https://artofproblemsolving.com/wiki/index.php?title=Special:NewPages&feed=atom&hideredirs=1&limit=50&offset=&namespace=0&username=&tagfilter=&size-mode=max&size=0 AoPS Wiki - New pages [en] 2020-10-29T00:14:59Z From AoPS Wiki MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_4 2021 AMC 10A Problems/Problem 4 2020-10-28T17:58:28Z <p>Virjoy2001: Problem 4</p> <hr /> <div>==Problem==</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php/IntelligentElephant2010 IntelligentElephant2010 2020-10-26T23:45:33Z <p>Intelligentelephant2010: Created page with &quot;H it's me! See this link: https://artofproblemsolving.com/wiki/index.php/User:IntelligentElephant2010#About_Me:&quot;</p> <hr /> <div>H it's me! See this link: https://artofproblemsolving.com/wiki/index.php/User:IntelligentElephant2010#About_Me:</div> Intelligentelephant2010 https://artofproblemsolving.com/wiki/index.php/Googol Googol 2020-10-25T14:46:09Z <p>El008: Created page with &quot;Googol is a huge number. It has a 1 followed by 100 zeroes or &lt;math&gt;10^{100}.&lt;/math&gt;&quot;</p> <hr /> <div>Googol is a huge number. It has a 1 followed by 100 zeroes or &lt;math&gt;10^{100}.&lt;/math&gt;</div> El008 https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Answer_Key 2020 AMC 8 Answer Key 2020-10-24T20:19:19Z <p>Zixuan12: Created page with &quot;1. Nobody cares 2. Can you PLEASE stop reading this 3. STOP!!!!!!!!!! 4. JUST STOP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! 5. GO AWAY YOU MONSTER!!!!!!!!!!!!...&quot;</p> <hr /> <div>1. Nobody cares<br /> 2. Can you PLEASE stop reading this<br /> 3. STOP!!!!!!!!!!<br /> 4. JUST STOP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!<br /> 5. GO AWAY YOU MONSTER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!<br /> 6. SHUT UP</div> Zixuan12 https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_24 2020 AMC 8 Problems/Problem 24 2020-10-24T20:17:12Z <p>Zixuan12: Created page with &quot;Go away. The problems are secret&quot;</p> <hr /> <div>Go away. The problems are secret</div> Zixuan12 https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems 2020 AMC 8 Problems 2020-10-24T20:16:21Z <p>Zixuan12: Created page with &quot;#Problem 1 #Problem 2 #Problem 3 #Problem 4 #Problem 5&quot;</p> <hr /> <div>#Problem 1<br /> #Problem 2<br /> #Problem 3<br /> #Problem 4<br /> #Problem 5</div> Zixuan12 https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_25 2020 AMC 8 Problems/Problem 25 2020-10-24T20:15:14Z <p>Zixuan12: Created page with &quot;so hard isnt it&quot;</p> <hr /> <div>so hard isnt it</div> Zixuan12 https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_16 2020 AMC 8 Problems/Problem 16 2020-10-24T20:14:43Z <p>Zixuan12: Created page with &quot;plz stop trying to cheat&quot;</p> <hr /> <div>plz stop trying to cheat</div> Zixuan12 https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_42 1955 AHSME Problems/Problem 42 2020-10-24T15:24:15Z <p>Thisusernameistaken: Created the page, not sure how to create LaTeX solution.</p> <hr /> <div>== Problem ==<br /> If &lt;math&gt;a, b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive integers, the radicals &lt;math&gt;\sqrt{a+\frac{b}{c}}&lt;/math&gt; and &lt;math&gt;a\sqrt{\frac{b}{c}}&lt;/math&gt; are equal when and only when: <br /> <br /> &lt;math&gt; \textbf{(A)}\ a=b=c=1\qquad\textbf{(B)}\ a=b\text{ and }c=a=1\qquad\textbf{(C)}\ c=\frac{b(a^2-1)}{a}\\ \textbf{(D)}\ a=b\text{ and }c\text{ is any value}\qquad\textbf{(E)}\ a=b\text{ and }c=a-1 &lt;/math&gt;<br /> <br /> <br /> == Solution ==<br /> <br /> <br /> == See Also ==<br /> <br /> {{AHSME 50p box|year=1955|num-b=41|num-a=43}}<br /> {{MAA Notice}}</div> Thisusernameistaken https://artofproblemsolving.com/wiki/index.php/2021_USAMO_Problems/Problem_1 2021 USAMO Problems/Problem 1 2020-10-23T21:24:18Z <p>Math31415926535: Created page with &quot;these problems will not be released until after the test date&quot;</p> <hr /> <div>these problems will not be released until after the test date</div> Math31415926535 https://artofproblemsolving.com/wiki/index.php/2020_USAMTS_Round_1_Problems/Problem_4 2020 USAMTS Round 1 Problems/Problem 4 2020-10-23T17:33:57Z <p>Icematrix2: Created page with &quot;Two beasts, Rosencrans and Gildenstern, play a game. They have a circle with &lt;math&gt;n&lt;/math&gt; points (&lt;math&gt;n \geq 5&lt;/math&gt;) on it. On their turn, each beast (starting with Rose...&quot;</p> <hr /> <div>Two beasts, Rosencrans and Gildenstern, play a game. They have a circle with &lt;math&gt;n&lt;/math&gt; points (&lt;math&gt;n \geq 5&lt;/math&gt;) on it. On their turn, each beast (starting with Rosencrans) draws a chord between a pair of points in such a way that any two chords have a shared point. (The chords either intersect or have a common endpoint).<br /> <br /> The game ends when a player cannot draw a chord. The last beast to draw a chord wins. For which &lt;math&gt;n&lt;/math&gt; does Rosencrans win?</div> Icematrix2 https://artofproblemsolving.com/wiki/index.php/2020_USAMTS_Round_1_Problems/Problem_5 2020 USAMTS Round 1 Problems/Problem 5 2020-10-23T03:08:44Z <p>Icematrix2: Created page with &quot;Find all pairs of rational numbers &lt;math&gt;(a,b)&lt;/math&gt; such that &lt;math&gt;0 &lt; a &lt; b&lt;/math&gt; and &lt;math&gt;a^a = b^b&lt;/math&gt;. =Solution 1= We claim that there is only one solution, whic...&quot;</p> <hr /> <div>Find all pairs of rational numbers &lt;math&gt;(a,b)&lt;/math&gt; such that &lt;math&gt;0 &lt; a &lt; b&lt;/math&gt; and &lt;math&gt;a^a = b^b&lt;/math&gt;.<br /> <br /> =Solution 1=<br /> We claim that there is only one solution, which is the solution pair &lt;math&gt;(a,b)=(\frac{1}{4},\frac{1}{2})&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;b=ta&lt;/math&gt;, where &lt;math&gt;t&lt;/math&gt; is a rational number greater than &lt;math&gt;1&lt;/math&gt;. Then when we substitute that into the equation &lt;math&gt;a^a=b^b&lt;/math&gt;, we get &lt;math&gt;a^a={ta}^{ta}&lt;/math&gt;. We want to separate the variables from each other, so we can multiply by &lt;math&gt;\frac{1}{a}&lt;/math&gt; to get &lt;math&gt;a={ta}^t \implies a = t^t \cdot a^t&lt;/math&gt;. Moving &lt;math&gt;a&lt;/math&gt; to the other side, and using the exponent rule, we get &lt;math&gt;a^{1-t}=t^t&lt;/math&gt;. Multiplying this by &lt;math&gt;\frac{1}{1-t}&lt;/math&gt;, we get &lt;math&gt;a=t^{\frac{t}{1-t}}&lt;/math&gt;. Substituting that into &lt;math&gt;b=ta&lt;/math&gt;, we get &lt;math&gt;b=t \cdot t^{\frac{t}{1-t}}&lt;/math&gt;, so &lt;math&gt;b=t^{\frac{1}{1-t}}&lt;/math&gt;. Therefore our solution set is &lt;math&gt;(t^{\frac{t}{1-t}},t^{\frac{1}{1-t}}&lt;/math&gt;, but wait! The solution set contains irrational numbers, and the question only asks for rational numbers. Therefore, &lt;math&gt;\frac{t}{1-t}&lt;/math&gt; and &lt;math&gt;\frac{1}{1-t}&lt;/math&gt; both need to be integers. By inspection, we can see that the only solution is &lt;math&gt;t=2&lt;/math&gt;, giving us the solution set &lt;math&gt;(\frac{1}{4},\frac{1}{2})&lt;/math&gt;.<br /> <br /> Solution and &lt;math&gt;\LaTeX&lt;/math&gt; by smartguy888<br /> <br /> {{MAA Notice}}</div> Icematrix2 https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_6 2020 AMC 8 Problems/Problem 6 2020-10-22T21:52:19Z <p>Math31415926535: Created page with &quot;These problems will not be released until after the AMC 8&quot;</p> <hr /> <div>These problems will not be released until after the AMC 8</div> Math31415926535 https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_5 2020 AMC 8 Problems/Problem 5 2020-10-22T21:37:31Z <p>Math31415926535: Created blank page</p> <hr /> <div></div> Math31415926535 https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10 2020 AMC 8 Problems/Problem 10 2020-10-22T21:35:55Z <p>Math31415926535: Created page with &quot;ssdsd&quot;</p> <hr /> <div>ssdsd</div> Math31415926535 https://artofproblemsolving.com/wiki/index.php/2020_USAMTS_Round_1_Problems/Problem_3 2020 USAMTS Round 1 Problems/Problem 3 2020-10-22T20:13:21Z <p>Aryabhata000: /* Solution 2 */</p> <hr /> <div>The bisectors of the internal angles of parallelogram &lt;math&gt;ABCD&lt;/math&gt; with &lt;math&gt;AB&gt;BC&lt;/math&gt; determine a quadrilateral with the same area as &lt;math&gt;ABCD&lt;/math&gt;. Determine, with proof, the value of &lt;math&gt;\frac{AB}{BC}&lt;/math&gt;.<br /> <br /> =Solution 1=<br /> We claim the answer is &lt;math&gt;2+\sqrt3.&lt;/math&gt; Let &lt;math&gt;HFGE&lt;/math&gt; be the new quadrilateral; that is, the quadrilateral determined by the internal bisectors of the angles of &lt;math&gt;ABCD&lt;/math&gt;.<br /> <br /> Lemma &lt;math&gt;1&lt;/math&gt; : &lt;math&gt;HFGE&lt;/math&gt; is a rectangle.<br /> &lt;math&gt;1.&lt;/math&gt; &lt;math&gt;ABCD&lt;/math&gt; is a parallelogram.<br /> &lt;math&gt;\angle DAB = \angle DCB,&lt;/math&gt; as &lt;math&gt;AE&lt;/math&gt; bisects &lt;math&gt;\angle DAB \Rightarrow \angle BAE = \frac{\angle DAB}{2}&lt;/math&gt; and &lt;math&gt;CE&lt;/math&gt; bisects &lt;math&gt;\angle DCB \Rightarrow \angle DCF = \frac{DCB}{2} \Rightarrow \angle DCF = \angle AJF \Rightarrow \angle BAE = \angle AJF \Rightarrow FG \parallel HE.&lt;/math&gt;<br /> By the same logic, &lt;math&gt;HF \parallel EG \Rightarrow GFHE&lt;/math&gt; is a parallelogram.<br /> 2. &lt;math&gt;\angle EAB = \frac{\angle DAB}{2}&lt;/math&gt; and &lt;math&gt;\angle ABE = \frac{\angle ABC}{2} \Rightarrow \angle EAB + \angle ABE = \frac{\angle DAB + \angle ABC}{2}&lt;/math&gt; and &lt;math&gt;\angle DAB + \angle ABC = 180^\circ \Rightarrow \angle EAB + \angle ABE = 90^\circ \Rightarrow \angle AEB = 90^\circ.&lt;/math&gt;<br /> By &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2,&lt;/math&gt; we can conclude that &lt;math&gt;HFGE&lt;/math&gt; is a rectangle.<br /> <br /> Now, knowing &lt;math&gt;HFGE&lt;/math&gt; is a rectangle, we can continue on.<br /> <br /> Let &lt;math&gt;AB = a, BC = b, &lt;/math&gt; and &lt;math&gt;\angle ABE = \alpha.&lt;/math&gt; Thus, &lt;math&gt;[ABCD] = ab\sin(2\alpha).&lt;/math&gt;<br /> &lt;math&gt;AD \parallel DC \Rightarrow \angle BJC = \angle JCD&lt;/math&gt; and &lt;math&gt;\angle JCD = \angle JCB \Rightarrow \angle BJC = \angle JCB \Rightarrow JB = BC =b.&lt;/math&gt;<br /> By the same logic, &lt;math&gt;AI = AD = b.&lt;/math&gt;<br /> &lt;math&gt;BE \parallel ED \Rightarrow \angle AIH = \angle ABE = \alpha.&lt;/math&gt;<br /> &lt;math&gt;HE = AE-AH = a\sin(\alpha) - b\sin(\alpha) = (\alpha - \beta)\sin(\alpha),&lt;/math&gt; and &lt;math&gt;EG = EB-GB = a\cos(\alpha) - b\cos(\alpha) = (a-b)\cos(\alpha).&lt;/math&gt;<br /> &lt;math&gt;[HFGE] = HE * EG = (a-b)^2\sin(\alpha)\cos(\alpha) \Rightarrow ab\sin(2\alpha) = (a-b)^2\text{sin}(\alpha)\text{cos}(\alpha) \Rightarrow ab\sin(2\alpha) = (a-b)^2\sin(\alpha)\cos(\alpha) \Rightarrow 2ab = (a-b)^2 \Rightarrow a^2 + b^2 -2ab = 2ab&lt;/math&gt; &lt;math&gt;\Rightarrow a^2 -4ab +b^2 = 0 \Rightarrow a = \frac{4b \pm \sqrt{16b^2 -4b^2}}{2} = 2b\pm b\sqrt{3}&lt;/math&gt; &lt;math&gt;\Rightarrow a=b(2\pm\sqrt{3}) \Rightarrow \frac{a}{b} = 2 \pm \sqrt{3}.&lt;/math&gt; Because &lt;math&gt;a&gt;b,&lt;/math&gt; we have &lt;math&gt;\frac{a}{b} = 2+\sqrt{3}.&lt;/math&gt;<br /> <br /> Solution and &lt;math&gt;\LaTeX&lt;/math&gt; by Sp3nc3r<br /> <br /> =Solution 2 (similar to Solution 1)=<br /> Let &lt;math&gt;P,Q,R,S&lt;/math&gt; be the intersections of the bisectors of &lt;math&gt;\angle C \text { and } \angle D, \angle B \text { and } \angle C, \angle A \text { and } \angle B, \angle A \text { and } \angle D&lt;/math&gt; respectively. <br /> <br /> Let &lt;math&gt; \angle BAD = \theta&lt;/math&gt; . Then &lt;math&gt;\angle SAD = \angle QCB= \frac{\theta}{2}&lt;/math&gt; and &lt;math&gt;\angle ADS = \angle QBC= \frac{180-\theta}{2}&lt;/math&gt;. So, &lt;math&gt;\angle ASD = \angle SRQ = \angle PQR = \angle 180 - (\frac{\theta}{2} + \frac{180-\theta}{2}) = 90&lt;/math&gt;. Therefore, &lt;math&gt;RSP = 90&lt;/math&gt;.<br /> <br /> Similarly, &lt;math&gt;\angle SPQ = \angle QRS = 180- (\frac{\theta}{2} + \frac{180-\theta}{2}) = 90&lt;/math&gt;.<br /> <br /> So, therefore, &lt;math&gt;PQRS&lt;/math&gt; must be a rectangle and &lt;math&gt;[PQRS] = SP \times RS&lt;/math&gt; <br /> <br /> Now, note that &lt;math&gt;SP= PD- SD = DC \sin(\frac{\theta}{2}) - AD \sin(\frac{\theta}{2})&lt;/math&gt;. Also, &lt;math&gt;RS = AR - AS - DC \cos(\frac{\theta}{2}) - AD \cos(\frac{\theta}{2})&lt;/math&gt;.<br /> <br /> So, we have &lt;cmath&gt; [PQRS] = (DC-AD)^2 \sin(\frac{\theta}{2}) \cos (\frac{\theta}{2})&lt;/cmath&gt; &lt;cmath&gt;[ABCD] = DC \times AD \sin{\theta} = DC \times AD \times 2 \sin(\frac{\theta}{2}) \cos (\frac{\theta}{2}). &lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;[PQRS] = [ABCD]&lt;/math&gt; :&lt;cmath&gt;(DC- AD)^2 = 2(DC)(AD) \implies r^2 - 4r + 1 = 0&lt;/cmath&gt; for &lt;math&gt;r = \frac{DC}{AD}&lt;/math&gt;.<br /> <br /> Therefore, by the Quadratic Formula, &lt;math&gt;r= 2 \pm \sqrt{3}&lt;/math&gt;. Since &lt;math&gt; AB &gt; BC&lt;/math&gt;, &lt;math&gt;r = \boxed{ 2+ \sqrt{3}}&lt;/math&gt;.<br /> <br /> <br /> {{USAMTS box|year=2020|round=1|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Icematrix2 https://artofproblemsolving.com/wiki/index.php/Https://artofproblemsolving.com/texer/zizfoozc Https://artofproblemsolving.com/texer/zizfoozc 2020-10-20T23:25:29Z <p>Piphi: </p> <hr /> <div>{{delete|does not help the wiki in any way}}</div> Medical ordnance https://artofproblemsolving.com/wiki/index.php/2020_SWMC_6_Problems 2020 SWMC 6 Problems 2020-10-20T10:59:38Z <p>Aops-g5-gethsemanea2: Created page with &quot;==Problem 1== Give the only positive value of &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;10x + 5 \leq 14 + x&lt;/math&gt;. ==Problem 2== An AoPS user is lining up the 5 names of his helpers in his...&quot;</p> <hr /> <div>==Problem 1==<br /> <br /> Give the only positive value of &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;10x + 5 \leq 14 + x&lt;/math&gt;. <br /> <br /> ==Problem 2==<br /> <br /> An AoPS user is lining up the 5 names of his helpers in his Mock Contest. However, he does not want aops-g5-gethsemanea2 to be first, second, or third in line. How many ways can he arrange the names to still keep himself happy?<br /> <br /> ==Problem 3==<br /> <br /> 30 students lined up to get food from one of the cafeteria booths in their school. How many terminating zeros (zeros at the end) are in the number of arrangements of these 30 students?</div> Aops-g5-gethsemanea2 https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_4 2020 AMC 8 Problems/Problem 4 2020-10-20T10:47:25Z <p>Aops-g5-gethsemanea2: Created page with &quot;the test did not come out yet. please wait...&quot;</p> <hr /> <div>the test did not come out yet. please wait...</div> Aops-g5-gethsemanea2 https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_5 2021 AMC 10A Problems/Problem 5 2020-10-19T20:53:02Z <p>Lnzhonglp: Ok</p> <hr /> <div>Ppkjb</div> Lnzhonglp https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems 2021 AMC 12B Problems 2020-10-18T19:41:19Z <p>Icematrix2: </p> <hr /> <div>{{AMC12 Problems|year=2021|ab=B}}<br /> ==Problem 1==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=B|before=[[2021 AMC 12A Problems]]|after=[[2022 AMC 12A Problems]]}}<br /> <br /> [[Category:AMC 12 Problems]]<br /> {{MAA Notice}}</div> Firebolt360 https://artofproblemsolving.com/wiki/index.php/Netherlands_mathematics_competitions Netherlands mathematics competitions 2020-10-17T08:43:35Z <p>Andyvanh1: Created page with &quot;Dutch Math Olympiad (Nederlandse Wiskundeolympiade)&quot;</p> <hr /> <div>Dutch Math Olympiad (Nederlandse Wiskundeolympiade)</div> Andyvanh1 https://artofproblemsolving.com/wiki/index.php/AoPSSheriff AoPSSheriff 2020-10-17T07:09:00Z <p>Aops-g5-gethsemanea2: </p> <hr /> <div>{{delete|this page does not tell who the AoPSSheriff is. The text here is against the AoPSSheriff}}<br /> THE AOPSSHERIFF IS VERY BAD HE IS SO BAD I DO NOT LIKE HIM HE BANS EVERYONE FOR NO REASON</div> Adolphhitler https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6 2021 AIME II Problems/Problem 6 2020-10-16T19:47:20Z <p>Icematrix2: Created page with &quot;These problems will not be available until the &lt;math&gt;2021&lt;/math&gt; AIME II is released on Thursday, March &lt;math&gt;25, 2021&lt;/math&gt;.&quot;</p> <hr /> <div>These problems will not be available until the &lt;math&gt;2021&lt;/math&gt; AIME II is released on Thursday, March &lt;math&gt;25, 2021&lt;/math&gt;.</div> Icematrix2 https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 2021 AIME II Problems/Problem 5 2020-10-16T19:47:10Z <p>Icematrix2: </p> <hr /> <div>==Problem==<br /> These problems will not be posted until the 2021 AIME II is released on Thursday, March 25, 2021.<br /> ==Solution==<br /> We can't have a solution without a problem.<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=II|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Icematrix2 https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 2021 AIME II Problems/Problem 4 2020-10-16T19:47:03Z <p>Icematrix2: </p> <hr /> <div>==Problem==<br /> These problems will not be posted until the 2021 AIME II is released on Thursday, March 25, 2021.<br /> ==Solution==<br /> We can't have a solution without a problem.<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=II|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Icematrix2 https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_3 2021 AIME II Problems/Problem 3 2020-10-16T19:46:49Z <p>Icematrix2: </p> <hr /> <div>==Problem==<br /> These problems will not be posted until the 2021 AIME II is released on Thursday, March 25, 2021.<br /> ==Solution==<br /> We can't have a solution without a problem.<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=II|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Icematrix2 https://artofproblemsolving.com/wiki/index.php/Homeomorphism Homeomorphism 2020-10-15T01:35:06Z <p>Duck master: replaced nonsense with definition</p> <hr /> <div>In [[abstract algebra]], a homeomorphism is a [[function]] from an object to another that preserves the structure of the two objects. An [[isomorphism]] is a homeomorphism that is also a [[bijection]].<br /> <br /> In [[category theory]], homeomorphisms are represented by arrows from one object to another.<br /> <br /> {{stub}}[[Category:Abstract algebra]]</div> Levizhao https://artofproblemsolving.com/wiki/index.php/Stereographic_projection Stereographic projection 2020-10-15T01:34:47Z <p>Duck master: replaced nonsense with content</p> <hr /> <div>A stereographic projection is a projection from a [[sphere]] to a tangent [[plane]]. Stereographic projections preserve [[angle|angles]].<br /> <br /> To stereographically project a point on a sphere to a plane tangent to its south pole, draw the line from the north pole of the sphere to the point in question. The stereographic projection of this point is then the intersection of this line with the plane. As such, the stereographic projection of the north pole will be undefined.<br /> <br /> {{stub}}[[Category:Geometry]]</div> Levizhao https://artofproblemsolving.com/wiki/index.php/2019_CIME_I_Problems/Problem_7 2019 CIME I Problems/Problem 7 2020-10-13T20:11:57Z <p>Icematrix2: </p> <hr /> <div>Albert, Bob, Carrie, and Douglas are travelling along a road at constant (but not necessarily equal) velocities&lt;math&gt;.&lt;/math&gt; Albert meets Bob at &lt;math&gt;12:00 \text{pm},&lt;/math&gt; Carrie at &lt;math&gt;12:20 \text{pm}&lt;/math&gt; and Douglas at &lt;math&gt;12:32\text{pm}.&lt;/math&gt; Later that same day, Douglas meets Carrie at &lt;math&gt;12:53\text{pm}&lt;/math&gt; and Bob at &lt;math&gt;1:17\text{pm}.&lt;/math&gt; If Bob and Carrie meet &lt;math&gt;m&lt;/math&gt; minutes after noon, compute &lt;math&gt;m&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{CIME box|year=2019|n=I|num-b=6|num-a=8}}<br /> {{MAC Notice}}</div> Icematrix2 https://artofproblemsolving.com/wiki/index.php/2020_SWMC_6_Problems/Problem_1 2020 SWMC 6 Problems/Problem 1 2020-10-13T07:12:46Z <p>Aops-g5-gethsemanea2: </p> <hr /> <div>==Problem==<br /> Give the only positive value of &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;10x + 5 \leq 14 + x&lt;/math&gt;. <br /> <br /> ==Solution==<br /> <br /> To figure out all possible values of &lt;math&gt;x&lt;/math&gt;, we have to solve the inequality.<br /> <br /> First, we have to make sure that only one side has the variable x. To do that, we subtract x from both sides so only the left side has x in it. We have<br /> <br /> &lt;cmath&gt;9x+5\leq14.&lt;/cmath&gt;<br /> <br /> Subtracting 5 from both sides gets &lt;math&gt;9x\leq9&lt;/math&gt;, so &lt;math&gt;x\leq1&lt;/math&gt;. Therefore the only positive value of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;\boxed1&lt;/math&gt;.</div> Aops-g5-gethsemanea2 https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_8 1957 AHSME Problems/Problem 8 2020-10-12T16:22:33Z <p>Angrybird029: </p> <hr /> <div>The numbers &lt;math&gt;x,\,y,\,z&lt;/math&gt; are proportional to &lt;math&gt;2,\,3,\,5&lt;/math&gt;. The sum of &lt;math&gt;x, y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;100&lt;/math&gt;. The number y is given by the equation &lt;math&gt;y = ax - 10&lt;/math&gt;. Then a is: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ \frac{3}{2}\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ \frac{5}{2}\qquad \textbf{(E)}\ 4 &lt;/math&gt;<br /> <br /> ==Solution==<br /> In order to solve the problem, we first need to find each of the three variables. We can use the proportions the problem gives us to find the value of one part, and, by extension, the values of the variables (as &lt;math&gt;x&lt;/math&gt; would have &lt;math&gt;2&lt;/math&gt; parts, &lt;math&gt;y&lt;/math&gt; would have &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; would have &lt;math&gt;5&lt;/math&gt;). One part, after some algebra, equals &lt;math&gt;10&lt;/math&gt;, so &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are &lt;math&gt;20&lt;/math&gt;, &lt;math&gt;30&lt;/math&gt;, and &lt;math&gt;50&lt;/math&gt;, respectively.<br /> <br /> We can plug &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; into the equation given to us: &lt;math&gt;30 = 20a-10&lt;/math&gt;, and then solve to get &lt;math&gt;a = \boxed{\textbf{(A)}2}&lt;/math&gt;.<br /> ==See Also==<br /> {{AHSME box|year=1957|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Angrybird029 https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_7 1957 AHSME Problems/Problem 7 2020-10-12T15:55:23Z <p>Angrybird029: /* See Also */</p> <hr /> <div>== Problem 7==<br /> <br /> The area of a circle inscribed in an equilateral triangle is &lt;math&gt;48\pi&lt;/math&gt;. The perimeter of this triangle is: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 72\sqrt{3} \qquad \textbf{(B)}\ 48\sqrt{3}\qquad \textbf{(C)}\ 36\qquad \textbf{(D)}\ 24\qquad \textbf{(E)}\ 72 &lt;/math&gt; <br /> ==Solution==<br /> &lt;asy&gt;<br /> draw((-3,-sqrt(3))--(3,-sqrt(3))--(0,2sqrt(3))--cycle);<br /> draw(circle((0,0),sqrt(3)));<br /> dot((0,0));<br /> draw((0,0)--(0,-sqrt(3)));<br /> &lt;/asy&gt;<br /> We can see that the radius of the circle is &lt;math&gt;4\sqrt{3}&lt;/math&gt;. We know that the radius is &lt;math&gt;\frac{1}{3}&lt;/math&gt; of each median line of the triangle; each median line is therefore &lt;math&gt;12\sqrt{3}&lt;/math&gt;. Since the median line completes a &lt;math&gt;30&lt;/math&gt;-&lt;math&gt;60&lt;/math&gt;-&lt;math&gt;90&lt;/math&gt; triangle, we can conclude that one of the sides of the triangle is &lt;math&gt;24&lt;/math&gt;. Triple the side length and we get our answer, &lt;math&gt;\boxed{\textbf{(E)}72}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AHSME box|year=1957|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Angrybird029 https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_16 1982 AHSME Problems/Problem 16 2020-10-12T15:26:55Z <p>Ab2024: Created page with &quot;== Solution for Problem 16 == Each exterior unit square which is removed exposes 4 faces of the unit interior squares, so the entire surface area in square meters is &lt;math&gt;6...&quot;</p> <hr /> <div>== Solution for Problem 16 ==<br /> <br /> Each exterior unit square which is removed exposes 4 faces of the unit interior squares, so the entire surface area in square meters is &lt;math&gt;6 \cdot 3^2 - 6 + 24=72.&lt;/math&gt;</div> Ab2024 https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_9 1982 AHSME Problems/Problem 9 2020-10-12T15:24:02Z <p>Ab2024: /* Solution for Problem 9 */</p> <hr /> <div>== Solution for Problem 9 ==<br /> <br /> &lt;asy&gt;<br /> size(350);<br /> defaultpen(fontsize(10));<br /> pair A=origin, O=(10,0), B=(3,0), N=(0,5), C=(3,5), P=(5,0), D=(1,1), G=(9,1), F=(1,0);<br /> draw(G--A--D--cycle, linewidth(0.7));<br /> draw(D--F, linewidth(0.6));<br /> draw(B--C, linewidth(0.5));<br /> draw(A--N, linewidth(0.4));<br /> draw(A--O, linewidth(0.4));<br /> dot(A^^B^^D^^G^^F);<br /> label(&quot;$A$&quot;, D, W);<br /> label(&quot;$B$&quot;, A, dir(0));<br /> label(&quot;$C$&quot;, G, dir(100));<br /> label(&quot;$E$&quot;, B, SE);<br /> label(&quot;$F$&quot;, F, SE);<br /> label(&quot;$(9,1)$&quot;, G, SE);<br /> label(&quot;$(0,0)$&quot;, A, SW);<br /> label(&quot;$8$&quot;, D--G, dir(40));<br /> label(&quot;$y=x/9$&quot;, A--G, SE);<br /> label(&quot;$x=a$&quot;, B--N, SE);<br /> label(&quot;$(1,1)$&quot;, D, NE);&lt;/asy&gt;<br /> The vertical line that divides &lt;math&gt;\triangle ABC&lt;/math&gt; into two equal regions has equation &lt;math&gt;x=a,&lt;/math&gt; as shown in the diagram.<br /> The area of &lt;math&gt;ABC&lt;/math&gt; is half of the height times &lt;math&gt;AC,&lt;/math&gt; so because the Y coordinate of A is 1 and &lt;math&gt;\overline{AF}&lt;/math&gt; is the height, because the difference of the x coordinates between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; is &lt;math&gt;8,&lt;/math&gt; we have &lt;math&gt;[ABC]=1/2 \cdot 8 \cdot 1 = 4.&lt;/math&gt; Thus the two regions must have area &lt;math&gt;2&lt;/math&gt; each.<br /> <br /> Since &lt;math&gt;\triangle ABF&lt;/math&gt; has area &lt;math&gt;1/2,&lt;/math&gt; we know that the portion of &lt;math&gt;\triangle ABC&lt;/math&gt; made by the points &lt;math&gt;A,&lt;/math&gt; &lt;math&gt;B&lt;/math&gt; and the intersection &lt;math&gt;\overline{AF}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; will be less than &lt;math&gt;1/2,&lt;/math&gt; which is less than half of the triangle's area, or 2. Therefore &lt;math&gt;\overline{AF}&lt;/math&gt; is to the left of vertical line &lt;math&gt;x=a&lt;/math&gt; (Passing through point &lt;math&gt;E&lt;/math&gt;).<br /> <br /> The equation of line BC is &lt;math&gt;y=x/9,&lt;/math&gt; and the vertical line &lt;math&gt;x=a&lt;/math&gt; intersects &lt;math&gt;\overline{BC}&lt;/math&gt; at the point &lt;math&gt;(a, a/9).&lt;/math&gt; Because the area of the portion of &lt;math&gt;\triangle ABC&lt;/math&gt; on the right is 2, we have &lt;cmath&gt;2=1/2(1-a/9)(9-a)&lt;/cmath&gt; or &lt;cmath&gt;(9-a)^2=36.&lt;/cmath&gt; Therefore &lt;math&gt;a&gt;0&lt;/math&gt; so &lt;math&gt;a=3=x.&lt;/math&gt;</div> Ab2024 https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25 2021 AMC 10A Problems/Problem 25 2020-10-12T13:58:31Z <p>Objectz: Created page with &quot;This article was made early. Please ignore it until 2021 AMC 10A Problems are released.&quot;</p> <hr /> <div>This article was made early. Please ignore it until 2021 AMC 10A Problems are released.</div> Objectz https://artofproblemsolving.com/wiki/index.php/Situsslotonlineqqmegah Situsslotonlineqqmegah 2020-10-12T09:03:50Z <p>Piphi: Replaced content with &quot;{{delete|advertisement}}&quot;</p> <hr /> <div>{{delete|advertisement}}</div> Situsslotonlineqqmegah https://artofproblemsolving.com/wiki/index.php/2019_CIME_I_Problems/Problem_8 2019 CIME I Problems/Problem 8 2020-10-12T04:28:44Z <p>Icematrix2: </p> <hr /> <div>In parallelogram &lt;math&gt;ABCD,&lt;/math&gt; the circumcircle of &lt;math&gt;\triangle BCD&lt;/math&gt; has center &lt;math&gt;O&lt;/math&gt; and intersects lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F,&lt;/math&gt; respectively&lt;math&gt;.&lt;/math&gt; Let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be the midpoints of &lt;math&gt;AO&lt;/math&gt; and &lt;math&gt;BD,&lt;/math&gt; respectively&lt;math&gt;.&lt;/math&gt; Suppose that &lt;math&gt;PQ=3&lt;/math&gt; and the height from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BD&lt;/math&gt; has length &lt;math&gt;7.&lt;/math&gt; Find the value of &lt;math&gt;BF \cdot DE.&lt;/math&gt;<br /> <br /> ==See also==<br /> {{CIME box|year=2019|n=I|num-b=7|num-a=9}}<br /> {{MAC Notice}}</div> Icematrix2 https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_8 1982 AHSME Problems/Problem 8 2020-10-11T21:47:15Z <p>Ab2024: /*Solutions*/</p> <hr /> <div>We know that &lt;math&gt;n \choose {2}&lt;/math&gt; &lt;math&gt;-&lt;/math&gt; &lt;math&gt; n \choose 1&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt; {n} \choose 3&lt;/math&gt; &lt;math&gt;-&lt;/math&gt; &lt;math&gt;{n} \choose 2,&lt;/math&gt; because they form an arithmetic sequence, and expanding, we have by the definitions in the problem: &lt;cmath&gt;\frac{n(n-1)(n-2)(n-3)...}{2(n-2)(n-3)...}-n=\frac{n(n-1)(n-2)(n-3)(n-4)...}{6((n-3)(n-4)...)}-\frac{n(n-1)(n-2)(n-3)...}{2(n-2)(n-3)...}.&lt;/cmath&gt;<br /> <br /> Canceling out the &lt;math&gt;(n-2)!&lt;/math&gt; and the &lt;math&gt;(n-3)!&lt;/math&gt; from each side of the equals sign, we have &lt;math&gt;\frac{n(n-1)}{2}-n = \frac {n(n-1)(n-2)}{6}-\frac{n(n-1)}{2}.&lt;/math&gt; Getting rid of the fractions by cross multiplication, and getting n on one side, we have &lt;math&gt;n^3 - 9n^2 + 14n = 0,&lt;/math&gt; and we can factor out the n, so n(n^2-9n+14)=0, and we are looking for two integers x and y such that &lt;math&gt;x+y=-9&lt;/math&gt; and &lt;math&gt;xy=14.&lt;/math&gt; By guess and check, our integers are -7 and -2, so &lt;math&gt;n(n-7)(n-2)=0!!!&lt;/math&gt; According to the problem, &lt;math&gt;n&gt;3,&lt;/math&gt; so we have n=7 or 2, thus &lt;math&gt;\boxed{\left(B\right)n=7}&lt;/math&gt;<br /> <br /> ~ab2024</div> Ab2024 https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_10 1982 AHSME Problems/Problem 10 2020-10-11T16:42:46Z <p>Ab2024: /* Problem 10 Solution */</p> <hr /> <div>== Problem 10 Solution ==<br /> <br /> Since &lt;math&gt;BO&lt;/math&gt; and &lt;math&gt;CO&lt;/math&gt; are angle bisectors of angles &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; respectively, &lt;math&gt;\angle MBO = \angle OBC&lt;/math&gt; and similarly &lt;math&gt;\angle NCO = \angle OCB&lt;/math&gt;. Because &lt;math&gt;MN&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; are parallel, &lt;math&gt;\angle OBC = \angle MOB&lt;/math&gt; and &lt;math&gt;\angle NOC = \angle OCB&lt;/math&gt; by corresponding angles. This relation makes &lt;math&gt;\triangle MOB&lt;/math&gt; and &lt;math&gt;\triangle NOC&lt;/math&gt; isosceles. This makes &lt;math&gt;MB = MO&lt;/math&gt; and &lt;math&gt;NO = NC&lt;/math&gt;. Therefore the perimeter of &lt;math&gt;\triangle AMN&lt;/math&gt; is &lt;math&gt;12 + 18 = 30&lt;/math&gt;.</div> Ab2024 https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_24 1982 AHSME Problems/Problem 24 2020-10-11T16:41:58Z <p>Ab2024: /* Solution 1 */</p> <hr /> <div>=== Solution 1===<br /> <br /> &lt;asy&gt; defaultpen(fontsize(10)); real r=sqrt(22); pair B=origin, A=16*dir(60), C=(16,0), D=(10-r,0), E=(10+r,0), F=C+1*dir(120), G=C+14*dir(120), H=13*dir(60), J=6*dir(60), O=circumcenter(G,H,J); dot(A^^B^^C^^D^^E^^F^^G^^H^^J); draw(Circle(O, abs(O-D))^^A--B--C--cycle, linewidth(0.7)); label(&quot;$A$&quot;, A, N); label(&quot;$B$&quot;, B, dir(210)); label(&quot;$C$&quot;, C, dir(330)); label(&quot;$D$&quot;, D, SW); label(&quot;$E$&quot;, E, SE); label(&quot;$F$&quot;, F, dir(170)); label(&quot;$G$&quot;, G, dir(250)); label(&quot;$H$&quot;, H, SE); label(&quot;$J$&quot;, J, dir(0)); label(&quot;2&quot;, A--G, dir(30)); label(&quot;13&quot;, F--G, dir(180+30)); label(&quot;1&quot;, F--C, dir(30)); label(&quot;7&quot;, H--J, dir(-30));&lt;/asy&gt;<br /> <br /> Let &lt;math&gt;AH=y, BD=a, DE=x,&lt;/math&gt; and &lt;math&gt;EC=b.&lt;/math&gt; Because &lt;math&gt;AG = 2, GF = 13, HJ = 7&lt;/math&gt; and &lt;math&gt;FC = 1&lt;/math&gt; and they sum to 16, the length of each side of the equilateral triangle will be 16. By Power of a Point on A, we have &lt;math&gt;y(y + 7) = 2(2 + 13),&lt;/math&gt; or expanding and factoring, &lt;math&gt;0 =y^2 + 7y - 30 = (y - 3)(y + 10).&lt;/math&gt; Therefore, y must equal 3 as this triangle must have positive side lengths.<br /> <br /> Hence, &lt;math&gt;BJ=16-(3+7)=6.&lt;/math&gt;<br /> <br /> Power of a point on C yields &lt;math&gt;b(b + x) = 1(1 + 13) = 14,&lt;/math&gt; and power of a point on B yields a(a + x) = 6(6 + 7) = 78. Also, we know that because a, b, and x are non-intersecting segments of the side of the triangle, a + b + x = 16. <br /> <br /> So, we have a system of equations:<br /> <br /> &lt;cmath&gt;a^2 + ax = 78,&lt;/cmath&gt;<br /> &lt;cmath&gt;b^2 + bx = 14,&lt;/cmath&gt;<br /> &lt;math&gt;a + b + x =16.&lt;/math&gt;<br /> <br /> If we subtract the second equation from the first, it appears to be a difference of squares! And we like differences of squares, because we can factor! Subtracting yields &lt;math&gt;a^2 - b^2 + (a - b)x = (a-b)(a+b)(a-b)x=(a - b)(a + b + x).&lt;/math&gt; Because we know that &lt;math&gt;a+b+x=16,&lt;/math&gt; inputting this we have that &lt;math&gt;(a - b)16 = 78 - 14 = 64.&lt;/math&gt; Therefore &lt;math&gt;a-b=4.&lt;/math&gt; Then adding this to the third equation, 2a + x = 20, so &lt;math&gt;a = 10 - (x/2).&lt;/math&gt; Substituting this into #1, we can now solve for x, and we have a difference of squares, or &lt;math&gt;100-x^2/4=78.&lt;/math&gt; This yields &lt;math&gt;x^2/4=22,&lt;/math&gt; so &lt;math&gt;x/2=\sqrt {22}&lt;/math&gt; and &lt;math&gt;x= \left(A\right)2\sqrt{22}!&lt;/math&gt;<br /> <br /> Further clarifications: If you are unsure why we didn't solve for a and b, it is because it is more efficient to use clever manipulations like the ones we used in this problem. We were looking for sums of values we already know, so we indirectly implied dividing &lt;math&gt;a^2 - b^2 + (a - b)x&lt;/math&gt; by &lt;math&gt;a+b+x&lt;/math&gt; to get &lt;math&gt;a-b,&lt;/math&gt; conveniently, then everything fell into place with substitution.<br /> <br /> <br /> ~*Solution by ab2024</div> Ab2024 https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_3 2020 AMC 8 Problems/Problem 3 2020-10-11T02:10:26Z <p>Samrocksnature: Created page with &quot;Btw, this isn't cheating and it's totally okay if you've come here on accident! The MAA cannot track your devices so don't feel threatened. But do remember that this test hasn...&quot;</p> <hr /> <div>Btw, this isn't cheating and it's totally okay if you've come here on accident! The MAA cannot track your devices so don't feel threatened. But do remember that this test hasn't even come out yet...</div> Samrocksnature https://artofproblemsolving.com/wiki/index.php/2017_Indonesia_MO_Problems/Problem_2 2017 Indonesia MO Problems/Problem 2 2020-10-10T05:18:37Z <p>Duck master: created page w/ solution</p> <hr /> <div>Five people are gathered in a meeting. Some pairs of people shakes hands. An ordered triple of people &lt;math&gt;(A,B,C)&lt;/math&gt; is a trio if one of the following is true:<br /> <br /> A shakes hands with B, and B shakes hands with C, or<br /> A doesn't shake hands with B, and B doesn't shake hands with C.<br /> If we consider &lt;math&gt;(A,B,C)&lt;/math&gt; and &lt;math&gt;(C,B,A)&lt;/math&gt; as the same trio, find the minimum possible number of trios.<br /> <br /> == Solution ==<br /> <br /> Note that the number of trios that &lt;math&gt;\{A, B, C\}&lt;/math&gt; constitute (including permutations) is one, plus two if all three shake hands with each other or all three do not shake hands with each other. However, we can arrange for no triple of people to all shake hands or all not shake hands (the minimum possible &quot;extras&quot;), by numbering the people by integers modulo &lt;math&gt;5&lt;/math&gt; and forcing &lt;math&gt;x&lt;/math&gt; to shake hands with &lt;math&gt;x + 1&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt;, and that only. Then there are &lt;math&gt;\binom{5}{3} = 10&lt;/math&gt; triples overall, the minimum possible, and we are done.<br /> <br /> == See also ==<br /> <br /> {{Indonesia MO box|year=2017|before=First Problem|num-a=2}}<br /> <br /> [[Category:Combinatorics Problems]]</div> Duck master https://artofproblemsolving.com/wiki/index.php/2000_Pan_African_MO_Problems/Problem_5 2000 Pan African MO Problems/Problem 5 2020-10-10T05:07:07Z <p>Duck master: created page with solution (add diagram?); NOTE: SOLUTION IS INCOMPLETE</p> <hr /> <div>Let &lt;math&gt;\gamma&lt;/math&gt; be circle and let &lt;math&gt;P&lt;/math&gt; be a point outside &lt;math&gt;\gamma&lt;/math&gt;. Let &lt;math&gt;PA&lt;/math&gt; and &lt;math&gt;PB&lt;/math&gt; be the tangents from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;\gamma&lt;/math&gt; (where &lt;math&gt;A, B \in \gamma&lt;/math&gt;). A line passing through &lt;math&gt;P&lt;/math&gt; intersects &lt;math&gt;\gamma&lt;/math&gt; at points &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;. Let &lt;math&gt;S&lt;/math&gt; be a point on &lt;math&gt;\gamma&lt;/math&gt; such that &lt;math&gt;BS \parallel QR&lt;/math&gt;. Prove that &lt;math&gt;SA&lt;/math&gt; bisects &lt;math&gt;QR&lt;/math&gt;.<br /> <br /> == Solution ==<br /> <br /> There is a projective transformation which maps &lt;math&gt;\gamma&lt;/math&gt; to a circle and that maps the midpoint of &lt;math&gt;QR&lt;/math&gt; to its center (EXPAND); therefore, we may assume without loss of generality that the midpoint of &lt;math&gt;QR&lt;/math&gt; is the center of &lt;math&gt;\gamma&lt;/math&gt;. But then &lt;math&gt;B&lt;/math&gt; is the reflection of &lt;math&gt;A&lt;/math&gt; across &lt;math&gt;QR&lt;/math&gt;, so that &lt;math&gt;S&lt;/math&gt; is the antipode of &lt;math&gt;A&lt;/math&gt; on &lt;math&gt;\gamma&lt;/math&gt;, and we are done.<br /> <br /> == See also ==<br /> <br /> {{Pan African MO box<br /> |year=2000<br /> |before=First Problem<br /> |num-a=2<br /> }}<br /> <br /> [[Category:Geometry Problems]] {{stub}}</div> Duck master https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Answer_Key 2021 AMC 10A Answer Key 2020-10-08T03:20:27Z <p>Soumyadeepbhattacharjee: Created page with &quot;1.A&quot;</p> <hr /> <div>1.A</div> Soumyadeepbhattacharjee https://artofproblemsolving.com/wiki/index.php/1998_JBMO_Problems/Problem_4 1998 JBMO Problems/Problem 4 2020-10-07T05:13:04Z <p>Duck master: deleted link to fifth problem</p> <hr /> <div>Do there exist 16 three digit numbers, using only three different digits in all, so that the all numbers give different residues when divided by 16?<br /> <br /> == Solution ==<br /> <br /> Let the three different digits be &lt;math&gt;a, b, c&lt;/math&gt;. <br /> <br /> If &lt;math&gt;a, b, c&lt;/math&gt; all have the same parity, then all sixteen numbers will also have the same parity. But then they can only cover at most 8 residues modulo 16, so they cannot have distinct residues by the [[pigeonhole principle]].<br /> <br /> Suppose that &lt;math&gt;a, b, c&lt;/math&gt; do not all have the same parity. If two are even and one is odd, then twelve of the eighteen possible three-digit numbers formed with &lt;math&gt;a, b, c&lt;/math&gt; will be even and six will be odd. But this means that there are not enough odd numbers to fill the 8 odd residues modulo 16, so it will not be possible to select 16 three digit numbers with this property in this case. <br /> <br /> Similarly, if two are odd and one is even, then twelve of the eighteen possible three-digit numbers formed with &lt;math&gt;a, b, c&lt;/math&gt; will be odd and six will be even. But this means that there are not enough even numbers to fill the 8 even residues modulo 16, so it will not be possible to select 16 three digit numbers with this property in this case.<br /> <br /> Therefore, we conclude that it is impossible to select 16 such numbers.<br /> <br /> == See also ==<br /> <br /> {{JBMO box|year=1998|num-b=3|after = Last Problem|five=}}<br /> <br /> [[Category:JBMO]][[Category: Number Theory Problems]]</div> Duck master https://artofproblemsolving.com/wiki/index.php/2019_CIME_I_Problems/Problem_9 2019 CIME I Problems/Problem 9 2020-10-06T20:25:51Z <p>Icematrix2: </p> <hr /> <div>Let &lt;math&gt;\text{N}&lt;/math&gt; denote the number of strictly increasing sequences of positive integers &lt;math&gt;a_1,a_2,\cdots, a_{19}&lt;/math&gt; satisfying the following two rules&lt;math&gt;:&lt;/math&gt;<br /> * &lt;math&gt;a_1=1&lt;/math&gt; and &lt;math&gt;a_{19}=361,&lt;/math&gt;<br /> * for any &lt;math&gt;i \neq j,&lt;/math&gt; if &lt;math&gt;b_{ij}&lt;/math&gt; is the &lt;math&gt;(i \cdot j)^{\text{th}}&lt;/math&gt; number not in the sequence&lt;math&gt;,&lt;/math&gt; then &lt;math&gt;(a_i-b_{ij})(a_j-b_{ij})&lt;0.&lt;/math&gt;<br /> Find the largest positive integer &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;2^k&lt;/math&gt; divides &lt;math&gt;\text{N}.&lt;/math&gt;<br /> <br /> =Solution 1=<br /> {{solution}}<br /> <br /> ==See also==<br /> {{CIME box|year=2019|n=I|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Icematrix2 https://artofproblemsolving.com/wiki/index.php/2019_CIME_I_Problems/Problem_10 2019 CIME I Problems/Problem 10 2020-10-06T20:21:51Z <p>Icematrix2: </p> <hr /> <div>Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; be real numbers such that &lt;math&gt;x^2=3-2yz&lt;/math&gt;, &lt;math&gt;y^2=4-2xz&lt;/math&gt;, and &lt;math&gt;z^2=5-2xy&lt;/math&gt;. The value of &lt;math&gt;x^2+y^2+z^2&lt;/math&gt; can be written as &lt;math&gt;a+\frac{b}{\sqrt c}&lt;/math&gt; for positive integers &lt;math&gt;a, b, c&lt;/math&gt;, where &lt;math&gt;c&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;a+b+c&lt;/math&gt;.<br /> <br /> =Solution 1=<br /> Let &lt;math&gt;n&lt;/math&gt; be our answer. Notice &lt;math&gt;(y-z)^2 = x^2 + y^2 + z^2 - (x2 + 2yz)= n-3&lt;/math&gt;. Similarly, &lt;math&gt;(x-z)^2 = n-4&lt;/math&gt; and &lt;math&gt;(y-x)^2 = n-5&lt;/math&gt;. Now notice that since &lt;math&gt;y-z = (x-z)+(y-x)&lt;/math&gt;, we have &lt;math&gt;\sqrt{n-3}=\sqrt{n-4}+\sqrt{n-5} \implies n^2-12n+36=4n^2-36n+80&lt;/math&gt; so &lt;math&gt;3n^2-24n+44=0&lt;/math&gt; and &lt;math&gt;n=4 \pm \frac{2}{\sqrt 3}&lt;/math&gt;. The answer is &lt;math&gt;\boxed 9&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{CIME box|year=2019|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAC Notice}}</div> Icematrix2 https://artofproblemsolving.com/wiki/index.php/2020_SWMC_6 2020 SWMC 6 2020-10-05T08:10:30Z <p>Aops-g5-gethsemanea2: </p> <hr /> <div>Here are problems and solutions for the 2020 SWMC 6 by aops-g5-gethsemanea2.<br /> <br /> <br /> [[2020 SWMC 6 Problems]]<br /> <br /> [[2020 SWMC 6 Answer Key]]<br /> <br /> [[2020 SWMC 6 Problems/Problem 1]]<br /> <br /> [[2020 SWMC 6 Problems/Problem 2]]<br /> <br /> [[2020 SWMC 6 Problems/Problem 3]]</div> Aops-g5-gethsemanea2 https://artofproblemsolving.com/wiki/index.php/SWMC_6_Problems_and_Solutions SWMC 6 Problems and Solutions 2020-10-05T08:09:21Z <p>Aops-g5-gethsemanea2: Created page with &quot;Here are the problems and solutions for the SWMC 6. 2020 SWMC 6&quot;</p> <hr /> <div>Here are the problems and solutions for the SWMC 6.<br /> <br /> [[2020 SWMC 6]]</div> Aops-g5-gethsemanea2 https://artofproblemsolving.com/wiki/index.php/Solution_Writing_Mathematics_Competitions Solution Writing Mathematics Competitions 2020-10-05T08:08:37Z <p>Aops-g5-gethsemanea2: Created page with &quot;The Solution Writing Mathematics Competitions (SWMC) is a series of competitions that helps students with their solution writing skills, since grading is not just based on get...&quot;</p> <hr /> <div>The Solution Writing Mathematics Competitions (SWMC) is a series of competitions that helps students with their solution writing skills, since grading is not just based on getting the correct answer.<br /> <br /> See the link below for sets of problems and answers so you can practice.<br /> <br /> [[SWMC 6 Problems and Solutions]]</div> Aops-g5-gethsemanea2