https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Kante314&feedformat=atom AoPS Wiki - User contributions [en] 2021-10-26T08:13:43Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_1&diff=163779 2014 AMC 10A Problems/Problem 1 2021-10-20T01:17:11Z <p>Kante314: </p> <hr /> <div>==Problem ==<br /> What is &lt;math&gt; 10\cdot\left(\tfrac{1}{2}+\tfrac{1}{5}+\tfrac{1}{10}\right)^{-1}? &lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{25}{2} \qquad\textbf{(D)}\ \frac{170}{3}\qquad\textbf{(E)}\ 170&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> We have &lt;cmath&gt;10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}&lt;/cmath&gt;<br /> Making the denominators equal gives<br /> &lt;cmath&gt;\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies 10\cdot\left(\frac{8}{10}\right)^{-1}&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies 10\cdot\left(\frac{4}{5}\right)^{-1}&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies 10\cdot\frac{5}{4}&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies \frac{50}{4}&lt;/cmath&gt;<br /> Finally, simplifying gives<br /> &lt;cmath&gt;\implies \boxed{\textbf{(C)}\ \frac{25}{2}}&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> We have<br /> &lt;cmath&gt;\left(\frac{1}{10}\right)^{-1}\cdot \left(\frac{1}{2} + \frac{1}{5} + \frac{1}{10}\right)^{-1}&lt;/cmath&gt;By Distributive Property,<br /> &lt;cmath&gt;\left(\frac{1}{20}+\frac{1}{50}+\frac{1}{100}\right)^{-1}&lt;/cmath&gt;Now, we want to find the least common multiple of &lt;math&gt;20, 50,&lt;/math&gt; and &lt;math&gt;100,&lt;/math&gt; so<br /> &lt;cmath&gt;\text{lcm}(20,50,100)=\text{lcm}(2^2 \cdot 5,2 \cdot 5^2,2^2 \cdot 5^2)=2^2 \cdot 5^2=100&lt;/cmath&gt;Converting everything to a denominator of &lt;math&gt;100,&lt;/math&gt;<br /> &lt;cmath&gt;\left(\frac{5}{100}+\frac{2}{100}+\frac{1}{100}\right)^{-1}=\left(\frac{8}{100}\right)^{-1}=\frac{100}{8}&lt;/cmath&gt;Now, we use Euclidean Algorithm, to find if this fraction is reducible, so<br /> &lt;cmath&gt;\gcd(100,8)=\gcd(12,8)=\gcd(4,8)=\gcd(4,4)&lt;/cmath&gt;Thus, both the numerator and denominator are divisible by &lt;math&gt;4,&lt;/math&gt; so<br /> &lt;cmath&gt;\frac{100}{8} \cdot \frac{4}{4}=\frac{100}{4} \cdot \frac{4}{8}=25 \cdot \frac{1}{2}=\boxed{\frac{25}{2}}&lt;/cmath&gt;<br /> <br /> - kante314<br /> <br /> ==Video Solution==<br /> https://youtu.be/QvkvhIMpXz8<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|before=First Problem|num-a=2}}<br /> {{AMC12 box|year=2014|ab=A|before=First Problem|num-a=2}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Prealgebra Problems]]</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_35&diff=162740 1950 AHSME Problems/Problem 35 2021-09-26T06:04:36Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AC=24&lt;/math&gt; inches, &lt;math&gt;BC=10&lt;/math&gt; inches, &lt;math&gt;AB=26&lt;/math&gt; inches. The radius of the inscribed circle is:<br /> <br /> &lt;math&gt;\textbf{(A)}\ 26\text{ in} \qquad<br /> \textbf{(B)}\ 4\text{ in} \qquad<br /> \textbf{(C)}\ 13\text{ in} \qquad<br /> \textbf{(D)}\ 8\text{ in} \qquad<br /> \textbf{(E)}\ \text{None of these}&lt;/math&gt;<br /> <br /> ==Solution==<br /> The inradius is equal to the area divided by semiperimeter. The area is &lt;math&gt;\frac{(10)(24)}{2} = 120&lt;/math&gt; because it's a right triangle, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is &lt;math&gt;30&lt;/math&gt;. Therefore the inradius is &lt;math&gt;\boxed{\textbf{(B)}\ 4}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Since this is a right triangle, we have<br /> &lt;cmath&gt;\frac{a+b-c}{2}=\boxed{4}&lt;/cmath&gt;<br /> <br /> - kante314<br /> <br /> ==See Also==<br /> {{AHSME 50p box|year=1950|num-b=34|num-a=36}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=User:Aops-g5-gethsemanea2&diff=161221 User:Aops-g5-gethsemanea2 2021-08-30T14:14:17Z <p>Kante314: /* User Count */</p> <hr /> <div><br /> {{Template:Aops-g5-gethsemanea2/Header}}<br /> <br /> {{User:CreativeHedgehog/Templates/SocialMedia}}<br /> <br /> {{shortcut|[[AGG]]}}<br /> <br /> Welcome to my user page! <br /> <br /> <br /> ==About Me==<br /> <br /> &lt;div style=&quot;padding-left:16px; padding-right:16px; border:15px #009fad; background-color:#eaeaea; padding-top:10px; padding-bottom:10px; max-height:200px; overflow:auto; display:block; width:100%;&quot;&gt;<br /> <br /> Aops-g5-gethsemanea2 is just an &lt;b&gt;Algebra A&lt;/b&gt; user &lt;s&gt;who is not as legendary as Piphi&lt;/s&gt;. <br /> <br /> Like CreativeHedgehog, Aops-g5-gethsemanea2 is making a lot of projects on this wiki including the Wiki Math Games and his very own main page.<br /> <br /> Aops-g5-gethsemanea2 got an 18 in the AMC 8 and a three-time national qualifier in mathleague.org.<br /> <br /> Aops-g5-gethsemanea2 is an International Countdown Champion in mathleague.org elementary 2021.<br /> <br /> ==Acknowledgements==<br /> Thanks to the following users!<br /> <br /> 1. Piphi for the inspiration<br /> <br /> 2. CreativeHedgehog for the social media buttons, the box code (the one used in his Viewer Count and in my User Count) and supporting this User Page<br /> &lt;/div&gt;<br /> <br /> ==User Count==<br /> &lt;div style=&quot;padding-left:16px; padding-right:16px; border:15px #1b365d dashed; background-color:white; padding-top:10px; padding-bottom:10px; max-height:200px; overflow:auto; display:block; width:100%;&quot;&gt;<br /> <br /> If this is the first time you visited this page, increase the following number by 1 and change the username written to yours:<br /> <br /> &lt;b&gt;&lt;i&gt;Latest user: kante314&lt;/i&gt;&lt;/b&gt;<br /> <br /> &lt;div style=&quot;font-size:10em; text-align:center;&quot;&gt;&lt;b&gt;24&lt;/b&gt;&lt;/div&gt;<br /> <br /> [https://artofproblemsolving.com/wiki/index.php?title=User:Aops-g5-gethsemanea2&amp;action=edit&amp;section=3 (Edit this User Count)]<br /> <br /> [https://artofproblemsolving.com/online/?login=1 (Log in Here)]<br /> [https://artofproblemsolving.com/community/logout (Log out Here)]<br /> <br /> &lt;/div&gt;<br /> <br /> <br /> <br /> &lt;div style=&quot;padding-left:16px; padding-right:16px; border:15px #009fad; background-color:#ffe4e1; padding-top:10px; padding-bottom:10px; max-height:200px; overflow:auto; display:block; width:100%;&quot;&gt;<br /> <br /> =WARNINGS!=<br /> <br /> ==WARNING 1==<br /> This wiki page will NOT work if you use a mobile device with the social media buttons.<br /> <br /> If you want it to work on a mobile device, you may put everything in a div and then align it to the center.<br /> <br /> Then make the width 50%.<br /> <br /> Example: Put the following text before every page:<br /> <br /> &lt;nowiki&gt;<br /> <br /> &lt;div style=&quot;position:fixed; left:50%; transform:translateX(-50%); width:50%;&quot;&gt;<br /> <br /> &lt;/nowiki&gt;<br /> <br /> And this after it:<br /> <br /> &lt;nowiki&gt;<br /> &lt;/div&gt;<br /> &lt;/nowiki&gt;<br /> <br /> I just didn't do that because I had 40 pages to edit that out with. [https://artofproblemsolving.com/wiki/index.php/User:CreativeHedgehog CreativeHedgehog] 22:32, 4 August 2020 (EDT)<br /> <br /> ==WARNING 2==<br /> If you followed the instruction above, then it will not work this time on a computer. Remove the div at the start and the end to make it back to normal.<br /> <br /> ==WARNING 3==<br /> This will only work on the Aops2 skin style. [https://artofproblemsolving.com/wiki/index.php/User:CreativeHedgehog CreativeHedgehog] 21:56, 8 August 2020 (EDT)<br /> <br /> =REMINDER=<br /> Most of these pages require you to sign in to edit the pages, especially the Wiki Math Games.<br /> &lt;/div&gt;</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_2&diff=161136 2010 AIME II Problems/Problem 2 2021-08-29T01:04:29Z <p>Kante314: /* Solution */</p> <hr /> <div>== Problem 2 ==<br /> A point &lt;math&gt;P&lt;/math&gt; is chosen at random in the interior of a unit square &lt;math&gt;S&lt;/math&gt;. Let &lt;math&gt;d(P)&lt;/math&gt; denote the distance from &lt;math&gt;P&lt;/math&gt; to the closest side of &lt;math&gt;S&lt;/math&gt;. The probability that &lt;math&gt;\frac{1}{5}\le d(P)\le\frac{1}{3}&lt;/math&gt; is equal to &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> <br /> == Solution ==<br /> <br /> Any point outside the square with side length &lt;math&gt;\frac{1}{3}&lt;/math&gt; that has the same center and orientation as the unit square and inside the square with side length &lt;math&gt;\frac{3}{5}&lt;/math&gt; that has the same center and orientation as the unit square has &lt;math&gt;\frac{1}{5}\le d(P)\le\frac{1}{3}&lt;/math&gt;.<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(1mm);<br /> defaultpen(linewidth(.8pt));<br /> <br /> draw((0,0)--(0,30)--(30,30)--(30,0)--cycle);<br /> draw((6,6)--(6,24)--(24,24)--(24,6)--cycle);<br /> draw((10,10)--(10,20)--(20,20)--(20,10)--cycle);<br /> fill((6,6)--(6,24)--(24,24)--(24,6)--cycle,gray);<br /> fill((10,10)--(10,20)--(20,20)--(20,10)--cycle,white);<br /> <br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Since the area of the unit square is &lt;math&gt;1&lt;/math&gt;, the probability of a point &lt;math&gt;P&lt;/math&gt; with &lt;math&gt;\frac{1}{5}\le d(P)\le\frac{1}{3}&lt;/math&gt; is the area of the shaded region, which is the difference of the area of two squares.<br /> <br /> &lt;math&gt;\left(\frac{3}{5}\right)^2-\left(\frac{1}{3}\right)^2=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}&lt;/math&gt;<br /> <br /> Thus, the answer is &lt;math&gt;56 + 225 = \boxed{281}.&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> First, let's figure out &lt;math&gt;d(P) \geq \frac{1}{3}&lt;/math&gt; which is&lt;cmath&gt;\left(\frac{3}{5}\right)^2=\frac{9}{25}.&lt;/cmath&gt;Then, &lt;math&gt;d(P) \geq \frac{1}{5}&lt;/math&gt; is a square inside &lt;math&gt;d(P) \geq \frac{1}{3}&lt;/math&gt;, so&lt;cmath&gt;\left(\frac{1}{3}\right)^2=\frac{1}{9}.&lt;/cmath&gt;Therefore, the probability that &lt;math&gt;\frac{1}{5}\le d(P)\le\frac{1}{3}&lt;/math&gt; is&lt;cmath&gt;\frac{9}{25}-\frac{1}{9}=\frac{56}{225}&lt;/cmath&gt;So, the answer is &lt;math&gt;56+225=\boxed{281}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2010|num-b=1|num-a=3|n=II}}<br /> {{MAA Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_2&diff=161135 2010 AIME II Problems/Problem 2 2021-08-29T01:03:50Z <p>Kante314: </p> <hr /> <div>== Problem 2 ==<br /> A point &lt;math&gt;P&lt;/math&gt; is chosen at random in the interior of a unit square &lt;math&gt;S&lt;/math&gt;. Let &lt;math&gt;d(P)&lt;/math&gt; denote the distance from &lt;math&gt;P&lt;/math&gt; to the closest side of &lt;math&gt;S&lt;/math&gt;. The probability that &lt;math&gt;\frac{1}{5}\le d(P)\le\frac{1}{3}&lt;/math&gt; is equal to &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> <br /> == Solution ==<br /> <br /> Any point outside the square with side length &lt;math&gt;\frac{1}{3}&lt;/math&gt; that has the same center and orientation as the unit square and inside the square with side length &lt;math&gt;\frac{3}{5}&lt;/math&gt; that has the same center and orientation as the unit square has &lt;math&gt;\frac{1}{5}\le d(P)\le\frac{1}{3}&lt;/math&gt;.<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(1mm);<br /> defaultpen(linewidth(.8pt));<br /> <br /> draw((0,0)--(0,30)--(30,30)--(30,0)--cycle);<br /> draw((6,6)--(6,24)--(24,24)--(24,6)--cycle);<br /> draw((10,10)--(10,20)--(20,20)--(20,10)--cycle);<br /> fill((6,6)--(6,24)--(24,24)--(24,6)--cycle,gray);<br /> fill((10,10)--(10,20)--(20,20)--(20,10)--cycle,white);<br /> <br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Since the area of the unit square is &lt;math&gt;1&lt;/math&gt;, the probability of a point &lt;math&gt;P&lt;/math&gt; with &lt;math&gt;\frac{1}{5}\le d(P)\le\frac{1}{3}&lt;/math&gt; is the area of the shaded region, which is the difference of the area of two squares.<br /> <br /> &lt;math&gt;\left(\frac{3}{5}\right)^2-\left(\frac{1}{3}\right)^2=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}&lt;/math&gt;<br /> <br /> Thus, the answer is &lt;math&gt;56 + 225 = \boxed{281}.&lt;/math&gt;<br /> <br /> == Solution ==<br /> First, let's figure out &lt;math&gt;d(P) \geq \frac{1}{3}&lt;/math&gt; which is&lt;cmath&gt;\left(\frac{3}{5}\right)^2=\frac{9}{25}.&lt;/cmath&gt;Then, &lt;math&gt;d(P) \geq \frac{1}{5}&lt;/math&gt; is a square inside &lt;math&gt;d(P) \geq \frac{1}{3}&lt;/math&gt;, so&lt;cmath&gt;\left(\frac{1}{3}\right)^2=\frac{1}{9}.&lt;/cmath&gt;Therefore, the probability that &lt;math&gt;\frac{1}{5}\le d(P)\le\frac{1}{3}&lt;/math&gt; is&lt;cmath&gt;\frac{9}{25}-\frac{1}{9}=\frac{56}{225}&lt;/cmath&gt;So, the answer is &lt;math&gt;56+225=\boxed{281}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2010|num-b=1|num-a=3|n=II}}<br /> {{MAA Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=1984_USAMO_Problems/Problem_1&diff=161134 1984 USAMO Problems/Problem 1 2021-08-29T00:58:36Z <p>Kante314: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> In the polynomial &lt;math&gt;x^4 - 18x^3 + kx^2 + 200x - 1984 = 0&lt;/math&gt;, the product of &lt;math&gt;2&lt;/math&gt; of its roots is &lt;math&gt;- 32&lt;/math&gt;. Find &lt;math&gt;k&lt;/math&gt;.<br /> <br /> === Solution 1 (ingenious)===<br /> <br /> <br /> Using Vieta's formulas, we have: <br /> <br /> &lt;cmath&gt;\begin{align*}a+b+c+d &amp;= 18,\\ ab+ac+ad+bc+bd+cd &amp;= k,\\ abc+abd+acd+bcd &amp;=-200,\\ abcd &amp;=-1984.\\ \end{align*}&lt;/cmath&gt; <br /> <br /> <br /> From the last of these equations, we see that &lt;math&gt;cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62&lt;/math&gt;. Thus, the second equation becomes &lt;math&gt;-32+ac+ad+bc+bd+62=k&lt;/math&gt;, and so &lt;math&gt;ac+ad+bc+bd=k-30&lt;/math&gt;. The key insight is now to factor the left-hand side as a product of two binomials: &lt;math&gt;(a+b)(c+d)=k-30&lt;/math&gt;, so that we now only need to determine &lt;math&gt;a+b&lt;/math&gt; and &lt;math&gt;c+d&lt;/math&gt; rather than all four of &lt;math&gt;a,b,c,d&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;p=a+b&lt;/math&gt; and &lt;math&gt;q=c+d&lt;/math&gt;. Plugging our known values for &lt;math&gt;ab&lt;/math&gt; and &lt;math&gt;cd&lt;/math&gt; into the third Vieta equation, &lt;math&gt;-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)&lt;/math&gt;, we have &lt;math&gt;-200 = -32(c+d) + 62(a+b) = 62p-32q&lt;/math&gt;. Moreover, the first Vieta equation, &lt;math&gt;a+b+c+d=18&lt;/math&gt;, gives &lt;math&gt;p+q=18&lt;/math&gt;. Thus we have two linear equations in &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;, which we solve to obtain &lt;math&gt;p=4&lt;/math&gt; and &lt;math&gt;q=14&lt;/math&gt;. <br /> <br /> Therefore, we have &lt;math&gt;(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30&lt;/math&gt;, yielding &lt;math&gt;k=4\cdot 14+30 = \boxed{86}&lt;/math&gt;.<br /> <br /> === Solution 2 (cool)===<br /> <br /> We start as before: &lt;math&gt;ab=-32&lt;/math&gt; and &lt;math&gt;cd=62&lt;/math&gt;. We now observe that a and b must be the roots of a quadratic, &lt;math&gt;x^2+rx-32&lt;/math&gt;, where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic &lt;math&gt;x^2+sx+62&lt;/math&gt;. <br /> <br /> Now <br /> <br /> &lt;cmath&gt; \begin{align*}x^4-18x^3+kx^2+200x-1984 =&amp; (x^2+rx-32)(x^2+sx+62)\\ =&amp; x^4+(r+s)x^3+(62-32+rs)x^2\\<br /> &amp;+(62s-32r)x-1984.\end{align*} &lt;/cmath&gt;<br /> <br /> Equating the coefficients of &lt;math&gt;x^3&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of &lt;math&gt;x^2&lt;/math&gt; and get &lt;math&gt;k=\boxed{86}.&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> Let the roots of the equation be &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;. By Vieta's,<br /> \begin{align*}<br /> a+b+c+d &amp;= 18\\<br /> ab+ac+ad+bc+bd+cd &amp;= k\\ <br /> abc+abd+acd+bcd &amp;=-200\\ <br /> abcd &amp;=-1984.\\ <br /> \end{align*}<br /> Since &lt;math&gt;abcd=-1984&lt;/math&gt; and &lt;math&gt;ab=-32&lt;/math&gt;, then, &lt;math&gt;cd=62&lt;/math&gt;. Notice that&lt;cmath&gt;abc + abd + acd + bcd = -200&lt;/cmath&gt;can be factored into&lt;cmath&gt;ab(c+d)+cd(a+b)=-32(c+d)+62(a+b).&lt;/cmath&gt;From the first equation, &lt;math&gt;c+d=18-a-b&lt;/math&gt;. Substituting it back into the equation,&lt;cmath&gt;-32(18-a-b)+62(a+b)=-200&lt;/cmath&gt;Expanding,&lt;cmath&gt;-576+32a+32b+62a+62b=-200 \implies 94a+94b=376&lt;/cmath&gt;So, &lt;math&gt;a+b=4&lt;/math&gt; and &lt;math&gt;c+d=14&lt;/math&gt;. Notice that&lt;cmath&gt;ab+ac+ad+bc+bd+cd=ab+cd+(a+b)(c+d)&lt;/cmath&gt;Plugging all our values in,&lt;cmath&gt;-32+62+4(14)=\boxed{86}.&lt;/cmath&gt;<br /> <br /> ~ kante314<br /> <br /> == Video Solution ==<br /> https://youtu.be/5QdPQ3__a7I?t=589<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{USAMO box|year=1984|before=First&lt;br&gt;Problem|num-a=2}}<br /> {{MAA Notice}}<br /> [[Category:Intermediate Algebra Problems]]</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=1984_USAMO_Problems/Problem_1&diff=161133 1984 USAMO Problems/Problem 1 2021-08-29T00:58:05Z <p>Kante314: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> In the polynomial &lt;math&gt;x^4 - 18x^3 + kx^2 + 200x - 1984 = 0&lt;/math&gt;, the product of &lt;math&gt;2&lt;/math&gt; of its roots is &lt;math&gt;- 32&lt;/math&gt;. Find &lt;math&gt;k&lt;/math&gt;.<br /> <br /> === Solution 1 (ingenious)===<br /> <br /> <br /> Using Vieta's formulas, we have: <br /> <br /> &lt;cmath&gt;\begin{align*}a+b+c+d &amp;= 18,\\ ab+ac+ad+bc+bd+cd &amp;= k,\\ abc+abd+acd+bcd &amp;=-200,\\ abcd &amp;=-1984.\\ \end{align*}&lt;/cmath&gt; <br /> <br /> <br /> From the last of these equations, we see that &lt;math&gt;cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62&lt;/math&gt;. Thus, the second equation becomes &lt;math&gt;-32+ac+ad+bc+bd+62=k&lt;/math&gt;, and so &lt;math&gt;ac+ad+bc+bd=k-30&lt;/math&gt;. The key insight is now to factor the left-hand side as a product of two binomials: &lt;math&gt;(a+b)(c+d)=k-30&lt;/math&gt;, so that we now only need to determine &lt;math&gt;a+b&lt;/math&gt; and &lt;math&gt;c+d&lt;/math&gt; rather than all four of &lt;math&gt;a,b,c,d&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;p=a+b&lt;/math&gt; and &lt;math&gt;q=c+d&lt;/math&gt;. Plugging our known values for &lt;math&gt;ab&lt;/math&gt; and &lt;math&gt;cd&lt;/math&gt; into the third Vieta equation, &lt;math&gt;-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)&lt;/math&gt;, we have &lt;math&gt;-200 = -32(c+d) + 62(a+b) = 62p-32q&lt;/math&gt;. Moreover, the first Vieta equation, &lt;math&gt;a+b+c+d=18&lt;/math&gt;, gives &lt;math&gt;p+q=18&lt;/math&gt;. Thus we have two linear equations in &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;, which we solve to obtain &lt;math&gt;p=4&lt;/math&gt; and &lt;math&gt;q=14&lt;/math&gt;. <br /> <br /> Therefore, we have &lt;math&gt;(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30&lt;/math&gt;, yielding &lt;math&gt;k=4\cdot 14+30 = \boxed{86}&lt;/math&gt;.<br /> <br /> === Solution 2 (cool)===<br /> <br /> We start as before: &lt;math&gt;ab=-32&lt;/math&gt; and &lt;math&gt;cd=62&lt;/math&gt;. We now observe that a and b must be the roots of a quadratic, &lt;math&gt;x^2+rx-32&lt;/math&gt;, where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic &lt;math&gt;x^2+sx+62&lt;/math&gt;. <br /> <br /> Now <br /> <br /> &lt;cmath&gt; \begin{align*}x^4-18x^3+kx^2+200x-1984 =&amp; (x^2+rx-32)(x^2+sx+62)\\ =&amp; x^4+(r+s)x^3+(62-32+rs)x^2\\<br /> &amp;+(62s-32r)x-1984.\end{align*} &lt;/cmath&gt;<br /> <br /> Equating the coefficients of &lt;math&gt;x^3&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of &lt;math&gt;x^2&lt;/math&gt; and get &lt;math&gt;k=\boxed{86}.&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> Let the roots of the equatoon be &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;. By Vieta's,<br /> \begin{align*}a+b+c+d &amp;= 18\\ ab+ac+ad+bc+bd+cd &amp;= k\\ abc+abd+acd+bcd &amp;=-200\\ abcd &amp;=-1984.\\ \end{align*}<br /> Since &lt;math&gt;abcd=-1984&lt;/math&gt; and &lt;math&gt;ab=-32&lt;/math&gt;, then, &lt;math&gt;cd=62&lt;/math&gt;. Notice that&lt;cmath&gt;abc + abd + acd + bcd = -200&lt;/cmath&gt;can be factored into&lt;cmath&gt;ab(c+d)+cd(a+b)=-32(c+d)+62(a+b).&lt;/cmath&gt;From the first equation, &lt;math&gt;c+d=18-a-b&lt;/math&gt;. Substituting it back into the equation,&lt;cmath&gt;-32(18-a-b)+62(a+b)=-200&lt;/cmath&gt;Expanding,&lt;cmath&gt;-576+32a+32b+62a+62b=-200 \implies 94a+94b=376&lt;/cmath&gt;So, &lt;math&gt;a+b=4&lt;/math&gt; and &lt;math&gt;c+d=14&lt;/math&gt;. Notice that&lt;cmath&gt;ab+ac+ad+bc+bd+cd=ab+cd+(a+b)(c+d)&lt;/cmath&gt;Plugging all our values in,&lt;cmath&gt;-32+62+4(14)=\boxed{86}.&lt;/cmath&gt;<br /> <br /> ~ kante314<br /> <br /> == Video Solution ==<br /> https://youtu.be/5QdPQ3__a7I?t=589<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{USAMO box|year=1984|before=First&lt;br&gt;Problem|num-a=2}}<br /> {{MAA Notice}}<br /> [[Category:Intermediate Algebra Problems]]</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=1984_USAMO_Problems/Problem_1&diff=161132 1984 USAMO Problems/Problem 1 2021-08-29T00:57:11Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> <br /> In the polynomial &lt;math&gt;x^4 - 18x^3 + kx^2 + 200x - 1984 = 0&lt;/math&gt;, the product of &lt;math&gt;2&lt;/math&gt; of its roots is &lt;math&gt;- 32&lt;/math&gt;. Find &lt;math&gt;k&lt;/math&gt;.<br /> <br /> === Solution 1 (ingenious)===<br /> <br /> <br /> Using Vieta's formulas, we have: <br /> <br /> &lt;cmath&gt;\begin{align*}a+b+c+d &amp;= 18,\\ ab+ac+ad+bc+bd+cd &amp;= k,\\ abc+abd+acd+bcd &amp;=-200,\\ abcd &amp;=-1984.\\ \end{align*}&lt;/cmath&gt; <br /> <br /> <br /> From the last of these equations, we see that &lt;math&gt;cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62&lt;/math&gt;. Thus, the second equation becomes &lt;math&gt;-32+ac+ad+bc+bd+62=k&lt;/math&gt;, and so &lt;math&gt;ac+ad+bc+bd=k-30&lt;/math&gt;. The key insight is now to factor the left-hand side as a product of two binomials: &lt;math&gt;(a+b)(c+d)=k-30&lt;/math&gt;, so that we now only need to determine &lt;math&gt;a+b&lt;/math&gt; and &lt;math&gt;c+d&lt;/math&gt; rather than all four of &lt;math&gt;a,b,c,d&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;p=a+b&lt;/math&gt; and &lt;math&gt;q=c+d&lt;/math&gt;. Plugging our known values for &lt;math&gt;ab&lt;/math&gt; and &lt;math&gt;cd&lt;/math&gt; into the third Vieta equation, &lt;math&gt;-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)&lt;/math&gt;, we have &lt;math&gt;-200 = -32(c+d) + 62(a+b) = 62p-32q&lt;/math&gt;. Moreover, the first Vieta equation, &lt;math&gt;a+b+c+d=18&lt;/math&gt;, gives &lt;math&gt;p+q=18&lt;/math&gt;. Thus we have two linear equations in &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;, which we solve to obtain &lt;math&gt;p=4&lt;/math&gt; and &lt;math&gt;q=14&lt;/math&gt;. <br /> <br /> Therefore, we have &lt;math&gt;(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30&lt;/math&gt;, yielding &lt;math&gt;k=4\cdot 14+30 = \boxed{86}&lt;/math&gt;.<br /> <br /> === Solution 2 (cool)===<br /> <br /> We start as before: &lt;math&gt;ab=-32&lt;/math&gt; and &lt;math&gt;cd=62&lt;/math&gt;. We now observe that a and b must be the roots of a quadratic, &lt;math&gt;x^2+rx-32&lt;/math&gt;, where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic &lt;math&gt;x^2+sx+62&lt;/math&gt;. <br /> <br /> Now <br /> <br /> &lt;cmath&gt; \begin{align*}x^4-18x^3+kx^2+200x-1984 =&amp; (x^2+rx-32)(x^2+sx+62)\\ =&amp; x^4+(r+s)x^3+(62-32+rs)x^2\\<br /> &amp;+(62s-32r)x-1984.\end{align*} &lt;/cmath&gt;<br /> <br /> Equating the coefficients of &lt;math&gt;x^3&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of &lt;math&gt;x^2&lt;/math&gt; and get &lt;math&gt;k=\boxed{86}.&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> Let the roots of the equatoon be &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;. By Vieta's,<br /> \begin{align*}<br /> a+b+c+d=18 \\<br /> ab+ac+ad+bc+bd+cd=k \\<br /> abc + abd + acd + bcd = -200 \\<br /> abcd=-1984<br /> \end{align*}Since &lt;math&gt;abcd=-1984&lt;/math&gt; and &lt;math&gt;ab=-32&lt;/math&gt;, then, &lt;math&gt;cd=62&lt;/math&gt;. Notice that&lt;cmath&gt;abc + abd + acd + bcd = -200&lt;/cmath&gt;can be factored into&lt;cmath&gt;ab(c+d)+cd(a+b)=-32(c+d)+62(a+b).&lt;/cmath&gt;From the first equation, &lt;math&gt;c+d=18-a-b&lt;/math&gt;. Substituting it back into the equation,&lt;cmath&gt;-32(18-a-b)+62(a+b)=-200&lt;/cmath&gt;Expanding,&lt;cmath&gt;-576+32a+32b+62a+62b=-200 \implies 94a+94b=376&lt;/cmath&gt;So, &lt;math&gt;a+b=4&lt;/math&gt; and &lt;math&gt;c+d=14&lt;/math&gt;. Notice that&lt;cmath&gt;ab+ac+ad+bc+bd+cd=ab+cd+(a+b)(c+d)&lt;/cmath&gt;Plugging all our values in,&lt;cmath&gt;-32+62+4(14)=\boxed{86}.&lt;/cmath&gt;<br /> <br /> ~ kante314<br /> <br /> == Video Solution ==<br /> https://youtu.be/5QdPQ3__a7I?t=589<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{USAMO box|year=1984|before=First&lt;br&gt;Problem|num-a=2}}<br /> {{MAA Notice}}<br /> [[Category:Intermediate Algebra Problems]]</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_13&diff=161108 2021 AMC 10A Problems/Problem 13 2021-08-28T22:45:56Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> What is the volume of tetrahedron &lt;math&gt;ABCD&lt;/math&gt; with edge lengths &lt;math&gt;AB = 2&lt;/math&gt;, &lt;math&gt;AC = 3&lt;/math&gt;, &lt;math&gt;AD = 4&lt;/math&gt;, &lt;math&gt;BC = \sqrt{13}&lt;/math&gt;, &lt;math&gt;BD = 2\sqrt{5}&lt;/math&gt;, and &lt;math&gt;CD = 5&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6&lt;/math&gt;<br /> <br /> ==Solution 1 (Three Right Triangles)==<br /> Drawing the tetrahedron out and testing side lengths, we realize that the &lt;math&gt;\triangle ACD, \triangle ABC,&lt;/math&gt; and &lt;math&gt;\triangle ABD&lt;/math&gt; are right triangles by the Converse of the Pythagorean Theorem. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid. If we take &lt;math&gt;\triangle ADC&lt;/math&gt; as the base, then &lt;math&gt;\overline{AB}&lt;/math&gt; must be the altitude. The volume of tetrahedron &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2=\boxed{\textbf{(C)} ~4}.&lt;/math&gt;<br /> <br /> ~Icewolf10 ~Bakedpotato66 ~MRENTHUSIASM<br /> <br /> ==Solution 2 (Bash: One Right Triangle)==<br /> We will place tetrahedron &lt;math&gt;ABCD&lt;/math&gt; in the &lt;math&gt;xyz&lt;/math&gt;-plane. By the Converse of the Pythagorean Theorem, we know that &lt;math&gt;\triangle ACD&lt;/math&gt; is a right triangle. Without the loss of generality, let &lt;math&gt;A=(0,0,0), C=(3,0,0), D=(0,4,0),&lt;/math&gt; and &lt;math&gt;B=(x,y,z).&lt;/math&gt;<br /> <br /> We apply the Distance Formula to &lt;math&gt;\overline{BA},\overline{BC},&lt;/math&gt; and &lt;math&gt;\overline{BD},&lt;/math&gt; respectively:<br /> &lt;cmath&gt;\begin{align*}<br /> x^2+y^2+z^2&amp;=2^2, &amp;(1) \\<br /> (x-3)^2+y^2+z^2&amp;=\sqrt{13}^2, &amp;(2) \\<br /> x^2+(y-4)^2+z^2&amp;=\left(2\sqrt5\right)^2. &amp;\hspace{1mm} (3)<br /> \end{align*}&lt;/cmath&gt;<br /> Subtracting &lt;math&gt;(1)&lt;/math&gt; from &lt;math&gt;(2)&lt;/math&gt; gives &lt;math&gt;-6x+9=9,&lt;/math&gt; from which &lt;math&gt;x=0.&lt;/math&gt;<br /> <br /> Subtracting &lt;math&gt;(1)&lt;/math&gt; from &lt;math&gt;(3)&lt;/math&gt; gives &lt;math&gt;-8y+16=16,&lt;/math&gt; from which &lt;math&gt;y=0.&lt;/math&gt;<br /> <br /> Substituting &lt;math&gt;(x,y)=(0,0)&lt;/math&gt; into &lt;math&gt;(1)&lt;/math&gt; produces &lt;math&gt;z^2=4,&lt;/math&gt; or &lt;math&gt;|z|=2.&lt;/math&gt;<br /> <br /> Let the brackets denote areas. Finally, we find the volume of tetrahedron &lt;math&gt;ABCD&lt;/math&gt; using &lt;math&gt;\triangle ACD&lt;/math&gt; as the base:<br /> &lt;cmath&gt;\begin{align*}<br /> V_{ABCD}&amp;=\frac13\cdot[ACD]\cdot h_B \\<br /> &amp;=\frac13\cdot\left(\frac12\cdot AC \cdot AD\right)\cdot |z| \\<br /> &amp;=\boxed{\textbf{(C)} ~4}.<br /> \end{align*}&lt;/cmath&gt;<br /> ~MRENTHUSIASM<br /> <br /> ==Solution 3 (Trirectangular Tetrahedron)==<br /> https://mathworld.wolfram.com/TrirectangularTetrahedron.html<br /> <br /> Given the observations from Solution 1, where &lt;math&gt;\triangle ACD, \triangle ABC,&lt;/math&gt; and &lt;math&gt;\triangle ABD&lt;/math&gt; are right triangles, the base is &lt;math&gt;\triangle ABD.&lt;/math&gt; We can apply the information about a trirectangular tetrahedron (all of the face angles are right angles), which states that the volume is<br /> &lt;cmath&gt;\begin{align*}<br /> V&amp;=\frac16\cdot AB\cdot AD\cdot BD \\<br /> &amp;=\frac16\cdot2\cdot4\cdot3 \\<br /> &amp;=\boxed{\textbf{(C)} ~4}.<br /> \end{align*}&lt;/cmath&gt;<br /> ~AMC60 (Solution)<br /> <br /> ~MRENTHUSIASM (Revision)<br /> <br /> ==Solution 4==<br /> Notice that &lt;math&gt;\angle ADC,\angle DAB,&lt;/math&gt; and &lt;math&gt;\angle BAC&lt;/math&gt; are right angles. Let the base of the tetrahedron be &lt;math&gt;\triangle DAC&lt;/math&gt;, so the height is &lt;math&gt;AB&lt;/math&gt;. &lt;cmath&gt;A_{\triangle CAD}=\frac{4 \cdot 3}{2}=6&lt;/cmath&gt;Therefore, &lt;cmath&gt;\frac{1}{3} \cdot 6 \cdot 2=\boxed{4}&lt;/cmath&gt;<br /> <br /> ~ kante314<br /> <br /> ==Remark==<br /> Here is a similar problem from another AMC test: [[2015_AMC_10A_Problems/Problem_21|2015 AMC 10A Problem 21]].<br /> <br /> ==Video Solution (Simple &amp; Quick)==<br /> https://youtu.be/bRrchiDCrKE<br /> <br /> ~ Education, the Study of Everything<br /> <br /> == Video Solution (Using Pythagorean Theorem, 3D Geometry: Tetrahedron) ==<br /> https://youtu.be/i4yUaXVUWKE<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution by TheBeautyofMath==<br /> https://youtu.be/t-EEP2V4nAE?t=813<br /> <br /> ~IceMatrix<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_21&diff=158228 2018 AMC 10A Problems/Problem 21 2021-07-12T15:13:09Z <p>Kante314: </p> <hr /> <div>{{duplicate|[[2018 AMC 12A Problems|2018 AMC 12A #16]] and [[2018 AMC 10A Problems|2018 AMC 10A #21]]}}<br /> <br /> == Problem ==<br /> <br /> Which of the following describes the set of values of &lt;math&gt;a&lt;/math&gt; for which the curves &lt;math&gt;x^2+y^2=a^2&lt;/math&gt; and &lt;math&gt;y=x^2-a&lt;/math&gt; in the real &lt;math&gt;xy&lt;/math&gt;-plane intersect at exactly &lt;math&gt;3&lt;/math&gt; points?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }a=\frac14 \qquad<br /> \textbf{(B) }\frac14 &lt; a &lt; \frac12 \qquad<br /> \textbf{(C) }a&gt;\frac14 \qquad<br /> \textbf{(D) }a=\frac12 \qquad<br /> \textbf{(E) }a&gt;\frac12 \qquad<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Substituting &lt;math&gt;y=x^2-a&lt;/math&gt; into &lt;math&gt;x^2+y^2=a^2&lt;/math&gt;, we get<br /> &lt;cmath&gt;<br /> x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0<br /> &lt;/cmath&gt;<br /> Since this is a quartic, there are 4 total roots (counting multiplicity). We see that &lt;math&gt;x=0&lt;/math&gt; always at least one intersection at &lt;math&gt;(0,-a)&lt;/math&gt; (and is in fact a double root). <br /> <br /> The other two intersection points have &lt;math&gt;x&lt;/math&gt; coordinates &lt;math&gt;\pm\sqrt{2a-1}&lt;/math&gt;. We must have &lt;math&gt;2a-1&gt; 0,&lt;/math&gt; otherwise we are in the case where the parabola lies entirely above the circle (tangent to it at the point &lt;math&gt;(0,a)&lt;/math&gt;). This only results in a single intersection point in the real coordinate plane. Thus, we see &lt;math&gt;\boxed{\textbf{(E) }a&gt;\frac12}&lt;/math&gt;.<br /> <br /> (projecteulerlover)<br /> <br /> == Solution 2 ==<br /> <br /> &lt;asy&gt;<br /> Label f; <br /> f.p=fontsize(6);<br /> xaxis(-2,2,Ticks(f, 0.2)); <br /> yaxis(-2,2,Ticks(f, 0.2)); <br /> real g(real x) <br /> { <br /> return x^2-1; <br /> } <br /> draw(graph(g, 1.7, -1.7));<br /> real h(real x) <br /> { <br /> return sqrt(1-x^2); <br /> } <br /> draw(graph(h, 1, -1));<br /> real j(real x) <br /> { <br /> return -sqrt(1-x^2); <br /> } <br /> draw(graph(j, 1, -1));<br /> &lt;/asy&gt;<br /> <br /> Looking at a graph, it is obvious that the two curves intersect at (0, -a). We also see that if the parabola goes 'in' the circle, then by going out of it (as it will), it will intersect five times, an impossibility. Thus we only look for cases where the parabola becomes externally tangent to the circle. We have &lt;math&gt;x^2 - a = -\sqrt{a^2 - x^2}&lt;/math&gt;. Squaring both sides and solving yields &lt;math&gt;x^4 - (2a - 1)x^2 = 0&lt;/math&gt;. Since &lt;math&gt;x = 0&lt;/math&gt; is already accounted for, we only need to find 1 solution for &lt;math&gt;x^2 = 2a - 1&lt;/math&gt;, where the right hand side portion is obviously increasing. Since &lt;math&gt;a = \frac{1}{2}&lt;/math&gt; begets &lt;math&gt;x = 0&lt;/math&gt; (an overcount), we have &lt;math&gt;\boxed{\textbf{(E) }a&gt;\frac12}&lt;/math&gt; is the right answer.<br /> <br /> Solution by JohnHankock<br /> <br /> == Solution 3 (The simple one, when you are running out of time) ==<br /> <br /> We can see that if &lt;math&gt;a = 1&lt;/math&gt;, we know that the points where the two curves intersect are &lt;math&gt;(0, -1), (1, 0)&lt;/math&gt; and &lt;math&gt;(-1, 0)&lt;/math&gt; .Because there are only 3 intersections and &lt;math&gt;a &gt; 1/2&lt;/math&gt;, we know that &lt;math&gt;\boxed{\textbf{(E) }a&gt;\frac12}&lt;/math&gt; is the correct answer.<br /> <br /> Solution by josephwidjaja<br /> <br /> == Solution 4 (Calculus Needed) ==<br /> <br /> In order to solve for the values of &lt;math&gt;a&lt;/math&gt;, we need to just count multiplicities of the roots when the equations are set equal to each other: in other words, take the derivative. We know that &lt;math&gt;\sqrt{a^2 - x^2} = x^2 - a&lt;/math&gt;. Now, we take square of both sides, and rearrange to obtain &lt;math&gt;x^4 - (2a - 1)x^2 = 0&lt;/math&gt;. Now, we may take the second derivative of the equation to obtain &lt;math&gt;6x^2 - (2a - 1) = 0&lt;/math&gt;. Now, we must take discriminant. Since we need the roots of that equation to be real and not repetitive (otherwise they would not intersect each other at three points), the discriminant must be greater than zero. Thus,<br /> <br /> &lt;math&gt;<br /> b^2 - 4ac &gt; 0 \rightarrow 0 - 4(6)(-(2a - 1)) &gt; 0 \rightarrow a &gt; \frac{1}{2}<br /> &lt;/math&gt;<br /> The answer is &lt;math&gt;\boxed{\textbf{(E) }a&gt;\frac12}&lt;/math&gt; and we are done. <br /> <br /> ~awesome1st <br /> <br /> (Edited by OlutosinNGA)<br /> <br /> == Solution 5 ==<br /> <br /> This describes a unit parabola, with a circle centered on the axis of symmetry and tangent to the vertex. As the curvature of the unit parabola at the vertex is 2, the radius of the circle that matches it has a radius of &lt;math&gt;\frac{1}{2}&lt;/math&gt;. This circle is tangent to an infinitesimally close pair of points, one on each side. Therefore, it is tangent to only 1 point. When a larger circle is used, it is tangent to 3 points because the points on either side are now separated from the vertex. Therefore, &lt;math&gt;\boxed{\textbf{(E) }a&gt;\frac12}&lt;/math&gt; is correct.<br /> <br /> &lt;math&gt;QED \blacksquare&lt;/math&gt;<br /> <br /> == Solution 6 ==<br /> <br /> Notice, the equations are of that of a circle of radius a centered at the origin and a parabola translated down by a units. They always intersect at the point &lt;math&gt;(0, a)&lt;/math&gt;, and they have symmetry across the y-axis, thus, for them to intersect at exactly 3 points, it suffices to find the y solution. <br /> <br /> First, rewrite the second equation to &lt;math&gt;y=x^2-a\implies x^2=y+a&lt;/math&gt;<br /> And substitute into the first equation: &lt;math&gt;y+a+y^2=a^2&lt;/math&gt; <br /> Since we're only interested in seeing the interval in which a can exist, we find the discriminant: &lt;math&gt;1-4a+4a^2&lt;/math&gt;. This value must not be less than 0 (It is the square root part of the quadratic formula). To find when it is 0, we find the roots: <br /> &lt;cmath&gt;4a^2-4a+1=0 \implies a=\frac{4\pm\sqrt{16-16}}{8}=\frac{1}{2}&lt;/cmath&gt;<br /> Since &lt;math&gt;\lim_{a\to \infty}(4a^2-4a+1)=\infty&lt;/math&gt;, our range is &lt;math&gt;\boxed{\textbf{(E) }a&gt;\frac12}&lt;/math&gt;.<br /> <br /> Solution by ktong<br /> <br /> == Solution 7 (Cheating with Answer Choices) ==<br /> Simply plug in &lt;math&gt;a = \frac{1}{2}, \frac{1}{4}, 1&lt;/math&gt; and solve the systems. (This shouldn't take too long.) And then realize that only &lt;math&gt;a=1&lt;/math&gt; yields three real solutions for &lt;math&gt;x&lt;/math&gt;, so we are done and the answer is &lt;math&gt;\boxed{\textbf{(E) }a&gt;\frac12}&lt;/math&gt;.<br /> <br /> ~ ccx09<br /> <br /> == Solution 8 ==<br /> Substituting &lt;math&gt;y = x^2 - a&lt;/math&gt; gives &lt;math&gt;x^2 + (x^2 - a)^2 = a^2&lt;/math&gt;, which simplifies to &lt;math&gt;x^2 + x^4 - 2x^2a + a^2 = a^2&lt;/math&gt;. This further simplifies to &lt;math&gt;x^2(1 + x^2 - 2a) = 0&lt;/math&gt;. Thus, either &lt;math&gt;x^2 = 0&lt;/math&gt;, or &lt;math&gt;x^2 - 2a + 1 = 0&lt;/math&gt;. Since we care about &lt;math&gt;a&lt;/math&gt;, we consider the second case. Then, we solve in terms of &lt;math&gt;a&lt;/math&gt; giving &lt;math&gt;a = \frac{x^2}{2} + \frac{1}{2}&lt;/math&gt;. We see that in order to find the range in which &lt;math&gt;a&lt;/math&gt; lies, we must find the vertex of the that equation, which turns out to be &lt;math&gt;(0, \frac{1}{2})&lt;/math&gt;, so we know that the minimum is &lt;math&gt;\frac{1}{2}&lt;/math&gt;, which further implies that &lt;math&gt;\boxed{\textbf{(E) }a &gt; \frac{1}{2}}&lt;/math&gt;<br /> <br /> ==Solution 9==<br /> Now, let's graph these two equations. We want the blue parabola to be inside this red circle.<br /> [asy]<br /> import graph;<br /> size(6cm);<br /> draw((0,0)--(0,10),EndArrow);<br /> draw((0,0)--(0,-10),EndArrow);<br /> draw((0,0)--(10,0),EndArrow);<br /> draw((0,0)--(-10,0),EndArrow);<br /> Label f;<br /> f.p=fontsize(6);<br /> xaxis(-10,10);<br /> yaxis(-10,10);<br /> real f(real x) <br /> { <br /> return x^2-5;<br /> }<br /> draw(graph(f,-4,4),blue+linewidth(1));<br /> draw(circle((0,0),5),red);<br /> dot(scale(.7)*&quot;&lt;math&gt;a&lt;/math&gt;&quot;,(0,5),NE);<br /> dot(scale(.7)*&quot;&lt;math&gt;-a&lt;/math&gt;&quot;,(0,-5),N);<br /> dot(scale(.7)*&quot;&lt;math&gt;a&lt;/math&gt;&quot;,(5,0),NE);<br /> dot(scale(.7)*&quot;&lt;math&gt;-a&lt;/math&gt;&quot;,(-5,0),SE);<br /> [/asy]<br /> Then we substitute &lt;math&gt;y&lt;/math&gt; into the first equation to get &lt;math&gt;x^2+(x^2-a)^2=a^2&lt;/math&gt;. Expanding, we get &lt;math&gt;x^4-2ax^2+x^2=0&lt;/math&gt;. Factoring out the &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;x^2(x^2-2a+1)=0&lt;/math&gt;. Then we find that &lt;math&gt;x=0&lt;/math&gt; or &lt;math&gt;x=\pm\sqrt{2a-1}&lt;/math&gt;. Therefore, &lt;math&gt;2a-1&gt;0&lt;/math&gt;, which means &lt;math&gt;\boxed{a&gt;\frac{1}{2}}&lt;/math&gt;.<br /> <br /> - kante314 -<br /> <br /> == Video Solution by Richard Rusczyk ==<br /> <br /> https://artofproblemsolving.com/videos/amc/2018amc10a/466<br /> <br /> ~ dolphin7<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2018|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_19&diff=158227 1955 AHSME Problems/Problem 19 2021-07-12T15:11:10Z <p>Kante314: </p> <hr /> <div>== Problem 19==<br /> <br /> Two numbers whose sum is &lt;math&gt;6&lt;/math&gt; and the absolute value of whose difference is &lt;math&gt;8&lt;/math&gt; are roots of the equation: <br /> <br /> &lt;math&gt; \textbf{(A)}\ x^2-6x+7=0\qquad\textbf{(B)}\ x^2-6x-7=0\qquad\textbf{(C)}\ x^2+6x-8=0\\ \textbf{(D)}\ x^2-6x+8=0\qquad\textbf{(E)}\ x^2+6x-7=0 &lt;/math&gt;<br /> ==Solution==<br /> The first two hints can be expressed as the following system of equations:<br /> &lt;cmath&gt;\begin{cases} (1) &amp; a + b = 6 \\ (2) &amp; a - b = 8 \end{cases}&lt;/cmath&gt;<br /> From this, we can clearly see that &lt;math&gt;a = 7&lt;/math&gt;, and that &lt;math&gt;b = -1&lt;/math&gt;.<br /> <br /> Since quadratic equations can generally be expressed in the form of &lt;math&gt;(x - a)(x - b) = 0&lt;/math&gt;, where a and b are roots, the correct quadratic, once factored, would look like &lt;math&gt;(x - 7)(x + 1) = 0&lt;/math&gt;<br /> <br /> Expanding the above equation gets us &lt;math&gt;\textbf{(B)} x^2 - 6x - 7 = 0&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let the roots of the equation be &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;. Therefore, we can set up a system of equation:<br /> &lt;cmath&gt;x+y=6&lt;/cmath&gt;&lt;cmath&gt;|x-y|=8&lt;/cmath&gt;Therefore, we get &lt;math&gt;x=7&lt;/math&gt; and &lt;math&gt;y=-1&lt;/math&gt;. So, &lt;math&gt;(x-7)(x+1)=\boxed{x^2-6x-7}&lt;/math&gt;<br /> <br /> - kante314 -<br /> <br /> == See Also ==<br /> {{AHSME box|year=1955|num-b=18|num-a=20}}<br /> <br /> {{MAA Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_19&diff=158226 1955 AHSME Problems/Problem 19 2021-07-12T15:10:57Z <p>Kante314: </p> <hr /> <div>== Problem 19==<br /> <br /> Two numbers whose sum is &lt;math&gt;6&lt;/math&gt; and the absolute value of whose difference is &lt;math&gt;8&lt;/math&gt; are roots of the equation: <br /> <br /> &lt;math&gt; \textbf{(A)}\ x^2-6x+7=0\qquad\textbf{(B)}\ x^2-6x-7=0\qquad\textbf{(C)}\ x^2+6x-8=0\\ \textbf{(D)}\ x^2-6x+8=0\qquad\textbf{(E)}\ x^2+6x-7=0 &lt;/math&gt;<br /> ==Solution==<br /> The first two hints can be expressed as the following system of equations:<br /> &lt;cmath&gt;\begin{cases} (1) &amp; a + b = 6 \\ (2) &amp; a - b = 8 \end{cases}&lt;/cmath&gt;<br /> From this, we can clearly see that &lt;math&gt;a = 7&lt;/math&gt;, and that &lt;math&gt;b = -1&lt;/math&gt;.<br /> <br /> Since quadratic equations can generally be expressed in the form of &lt;math&gt;(x - a)(x - b) = 0&lt;/math&gt;, where a and b are roots, the correct quadratic, once factored, would look like &lt;math&gt;(x - 7)(x + 1) = 0&lt;/math&gt;<br /> <br /> Expanding the above equation gets us &lt;math&gt;\textbf{(B)} x^2 - 6x - 7 = 0&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let the roots of the equation be &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;. Therefore, we can set up a system of equation:<br /> &lt;cmath&gt;x+y=6&lt;/cmath&gt;&lt;cmath&gt;|x-y|=8&lt;/cmath&gt;Therefore, we get &lt;math&gt;x=7&lt;/math&gt; and &lt;math&gt;y=-1&lt;/math&gt;. So, &lt;math&gt;(x-7)(x+1)=\boxed{x^2-6x-7}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AHSME box|year=1955|num-b=18|num-a=20}}<br /> <br /> {{MAA Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_13&diff=158222 2021 JMPSC Sprint Problems/Problem 13 2021-07-12T14:47:50Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> Grace places a pencil in a cylindrical cup and is surprised to see that it fits diagonally. The pencil is &lt;math&gt;17&lt;/math&gt; units long and of negligible thickness. The cup is &lt;math&gt;8&lt;/math&gt; units tall. The volume of the cup can be written as &lt;math&gt;k \pi&lt;/math&gt; cubic units. Find &lt;math&gt;k&lt;/math&gt;.<br /> &lt;center&gt;<br /> [[File:Sprint14.jpg|200px]]<br /> &lt;/center&gt;<br /> <br /> ==Solution==<br /> By the [[Pythagorean Theorem]], we have that the diameter of the cylinder's base is 15 units long. Thus, the cylinder's base has radius &lt;math&gt;\frac{15}{2}&lt;/math&gt; units. Thus, the volume of the cylinder is &lt;math&gt;\left(\frac{15}{2}\right)^2\cdot8\pi=\boxed{450}\pi.&lt;/math&gt;<br /> <br /> ~Lamboreghini<br /> <br /> == Solution 2==<br /> The diameter is &lt;math&gt;15&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;\pi \left(\frac{15}{2}\right)^2 \cdot 8=450 \pi&lt;/cmath&gt; So, &lt;math&gt;k=\boxed{450}&lt;/math&gt;<br /> <br /> - kante314 -<br /> <br /> ==See also==<br /> #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]<br /> #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_13&diff=158221 2021 JMPSC Sprint Problems/Problem 13 2021-07-12T14:47:29Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> Grace places a pencil in a cylindrical cup and is surprised to see that it fits diagonally. The pencil is &lt;math&gt;17&lt;/math&gt; units long and of negligible thickness. The cup is &lt;math&gt;8&lt;/math&gt; units tall. The volume of the cup can be written as &lt;math&gt;k \pi&lt;/math&gt; cubic units. Find &lt;math&gt;k&lt;/math&gt;.<br /> &lt;center&gt;<br /> [[File:Sprint14.jpg|200px]]<br /> &lt;/center&gt;<br /> <br /> ==Solution==<br /> By the [[Pythagorean Theorem]], we have that the diameter of the cylinder's base is 15 units long. Thus, the cylinder's base has radius &lt;math&gt;\frac{15}{2}&lt;/math&gt; units. Thus, the volume of the cylinder is &lt;math&gt;\left(\frac{15}{2}\right)^2\cdot8\pi=\boxed{450}\pi.&lt;/math&gt;<br /> <br /> ~Lamboreghini<br /> <br /> == Solution 2==<br /> The diameter is &lt;math&gt;15&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;\pi \left(\frac{15}{2}\right)^2 \cdot 8=450 \pi&lt;/cmath&gt;SO, &lt;math&gt;k=\boxed{450}&lt;/math&gt;<br /> <br /> - kante314 -<br /> <br /> ==See also==<br /> #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]<br /> #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_13&diff=158220 2021 JMPSC Sprint Problems/Problem 13 2021-07-12T14:46:39Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> Grace places a pencil in a cylindrical cup and is surprised to see that it fits diagonally. The pencil is &lt;math&gt;17&lt;/math&gt; units long and of negligible thickness. The cup is &lt;math&gt;8&lt;/math&gt; units tall. The volume of the cup can be written as &lt;math&gt;k \pi&lt;/math&gt; cubic units. Find &lt;math&gt;k&lt;/math&gt;.<br /> &lt;center&gt;<br /> [[File:Sprint14.jpg|200px]]<br /> &lt;/center&gt;<br /> <br /> ==Solution==<br /> By the [[Pythagorean Theorem]], we have that the diameter of the cylinder's base is 15 units long. Thus, the cylinder's base has radius &lt;math&gt;\frac{15}{2}&lt;/math&gt; units. Thus, the volume of the cylinder is &lt;math&gt;\left(\frac{15}{2}\right)^2\cdot8\pi=\boxed{450}\pi.&lt;/math&gt;<br /> <br /> ~Lamboreghini<br /> <br /> == Solution 2==<br /> The diameter is &lt;math&gt;15&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;\pi \left(\frac{15}{2}\right)^2 \cdot 8=450 \pi&lt;/cmath&gt;SO, &lt;math&gt;k=\boxed{450}&lt;/math&gt;<br /> <br /> ==See also==<br /> #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]<br /> #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_15&diff=158219 2021 JMPSC Sprint Problems/Problem 15 2021-07-12T14:45:44Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> Find the last two digits of &lt;math&gt;10^{10}-5^{10}.&lt;/math&gt;<br /> <br /> ==Solution==<br /> Note that &lt;math&gt;10^{10}\equiv0\pmod{100}&lt;/math&gt; and &lt;math&gt;5^{10}\equiv25\pmod{100}&lt;/math&gt;. <br /> <br /> &lt;math&gt;0-25=-25&lt;/math&gt;. &lt;math&gt;-25\equiv\boxed{75}\pmod{100}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> By multiplying out several powers of &lt;math&gt;5&lt;/math&gt;, we can observe that the last &lt;math&gt;2&lt;/math&gt; digits are always &lt;math&gt;25&lt;/math&gt; (with the exception of &lt;math&gt;5^n&lt;/math&gt; where &lt;math&gt;n \le 1&lt;/math&gt;). Also, &lt;math&gt;10^10&lt;/math&gt; ends with several zeros, so the answer is &lt;math&gt;100...00 - 25 = 99...99 - 24 = 999...75&lt;/math&gt;.<br /> <br /> ~Mathdreams<br /> <br /> == Solution 3 ==<br /> &lt;cmath&gt;100^{10} \equiv 0 \mod 100&lt;/cmath&gt;&lt;cmath&gt;5^{10} \equiv 25 \mod 100&lt;/cmath&gt;Therefore, the answer is &lt;math&gt;75&lt;/math&gt;<br /> <br /> - kante314 -<br /> <br /> ==See also==<br /> #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]<br /> #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_8&diff=158218 2021 JMPSC Sprint Problems/Problem 8 2021-07-12T14:44:53Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> How many positive two-digit numbers exist such that the product of its digits is not zero?<br /> <br /> ==Solution==<br /> Rather than counting all the two-digit numbers that exist with those characteristics, we should do complementary counting to find the numbers with the product of its digits as 0. <br /> <br /> The only numbers with &lt;math&gt;0&lt;/math&gt;'s in their digits are the multiples of &lt;math&gt;10&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;10, 20, 30, 40, 50, 60, 70, 80, 90&lt;/cmath&gt;<br /> <br /> Therefore, there are only &lt;math&gt;9&lt;/math&gt; two-digit numbers that do not satisfy the requirements. There are &lt;math&gt;100-11+1=90&lt;/math&gt; two-digit numbers total, so there are &lt;math&gt;90-9=\boxed{81}&lt;/math&gt; numbers.<br /> <br /> -OofPirate<br /> <br /> == Solution 2 ==<br /> You don't want a digit in this number to contain &lt;math&gt;0&lt;/math&gt;. Therefore, the answer is &lt;math&gt;9 \cdot 9=\boxed{81}&lt;/math&gt;<br /> <br /> - kante314 -<br /> <br /> ==See also==<br /> #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]<br /> #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_5&diff=158217 2021 JMPSC Sprint Problems/Problem 5 2021-07-12T14:43:17Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> What two-digit even number has digits that sum to &lt;math&gt;17&lt;/math&gt;?<br /> <br /> ==Solution==<br /> There exists a &lt;math&gt;2&lt;/math&gt; digit even number that has digits that sum to &lt;math&gt;17&lt;/math&gt;. Pertaining to the assumption that this operation is in base &lt;math&gt;10&lt;/math&gt;, there exists only &lt;math&gt;10&lt;/math&gt; digits to be used, specifically only &lt;math&gt;5&lt;/math&gt; for the first digit. Only &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt; may be used, as there isn't other pair of digits which sum to &lt;math&gt;17&lt;/math&gt;<br /> <br /> The only two numbers in which satisfy the fact that the digits sum to &lt;math&gt;17&lt;/math&gt; are &lt;math&gt;98&lt;/math&gt; and &lt;math&gt;89&lt;/math&gt;. Yet, only &lt;math&gt;98&lt;/math&gt; works because it is the only one in which satisfies the condition that the number must be even. <br /> <br /> Therefore, &lt;math&gt;\boxed{98}&lt;/math&gt; is the only two-digit even number that has digits that sum to &lt;math&gt;17&lt;/math&gt;.<br /> <br /> -OofPirate<br /> <br /> == Solution 2 ==<br /> The number must have digits &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt;. &lt;math&gt;89&lt;/math&gt; is odd but &lt;math&gt;\boxed{98}&lt;/math&gt; is even.<br /> <br /> - kante314 -<br /> <br /> ==See also==<br /> #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]<br /> #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_3&diff=158216 2021 JMPSC Sprint Problems/Problem 3 2021-07-12T14:41:11Z <p>Kante314: </p> <hr /> <div>==Problem== <br /> If all angles marked with a red square are &lt;math&gt;90^\circ&lt;/math&gt; and all angles marked with one black curve are equal, find the measure of the angle with a question mark.<br /> &lt;center&gt;<br /> [[File:Sprint4.png|400px]]<br /> &lt;/center&gt;<br /> <br /> ==Solution==<br /> <br /> It is given that the right angles are &lt;math&gt;90&lt;/math&gt; degrees, and that all the angles in the two triangles are all equal. We can already infer that the black angles are all &lt;math&gt;60&lt;/math&gt; degrees, since they are equilateral triangles. <br /> <br /> There are &lt;math&gt;360&lt;/math&gt; degrees in a whole circle. We are given two of the black curves, and a &lt;math&gt;90&lt;/math&gt; degree angle, in which all three of them add up to &lt;math&gt;210&lt;/math&gt; degrees.<br /> <br /> &lt;math&gt;360-210=150&lt;/math&gt;. Therefore, the angle marked with a question mark has a measure of &lt;math&gt;150&lt;/math&gt; degrees.<br /> <br /> -OofPirate<br /> <br /> == Solution 2 ==<br /> &lt;math&gt;360^{\circ}-90^{\circ}-60^{\circ}-60^{\circ}=\boxed{150^{\circ}}&lt;/math&gt; <br /> <br /> - kante314 -<br /> <br /> ==See also==<br /> #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]<br /> #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_3&diff=158215 2021 JMPSC Sprint Problems/Problem 3 2021-07-12T14:40:54Z <p>Kante314: </p> <hr /> <div>==Problem== <br /> If all angles marked with a red square are &lt;math&gt;90^\circ&lt;/math&gt; and all angles marked with one black curve are equal, find the measure of the angle with a question mark.<br /> &lt;center&gt;<br /> [[File:Sprint4.png|400px]]<br /> &lt;/center&gt;<br /> <br /> ==Solution==<br /> <br /> It is given that the right angles are &lt;math&gt;90&lt;/math&gt; degrees, and that all the angles in the two triangles are all equal. We can already infer that the black angles are all &lt;math&gt;60&lt;/math&gt; degrees, since they are equilateral triangles. <br /> <br /> There are &lt;math&gt;360&lt;/math&gt; degrees in a whole circle. We are given two of the black curves, and a &lt;math&gt;90&lt;/math&gt; degree angle, in which all three of them add up to &lt;math&gt;210&lt;/math&gt; degrees.<br /> <br /> &lt;math&gt;360-210=150&lt;/math&gt;. Therefore, the angle marked with a question mark has a measure of &lt;math&gt;150&lt;/math&gt; degrees.<br /> <br /> -OofPirate<br /> <br /> == Solution 2 ==<br /> &lt;math&gt;360^{\circ}-90^{\circ}-60^{\circ}-60^{\circ}=\boxed{150^{\circ}}&lt;/math&gt; <br /> <br /> ==See also==<br /> #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]<br /> #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_16&diff=158214 2021 JMPSC Sprint Problems/Problem 16 2021-07-12T14:39:05Z <p>Kante314: </p> <hr /> <div>==Problem== <br /> &lt;math&gt;ABCD&lt;/math&gt; is a concave quadrilateral with &lt;math&gt;AB = 12&lt;/math&gt;, &lt;math&gt;BC = 16&lt;/math&gt;, &lt;math&gt;AD = CD = 26&lt;/math&gt;, and &lt;math&gt;\angle ABC=90^\circ&lt;/math&gt;. Find the area of &lt;math&gt;ABCD&lt;/math&gt;.<br /> &lt;center&gt;<br /> [[File:Sprint16.jpg|350px]]<br /> &lt;/center&gt;<br /> <br /> ==Solution==<br /> <br /> Notice that &lt;math&gt;[ABCD] = [ADC] - [ABC]&lt;/math&gt; and &lt;math&gt;AC = \sqrt{12^2 + 16^2} = 20&lt;/math&gt; by the Pythagorean Thereom. We then have that the area of triangle of &lt;math&gt;ADC&lt;/math&gt; is &lt;math&gt;\frac{20 \cdot \sqrt{26^2 - 10^2}}{2} = 240&lt;/math&gt;, and the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{12 \cdot 16}{2} = 96&lt;/math&gt;, so the area of quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;240 - 96 = 144&lt;/math&gt;. <br /> <br /> ~Mathdreams<br /> <br /> == Solution 2 ==<br /> &lt;cmath&gt;[ACD] = \frac{24 \cdot 20}{2}=240&lt;/cmath&gt;<br /> &lt;cmath&gt;[ABC] = \frac{12 \cdot 16}{2}=96&lt;/cmath&gt; <br /> Therefore, &lt;math&gt;[ABCD] = 240-96=144&lt;/math&gt;<br /> <br /> - kante314 -<br /> <br /> ==See also==<br /> #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]<br /> #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_5&diff=158213 2021 JMPSC Sprint Problems/Problem 5 2021-07-12T14:35:00Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> What two-digit even number has digits that sum to &lt;math&gt;17&lt;/math&gt;?<br /> <br /> ==Solution==<br /> There exists a &lt;math&gt;2&lt;/math&gt; digit even number that has digits that sum to &lt;math&gt;17&lt;/math&gt;. Pertaining to the assumption that this operation is in base &lt;math&gt;10&lt;/math&gt;, there exists only &lt;math&gt;10&lt;/math&gt; digits to be used, specifically only &lt;math&gt;5&lt;/math&gt; for the first digit. Only &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt; may be used, as there isn't other pair of digits which sum to &lt;math&gt;17&lt;/math&gt;<br /> <br /> The only two numbers in which satisfy the fact that the digits sum to &lt;math&gt;17&lt;/math&gt; are &lt;math&gt;98&lt;/math&gt; and &lt;math&gt;89&lt;/math&gt;. Yet, only &lt;math&gt;98&lt;/math&gt; works because it is the only one in which satisfies the condition that the number must be even. <br /> <br /> Therefore, &lt;math&gt;\boxed{98}&lt;/math&gt; is the only two-digit even number that has digits that sum to &lt;math&gt;17&lt;/math&gt;.<br /> <br /> -OofPirate<br /> <br /> == Solution 2 ==<br /> The number must have digits &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt;. &lt;math&gt;89&lt;/math&gt; is odd but &lt;math&gt;\boxed{98}&lt;/math&gt; is even.<br /> <br /> ==See also==<br /> #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]<br /> #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_2&diff=158212 2021 JMPSC Sprint Problems/Problem 2 2021-07-12T14:09:56Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> Brady has an unlimited supply of quarters (\$0.25), dimes (\$0.10), nickels (\$0.05), and pennies (\$0.01). What is the least number (quantity, not type) of coins Brady can use to pay off \$&lt;math&gt;2.78&lt;/math&gt;?<br /> <br /> ==Solution==<br /> It is generally best to use the smallest number of coins with the most value, specifically the quarters, for taking away a big chunk of the problem. We are able to fit &lt;math&gt;11&lt;/math&gt; quarters, or &lt;math&gt;\$2.75&lt;/math&gt; into &lt;math&gt;\$2.78&lt;/math&gt;. That only leaves &lt;math&gt;3&lt;/math&gt; cents. We cannot put any nickels nor dimes, therefore we require three pennies to get a total of &lt;math&gt;\$2.78&lt;/math&gt;.<br /> <br /> The least number of coins Brady can use to pay off &lt;math&gt;\$2.78&lt;/math&gt; will be &lt;math&gt;14&lt;/math&gt; coins.<br /> <br /> -OofPirate<br /> <br /> == Solution 2 ==<br /> You want as many quarters in order to cut down on the number of coins. The most amount of quarters you can have is &lt;math&gt;11&lt;/math&gt;. Since you can't use three cents on anything other than pennies, the remaining coins are &lt;math&gt;3&lt;/math&gt; pennies. Therefore &lt;math&gt;11+3=14&lt;/math&gt;<br /> <br /> - kante314 -<br /> <br /> ==See also==<br /> #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]<br /> #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_2&diff=158211 2021 JMPSC Sprint Problems/Problem 2 2021-07-12T14:09:40Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> Brady has an unlimited supply of quarters (\$0.25), dimes (\$0.10), nickels (\$0.05), and pennies (\$0.01). What is the least number (quantity, not type) of coins Brady can use to pay off \$&lt;math&gt;2.78&lt;/math&gt;?<br /> <br /> ==Solution==<br /> It is generally best to use the smallest number of coins with the most value, specifically the quarters, for taking away a big chunk of the problem. We are able to fit &lt;math&gt;11&lt;/math&gt; quarters, or &lt;math&gt;\$2.75&lt;/math&gt; into &lt;math&gt;\$2.78&lt;/math&gt;. That only leaves &lt;math&gt;3&lt;/math&gt; cents. We cannot put any nickels nor dimes, therefore we require three pennies to get a total of &lt;math&gt;\$2.78&lt;/math&gt;.<br /> <br /> The least number of coins Brady can use to pay off &lt;math&gt;\$2.78&lt;/math&gt; will be &lt;math&gt;14&lt;/math&gt; coins.<br /> <br /> -OofPirate<br /> <br /> == Solution 2 ==<br /> You want as many quarters in order to cut down on the number of coins. The most amount of quarters you can have is &lt;math&gt;11&lt;/math&gt;. Since you can't use three cents on anything other than pennies, the remaining coins are &lt;math&gt;3&lt;/math&gt; pennies. Therefore &lt;math&gt;11+3=14&lt;/math&gt;<br /> <br /> ==See also==<br /> #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]<br /> #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Invitationals_Problems/Problem_8&diff=158210 2021 JMPSC Invitationals Problems/Problem 8 2021-07-12T14:07:48Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be real numbers that satisfy<br /> &lt;cmath&gt;(x+y)^2(20x+21y) = 12&lt;/cmath&gt;<br /> &lt;cmath&gt;(x+y)(20x+21y)^2 = 18.&lt;/cmath&gt;<br /> Find &lt;math&gt;21x+20y&lt;/math&gt;.<br /> <br /> ==Solution==<br /> We let &lt;math&gt;a=(x+y)&lt;/math&gt; and &lt;math&gt;b=(20x+21y)&lt;/math&gt; to get the new system of equations &lt;cmath&gt;a^2b=12 \qquad (1)&lt;/cmath&gt; &lt;cmath&gt;ab^2=18 \qquad(2).&lt;/cmath&gt; Multiplying these two, we have &lt;math&gt;(ab)^3=12 \cdot 18&lt;/math&gt; or &lt;cmath&gt;ab=6 \qquad (3).&lt;/cmath&gt; We divide &lt;math&gt;(3)&lt;/math&gt; by &lt;math&gt;(1)&lt;/math&gt; to get &lt;math&gt;a=2&lt;/math&gt; and divide &lt;math&gt;(2)&lt;/math&gt; by &lt;math&gt;(1)&lt;/math&gt; to get &lt;math&gt;b=3&lt;/math&gt;. Recall that &lt;math&gt;a=x+y=2&lt;/math&gt; and &lt;math&gt;b=20x+21y=3&lt;/math&gt;. Solving the system of equations &lt;cmath&gt;x+y=2&lt;/cmath&gt; &lt;cmath&gt;20x+21y=3,&lt;/cmath&gt; we get &lt;math&gt;y=-37&lt;/math&gt; and &lt;math&gt;x=39&lt;/math&gt;. This means that &lt;cmath&gt;21x+20y=20x+21y+x-y=3+39-(-37)=\boxed{79}.&lt;/cmath&gt; ~samrocksnature<br /> <br /> ==Solution 2==<br /> Each number shares are factor of &lt;math&gt;6&lt;/math&gt;, which means &lt;math&gt;(x+y)(20x+21y)=6&lt;/math&gt;, or &lt;math&gt;x+y=2&lt;/math&gt; and &lt;math&gt;20x+21y=3&lt;/math&gt;. We see &lt;math&gt;y=-37&lt;/math&gt; and &lt;math&gt;x=39&lt;/math&gt;, so &lt;math&gt;39(21)-20(37)=\boxed{79}&lt;/math&gt;<br /> <br /> ~Geometry285<br /> <br /> == Solution 3 ==<br /> Multiplying the equations together, we get<br /> &lt;cmath&gt;(x+y)^3(20x+21y)^3=2^3 \cdot 3^3 \implies (x+y)(20x+21y)=6&lt;/cmath&gt;Therefore,<br /> &lt;cmath&gt;x+y=2 \implies 20x+20y=40&lt;/cmath&gt;&lt;cmath&gt;20x+21y=3&lt;/cmath&gt;Subtracting the equations, we get &lt;math&gt;y=-37&lt;/math&gt; and &lt;math&gt;x=39&lt;/math&gt;, therefore, &lt;math&gt;21 (39) - 20 (37) =\boxed{79}&lt;/math&gt;<br /> <br /> - kante314 -<br /> <br /> ==See also==<br /> #[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]<br /> #[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Invitationals_Problems/Problem_1&diff=158209 2021 JMPSC Invitationals Problems/Problem 1 2021-07-12T14:06:31Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> The equation &lt;math&gt;ax^2 + 5x = 4,&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; is some constant, has &lt;math&gt;x = 1&lt;/math&gt; as a solution. What is the other solution?<br /> <br /> ==Solution==<br /> Since &lt;math&gt;x=1&lt;/math&gt; must be a solution, &lt;math&gt;a+5=4&lt;/math&gt; must be true. Therefore, &lt;math&gt;a = -1&lt;/math&gt;. We plug this back in to the original quadratic to get &lt;math&gt;5x-x^2=4&lt;/math&gt;. We can solve this quadratic to get &lt;math&gt;1,4&lt;/math&gt;. We are asked to find the 2nd solution so our answer is &lt;math&gt;\boxed{4}&lt;/math&gt;<br /> <br /> ~Grisham<br /> <br /> ==Solution 2==<br /> Plug &lt;math&gt;x=1&lt;/math&gt; to get &lt;math&gt;a=-1&lt;/math&gt;, so &lt;math&gt;x^2-5x+4=0&lt;/math&gt;, or &lt;math&gt;(x-4)(x-1)=0&lt;/math&gt;, meaning the other solution is &lt;math&gt;x=\boxed{4}&lt;/math&gt;<br /> &lt;math&gt;\linebreak&lt;/math&gt;<br /> ~Geometry285<br /> <br /> == Solution 3 ==<br /> &lt;cmath&gt;ax^2+5x-4=0&lt;/cmath&gt;Plugging in &lt;math&gt;1&lt;/math&gt;, we get &lt;math&gt;a+5-4=0 \implies a+1=0 \implies a=-1&lt;/math&gt;, therefore,<br /> &lt;cmath&gt;-x^2+5x-4=0 \implies (x-4)(x-1)=0&lt;/cmath&gt;Finally, we get the other root is &lt;math&gt;4&lt;/math&gt;.<br /> <br /> - kante314 -<br /> <br /> ==See also==<br /> #[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]<br /> #[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Accuracy_Problems/Problem_5&diff=158208 2021 JMPSC Accuracy Problems/Problem 5 2021-07-12T14:04:21Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;n!=n \cdot (n-1) \cdot (n-2) \cdots 2 \cdot 1&lt;/math&gt; for all positive integers &lt;math&gt;n&lt;/math&gt;. Find the value of &lt;math&gt;x&lt;/math&gt; that satisfies &lt;cmath&gt;\frac{5!x}{2022!}=\frac{20}{2021!}.&lt;/cmath&gt;<br /> <br /> ==Solution==<br /> We can multiply both sides by &lt;math&gt;2022!&lt;/math&gt; to get rid of the fractions<br /> &lt;cmath&gt;\frac{5!x}{2022!}=\frac{20}{2021!}&lt;/cmath&gt;<br /> &lt;cmath&gt;5!x=20 \cdot 2022&lt;/cmath&gt;<br /> &lt;cmath&gt;120x=(120)(337)&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\boxed{337}&lt;/cmath&gt;<br /> <br /> ~Bradygho<br /> <br /> == Solution 2 ==<br /> &lt;cmath&gt;\frac{120x}{2022}=20 \implies \frac{6x}{2022}=1 \implies x=337&lt;/cmath&gt;<br /> <br /> - kante314 -<br /> <br /> ==See also==<br /> #[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]<br /> #[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Invitationals_Problems/Problem_6&diff=158207 2021 JMPSC Invitationals Problems/Problem 6 2021-07-12T14:01:22Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> Five friends decide to meet together for a party. However, they did not plan the party well, and at noon, every friend leaves their own house and travels to one of the other four friends' houses, chosen uniformly at random. The probability that every friend sees another friend in the house they chose can be expressed in the form &lt;math&gt;\frac{m}{n}&lt;/math&gt;. If &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> There are &lt;math&gt;4^5&lt;/math&gt; ways for the friends to choose houses. Now, we have friends can meet in pairs, so &lt;math&gt;(2,3)&lt;/math&gt; is the only way they can meet. There are &lt;math&gt;20&lt;/math&gt; ways for them to choose the &lt;math&gt;2&lt;/math&gt; houses to meet, and &lt;math&gt;3&lt;/math&gt; ways to rotate the &quot;extra&quot; partner (2 houses are already selected) The answer is &lt;cmath&gt;\frac{2^2 \cdot 3 \cdot 5}{2^{10}} \implies \frac{15}{256} \implies 256+15=\boxed{271}.&lt;/cmath&gt; ~Geometry285<br /> <br /> == Solution 2 ==<br /> &lt;cmath&gt;\frac{2^2 \cdot 3 \cdot 5}{4^{5}} \implies \frac{15}{256}&lt;/cmath&gt;Therefore, &lt;math&gt;15+256=\boxed{271}&lt;/math&gt;<br /> <br /> - kante314 -<br /> <br /> ==See also==<br /> #[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]<br /> #[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Invitationals_Problems/Problem_6&diff=158206 2021 JMPSC Invitationals Problems/Problem 6 2021-07-12T14:01:08Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> Five friends decide to meet together for a party. However, they did not plan the party well, and at noon, every friend leaves their own house and travels to one of the other four friends' houses, chosen uniformly at random. The probability that every friend sees another friend in the house they chose can be expressed in the form &lt;math&gt;\frac{m}{n}&lt;/math&gt;. If &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> There are &lt;math&gt;4^5&lt;/math&gt; ways for the friends to choose houses. Now, we have friends can meet in pairs, so &lt;math&gt;(2,3)&lt;/math&gt; is the only way they can meet. There are &lt;math&gt;20&lt;/math&gt; ways for them to choose the &lt;math&gt;2&lt;/math&gt; houses to meet, and &lt;math&gt;3&lt;/math&gt; ways to rotate the &quot;extra&quot; partner (2 houses are already selected) The answer is &lt;cmath&gt;\frac{2^2 \cdot 3 \cdot 5}{2^{10}} \implies \frac{15}{256} \implies 256+15=\boxed{271}.&lt;/cmath&gt; ~Geometry285<br /> <br /> == Solution 2 ==<br /> &lt;cmath&gt;\frac{2^2 \cdot 3 \cdot 5}{4^{5}} \implies \frac{15}{256}&lt;/cmath&gt;Therefore, &lt;math&gt;15+256=\boxed{271}&lt;/math&gt;<br /> <br /> ==See also==<br /> #[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]<br /> #[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_20&diff=158205 2021 JMPSC Sprint Problems/Problem 20 2021-07-12T14:00:08Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> For all integers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, define the operation &lt;math&gt;\Delta&lt;/math&gt; as &lt;cmath&gt;x \Delta y = x^3+y^2+x+y.&lt;/cmath&gt; Find &lt;cmath&gt;\sqrt{\dfrac{257 \Delta 256}{258}}.&lt;/cmath&gt; <br /> <br /> ==Solution==<br /> Let &lt;math&gt;258=a&lt;/math&gt;. Then, &lt;math&gt;257=a-1&lt;/math&gt; and &lt;math&gt;256=a-2&lt;/math&gt;. We substitute these values into expression &lt;math&gt;(1)&lt;/math&gt; to get &lt;cmath&gt;\sqrt{\frac{(a-1) \Delta (a-2)}{a}}.&lt;/cmath&gt; Recall the definition for the operation &lt;math&gt;\Delta&lt;/math&gt;; using this, we simplify our expression to &lt;cmath&gt;\sqrt{\frac{(a-1)^3+(a-2)^2+(a-1)+(a-2)}{a}}.&lt;/cmath&gt; We have &lt;math&gt;(a-1)^3=a^3-3a^2+3a-1&lt;/math&gt; and &lt;math&gt;(a-2)^2=a^2-4a+4&lt;/math&gt;, so we can expand the numerator of the fraction within the square root as &lt;math&gt;a^3-3a^2+3a-1+a^2-4a+4+a-1+a-2=a^3-2a^2+a&lt;/math&gt; to get &lt;cmath&gt;\sqrt{\frac{a^3-2a^2+a}{a}}=\sqrt{a^2-2a+1}=\sqrt{(a-1)^2}=a-1=\boxed{257}.&lt;/cmath&gt; ~samrocksnature<br /> <br /> <br /> ==Solution 2==<br /> <br /> Basically the same as above, but instead we can let &lt;math&gt;257 = 256 + 1&lt;/math&gt;. Then we have<br /> &lt;cmath&gt;\sqrt{\frac{(256+1)(256^2 + 256 + 1) + 1(256^2 + 257) + 256}{258}},&lt;/cmath&gt;<br /> &lt;cmath&gt;\sqrt{\frac{258(256^2 + 257) + 256}{258}},&lt;/cmath&gt;<br /> &lt;cmath&gt;\sqrt{256^2 + 256 + 256 + 1} =&lt;/cmath&gt; &lt;cmath&gt;\sqrt{256^2 + 2\cdot256 + 1} =&lt;/cmath&gt; &lt;cmath&gt;\sqrt{(256+1)^2} =&lt;/cmath&gt; &lt;cmath&gt;\sqrt{(257^2)}&lt;/cmath&gt;<br /> <br /> which equals &lt;math&gt;\boxed{257}&lt;/math&gt;.<br /> <br /> <br /> ~~abhinavg0627<br /> <br /> == Note: ==<br /> <br /> &lt;math&gt;257^3 = 16974593&lt;/math&gt;, &lt;math&gt;256^2 = 65536&lt;/math&gt;, and &lt;math&gt;257^2 = 66049&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> Notice that &lt;math&gt;x=y+1&lt;/math&gt;, substituting this in, we get &lt;math&gt;x^2(x+1)&lt;/math&gt;. Therefore, &lt;math&gt;\sqrt{\frac{257^2(258)}{258}}=\boxed{257}&lt;/math&gt;<br /> <br /> - kante314 -<br /> <br /> ==See also==<br /> #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]<br /> #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_20&diff=158204 2021 JMPSC Sprint Problems/Problem 20 2021-07-12T13:59:50Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> For all integers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, define the operation &lt;math&gt;\Delta&lt;/math&gt; as &lt;cmath&gt;x \Delta y = x^3+y^2+x+y.&lt;/cmath&gt; Find &lt;cmath&gt;\sqrt{\dfrac{257 \Delta 256}{258}}.&lt;/cmath&gt; <br /> <br /> ==Solution==<br /> Let &lt;math&gt;258=a&lt;/math&gt;. Then, &lt;math&gt;257=a-1&lt;/math&gt; and &lt;math&gt;256=a-2&lt;/math&gt;. We substitute these values into expression &lt;math&gt;(1)&lt;/math&gt; to get &lt;cmath&gt;\sqrt{\frac{(a-1) \Delta (a-2)}{a}}.&lt;/cmath&gt; Recall the definition for the operation &lt;math&gt;\Delta&lt;/math&gt;; using this, we simplify our expression to &lt;cmath&gt;\sqrt{\frac{(a-1)^3+(a-2)^2+(a-1)+(a-2)}{a}}.&lt;/cmath&gt; We have &lt;math&gt;(a-1)^3=a^3-3a^2+3a-1&lt;/math&gt; and &lt;math&gt;(a-2)^2=a^2-4a+4&lt;/math&gt;, so we can expand the numerator of the fraction within the square root as &lt;math&gt;a^3-3a^2+3a-1+a^2-4a+4+a-1+a-2=a^3-2a^2+a&lt;/math&gt; to get &lt;cmath&gt;\sqrt{\frac{a^3-2a^2+a}{a}}=\sqrt{a^2-2a+1}=\sqrt{(a-1)^2}=a-1=\boxed{257}.&lt;/cmath&gt; ~samrocksnature<br /> <br /> <br /> ==Solution 2==<br /> <br /> Basically the same as above, but instead we can let &lt;math&gt;257 = 256 + 1&lt;/math&gt;. Then we have<br /> &lt;cmath&gt;\sqrt{\frac{(256+1)(256^2 + 256 + 1) + 1(256^2 + 257) + 256}{258}},&lt;/cmath&gt;<br /> &lt;cmath&gt;\sqrt{\frac{258(256^2 + 257) + 256}{258}},&lt;/cmath&gt;<br /> &lt;cmath&gt;\sqrt{256^2 + 256 + 256 + 1} =&lt;/cmath&gt; &lt;cmath&gt;\sqrt{256^2 + 2\cdot256 + 1} =&lt;/cmath&gt; &lt;cmath&gt;\sqrt{(256+1)^2} =&lt;/cmath&gt; &lt;cmath&gt;\sqrt{(257^2)}&lt;/cmath&gt;<br /> <br /> which equals &lt;math&gt;\boxed{257}&lt;/math&gt;.<br /> <br /> <br /> ~~abhinavg0627<br /> <br /> == Note: ==<br /> <br /> &lt;math&gt;257^3 = 16974593&lt;/math&gt;, &lt;math&gt;256^2 = 65536&lt;/math&gt;, and &lt;math&gt;257^2 = 66049&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> Notice that &lt;math&gt;x=y+1&lt;/math&gt;, substituting this in, we get &lt;math&gt;x^2(x+1)&lt;/math&gt;. Therefore, &lt;math&gt;\sqrt{\frac{257^2(258)}{258}}=\boxed{257}&lt;/math&gt;<br /> <br /> ==See also==<br /> #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]<br /> #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Accuracy_Problems/Problem_7&diff=158203 2021 JMPSC Accuracy Problems/Problem 7 2021-07-12T13:57:37Z <p>Kante314: </p> <hr /> <div>==Problem==<br /> If &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; each represent a single digit and they satisfy the equation &lt;cmath&gt;\begin{array}{cccc}&amp; A &amp; B &amp; C \\ \times &amp; &amp; &amp;3 \\ \hline &amp; 7 &amp; 9 &amp; C\end{array},&lt;/cmath&gt; find &lt;math&gt;3A+2B+C&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Notice that &lt;math&gt;C&lt;/math&gt; can only be &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;. However, &lt;math&gt;790&lt;/math&gt; is not divisible by &lt;math&gt;3&lt;/math&gt;, so &lt;cmath&gt;3 \times ABC = 795&lt;/cmath&gt; &lt;cmath&gt;ABC = 265&lt;/cmath&gt; Thus, &lt;math&gt;3A + 2B + C = \boxed{23}&lt;/math&gt;<br /> <br /> ~Bradygho<br /> <br /> ==Solution 2==<br /> Clearly we see &lt;math&gt;C=1&lt;/math&gt; does not work, but &lt;math&gt;C=5&lt;/math&gt; works with simple guess-and-check. We have &lt;math&gt;AB5=\frac{795}{3}=265&lt;/math&gt;, so &lt;math&gt;A=2&lt;/math&gt; and &lt;math&gt;B=6&lt;/math&gt;. The answer is &lt;math&gt;3(2)+6(2)+1(5)=\boxed{23}&lt;/math&gt;<br /> <br /> ~Geometry285<br /> <br /> == Solution 3 ==<br /> Easily, we can see that &lt;math&gt;A=2&lt;/math&gt;. Therefore,&lt;cmath&gt;\overline{BC} \cdot 3 = \overline{19C}.&lt;/cmath&gt;We can see that &lt;math&gt;C&lt;/math&gt; must be &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt;. If &lt;math&gt;C=1&lt;/math&gt;, then&lt;cmath&gt;\overline{B1} \cdot 3 = 191.&lt;/cmath&gt;This doesn't work because &lt;math&gt;191&lt;/math&gt; isn't divisible by &lt;math&gt;3&lt;/math&gt;. If &lt;math&gt;C=5&lt;/math&gt;, then&lt;cmath&gt;\overline{B5} \cdot 3 = 195.&lt;/cmath&gt;Therefore, &lt;math&gt;B=6&lt;/math&gt;. So, we have &lt;math&gt;3(2) + 2(6) + 5=6+12+5=18+5=\boxed{23}&lt;/math&gt;.<br /> <br /> - kante314 -<br /> <br /> ==See also==<br /> #[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]<br /> #[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_1&diff=158202 2021 JMPSC Sprint Problems/Problem 1 2021-07-12T13:56:09Z <p>Kante314: </p> <hr /> <div>== Problem ==<br /> <br /> Compute &lt;math&gt;\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{9}\right)(3+6+9)&lt;/math&gt;. <br /> <br /> == Solution ==<br /> <br /> Solving the right side gives &lt;math&gt;3 + 6 + 9 = 18&lt;/math&gt;. Distributing into the left side gives &lt;math&gt;\frac{18}{3}+\frac{18}{6}+\frac{18}{9}&lt;/math&gt;, so the answer is &lt;math&gt;6 + 3 + 2 = \boxed{11}&lt;/math&gt;.<br /> <br /> <br /> == Solution 2 ==<br /> <br /> &lt;math&gt;\frac{1}{3}+\frac{1}{6}+\frac{1}{9}=\frac{6}{18}+\frac{3}{18}+\frac{2}{18}=\frac{11}{18}&lt;/math&gt; and &lt;math&gt;3+6+9=18&lt;/math&gt;, so the answer is &lt;math&gt;\frac{11}{18}\cdot18=\boxed{11}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> &lt;cmath&gt;3 \left(\frac{1}{3}\right) + 3 \left(\frac{1}{6}\right) + 3 \left(\frac{1}{9}\right)=1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}&lt;/cmath&gt;&lt;cmath&gt;6 \left(\frac{1}{3}\right) + 6 \left(\frac{1}{6}\right) + 6 \left(\frac{1}{9}\right)=2+1+\frac{2}{3}=\frac{22}{6}&lt;/cmath&gt;&lt;cmath&gt;9 \left(\frac{1}{3}\right) + 9 \left(\frac{1}{6}\right) + 9 \left(\frac{1}{9}\right)=3+\frac{3}{2}+1=\frac{33}{6}&lt;/cmath&gt;<br /> Therefore, the answer is &lt;math&gt;\frac{11}{6}+\frac{22}{6}+\frac{33}{6}=11&lt;/math&gt;<br /> <br /> - kante314 -<br /> <br /> ==See also==<br /> #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]<br /> #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]<br /> #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]<br /> {{JMPSC Notice}}</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Invitationals_Problems/Problem_15&diff=157952 2021 JMPSC Invitationals Problems/Problem 15 2021-07-11T19:26:32Z <p>Kante314: /* Solution */</p> <hr /> <div>==Problem==<br /> Abhishek is choosing positive integer factors of &lt;math&gt;2021 \times 2^{2021}&lt;/math&gt; with replacement. After a minute passes, he chooses a random factor and writes it down. Abhishek repeats this process until the first time the product of all numbers written down is a perfect square. Find the expected number of minutes it takes for him to stop. <br /> <br /> ==Solution==<br /> asdf orz</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Accuracy_Problems/Problem_4&diff=157814 2021 JMPSC Accuracy Problems/Problem 4 2021-07-11T04:20:05Z <p>Kante314: /* Solution */</p> <hr /> <div>==Problem==<br /> If &lt;math&gt;\frac{x+2}{6}&lt;/math&gt; is its own reciprocal, find the product of all possible values of &lt;math&gt;x.&lt;/math&gt;<br /> <br /> ==Solution==<br /> From the problem, we know that <br /> &lt;cmath&gt;\frac{x+2}{6} = \frac{6}{x+2}&lt;/cmath&gt;<br /> &lt;cmath&gt;(x+2)^2 = 6^2&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2+ 4x + 4 = 36&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2 + 4x - 32 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;(x-8)(x+4) = 0&lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;x = 8&lt;/math&gt; or &lt;math&gt;x = -4&lt;/math&gt;. Our answer is &lt;math&gt;8 \cdot(-4)=\boxed{-32}&lt;/math&gt;<br /> <br /> ~Bradygho<br /> <br /> &lt;math&gt;\frac{x+2}{6}=\frac{6}{x+2} \implies x^2+4x-32&lt;/math&gt; Therefore, the product of the root is &lt;math&gt;-32&lt;/math&gt; ~ kante314</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Accuracy_Problems/Problem_13&diff=157813 2021 JMPSC Accuracy Problems/Problem 13 2021-07-11T04:19:12Z <p>Kante314: /* Solution */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be nonnegative integers such that &lt;math&gt;(x+y)^2+(xy)^2=25.&lt;/math&gt; Find the sum of all possible values of &lt;math&gt;x.&lt;/math&gt;<br /> <br /> ==Solution==<br /> asdf<br /> <br /> Case 1: &lt;math&gt;x+y=4,3&lt;/math&gt;.<br /> There are no possible answer when &lt;math&gt;x+y=3&lt;/math&gt;, but when &lt;math&gt;x+y=4&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can equal &lt;math&gt;3&lt;/math&gt; or &lt;math&gt;1&lt;/math&gt;.<br /> Case 2: &lt;math&gt;x+y=5,0&lt;/math&gt;<br /> This works when &lt;math&gt;x=0,5&lt;/math&gt;.<br /> Therefore, the answer is &lt;math&gt;9&lt;/math&gt;. ~ kante314</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Accuracy_Problems/Problem_13&diff=157812 2021 JMPSC Accuracy Problems/Problem 13 2021-07-11T04:19:00Z <p>Kante314: /* Solution */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be nonnegative integers such that &lt;math&gt;(x+y)^2+(xy)^2=25.&lt;/math&gt; Find the sum of all possible values of &lt;math&gt;x.&lt;/math&gt;<br /> <br /> ==Solution==<br /> asdf<br /> <br /> Easily, we can see that &lt;math&gt;C=3,5&lt;/math&gt;. &lt;math&gt;C&lt;/math&gt; must be &lt;math&gt;5&lt;/math&gt; because &lt;math&gt;795&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;793&lt;/math&gt; is not divisible by &lt;math&gt;3&lt;/math&gt;. Therefore &lt;math&gt;\frac{795}{3}=265&lt;/math&gt;. So, &lt;math&gt;3A+2B+C=6+12+5=23&lt;/math&gt;<br /> <br /> 13.<br /> Case 1: &lt;math&gt;x+y=4,3&lt;/math&gt;.<br /> There are no possible answer when &lt;math&gt;x+y=3&lt;/math&gt;, but when &lt;math&gt;x+y=4&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can equal &lt;math&gt;3&lt;/math&gt; or &lt;math&gt;1&lt;/math&gt;.<br /> Case 2: &lt;math&gt;x+y=5,0&lt;/math&gt;<br /> This works when &lt;math&gt;x=0,5&lt;/math&gt;.<br /> Therefore, the answer is &lt;math&gt;9&lt;/math&gt;. ~ kante314</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=154024 User:Piphi 2021-05-20T22:01:21Z <p>Kante314: /* User Count */</p> <hr /> <div>{{User:Piphi/Template:Header}}<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;100px&quot;&gt;494&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#919293;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> &lt;!-- &lt;div&gt;Q: How to annoy a web developer?&lt;/span&gt; :P --&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;&lt;font color=&quot;black&quot;&gt;<br /> Piphi created [[User:Piphi/Games|AoPS Wiki Games by Piphi]].&lt;br&gt;<br /> <br /> Piphi has been very close to winning multiple [[Greed Control]] games, piphi placed 5th in game #18 and 2nd in game #19. Thanks to piphi, Greed Control games have started to be kept track of. Piphi made a spreadsheet that has all of Greed Control history [https://artofproblemsolving.com/community/c19451h2126208p15569802 here].&lt;br&gt;<br /> <br /> Piphi has a side-project that is making the Wiki's [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]].&lt;br&gt;<br /> <br /> Piphi is a proud member of [https://artofproblemsolving.com/community/c562043 The Interuniversal GMAAS Society].<br /> <br /> Piphi is part of the [https://artofproblemsolving.com/community/c1124279 Asymptote Competition] staff.<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#333333;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;#f0f2f3&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;#f0f2f3&quot;&gt;<br /> You can check out more goals/statistics [[User:Piphi/Statistics|here]].<br /> <br /> A User Count of 500<br /> {{User:Piphi/Template:Progress_Bar|98|width=100%}}<br /> <br /> 200 subpages of [[User:Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|99|width=100%}}<br /> <br /> 200 signups for [[User:Piphi/Games|AoPS Wiki Games by Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|54|width=100%}}<br /> <br /> Make 10,000 edits<br /> {{User:Piphi/Template:Progress_Bar|22.99|width=100%}}&lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=User:Rusczyk&diff=147876 User:Rusczyk 2021-02-24T22:13:38Z <p>Kante314: /* User Count */</p> <hr /> <div>Rusczyk's Page:<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#4EC284&quot;&gt;<br /> ==&lt;font color=&quot;white&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;white&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;&lt;font color=&quot;white&quot;&gt;<br /> &lt;center&gt;&lt;font size=&quot;101px&quot;&gt;86&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#4EC284;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;white&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;white&quot;&gt;Rusczyk is currently borderline AIME.&lt;br&gt;<br /> <br /> Rusczyk just turned 13 years old.&lt;br&gt;<br /> <br /> Rusczyk scored 44/46 when mocking the 2018 MATHCOUNTS State test, and got silver on the 2020 online MATHCOUNTS State held on AoPS.&lt;br&gt;<br /> <br /> Rusczyk is a pro at maths and physics<br /> <br /> Rusczyk has come world and country #1 in various international tournaments and competitions starting from 2017<br /> <br /> I am better than Rusczyk at math<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#4EC284;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;white&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;white&quot;&gt; A User Count of &lt;math&gt;\color{white}{\infty}&lt;/math&gt;<br /> <br /> Make AIME 2021 (Currently borderline)<br /> <br /> Pass AP Calculus AB, BC and AP Physics exam<br /> <br /> Get in the Alcumus HoF in the next 6 months<br /> <br /> Convince OlympusHero that he is better than Rusczyk (at Math)<br /> <br /> Get &lt;math&gt;\color{white}{2 \times}&lt;/math&gt; medals this year as compared to what they did last year. That is &lt;math&gt;\color{white}{2 \times 14 = \boxed{28}}&lt;/math&gt; which is nearly impossible.<br /> &lt;/div&gt;<br /> &lt;/div&gt;</div> Kante314 https://artofproblemsolving.com/wiki/index.php?title=User:Rusczyk&diff=147875 User:Rusczyk 2021-02-24T22:12:33Z <p>Kante314: /* User Count */</p> <hr /> <div>Rusczyk's Page:<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#4EC284&quot;&gt;<br /> ==&lt;font color=&quot;white&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;white&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;&lt;font color=&quot;white&quot;&gt;<br /> &lt;center&gt;&lt;font size=&quot;101px&quot;&gt;65&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#4EC284;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;white&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;white&quot;&gt;Rusczyk is currently borderline AIME.&lt;br&gt;<br /> <br /> Rusczyk just turned 13 years old.&lt;br&gt;<br /> <br /> Rusczyk scored 44/46 when mocking the 2018 MATHCOUNTS State test, and got silver on the 2020 online MATHCOUNTS State held on AoPS.&lt;br&gt;<br /> <br /> Rusczyk is a pro at maths and physics<br /> <br /> Rusczyk has come world and country #1 in various international tournaments and competitions starting from 2017<br /> <br /> I am better than Rusczyk at math<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#4EC284;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;white&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;white&quot;&gt; A User Count of &lt;math&gt;\color{white}{\infty}&lt;/math&gt;<br /> <br /> Make AIME 2021 (Currently borderline)<br /> <br /> Pass AP Calculus AB, BC and AP Physics exam<br /> <br /> Get in the Alcumus HoF in the next 6 months<br /> <br /> Convince OlympusHero that he is better than Rusczyk (at Math)<br /> <br /> Get &lt;math&gt;\color{white}{2 \times}&lt;/math&gt; medals this year as compared to what they did last year. That is &lt;math&gt;\color{white}{2 \times 14 = \boxed{28}}&lt;/math&gt; which is nearly impossible.<br /> &lt;/div&gt;<br /> &lt;/div&gt;</div> Kante314