The volume of the new rectangular solid is &lt;math&gt;5&lt;/math&gt; less than that of the cube. What was the volume of the cube?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 125 \qquad \textbf{(E)}\ 216&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let the original cube have edge length &lt;math&gt;a&lt;/math&gt;. Then its volume is &lt;math&gt;a^3&lt;/math&gt;.<br /> The new box has dimensions &lt;math&gt;a-1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;, and &lt;math&gt;a+1&lt;/math&gt;, hence its volume is &lt;math&gt;(a-1)a(a+1) = a^3-a&lt;/math&gt;. <br /> The difference between the two volumes is &lt;math&gt;a&lt;/math&gt;. As we are given that the difference is &lt;math&gt;5&lt;/math&gt;, we have &lt;math&gt;a=5&lt;/math&gt;, and the volume of the original cube was &lt;math&gt;5^3 = 125\Rightarrow\boxed{\text{(D)}}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{AMC10 box|year=2009|ab=A|num-b=10|num-a=12}}</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_5&diff=49048 2009 AMC 12A Problems/Problem 5 2012-11-03T00:24:39Z <p>Williamhu888: /* Solution */</p> <hr /> <div>{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #5]] and [[2009 AMC 10A Problems|2009 AMC 10A #11]]}}<br /> <br /> == Problem ==<br /> One dimension of a cube is increased by &lt;math&gt;1&lt;/math&gt;, another is decreased by &lt;math&gt;1&lt;/math&gt;, and the third is left unchanged. The volume of the new rectangular solid is &lt;math&gt;5&lt;/math&gt; less than that of the cube. What was the volume of the cube?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 125 \qquad \textbf{(E)}\ 216&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let the original cube have edge length &lt;math&gt;a&lt;/math&gt;. Then its volume is &lt;math&gt;a^3&lt;/math&gt;.<br /> The new box has dimensions &lt;math&gt;a-1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;, and &lt;math&gt;a+1&lt;/math&gt;, hence its volume is &lt;math&gt;(a-1)a(a+1) = a^3-a&lt;/math&gt;. <br /> The difference between the two volumes is &lt;math&gt;a&lt;/math&gt;. As we are given that the difference is &lt;math&gt;5&lt;/math&gt;, we have &lt;math&gt;a=5&lt;/math&gt;, and the volume of the original cube was &lt;math&gt;5^3 = 125\Rightarrow\boxed{(D)}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{AMC10 box|year=2009|ab=A|num-b=10|num-a=12}}</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_10&diff=49047 2009 AMC 10A Problems/Problem 10 2012-11-03T00:24:02Z <p>Williamhu888: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has a right angle at &lt;math&gt;B&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; is the foot of the altitude from &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;AD=3&lt;/math&gt;, and &lt;math&gt;DC=4&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(5mm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> dotfactor=4;<br /> <br /> pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21));<br /> pair D=foot(B,A,C);<br /> pair[] ps={B,C,A,D};<br /> <br /> draw(A--B--C--cycle);<br /> draw(B--D);<br /> draw(rightanglemark(B,D,C));<br /> <br /> dot(ps);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$3$&quot;,midpoint(A--D),NE);<br /> label(&quot;$4$&quot;,midpoint(D--C),NE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 4\sqrt3<br /> \qquad<br /> \mathrm{(B)}\ 7\sqrt3<br /> \qquad<br /> \mathrm{(C)}\ 21<br /> \qquad<br /> \mathrm{(D)}\ 14\sqrt3 <br /> \qquad<br /> \mathrm{(E)}\ 42<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> It is a well-known fact that in any right triangle &lt;math&gt;ABC&lt;/math&gt; with the right angle at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; the foot of the altitude from &lt;math&gt;B&lt;/math&gt; onto &lt;math&gt;AC&lt;/math&gt; we have &lt;math&gt;BD^2 = AD\cdot CD&lt;/math&gt;. (See below for a proof.) Then &lt;math&gt;BD = \sqrt{ 3\cdot 4 } = 2\sqrt 3&lt;/math&gt;, and the area of the triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{AC\cdot BD}2 = 7\sqrt3\Rightarrow\boxed{\text{(B)}}&lt;/math&gt;.<br /> <br /> ''Proof'': Consider the [[Pythagorean theorem]] for each of the triangles &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;ABD&lt;/math&gt;, and &lt;math&gt;CBD&lt;/math&gt;. We get:<br /> # &lt;math&gt;AB^2 + BC^2 = AC^2 = (AD+DC)^2 = AD^2 + DC^2 + 2 \cdot AD \cdot DC&lt;/math&gt;.<br /> # &lt;math&gt;AB^2 = AD^2 + BD^2&lt;/math&gt;<br /> # &lt;math&gt;BC^2 = BD^2 + CD^2&lt;/math&gt;<br /> <br /> Substituting equations 2 and 3 into the left hand side of equation 1, we get &lt;math&gt;BD^2 = AD \cdot DC&lt;/math&gt;. <br /> <br /> Alternatively, note that &lt;math&gt;\triangle ABD \sim \triangle BCD \Longrightarrow \frac{AD}{BD} = \frac{BD}{CD}&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=A|num-b=9|num-a=11}}</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_9&diff=49046 2009 AMC 10A Problems/Problem 9 2012-11-03T00:23:14Z <p>Williamhu888: </p> <hr /> <div>== Problem ==<br /> <br /> Positive integers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;2009&lt;/math&gt;, with &lt;math&gt;a&lt;b&lt;2009&lt;/math&gt;, form a geometric sequence with an integer ratio. What is &lt;math&gt;a&lt;/math&gt;? <br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 7<br /> \qquad<br /> \mathrm{(B)}\ 41<br /> \qquad<br /> \mathrm{(C)}\ 49<br /> \qquad<br /> \mathrm{(D)}\ 289<br /> \qquad<br /> \mathrm{(E)}\ 2009<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> The prime factorization of &lt;math&gt;2009&lt;/math&gt; is &lt;math&gt;2009 = 7\cdot 7\cdot 41&lt;/math&gt;. As &lt;math&gt;a&lt;b&lt;2009&lt;/math&gt;, the ratio must be positive and larger than &lt;math&gt;1&lt;/math&gt;, hence there is only one possibility: the ratio must be &lt;math&gt;7&lt;/math&gt;, and then &lt;math&gt;b=7\cdot 41&lt;/math&gt;, and &lt;math&gt;a=41\Rightarrow\text{(B)}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=A|num-b=8|num-a=10}}</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_24&diff=49045 2011 AMC 10A Problems/Problem 24 2012-11-02T02:52:45Z <p>Williamhu888: /* Solution */</p> <hr /> <div>==Problem 24==<br /> Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?<br /> <br /> &lt;math&gt;\text{(A)}\,\frac{1}{12} \qquad\text{(B)}\,\frac{\sqrt2}{12} \qquad\text{(C)}\,\frac{\sqrt3}{12} \qquad\text{(D)}\,\frac{1}{6} \qquad\text{(E)}\,\frac{\sqrt2}{6}&lt;/math&gt;<br /> <br /> ==Solution==<br /> The two tetrahedra look somewhat like this.[[File:CubeAndStel.gif]]<br /> <br /> A regular unit tetrahedron can be split into eight tetrahedra that have lengths of &lt;math&gt;\frac{1}{2}&lt;/math&gt;. <br /> The volume of a regular tetrahedron can be found using base area and height:<br /> <br /> For a tetrahedron of side length 1, its base area is &lt;math&gt;\frac{\sqrt{3}}{4}&lt;/math&gt;, and its height can be found using Pythagoras' Theorem. Its height is &lt;math&gt;\sqrt{1^2-\left(\frac{\sqrt3}{3}\right)^2}=\frac{\sqrt2}{\sqrt3}&lt;/math&gt;. Its volume is &lt;math&gt;\frac13\times\frac{\sqrt{3}}{4}\times\frac{\sqrt2}{\sqrt3}=\frac{\sqrt{2}}{12}&lt;/math&gt;.<br /> <br /> The tetrahedron actually has side length &lt;math&gt;\sqrt2&lt;/math&gt;, so the actual volume is &lt;math&gt;\frac{\sqrt{2}}{12}\times\sqrt2^3=\frac13&lt;/math&gt;.<br /> <br /> On the eight small tetrahedra, the four tetrahedra on the corners of the large tetrahedra are not inside the other large tetrahedra. Thus, &lt;math&gt;\frac{4}{8}=\frac{1}{2}&lt;/math&gt; of the large tetrahedra will not be inside the other large tetrahedra.<br /> <br /> The intersection of the two tetrahedra is thus &lt;math&gt;\frac12\times\frac13=\frac{1}{6}=\boxed{\text{(D)}}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=23|num-a=25}}</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_24&diff=49044 2011 AMC 10A Problems/Problem 24 2012-11-02T02:52:01Z <p>Williamhu888: /* Solution */</p> <hr /> <div>==Problem 24==<br /> Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?<br /> <br /> &lt;math&gt;\text{(A)}\,\frac{1}{12} \qquad\text{(B)}\,\frac{\sqrt2}{12} \qquad\text{(C)}\,\frac{\sqrt3}{12} \qquad\text{(D)}\,\frac{1}{6} \qquad\text{(E)}\,\frac{\sqrt2}{6}&lt;/math&gt;<br /> <br /> ==Solution==<br /> The two tetrahedra look somewhat like this.[[File:CubeAndStel.gif]]<br /> <br /> A regular unit tetrahedron can be split into eight tetrahedra that have lengths of &lt;math&gt;\frac{1}{2}&lt;/math&gt;. <br /> The volume of a regular tetrahedron can be found using base area and height:<br /> <br /> For a tetrahedron of side length 1, its base area is &lt;math&gt;\frac{\sqrt{3}}{4}&lt;/math&gt;, and its height can be found using Pythagoras' Theorem. Its height is &lt;math&gt;\sqrt{1^2-\left(\frac{\sqrt3}{3}\right)^2}=\frac{\sqrt2}{\sqrt3}&lt;/math&gt;. Its volume is &lt;math&gt;\frac13\times\frac{\sqrt{3}}{4}\times\frac{\sqrt2}{\sqrt3}=\frac{\sqrt{2}}{12}&lt;/math&gt;.<br /> <br /> The tetrahedron actually has side length &lt;math&gt;\sqrt2&lt;/math&gt;, so the actual volume is &lt;math&gt;\frac{\sqrt{2}}{12}\times\sqrt2^3=\frac13&lt;/math&gt;.<br /> <br /> On the eight small tetrahedra, the four tetrahedra on the corners of the large tetrahedra are not inside the other large tetrahedra. Thus, &lt;math&gt;\frac{4}{8}=\frac{1}{2}&lt;/math&gt; of the large tetrahedra will not be inside the other large tetrahedra.<br /> <br /> The intersection of the two tetrahedra is thus &lt;math&gt;\frac12\times\frac13=\frac{1}{6}=\boxed{\text{D}}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=23|num-a=25}}</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=Right_triangle&diff=48949 Right triangle 2012-10-29T02:23:02Z <p>Williamhu888: /* Special right triangles */</p> <hr /> <div>A '''right triangle''' is any [[triangle]] with an angle of 90 degrees (that is, a [[right angle]]).<br /> <br /> [[Image:Righttriangle.png]]<br /> <br /> In the image above, you see that in triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, angle C has a measure of 90 degrees, so &lt;math&gt;\triangle ABC&lt;/math&gt; is a right triangle. The sides of a right triangle have different names: The longest side, opposite the right angle, is called the [[hypotenuse]]. In the diagram, the hypotenuse is labelled c. The other two sides are called the legs of the triangle.<br /> <br /> Right triangles are very useful in [[geometry]] and for finding the [[area]]s of [[polygon]]s. The most important relationship for right triangles is the [[Pythagorean Theorem]]. In addition, the field of [[trigonometry]] arises from the study of right triangles, and nearly all [[trigonometric identities]] can be deduced from them.<br /> <br /> == Special right triangles ==<br /> There are several well-known right triangles which are easy to solve. These include the [[isosceles triangle|isosceles]] &lt;math&gt;45-45-90&lt;/math&gt;, where the hypotenuse is equal to &lt;math&gt;\sqrt{2}&lt;/math&gt; times the length of either of the legs. The &lt;math&gt;30-60-90&lt;/math&gt; has sides in the ratio of &lt;math&gt;x, x\sqrt{3}, 2x&lt;/math&gt;.<br /> <br /> If the lengths of the legs and hypotenuse are integral, then they form a [[Pythagorean triple]].<br /> <br /> Some well known Pythagorean triples include 3-4-5 ,5-12-13, 7-24-25, and others.<br /> <br /> == Properties ==<br /> The [[area]] of the triangle can be calculated using half of the product of the lengths of the legs. It can also be calculated using half of the product of the [[median of a triangle|median]] to the hypotenuse and the hypotenuse. Using similarity, it is possible to derive several formulas relating the sides, the hypotenuse, and median.<br /> <br /> The [[circumradius]] is equal to half of the hypotenuse, or the median to the hypotenuse.<br /> <br /> == Problems ==<br /> [[2007 AMC 12A Problems/Problem 10 | 2007 AMC 12A Problem 10]]<br /> <br /> == See also ==<br /> *[[Acute triangle]]<br /> *[[Obtuse triangle]]<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=Right_triangle&diff=48948 Right triangle 2012-10-29T02:22:40Z <p>Williamhu888: /* Special right triangles */</p> <hr /> <div>A '''right triangle''' is any [[triangle]] with an angle of 90 degrees (that is, a [[right angle]]).<br /> <br /> [[Image:Righttriangle.png]]<br /> <br /> In the image above, you see that in triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, angle C has a measure of 90 degrees, so &lt;math&gt;\triangle ABC&lt;/math&gt; is a right triangle. The sides of a right triangle have different names: The longest side, opposite the right angle, is called the [[hypotenuse]]. In the diagram, the hypotenuse is labelled c. The other two sides are called the legs of the triangle.<br /> <br /> Right triangles are very useful in [[geometry]] and for finding the [[area]]s of [[polygon]]s. The most important relationship for right triangles is the [[Pythagorean Theorem]]. In addition, the field of [[trigonometry]] arises from the study of right triangles, and nearly all [[trigonometric identities]] can be deduced from them.<br /> <br /> == Special right triangles ==<br /> There are several well-known right triangles which are easy to solve. These include the [[isosceles triangle|isosceles]] &lt;math&gt;45-45-90&lt;/math&gt;, where the hypotenuse is equal to &lt;math&gt;\sqrt{2}&lt;/math&gt; times the length of either of the legs. The &lt;math&gt;30-60-90&lt;/math&gt; has sides in the ratio of &lt;math&gt;x, x\sqrt{3}, 2x&lt;/math&gt;.<br /> <br /> If the lengths of the legs and hypotenuse are integral, then they form a [[Pythagorean triple]].<br /> <br /> Some well known Pythagorean triples include 3-4-5,5-12-13,7-24-25, and others.<br /> <br /> == Properties ==<br /> The [[area]] of the triangle can be calculated using half of the product of the lengths of the legs. It can also be calculated using half of the product of the [[median of a triangle|median]] to the hypotenuse and the hypotenuse. Using similarity, it is possible to derive several formulas relating the sides, the hypotenuse, and median.<br /> <br /> The [[circumradius]] is equal to half of the hypotenuse, or the median to the hypotenuse.<br /> <br /> == Problems ==<br /> [[2007 AMC 12A Problems/Problem 10 | 2007 AMC 12A Problem 10]]<br /> <br /> == See also ==<br /> *[[Acute triangle]]<br /> *[[Obtuse triangle]]<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_11&diff=47957 1997 AIME Problems/Problem 11 2012-08-20T00:29:25Z <p>Williamhu888: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;x=\frac{\sum_{n=1}^{44} \cos n^\circ}{\sum_{n=1}^{44} \sin n^\circ}&lt;/math&gt;. What is the [[greatest integer function|greatest integer]] that does not exceed &lt;math&gt;100x&lt;/math&gt;?<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> &lt;cmath&gt;\begin{eqnarray*} x &amp;=&amp; \frac {\sum_{n = 1}^{44} \cos n^\circ}{\sum_{n = 1}^{44} \sin n^\circ} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\\<br /> &amp;=&amp; \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44}\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Using the identity &lt;math&gt;\sin a + \sin b = 2\sin \frac{a+b}2 \cos \frac{a-b}{2}&lt;/math&gt; &lt;math&gt;\Longrightarrow \sin x + \cos x&lt;/math&gt; &lt;math&gt;= \sin x + \sin (90-x)&lt;/math&gt; &lt;math&gt;= 2 \sin 45 \cos (45-x)&lt;/math&gt; &lt;math&gt;= \sqrt{2} \cos (45-x)&lt;/math&gt;, that [[summation]] reduces to<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}x &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(\frac {(\cos 1 + \cos2 + \dots + \cos44) + (\sin1 + \sin2 + \dots + \sin44)}{\sin1 + \sin2 + \dots + \sin44}\right)\\<br /> &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(1 + \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\right)<br /> &lt;/cmath&gt;<br /> <br /> This fraction is equivalent to &lt;math&gt;x&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;\begin{eqnarray*}<br /> x &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(1 + x\right)\\<br /> \frac {1}{\sqrt {2}} &amp;=&amp; x\left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\\<br /> x &amp;=&amp; \frac {1}{\sqrt {2} - 1} = 1 + \sqrt {2}\\<br /> \lfloor 100x \rfloor &amp;=&amp; \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}\\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> === Solution 2 ===<br /> A slight variant of the above solution, note that <br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &amp;=&amp; \sum_{n=1}^{44} \sin n + \sin(90-n)\\<br /> &amp;=&amp; \sqrt{2}\sum_{n=1}^{44} \cos(45-n) = \sqrt{2}\sum_{n=1}^{44} \cos n\\<br /> \sum_{n=1}^{44} \sin n &amp;=&amp; (\sqrt{2}-1)\sum_{n=1}^{44} \cos n&lt;/cmath&gt;<br /> <br /> This is the [[ratio]] we are looking for. &lt;math&gt;x&lt;/math&gt; reduces to &lt;math&gt;\frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1&lt;/math&gt;, and &lt;math&gt;\lfloor 100(\sqrt{2} + 1)\rfloor = \boxed{241}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Consider the sum &lt;math&gt;\sum_{n = 1}^{44} \text{cis } n^\circ&lt;/math&gt;. The fraction is given by the real part divided by the imaginary part.<br /> <br /> The sum can be written &lt;math&gt;- 1 + \sum_{n = 0}^{44} \text{cis } n^\circ = - 1 + \frac {\text{cis } 45^\circ - 1}{\text{cis } 1^\circ - 1}&lt;/math&gt; (by [[De Moivre's Theorem]] with geometric series)<br /> <br /> &lt;math&gt;= - 1 + \frac {\frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2}}{\text{cis } 1^\circ - 1} = - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2} \right) (\text{cis } ( - 1^\circ) - 1)}{(\cos 1^\circ - 1)^2 + \sin^2 1^\circ}&lt;/math&gt; (after multiplying by [[complex conjugate]])<br /> <br /> &lt;math&gt;= - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 \right) (\cos 1^\circ - 1) + \frac {\sqrt {2}}{2}\sin 1^\circ + i\left( \left(1 - \frac {\sqrt {2}}{2} \right) \sin 1^\circ + \frac {\sqrt {2}}{2} (\cos 1^\circ - 1)\right)}{2(1 - \cos 1^\circ)}&lt;/math&gt;<br /> <br /> &lt;math&gt;= - \frac {1}{2} - \frac {\sqrt {2}}{4} - \frac {i\sqrt {2}}{4} + \frac {\sin 1^\circ \left( \frac {\sqrt {2}}{2} + i\left( 1 - \frac {\sqrt {2}}{2} \right) \right)}{2(1 - \cos 1^\circ)}&lt;/math&gt;<br /> <br /> Using the [[trigonometric identities|tangent half-angle formula]], this becomes &lt;math&gt;\left( - \frac {1}{2} + \frac {\sqrt {2}}{4}[\cot (1/2^\circ) - 1] \right) + i\left( \frac {1}{2}\cot (1/2^\circ) - \frac {\sqrt {2}}{4}[\cot (1/2^\circ) + 1] \right)&lt;/math&gt;.<br /> <br /> Dividing the two parts and multiplying each part by 4, the fraction is &lt;math&gt;\frac { - 2 + \sqrt {2}[\cot (1/2^\circ) - 1]}{2\cot (1/2^\circ) - \sqrt {2}[\cot (1/2^\circ) + 1]}&lt;/math&gt;.<br /> <br /> Although an exact value for &lt;math&gt;\cot (1/2^\circ)&lt;/math&gt; in terms of radicals will be difficult, this is easily known: it is really large!<br /> <br /> So treat it as though it were &lt;math&gt;\infty&lt;/math&gt;. The fraction is approximated by &lt;math&gt;\frac {\sqrt {2}}{2 - \sqrt {2}} = \frac {\sqrt {2}(2 + \sqrt {2})}{2} = 1 + \sqrt {2}\Rightarrow \lfloor 100(1+\sqrt2)\rfloor=\boxed{241}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_9&diff=47594 2012 AIME II Problems/Problem 9 2012-07-08T03:41:00Z <p>Williamhu888: /* The Second Term */</p> <hr /> <div>== Problem 9 ==<br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be real numbers such that &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; and &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt;. The value of &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac pq&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> <br /> == Solution ==<br /> Examine the first term in the expression we want to evaluate, &lt;math&gt;\frac{\sin 2x}{\sin 2y}&lt;/math&gt;, separately from the second term, &lt;math&gt;\frac{\cos 2x}{\cos 2y}&lt;/math&gt;.<br /> <br /> === The First Term ===<br /> <br /> Using the identity &lt;math&gt;\sin 2\theta = 2\sin\theta\cos\theta&lt;/math&gt;, we have:<br /> <br /> &lt;math&gt;\frac{2\sin x \cos x}{2\sin y \cos y} = \frac{\sin x \cos x}{\sin y \cos y} = \frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}=3\cdot\frac{1}{2} = \frac{3}{2}&lt;/math&gt;<br /> <br /> === The Second Term ===<br /> <br /> Let the equation &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; be equation 1, and let the equation &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt; be equation 2.<br /> Hungry for the widely-used identity &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt;, we cross multiply equation 1 by &lt;math&gt;\sin y&lt;/math&gt; and multiply equation 2 by &lt;math&gt;\cos y&lt;/math&gt;.<br /> <br /> Equation 1 then becomes:<br /> <br /> &lt;math&gt;\sin x = 3\sin y&lt;/math&gt;.<br /> <br /> Equation 2 then becomes:<br /> <br /> &lt;math&gt;\cos x = \frac{1}{2} \cos y&lt;/math&gt;<br /> <br /> Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:<br /> <br /> &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} \cos^2 y&lt;/math&gt;<br /> <br /> Applying the identity &lt;math&gt;\cos^2 y = 1 - \sin^2 y&lt;/math&gt; (which is similar to &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt; but a bit different), we can change &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} \cos^2 y&lt;/math&gt; into:<br /> <br /> &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y&lt;/math&gt;<br /> <br /> Rearranging, we get &lt;math&gt;\frac{3}{4} = \frac{35}{4} \sin^2 y &lt;/math&gt;.<br /> <br /> So, &lt;math&gt;\sin^2 y = \frac{3}{35}&lt;/math&gt;.<br /> <br /> Squaring Equation 1 (leading to &lt;math&gt;\sin^2 x = 9\sin^2 y&lt;/math&gt;), we can solve for &lt;math&gt;\sin^2 x&lt;/math&gt;:<br /> <br /> &lt;math&gt;\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}&lt;/math&gt;<br /> <br /> Using the identity &lt;math&gt;\cos 2\theta = 1 - 2\sin^2\theta&lt;/math&gt;, we can solve for &lt;math&gt;\frac{\cos 2x}{\cos 2y}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos 2y = 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}&lt;/math&gt;.<br /> <br /> === Now Back to the Solution! ===<br /> <br /> Finally, &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}&lt;/math&gt;.<br /> <br /> So, the answer is &lt;math&gt;49+58=\boxed{107}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=8|num-a=10}}</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=Vector&diff=47298 Vector 2012-05-30T23:36:50Z <p>Williamhu888: </p> <hr /> <div>The word '''vector''' has many different definitions, depending on who is defining it and in what context. Physicists will often refer to a vector as &quot;a quantity with a direction and magnitude.&quot; For Euclidean geometers, a vector is essentially a directed line segment. In many situations, a vector is best considered as an n-tuple of numbers (often real or complex). Most generally, but also most abstractly, a vector is any object which is an element of a given vector space. <br /> <br /> A vector is usually graphically represented as an arrow. Vectors can be uniquely described in many ways. The two most common is (for 2-dimensional vectors) by describing it with its length (or magnitude) and the angle it makes with some fixed line (usually the x-axis) or by describing it as an arrow beginning at the origin and ending at the point &lt;math&gt;(x,y)&lt;/math&gt;. An &lt;math&gt;n&lt;/math&gt;-dimensional vector can be described in this coordinate form as an ordered &lt;math&gt;n&lt;/math&gt;-tuple of numbers within angle brackets or parentheses, &lt;math&gt;(x\,\,y\,\,z\,\,...)&lt;/math&gt;. The set of vectors over a [[field]] is called a [[vector space]].<br /> <br /> == Description ==<br /> Every vector &lt;math&gt;\overrightarrow{PQ}&lt;/math&gt; has a starting point &lt;math&gt;P\langle x_1, y_1\rangle&lt;/math&gt; and an endpoint &lt;math&gt;Q\langle x_2, y_2\rangle&lt;/math&gt;. Since the only thing that distinguishes one vector from another is its magnitude or length, and direction, vectors can be freely translated about a plane without changing. Hence, it is convenient to consider a vector as originating from the origin. This way, two vectors can be compared by only looking at their endpoints. This is why we only require &lt;math&gt;n&lt;/math&gt; values for an &lt;math&gt;n&lt;/math&gt; dimensional vector written in the form &lt;math&gt;(x\,\,y\,\,z\,\,...)&lt;/math&gt;. The magnitude of a vector, denoted &lt;math&gt;\|\vec{v}\|&lt;/math&gt;, is found simply by <br /> using the distance formula.<br /> <br /> == Addition of Vectors ==<br /> For vectors &lt;math&gt;\vec{v}&lt;/math&gt; and &lt;math&gt;\vec{w}&lt;/math&gt;, with angle &lt;math&gt;\theta&lt;/math&gt; formed by them, &lt;math&gt;\|\vec{v}+\vec{w}\|^2=\|\vec{v}\|^2+\|\vec{w}\|^2+2\|\vec{v}\|\|\vec{w}\|\cos\theta&lt;/math&gt;.<br /> {{asy image|&lt;asy&gt; <br /> size(150);<br /> pen p=linewidth(1);<br /> MA(&quot;\theta&quot;,(5,-1),(2,3),(4,6),0.3,9,yellow);<br /> MC(&quot;\vec v&quot;,D((0,0)--(2,3),orange+p,Arrow),NW);<br /> D((2,3)--(3,4.5));<br /> MC(&quot;\vec w&quot;,D((2,3)--(5,-1),green+p,Arrow),NE);<br /> MC(-10,&quot;\vec{v}+\vec{w}&quot;,D((0,0)--(5,-1),red+p,Arrow),S);<br /> &lt;/asy&gt;|right|Addition of vectors}}<br /> <br /> From this it is simple to derive that for a real number &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;c\vec{v}&lt;/math&gt; is the vector &lt;math&gt;\vec{v}&lt;/math&gt; with magnitude multiplied by &lt;math&gt;c&lt;/math&gt;. Negative &lt;math&gt;c&lt;/math&gt; corresponds to opposite directions.<br /> <br /> == Properties of Vectors ==<br /> Since a [[vector space]] is defined over a [[field]] &lt;math&gt;K&lt;/math&gt;, it is logically inherent that vectors have the same properties as those elements in a field.<br /> <br /> For any vectors &lt;math&gt;\vec{x}&lt;/math&gt;, &lt;math&gt;\vec{y}&lt;/math&gt;, &lt;math&gt;\vec{z}&lt;/math&gt;, and real numbers &lt;math&gt;a,b&lt;/math&gt;,<br /> <br /> #&lt;math&gt;\vec{x}+\vec{y}=\vec{y}+\vec{x}&lt;/math&gt; ([[Commutative]] in +)<br /> #&lt;math&gt;(\vec{x}+\vec{y})+\vec{z}=\vec{x}+(\vec{y}+\vec{z})&lt;/math&gt; ([[Associative]] in +)<br /> #There exists the zero vector &lt;math&gt;\vec{0}&lt;/math&gt; such that &lt;math&gt;\vec{x}+\vec{0}=\vec{x}&lt;/math&gt; ([[Additive identity]])<br /> #For each &lt;math&gt;\vec{x}&lt;/math&gt;, there is a vector &lt;math&gt;\vec{y}&lt;/math&gt; such that &lt;math&gt;\vec{x}+\vec{y}=\vec{0}&lt;/math&gt; ([[Additive inverse]])<br /> #&lt;math&gt;1\vec{x}=\vec{x}&lt;/math&gt; (Unit scalar identity)<br /> #&lt;math&gt;(ab)\vec{x}=a(b\vec{x})&lt;/math&gt; ([[Associative]] in scalar)<br /> #&lt;math&gt;a(\vec{x}+\vec{y})=a\vec{x}+a\vec{y}&lt;/math&gt; ([[Distributive]] on vectors)<br /> #&lt;math&gt;(a+b)\vec{x}=a\vec{x}+b\vec{x}&lt;/math&gt; ([[Distributive]] on scalars)<br /> <br /> == Vector Operations ==<br /> ===Dot (Scalar) Product===<br /> Consider two vectors &lt;math&gt;\bold{a}=\langle a_1,a_2,\ldots,a_n\rangle&lt;/math&gt; and &lt;math&gt;\bold{b}=\langle b_1, b_2,\ldots,b_n\rangle&lt;/math&gt; in &lt;math&gt;\mathbb{R}^n&lt;/math&gt;. The dot product is defined as &lt;math&gt;\bold{a}\cdot\bold{b}=\bold{b}\cdot\bold{a}=|\bold{a}| |\bold{b}|\cos\theta=a_1b_1+a_2b_2+\cdots+a_nb_n&lt;/math&gt;, where &lt;math&gt;\theta&lt;/math&gt; is the angle formed by the two vectors. This also yields the geometric interpretation of the dot product: from basic right triangle trigonometry, it follows that the dot product is equal to the length of the [[projection]] (i.e. the distance from the origin to the foot of the head of &lt;math&gt;\bold{a}&lt;/math&gt; to &lt;math&gt;\bold{b}&lt;/math&gt;) of &lt;math&gt;\bold{a}&lt;/math&gt; onto &lt;math&gt;\bold{b}&lt;/math&gt; times the length of &lt;math&gt;\bold{b}&lt;/math&gt;. Note that the dot product is &lt;math&gt;0&lt;/math&gt; if and only if the two vectors are perpendicular.<br /> <br /> ===Cross (Vector) Product===<br /> The cross product between two vectors &lt;math&gt;\bold{a}&lt;/math&gt; and &lt;math&gt;\bold{b}&lt;/math&gt; in &lt;math&gt;\mathbb{R}^3&lt;/math&gt; is defined as the vector whose length is equal to the area of the parallelogram spanned by &lt;math&gt;\bold{a}&lt;/math&gt; and &lt;math&gt;\bold{b}&lt;/math&gt; and whose direction is in accordance with the [[right-hand rule]]. Because of this, &lt;math&gt;|\bold{a}\times\bold{b}|=|\bold{a}| |\bold{b}|\sin\theta&lt;/math&gt;, where &lt;math&gt;\theta&lt;/math&gt; is the angle formed by the two vectors, and from the [[right-hand rule]] condition, &lt;math&gt;\bold{a}\times\bold{b}=-\bold{b}\times\bold{a}&lt;/math&gt;. Also, &lt;math&gt;\sin^2\theta+\cos^2\theta=1&lt;/math&gt; gives that &lt;math&gt;|\bold{a}|^2|\bold{b}|^2=|\bold{a}\cdot\bold{b}|^2+|\bold{a}\times\bold{b}|^2&lt;/math&gt;.<br /> <br /> If &lt;math&gt;\bold{a}=\langle a_1,a_2,a_3\rangle&lt;/math&gt; and &lt;math&gt;\bold{b}=\langle b_1,b_2,b_3\rangle&lt;/math&gt;, then the cross product of &lt;math&gt;\bold{a}&lt;/math&gt; and &lt;math&gt;\bold{b}&lt;/math&gt; is given by <br /> &lt;center&gt;&lt;math&gt;\bold{a}\times\bold{b}=\begin{vmatrix} \hat{i} &amp; \hat{j} &amp; \hat{k} \\ a_1 &amp; a_2 &amp; a_3 \\ b_1 &amp; b_2 &amp; b_3\end{vmatrix}.&lt;/math&gt;&lt;/center&gt;<br /> where &lt;math&gt;\hat{i},\hat{j},\hat{k}&lt;/math&gt; are [[unit vector]]s along the coordinate axes, or equivalently, &lt;math&gt;\bold{a}\times\bold{b}=\langle a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1\rangle&lt;/math&gt;. Also, &lt;math&gt;\bold{a}\times\bold{a}=\bold{0}&lt;/math&gt;<br /> <br /> ===Triple Scalar Product===<br /> The triple scalar product of three vectors &lt;math&gt;\bold{a,b,c}&lt;/math&gt; is defined as &lt;math&gt;(\bold{a}\times\bold{b})\cdot \bold{c}&lt;/math&gt;. Geometrically, the triple scalar product gives the signed volume of the [[parallelepiped]] determined by &lt;math&gt;\bold{a,b}&lt;/math&gt; and &lt;math&gt;\bold{c}&lt;/math&gt;. It follows that <br /> <br /> &lt;center&gt;&lt;math&gt;(\bold{a}\times\bold{b})\cdot \bold{c} = (\bold{c}\times\bold{a})\cdot \bold{b} = (\bold{b}\times\bold{c})\cdot \bold{a}.&lt;/math&gt;&lt;/center&gt;<br /> <br /> It can also be shown that <br /> <br /> &lt;center&gt;&lt;math&gt;(\bold{a}\times\bold{b})\cdot \bold{c} = \begin{vmatrix} a_1 &amp; a_2 &amp; a_3 \\ b_1 &amp; b_2 &amp; b_3 \\ c_1 &amp; c_2 &amp; c_3 \end{vmatrix}.&lt;/math&gt;&lt;/center&gt;<br /> <br /> ===Triple Vector Product===<br /> The vector triple product of &lt;math&gt;\bold{a},\bold{b},\bold{c}&lt;/math&gt; is defined as the cross product of one vector, so that &lt;math&gt;\bold{a}\times(\bold{b}\times\bold{c})=\bold{b}(\bold{a}\cdot\bold{c})-\bold{c}(\bold{a}\cdot\bold{b})&lt;/math&gt;, which can be remembered by the mnemonic &quot;BAC-CAB&quot; (this relationship between the cross product and dot product is called the triple product expansion, or Lagrange's formula).<br /> <br /> == See Also ==<br /> *[[Linear Algebra]]<br /> *[[Matrix]]<br /> *[http://www.artofproblemsolving.com/Forum/index.php?f=346\ Matrix-Linear Algebra AOPS forum]<br /> <br /> == Discussion ==<br /> *[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89911\ This is a thread about what vectors are.]</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=MATHCOUNTS&diff=46927 MATHCOUNTS 2012-05-11T22:37:27Z <p>Williamhu888: </p> <hr /> <div>'''MATHCOUNTS''' is a large national [[mathematics competition]] and [[mathematics coaching]] program that has served millions of middle school students since 1984. Sponsored by the [[CNA Foundation]], [[National Society of Professional Engineers]], the [[National Council of Teachers of Mathematics]], and others, the focus of MATHCOUNTS is on [[mathematical problem solving]]. Students are eligible for up to three years, but cannot compete beyond their eighth grade year.<br /> <br /> == MATHCOUNTS Curriculum ==<br /> MATHCOUNTS curriculum includes [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], [[probability]], and [[statistics]]. The focus of MATHCOUNTS curriculum is in developing [[mathematical problem solving]] skills.<br /> <br /> Before 1990, MATHCOUNTS chose particular areas of mathematics to highlight each year before changing the focus of the competition more broadly to problem solving.<br /> <br /> ==Past Winners==<br /> * 1984: Michael Edwards, Texas<br /> * 1985: Timothy Kokesh, Oklahoma<br /> * 1986: Brian David Ewald, Florida<br /> * 1987: Russell Mann, Tennessee<br /> * 1988: Andrew Schultz, Illinois<br /> * 1989: Albert Kurz, Pennsylvania<br /> * 1990: Brian Jenkins, Arkansas<br /> * 1991: Jonathan L. Weinstein, Massachusetts<br /> * 1992: Andrei C. Gnepp, Ohio<br /> * 1993: Carleton Bosley, Kansas<br /> * 1994: William O. Engel, Illinois<br /> * 1995: Richard Reifsnyder, Kentucky<br /> * 1996: Alexander Schwartz, Pennsylvania<br /> * 1997: Zhihao Liu, Wisconsin<br /> * 1998: Ricky Liu, Massachusetts<br /> * 1999: Po-Ru Loh, Wisconsin<br /> * 2000: Ruozhou Jia, Illinois<br /> * 2001: Ryan Ko, New Jersey<br /> * 2002: Albert Ni, Illinois<br /> * 2003: Adam Hesterberg, Washington<br /> * 2004: Gregory Gauthier, Illinois<br /> * 2005: Neal Wu, Louisiana<br /> * 2006: Daesun Yim, New Jersey<br /> * 2007: Kevin Chen, Texas<br /> * 2008: Darryl Wu, Washington<br /> * 2009: Bobby Shen, Texas<br /> * 2010: Mark Sellke, Indiana<br /> * 2011: Scott Wu, Louisiana<br /> * 2012: Chad Qian, Indiana<br /> <br /> == Past State Team Winners ==<br /> * 1984: Virginia<br /> * 1985: Florida<br /> * 1986: California<br /> * 1987: New York<br /> * 1988: New York<br /> * 1989: North Carolina<br /> * 1990: Ohio<br /> * 1991: Alabama<br /> * 1992: California<br /> * 1993: Kansas<br /> * 1994: Pennsylvania<br /> * 1995: Massachusetts<br /> * 1996: Wisconsin<br /> * 1997: Massachusetts<br /> * 1998: Wisconsin<br /> * 1999: Massachusetts<br /> * 2000: California<br /> * 2001: Virginia<br /> * 2002: California<br /> * 2003: California<br /> * 2004: Illinois<br /> * 2005: Texas<br /> * 2006: Virginia<br /> * 2007: Texas<br /> * 2008: Texas<br /> * 2009: Texas<br /> * 2010: California<br /> * 2011: California<br /> * 2012: Massachusetts<br /> <br /> == MATHCOUNTS Competition Structure ==<br /> <br /> === Sprint Round ===<br /> <br /> 30 problems in 40 minutes. This round is generally made up questions ranging from relatively easy to relatively difficult. Some of the difficult problems are only difficult because calculators are not allowed in this round.<br /> <br /> === Target Round ===<br /> 8 problems given 2 at a time. Each set of two problems is given six minutes. Students may not go back to previous rounds (or forwards to future rounds) even if they finish before time is called. Students may use calculators.<br /> <br /> === Team Round ===<br /> <br /> 10 problems in 20 minutes for a team of 4 students. These problems typically include some of the most difficult problems of the competition. Use of a calculator is allowed (and required for some questions).<br /> <br /> === Countdown Round ===<br /> High scoring individuals compete head-to-head until a champion is crowned. People compete from off a screen taking 45 seconds or less to finish the problem. The Countdown round is run differently in various different chapter, state, and national competitions. In the national competitions, it is the round that determines the champion. Calculators are not allowed.<br /> <br /> <br /> ====Chapter and State Competitions====<br /> <br /> In the chapter and state competitions, the countdown round is not mandatory. However, if it is deemed official by the chapter or state, the following format must be used:<br /> <br /> *The 10th place written finisher competes against the 9th place written finisher. A problem is displayed, and both competitors have 45 seconds to answer the question, and the first competitor to correctly answer the question receives one point. The person who gets the most correct out of three questions (not necessarily two out of three) is the winner.<br /> <br /> *The winner of the first round goes up against the 8th place finisher.<br /> <br /> *The winner of the second round goes up against the 7th place finisher.<br /> <br /> This process is continued until the countdown round reaches the top four written competitors. Starting then, the first person to get three questions correct wins (as opposed to the best-out-of-three rule).<br /> <br /> If the countdown round is unofficial, any format may be used. Single-elimination bracket-style tournaments are common.<br /> <br /> ====National Competition====<br /> <br /> At the national competition, there are some structural changes to the countdown round. The top 12 (not the top 10) written finishers make it to the countdown round, and the format is changed from a ladder competition to a single elimination tournament where the top four written competitors get a bye. This setup makes it far more likely for a 12th place finisher to become champion, and it makes it less likely for a first place written finisher to become champion, equalizing the field. But even then, a 12th place written competitor will have less of a chance to become champion than the top 4, because the top 4 get a bye. Until the semi-finals, the scoring is best out of five advances.<br /> <br /> At the first round and the second round, the person to correctly answer the most out of 5 questions wins. However, at the semifinals, the rules slightly change&amp;mdash;the first person to correctly answer four questions wins.<br /> <br /> === Ciphering Round ===<br /> In some states, (most notably Florida) there is an optional ciphering round. Very similar to countdown (in both difficulty and layout), a team sends up a representative to go against all representatives from the other teams. A problem is shown on a screen and students work fast to answer the problem. The students give their answer and after 45 seconds the answer is shown and the answers are checked to see if they are right. The fastest correct answer gets five points, the next fastest gets 4, etc. There are 4 questions per individual and teams send up 4 people. A perfect score is then 80. Often times the questions take clever reading skills. For example, one question was &quot;How much dirt is in a 3 ft by 3 ft by 4 ft hole?&quot; The answer was 0 because there is no dirt in a hole.<br /> <br /> === Masters Round ===<br /> Top students give in-depth explanations to challenging problems. This round is optional at the state level competition and is mandatory at the national competition. At nationals the top two on the written and countdown participate.<br /> <br /> === Scoring and Ranking ===<br /> An individual's score is their total number of correct sprint round answers plus 2 times their total number of correct target round answers. This total is out of a maximum of &lt;math&gt;30 + 2(8) = 46&lt;/math&gt; points.<br /> <br /> A team's score is the average of the individual scores of its four members plus 2 points for every correct team round answer, making a team's maximum possible score 66 points. Therefore, it is possible to win with a relatively low team score and a phenomenal individual score, as the team score is only roughly 30% of the total team score.<br /> <br /> == MATHCOUNTS Competition Levels ==<br /> === School Competition ===<br /> Students vie for the chance to make their school teams. Problems at this level are generally the easiest and most basic in curriculum.<br /> <br /> === Chapter Competition ===<br /> Chapter competitions serve as a selection filter for state competitions. A few states don't need to host chapter competitions due to a small population size.<br /> <br /> === State Competition ===<br /> The top 4 students in each state form the state team for the national competition. The coach of the top school team at the state level is invited to coach the state team at the national competition. Interestingly, the coach of a state team is not necessarily the coach of any of the state's team members.<br /> <br /> === National Competition ===<br /> ==== National Competition Sites ====<br /> For many years, the National MATHCOUNTS competition was held in Washington, D.C. More recently, the competition has changed venues often.<br /> <br /> * The 2012 competition was held in Orlando, Florida.<br /> * The 2011 competition was held in Washington, D.C.<br /> * The 2009 and 2010 competitions was held in Orlando, Florida.<br /> * The 2008 competition was held in Denver, Colorado.<br /> * The 2007 competition was held in Fort Worth, Texas.<br /> * The 2006 competition was held in Arlington, Virginia.<br /> * The 2005 competition was held in Detroit, Michigan.<br /> * The 2004 competition was held in Washington, D.C.<br /> * The 2002 and 2003 competitions were held in Chicago, Illinois.<br /> <br /> ==== Rewards ====<br /> <br /> Every competitor at the national competition receives a graphing calculator that varies by year - for example, in 2006 it was a TI-84 Plus Silver Edition with the MATHCOUNTS logo on the back. In 2007, MATHCOUNTS took the logo off. In 2008 to 2011, they gave TI-&lt;math&gt;n&lt;/math&gt;spires to everyone. They also give out a laptop and an $8000 scholarship to the winner.<br /> <br /> == MATHCOUNTS Resources ==<br /> === MATHCOUNTS Books ===<br /> * [http://www.artofproblemsolving.com/Books/AoPS_B_CP_MC.php MATHCOUNTS books] at the [http://www.artofproblemsolving.com/Books/AoPS_B_About.php AoPS Bookstore]<br /> * [[Art of Problem Solving]]'s [http://www.artofproblemsolving.com/Books/AoPS_B_Rec_Middle.php Introductory subject textbooks] are ideal for students preparing for MATHCOUNTS.<br /> <br /> === MATHCOUNTS Classes ===<br /> * [[Art of Problem Solving]] hosts [http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesP.php#mc MATHCOUNTS preparation classes].<br /> * [[Art of Problem Solving]] hosts many free MATHCOUNTS [[Math Jams]]. [http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php Math Jam Schedule]. [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php Math Jam Transcript Archive].<br /> <br /> === MATHCOUNTS Online ===<br /> * [http://www.mathcounts.org MATHCOUNTS Homepage]<br /> * [[Art of Problem Solving]] hosts a large [http://www.artofproblemsolving.com/Forum/index.php?f=132 MATHCOUNTS Forum] as well as a private [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23209 MATHCOUNTS Coaches Forum].<br /> * [http://mathcounts.saab.org/ Elias Saab's MATHCOUNTS Preparation Homepage]<br /> * [http://www.unidata.ucar.edu/staff/russ/mathcounts/diaz.html The MATHCOUNTS Bible According to Mr. Diaz]<br /> *[http://www.artofproblemsolving.com/Resources/AoPS_R_A_MATHCOUNTS.php/ Building a Successful MATHCOUNTS Program] by [[Jeff Boyd]], who coached the 2005, 2007, and 2008 National Champion [[Texas MathCounts]] team.<br /> <br /> == What comes after MATHCOUNTS? ==<br /> <br /> Give the following competitions a try and take a look at the [[List of United States high school mathematics competitions]].<br /> * [[American Mathematics Competitions]]<br /> * [[American Regions Math League]]<br /> * [[Mandelbrot Competition]]<br /> * [[Mu Alpha Theta]]<br /> <br /> [[Category:Mathematics competitions]]<br /> <br /> == See also ==<br /> * [[List of national MATHCOUNTS teams]]<br /> * [[MATHCOUNTS historical results]]<br /> * [[Mathematics competition resources]]<br /> * [[Math contest books]]<br /> * [[Math books]]<br /> * [[List of United States middle school mathematics competitions]]<br /> * [[List of United States high school mathematics competitions]]<br /> * [http://www.mathcounts.org/webarticles/anmviewer.asp?a=921&amp;z=71 2006 MATHCOUNTS Countdown Video]</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=MATHCOUNTS&diff=46926 MATHCOUNTS 2012-05-11T22:36:44Z <p>Williamhu888: /* Past State Team Winners */</p> <hr /> <div>'''MATHCOUNTS''' is a large national [[mathematics competition]] and [[mathematics coaching]] program that has served millions of middle school students since 1984. Sponsored by the [[CNA Foundation]], [[National Society of Professional Engineers]], the [[National Council of Teachers of Mathematics]], and others, the focus of MATHCOUNTS is on [[mathematical problem solving]]. Students are eligible for up to three years, but cannot compete beyond their eighth grade year.<br /> <br /> == MATHCOUNTS Curriculum ==<br /> MATHCOUNTS curriculum includes [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], [[probability]], and [[statistics]]. The focus of MATHCOUNTS curriculum is in developing [[mathematical problem solving]] skills.<br /> <br /> Before 1990, MATHCOUNTS chose particular areas of mathematics to highlight each year before changing the focus of the competition more broadly to problem solving.<br /> <br /> ==Past Winners==<br /> * 1984: Michael Edwards, Texas<br /> * 1985: Timothy Kokesh, Oklahoma<br /> * 1986: Brian David Ewald, Florida<br /> * 1987: Russell Mann, Tennessee<br /> * 1988: Andrew Schultz, Illinois<br /> * 1989: Albert Kurz, Pennsylvania<br /> * 1990: Brian Jenkins, Arkansas<br /> * 1991: Jonathan L. Weinstein, Massachusetts<br /> * 1992: Andrei C. Gnepp, Ohio<br /> * 1993: Carleton Bosley, Kansas<br /> * 1994: William O. Engel, Illinois<br /> * 1995: Richard Reifsnyder, Kentucky<br /> * 1996: Alexander Schwartz, Pennsylvania<br /> * 1997: Zhihao Liu, Wisconsin<br /> * 1998: Ricky Liu, Massachusetts<br /> * 1999: Po-Ru Loh, Wisconsin<br /> * 2000: Ruozhou Jia, Illinois<br /> * 2001: Ryan Ko, New Jersey<br /> * 2002: Albert Ni, Illinois<br /> * 2003: Adam Hesterberg, Washington<br /> * 2004: Gregory Gauthier, Illinois<br /> * 2005: Neal Wu, Louisiana<br /> * 2006: Daesun Yim, New Jersey<br /> * 2007: Kevin Chen, Texas<br /> * 2008: Darryl Wu, Washington<br /> * 2009: Bobby Shen, Texas<br /> * 2010: Mark Sellke, Indiana<br /> * 2011: Scott Wu, Louisiana<br /> * 2012: Chad Qian, Indiana<br /> <br /> == Past State Team Winners ==<br /> * 1984: Virginia<br /> * 1985: Florida<br /> * 1986: California<br /> * 1987: New York<br /> * 1988: New York<br /> * 1989: North Carolina<br /> * 1990: Ohio<br /> * 1991: Alabama<br /> * 1992: California<br /> * 1993: Kansas<br /> * 1994: Pennsylvania<br /> * 1995: Massachusetts<br /> * 1996: Wisconsin<br /> * 1997: Massachusetts<br /> * 1998: Wisconsin<br /> * 1999: Massachusetts<br /> * 2000: California<br /> * 2001: Virginia<br /> * 2002: California<br /> * 2003: California<br /> * 2004: Illinois<br /> * 2005: Texas<br /> * 2006: Virginia<br /> * 2007: Texas<br /> * 2008: Texas<br /> * 2009: Texas<br /> * 2010: California<br /> * 2011: California<br /> * 2012: Massachusetts<br /> <br /> == MATHCOUNTS Competition Structure ==<br /> <br /> === Sprint Round ===<br /> <br /> 30 problems in 40 minutes. This round is generally made up questions ranging from relatively easy to relatively difficult. Some of the difficult problems are only difficult because calculators are not allowed in this round.<br /> <br /> === Target Round ===<br /> 8 problems given 2 at a time. Each set of two problems is given six minutes. Students may not go back to previous rounds (or forwards to future rounds) even if they finish before time is called. Students may use calculators.<br /> <br /> === Team Round ===<br /> <br /> 10 problems in 20 minutes for a team of 4 students. These problems typically include some of the most difficult problems of the competition. Use of a calculator is allowed (and required for some questions).<br /> <br /> === Countdown Round ===<br /> High scoring individuals compete head-to-head until a champion is crowned. People compete from off a screen taking 45 seconds or less to finish the problem. The Countdown round is run differently in various different chapter, state, and national competitions. In the national competitions, it is the round that determines the champion. Calculators are not allowed.<br /> <br /> <br /> ====Chapter and State Competitions====<br /> <br /> In the chapter and state competitions, the countdown round is not mandatory. However, if it is deemed official by the chapter or state, the following format must be used:<br /> <br /> *The 10th place written finisher competes against the 9th place written finisher. A problem is displayed, and both competitors have 45 seconds to answer the question, and the first competitor to correctly answer the question receives one point. The person who gets the most correct out of three questions (not necessarily two out of three) is the winner.<br /> <br /> *The winner of the first round goes up against the 8th place finisher.<br /> <br /> *The winner of the second round goes up against the 7th place finisher.<br /> <br /> This process is continued until the countdown round reaches the top four written competitors. Starting then, the first person to get three questions correct wins (as opposed to the best-out-of-three rule).<br /> <br /> If the countdown round is unofficial, any format may be used. Single-elimination bracket-style tournaments are common.<br /> <br /> ====National Competition====<br /> <br /> At the national competition, there are some structural changes to the countdown round. The top 12 (not the top 10) written finishers make it to the countdown round, and the format is changed from a ladder competition to a single elimination tournament where the top four written competitors get a bye. This setup makes it far more likely for a 12th place finisher to become champion, and it makes it less likely for a first place written finisher to become champion, equalizing the field. But even then, a 12th place written competitor will have less of a chance to become champion than the top 4, because the top 4 get a bye. Until the semi-finals, the scoring is best out of five advances.<br /> <br /> At the first round and the second round, the person to correctly answer the most out of 5 questions wins. However, at the semifinals, the rules slightly change&amp;mdash;the first person to correctly answer four questions wins.<br /> <br /> === Ciphering Round ===<br /> In some states, (most notably Florida) there is an optional ciphering round. Very similar to countdown (in both difficulty and layout), a team sends up a representative to go against all representatives from the other teams. A problem is shown on a screen and students work fast to answer the problem. The students give their answer and after 45 seconds the answer is shown and the answers are checked to see if they are right. The fastest correct answer gets five points, the next fastest gets 4, etc. There are 4 questions per individual and teams send up 4 people. A perfect score is then 80. Often times the questions take clever reading skills. For example, one question was &quot;How much dirt is in a 3 ft by 3 ft by 4 ft hole?&quot; The answer was 0 because there is no dirt in a hole.<br /> <br /> === Masters Round ===<br /> Top students give in-depth explanations to challenging problems. This round is optional at the state level competition and is mandatory at the national competition. At nationals the top two on the written and countdown participate.<br /> <br /> === Scoring and Ranking ===<br /> An individual's score is their total number of correct sprint round answers plus 2 times their total number of correct target round answers. This total is out of a maximum of &lt;math&gt;30 + 2(8) = 46&lt;/math&gt; points.<br /> <br /> A team's score is the average of the individual scores of its four members plus 2 points for every correct team round answer, making a team's maximum possible score 66 points. Therefore, it is possible to win with a relatively low team score and a phenomenal individual score, as the team score is only roughly 30% of the total team score.<br /> <br /> == MATHCOUNTS Competition Levels ==<br /> === School Competition ===<br /> Students vie for the chance to make their school teams. Problems at this level are generally the easiest and most basic in curriculum.<br /> <br /> === Chapter Competition ===<br /> Chapter competitions serve as a selection filter for state competitions. A few states don't need to host chapter competitions due to a small population size.<br /> <br /> === State Competition ===<br /> The top 4 students in each state form the state team for the national competition. The coach of the top school team at the state level is invited to coach the state team at the national competition. Interestingly, the coach of a state team is not necessarily the coach of any of the state's team members.<br /> <br /> === National Competition ===<br /> ==== National Competition Sites ====<br /> For many years, the National MATHCOUNTS competition was held in Washington, D.C. More recently, the competition has changed venues often.<br /> <br /> * The 2011 competition was held in Washington, D.C.<br /> * The 2009 and 2010 competitions was held in Orlando, Florida.<br /> * The 2008 competition was held in Denver, Colorado.<br /> * The 2007 competition was held in Fort Worth, Texas.<br /> * The 2006 competition was held in Arlington, Virginia.<br /> * The 2005 competition was held in Detroit, Michigan.<br /> * The 2004 competition was held in Washington, D.C.<br /> * The 2002 and 2003 competitions were held in Chicago, Illinois.<br /> <br /> ==== Rewards ====<br /> <br /> Every competitor at the national competition receives a graphing calculator that varies by year - for example, in 2006 it was a TI-84 Plus Silver Edition with the MATHCOUNTS logo on the back. In 2007, MATHCOUNTS took the logo off. In 2008 to 2011, they gave TI-&lt;math&gt;n&lt;/math&gt;spires to everyone. They also give out a laptop and an$8000 scholarship to the winner.<br /> <br /> == MATHCOUNTS Resources ==<br /> === MATHCOUNTS Books ===<br /> * [http://www.artofproblemsolving.com/Books/AoPS_B_CP_MC.php MATHCOUNTS books] at the [http://www.artofproblemsolving.com/Books/AoPS_B_About.php AoPS Bookstore]<br /> * [[Art of Problem Solving]]'s [http://www.artofproblemsolving.com/Books/AoPS_B_Rec_Middle.php Introductory subject textbooks] are ideal for students preparing for MATHCOUNTS.<br /> <br /> === MATHCOUNTS Classes ===<br /> * [[Art of Problem Solving]] hosts [http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesP.php#mc MATHCOUNTS preparation classes].<br /> * [[Art of Problem Solving]] hosts many free MATHCOUNTS [[Math Jams]]. [http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php Math Jam Schedule]. [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php Math Jam Transcript Archive].<br /> <br /> === MATHCOUNTS Online ===<br /> * [http://www.mathcounts.org MATHCOUNTS Homepage]<br /> * [[Art of Problem Solving]] hosts a large [http://www.artofproblemsolving.com/Forum/index.php?f=132 MATHCOUNTS Forum] as well as a private [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23209 MATHCOUNTS Coaches Forum].<br /> * [http://mathcounts.saab.org/ Elias Saab's MATHCOUNTS Preparation Homepage]<br /> * [http://www.unidata.ucar.edu/staff/russ/mathcounts/diaz.html The MATHCOUNTS Bible According to Mr. Diaz]<br /> *[http://www.artofproblemsolving.com/Resources/AoPS_R_A_MATHCOUNTS.php/ Building a Successful MATHCOUNTS Program] by [[Jeff Boyd]], who coached the 2005, 2007, and 2008 National Champion [[Texas MathCounts]] team.<br /> <br /> == What comes after MATHCOUNTS? ==<br /> <br /> Give the following competitions a try and take a look at the [[List of United States high school mathematics competitions]].<br /> * [[American Mathematics Competitions]]<br /> * [[American Regions Math League]]<br /> * [[Mandelbrot Competition]]<br /> * [[Mu Alpha Theta]]<br /> <br /> [[Category:Mathematics competitions]]<br /> <br /> == See also ==<br /> * [[List of national MATHCOUNTS teams]]<br /> * [[MATHCOUNTS historical results]]<br /> * [[Mathematics competition resources]]<br /> * [[Math contest books]]<br /> * [[Math books]]<br /> * [[List of United States middle school mathematics competitions]]<br /> * [[List of United States high school mathematics competitions]]<br /> * [http://www.mathcounts.org/webarticles/anmviewer.asp?a=921&amp;z=71 2006 MATHCOUNTS Countdown Video]</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=MATHCOUNTS&diff=46925 MATHCOUNTS 2012-05-11T22:36:14Z <p>Williamhu888: /* Past Winners */</p> <hr /> <div>'''MATHCOUNTS''' is a large national [[mathematics competition]] and [[mathematics coaching]] program that has served millions of middle school students since 1984. Sponsored by the [[CNA Foundation]], [[National Society of Professional Engineers]], the [[National Council of Teachers of Mathematics]], and others, the focus of MATHCOUNTS is on [[mathematical problem solving]]. Students are eligible for up to three years, but cannot compete beyond their eighth grade year.<br /> <br /> == MATHCOUNTS Curriculum ==<br /> MATHCOUNTS curriculum includes [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], [[probability]], and [[statistics]]. The focus of MATHCOUNTS curriculum is in developing [[mathematical problem solving]] skills.<br /> <br /> Before 1990, MATHCOUNTS chose particular areas of mathematics to highlight each year before changing the focus of the competition more broadly to problem solving.<br /> <br /> ==Past Winners==<br /> * 1984: Michael Edwards, Texas<br /> * 1985: Timothy Kokesh, Oklahoma<br /> * 1986: Brian David Ewald, Florida<br /> * 1987: Russell Mann, Tennessee<br /> * 1988: Andrew Schultz, Illinois<br /> * 1989: Albert Kurz, Pennsylvania<br /> * 1990: Brian Jenkins, Arkansas<br /> * 1991: Jonathan L. Weinstein, Massachusetts<br /> * 1992: Andrei C. Gnepp, Ohio<br /> * 1993: Carleton Bosley, Kansas<br /> * 1994: William O. Engel, Illinois<br /> * 1995: Richard Reifsnyder, Kentucky<br /> * 1996: Alexander Schwartz, Pennsylvania<br /> * 1997: Zhihao Liu, Wisconsin<br /> * 1998: Ricky Liu, Massachusetts<br /> * 1999: Po-Ru Loh, Wisconsin<br /> * 2000: Ruozhou Jia, Illinois<br /> * 2001: Ryan Ko, New Jersey<br /> * 2002: Albert Ni, Illinois<br /> * 2003: Adam Hesterberg, Washington<br /> * 2004: Gregory Gauthier, Illinois<br /> * 2005: Neal Wu, Louisiana<br /> * 2006: Daesun Yim, New Jersey<br /> * 2007: Kevin Chen, Texas<br /> * 2008: Darryl Wu, Washington<br /> * 2009: Bobby Shen, Texas<br /> * 2010: Mark Sellke, Indiana<br /> * 2011: Scott Wu, Louisiana<br /> * 2012: Chad Qian, Indiana<br /> <br /> == Past State Team Winners ==<br /> * 1984: Virginia<br /> * 1985: Florida<br /> * 1986: California<br /> * 1987: New York<br /> * 1988: New York<br /> * 1989: North Carolina<br /> * 1990: Ohio<br /> * 1991: Alabama<br /> * 1992: California<br /> * 1993: Kansas<br /> * 1994: Pennsylvania<br /> * 1995: Massachusetts<br /> * 1996: Wisconsin<br /> * 1997: Massachusetts<br /> * 1998: Wisconsin<br /> * 1999: Massachusetts<br /> * 2000: California<br /> * 2001: Virginia<br /> * 2002: California<br /> * 2003: California<br /> * 2004: Illinois<br /> * 2005: Texas<br /> * 2006: Virginia<br /> * 2007: Texas<br /> * 2008: Texas<br /> * 2009: Texas<br /> * 2010: California<br /> * 2011: California<br /> <br /> == MATHCOUNTS Competition Structure ==<br /> <br /> === Sprint Round ===<br /> <br /> 30 problems in 40 minutes. This round is generally made up questions ranging from relatively easy to relatively difficult. Some of the difficult problems are only difficult because calculators are not allowed in this round.<br /> <br /> === Target Round ===<br /> 8 problems given 2 at a time. Each set of two problems is given six minutes. Students may not go back to previous rounds (or forwards to future rounds) even if they finish before time is called. Students may use calculators.<br /> <br /> === Team Round ===<br /> <br /> 10 problems in 20 minutes for a team of 4 students. These problems typically include some of the most difficult problems of the competition. Use of a calculator is allowed (and required for some questions).<br /> <br /> === Countdown Round ===<br /> High scoring individuals compete head-to-head until a champion is crowned. People compete from off a screen taking 45 seconds or less to finish the problem. The Countdown round is run differently in various different chapter, state, and national competitions. In the national competitions, it is the round that determines the champion. Calculators are not allowed.<br /> <br /> <br /> ====Chapter and State Competitions====<br /> <br /> In the chapter and state competitions, the countdown round is not mandatory. However, if it is deemed official by the chapter or state, the following format must be used:<br /> <br /> *The 10th place written finisher competes against the 9th place written finisher. A problem is displayed, and both competitors have 45 seconds to answer the question, and the first competitor to correctly answer the question receives one point. The person who gets the most correct out of three questions (not necessarily two out of three) is the winner.<br /> <br /> *The winner of the first round goes up against the 8th place finisher.<br /> <br /> *The winner of the second round goes up against the 7th place finisher.<br /> <br /> This process is continued until the countdown round reaches the top four written competitors. Starting then, the first person to get three questions correct wins (as opposed to the best-out-of-three rule).<br /> <br /> If the countdown round is unofficial, any format may be used. Single-elimination bracket-style tournaments are common.<br /> <br /> ====National Competition====<br /> <br /> At the national competition, there are some structural changes to the countdown round. The top 12 (not the top 10) written finishers make it to the countdown round, and the format is changed from a ladder competition to a single elimination tournament where the top four written competitors get a bye. This setup makes it far more likely for a 12th place finisher to become champion, and it makes it less likely for a first place written finisher to become champion, equalizing the field. But even then, a 12th place written competitor will have less of a chance to become champion than the top 4, because the top 4 get a bye. Until the semi-finals, the scoring is best out of five advances.<br /> <br /> At the first round and the second round, the person to correctly answer the most out of 5 questions wins. However, at the semifinals, the rules slightly change&amp;mdash;the first person to correctly answer four questions wins.<br /> <br /> === Ciphering Round ===<br /> In some states, (most notably Florida) there is an optional ciphering round. Very similar to countdown (in both difficulty and layout), a team sends up a representative to go against all representatives from the other teams. A problem is shown on a screen and students work fast to answer the problem. The students give their answer and after 45 seconds the answer is shown and the answers are checked to see if they are right. The fastest correct answer gets five points, the next fastest gets 4, etc. There are 4 questions per individual and teams send up 4 people. A perfect score is then 80. Often times the questions take clever reading skills. For example, one question was &quot;How much dirt is in a 3 ft by 3 ft by 4 ft hole?&quot; The answer was 0 because there is no dirt in a hole.<br /> <br /> === Masters Round ===<br /> Top students give in-depth explanations to challenging problems. This round is optional at the state level competition and is mandatory at the national competition. At nationals the top two on the written and countdown participate.<br /> <br /> === Scoring and Ranking ===<br /> An individual's score is their total number of correct sprint round answers plus 2 times their total number of correct target round answers. This total is out of a maximum of &lt;math&gt;30 + 2(8) = 46&lt;/math&gt; points.<br /> <br /> A team's score is the average of the individual scores of its four members plus 2 points for every correct team round answer, making a team's maximum possible score 66 points. Therefore, it is possible to win with a relatively low team score and a phenomenal individual score, as the team score is only roughly 30% of the total team score.<br /> <br /> == MATHCOUNTS Competition Levels ==<br /> === School Competition ===<br /> Students vie for the chance to make their school teams. Problems at this level are generally the easiest and most basic in curriculum.<br /> <br /> === Chapter Competition ===<br /> Chapter competitions serve as a selection filter for state competitions. A few states don't need to host chapter competitions due to a small population size.<br /> <br /> === State Competition ===<br /> The top 4 students in each state form the state team for the national competition. The coach of the top school team at the state level is invited to coach the state team at the national competition. Interestingly, the coach of a state team is not necessarily the coach of any of the state's team members.<br /> <br /> === National Competition ===<br /> ==== National Competition Sites ====<br /> For many years, the National MATHCOUNTS competition was held in Washington, D.C. More recently, the competition has changed venues often.<br /> <br /> * The 2011 competition was held in Washington, D.C.<br /> * The 2009 and 2010 competitions was held in Orlando, Florida.<br /> * The 2008 competition was held in Denver, Colorado.<br /> * The 2007 competition was held in Fort Worth, Texas.<br /> * The 2006 competition was held in Arlington, Virginia.<br /> * The 2005 competition was held in Detroit, Michigan.<br /> * The 2004 competition was held in Washington, D.C.<br /> * The 2002 and 2003 competitions were held in Chicago, Illinois.<br /> <br /> ==== Rewards ====<br /> <br /> Every competitor at the national competition receives a graphing calculator that varies by year - for example, in 2006 it was a TI-84 Plus Silver Edition with the MATHCOUNTS logo on the back. In 2007, MATHCOUNTS took the logo off. In 2008 to 2011, they gave TI-&lt;math&gt;n&lt;/math&gt;spires to everyone. They also give out a laptop and an \$8000 scholarship to the winner.<br /> <br /> == MATHCOUNTS Resources ==<br /> === MATHCOUNTS Books ===<br /> * [http://www.artofproblemsolving.com/Books/AoPS_B_CP_MC.php MATHCOUNTS books] at the [http://www.artofproblemsolving.com/Books/AoPS_B_About.php AoPS Bookstore]<br /> * [[Art of Problem Solving]]'s [http://www.artofproblemsolving.com/Books/AoPS_B_Rec_Middle.php Introductory subject textbooks] are ideal for students preparing for MATHCOUNTS.<br /> <br /> === MATHCOUNTS Classes ===<br /> * [[Art of Problem Solving]] hosts [http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesP.php#mc MATHCOUNTS preparation classes].<br /> * [[Art of Problem Solving]] hosts many free MATHCOUNTS [[Math Jams]]. [http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php Math Jam Schedule]. [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php Math Jam Transcript Archive].<br /> <br /> === MATHCOUNTS Online ===<br /> * [http://www.mathcounts.org MATHCOUNTS Homepage]<br /> * [[Art of Problem Solving]] hosts a large [http://www.artofproblemsolving.com/Forum/index.php?f=132 MATHCOUNTS Forum] as well as a private [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23209 MATHCOUNTS Coaches Forum].<br /> * [http://mathcounts.saab.org/ Elias Saab's MATHCOUNTS Preparation Homepage]<br /> * [http://www.unidata.ucar.edu/staff/russ/mathcounts/diaz.html The MATHCOUNTS Bible According to Mr. Diaz]<br /> *[http://www.artofproblemsolving.com/Resources/AoPS_R_A_MATHCOUNTS.php/ Building a Successful MATHCOUNTS Program] by [[Jeff Boyd]], who coached the 2005, 2007, and 2008 National Champion [[Texas MathCounts]] team.<br /> <br /> == What comes after MATHCOUNTS? ==<br /> <br /> Give the following competitions a try and take a look at the [[List of United States high school mathematics competitions]].<br /> * [[American Mathematics Competitions]]<br /> * [[American Regions Math League]]<br /> * [[Mandelbrot Competition]]<br /> * [[Mu Alpha Theta]]<br /> <br /> [[Category:Mathematics competitions]]<br /> <br /> == See also ==<br /> * [[List of national MATHCOUNTS teams]]<br /> * [[MATHCOUNTS historical results]]<br /> * [[Mathematics competition resources]]<br /> * [[Math contest books]]<br /> * [[Math books]]<br /> * [[List of United States middle school mathematics competitions]]<br /> * [[List of United States high school mathematics competitions]]<br /> * [http://www.mathcounts.org/webarticles/anmviewer.asp?a=921&amp;z=71 2006 MATHCOUNTS Countdown Video]</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_9&diff=46049 2012 AIME II Problems/Problem 9 2012-04-04T00:26:45Z <p>Williamhu888: /* Solution */</p> <hr /> <div>== Problem 9 ==<br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be real numbers such that &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; and &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt;. The value of &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac pq&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> <br /> == Solution ==<br /> Examine the first term in the expression we want to evaluate, &lt;math&gt;\frac{\sin 2x}{\sin 2y}&lt;/math&gt;, separately from the second term, &lt;math&gt;\frac{\cos 2x}{\cos 2y}&lt;/math&gt;.<br /> <br /> == The First Term ==<br /> <br /> Using the identity &lt;math&gt;\sin 2\theta = 2\sin\theta\cos\theta&lt;/math&gt;, we have:<br /> <br /> &lt;math&gt;\frac{2\sin x \cos x}{2\sin y \cos y} = \frac{\sin x \cos x}{\sin y \cos y} = \frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}=3\cdot\frac{1}{2} = \frac{3}{2}&lt;/math&gt;<br /> <br /> == The Second Term ==<br /> <br /> Let the equation &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; be equation 1, and let the equation &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt; be equation 2.<br /> Hungry for the widely-used identity &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt;, we cross multiply equation 1 by &lt;math&gt;\sin y&lt;/math&gt; and multiply equation 2 by &lt;math&gt;\cos y&lt;/math&gt;.<br /> <br /> Equation 1 then becomes:<br /> <br /> &lt;math&gt;\sin x = 3\sin y&lt;/math&gt;.<br /> <br /> Equation 2 then becomes:<br /> <br /> &lt;math&gt;\cos x = \frac{1}{2} \cos y&lt;/math&gt;<br /> <br /> Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:<br /> <br /> &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} \cos^2 y&lt;/math&gt;<br /> <br /> Applying the identity &lt;math&gt;\cos^2 y = 1 - \sin^2 y&lt;/math&gt; (which is similar to &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt; but a bit different), we can change &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} \cos^2 y&lt;/math&gt; into:<br /> <br /> &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y&lt;/math&gt;<br /> <br /> Rearranging, we get &lt;math&gt;\frac{3}{4} = \frac{35}{4} \sin^2 y &lt;/math&gt;.<br /> <br /> So, &lt;math&gt;\sin^2 y = \frac{3}{35}&lt;/math&gt;.<br /> <br /> Squaring Equation 1 (leading to &lt;math&gt;\sin^2 x = 9\sin^2 y&lt;/math&gt;), we can solve for &lt;math&gt;\sin^2 y&lt;/math&gt;:<br /> <br /> &lt;math&gt;\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}&lt;/math&gt;<br /> <br /> Using the identity &lt;math&gt;\cos 2\theta = 1 - 2\sin^2\theta&lt;/math&gt;, we can solve for &lt;math&gt;\frac{\cos 2x}{\cos 2y}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos 2y = 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}&lt;/math&gt;.<br /> <br /> == Now Back to the Solution! ==<br /> <br /> Finally, &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}&lt;/math&gt;.<br /> <br /> So, the answer is &lt;math&gt;49+58=\boxed{107}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=8|num-a=10}}</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_15&diff=45813 2012 AIME II Problems/Problem 15 2012-03-31T20:13:43Z <p>Williamhu888: Created page with &quot;== Problem 15 == Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=7&lt;/math&gt;, and &lt;math&gt;AC=3&lt;/math&gt;. The bisector of angle &lt;mat...&quot;</p> <hr /> <div>== Problem 15 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=7&lt;/math&gt;, and &lt;math&gt;AC=3&lt;/math&gt;. The bisector of angle &lt;math&gt;A&lt;/math&gt; meets side &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and circle &lt;math&gt;\omega&lt;/math&gt; at a second point &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;\gamma&lt;/math&gt; be the circle with diameter &lt;math&gt;\overline{DE}&lt;/math&gt;. Circles &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; meet at &lt;math&gt;E&lt;/math&gt; and a second point &lt;math&gt;F&lt;/math&gt;. Then &lt;math&gt;AF^2 = \frac mn&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_14&diff=45812 2012 AIME II Problems/Problem 14 2012-03-31T20:12:52Z <p>Williamhu888: Created page with &quot;== Problem 14 == In a group of nine people each person shakes hands with exactly two of the other people from the group. Let &lt;math&gt;N&lt;/math&gt; be the number of ways this handshaking...&quot;</p> <hr /> <div>== Problem 14 ==<br /> In a group of nine people each person shakes hands with exactly two of the other people from the group. Let &lt;math&gt;N&lt;/math&gt; be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_13&diff=45811 2012 AIME II Problems/Problem 13 2012-03-31T20:11:47Z <p>Williamhu888: Created page with &quot;== Problem 13 == Equilateral &lt;math&gt;\triangle ABC&lt;/math&gt; has side length &lt;math&gt;\sqrt{111}&lt;/math&gt;. There are four distinct triangles &lt;math&gt;AD_1E_1&lt;/math&gt;, &lt;math&gt;AD_1E_2&lt;/math&gt;, &lt;ma...&quot;</p> <hr /> <div>== Problem 13 ==<br /> Equilateral &lt;math&gt;\triangle ABC&lt;/math&gt; has side length &lt;math&gt;\sqrt{111}&lt;/math&gt;. There are four distinct triangles &lt;math&gt;AD_1E_1&lt;/math&gt;, &lt;math&gt;AD_1E_2&lt;/math&gt;, &lt;math&gt;AD_2E_3&lt;/math&gt;, and &lt;math&gt;AD_2E_4&lt;/math&gt;, each congruent to &lt;math&gt;\triangle ABC&lt;/math&gt;,<br /> with &lt;math&gt;BD_1 = BD_2 = \sqrt{11}&lt;/math&gt;. Find &lt;math&gt;\sum_{k=1}^4(CE_k)^2&lt;/math&gt;.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_12&diff=45810 2012 AIME II Problems/Problem 12 2012-03-31T20:11:10Z <p>Williamhu888: Created page with &quot;== Problem 12 == For a positive integer &lt;math&gt;p&lt;/math&gt;, define the positive integer &lt;math&gt;n&lt;/math&gt; to be &lt;math&gt;p&lt;/math&gt;''-safe'' if &lt;math&gt;n&lt;/math&gt; differs in absolute value by mo...&quot;</p> <hr /> <div>== Problem 12 ==<br /> For a positive integer &lt;math&gt;p&lt;/math&gt;, define the positive integer &lt;math&gt;n&lt;/math&gt; to be &lt;math&gt;p&lt;/math&gt;''-safe'' if &lt;math&gt;n&lt;/math&gt; differs in absolute value by more than &lt;math&gt;2&lt;/math&gt; from all multiples of &lt;math&gt;p&lt;/math&gt;. For example, the set of &lt;math&gt;10&lt;/math&gt;-safe numbers is &lt;math&gt;\{ 3, 4, 5, 6, 7, 13, 14, 15, 16, 17, 23, \ldots\}&lt;/math&gt;. Find the number of positive integers less than or equal to &lt;math&gt;10,000&lt;/math&gt; which are simultaneously &lt;math&gt;7&lt;/math&gt;-safe, &lt;math&gt;11&lt;/math&gt;-safe, and &lt;math&gt;13&lt;/math&gt;-safe.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_11&diff=45809 2012 AIME II Problems/Problem 11 2012-03-31T20:10:22Z <p>Williamhu888: Created page with &quot;== Problem 11 == Let &lt;math&gt;f_1(x) = \frac23 - \frac3{3x+1}&lt;/math&gt;, and for &lt;math&gt;n \ge 2&lt;/math&gt;, define &lt;math&gt;f_n(x) = f_1(f_{n-1}(x))&lt;/math&gt;. The value of &lt;math&gt;x&lt;/math&gt; that sa...&quot;</p> <hr /> <div>== Problem 11 ==<br /> Let &lt;math&gt;f_1(x) = \frac23 - \frac3{3x+1}&lt;/math&gt;, and for &lt;math&gt;n \ge 2&lt;/math&gt;, define &lt;math&gt;f_n(x) = f_1(f_{n-1}(x))&lt;/math&gt;. The value of &lt;math&gt;x&lt;/math&gt; that satisfies &lt;math&gt;f_{1001}(x) = x-3&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac mn&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_10&diff=45808 2012 AIME II Problems/Problem 10 2012-03-31T20:09:47Z <p>Williamhu888: Created page with &quot;== Problem 10 == Find the number of positive integers &lt;math&gt;n&lt;/math&gt; less than &lt;math&gt;1000&lt;/math&gt; for which there exists a positive real number &lt;math&gt;x&lt;/math&gt; such that &lt;math&gt;n=x\...&quot;</p> <hr /> <div>== Problem 10 ==<br /> Find the number of positive integers &lt;math&gt;n&lt;/math&gt; less than &lt;math&gt;1000&lt;/math&gt; for which there exists a positive real number &lt;math&gt;x&lt;/math&gt; such that &lt;math&gt;n=x\lfloor x \rfloor&lt;/math&gt;.<br /> <br /> Note: &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_9&diff=45807 2012 AIME II Problems/Problem 9 2012-03-31T20:09:13Z <p>Williamhu888: Created page with &quot;== Problem 9 == Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be real numbers such that &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; and &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt;. The value...&quot;</p> <hr /> <div>== Problem 9 ==<br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be real numbers such that &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; and &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt;. The value of &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac pq&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p+q&lt;/math&gt;.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_8&diff=45806 2012 AIME II Problems/Problem 8 2012-03-31T20:08:38Z <p>Williamhu888: Created page with &quot;== Problem 8 == The complex numbers &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt; satisfy the system &lt;cmath&gt; z + \frac{20i}w = 5+i \\ \\ w+\frac{12i}z = -4+10i &lt;/cmath&gt; Find the smallest pos...&quot;</p> <hr /> <div>== Problem 8 ==<br /> The complex numbers &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt; satisfy the system &lt;cmath&gt; z + \frac{20i}w = 5+i \\ \\<br /> w+\frac{12i}z = -4+10i &lt;/cmath&gt; Find the smallest possible value of &lt;math&gt;\vert zw\vert^2&lt;/math&gt;.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_7&diff=45805 2012 AIME II Problems/Problem 7 2012-03-31T20:07:39Z <p>Williamhu888: Created page with &quot;== Problem 7 == Let &lt;math&gt;S&lt;/math&gt; be the increasing sequence of positive integers whose binary representation has exactly &lt;math&gt;8&lt;/math&gt; ones. Let &lt;math&gt;N&lt;/math&gt; be the 1000th n...&quot;</p> <hr /> <div>== Problem 7 ==<br /> Let &lt;math&gt;S&lt;/math&gt; be the increasing sequence of positive integers whose binary representation has exactly &lt;math&gt;8&lt;/math&gt; ones. Let &lt;math&gt;N&lt;/math&gt; be the 1000th number in &lt;math&gt;S&lt;/math&gt;. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_6&diff=45804 2012 AIME II Problems/Problem 6 2012-03-31T20:07:01Z <p>Williamhu888: </p> <hr /> <div>== Problem 6 ==<br /> Let &lt;math&gt;z=a+bi&lt;/math&gt; be the complex number with &lt;math&gt;\vert z \vert = 5&lt;/math&gt; and &lt;math&gt;b &gt; 0&lt;/math&gt; such that the distance between &lt;math&gt;(1+2i)z^3&lt;/math&gt; and &lt;math&gt;z^5&lt;/math&gt; is maximized, and let &lt;math&gt;z^4 = c+di&lt;/math&gt;. Find &lt;math&gt;c+d&lt;/math&gt;.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_5&diff=45803 2012 AIME II Problems/Problem 5 2012-03-31T20:05:29Z <p>Williamhu888: Created page with &quot;== Problem 5 == In the accompanying figure, the outer square &lt;math&gt;S&lt;/math&gt; has side length &lt;math&gt;40&lt;/math&gt;. A second square &lt;math&gt;S'&lt;/math&gt; of side length &lt;math&gt;15&lt;/math&gt; is con...&quot;</p> <hr /> <div>== Problem 5 ==<br /> In the accompanying figure, the outer square &lt;math&gt;S&lt;/math&gt; has side length &lt;math&gt;40&lt;/math&gt;. A second square &lt;math&gt;S'&lt;/math&gt; of side length &lt;math&gt;15&lt;/math&gt; is constructed inside &lt;math&gt;S&lt;/math&gt; with the same center as &lt;math&gt;S&lt;/math&gt; and with sides parallel to those of &lt;math&gt;S&lt;/math&gt;. From each midpoint of a side of &lt;math&gt;S&lt;/math&gt;, segments are drawn to the two closest vertices of &lt;math&gt;S'&lt;/math&gt;. The result is a four-pointed starlike figure inscribed in &lt;math&gt;S&lt;/math&gt;. The star figure is cut out and then folded to form a pyramid with base &lt;math&gt;S'&lt;/math&gt;. Find the volume of this pyramid.<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> <br /> pair S1 = (20, 20), S2 = (-20, 20), S3 = (-20, -20), S4 = (20, -20);<br /> pair M1 = (S1+S2)/2, M2 = (S2+S3)/2, M3=(S3+S4)/2, M4=(S4+S1)/2;<br /> pair Sp1 = (7.5, 7.5), Sp2=(-7.5, 7.5), Sp3 = (-7.5, -7.5), Sp4 = (7.5, -7.5);<br /> <br /> draw(S1--S2--S3--S4--cycle);<br /> draw(Sp1--Sp2--Sp3--Sp4--cycle);<br /> draw(Sp1--M1--Sp2--M2--Sp3--M3--Sp4--M4--cycle);<br /> &lt;/asy&gt;&lt;/center&gt;</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_4&diff=45802 2012 AIME II Problems/Problem 4 2012-03-31T20:04:31Z <p>Williamhu888: Created page with &quot;== Problem 4 == Ana, Bob, and Cao bike at constant rates of &lt;math&gt;8.6&lt;/math&gt; meters per second, &lt;math&gt;6.2&lt;/math&gt; meters per second, and &lt;math&gt;5&lt;/math&gt; meters per second, respecti...&quot;</p> <hr /> <div>== Problem 4 ==<br /> Ana, Bob, and Cao bike at constant rates of &lt;math&gt;8.6&lt;/math&gt; meters per second, &lt;math&gt;6.2&lt;/math&gt; meters per second, and &lt;math&gt;5&lt;/math&gt; meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point &lt;math&gt;D&lt;/math&gt; on the south edge of the field. Cao arrives at point &lt;math&gt;D&lt;/math&gt; at the same time that Ana and Bob arrive at &lt;math&gt;D&lt;/math&gt; for the first time. The ratio of the field's length to the field's width to the distance from point &lt;math&gt;D&lt;/math&gt; to the southeast corner of the field can be represented as &lt;math&gt;p : q : r&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; are positive integers with &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; relatively prime. Find &lt;math&gt;p+q+r&lt;/math&gt;.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_3&diff=45801 2012 AIME II Problems/Problem 3 2012-03-31T20:03:00Z <p>Williamhu888: Created page with &quot;== Problem 3 == At a certain university, the division of mathematical sciences consists of the departments of mathematics, statistics, and computer science. There are two male an...&quot;</p> <hr /> <div>== Problem 3 ==<br /> At a certain university, the division of mathematical sciences consists of the departments of mathematics, statistics, and computer science. There are two male and two female professors in each department. A committee of six professors is to contain three men and three women and must also contain two professors from each of the three departments. Find the number of possible committees that can be formed subject to these requirements.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_2&diff=45800 2012 AIME II Problems/Problem 2 2012-03-31T20:02:04Z <p>Williamhu888: Created page with &quot;== Problem 2 == Two geometric sequences &lt;math&gt;a_1, a_2, a_3, \ldots&lt;/math&gt; and &lt;math&gt;b_1, b_2, b_3, \ldots&lt;/math&gt; have the same common ratio, with &lt;math&gt;a_1 = 27&lt;/math&gt;, &lt;math&gt;b_...&quot;</p> <hr /> <div>== Problem 2 ==<br /> Two geometric sequences &lt;math&gt;a_1, a_2, a_3, \ldots&lt;/math&gt; and &lt;math&gt;b_1, b_2, b_3, \ldots&lt;/math&gt; have the same common ratio, with &lt;math&gt;a_1 = 27&lt;/math&gt;, &lt;math&gt;b_1=99&lt;/math&gt;, and &lt;math&gt;a_{15}=b_{11}&lt;/math&gt;. Find &lt;math&gt;a_9&lt;/math&gt;.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_1&diff=45799 2012 AIME II Problems/Problem 1 2012-03-31T20:00:31Z <p>Williamhu888: Created page with &quot;== Problem 1 == Find the number of ordered pairs of positive integer solutions &lt;math&gt;(m, n)&lt;/math&gt; to the equation &lt;math&gt;20m + 12n = 2012&lt;/math&gt;.&quot;</p> <hr /> <div>== Problem 1 ==<br /> Find the number of ordered pairs of positive integer solutions &lt;math&gt;(m, n)&lt;/math&gt; to the equation &lt;math&gt;20m + 12n = 2012&lt;/math&gt;.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Answer_Key&diff=45790 2012 AIME II Answer Key 2012-03-30T22:46:41Z <p>Williamhu888: </p> <hr /> <div>1. 034<br /> <br /> 2. 363<br /> <br /> 3. 088<br /> <br /> 4. 061<br /> <br /> 5. 750<br /> <br /> 6. 125<br /> <br /> 7. 032<br /> <br /> 8. 040<br /> <br /> 9. 107<br /> <br /> 10. 496<br /> <br /> 11. 008<br /> <br /> 12. 958<br /> <br /> 13. 677<br /> <br /> 14. 016<br /> <br /> 15. 919</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_11&diff=45755 2009 AIME I Problems/Problem 11 2012-03-26T00:29:49Z <p>Williamhu888: /* Solution 2 */</p> <hr /> <div><br /> == Problem ==<br /> Consider the set of all triangles &lt;math&gt;OPQ&lt;/math&gt; where &lt;math&gt;O&lt;/math&gt; is the origin and &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; are distinct points in the plane with nonnegative integer coordinates &lt;math&gt;(x,y)&lt;/math&gt; such that &lt;math&gt;41x + y = 2009&lt;/math&gt;. Find the number of such distinct triangles whose area is a positive integer.<br /> <br /> == Solution 1 ==<br /> Let the two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be defined with coordinates; &lt;math&gt;P=(x_1,y_1)&lt;/math&gt; and &lt;math&gt;Q=(x_2,y_2)&lt;/math&gt;<br /> <br /> We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem). <br /> <br /> &lt;math&gt;\det \left({\matrix {P \above Q}}\right)=\det \left({\matrix {x_1 \above x_2}\matrix {y_1 \above y_2}\right).&lt;/math&gt;<br /> <br /> Since the triangle has half the area of the parallelogram, we just need the determinant to be even.<br /> <br /> The determinant is<br /> <br /> &lt;cmath&gt;(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;2009&lt;/math&gt; is not even, &lt;math&gt;((x_1)-(x_2))&lt;/math&gt; must be even, thus the two &lt;math&gt;x&lt;/math&gt;'s must be of the same parity. Also note that the maximum value for &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;49&lt;/math&gt; and the minimum is &lt;math&gt;0&lt;/math&gt;. There are &lt;math&gt;25&lt;/math&gt; even and &lt;math&gt;25&lt;/math&gt; odd numbers available for use as coordinates and thus there are &lt;math&gt;(_{25}C_2)+(_{25}C_2)=\boxed{600}&lt;/math&gt; such triangles.<br /> <br /> == Solution 2 ==<br /> As in the solution above, let the two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be defined with coordinates; &lt;math&gt;P=(x_1,y_1)&lt;/math&gt; and &lt;math&gt;Q=(x_2,y_2)&lt;/math&gt;.<br /> <br /> <br /> If the coordinates of &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; have nonnegative integer coordinates, &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; must be lattice points either<br /> <br /> *on the nonnegative x-axis<br /> <br /> *on the nonnegative y-axis<br /> <br /> *in the first quadrant<br /> <br /> We can calculate the y-intercept of the line &lt;math&gt;41x+y=2009&lt;/math&gt; to be &lt;math&gt;(0,2009)&lt;/math&gt; and the x-intercept to be &lt;math&gt;(49,0)&lt;/math&gt;.<br /> <br /> Using the point-to-line distance formula, we can calculate the height of &lt;math&gt;\triangle OPQ&lt;/math&gt; from vertex &lt;math&gt;O&lt;/math&gt; (the origin) to be:<br /> <br /> &lt;math&gt;\dfrac{|41(0) + 1(0) - 2009|}{\sqrt{41^2 + 1^2}} = \dfrac{2009}{\sqrt{1682}} = \dfrac{2009}{29\sqrt2}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;b&lt;/math&gt; be the base of the triangle that is part of the line &lt;math&gt;41x+y=2009&lt;/math&gt;.<br /> <br /> The area is calculated as: <br /> &lt;math&gt;\dfrac{1}{2}\times b \times \dfrac{2009}{29\sqrt2} = \dfrac{2009}{58\sqrt2}\times b&lt;/math&gt;<br /> <br /> Let the numerical area of the triangle be &lt;math&gt;k&lt;/math&gt;. <br /> <br /> So, &lt;math&gt;k = \dfrac{2009}{58\sqrt2}\times b&lt;/math&gt;<br /> <br /> We know that &lt;math&gt;k&lt;/math&gt; is an integer. So, &lt;math&gt;b = 58\sqrt2 \times z&lt;/math&gt;, where &lt;math&gt;z&lt;/math&gt; is also an integer.<br /> <br /> We defined the points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; as &lt;math&gt;P=(x_1,y_1)&lt;/math&gt; and &lt;math&gt;Q=(x_2,y_2)&lt;/math&gt;.<br /> <br /> Changing the y-coordinates to be in terms of x, we get:<br /> <br /> &lt;math&gt;P=(x_1,2009-41x_1)&lt;/math&gt; and &lt;math&gt;Q=(x_2,2009-41x_2)&lt;/math&gt;.<br /> <br /> The distance between them equals &lt;math&gt;b&lt;/math&gt;. <br /> <br /> Using the distance formula, we get<br /> <br /> &lt;math&gt;PQ = b = \sqrt{(-41x_2+ 41x_1)^2 + (x_2 - x_1)^2} = 29\sqrt2 \times |x_2 - x_1| = 58\sqrt2\times z&lt;/math&gt; &lt;math&gt;(*)&lt;/math&gt;<br /> <br /> WLOG, we can assume that &lt;math&gt;x_2 &gt; x_1&lt;/math&gt;.<br /> <br /> Taking the last two equalities from the &lt;math&gt;(*)&lt;/math&gt; string of equalities and putting in our assumption that &lt;math&gt;x_2&gt;x_1&lt;/math&gt;, we get<br /> <br /> &lt;math&gt;29\sqrt2\times (x_2-x_1) = 58\sqrt2\times z&lt;/math&gt;.<br /> <br /> Dividing both sides by &lt;math&gt;29\sqrt2&lt;/math&gt;, we get<br /> <br /> &lt;math&gt;x_2-x_1 = 2z&lt;/math&gt;<br /> <br /> As we mentioned, &lt;math&gt;z&lt;/math&gt; is an integer, so &lt;math&gt;x_2-x_1&lt;/math&gt; is an even integer. Also, &lt;math&gt;x_2&lt;/math&gt; and &lt;math&gt;x_1&lt;/math&gt; are both positive integers. So, &lt;math&gt;x_2&lt;/math&gt; and &lt;math&gt;x_1&lt;/math&gt; are between 0 and 49, inclusive. Remember, &lt;math&gt;x_2&gt;x_1&lt;/math&gt; as well.<br /> <br /> *There are 48 ordered pairs &lt;math&gt;(x_2,x_1)&lt;/math&gt; such that their positive difference is 2.<br /> <br /> *There are 46 ordered pairs &lt;math&gt;(x_2,x_1)&lt;/math&gt; such that their positive difference is 4.<br /> <br /> ...<br /> <br /> *Finally, there are 2 ordered pairs &lt;math&gt;(x_2,x_1)&lt;/math&gt; such that their positive difference is 48.<br /> <br /> Summing them up, we get that there are &lt;math&gt;2+4+\dots + 48 = \boxed{600}&lt;/math&gt; triangles.<br /> <br /> == See also ==<br /> {{AIME box|year=2009|n=I|num-b=10|num-a=12}}</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_11&diff=45750 2009 AIME I Problems/Problem 11 2012-03-25T20:23:45Z <p>Williamhu888: /* Solution 2 */</p> <hr /> <div><br /> == Problem ==<br /> Consider the set of all triangles &lt;math&gt;OPQ&lt;/math&gt; where &lt;math&gt;O&lt;/math&gt; is the origin and &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; are distinct points in the plane with nonnegative integer coordinates &lt;math&gt;(x,y)&lt;/math&gt; such that &lt;math&gt;41x + y = 2009&lt;/math&gt;. Find the number of such distinct triangles whose area is a positive integer.<br /> <br /> == Solution 1 ==<br /> Let the two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be defined with coordinates; &lt;math&gt;P=(x_1,y_1)&lt;/math&gt; and &lt;math&gt;Q=(x_2,y_2)&lt;/math&gt;<br /> <br /> We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem). <br /> <br /> &lt;math&gt;\det \left({\matrix {P \above Q}}\right)=\det \left({\matrix {x_1 \above x_2}\matrix {y_1 \above y_2}\right).&lt;/math&gt;<br /> <br /> Since the triangle has half the area of the parallelogram, we just need the determinant to be even.<br /> <br /> The determinant is<br /> <br /> &lt;cmath&gt;(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;2009&lt;/math&gt; is not even, &lt;math&gt;((x_1)-(x_2))&lt;/math&gt; must be even, thus the two &lt;math&gt;x&lt;/math&gt;'s must be of the same parity. Also note that the maximum value for &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;49&lt;/math&gt; and the minimum is &lt;math&gt;0&lt;/math&gt;. There are &lt;math&gt;25&lt;/math&gt; even and &lt;math&gt;25&lt;/math&gt; odd numbers available for use as coordinates and thus there are &lt;math&gt;(_{25}C_2)+(_{25}C_2)=\boxed{600}&lt;/math&gt; such triangles.<br /> <br /> == Solution 2 ==<br /> As in the solution above, let the two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be defined with coordinates; &lt;math&gt;P=(x_1,y_1)&lt;/math&gt; and &lt;math&gt;Q=(x_2,y_2)&lt;/math&gt;.<br /> <br /> <br /> If the coordinates of &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; have nonnegative integer coordinates, &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; must be lattice points either<br /> <br /> *on the nonnegative x-axis<br /> <br /> *on the nonnegative y-axis<br /> <br /> *in the first quadrant<br /> <br /> We can calculate the y-intercept of the line &lt;math&gt;41x+y=2009&lt;/math&gt; to be &lt;math&gt;(0,2009)&lt;/math&gt; and the x-intercept to be &lt;math&gt;(49,0)&lt;/math&gt;.<br /> <br /> Using the point-to-line distance formula, we can calculate the height of &lt;math&gt;\triangle OPQ&lt;/math&gt; from vertex &lt;math&gt;O&lt;/math&gt; (the origin) to be:<br /> <br /> &lt;math&gt;\dfrac{|41(0) + 1(0) - 2009|}{\sqrt{41^2 + 1^2}} = \dfrac{2009}{\sqrt{1682}} = \dfrac{2009}{29\sqrt2}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;b&lt;/math&gt; be the base of the triangle that is part of the line &lt;math&gt;41x+y=2009&lt;/math&gt;.<br /> <br /> The area is calculated as: <br /> &lt;math&gt;\dfrac{1}{2}\times b \times \dfrac{2009}{29\sqrt2} = \dfrac{2009}{58\sqrt2}\times b&lt;/math&gt;<br /> <br /> Let the numerical area of the triangle be &lt;math&gt;k&lt;/math&gt;. <br /> <br /> So, &lt;math&gt;k = \dfrac{2009}{58\sqrt2}\times b&lt;/math&gt;<br /> <br /> We know that &lt;math&gt;k&lt;/math&gt; is an integer. So, &lt;math&gt;b = 58\sqrt2 \times z&lt;/math&gt;, where &lt;math&gt;z&lt;/math&gt; is also an integer.<br /> <br /> We defined the points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; as &lt;math&gt;P=(x_1,y_1)&lt;/math&gt; and &lt;math&gt;Q=(x_2,y_2)&lt;/math&gt;.<br /> <br /> Changing the y-coordinates to be in terms of x, we get:<br /> <br /> &lt;math&gt;P=(x_1,2009-41x_1)&lt;/math&gt; and &lt;math&gt;Q=(x_2,2009-41x_2)&lt;/math&gt;.<br /> <br /> The distance between them equals &lt;math&gt;b&lt;/math&gt;. <br /> <br /> Using the distance formula, we get<br /> <br /> &lt;math&gt;PQ = b = \sqrt{(-41x_2+ 41x_1)^2 + (x_2 - x_1)^2} = 29\sqrt2 \times |x_2 - x_1| = 58\sqrt2\times z&lt;/math&gt; (*)<br /> <br /> WLOG, we can assume that &lt;math&gt;x_2 &gt; x_1&lt;/math&gt;.<br /> <br /> Taking the last two equalities from the &lt;math&gt;(*)&lt;/math&gt; string of equalities and putting in our assumption that &lt;math&gt;x_2&gt;x_1&lt;/math&gt;, we get<br /> <br /> &lt;math&gt;29\sqrt2\times (x_2-x_1) = 58\sqrt2\times z&lt;/math&gt;.<br /> <br /> Dividing both sides by &lt;math&gt;29\sqrt2&lt;/math&gt;, we get<br /> <br /> &lt;math&gt;x_2-x_1 = 2z&lt;/math&gt;<br /> <br /> As we mentioned, &lt;math&gt;z&lt;/math&gt; is an integer, so &lt;math&gt;x_2-x_1&lt;/math&gt; is an even integer. Also, &lt;math&gt;x_2&lt;/math&gt; and &lt;math&gt;x_1&lt;/math&gt; are both positive integers. So, &lt;math&gt;x_2&lt;/math&gt; and &lt;math&gt;x_1&lt;/math&gt; are between 0 and 49, inclusive. Remember, &lt;math&gt;x_2&gt;x_1&lt;/math&gt; as well.<br /> <br /> *There are 48 ordered pairs &lt;math&gt;(x_2,x_1)&lt;/math&gt; such that their positive difference is 2.<br /> <br /> *There are 46 ordered pairs &lt;math&gt;(x_2,x_1)&lt;/math&gt; such that their positive difference is 4.<br /> <br /> ...<br /> <br /> *Finally, there are 2 ordered pairs &lt;math&gt;(x_2,x_1)&lt;/math&gt; such that their positive difference is 48.<br /> <br /> Summing them up, we get that there are &lt;math&gt;2+4+\dots + 48 = \boxed{600}&lt;/math&gt; triangles.<br /> <br /> == See also ==<br /> {{AIME box|year=2009|n=I|num-b=10|num-a=12}}</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_11&diff=45749 2009 AIME I Problems/Problem 11 2012-03-25T20:07:20Z <p>Williamhu888: /* Solution */</p> <hr /> <div><br /> == Problem ==<br /> Consider the set of all triangles &lt;math&gt;OPQ&lt;/math&gt; where &lt;math&gt;O&lt;/math&gt; is the origin and &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; are distinct points in the plane with nonnegative integer coordinates &lt;math&gt;(x,y)&lt;/math&gt; such that &lt;math&gt;41x + y = 2009&lt;/math&gt;. Find the number of such distinct triangles whose area is a positive integer.<br /> <br /> == Solution 1 ==<br /> Let the two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be defined with coordinates; &lt;math&gt;P=(x_1,y_1)&lt;/math&gt; and &lt;math&gt;Q=(x_2,y_2)&lt;/math&gt;<br /> <br /> We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem). <br /> <br /> &lt;math&gt;\det \left({\matrix {P \above Q}}\right)=\det \left({\matrix {x_1 \above x_2}\matrix {y_1 \above y_2}\right).&lt;/math&gt;<br /> <br /> Since the triangle has half the area of the parallelogram, we just need the determinant to be even.<br /> <br /> The determinant is<br /> <br /> &lt;cmath&gt;(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;2009&lt;/math&gt; is not even, &lt;math&gt;((x_1)-(x_2))&lt;/math&gt; must be even, thus the two &lt;math&gt;x&lt;/math&gt;'s must be of the same parity. Also note that the maximum value for &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;49&lt;/math&gt; and the minimum is &lt;math&gt;0&lt;/math&gt;. There are &lt;math&gt;25&lt;/math&gt; even and &lt;math&gt;25&lt;/math&gt; odd numbers available for use as coordinates and thus there are &lt;math&gt;(_{25}C_2)+(_{25}C_2)=\boxed{600}&lt;/math&gt; such triangles.<br /> <br /> == Solution 2 ==<br /> As in the solution above, let the two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be defined with coordinates; &lt;math&gt;P=(x_1,y_1)&lt;/math&gt; and &lt;math&gt;Q=(x_2,y_2)&lt;/math&gt;.<br /> <br /> <br /> If the coordinates of &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; have nonnegative integer coordinates, &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; must be lattice points either<br /> <br /> *on the nonnegative x-axis<br /> <br /> *on the nonnegative y-axis<br /> <br /> *in the first quadrant<br /> <br /> We can calculate the y-intercept of the line &lt;math&gt;41x+y=2009&lt;/math&gt; to be &lt;math&gt;(0,2009)&lt;/math&gt; and the x-intercept to be &lt;math&gt;(49,0)&lt;/math&gt;.<br /> <br /> Using the point-to-line distance formula, we can calculate the height of &lt;math&gt;\triangle OPQ&lt;/math&gt; from vertex &lt;math&gt;O&lt;/math&gt; (the origin) to be:<br /> <br /> &lt;math&gt;\dfrac{|41(0) + 1(0) - 2009|}{\sqrt{41^2 + 1^2}} = \dfrac{2009}{\sqrt{1682}} = \dfrac{2009}{29\sqrt2}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;b&lt;/math&gt; be the base of the triangle that is part of the line &lt;math&gt;41x+y=2009&lt;/math&gt;.<br /> <br /> The area is calculated as: <br /> &lt;math&gt;\dfrac{1}{2}\times b \times \dfrac{2009}{29\sqrt2} = \dfrac{2009}{58\sqrt2}\times b&lt;/math&gt;<br /> <br /> Let the numerical area of the triangle be &lt;math&gt;k&lt;/math&gt;. <br /> <br /> So, &lt;math&gt;k = \dfrac{2009}{58\sqrt2}\times b&lt;/math&gt;<br /> <br /> We know that &lt;math&gt;k&lt;/math&gt; is an integer. So, &lt;math&gt;b = 58\sqrt2 \times z&lt;/math&gt;, where &lt;math&gt;z&lt;/math&gt; is also an integer.<br /> <br /> We defined the points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; as &lt;math&gt;P=(x_1,y_1)&lt;/math&gt; and &lt;math&gt;Q=(x_2,y_2)&lt;/math&gt;.<br /> <br /> Changing the y-coordinates to be in terms of x, we get:<br /> <br /> &lt;math&gt;P=(x_1,2009-41x_1)&lt;/math&gt; and &lt;math&gt;Q=(x_2,2009-41x_2)&lt;/math&gt;.<br /> <br /> The distance between them equals &lt;math&gt;b&lt;/math&gt;. <br /> <br /> Using the distance formula, we get<br /> <br /> &lt;math&gt;PQ = b = \sqrt{(-41x_2+ 41x_1)^2 + (x_2 - x_1)^2} = 29\sqrt2 \times |x_2 - x_1| = 58\sqrt2\times z&lt;/math&gt; (*)<br /> <br /> WLOG, we can assume that &lt;math&gt;x_2 &gt; x_1&lt;/math&gt;.<br /> <br /> Taking the last two equalities from the &lt;math&gt;(*)&lt;/math&gt; string of equalities and putting in our assumption that &lt;math&gt;x_2&gt;x_1&lt;/math&gt;, we get<br /> <br /> &lt;math&gt;29\sqrt2\times (x_2-x_1) = 58\sqrt2\times z&lt;/math&gt;.<br /> <br /> Dividing both sides by &lt;math&gt;29\sqrt2&lt;/math&gt;, we get<br /> <br /> &lt;math&gt;x_2-x_1 = 2z&lt;/math&gt;<br /> <br /> As we mentioned, &lt;math&gt;z&lt;/math&gt; is an integer, so &lt;math&gt;x_2-x_1&lt;/math&gt; is an even integer. Also, &lt;math&gt;x_2&lt;/math&gt; and &lt;math&gt;x_1&lt;/math&gt; are both positive integers. So, &lt;math&gt;x_2&lt;/math&gt; and &lt;math&gt;x_1&lt;/math&gt; are between 0 and 49, inclusive. Also, &lt;math&gt;x_2&gt;x_1&lt;/math&gt;.<br /> <br /> *There are 48 ordered pairs &lt;math&gt;(x_2,x_1)&lt;/math&gt; such that their positive difference is 2.<br /> <br /> *There are 46 ordered pairs &lt;math&gt;(x_2,x_1)&lt;/math&gt; such that their positive difference is 4.<br /> <br /> ...<br /> <br /> *Finally, there are 2 ordered pairs &lt;math&gt;(x_2,x_1)&lt;/math&gt; such that their positive difference is 48.<br /> <br /> Summing them up, we get that there are &lt;math&gt;2+4+\dots + 48 = \boxed{600}&lt;/math&gt; triangles.<br /> <br /> == See also ==<br /> {{AIME box|year=2009|n=I|num-b=10|num-a=12}}</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=Rational_Root_Theorem&diff=45496 Rational Root Theorem 2012-03-16T03:51:28Z <p>Williamhu888: /* Proof */</p> <hr /> <div>{{stub}}<br /> <br /> <br /> Given a [[polynomial]] &lt;math&gt;P(x) = a_n x^n + a_{n - 1}x^{n - 1} + \ldots + a_1 x + a_0&lt;/math&gt; with [[integer | integral]] [[coefficient]]s, &lt;math&gt;a_n \neq 0&lt;/math&gt;. The '''Rational Root Theorem''' states that if &lt;math&gt;P(x)&lt;/math&gt; has a [[rational number| rational]] [[root]] &lt;math&gt;r = \pm\frac pq&lt;/math&gt; with &lt;math&gt;p, q&lt;/math&gt; [[relatively prime]] [[positive integer]]s, &lt;math&gt;p&lt;/math&gt; is a [[divisor]] of &lt;math&gt;a_0&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; is a divisor of &lt;math&gt;a_n&lt;/math&gt;.<br /> <br /> As a consequence, every rational root of a [[monic polynomial]] with integral coefficients must be integral.<br /> <br /> This gives us a relatively quick process to find all &quot;nice&quot; roots of a given polynomial, since given the coefficients we have only a finite number of rational numbers to check.<br /> <br /> == Proof ==<br /> <br /> Given &lt;math&gt;\frac{p}{q}&lt;/math&gt; is a rational root of a polynomial &lt;math&gt;f(x)=a_nx^n+x_{n-1}x^{n-1}+\cdots +a_0&lt;/math&gt;, where the &lt;math&gt;a_n&lt;/math&gt;'s are integers, we wish to show that &lt;math&gt;p|a_0&lt;/math&gt; and &lt;math&gt;q|a_n&lt;/math&gt;. Since &lt;math&gt;\frac{p}{q}&lt;/math&gt; is a root, &lt;cmath&gt;0=a_n\left(\frac{p}{q}\right)^n+\cdots +a_0&lt;/cmath&gt; Multiplying by &lt;math&gt;q^n&lt;/math&gt;, we have: &lt;cmath&gt;0=a_np^n+a_{n-1}p^{n-1}q+\cdots+a_0q^n&lt;/cmath&gt; Examining this in modulo &lt;math&gt;p&lt;/math&gt;, we have &lt;math&gt;a_0q^n\equiv 0\pmod p&lt;/math&gt;. As &lt;math&gt;q&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are relatively prime, &lt;math&gt;p|a_0&lt;/math&gt;. With the same logic, but with modulo &lt;math&gt;q&lt;/math&gt;, we have &lt;math&gt;q|a_n&lt;/math&gt;, and we are done.<br /> <br /> ==Problems==<br /> <br /> ===Easy===<br /> <br /> Factor the polynomial &lt;math&gt;x^3-5x^2+2x+8&lt;/math&gt;.<br /> <br /> ===Intermediate===<br /> <br /> Find all rational roots of the polynomial &lt;math&gt;x^4-x^3-x^2+x+57&lt;/math&gt;.<br /> <br /> Prove that &lt;math&gt;\sqrt{2}&lt;/math&gt; is irrational, using the Rational Root Theorem.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=Rational_Root_Theorem&diff=45495 Rational Root Theorem 2012-03-16T03:51:03Z <p>Williamhu888: /* Proof */</p> <hr /> <div>{{stub}}<br /> <br /> <br /> Given a [[polynomial]] &lt;math&gt;P(x) = a_n x^n + a_{n - 1}x^{n - 1} + \ldots + a_1 x + a_0&lt;/math&gt; with [[integer | integral]] [[coefficient]]s, &lt;math&gt;a_n \neq 0&lt;/math&gt;. The '''Rational Root Theorem''' states that if &lt;math&gt;P(x)&lt;/math&gt; has a [[rational number| rational]] [[root]] &lt;math&gt;r = \pm\frac pq&lt;/math&gt; with &lt;math&gt;p, q&lt;/math&gt; [[relatively prime]] [[positive integer]]s, &lt;math&gt;p&lt;/math&gt; is a [[divisor]] of &lt;math&gt;a_0&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; is a divisor of &lt;math&gt;a_n&lt;/math&gt;.<br /> <br /> As a consequence, every rational root of a [[monic polynomial]] with integral coefficients must be integral.<br /> <br /> This gives us a relatively quick process to find all &quot;nice&quot; roots of a given polynomial, since given the coefficients we have only a finite number of rational numbers to check.<br /> <br /> == Proof ==<br /> <br /> Given &lt;math&gt;\frac{p}{q}&lt;/math&gt; is a rational root of a polynomial &lt;math&gt;f(x)=a_nx^n+x_{n-1}x^{n-1}+\cdots +a_0&lt;/math&gt;, where the coefficients are integers, we wish to show that &lt;math&gt;p|a_0&lt;/math&gt; and &lt;math&gt;q|a_n&lt;/math&gt;. Since &lt;math&gt;\frac{p}{q}&lt;/math&gt; is a root, &lt;cmath&gt;0=a_n\left(\frac{p}{q}\right)^n+\cdots +a_0&lt;/cmath&gt; Multiplying by &lt;math&gt;q^n&lt;/math&gt;, we have: &lt;cmath&gt;0=a_np^n+a_{n-1}p^{n-1}q+\cdots+a_0q^n&lt;/cmath&gt; Examining this in modulo &lt;math&gt;p&lt;/math&gt;, we have &lt;math&gt;a_0q^n\equiv 0\pmod p&lt;/math&gt;. As &lt;math&gt;q&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are relatively prime, &lt;math&gt;p|a_0&lt;/math&gt;. With the same logic, but with modulo &lt;math&gt;q&lt;/math&gt;, we have &lt;math&gt;q|a_n&lt;/math&gt;, and we are done.<br /> <br /> ==Problems==<br /> <br /> ===Easy===<br /> <br /> Factor the polynomial &lt;math&gt;x^3-5x^2+2x+8&lt;/math&gt;.<br /> <br /> ===Intermediate===<br /> <br /> Find all rational roots of the polynomial &lt;math&gt;x^4-x^3-x^2+x+57&lt;/math&gt;.<br /> <br /> Prove that &lt;math&gt;\sqrt{2}&lt;/math&gt; is irrational, using the Rational Root Theorem.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_12&diff=45382 2012 AMC 12B Problems/Problem 12 2012-03-12T03:00:33Z <p>Williamhu888: /* Solution 1 */</p> <hr /> <div>==Solution 1==<br /> <br /> There are 20 Choose 2 selections however, we count these twice therefore<br /> <br /> 2* 20 C 2 = 380. The wording of the question implies D not E.<br /> <br /> MAA decided to accept both D and E, however.<br /> <br /> ==Solution 2==<br /> <br /> Consider the 20 term sequence of 0's and 1's. Keeping all other terms 1, a sequence of &lt;math&gt;k&gt;0&lt;/math&gt; consecutive 0's can be placed in &lt;math&gt;21-k&lt;/math&gt; locations. That is, there are 20 strings with 1 zero, 19 strings with 2 consecutive zeros, 18 strings with 3 consecutive zeros, ..., 1 string with 20 consecutive zeros. Hence there are &lt;math&gt;20+19+\cdots+1=\binom{21}{2}&lt;/math&gt; strings with consecutive zeros. The same argument shows there are &lt;math&gt;\binom{21}{2}&lt;/math&gt; strings with consecutive 1's. This yields &lt;math&gt;2\binom{21}{2}&lt;/math&gt; strings in all. However, we have counted twice those strings in which all the 1's and all the 0's are consecutive. These are the cases &lt;math&gt;01111...&lt;/math&gt;, &lt;math&gt;00111...&lt;/math&gt;, &lt;math&gt;000111...&lt;/math&gt;, ..., &lt;math&gt;000...0001&lt;/math&gt; (of which there are 19) as well as the cases &lt;math&gt;10000...&lt;/math&gt;, &lt;math&gt;11000...&lt;/math&gt;, &lt;math&gt;111000...&lt;/math&gt;, ..., &lt;math&gt;111...110&lt;/math&gt; (of which there are 19 as well). This yields &lt;math&gt;2\binom{21}{2}-2\cdot19=382&lt;/math&gt; so that the answer is &lt;math&gt;\framebox{E}&lt;/math&gt;.</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=User:Williamhu888&diff=45365 User:Williamhu888 2012-03-11T18:56:31Z <p>Williamhu888: Blanked the page</p> <hr /> <div></div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_8&diff=45335 2006 AIME I Problems/Problem 8 2012-03-09T04:03:46Z <p>Williamhu888: /* Solution */</p> <hr /> <div>== Problem ==<br /> [[Hexagon]] &lt;math&gt; ABCDEF &lt;/math&gt; is divided into five [[rhombus]]es, &lt;math&gt; \mathcal{P, Q, R, S,} &lt;/math&gt; and &lt;math&gt; \mathcal{T,} &lt;/math&gt; as shown. Rhombuses &lt;math&gt; \mathcal{P, Q, R,} &lt;/math&gt; and &lt;math&gt; \mathcal{S} &lt;/math&gt; are [[congruent (geometry) | congruent]], and each has [[area]] &lt;math&gt; \sqrt{2006}. &lt;/math&gt; Let &lt;math&gt; K &lt;/math&gt; be the area of rhombus &lt;math&gt; \mathcal{T}&lt;/math&gt;. Given that &lt;math&gt; K &lt;/math&gt; is a [[positive integer]], find the number of possible values for &lt;math&gt; K&lt;/math&gt;.<br /> <br /> [[Image:2006AimeA8.PNG]]<br /> <br /> == Solution ==<br /> Let &lt;math&gt;x&lt;/math&gt; denote the common side length of the rhombi.<br /> Let &lt;math&gt;y&lt;/math&gt; denote one of the smaller interior [[angle]]s of rhombus &lt;math&gt; \mathcal{P} &lt;/math&gt;. Then &lt;math&gt;x^2\sin(y)=\sqrt{2006}&lt;/math&gt;. We also see that &lt;math&gt;K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y&lt;/math&gt;. Thus &lt;math&gt;K&lt;/math&gt; can be any positive integer in the [[interval]] &lt;math&gt;(0, 2\sqrt{2006})&lt;/math&gt;. <br /> &lt;math&gt;2\sqrt{2006} = \sqrt{8024}&lt;/math&gt; and &lt;math&gt;89^2 = 7921 &lt; 8024 &lt; 8100 = 90^2&lt;/math&gt;, so &lt;math&gt;K&lt;/math&gt; can be any [[integer]] between 1 and 89, inclusive. Thus the number of positive values for &lt;math&gt;K&lt;/math&gt; is &lt;math&gt;\boxed{089}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Trigonometry Problems]]</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_5&diff=45265 2005 AIME II Problems/Problem 5 2012-03-03T05:27:29Z <p>Williamhu888: /* Solution */</p> <hr /> <div>== Problem ==<br /> Determine the number of [[ordered pair]]s &lt;math&gt; (a,b) &lt;/math&gt; of [[integer]]s such that &lt;math&gt; \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, &lt;/math&gt; and &lt;math&gt; 2 \leq b \leq 2005. &lt;/math&gt;<br /> <br /> == Solution I ==<br /> The equation can be rewritten as &lt;math&gt; \frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \frac{(\log b)^2+6(\log a)^2}{\log a \log b}=5 &lt;/math&gt; Multiplying through by &lt;math&gt;\log a \log b &lt;/math&gt; and factoring yields &lt;math&gt;(\log b - 3\log a)(\log b - 2\log a)=0 &lt;/math&gt;. Therefore, &lt;math&gt;\log b=3\log a &lt;/math&gt; or &lt;math&gt;\log b=2\log a &lt;/math&gt;, so either &lt;math&gt; b=a^3 &lt;/math&gt; or &lt;math&gt; b=a^2 &lt;/math&gt;. <br /> *For the case &lt;math&gt; b=a^2 &lt;/math&gt;, note that &lt;math&gt; 44^2=1936 &lt;/math&gt; and &lt;math&gt; 45^2=2025 &lt;/math&gt;. Thus, all values of &lt;math&gt;a&lt;/math&gt; from &lt;math&gt;2&lt;/math&gt; to &lt;math&gt;44&lt;/math&gt; will work. <br /> *For the case &lt;math&gt; b=a^3 &lt;/math&gt;, note that &lt;math&gt; 12^3=1728 &lt;/math&gt; while &lt;math&gt; 13^3=2197 &lt;/math&gt;. Therefore, for this case, all values of &lt;math&gt;a&lt;/math&gt; from &lt;math&gt;2&lt;/math&gt; to &lt;math&gt;12&lt;/math&gt; work. <br /> There are &lt;math&gt; 44-2+1=43 &lt;/math&gt; possibilities for the square case and &lt;math&gt; 12-2+1=11 &lt;/math&gt; possibilities for the cube case. Thus, the answer is &lt;math&gt; 43+11= \boxed{054}&lt;/math&gt;.<br /> <br /> Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs &lt;math&gt;(a,b)&lt;/math&gt;, and not for the number of possible values of &lt;math&gt;b&lt;/math&gt;. Were the problem to ask for the number of possible values of &lt;math&gt;b&lt;/math&gt;, the values of &lt;math&gt;b^6&lt;/math&gt; under &lt;math&gt;2005&lt;/math&gt; would have to be subtracted, which would just be &lt;math&gt;2&lt;/math&gt; values: &lt;math&gt;2^6&lt;/math&gt; and &lt;math&gt;3^6&lt;/math&gt;.<br /> <br /> ==Solution II ==<br /> Let &lt;math&gt;k=\log_a b&lt;/math&gt;. Then our equation becomes &lt;math&gt;k+\frac{6}{k}=5&lt;/math&gt;. Multiplying through by &lt;math&gt;k&lt;/math&gt; and solving the quadratic gives us &lt;math&gt;k=2&lt;/math&gt; or &lt;math&gt;k=3&lt;/math&gt;. Hence &lt;math&gt;a^2=b&lt;/math&gt; or &lt;math&gt;a^3=b&lt;/math&gt;. <br /> <br /> For the first case &lt;math&gt;a^2=b&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt; can range from 2 to 44, a total of 43 values.<br /> For the second case &lt;math&gt;a^3=b&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt; can range from 2 to 12, a total of 11 values.<br /> <br /> <br /> Thus the total number of possible values is &lt;math&gt;43+11=\boxed{54}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_II_Problems/Problem_1&diff=45147 2004 AIME II Problems/Problem 1 2012-02-25T19:02:32Z <p>Williamhu888: /* Solution */</p> <hr /> <div>== Problem ==<br /> A [[chord]] of a [[circle]] is [[perpendicular]] to a [[radius]] at the [[midpoint]] of the radius. The [[ratio]] of the [[area]] of the larger of the two regions into which the chord divides the circle to the smaller can be expressed in the form &lt;math&gt; \frac{a\pi+b\sqrt{c}}{d\pi-e\sqrt{f}}, &lt;/math&gt; where &lt;math&gt; a, b, c, d, e, &lt;/math&gt; and &lt;math&gt; f &lt;/math&gt; are [[positive integer]]s, &lt;math&gt; a &lt;/math&gt; and &lt;math&gt; e &lt;/math&gt; are [[relatively prime]], and neither &lt;math&gt; c &lt;/math&gt; nor &lt;math&gt; f &lt;/math&gt; is [[divisibility | divisible]] by the [[square]] of any [[prime]]. Find the [[remainder]] when the product &lt;math&gt; abcdef &lt;/math&gt; is divided by 1000.<br /> <br /> [[Image:2004 AIME II Problem 1.png]]<br /> <br /> == Solution ==<br /> Let &lt;math&gt;r&lt;/math&gt; be the [[length]] of the radius of the circle. A [[right triangle]] is formed by half of the chord, half of the radius (since the chord [[bisect]]s it), and the radius. Thus, it is a &lt;math&gt;30^\circ&lt;/math&gt; - &lt;math&gt;60^\circ&lt;/math&gt; - &lt;math&gt;90^\circ&lt;/math&gt; [[triangle]], and the area of two such triangles is &lt;math&gt;2 \cdot \frac{1}{2} \cdot \frac{r}{2} \cdot \frac{r\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}&lt;/math&gt;. The [[central angle]] which contains the entire chord is &lt;math&gt;60 \cdot 2 = 120&lt;/math&gt; [[degree]]s, so the area of the [[sector]] is &lt;math&gt;\frac{1}{3}r^2\pi&lt;/math&gt;; the rest of the area of the circle is then equal to &lt;math&gt;\frac{2}{3}r^2\pi&lt;/math&gt;.<br /> <br /> The smaller area cut off by the chord is equal to the area of the sector minus the area of the triangle. The larger area is equal to the area of the circle not within the sector and the area of the triangle. Thus, the desired ratio is &lt;math&gt;\frac{\frac{2}{3}r^2\pi + \frac{r^2\sqrt{3}}{4}}{\frac{1}{3}r^2\pi - \frac{r^2\sqrt{3}}{4}} = \frac{8\pi + 3\sqrt{3}}{4\pi - 3\sqrt{3}}&lt;/math&gt;<br /> <br /> Therefore, &lt;math&gt;abcdef = 2^53^4 = 2592 \Longrightarrow \boxed{592}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2004|n=II|before=First Question|num-a=2}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_7&diff=44895 2003 AIME II Problems/Problem 7 2012-02-19T19:31:53Z <p>Williamhu888: </p> <hr /> <div>== Problem ==<br /> Find the area of rhombus &lt;math&gt;ABCD&lt;/math&gt; given that the radii of the circles circumscribed around triangles &lt;math&gt;ABD&lt;/math&gt; and &lt;math&gt;ACD&lt;/math&gt; are &lt;math&gt;12.5&lt;/math&gt; and &lt;math&gt;25&lt;/math&gt;, respectively.<br /> <br /> == Solution ==<br /> The diagonals of the rhombus perpendicularly bisect each other. Call half of diagonal BD &lt;math&gt;a&lt;/math&gt; and half of diagonal AC &lt;math&gt;b&lt;/math&gt;. The length of the four sides of the rhombus is &lt;math&gt;\sqrt{a^2+b^2}&lt;/math&gt;. <br /> <br /> The area of any triangle can be expressed as &lt;math&gt;\frac{a\cdot b\cdot c}{4R}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are the sides and &lt;math&gt;R&lt;/math&gt; is the circumradius. Thus, the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;ab=2a(a^2+b^2)/(4\cdot12.5)&lt;/math&gt;. Also, the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;ab=2b(a^2+b^2)/(4\cdot25)&lt;/math&gt;. Setting these two expressions equal to each other and simplifying gives &lt;math&gt;b=2a&lt;/math&gt;. Substitution yields &lt;math&gt;a=10&lt;/math&gt; and &lt;math&gt;b=20&lt;/math&gt;, so the area of the rhombus is &lt;math&gt;20\cdot40/2=\boxed{400}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2003|n=II|num-b=6|num-a=8}}</div> Williamhu888 https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems/Problem_3&diff=44821 2003 AIME I Problems/Problem 3 2012-02-18T23:28:04Z <p>Williamhu888: </p> <hr /> <div>== Problem ==<br /> Let the [[set]] &lt;math&gt; \mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}. &lt;/math&gt; Susan makes a list as follows: for each two-element subset of &lt;math&gt; \mathcal{S}, &lt;/math&gt; she writes on her list the greater of the set's two elements. Find the sum of the numbers on the list.<br /> <br /> == Solution ==<br /> Order the numbers in the set from greatest to least to reduce error: &lt;math&gt;\{34, 21, 13, 8, 5, 3, 2, 1\}.&lt;/math&gt;<br /> Each [[element]] of the [[set]] will appear in &lt;math&gt;7&lt;/math&gt; two-element [[subset]]s, once with each other number.<br /> <br /> *&lt;math&gt;34&lt;/math&gt; will be the greater number in &lt;math&gt;7&lt;/math&gt; subsets. <br /> *&lt;math&gt;21&lt;/math&gt; will be the greater number in &lt;math&gt;6&lt;/math&gt; subsets. <br /> *&lt;math&gt;13&lt;/math&gt; will be the greater number in &lt;math&gt;5&lt;/math&gt; subsets. <br /> *&lt;math&gt;8&lt;/math&gt; will be the greater number in &lt;math&gt;4&lt;/math&gt; subsets. <br /> *&lt;math&gt;5&lt;/math&gt; will be the greater number in &lt;math&gt;3&lt;/math&gt; subsets. <br /> *&lt;math&gt;3&lt;/math&gt; will be the greater number in &lt;math&gt;2&lt;/math&gt; subsets. <br /> *&lt;math&gt;2&lt;/math&gt; will be the greater number in &lt;math&gt;1&lt;/math&gt; subsets. <br /> *&lt;math&gt;1&lt;/math&gt; will be the greater number in &lt;math&gt;0&lt;/math&gt; subsets. <br /> <br /> Therefore the desired sum is &lt;math&gt;34\cdot7+21\cdot6+13\cdot5+8\cdot4+5\cdot3+3 \cdot2+2\cdot1+1\cdot0=\boxed{484}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2003|n=I|num-b=2|num-a=4}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]</div> Williamhu888