https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=CoolCarsOnTheRun&feedformat=atom AoPS Wiki - User contributions [en] 2021-07-24T19:50:26Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_10&diff=153628 2020 AIME II Problems/Problem 10 2021-05-12T22:32:55Z <p>CoolCarsOnTheRun: improved some explanation and readability</p> <hr /> <div>== Problem ==<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that when &lt;math&gt;1^3+2^3+3^3+\cdots +n^3&lt;/math&gt; is divided by &lt;math&gt;n+5&lt;/math&gt;, the remainder is &lt;math&gt;17&lt;/math&gt;.<br /> <br /> == Solution 1 (If you don't remember the formula for sum of cubes) ==<br /> We first note that since the remainder is &lt;math&gt;17&lt;/math&gt; and we are dividing by &lt;math&gt;n+5&lt;/math&gt;, &lt;math&gt;n+5&lt;/math&gt; must be greater than &lt;math&gt;17&lt;/math&gt;, meaning that &lt;math&gt;n&lt;/math&gt; has to be at least &lt;math&gt;13&lt;/math&gt;.<br /> <br /> We then notice that we can pair the &lt;math&gt;5^3&lt;/math&gt; term with the &lt;math&gt;n^3&lt;/math&gt; term to factor it into &lt;math&gt;(n+5)(n^2-5n+25)&lt;/math&gt; using the sum of two cubes factorization (&lt;math&gt;a^3+b^3=(a+b)(a^2-ab+b^2)&lt;/math&gt;), which is divisible by &lt;math&gt;n+5&lt;/math&gt;. We can do the same for the &lt;math&gt;6^3&lt;/math&gt; term with the &lt;math&gt;(n-1)^3&lt;/math&gt; term, the &lt;math&gt;7^3&lt;/math&gt; term with the &lt;math&gt;(n-2)^3&lt;/math&gt;, and so on, which are all divisible by &lt;math&gt;n+5&lt;/math&gt;. However, when &lt;math&gt;n&lt;/math&gt; is odd, we will have a middle term that is not paired with any other terms, which is not necessarily divisible by &lt;math&gt;n+5&lt;/math&gt;. Thus, we have two cases:<br /> &lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;\textbf{CASE 1: } n&lt;/math&gt; is even &lt;br&gt;<br /> If &lt;math&gt;n&lt;/math&gt; is even, all terms that are greater than &lt;math&gt;4^3&lt;/math&gt; pair, as there are an even number of terms that are greater than &lt;math&gt;4^3&lt;/math&gt;. Therefore, all we need in order for the entire sequence to have a remainder of &lt;math&gt;17&lt;/math&gt; when divided by &lt;math&gt;n+5&lt;/math&gt; is &lt;math&gt;1^3+2^3+3^3+4^3&lt;/math&gt; to have a remainder of &lt;math&gt;17&lt;/math&gt; when divided by &lt;math&gt;n+5&lt;/math&gt;.<br /> <br /> Evaluating &lt;math&gt;1^3+2^3+3^3+4^3&lt;/math&gt; as &lt;math&gt;100&lt;/math&gt;, all we need to be true is <br /> &lt;cmath&gt;100\equiv 17\pmod {n+5},&lt;/cmath&gt;<br /> or that<br /> &lt;cmath&gt;83\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> Thus, &lt;math&gt;83&lt;/math&gt; will be divisible by &lt;math&gt;n+5&lt;/math&gt; where &lt;math&gt;n\geq 13&lt;/math&gt;. As &lt;math&gt;83&lt;/math&gt; is prime, &lt;math&gt;n+5&lt;/math&gt; must be equal to either &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;83&lt;/math&gt;. If &lt;math&gt;n+5=1&lt;/math&gt;, we have that &lt;math&gt;n=-4&lt;/math&gt;, which is not greater than or equal to &lt;math&gt;13&lt;/math&gt;, so that solution is extraneous.<br /> <br /> If &lt;math&gt;n+5=83&lt;/math&gt;, we have that &lt;math&gt;n=78&lt;/math&gt;, which is &lt;math&gt;\geq 13&lt;/math&gt;, so one of our solutions is &lt;math&gt;n=78&lt;/math&gt;, and we are done with our first case.<br /> &lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;\textbf{CASE 2: } n&lt;/math&gt; is odd &lt;br&gt;<br /> If &lt;math&gt;n&lt;/math&gt; is odd, the only term that does not pair is the arithmetic mean of the numbers under the cube of the largest and smallest terms that would pair, or &lt;math&gt;\left(\frac{n+5}{2}\right)^3&lt;/math&gt;. Therefore, as all other terms that are &lt;math&gt;\geq 5^3&lt;/math&gt; pair, the requirement that we have is<br /> &lt;cmath&gt;1^3+2^3+3^3+4^3+\left(\frac{n+5}{2}\right)^3\equiv 17\pmod {n+5}.&lt;/cmath&gt;<br /> Calculating and simplifying, we have that<br /> &lt;cmath&gt;83+\left(\frac{n+5}{2}\right)^3\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> Now, we multiply both sides by &lt;math&gt;8&lt;/math&gt;. However, since multiplication is not reversible in modular arithmetic, we need to check whether any solutions are extraneous after solving. The congruence that we now have is<br /> &lt;cmath&gt;8\cdot 83+(n+5)^3\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> As we know that &lt;math&gt;(n+5)^3&lt;/math&gt; is divisible by &lt;math&gt;n+5&lt;/math&gt;, what we need now is<br /> &lt;cmath&gt;8\cdot 83\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> We now check each solution to see whether it works.<br /> <br /> If &lt;math&gt;n+5=1, 2, 4, 8&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would be less than &lt;math&gt;13&lt;/math&gt;, so none of these solutions work. If &lt;math&gt;n+5=83&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would be even, so that solution does not work for this case. Therefore, the only three solutions we need to check for this case are when &lt;math&gt;n+5=166&lt;/math&gt;, &lt;math&gt;n+5=332&lt;/math&gt;, or &lt;math&gt;n+5=664&lt;/math&gt;. We plug these values into the congruence before we multiplied both sides by &lt;math&gt;8&lt;/math&gt; to check.<br /> <br /> If &lt;math&gt;n+5=166&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{166}{2}\right)^3\equiv 0\pmod {166}.&lt;/cmath&gt;<br /> Calculating and factoring out &lt;math&gt;83&lt;/math&gt;, we have that<br /> &lt;cmath&gt;83(1+83\cdot 83)\equiv 0\pmod {166}.&lt;/cmath&gt;<br /> As the right parenthesis is odd and &lt;math&gt;166=83\cdot 2&lt;/math&gt;, we know that this solution works, so we have another solution: &lt;math&gt;n=166-5=161&lt;/math&gt;.<br /> <br /> If &lt;math&gt;n+5=332&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{332}{2}\right)^3=83+166^3\equiv 0\pmod {322}.&lt;/cmath&gt;<br /> As the left hand side is odd, but all multiples of &lt;math&gt;322&lt;/math&gt; is even, this solution is therefore extraneous.<br /> <br /> If &lt;math&gt;n+5=664&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{664}{2}\right)^3=83+332^3\equiv 0\pmod {664}.&lt;/cmath&gt;<br /> Again, the left hand side is odd, and all multiples of &lt;math&gt;664&lt;/math&gt; are even, so this solution is extraneous.<br /> &lt;br&gt;&lt;br&gt;<br /> Therefore, our final answer is &lt;math&gt;78+161=\boxed{239}&lt;/math&gt;.<br /> &lt;br&gt;&lt;br&gt;<br /> ~ CoolCarsOnTheRun<br /> <br /> ==Solution 2 (w/ formula)==<br /> Let &lt;math&gt;m=n+5&lt;/math&gt;. Then we have<br /> &lt;cmath&gt;1^3+2^3+3^3+\cdots+(m-5)^3\equiv 17 \mod m&lt;/cmath&gt;<br /> &lt;cmath&gt;\left(\frac{(m-5)(m-4)}{2}\right)^2\equiv 17 \mod m &lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{400}{4}\equiv 17 \mod m &lt;/cmath&gt;<br /> &lt;cmath&gt;332 \equiv 0 \mod m &lt;/cmath&gt;<br /> So, &lt;math&gt;m\in\{83,166,332\}&lt;/math&gt;. Testing, the cases, only &lt;math&gt;332&lt;/math&gt; fails. This leaves &lt;math&gt;78+161=\boxed{239}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Official MAA 1)==<br /> The sum of the cubes from 1 to &lt;math&gt;n&lt;/math&gt; is<br /> &lt;cmath&gt;1^3+2^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}.&lt;/cmath&gt;For this to be equal to &lt;math&gt;(n+5)q+17&lt;/math&gt; for some integer &lt;math&gt;q&lt;/math&gt;, it must be that&lt;cmath&gt;n^2(n+1)^2=4(n+5)q+4\cdot 17,&lt;/cmath&gt;so&lt;cmath&gt;n^2(n+1)^2 \equiv 4 \cdot 17= 68\hskip-.2cm \pmod{n+5}.&lt;/cmath&gt;But &lt;math&gt;n^2(n+1)^2 \equiv (-5)^2(-4)^2 = 400 \pmod{n+5}.&lt;/math&gt; Thus &lt;math&gt;n^2(n+1)^2&lt;/math&gt; is congruent to both &lt;math&gt;68&lt;/math&gt; and &lt;math&gt;400,&lt;/math&gt; which implies that &lt;math&gt;n+5&lt;/math&gt; divides &lt;math&gt;400-68 = 332=2^2 \cdot 83&lt;/math&gt;. Because &lt;math&gt;n+5 &gt; 17&lt;/math&gt;, the only choices for &lt;math&gt;n+5&lt;/math&gt; are &lt;math&gt;83, 166,&lt;/math&gt; and &lt;math&gt;332.&lt;/math&gt; Checking all three cases verifies that &lt;math&gt;n=78&lt;/math&gt; and &lt;math&gt;n=161&lt;/math&gt; work, but &lt;math&gt;n=327&lt;/math&gt; does not. The requested sum is &lt;math&gt;78+161 = 239&lt;/math&gt;.<br /> <br /> ==Solution 4 (Official MAA 2)==<br /> The sum of the cubes of the integers from &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;n&lt;/math&gt; is&lt;cmath&gt;\frac{n^2(n+1)^2}{4},&lt;/cmath&gt;which, when divided by &lt;math&gt;n+5&lt;/math&gt;, has quotient&lt;cmath&gt;Q=\frac14n^3 -\frac34n^2+4n-20 = \frac{n^2(n-3)}4+4n-20&lt;/cmath&gt;with remainder &lt;math&gt;100.&lt;/math&gt; If &lt;math&gt;n&lt;/math&gt; is not congruent to &lt;math&gt;1\pmod4&lt;/math&gt;, then &lt;math&gt;Q&lt;/math&gt; is an integer, and&lt;cmath&gt;\frac{n^2(n+1)^2}{4} = (n+5)Q + 100 \equiv 17\pmod{n+5},&lt;/cmath&gt;so &lt;math&gt;n+5&lt;/math&gt; divides &lt;math&gt;100 - 17 =83&lt;/math&gt;, and &lt;math&gt;n = 78&lt;/math&gt;. If &lt;math&gt;n \equiv 1 \pmod4&lt;/math&gt;, then &lt;math&gt;Q&lt;/math&gt; is half of an integer, and letting &lt;math&gt;n = 4k+1&lt;/math&gt; for some integer &lt;math&gt;k&lt;/math&gt; gives&lt;cmath&gt;\frac{n^2(n+1)^2}{4} = 2(2k+3)Q + 100 \equiv 17\pmod{n+5}.&lt;/cmath&gt;Thus &lt;math&gt;2k+3&lt;/math&gt; divides &lt;math&gt;100-17 = 83&lt;/math&gt;. It follows that &lt;math&gt;k=40&lt;/math&gt;, and &lt;math&gt;n = 161&lt;/math&gt;. The requested sum is &lt;math&gt;161 + 78 = 239&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Using the formula for &lt;math&gt;\sum_{k=1}^n k^3&lt;/math&gt;,<br /> &lt;cmath&gt;1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2(n+1)^2}{4}&lt;/cmath&gt;<br /> Since &lt;math&gt;1^3 + 2^3 + 3^3 + ... + n^3&lt;/math&gt; divided by &lt;math&gt;n + 5&lt;/math&gt; has a remainder of &lt;math&gt;17&lt;/math&gt;,<br /> &lt;cmath&gt;\frac{n^2(n+1)^2}{4} \equiv 17\pmod {n + 5}&lt;/cmath&gt;<br /> Using the rules of modular arithmetic,<br /> &lt;cmath&gt;n^2(n+1)^2 \equiv 68\pmod {n + 5}&lt;/cmath&gt;&lt;cmath&gt;n^2(n+1)^2 - 68\equiv 0\pmod {n + 5}&lt;/cmath&gt;<br /> Expanding the left hand side,<br /> &lt;cmath&gt;n^4 + 2 n^3 + n^2 - 68\equiv 0\pmod {n + 5}&lt;/cmath&gt;<br /> This means that <br /> &lt;math&gt;n^4 + 2 n^3 + n^2 - 68&lt;/math&gt; is divisible by &lt;math&gt;{n + 5}&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;(n + 5) | (n^4 + 2 n^3 + n^2 - 68)&lt;/cmath&gt;<br /> Dividing polynomials,<br /> &lt;cmath&gt;\frac{n^4 + 2 n^3 + n^2 - 68}{n + 5}&lt;/cmath&gt;<br /> &lt;cmath&gt;= n^3 - 3 n^2 + 16n - 80 + \frac{332}{(n + 5)}&lt;/cmath&gt;<br /> &lt;math&gt;(n + 5)&lt;/math&gt; &lt;math&gt;|&lt;/math&gt; &lt;math&gt;(n^4 + 2 n^3 + n^2 - 68)&lt;/math&gt; &lt;math&gt;\iff&lt;/math&gt; &lt;math&gt;\frac{332}{(n + 5)}&lt;/math&gt; &lt;math&gt;\in&lt;/math&gt; &lt;math&gt;\mathbb{Z}&lt;/math&gt;<br /> &lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;\frac{332}{(n + 5)}&lt;/math&gt; &lt;math&gt;\in&lt;/math&gt;<br /> &lt;math&gt;\mathbb{Z}&lt;/math&gt; &lt;math&gt;\iff&lt;/math&gt; &lt;math&gt;(n + 5) = \pm 1, \pm 2, \pm 4, \pm 83, \pm 166, \pm 332&lt;/math&gt;<br /> &lt;br&gt;&lt;br&gt;<br /> Note that &lt;math&gt;n&lt;/math&gt; &lt;math&gt;\in&lt;/math&gt; &lt;math&gt;\mathbb{N}&lt;/math&gt; and &lt;math&gt;n + 5 &gt; 17&lt;/math&gt; (because the remainder when dividing by &lt;math&gt;n + 5&lt;/math&gt; is &lt;math&gt;17&lt;/math&gt;, so &lt;math&gt;n + 5&lt;/math&gt; must be greater than &lt;math&gt;17&lt;/math&gt;), so all options &lt;math&gt;\leq 17&lt;/math&gt; can be eliminated.<br /> &lt;cmath&gt;(n + 5) = 83, 166, 332&lt;/cmath&gt;<br /> &lt;cmath&gt;n = 78, 161, 327&lt;/cmath&gt;<br /> Checking all 3 cases, &lt;math&gt;n = 78&lt;/math&gt; and &lt;math&gt;n = 161&lt;/math&gt; work; &lt;math&gt;n = 327&lt;/math&gt; fails.<br /> &lt;br&gt;&lt;br&gt;<br /> Therefore, the answer is &lt;math&gt;78 + 161 = \boxed{239}&lt;/math&gt;.<br /> &lt;br&gt;&lt;br&gt;<br /> ~ {TSun} ~<br /> <br /> == Solution 6 (similar ideas to Solution 1, but faster) ==<br /> As before, we note that &lt;math&gt;(5+a)^3 + (n-a)^3 \equiv (5+a)^3 - (n+5 - (n-a))^3 \equiv 0 \pmod {n+5}.&lt;/math&gt;<br /> Thus, we can pair up the terms from &lt;math&gt;5^3&lt;/math&gt; to &lt;math&gt;n^3&lt;/math&gt; and cancel them. We have to deal with two cases:<br /> <br /> If &lt;math&gt;n&lt;/math&gt; is even, then &lt;math&gt;5^3+6^3 + \cdots + n^3 \equiv 0 \pmod {n+5},&lt;/math&gt; as there are an even number of terms and they pair and cancel. We thus get &lt;math&gt;1^2+2^3+3^3+4^3 = 100 \equiv 17 \pmod {n+5},&lt;/math&gt; or &lt;math&gt;(n+5) | 83,&lt;/math&gt; which yields &lt;math&gt;n=78.&lt;/math&gt;<br /> <br /> If &lt;math&gt;n&lt;/math&gt; is odd, then &lt;math&gt;1^3+2^3+\cdots + n^3 \equiv 1^3+2^3+3^3+4^3+\left( \frac{n+5}{2} \right)^3 \equiv 17 \pmod {n+5}.&lt;/math&gt;<br /> Letting &lt;math&gt;k = \frac{n+5}{2}&lt;/math&gt; yields &lt;math&gt;k^2 + 83 \equiv 0 \pmod {2k}.&lt;/math&gt; However, this means that &lt;math&gt;83&lt;/math&gt; is divisible by &lt;math&gt;k,&lt;/math&gt; so &lt;math&gt;k=1,83.&lt;/math&gt;<br /> Plugging this back into &lt;math&gt;n&lt;/math&gt; yields &lt;math&gt;n=2(83)-5 = 161&lt;/math&gt; in the latter case.<br /> <br /> Thus, the sum of all possible &lt;math&gt;n&lt;/math&gt; is just &lt;math&gt;78+161 = \boxed{239}.&lt;/math&gt;<br /> <br /> - ccx09<br /> <br /> ==Video solution==<br /> https://www.youtube.com/watch?v=87Mp0cdUtCU<br /> ~ North America Math Contest Go Go Go <br /> <br /> ==Video Solution==<br /> https://youtu.be/bz5N-jI2e0U?t=201<br /> <br /> ==Video Solution==<br /> https://youtu.be/bz5N-jI2e0U?t=201<br /> <br /> ==See Also==<br /> {{AIME box|year=2020|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_10&diff=153627 2020 AIME II Problems/Problem 10 2021-05-12T22:31:45Z <p>CoolCarsOnTheRun: added line break to improve readability</p> <hr /> <div>== Problem ==<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that when &lt;math&gt;1^3+2^3+3^3+\cdots +n^3&lt;/math&gt; is divided by &lt;math&gt;n+5&lt;/math&gt;, the remainder is &lt;math&gt;17&lt;/math&gt;.<br /> <br /> == Solution 1 (If you don't remember the formula for sum of cubes) ==<br /> We first note that since the remainder is &lt;math&gt;17&lt;/math&gt; and we are dividing by &lt;math&gt;n+5&lt;/math&gt;, &lt;math&gt;n+5&lt;/math&gt; must be greater than &lt;math&gt;17&lt;/math&gt;, meaning that &lt;math&gt;n&lt;/math&gt; has to be at least &lt;math&gt;13&lt;/math&gt;.<br /> <br /> We then notice that we can pair the &lt;math&gt;5^3&lt;/math&gt; term with the &lt;math&gt;n^3&lt;/math&gt; term to factor it into &lt;math&gt;(n+5)(n^2-5n+25)&lt;/math&gt; using the sum of two cubes formula (&lt;math&gt;a^3+b^3=(a+b)(a^2-ab+b^2)&lt;/math&gt;), which is divisible by &lt;math&gt;n+5&lt;/math&gt;. We can do the same for the &lt;math&gt;6^3&lt;/math&gt; term with the &lt;math&gt;(n-1)^3&lt;/math&gt; term, the &lt;math&gt;7^3&lt;/math&gt; term with the &lt;math&gt;(n-2)^3&lt;/math&gt;, and so on, which are all divisible by &lt;math&gt;n+5&lt;/math&gt;. However, when &lt;math&gt;n&lt;/math&gt; is odd, we will have a middle term that is not paired with any other terms, which is not necessarily divisible by &lt;math&gt;n+5&lt;/math&gt;. Thus, we have two cases:<br /> &lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;\textbf{CASE 1: } n&lt;/math&gt; is even &lt;br&gt;<br /> If &lt;math&gt;n&lt;/math&gt; is even, all terms that are greater than &lt;math&gt;4^3&lt;/math&gt; pair, as there are an even number of terms that are greater than &lt;math&gt;4^3&lt;/math&gt;. Therefore, all we need in order for the entire sequence to have a remainder of &lt;math&gt;17&lt;/math&gt; when divided by &lt;math&gt;n+5&lt;/math&gt; is &lt;math&gt;1^3+2^3+3^3+4^3&lt;/math&gt; to have a remainder of &lt;math&gt;17&lt;/math&gt; when divided by &lt;math&gt;n+5&lt;/math&gt;.<br /> <br /> Evaluating &lt;math&gt;1^3+2^3+3^3+4^3&lt;/math&gt; as &lt;math&gt;100&lt;/math&gt;, all we need to be true is <br /> &lt;cmath&gt;100\equiv 17\pmod {n+5},&lt;/cmath&gt;<br /> or that<br /> &lt;cmath&gt;83\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> Thus, &lt;math&gt;83&lt;/math&gt; will be divisible by &lt;math&gt;n+5&lt;/math&gt; where &lt;math&gt;n\geq 13&lt;/math&gt;. As &lt;math&gt;83&lt;/math&gt; is prime, &lt;math&gt;n+5&lt;/math&gt; must be equal to either &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;83&lt;/math&gt;. If &lt;math&gt;n+5=1&lt;/math&gt;, we have that &lt;math&gt;n=-4&lt;/math&gt;, which is not greater than or equal to &lt;math&gt;13&lt;/math&gt;, so that solution is extraneous.<br /> <br /> If &lt;math&gt;n+5=83&lt;/math&gt;, we have that &lt;math&gt;n=78&lt;/math&gt;, which is &lt;math&gt;\geq 13&lt;/math&gt;, so one of our solutions is &lt;math&gt;n=78&lt;/math&gt;, and we are done with our first case.<br /> &lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;\textbf{CASE 2: } n&lt;/math&gt; is odd &lt;br&gt;<br /> If &lt;math&gt;n&lt;/math&gt; is odd, the only term that does not pair is the arithmetic mean of the numbers under the cube of the largest and smallest terms that would pair, or &lt;math&gt;(\frac{n+5}{2})^3&lt;/math&gt;. Therefore, as all other terms that are &lt;math&gt;\geq 5^3&lt;/math&gt; pair, the requirement that we have is<br /> &lt;cmath&gt;1^3+2^3+3^3+4^3+\left(\frac{n+5}{2}\right)^3\equiv 17\pmod {n+5}.&lt;/cmath&gt;<br /> Calculating and simplifying, we have that<br /> &lt;cmath&gt;83+\left(\frac{n+5}{2}\right)^3\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> Now, we multiply both sides by &lt;math&gt;8&lt;/math&gt;. However, since multiplication is not reversible in modular arithmetic, we need to check whether any solutions are extraneous after solving. The congruence that we now have is<br /> &lt;cmath&gt;8\cdot 83+(n+5)^3\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> As we know that &lt;math&gt;(n+5)^3&lt;/math&gt; is divisible by &lt;math&gt;n+5&lt;/math&gt;, what we need now is<br /> &lt;cmath&gt;8\cdot 83\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> We now check each solution to see whether it works.<br /> <br /> If &lt;math&gt;n+5=1, 2, 4, 8&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would be less than &lt;math&gt;13&lt;/math&gt;, so none of these solutions work. If &lt;math&gt;n+5=83&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would be even, so that solution does not work for this case. Therefore, the only three solutions we need to check for this case are when &lt;math&gt;n+5=166&lt;/math&gt;, &lt;math&gt;n+5=332&lt;/math&gt;, or &lt;math&gt;n+5=664&lt;/math&gt;. We plug these values into the congruence before we multiplied both sides by &lt;math&gt;8&lt;/math&gt; to check.<br /> <br /> If &lt;math&gt;n+5=166&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{166}{2}\right)^3\equiv 0\pmod {166}.&lt;/cmath&gt;<br /> Calculating and factoring out &lt;math&gt;83&lt;/math&gt;, we have that<br /> &lt;cmath&gt;83(1+83\cdot 83)\equiv 0\pmod {166}.&lt;/cmath&gt;<br /> As the right parenthesis is odd and &lt;math&gt;166=83\cdot 2&lt;/math&gt;, we know that this solution works, so we have another solution: &lt;math&gt;n=166-5=161&lt;/math&gt;.<br /> <br /> If &lt;math&gt;n+5=332&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{332}{2}\right)^3=83+166^3\equiv 0\pmod {322}.&lt;/cmath&gt;<br /> As the left hand side is odd, but all multiples of &lt;math&gt;322&lt;/math&gt; is even, this solution is therefore extraneous.<br /> <br /> If &lt;math&gt;n+5=664&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{664}{2}\right)^3=83+332^3\equiv 0\pmod {664}.&lt;/cmath&gt;<br /> Again, the left hand side is odd, and all multiples of &lt;math&gt;664&lt;/math&gt; are even, so this solution is extraneous.<br /> &lt;br&gt;&lt;br&gt;<br /> Therefore, our final answer is &lt;math&gt;78+161=\boxed{239}&lt;/math&gt;.<br /> &lt;br&gt;&lt;br&gt;<br /> ~ CoolCarsOnTheRun<br /> <br /> ==Solution 2 (w/ formula)==<br /> Let &lt;math&gt;m=n+5&lt;/math&gt;. Then we have<br /> &lt;cmath&gt;1^3+2^3+3^3+\cdots+(m-5)^3\equiv 17 \mod m&lt;/cmath&gt;<br /> &lt;cmath&gt;\left(\frac{(m-5)(m-4)}{2}\right)^2\equiv 17 \mod m &lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{400}{4}\equiv 17 \mod m &lt;/cmath&gt;<br /> &lt;cmath&gt;332 \equiv 0 \mod m &lt;/cmath&gt;<br /> So, &lt;math&gt;m\in\{83,166,332\}&lt;/math&gt;. Testing, the cases, only &lt;math&gt;332&lt;/math&gt; fails. This leaves &lt;math&gt;78+161=\boxed{239}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Official MAA 1)==<br /> The sum of the cubes from 1 to &lt;math&gt;n&lt;/math&gt; is<br /> &lt;cmath&gt;1^3+2^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}.&lt;/cmath&gt;For this to be equal to &lt;math&gt;(n+5)q+17&lt;/math&gt; for some integer &lt;math&gt;q&lt;/math&gt;, it must be that&lt;cmath&gt;n^2(n+1)^2=4(n+5)q+4\cdot 17,&lt;/cmath&gt;so&lt;cmath&gt;n^2(n+1)^2 \equiv 4 \cdot 17= 68\hskip-.2cm \pmod{n+5}.&lt;/cmath&gt;But &lt;math&gt;n^2(n+1)^2 \equiv (-5)^2(-4)^2 = 400 \pmod{n+5}.&lt;/math&gt; Thus &lt;math&gt;n^2(n+1)^2&lt;/math&gt; is congruent to both &lt;math&gt;68&lt;/math&gt; and &lt;math&gt;400,&lt;/math&gt; which implies that &lt;math&gt;n+5&lt;/math&gt; divides &lt;math&gt;400-68 = 332=2^2 \cdot 83&lt;/math&gt;. Because &lt;math&gt;n+5 &gt; 17&lt;/math&gt;, the only choices for &lt;math&gt;n+5&lt;/math&gt; are &lt;math&gt;83, 166,&lt;/math&gt; and &lt;math&gt;332.&lt;/math&gt; Checking all three cases verifies that &lt;math&gt;n=78&lt;/math&gt; and &lt;math&gt;n=161&lt;/math&gt; work, but &lt;math&gt;n=327&lt;/math&gt; does not. The requested sum is &lt;math&gt;78+161 = 239&lt;/math&gt;.<br /> <br /> ==Solution 4 (Official MAA 2)==<br /> The sum of the cubes of the integers from &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;n&lt;/math&gt; is&lt;cmath&gt;\frac{n^2(n+1)^2}{4},&lt;/cmath&gt;which, when divided by &lt;math&gt;n+5&lt;/math&gt;, has quotient&lt;cmath&gt;Q=\frac14n^3 -\frac34n^2+4n-20 = \frac{n^2(n-3)}4+4n-20&lt;/cmath&gt;with remainder &lt;math&gt;100.&lt;/math&gt; If &lt;math&gt;n&lt;/math&gt; is not congruent to &lt;math&gt;1\pmod4&lt;/math&gt;, then &lt;math&gt;Q&lt;/math&gt; is an integer, and&lt;cmath&gt;\frac{n^2(n+1)^2}{4} = (n+5)Q + 100 \equiv 17\pmod{n+5},&lt;/cmath&gt;so &lt;math&gt;n+5&lt;/math&gt; divides &lt;math&gt;100 - 17 =83&lt;/math&gt;, and &lt;math&gt;n = 78&lt;/math&gt;. If &lt;math&gt;n \equiv 1 \pmod4&lt;/math&gt;, then &lt;math&gt;Q&lt;/math&gt; is half of an integer, and letting &lt;math&gt;n = 4k+1&lt;/math&gt; for some integer &lt;math&gt;k&lt;/math&gt; gives&lt;cmath&gt;\frac{n^2(n+1)^2}{4} = 2(2k+3)Q + 100 \equiv 17\pmod{n+5}.&lt;/cmath&gt;Thus &lt;math&gt;2k+3&lt;/math&gt; divides &lt;math&gt;100-17 = 83&lt;/math&gt;. It follows that &lt;math&gt;k=40&lt;/math&gt;, and &lt;math&gt;n = 161&lt;/math&gt;. The requested sum is &lt;math&gt;161 + 78 = 239&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Using the formula for &lt;math&gt;\sum_{k=1}^n k^3&lt;/math&gt;,<br /> &lt;cmath&gt;1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2(n+1)^2}{4}&lt;/cmath&gt;<br /> Since &lt;math&gt;1^3 + 2^3 + 3^3 + ... + n^3&lt;/math&gt; divided by &lt;math&gt;n + 5&lt;/math&gt; has a remainder of &lt;math&gt;17&lt;/math&gt;,<br /> &lt;cmath&gt;\frac{n^2(n+1)^2}{4} \equiv 17\pmod {n + 5}&lt;/cmath&gt;<br /> Using the rules of modular arithmetic,<br /> &lt;cmath&gt;n^2(n+1)^2 \equiv 68\pmod {n + 5}&lt;/cmath&gt;&lt;cmath&gt;n^2(n+1)^2 - 68\equiv 0\pmod {n + 5}&lt;/cmath&gt;<br /> Expanding the left hand side,<br /> &lt;cmath&gt;n^4 + 2 n^3 + n^2 - 68\equiv 0\pmod {n + 5}&lt;/cmath&gt;<br /> This means that <br /> &lt;math&gt;n^4 + 2 n^3 + n^2 - 68&lt;/math&gt; is divisible by &lt;math&gt;{n + 5}&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;(n + 5) | (n^4 + 2 n^3 + n^2 - 68)&lt;/cmath&gt;<br /> Dividing polynomials,<br /> &lt;cmath&gt;\frac{n^4 + 2 n^3 + n^2 - 68}{n + 5}&lt;/cmath&gt;<br /> &lt;cmath&gt;= n^3 - 3 n^2 + 16n - 80 + \frac{332}{(n + 5)}&lt;/cmath&gt;<br /> &lt;math&gt;(n + 5)&lt;/math&gt; &lt;math&gt;|&lt;/math&gt; &lt;math&gt;(n^4 + 2 n^3 + n^2 - 68)&lt;/math&gt; &lt;math&gt;\iff&lt;/math&gt; &lt;math&gt;\frac{332}{(n + 5)}&lt;/math&gt; &lt;math&gt;\in&lt;/math&gt; &lt;math&gt;\mathbb{Z}&lt;/math&gt;<br /> &lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;\frac{332}{(n + 5)}&lt;/math&gt; &lt;math&gt;\in&lt;/math&gt;<br /> &lt;math&gt;\mathbb{Z}&lt;/math&gt; &lt;math&gt;\iff&lt;/math&gt; &lt;math&gt;(n + 5) = \pm 1, \pm 2, \pm 4, \pm 83, \pm 166, \pm 332&lt;/math&gt;<br /> &lt;br&gt;&lt;br&gt;<br /> Note that &lt;math&gt;n&lt;/math&gt; &lt;math&gt;\in&lt;/math&gt; &lt;math&gt;\mathbb{N}&lt;/math&gt; and &lt;math&gt;n + 5 &gt; 17&lt;/math&gt; (because the remainder when dividing by &lt;math&gt;n + 5&lt;/math&gt; is &lt;math&gt;17&lt;/math&gt;, so &lt;math&gt;n + 5&lt;/math&gt; must be greater than &lt;math&gt;17&lt;/math&gt;), so all options &lt;math&gt;\leq 17&lt;/math&gt; can be eliminated.<br /> &lt;cmath&gt;(n + 5) = 83, 166, 332&lt;/cmath&gt;<br /> &lt;cmath&gt;n = 78, 161, 327&lt;/cmath&gt;<br /> Checking all 3 cases, &lt;math&gt;n = 78&lt;/math&gt; and &lt;math&gt;n = 161&lt;/math&gt; work; &lt;math&gt;n = 327&lt;/math&gt; fails.<br /> &lt;br&gt;&lt;br&gt;<br /> Therefore, the answer is &lt;math&gt;78 + 161 = \boxed{239}&lt;/math&gt;.<br /> &lt;br&gt;&lt;br&gt;<br /> ~ {TSun} ~<br /> <br /> == Solution 6 (similar ideas to Solution 1, but faster) ==<br /> As before, we note that &lt;math&gt;(5+a)^3 + (n-a)^3 \equiv (5+a)^3 - (n+5 - (n-a))^3 \equiv 0 \pmod {n+5}.&lt;/math&gt;<br /> Thus, we can pair up the terms from &lt;math&gt;5^3&lt;/math&gt; to &lt;math&gt;n^3&lt;/math&gt; and cancel them. We have to deal with two cases:<br /> <br /> If &lt;math&gt;n&lt;/math&gt; is even, then &lt;math&gt;5^3+6^3 + \cdots + n^3 \equiv 0 \pmod {n+5},&lt;/math&gt; as there are an even number of terms and they pair and cancel. We thus get &lt;math&gt;1^2+2^3+3^3+4^3 = 100 \equiv 17 \pmod {n+5},&lt;/math&gt; or &lt;math&gt;(n+5) | 83,&lt;/math&gt; which yields &lt;math&gt;n=78.&lt;/math&gt;<br /> <br /> If &lt;math&gt;n&lt;/math&gt; is odd, then &lt;math&gt;1^3+2^3+\cdots + n^3 \equiv 1^3+2^3+3^3+4^3+\left( \frac{n+5}{2} \right)^3 \equiv 17 \pmod {n+5}.&lt;/math&gt;<br /> Letting &lt;math&gt;k = \frac{n+5}{2}&lt;/math&gt; yields &lt;math&gt;k^2 + 83 \equiv 0 \pmod {2k}.&lt;/math&gt; However, this means that &lt;math&gt;83&lt;/math&gt; is divisible by &lt;math&gt;k,&lt;/math&gt; so &lt;math&gt;k=1,83.&lt;/math&gt;<br /> Plugging this back into &lt;math&gt;n&lt;/math&gt; yields &lt;math&gt;n=2(83)-5 = 161&lt;/math&gt; in the latter case.<br /> <br /> Thus, the sum of all possible &lt;math&gt;n&lt;/math&gt; is just &lt;math&gt;78+161 = \boxed{239}.&lt;/math&gt;<br /> <br /> - ccx09<br /> <br /> ==Video solution==<br /> https://www.youtube.com/watch?v=87Mp0cdUtCU<br /> ~ North America Math Contest Go Go Go <br /> <br /> ==Video Solution==<br /> https://youtu.be/bz5N-jI2e0U?t=201<br /> <br /> ==Video Solution==<br /> https://youtu.be/bz5N-jI2e0U?t=201<br /> <br /> ==See Also==<br /> {{AIME box|year=2020|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_20&diff=153004 2007 AMC 8 Problems/Problem 20 2021-04-30T20:48:54Z <p>CoolCarsOnTheRun: </p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;Before the district play, the Unicorns had won &lt;math&gt;45&lt;/math&gt;% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> At the beginning of the problem, the Unicorns had played &lt;math&gt;y&lt;/math&gt; games and they had won &lt;math&gt;x&lt;/math&gt; of these games. From the information given in the problem, we can say that &lt;math&gt;\frac{x}{y}=0.45.&lt;/math&gt; Next, the Unicorns win 6 more games and lose 2 more, for a total of &lt;math&gt;6+2=8&lt;/math&gt; games played during district play. We are told that they end the season having won half of their games, or &lt;math&gt;0.5 &lt;/math&gt; of their games. We can write another equation: &lt;math&gt;\frac{x+6}{y+8}=0.5.&lt;/math&gt; This gives us a system of equations:<br /> &lt;math&gt;\frac{x}{y}=0.45&lt;/math&gt; and &lt;math&gt;\frac{x+6}{y+8}=0.5.&lt;/math&gt;<br /> We first multiply both sides of the first equation by &lt;math&gt;y&lt;/math&gt; to get &lt;math&gt;x=0.45y.&lt;/math&gt; Then, we multiply both sides of the second equation by &lt;math&gt;(y+8)&lt;/math&gt; to get &lt;math&gt;x+6=0.5(y+8).&lt;/math&gt; Applying the Distributive Property gives yields &lt;math&gt;x+6=0.5y+4.&lt;/math&gt; Now we substitute &lt;math&gt;0.45y&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; to get &lt;math&gt;0.45y+6=0.5y+4.&lt;/math&gt; Solving gives us &lt;math&gt;y=40.&lt;/math&gt; Since the problem asks for the total number of games, we add on the last 8 games to get the solution &lt;math&gt;\boxed{\textbf{(A)}\ 48}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Simplifying 45% to &lt;math&gt;\frac{9}{20}&lt;/math&gt;, we see that the numbers of games are a multiple of 20. After that the Unicorns played 8 more games to the total number of games is in the form of 20x+8 where x is any positive integer. The only answer choice is &lt;math&gt;\boxed{48}&lt;/math&gt;, which is 20(2)+8.<br /> <br /> -harsha12345<br /> <br /> ==Solution 3==<br /> First we simplify 45% to &lt;math&gt;\frac{9}{20}&lt;/math&gt;. After they won 6 more games and lost 2 more games the number of games they won is &lt;math&gt;9x+6&lt;/math&gt;, and the total number of games is &lt;math&gt;20x+8&lt;/math&gt;. Turning it into a fraction we get &lt;math&gt;\frac{9x+6}{20x+8}=\frac{1}{2}&lt;/math&gt;, so solving for &lt;math&gt;x&lt;/math&gt; we get &lt;math&gt;x=2.&lt;/math&gt; Plugging in 2 for &lt;math&gt;x&lt;/math&gt; we get &lt;math&gt;20(2)+8=\boxed{48}&lt;/math&gt;.<br /> <br /> -harsha12345<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2007|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_20&diff=153003 2007 AMC 8 Problems/Problem 20 2021-04-30T20:48:39Z <p>CoolCarsOnTheRun: </p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;Before the district play, the Unicorns had won &lt;math&gt;45&lt;/math&gt;% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> At the beginning of the problem, the avengers had played &lt;math&gt;y&lt;/math&gt; games and they had won &lt;math&gt;x&lt;/math&gt; of these games. From the information given in the problem, we can say that &lt;math&gt;\frac{x}{y}=0.45.&lt;/math&gt; Next, the Unicorns win 6 more games and lose 2 more, for a total of &lt;math&gt;6+2=8&lt;/math&gt; games played during district play. We are told that they end the season having won half of their games, or &lt;math&gt;0.5 &lt;/math&gt; of their games. We can write another equation: &lt;math&gt;\frac{x+6}{y+8}=0.5.&lt;/math&gt; This gives us a system of equations:<br /> &lt;math&gt;\frac{x}{y}=0.45&lt;/math&gt; and &lt;math&gt;\frac{x+6}{y+8}=0.5.&lt;/math&gt;<br /> We first multiply both sides of the first equation by &lt;math&gt;y&lt;/math&gt; to get &lt;math&gt;x=0.45y.&lt;/math&gt; Then, we multiply both sides of the second equation by &lt;math&gt;(y+8)&lt;/math&gt; to get &lt;math&gt;x+6=0.5(y+8).&lt;/math&gt; Applying the Distributive Property gives yields &lt;math&gt;x+6=0.5y+4.&lt;/math&gt; Now we substitute &lt;math&gt;0.45y&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; to get &lt;math&gt;0.45y+6=0.5y+4.&lt;/math&gt; Solving gives us &lt;math&gt;y=40.&lt;/math&gt; Since the problem asks for the total number of games, we add on the last 8 games to get the solution &lt;math&gt;\boxed{\textbf{(A)}\ 48}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Simplifying 45% to &lt;math&gt;\frac{9}{20}&lt;/math&gt;, we see that the numbers of games are a multiple of 20. After that the Unicorns played 8 more games to the total number of games is in the form of 20x+8 where x is any positive integer. The only answer choice is &lt;math&gt;\boxed{48}&lt;/math&gt;, which is 20(2)+8.<br /> <br /> -harsha12345<br /> <br /> ==Solution 3==<br /> First we simplify 45% to &lt;math&gt;\frac{9}{20}&lt;/math&gt;. After they won 6 more games and lost 2 more games the number of games they won is &lt;math&gt;9x+6&lt;/math&gt;, and the total number of games is &lt;math&gt;20x+8&lt;/math&gt;. Turning it into a fraction we get &lt;math&gt;\frac{9x+6}{20x+8}=\frac{1}{2}&lt;/math&gt;, so solving for &lt;math&gt;x&lt;/math&gt; we get &lt;math&gt;x=2.&lt;/math&gt; Plugging in 2 for &lt;math&gt;x&lt;/math&gt; we get &lt;math&gt;20(2)+8=\boxed{48}&lt;/math&gt;.<br /> <br /> -harsha12345<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2007|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=Wooga_Looga_Theorem&diff=152781 Wooga Looga Theorem 2021-04-27T00:46:54Z <p>CoolCarsOnTheRun: </p> <hr /> <div>=Definition=<br /> If there is &lt;math&gt;\triangle ABC&lt;/math&gt; and points &lt;math&gt;D,E,F&lt;/math&gt; on the sides &lt;math&gt;BC,CA,AB&lt;/math&gt; respectively such that &lt;math&gt;\frac{DB}{DC}=\frac{EC}{EA}=\frac{FA}{FB}=r&lt;/math&gt;, then the ratio &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{r^2-r+1}{(r+1)^2}&lt;/math&gt;.<br /> <br /> <br /> Created by Foogle and Hoogle of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> =Proofs=<br /> ==Proof 1==<br /> Proof by Gogobao:<br /> <br /> We have: &lt;math&gt;\frac{DB}{BC} = \frac{r}{r+1}, \frac{DC}{BC} = \frac{1}{r+1}, \frac{EC}{AC} = \frac{r}{r+1}, \frac{EA}{AC} = \frac{1}{r+1}, \frac{FA}{BA} = \frac{r}{r+1}, \frac{FB}{BA} = \frac{1}{r+1} &lt;/math&gt;<br /> <br /> We have: &lt;math&gt;[DEF] = [ABC] - [DCE] - [FAE] - [FBD]&lt;/math&gt;<br /> <br /> &lt;math&gt;[DCE] = [ABC] \cdot \frac{DC}{CB} \cdot \frac{CE}{CA} = [ABC] \cdot \frac{r}{(r+1)^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;[EAF] = [ABC] \cdot \frac{EA}{CA} \cdot \frac{FA}{BA} = [ABC] \cdot \frac{r}{(r+1)^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;[FBD] = [ABC] \cdot \frac{FB}{AB} \cdot \frac{BD}{CB} = [ABC] \cdot \frac{r}{(r+1)^2}&lt;/math&gt;<br /> <br /> Therefore &lt;math&gt;[DEF] = [ABC] (1-\frac{3r}{(r+1)^2})&lt;/math&gt;<br /> <br /> So we have &lt;math&gt;\frac{[DEF]}{[ABC]} = \frac{r^2-r+1}{(r+1)^2}&lt;/math&gt;<br /> <br /> ==Proof 2==<br /> Proof by franzliszt<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Then &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. We can also find that &lt;math&gt;D=\left(0,\tfrac {1}{r+1},\tfrac {r}{r+1}\right),E=\left(\tfrac {r}{r+1},0,\tfrac {1}{r+1}\right),F=\left(\tfrac {1}{r+1},\tfrac {r}{r+1},0\right)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that &lt;cmath&gt;\frac{[DEF]}{[ABC]}= \begin{vmatrix} 0&amp;\tfrac {1}{r+1}&amp;\tfrac {r}{r+1} \\ \tfrac {r}{r+1}&amp;0&amp;\tfrac {1}{r+1}\\ \tfrac {1}{r+1}&amp;\tfrac {r}{r+1}&amp;0 \end{vmatrix}=\frac{r^2-r+1}{(r+1)^2}.&lt;/cmath&gt;<br /> <br /> ==Proof 3==<br /> Proof by [[User:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;RedFireTruck&lt;/font&gt;]] ([[User talk:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;talk&lt;/font&gt;]]) 12:11, 1 February 2021 (EST):<br /> <br /> WLOG we let &lt;math&gt;A=(0, 0)&lt;/math&gt;, &lt;math&gt;B=(1, 0)&lt;/math&gt;, &lt;math&gt;C=(x, y)&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y\in\mathbb{R}&lt;/math&gt;. We then use Shoelace Forumla to get &lt;math&gt;[ABC]=\frac12|y|&lt;/math&gt;. We then figure out that &lt;math&gt;D=\left(\frac{rx+1}{r+1}, \frac{ry}{r+1}\right)&lt;/math&gt;, &lt;math&gt;E=\left(\frac{x}{r+1}, \frac{y}{r+1}\right)&lt;/math&gt;, and &lt;math&gt;F=\left(\frac{r}{r+1}, 0\right)&lt;/math&gt; so we know that by Shoelace Formula &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{\frac12\left|\frac{r^2y-ry+y}{(r+1)^2}\right|}{\frac12|y|}=\left|\frac{r^2-r+1}{(r+1)^2}\right|&lt;/math&gt;. We know that &lt;math&gt;\frac{r^2-r+1}{(r+1)^2}\ge0&lt;/math&gt; for all &lt;math&gt;r\in\mathbb{R}&lt;/math&gt; so &lt;math&gt;\left|\frac{r^2-r+1}{(r+1)^2}\right|=\frac{r^2-r+1}{(r+1)^2}&lt;/math&gt;.<br /> <br /> ==Proof 4==<br /> Proof by ishanvannadil2008:<br /> <br /> Just use jayasharmaramankumarguptareddybavarajugopal's lemma. (Thanks to tenebrine)<br /> <br /> =Application 1=<br /> ==Problem==<br /> The Wooga Looga Theorem states that the solution to this problem by franzliszt:<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; points &lt;math&gt;X,Y,Z&lt;/math&gt; are on sides &lt;math&gt;BC,CA,AB&lt;/math&gt; such that &lt;math&gt;\frac{XB}{XC}=\frac{YC}{YA}=\frac{ZA}{ZB}=\frac 71&lt;/math&gt;. Find the ratio of &lt;math&gt;[XYZ]&lt;/math&gt; to &lt;math&gt;[ABC]&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> One solution is this one by [[User:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;RedFireTruck&lt;/font&gt;]] ([[User talk:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;talk&lt;/font&gt;]]) 12:11, 1 February 2021 (EST):<br /> <br /> WLOG let &lt;math&gt;A=(0, 0)&lt;/math&gt;, &lt;math&gt;B=(1, 0)&lt;/math&gt;, &lt;math&gt;C=(x, y)&lt;/math&gt;. Then &lt;math&gt;[ABC]=\frac12|y|&lt;/math&gt; by Shoelace Theorem and &lt;math&gt;X=\left(\frac{7x+1}{8}, \frac{7y}{8}\right)&lt;/math&gt;, &lt;math&gt;Y=\left(\frac{x}{8}, \frac{y}{8}\right)&lt;/math&gt;, &lt;math&gt;Z=\left(\frac78, 0\right)&lt;/math&gt;. Then &lt;math&gt;[XYZ]=\frac12\left|\frac{43y}{64}\right|&lt;/math&gt; by Shoelace Theorem. Therefore the answer is &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> or this solution by franzliszt:<br /> <br /> We apply Barycentric Coordinates w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Let &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. Then we find that &lt;math&gt;X=\left(0,\tfrac 18,\tfrac 78\right),Y=\left(\tfrac 78,0,\tfrac 18\right),Z=\left(\tfrac18,\tfrac78,0\right)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix}<br /> x_{1} &amp;y_{1} &amp;z_{1} \\<br /> x_{2} &amp;y_{2} &amp;z_{2} \\ <br /> x_{3}&amp; y_{3} &amp; z_{3}<br /> \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that &lt;cmath&gt;\frac{[XYZ]}{[ABC]}=\begin{vmatrix}<br /> 0&amp;\tfrac 18&amp;\tfrac 78\\<br /> \tfrac 78&amp;0&amp;\tfrac 18\\<br /> \tfrac18&amp;\tfrac78&amp;0<br /> \end{vmatrix}=\frac{43}{64}.&lt;/cmath&gt; &lt;math&gt;\blacksquare&lt;/math&gt;<br /> ==Solution 3==<br /> or this solution by aaja3427:<br /> <br /> According the the Wooga Looga Theorem, It is &lt;math&gt;\frac{49-7+1}{8^2}&lt;/math&gt;. This is &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> <br /> or this solution by AoPS user ilovepizza2020:<br /> <br /> We use the &lt;math&gt;\mathbf{FUNDAMENTAL~THEOREM~OF~GEOGEBRA}&lt;/math&gt; to instantly get &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. (Note: You can only use this method when you are not in a contest, as this method is so overpowered that the people behind mathematics examinations decided to ban it.)<br /> <br /> ==Solution 5==<br /> or this solution by eduD_looC:<br /> <br /> This is a perfect application of the Adihaya Jayasharmaramankumarguptareddybavarajugopal's Lemma, which results in the answer being &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. A very beautiful application, which leaves graders and readers speechless. Great for math contests with proofs.<br /> <br /> ==Solution 6==<br /> or this solution by CoolJupiter:<br /> <br /> Wow. All of your solutions are slow, compared to my sol:<br /> <br /> By math, we have &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;.<br /> <br /> ~CoolJupiter<br /> ^<br /> |<br /> EVERYONE USE THIS SOLUTION IT'S BRILLIANT <br /> ~bsu1<br /> Yes, very BRILLIANT!<br /> ~ TheAoPSLebron<br /> <br /> ==The Best Solution==<br /> <br /> By the &lt;math&gt;1+1=\text{BREAD}&lt;/math&gt; principle, we get &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. Definitely the best method. When asked, please say that OlympusHero taught you this method. Because he did.<br /> <br /> ==Easiest Solution==<br /> <br /> The answer is clearly &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. We leave the proof and intermediate steps to the reader as an exercise.<br /> <br /> =Application 2=<br /> ==Problem==<br /> The Wooga Looga Theorem states that the solution to this problem by Matholic:<br /> <br /> The figure below shows a triangle ABC whose area is &lt;math&gt;72 \text{cm}^2&lt;/math&gt;. If &lt;math&gt;\dfrac{AD}{DB}=\dfrac{BE}{EC}=\dfrac{CF}{FA}=\dfrac{1}{5}&lt;/math&gt;, find &lt;math&gt;[DEF].&lt;/math&gt;<br /> <br /> ~LaTeX-ifyed by RP3.1415<br /> <br /> ==Solution 1==<br /> is this solution by franzliszt:<br /> <br /> We apply Barycentric Coordinates w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Let &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. Then we find that &lt;math&gt;D=(\tfrac 56,\tfrac 16,0),E=(0,\tfrac 56,\tfrac 16),F=(\tfrac16,0,\tfrac56)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix}<br /> x_{1} &amp;y_{1} &amp;z_{1} \\ <br /> x_{2} &amp;y_{2} &amp;z_{2} \\ <br /> x_{3}&amp; y_{3} &amp; z_{3}<br /> \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[DEF]}{}=\begin{vmatrix}<br /> \tfrac 56&amp;\tfrac 16&amp;0\\<br /> 0&amp;\tfrac 56&amp;\tfrac 16\\<br /> \tfrac16&amp;0&amp;\tfrac56<br /> \end{vmatrix}=\frac{7}{12}&lt;/cmath&gt; so &lt;math&gt;[DEF]=42&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> ==Solution 2==<br /> or this solution by [[User:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;RedFireTruck&lt;/font&gt;]] ([[User talk:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;talk&lt;/font&gt;]]) 12:11, 1 February 2021 (EST):<br /> <br /> By the Wooga Looga Theorem, &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{5^2-5+1}{(5+1)^2}=\frac{21}{36}=\frac{7}{12}&lt;/math&gt;. We are given that &lt;math&gt;[ABC]=72&lt;/math&gt; so &lt;math&gt;[DEF]=\frac{7}{12}\cdot72=\boxed{42}&lt;/math&gt;<br /> <br /> =Application 3=<br /> ==Problem==<br /> The Wooga Looga Theorem states that the solution to this problem by RedFireTruck:<br /> <br /> Find the ratio &lt;math&gt;\frac{[GHI]}{[ABC]}&lt;/math&gt; if &lt;math&gt;\frac{AD}{DB}=\frac{BE}{EC}=\frac{CF}{FA}=\frac12&lt;/math&gt; and &lt;math&gt;\frac{DG}{GE}=\frac{EH}{HF}=\frac{FI}{ID}=1&lt;/math&gt; in the diagram below.&lt;asy&gt;<br /> draw((0, 0)--(6, 0)--(4, 3)--cycle);<br /> draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle);<br /> draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle);<br /> label(&quot;$A$&quot;, (0, 0), SW);<br /> label(&quot;$B$&quot;, (6, 0), SE);<br /> label(&quot;$C$&quot;, (4, 3), N);<br /> label(&quot;$D$&quot;, (2, 0), S);<br /> label(&quot;$E$&quot;, (16/3, 1), NE);<br /> label(&quot;$F$&quot;, (8/3, 2), NW);<br /> label(&quot;$G$&quot;, (11/3, 1/2), SE);<br /> label(&quot;$H$&quot;, (4, 3/2), NE);<br /> label(&quot;$I$&quot;, (7/3, 1), W);<br /> &lt;/asy&gt;<br /> ==Solution 1==<br /> is this solution by franzliszt:<br /> <br /> By the Wooga Looga Theorem, &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{2^2-2+1}{(1+2)^2}=\frac 13&lt;/math&gt;. Notice that &lt;math&gt;\triangle GHI&lt;/math&gt; is the medial triangle of '''Wooga Looga Triangle ''' of &lt;math&gt;\triangle ABC&lt;/math&gt;. So &lt;math&gt;\frac{[GHI]}{[DEF]}=\frac 14&lt;/math&gt; and &lt;math&gt;\frac{[GHI]}{[ABD]}=\frac{[DEF]}{[ABC]}\cdot\frac{[GHI]}{[DEF]}=\frac 13 \cdot \frac 14 = \frac {1}{12}&lt;/math&gt; by Chain Rule ideas.<br /> <br /> ==Solution 2==<br /> or this solution by franzliszt:<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt; so that &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. Then &lt;math&gt;D=(\tfrac 23,\tfrac 13,0),E=(0,\tfrac 23,\tfrac 13),F=(\tfrac 13,0,\tfrac 23)&lt;/math&gt;. And &lt;math&gt;G=(\tfrac 13,\tfrac 12,\tfrac 16),H=(\tfrac 16,\tfrac 13,\tfrac 12),I=(\tfrac 12,\tfrac 16,\tfrac 13)&lt;/math&gt;.<br /> <br /> In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[GHI]}{[ABC]}=\begin{vmatrix} \tfrac 13&amp;\tfrac 12&amp;\tfrac 16\\ \tfrac 16&amp;\tfrac 13&amp;\tfrac 12\\ \tfrac 12&amp;\tfrac 16&amp;\tfrac 13 \end{vmatrix}=\frac{1}{12}.&lt;/cmath&gt;<br /> <br /> =Application 4=<br /> ==Problem==<br /> <br /> Let &lt;math&gt;ABC&lt;/math&gt; be a triangle and &lt;math&gt;D,E,F&lt;/math&gt; be points on sides &lt;math&gt;BC,AC,&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; respectively. We have that &lt;math&gt;\frac{BD}{DC} = 3&lt;/math&gt; and similar for the other sides. If the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;, then what is the area of triangle &lt;math&gt;DEF&lt;/math&gt;? (By ilovepizza2020)<br /> <br /> ==Solution 1==<br /> <br /> By Franzliszt<br /> <br /> By Wooga Looga, &lt;math&gt;\frac{[DEF]}{16} = \frac{3^2-3+1}{(3+1)^2}=\frac{7}{16}&lt;/math&gt; so the answer is &lt;math&gt;\boxed7&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> By franzliszt<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Then &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. We can also find that &lt;math&gt;D=(0,\tfrac 14,\tfrac 34),E=(\tfrac 34,0,\tfrac 14),F=(\tfrac 14,\tfrac 34,0)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[DEF]}{[ABC]}=\begin{vmatrix} 0&amp;\tfrac 14&amp;\tfrac 34\\ \tfrac 34&amp;0&amp;\tfrac 14\\ \tfrac 14&amp;\tfrac 34&amp;0 \end{vmatrix}=\frac{7}{16}.&lt;/cmath&gt;So the answer is &lt;math&gt;\boxed{7}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> A long story short, the answer must be &lt;math&gt;\boxed{7}&lt;/math&gt; by the inverse of the Inverse Wooga Looga Theorem<br /> <br /> =Testimonials=<br /> <br /> Pogpr0 = wooga looga - Ladka13<br /> The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook.<br /> ~ilp2020<br /> <br /> Thanks for rediscovering our theorem RedFireTruck - Foogle and Hoogle of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> Franzlist is wooga looga howsopro - volkie boy<br /> <br /> this is in fact a pretty sensible theorem. Nothing to be so excited about, though. ~DofL<br /> <br /> The Wooga Looga Theorem is EPIC POGGERS WHOLESOME 100 KEANU CHUNGUS AMAZING SKILL THEOREM!!!!!1!!!111111 -centslordm<br /> <br /> The Wooga Looga Theorem is amazing and can be applied to so many problems and should be taught in every school. - [[User:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;RedFireTruck&lt;/font&gt;]] ([[User talk:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;talk&lt;/font&gt;]]) 11:00, 1 February 2021 (EST)<br /> <br /> The Wooga Looga Theorem is the best. -aaja3427<br /> <br /> The Wooga Looga Theorem is needed for everything and it is great-hi..<br /> <br /> The Wooga Looga Theorem was made by the author of the 5th Testimonial, RedFireTruck, which means they are the ooga booga tribe... proof: go to https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ and click &quot;about&quot;. now copy and paste the aops URL. you got RedFireTruck! Great Job! now go check out his thread for post milestones, https://artofproblemsolving.com/community/c3h2319596, and give him a friend request! -FPT<br /> <br /> This theorem has helped me with school and I am no longer failing my math class. -mchang<br /> <br /> I, who [u]rarely[\u] edits the AoPS Wiki, has edited this to show how amazing this theorem is! The Wooga Looga theorem has actually helped me on a school test, a math competition, and more! My teacher got upset at me for not doing it the way I she taught it though - ChrisalonaLiverspur<br /> <br /> &quot;I can't believe AoPS books don't have this amazing theorem. If you need help with math, you can depend on caveman.&quot; ~CoolJupiter<br /> <br /> Before the Wooga Looga Theorem, I had NO IDEA how to solve any hard geo. But, now that I've learned it, I can solve hard geo in 7 seconds ~ ilp2020 (2nd testimonial by me)<br /> <br /> Too powerful... ~franzliszt<br /> <br /> The Wooga Looga Theorem is so pro ~ ac142931<br /> <br /> It is so epic and awesome that it will blow the minds of people if they saw this ~ ac142931(2nd testimonial by me)<br /> <br /> This theorem changed my life... ~ samrocksnature<br /> <br /> Math competitions need to ban the use of the Wooga Looga Theorem, it's just too good. ~ jasperE3<br /> <br /> It actually can be. I never thought I'd say this, but the Wooga Looga theorem is a legit theorem. ~ jasperE3<br /> <br /> This is franzliszt and I endorse this theorem. ~franzliszt<br /> <br /> This theorem is too OP. ~bestzack66<br /> <br /> This is amazing! However much it looks like a joke, it is a legitimate - and powerful - theorem. -Supernova283<br /> <br /> Wooga Looga Theorem is extremely useful. Someone needs to make a handout on this so everyone can obtain the power of Wooga Looga ~RP3.1415<br /> <br /> The Wooga Looga cavemen were way ahead of their time. Good job (dead) guys! -HIA2020<br /> <br /> It's like the Ooga Booga Theorem (also OP), but better!!! - BobDBuilder321<br /> <br /> The Wooga Looga Theorem is a special case of [https://en.wikipedia.org/wiki/Routh%27s_theorem Routh's Theorem.] So this wiki article is DEFINITELY needed. -peace<br /> <br /> I actually thought this was a joke theorem until I read this page - HumanCalculator9<br /> <br /> I endorse the Wooga Looga theorem for its utter usefulness and seriousness. -HamstPan38825<br /> <br /> &lt;s&gt;I ReAlLy don't get it - Senguamar&lt;/s&gt; HOW DARE YOU!!!!<br /> <br /> The Wooga Looga Theorem is the base of all geometry. It is so OP that even I don't understand how to use it.<br /> <br /> You know what, this is jayasharmaramankumarguptareddybavarajugopal's lemma - Ishan<br /> <br /> If only I knew this on some contests that I had done previously... - JacobJB<br /> <br /> The Wooga Looga Theorem is so pr0 that it needs to be nerfed. - rocketsri<br /> <br /> &quot;The Wooga Looga Theorem should be used in contests and should be part of geometry books.&quot; ~ [[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 21:56, 21 December 2020 (EST)<br /> <br /> The Wooga Looga Theorem is so OP BRUH<br /> <br /> thank for the theorem it is trivial by 1/2 ab sin(C) formula but very helpful I have used it zero times so far in competitions so it is of great use thank - bussie<br /> <br /> I have no idea what is going on here - awesomeguy856<br /> <br /> fuzimiao2013 waz hear<br /> <br /> this theorem is bad<br /> <br /> poggers theorem - awesomeming327<br /> <br /> The Wooga Looga theorem is very OP and not to be frowned upon - Yelly314<br /> <br /> person who invented Wooga Looga theorem is orz orz orz wooga looga theorem OP, citing it on ANY olympiad test = instant full marks -awesomeness_in_a_bun<br /> <br /> Wooga Looga Theorem is TRASH.<br /> <br /> HOW DARE YOU @above and @5above DISRESPECT THE WOOGA LOOGA THEOREM. THIS IS THE MOST OP THEOREM EVER AND CAN BE USED TO SOLVE EVERY PROBLEM. BECAUSE OF THIS THEOREM, I GOT A 187 ON AMC 10A AND 10B, a 665 on the AIME, AND A 420 ON THE USAJMO THIS YEAR. I HAVE ALSO USED THIS TO PROVE THE GOLDBACH CONJECTURE, THE TWIN PRIME CONJECTURE, EVERY SINGLE IMO PROBLEM, AND I HAVE PROVED THE RIEMANN HYPOTHESIS. I ALSO INVENTED HAND SANITIZER THAT KILLS 100% OF BACTERIA, MILK THAT IS 0% MILK, AND A TIME MACHINE WITH THIS THEOREM. THANK YOU SO MUCH @RedFireTruck FOR THIS LEGENDARY THEOREM. I WILL ALWAYS BE INDEBTED TO YOU.</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_10&diff=141678 2020 AIME II Problems/Problem 10 2021-01-07T16:30:07Z <p>CoolCarsOnTheRun: /* Solution 1 (If you don't remember the formula for sum of cubes) */</p> <hr /> <div>= Problem =<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that when &lt;math&gt;1^3+2^3+3^3+\cdots +n^3&lt;/math&gt; is divided by &lt;math&gt;n+5&lt;/math&gt;, the remainder is &lt;math&gt;17&lt;/math&gt;.<br /> <br /> == Solution 1 (If you don't remember the formula for sum of cubes) ==<br /> We first note that since the remainder is &lt;math&gt;17&lt;/math&gt; and we are dividing by &lt;math&gt;n+5&lt;/math&gt;, &lt;math&gt;n+5&lt;/math&gt; must be greater than &lt;math&gt;17&lt;/math&gt;, meaning that &lt;math&gt;n&lt;/math&gt; has to be at least &lt;math&gt;13&lt;/math&gt;.<br /> <br /> We then notice that we can pair the &lt;math&gt;5^3&lt;/math&gt; term with the &lt;math&gt;n^3&lt;/math&gt; term to factor it into &lt;math&gt;(n+5)(n^2-5n+25)&lt;/math&gt; using the sum of two cubes formula (&lt;math&gt;a^3+b^3=(a+b)(a^2-ab+b^2)&lt;/math&gt;), which is divisible by &lt;math&gt;n+5&lt;/math&gt;. We can do the same for the &lt;math&gt;6^3&lt;/math&gt; term with the &lt;math&gt;(n-1)^3&lt;/math&gt; term, the &lt;math&gt;7^3&lt;/math&gt; term with the &lt;math&gt;(n-2)^3&lt;/math&gt;, and so on, which are all divisible by &lt;math&gt;n+5&lt;/math&gt;. However, when &lt;math&gt;n&lt;/math&gt; is odd, we will have a middle term that is not paired with any other terms, which is not necessarily divisible by &lt;math&gt;n+5&lt;/math&gt;. Thus, we have two cases:<br /> &lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;\textbf{CASE 1: } n&lt;/math&gt; is even &lt;br&gt;<br /> If &lt;math&gt;n&lt;/math&gt; is even, all terms that are greater than &lt;math&gt;4^3&lt;/math&gt; pair, as there are an even number of terms that are greater than &lt;math&gt;4^3&lt;/math&gt;. Therefore, all we need in order for the entire sequence to have a remainder of &lt;math&gt;17&lt;/math&gt; when divided by &lt;math&gt;n+5&lt;/math&gt; is &lt;math&gt;1^3+2^3+3^3+4^3&lt;/math&gt; to have a remainder of &lt;math&gt;17&lt;/math&gt; when divided by &lt;math&gt;n+5&lt;/math&gt;.<br /> <br /> Evaluating &lt;math&gt;1^3+2^3+3^3+4^3&lt;/math&gt; as &lt;math&gt;100&lt;/math&gt;, all we need to be true is <br /> &lt;cmath&gt;100\equiv 17\pmod {n+5},&lt;/cmath&gt;<br /> or that<br /> &lt;cmath&gt;83\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> Thus, &lt;math&gt;83&lt;/math&gt; will be divisible by &lt;math&gt;n+5&lt;/math&gt; where &lt;math&gt;n\geq 13&lt;/math&gt;. As &lt;math&gt;83&lt;/math&gt; is prime, &lt;math&gt;n+5&lt;/math&gt; must be equal to either &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;83&lt;/math&gt;. If &lt;math&gt;n+5=1&lt;/math&gt;, we have that &lt;math&gt;n=-4&lt;/math&gt;, which is not greater than or equal to &lt;math&gt;13&lt;/math&gt;, so that solution is extraneous.<br /> <br /> If &lt;math&gt;n+5=83&lt;/math&gt;, we have that &lt;math&gt;n=78&lt;/math&gt;, which is &lt;math&gt;\geq 13&lt;/math&gt;, so one of our solutions is &lt;math&gt;n=78&lt;/math&gt;, and we are done with our first case.<br /> &lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;\textbf{CASE 2: } n&lt;/math&gt; is odd<br /> If &lt;math&gt;n&lt;/math&gt; is odd, the only term that does not pair is the arithmetic mean of the numbers under the cube of the largest and smallest terms that would pair, or &lt;math&gt;(\frac{n+5}{2})^3&lt;/math&gt;. Therefore, as all other terms that are &lt;math&gt;\geq 5^3&lt;/math&gt; pair, the requirement that we have is<br /> &lt;cmath&gt;1^3+2^3+3^3+4^3+\left(\frac{n+5}{2}\right)^3\equiv 17\pmod {n+5}.&lt;/cmath&gt;<br /> Calculating and simplifying, we have that<br /> &lt;cmath&gt;83+\left(\frac{n+5}{2}\right)^3\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> Now, we multiply both sides by &lt;math&gt;8&lt;/math&gt;. However, since multiplication is not reversible in modular arithmetic, we need to check whether any solutions are extraneous after solving. The congruence that we now have is<br /> &lt;cmath&gt;8\cdot 83+(n+5)^3\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> As we know that &lt;math&gt;(n+5)^3&lt;/math&gt; is divisible by &lt;math&gt;n+5&lt;/math&gt;, what we need now is<br /> &lt;cmath&gt;8\cdot 83\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> We now check each solution to see whether it works.<br /> <br /> If &lt;math&gt;n+5=1, 2, 4, 8&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would be less than &lt;math&gt;13&lt;/math&gt;, so none of these solutions work. If &lt;math&gt;n+5=83&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would be even, so that solution does not work for this case. Therefore, the only three solutions we need to check for this case are when &lt;math&gt;n+5=166&lt;/math&gt;, &lt;math&gt;n+5=332&lt;/math&gt;, or &lt;math&gt;n+5=664&lt;/math&gt;. We plug these values into the congruence before we multiplied both sides by &lt;math&gt;8&lt;/math&gt; to check.<br /> <br /> If &lt;math&gt;n+5=166&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{166}{2}\right)^3\equiv 0\pmod {166}.&lt;/cmath&gt;<br /> Calculating and factoring out &lt;math&gt;83&lt;/math&gt;, we have that<br /> &lt;cmath&gt;83(1+83\cdot 83)\equiv 0\pmod {166}.&lt;/cmath&gt;<br /> As the right parenthesis is odd and &lt;math&gt;166=83\cdot 2&lt;/math&gt;, we know that this solution works, so we have another solution: &lt;math&gt;n=166-5=161&lt;/math&gt;.<br /> <br /> If &lt;math&gt;n+5=332&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{332}{2}\right)^3=83+166^3\equiv 0\pmod {322}.&lt;/cmath&gt;<br /> As the left hand side is odd, but all multiples of &lt;math&gt;322&lt;/math&gt; is even, this solution is therefore extraneous.<br /> <br /> If &lt;math&gt;n+5=664&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{664}{2}\right)^3=83+332^3\equiv 0\pmod {664}.&lt;/cmath&gt;<br /> Again, the left hand side is odd, and all multiples of &lt;math&gt;664&lt;/math&gt; are even, so this solution is extraneous.<br /> &lt;br&gt;&lt;br&gt;<br /> Therefore, our final answer is &lt;math&gt;78+161=\boxed{239}&lt;/math&gt;.<br /> &lt;br&gt;&lt;br&gt;<br /> ~ CoolCarsOnTheRun<br /> <br /> ==Solution 2 (w/ formula)==<br /> Let &lt;math&gt;m=n+5&lt;/math&gt;. Then we have<br /> &lt;cmath&gt;1^3+2^3+3^3+\cdots+(m-5)^3\equiv 17 \mod m&lt;/cmath&gt;<br /> &lt;cmath&gt;\left(\frac{(m-5)(m-4)}{2}\right)^2\equiv 17 \mod m &lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{400}{4}\equiv 17 \mod m &lt;/cmath&gt;<br /> &lt;cmath&gt;332 \equiv 0 \mod m &lt;/cmath&gt;<br /> So, &lt;math&gt;m\in\{83,166,332\}&lt;/math&gt;. Testing, the cases, only &lt;math&gt;332&lt;/math&gt; fails. This leaves &lt;math&gt;78+161=\boxed{239}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Official MAA 1)==<br /> The sum of the cubes from 1 to &lt;math&gt;n&lt;/math&gt; is<br /> &lt;cmath&gt;1^3+2^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}.&lt;/cmath&gt;For this to be equal to &lt;math&gt;(n+5)q+17&lt;/math&gt; for some integer &lt;math&gt;q&lt;/math&gt;, it must be that&lt;cmath&gt;n^2(n+1)^2=4(n+5)q+4\cdot 17,&lt;/cmath&gt;so&lt;cmath&gt;n^2(n+1)^2 \equiv 4 \cdot 17= 68\hskip-.2cm \pmod{n+5}.&lt;/cmath&gt;But &lt;math&gt;n^2(n+1)^2 \equiv (-5)^2(-4)^2 = 400 \pmod{n+5}.&lt;/math&gt; Thus &lt;math&gt;n^2(n+1)^2&lt;/math&gt; is congruent to both &lt;math&gt;68&lt;/math&gt; and &lt;math&gt;400,&lt;/math&gt; which implies that &lt;math&gt;n+5&lt;/math&gt; divides &lt;math&gt;400-68 = 332=2^2 \cdot 83&lt;/math&gt;. Because &lt;math&gt;n+5 &gt; 17&lt;/math&gt;, the only choices for &lt;math&gt;n+5&lt;/math&gt; are &lt;math&gt;83, 166,&lt;/math&gt; and &lt;math&gt;332.&lt;/math&gt; Checking all three cases verifies that &lt;math&gt;n=78&lt;/math&gt; and &lt;math&gt;n=161&lt;/math&gt; work, but &lt;math&gt;n=327&lt;/math&gt; does not. The requested sum is &lt;math&gt;78+161 = 239&lt;/math&gt;.<br /> <br /> ==Solution 4 (Official MAA 2)==<br /> The sum of the cubes of the integers from &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;n&lt;/math&gt; is&lt;cmath&gt;\frac{n^2(n+1)^2}{4},&lt;/cmath&gt;which, when divided by &lt;math&gt;n+5&lt;/math&gt;, has quotient&lt;cmath&gt;Q=\frac14n^3 -\frac34n^2+4n-20 = \frac{n^2(n-3)}4+4n-20&lt;/cmath&gt;with remainder &lt;math&gt;100.&lt;/math&gt; If &lt;math&gt;n&lt;/math&gt; is not congruent to &lt;math&gt;1\pmod4&lt;/math&gt;, then &lt;math&gt;Q&lt;/math&gt; is an integer, and&lt;cmath&gt;\frac{n^2(n+1)^2}{4} = (n+5)Q + 100 \equiv 17\pmod{n+5},&lt;/cmath&gt;so &lt;math&gt;n+5&lt;/math&gt; divides &lt;math&gt;100 - 17 =83&lt;/math&gt;, and &lt;math&gt;n = 78&lt;/math&gt;. If &lt;math&gt;n \equiv 1 \pmod4&lt;/math&gt;, then &lt;math&gt;Q&lt;/math&gt; is half of an integer, and letting &lt;math&gt;n = 4k+1&lt;/math&gt; for some integer &lt;math&gt;k&lt;/math&gt; gives&lt;cmath&gt;\frac{n^2(n+1)^2}{4} = 2(2k+3)Q + 100 \equiv 17\pmod{n+5}.&lt;/cmath&gt;Thus &lt;math&gt;2k+3&lt;/math&gt; divides &lt;math&gt;100-17 = 83&lt;/math&gt;. It follows that &lt;math&gt;k=40&lt;/math&gt;, and &lt;math&gt;n = 161&lt;/math&gt;. The requested sum is &lt;math&gt;161 + 78 = 239&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Using the formula for &lt;math&gt;\sum_{k=1}^n n^3&lt;/math&gt;,<br /> &lt;cmath&gt;1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2(n+1)^2}{4}&lt;/cmath&gt;<br /> Since &lt;math&gt;1^3 + 2^3 + 3^3 + ... + n^3&lt;/math&gt; divided by &lt;math&gt;n + 5&lt;/math&gt; has a remainder of &lt;math&gt;17&lt;/math&gt;,<br /> &lt;cmath&gt;\frac{n^2(n+1)^2}{4} \equiv 17\pmod {n + 5}&lt;/cmath&gt;<br /> Using the rules of modular arithmetic,<br /> &lt;cmath&gt;n^2(n+1)^2 \equiv 68\pmod {n + 5}&lt;/cmath&gt;&lt;cmath&gt;n^2(n+1)^2 - 68\equiv 0\pmod {n + 5}&lt;/cmath&gt;<br /> Expanding the left hand side,<br /> &lt;cmath&gt;n^4 + 2 n^3 + n^2 - 68\equiv 0\pmod {n + 5}&lt;/cmath&gt;<br /> This means that <br /> &lt;math&gt;n^4 + 2 n^3 + n^2 - 68&lt;/math&gt; is divisible by &lt;math&gt;{n + 5}&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;(n + 5) | (n^4 + 2 n^3 + n^2 - 68)&lt;/cmath&gt;<br /> Dividing polynomials,<br /> &lt;cmath&gt;\frac{n^4 + 2 n^3 + n^2 - 68}{n + 5}&lt;/cmath&gt;<br /> &lt;cmath&gt;= n^3 - 3 n^2 + 16n - 80 + \frac{332}{(n + 5)}&lt;/cmath&gt;<br /> &lt;math&gt;(n + 5)&lt;/math&gt; &lt;math&gt;|&lt;/math&gt; &lt;math&gt;(n^4 + 2 n^3 + n^2 - 68)&lt;/math&gt; &lt;math&gt;\iff&lt;/math&gt; &lt;math&gt;\frac{332}{(n + 5)}&lt;/math&gt; &lt;math&gt;\in&lt;/math&gt; &lt;math&gt;\mathbb{Z}&lt;/math&gt;<br /> &lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;\frac{332}{(n + 5)}&lt;/math&gt; &lt;math&gt;\in&lt;/math&gt;<br /> &lt;math&gt;\mathbb{Z}&lt;/math&gt; &lt;math&gt;\iff&lt;/math&gt; &lt;math&gt;(n + 5) = \pm 1, \pm 2, \pm 4, \pm 83, \pm 166, \pm 332&lt;/math&gt;<br /> &lt;br&gt;&lt;br&gt;<br /> Note that &lt;math&gt;n&lt;/math&gt; &lt;math&gt;\in&lt;/math&gt; &lt;math&gt;\mathbb{N}&lt;/math&gt; and &lt;math&gt;n + 5 &gt; 17&lt;/math&gt; (because the remainder when dividing by &lt;math&gt;n + 5&lt;/math&gt; is &lt;math&gt;17&lt;/math&gt;, so &lt;math&gt;n + 5&lt;/math&gt; must be greater than &lt;math&gt;17&lt;/math&gt;), so all options &lt;math&gt;\leq 17&lt;/math&gt; can be eliminated.<br /> &lt;cmath&gt;(n + 5) = 83, 166, 332&lt;/cmath&gt;<br /> &lt;cmath&gt;n = 78, 161, 327&lt;/cmath&gt;<br /> Checking all 3 cases, &lt;math&gt;n = 78&lt;/math&gt; and &lt;math&gt;n = 161&lt;/math&gt; work; &lt;math&gt;n = 327&lt;/math&gt; fails.<br /> &lt;br&gt;&lt;br&gt;<br /> Therefore, the answer is &lt;math&gt;78 + 161 = \boxed{239}&lt;/math&gt;.<br /> &lt;br&gt;&lt;br&gt;<br /> ~ {TSun} ~<br /> ==Video Solution==<br /> https://youtu.be/bz5N-jI2e0U?t=201<br /> <br /> ==See Also==<br /> {{AIME box|year=2020|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=Wooga_Looga_Theorem&diff=136863 Wooga Looga Theorem 2020-11-07T05:16:33Z <p>CoolCarsOnTheRun: /* Testimonials */</p> <hr /> <div>=Definition=<br /> If there is &lt;math&gt;\triangle ABC&lt;/math&gt; and points &lt;math&gt;D,E,F&lt;/math&gt; on the sides &lt;math&gt;BC,CA,AB&lt;/math&gt; respectively such that &lt;math&gt;\frac{DB}{DC}=\frac{EC}{EA}=\frac{FA}{FB}=r&lt;/math&gt;, then the ratio &lt;math&gt;\frac{[DEF]}{[ABC]}=\left|\frac{r^2-r+1}{(r+1)^2}\right|&lt;/math&gt;.<br /> <br /> Created by the Ooga Booga Tribe of the Caveman Society, https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ<br /> <br /> =Proofs=<br /> ==Proof 1==<br /> Proof by Gogobao:<br /> <br /> We have: &lt;math&gt;\frac{DB}{BC} = \frac{r}{r+1}, \frac{DC}{BC} = \frac{1}{r+1}, \frac{EC}{AC} = \frac{r}{r+1}, \frac{EA}{AC} = \frac{1}{r+1}, \frac{FA}{BA} = \frac{r}{r+1}, \frac{FB}{BA} = \frac{1}{r+1} &lt;/math&gt;<br /> <br /> We have: &lt;math&gt;[DEF] = [ABC] - [DCE] - [FAE] - [FBD]&lt;/math&gt;<br /> <br /> &lt;math&gt;[DCE] = [ABC] \cdot \frac{DC}{CB} \cdot \frac{CE}{CA} = [ABC] \cdot \frac{r}{(r+1)^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;[EAF] = [ABC] \cdot \frac{EA}{CA} \cdot \frac{FA}{BA} = [ABC] \cdot \frac{r}{(r+1)^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;[FBD] = [ABC] \cdot \frac{FB}{AB} \cdot \frac{BD}{CB} = [ABC] \cdot \frac{r}{(r+1)^2}&lt;/math&gt;<br /> <br /> Therefore &lt;math&gt;[DEF] = [ABC] (1-\frac{3r}{(r+1)^2})&lt;/math&gt;<br /> <br /> So we have &lt;math&gt;\frac{[DEF]}{[ABC]} = \frac{r^2-r+1}{(r+1)^2}&lt;/math&gt;<br /> <br /> ==Proof 2==<br /> Proof by franzliszt<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Then &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. We can also find that &lt;math&gt;D=\left(0,\tfrac {1}{r+1},\tfrac {r}{r+1}\right),E=\left(\tfrac {r}{r+1},0,\tfrac {1}{r+1}\right),F=\left(\tfrac {1}{r+1},\tfrac {r}{r+1},0\right)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that &lt;cmath&gt;\frac{[DEF]}{[ABC]}= \begin{vmatrix} 0&amp;\tfrac {1}{r+1}&amp;\tfrac {r}{r+1} \\ \tfrac {r}{r+1}&amp;0&amp;\tfrac {1}{r+1}\\ \tfrac {1}{r+1}&amp;\tfrac {r}{r+1}&amp;0 \end{vmatrix}=\frac{r^2-r+1}{(r+1)^2}.&lt;/cmath&gt;<br /> <br /> ==Proof 3==<br /> Proof by RedFireTruck:<br /> <br /> WLOG we let &lt;math&gt;A=(0, 0)&lt;/math&gt;, &lt;math&gt;B=(1, 0)&lt;/math&gt;, &lt;math&gt;C=(x, y)&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y\in\mathbb{R}&lt;/math&gt;. We then use Shoelace Forumla to get &lt;math&gt;[ABC]=\frac12|y|&lt;/math&gt;. We then figure out that &lt;math&gt;D=\left(\frac{rx+1}{r+1}, \frac{ry}{r+1}\right)&lt;/math&gt;, &lt;math&gt;E=\left(\frac{x}{r+1}, \frac{y}{r+1}\right)&lt;/math&gt;, and &lt;math&gt;F=\left(\frac{r}{r+1}, 0\right)&lt;/math&gt; so we know that by Shoelace Formula &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{\frac12\left|\frac{r^2y-ry+y}{(r+1)^2}\right|}{\frac12|y|}=\left|\frac{r^2-r+1}{(r+1)^2}\right|&lt;/math&gt;. We know that &lt;math&gt;\frac{r^2-r+1}{(r+1)^2}\ge0&lt;/math&gt; for all &lt;math&gt;r\in\mathbb{R}&lt;/math&gt; so &lt;math&gt;\left|\frac{r^2-r+1}{(r+1)^2}\right|=\frac{r^2-r+1}{(r+1)^2}&lt;/math&gt;.<br /> <br /> ==Proof 4==<br /> Proof by jasperE3:<br /> The Wooga Looga theorem is a direct application of Routh's Theorem when a=b=c.<br /> <br /> =Application 1=<br /> ==Problem==<br /> The Wooga Looga Theorem states that the solution to this problem by franzliszt:<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; points &lt;math&gt;X,Y,Z&lt;/math&gt; are on sides &lt;math&gt;BC,CA,AB&lt;/math&gt; such that &lt;math&gt;\frac{XB}{XC}=\frac{YC}{YA}=\frac{ZA}{ZB}=\frac 71&lt;/math&gt;. Find the ratio of &lt;math&gt;[XYZ]&lt;/math&gt; to &lt;math&gt;[ABC]&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> is this solution by RedFireTruck:<br /> <br /> WLOG let &lt;math&gt;A=(0, 0)&lt;/math&gt;, &lt;math&gt;B=(1, 0)&lt;/math&gt;, &lt;math&gt;C=(x, y)&lt;/math&gt;. Then &lt;math&gt;[ABC]=\frac12|y|&lt;/math&gt; by Shoelace Theorem and &lt;math&gt;X=\left(\frac{7x+1}{8}, \frac{7y}{8}\right)&lt;/math&gt;, &lt;math&gt;Y=\left(\frac{x}{8}, \frac{y}{8}\right)&lt;/math&gt;, &lt;math&gt;Z=\left(\frac78, 0\right)&lt;/math&gt;. Then &lt;math&gt;[XYZ]=\frac12\left|\frac{43y}{64}\right|&lt;/math&gt; by Shoelace Theorem. Therefore the answer is &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;.<br /> ==Solution 2==<br /> or this solution by franzliszt:<br /> <br /> We apply Barycentric Coordinates w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Let &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. Then we find that &lt;math&gt;X=\left(0,\tfrac 18,\tfrac 78\right),Y=\left(\tfrac 78,0,\tfrac 18\right),Z=\left(\tfrac18,\tfrac78,0\right)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix}<br /> x_{1} &amp;y_{1} &amp;z_{1} \\<br /> x_{2} &amp;y_{2} &amp;z_{2} \\ <br /> x_{3}&amp; y_{3} &amp; z_{3}<br /> \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that &lt;cmath&gt;\frac{[XYZ]}{[ABC]}=\begin{vmatrix}<br /> 0&amp;\tfrac 18&amp;\tfrac 78\\<br /> \tfrac 78&amp;0&amp;\tfrac 18\\<br /> \tfrac18&amp;\tfrac78&amp;0<br /> \end{vmatrix}=\frac{43}{64}.&lt;/cmath&gt; &lt;math&gt;\blacksquare&lt;/math&gt;<br /> ==Solution 3==<br /> or this solution by aaja3427:<br /> <br /> According the the Wooga Looga Theorem, It is &lt;math&gt;\frac{49-7+1}{8^2}&lt;/math&gt;. This is &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> or this solution by eduD_looC:<br /> <br /> This is a perfect application of the Adihaya Jayasharmaramankumarguptareddybavarajugopal's Lemma, which results in the answer being &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. A very beautiful application, which leaves graders and readers speechless.<br /> <br /> ==Solution 6==<br /> or this solution by CoolJupiter:<br /> <br /> Wow. All of your solutions are slow, compared to my sol:<br /> <br /> By math, we have &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;.<br /> <br /> ~CoolJupiter<br /> ^<br /> |<br /> EVERYONE USE THIS SOLUTION IT'S BRILLIANT <br /> ~bsu1<br /> Yes, very BRILLIANT!<br /> ~ TheAoPSLebron<br /> <br /> ==The Best Solution==<br /> <br /> By the 1+1=2 principle, we get &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. Definitely the best method. When asked, please say that OlympusHero taught you this method. Cuz he did.<br /> <br /> =Application 2=<br /> ==Problem==<br /> The Wooga Looga Theorem states that the solution to this problem by Matholic:<br /> <br /> The figure below shows a triangle ABC whose area is &lt;math&gt;72 \text{cm}^2&lt;/math&gt;. If &lt;math&gt;\dfrac{AD}{DB}=\dfrac{BE}{EC}=\dfrac{CF}{FA}=\dfrac{1}{5}&lt;/math&gt;, find &lt;math&gt;[DEF].&lt;/math&gt;<br /> <br /> ~LaTeX-ifyed by RP3.1415<br /> <br /> ==Solution 1==<br /> is this solution by franzliszt:<br /> <br /> We apply Barycentric Coordinates w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Let &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. Then we find that &lt;math&gt;D=(\tfrac 56,\tfrac 16,0),E=(0,\tfrac 56,\tfrac 16),F=(\tfrac16,0,\tfrac56)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix}<br /> x_{1} &amp;y_{1} &amp;z_{1} \\ <br /> x_{2} &amp;y_{2} &amp;z_{2} \\ <br /> x_{3}&amp; y_{3} &amp; z_{3}<br /> \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[DEF]}{}=\begin{vmatrix}<br /> \tfrac 56&amp;\tfrac 16&amp;0\\<br /> 0&amp;\tfrac 56&amp;\tfrac 16\\<br /> \tfrac16&amp;0&amp;\tfrac56<br /> \end{vmatrix}=\frac{7}{12}&lt;/cmath&gt; so &lt;math&gt;[DEF]=42&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> ==Solution 2==<br /> or this solution by RedFireTruck:<br /> <br /> By the Wooga Looga Theorem, &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{5^2-5+1}{(5+1)^2}=\frac{21}{36}=\frac{7}{12}&lt;/math&gt;. We are given that &lt;math&gt;[ABC]=72&lt;/math&gt; so &lt;math&gt;[DEF]=\frac{7}{12}\cdot72=\boxed{42}&lt;/math&gt;<br /> <br /> =Application 3=<br /> ==Problem==<br /> The Wooga Looga Theorem states that the solution to this problem by RedFireTruck:<br /> <br /> Find the ratio &lt;math&gt;\frac{[GHI]}{[ABC]}&lt;/math&gt; if &lt;math&gt;\frac{AD}{DB}=\frac{BE}{EC}=\frac{CF}{FA}=\frac12&lt;/math&gt; and &lt;math&gt;\frac{DG}{GE}=\frac{EH}{HF}=\frac{FI}{ID}=1&lt;/math&gt; in the diagram below.&lt;asy&gt;<br /> draw((0, 0)--(6, 0)--(4, 3)--cycle);<br /> draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle);<br /> draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle);<br /> label(&quot;$A$&quot;, (0, 0), SW);<br /> label(&quot;$B$&quot;, (6, 0), SE);<br /> label(&quot;$C$&quot;, (4, 3), N);<br /> label(&quot;$D$&quot;, (2, 0), S);<br /> label(&quot;$E$&quot;, (16/3, 1), NE);<br /> label(&quot;$F$&quot;, (8/3, 2), NW);<br /> label(&quot;$G$&quot;, (11/3, 1/2), SE);<br /> label(&quot;$H$&quot;, (4, 3/2), NE);<br /> label(&quot;$I$&quot;, (7/3, 1), W);<br /> &lt;/asy&gt;<br /> ==Solution 1==<br /> is this solution by franzliszt:<br /> <br /> By the Wooga Looga Theorem, &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{2^2-2+1}{(1+2)^2}=\frac 13&lt;/math&gt;. Notice that &lt;math&gt;\triangle GHI&lt;/math&gt; is the medial triangle of '''Wooga Looga Triangle ''' of &lt;math&gt;\triangle ABC&lt;/math&gt;. So &lt;math&gt;\frac{[GHI]}{[DEF]}=\frac 14&lt;/math&gt; and &lt;math&gt;\frac{[GHI]}{[ABD]}=\frac{[DEF]}{[ABC]}\cdot\frac{[GHI]}{[DEF]}=\frac 13 \cdot \frac 14 = \frac {1}{12}&lt;/math&gt; by Chain Rule ideas.<br /> <br /> ==Solution 2==<br /> or this solution by franzliszt:<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt; so that &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. Then &lt;math&gt;D=(\tfrac 23,\tfrac 13,0),E=(0,\tfrac 23,\tfrac 13),F=(\tfrac 13,0,\tfrac 23)&lt;/math&gt;. And &lt;math&gt;G=(\tfrac 13,\tfrac 12,\tfrac 16),H=(\tfrac 16,\tfrac 13,\tfrac 12),I=(\tfrac 12,\tfrac 16,\tfrac 13)&lt;/math&gt;.<br /> <br /> In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[GHI]}{[ABC]}=\begin{vmatrix} \tfrac 13&amp;\tfrac 12&amp;\tfrac 16\\ \tfrac 16&amp;\tfrac 13&amp;\tfrac 12\\ \tfrac 12&amp;\tfrac 16&amp;\tfrac 13 \end{vmatrix}=\frac{1}{12}.&lt;/cmath&gt;<br /> <br /> =Application 4=<br /> ==Problem==<br /> <br /> Let &lt;math&gt;ABC&lt;/math&gt; be a triangle and &lt;math&gt;D,E,F&lt;/math&gt; be points on sides &lt;math&gt;BC,AC,&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; respectively. We have that &lt;math&gt;\frac{BD}{DC} = 3&lt;/math&gt; and similar for the other sides. If the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;, then what is the area of triangle &lt;math&gt;DEF&lt;/math&gt;? (By ilovepizza2020)<br /> <br /> ==Solution 1==<br /> <br /> By Franzliszt<br /> <br /> By Wooga Looga, &lt;math&gt;\frac{[DEF]}{16} = \frac{3^2-3+1}{(3+1)^2}=\frac{7}{16}&lt;/math&gt; so the answer is &lt;math&gt;7&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> By franzliszt<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Then &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. We can also find that &lt;math&gt;D=(0,\tfrac 14,\tfrac 34),E=(\tfrac 34,0,\tfrac 14),F=(\tfrac 14,\tfrac 34,0)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[DEF]}{[ABC]}=\begin{vmatrix} 0&amp;\tfrac 14&amp;\tfrac 34\\ \tfrac 34&amp;0&amp;\tfrac 14\\ \tfrac 14&amp;\tfrac 34&amp;0 \end{vmatrix}=\frac{7}{16}.&lt;/cmath&gt;So the answer is &lt;math&gt;\boxed{7}&lt;/math&gt;.<br /> <br /> =Testimonials=<br /> Franzlist is wooga looga howsopro - volkie boy<br /> <br /> The Wooga Looga Theorem is EPIC POGGERS WHOLESOME 100 KEANU CHUNGUS AMAZING SKILL THEOREM!!!!!1!!!111111 -centslordm<br /> <br /> The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook.<br /> ~ilp2020<br /> <br /> The Wooga Looga Theorem is amazing and can be applied to so many problems and should be taught in every school. - RedFireTruck<br /> <br /> The Wooga Looga Theorem is the best. -aaja3427<br /> <br /> The Wooga Looga Theorem is needed for everything and it is great-hi..<br /> <br /> The Wooga Looga Theorem was made by the author of the 4th Testimonial, RedFireTruck, which means they are the ooga booga tribe... proof: go to https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ and click &quot;about&quot;. now copy and paste the aops URL. you got RedFireTruck! Great Job! now go check out his thread for post milestones, https://artofproblemsolving.com/community/c3h2319596, and give him a friend request! -FPT<br /> <br /> This theorem has helped me with school and I am no longer failing my math class. -mchang<br /> <br /> &quot;I can't believe AoPS books don't have this amazing theorem. If you need help with math, you can depend on caveman.&quot; ~CoolJupiter<br /> <br /> Before the Wooga Looga Theorem, I had NO IDEA how to solve any hard geo. But, now that I've learned it, I can solve hard geo in 7 seconds ~ ilp2020 (2nd testimonial by me)<br /> <br /> Too powerful... ~franzliszt<br /> <br /> The Wooga Looga Theorem is so pro ~ ac142931<br /> <br /> It is so epic and awesome that it will blow the minds of people if they saw this ~ ac142931(2nd testimonial by me)<br /> <br /> This theorem changed my life... ~ samrocksnature<br /> <br /> Math competitions need to ban the use of the Wooga Looga Theorem, it's just too good. ~ jasperE3<br /> <br /> It actually can be. I never thought I'd say this, but the Wooga Looga theorem is a legit theorem. ~ jasperE3<br /> <br /> This is franzliszt and I endorse this theorem. ~franzliszt<br /> <br /> This theorem is too OP. ~bestzack66<br /> <br /> This is amazing! It literally changed my geo career. Before Wooga Looga, I was awful at geo. Now, with this weapon in my hands, I am much better. Thank the cavemen! -Supernova283<br /> <br /> Wooga Looga Theorem is extremely useful. Someone needs to make a handout on this so everyone can obtain the power of Wooga Looga ~RP3.1415<br /> <br /> The Wooga Looga cavemen were way ahead of their time. Good job (dead) guys! -HIA2020<br /> <br /> It's like the Ooga Booga Theorem (also OP), but better!!! - BobDBuilder321<br /> <br /> The Wooga Looga Theorem is a special case of [url=https://en.wikipedia.org/wiki/Routh%27s_theorem]Routh's Theorem.[/url] So this wiki article is DEFINITELY needed. -peace<br /> <br /> I actually thought this was a joke theorem until I read this page - HumanCalculator9<br /> <br /> I endorse the Wooga Looga theorem for its utter usefulness and seriousness. -HamstPan38825<br /> <br /> This is &lt;i&gt;almost&lt;/i&gt; as OP as the Adihaya Jayasharmaramankumarguptareddybavarajugopal Lemma. Needs to be nerfed. -CoolCarsonTheRun</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=AoPS_Blog_Layout&diff=136741 AoPS Blog Layout 2020-11-06T16:31:48Z <p>CoolCarsOnTheRun: /* Layout */</p> <hr /> <div>==How to use blog layout in your CSS==<br /> <br /> Blogs hosted by AoPS follow a certain layout. You can take advantage of this by specifying the &lt;code&gt;div&lt;/code&gt; you want your CSS to affect. <br /> <br /> ===Specifying sections===<br /> <br /> You can specify a &lt;code&gt;div&lt;/code&gt; with a specific ID (specified below) with:<br /> <br /> &lt;code&gt; div #id_goes_here {&lt;/code&gt;<br /> &lt;code&gt; CSS goes here&lt;/code&gt;<br /> &lt;code&gt; }&lt;/code&gt;<br /> <br /> IDs are used with &lt;code&gt;div&lt;/code&gt; that only appears once in the code. On the other hand, &lt;code&gt;div&lt;/code&gt;s that occur multiple times, like a blog entry, have a class. You can specify a class like this:<br /> <br /> &lt;code&gt; div .class_goes_here {&lt;/code&gt;<br /> &lt;code&gt; CSS goes here&lt;/code&gt;<br /> &lt;code&gt; }&lt;/code&gt;<br /> <br /> If you want to specify a specific HTML element, like &lt;code&gt;p&lt;/code&gt;, the syntax is like this:<br /> <br /> &lt;code&gt; div #id__goes_here p {&lt;/code&gt;<br /> &lt;code&gt; CSS goes here&lt;/code&gt;<br /> &lt;code&gt; }&lt;/code&gt;<br /> <br /> &lt;code&gt; div .class_goes_here p {&lt;/code&gt;<br /> &lt;code&gt; CSS goes here&lt;/code&gt;<br /> &lt;code&gt; }&lt;/code&gt;<br /> <br /> If you want to specify all HTML elements, use &lt;code&gt;*&lt;/code&gt;. Here is an example:<br /> <br /> &lt;code&gt; * {&lt;/code&gt;<br /> &lt;code&gt; CSS goes here&lt;/code&gt;<br /> &lt;code&gt; }&lt;/code&gt;<br /> <br /> If you just want to specify for one HTML element, use:<br /> <br /> &lt;code&gt; p {&lt;/code&gt;<br /> &lt;code&gt; CSS;&lt;/code&gt;<br /> &lt;code&gt; }&lt;/code&gt;<br /> <br /> All items listed below are HTML &lt;code&gt;div&lt;/code&gt;s unless otherwise specified.<br /> <br /> ==Layout==<br /> <br /> * '''ID: page-wrapper''' Wraps the entire page.<br /> ** '''ID: navigation_box''' Wraps the navigation bar, on the top of the page<br /> *** '''ID: left_navigation_box '''Holds the links to the<br /> **** community<br /> **** AoPS Blogroll<br /> **** Blog Homepage<br /> *** '''ID: right_navigation_box''' Holds the links to the<br /> **** subscribe link<br /> **** logout<br /> **** Blog Information<br /> *** '''ID: clear '''Holds nothing.<br /> ** '''ID: wrapper''' Wraps all the blog's content.<br /> *** '''ID: header''' Holds the header space and the title link, which is a &lt;code&gt;h1&lt;/code&gt; inside a HTML link<br /> *** '''ID: content''' Holds the rest of the page<br /> **** '''ID: main''' Holds the blog entries on the left side of the page<br /> ***** '''p: post-new-entry-p''' Holds the '''post-new-entry-img''', the pencil image, and the '''post-new-entry text''', a &quot;a&quot;<br /> ***** Unedited Entries: are held in '''unnamed divs'''<br /> ***** '''CLASS: cmty-post-edited''' Holds all edited posts. Content same as CLASS: entry but cmty-post-edit-info is filled with text for edit status<br /> ****** '''CLASS: entry''' Wraps the entry and title<br /> ******* '''h1: Title of entry'''<br /> ******* '''h2: poster info and time posted'''<br /> ******* '''CLASS: entrywrap''' Wraps the entry text and widgets (comments, etc.)<br /> ******** '''CLASS: message''' Holds the actual entry text.<br /> ******** '''CLASS: cmty-post-attachments''' Unknown. Most likely images.<br /> ******** '''CLASS: cmty-post-edit-info''' Empty on unedited entries (Unnamed div) but filled up in in divs with CLASS: cmty-post-edited<br /> ******** '''CLASS: cmty-tags-itembox-wrapper''' Wraps the edit tags box.<br /> ********* '''CLASS: cmty-itembox''' Holds item.<br /> ********** '''TITLE: Edit tags''' Holds the edit tag itself.<br /> ******** '''CLASS: efooter''' Holds comments area and mod actions, such as edit post.<br /> ********* '''CLASS: actions''' Holds comments area. Holds an unordered list, with two &quot;a&quot;s: one of '''CLASS: post-replies''' and another '''CLASS: post-comment'''<br /> ********* '''CLASS: modactions '''Holds moderator actions inside an unordered list. Two &quot;a&quot;s are inside: one '''CLASS: blog-edit-post''' for editing, and the other '''CLASS: blog-moderate-topic''' for &quot;Moderate&quot;.<br /> ******** '''CLASS: clear''' Empty.<br /> **** '''ID: side''' Holds all side content, including profile, shout box, and blog stats. All divs below also belong to CLASS: block widget (with space in between)<br /> ***** '''ID: user-menu-widget '''<br /> ***** '''ID: archive-widget'''<br /> ***** '''ID: shouts-widget'''<br /> ***** '''Unnamed div.'''<br /> ***** '''ID: about-owner-widget'''<br /> ***** '''ID: stats-widget'''<br /> ***** '''ID: search-widget'''<br /> **** '''CLASS: clear''' Holds nothing<br /> * '''ID: login-form''' Wraps the login form.<br /> <br /> *'''CLASS: aops-modal-wrapper''' Wraps AoPS Modals<br /> **'''CLASS: aops-modal-frame''' Frame border for aops modals<br /> ***'''CLASS: aops-modal-content-wrapper''' Wrapps modals<br /> ****'''CLASS: aops-modal-nonscroll''' Wrapps Content<br /> *****'''CLASS: aops-modal-body''' Text for modal<br /> *'''CLASS: aops-modal-mask''' Holds AoPS Background<br /> <br /> ==Web Inspector==<br /> <br /> If you wish to see the HTML code of your blog in more detail, use a code inspector.<br /> * '''Macs with Safari''': Click on &quot;Web Inspector&quot; under the Develop menu (top navigation bar). If you do not see the Develop Menu, go to Safari&gt;Preferences&gt;Advanced&gt; and make sure Show Develop menu in hot bar.<br /> * '''Macs with Google Chrome''': control-click an element. Click inspect.<br /> * '''PCs with Google Chrome''': right-click and click on &quot;Inspect Element&quot;.<br /> <br /> ==AoPS Blog Design==<br /> <br /> {{Blog Design}}<br /> <br /> [[Category:CSS]]</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_1&diff=124329 2020 AIME II Problems/Problem 1 2020-06-08T00:38:36Z <p>CoolCarsOnTheRun: /* Solution */</p> <hr /> <div>==Problem==<br /> Find the number of ordered pairs of positive integers &lt;math&gt;(m,n)&lt;/math&gt; such that &lt;math&gt;{m^2n = 20 ^{20}}&lt;/math&gt;.<br /> <br /> ==Solution==<br /> First, we find the prime factorization of &lt;math&gt;20^{20}&lt;/math&gt;, which is &lt;math&gt;2^{40}\times5^{20}&lt;/math&gt;. The equation &lt;math&gt;{m^2n = 20 ^{20}}&lt;/math&gt; tells us that we want to select a perfect square factor of &lt;math&gt;20^{20}&lt;/math&gt;, &lt;math&gt;m^2&lt;/math&gt;. &lt;math&gt;n&lt;/math&gt; will be assigned by default. There are &lt;math&gt;21\times11=231&lt;/math&gt; ways to select a perfect square factor of &lt;math&gt;20^{20}&lt;/math&gt;, thus our answer is &lt;math&gt;\mbox{231}&lt;/math&gt;.<br /> <br /> ~superagh<br /> &lt;br&gt;&lt;br&gt;<br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_10&diff=124279 2020 AIME II Problems/Problem 10 2020-06-07T20:09:16Z <p>CoolCarsOnTheRun: /* Solution 1 (If you don't remember formula for sum of cubes) */</p> <hr /> <div>= Problem =<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that when &lt;math&gt;1^3+2^3+3^3+\cdots +n^3&lt;/math&gt; is divided by &lt;math&gt;n+5&lt;/math&gt;, the remainder is &lt;math&gt;17&lt;/math&gt;.<br /> <br /> == Solution 1 (If you don't remember the formula for sum of cubes) ==<br /> We first note that since the remainder is &lt;math&gt;17&lt;/math&gt; and we are dividing by &lt;math&gt;n+5&lt;/math&gt;, &lt;math&gt;n+5&lt;/math&gt; must be greater than &lt;math&gt;17&lt;/math&gt;, meaning that &lt;math&gt;n&lt;/math&gt; has to be at least &lt;math&gt;13&lt;/math&gt;.<br /> <br /> We then notice that we can pair the &lt;math&gt;5^3&lt;/math&gt; term with the &lt;math&gt;n^3&lt;/math&gt; term to factor it into &lt;math&gt;(n+5)(n^2-5n+25)&lt;/math&gt; using the sum of cubes formula, which is divisible by &lt;math&gt;n+5&lt;/math&gt;. We can do the same for the &lt;math&gt;6^3&lt;/math&gt; term with the &lt;math&gt;(n-1)^3&lt;/math&gt; term, the &lt;math&gt;7^3&lt;/math&gt; term with the &lt;math&gt;(n-2)^3&lt;/math&gt;, and so on, which are all divisible by &lt;math&gt;n+5&lt;/math&gt;. However, when &lt;math&gt;n&lt;/math&gt; is odd, we will have a middle term that is not paired with any other terms, which is not necessarily divisible by &lt;math&gt;n+5&lt;/math&gt;. Thus, we have two cases:<br /> &lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;\textbf{CASE 1: } n&lt;/math&gt; is even &lt;br&gt;<br /> If &lt;math&gt;n&lt;/math&gt; is even, all terms that are greater than &lt;math&gt;4^3&lt;/math&gt; pair, as there are an even number of terms that are greater than &lt;math&gt;4^3&lt;/math&gt;. Therefore, all we need in order for the entire sequence to have a remainder of &lt;math&gt;17&lt;/math&gt; when divided by &lt;math&gt;n+5&lt;/math&gt; is &lt;math&gt;1^3+2^3+3^3+4^3&lt;/math&gt; to have a remainder of &lt;math&gt;17&lt;/math&gt; when divided by &lt;math&gt;n+5&lt;/math&gt;.<br /> <br /> Evaluating &lt;math&gt;1^3+2^3+3^3+4^3&lt;/math&gt; as &lt;math&gt;100&lt;/math&gt;, all we need to be true is <br /> &lt;cmath&gt;100\equiv 17\pmod {n+5},&lt;/cmath&gt;<br /> or that<br /> &lt;cmath&gt;83\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> Thus, &lt;math&gt;83&lt;/math&gt; will be divisible by &lt;math&gt;n+5&lt;/math&gt; where &lt;math&gt;n\geq 13&lt;/math&gt;. As &lt;math&gt;83&lt;/math&gt; is prime, &lt;math&gt;n+5&lt;/math&gt; must be equal to either &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;83&lt;/math&gt;. If &lt;math&gt;n+5=1&lt;/math&gt;, we have that &lt;math&gt;n=-4&lt;/math&gt;, which is not greater than or equal to &lt;math&gt;13&lt;/math&gt;, so that solution is extraneous.<br /> <br /> If &lt;math&gt;n+5=83&lt;/math&gt;, we have that &lt;math&gt;n=78&lt;/math&gt;, which is &lt;math&gt;\geq 13&lt;/math&gt;, so one of our solutions is &lt;math&gt;n=78&lt;/math&gt;, and we are done with our first case.<br /> &lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;\textbf{CASE 2: } n&lt;/math&gt; is odd<br /> If &lt;math&gt;n&lt;/math&gt; is odd, the only term that does not pair is the arithmetic mean of the numbers under the cube of the largest and smallest terms that would pair, or &lt;math&gt;(\frac{n+5}{2})^3&lt;/math&gt;. Therefore, as all other terms that are &lt;math&gt;\geq 5^3&lt;/math&gt; pair, the requirement that we have is<br /> &lt;cmath&gt;1^3+2^3+3^3+4^3+\left(\frac{n+5}{2}\right)^3\equiv 17\pmod {n+5}.&lt;/cmath&gt;<br /> Calculating and simplifying, we have that<br /> &lt;cmath&gt;83+\left(\frac{n+5}{2}\right)^3\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> Now, we multiply both sides by &lt;math&gt;8&lt;/math&gt;. However, since multiplication is not reversible in modular arithmetic, we need to check whether any solutions are extraneous after solving. The congruence that we now have is<br /> &lt;cmath&gt;8\cdot 83+(n+5)^3\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> As we know that &lt;math&gt;(n+5)^3&lt;/math&gt; is divisible by &lt;math&gt;n+5&lt;/math&gt;, what we need now is<br /> &lt;cmath&gt;8\cdot 83\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> We now check each solution to see whether it works.<br /> <br /> If &lt;math&gt;n+5=1, 2, 4, 8&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would be less than &lt;math&gt;13&lt;/math&gt;, so none of these solutions work. If &lt;math&gt;n+5=83&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would be even, so that solution does not work for this case. Therefore, the only three solutions we need to check for this case are when &lt;math&gt;n+5=166&lt;/math&gt;, &lt;math&gt;n+5=332&lt;/math&gt;, or &lt;math&gt;n+5=664&lt;/math&gt;. We plug these values into the congruence before we multiplied both sides by &lt;math&gt;8&lt;/math&gt; to check.<br /> <br /> If &lt;math&gt;n+5=166&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{166}{2}\right)^3\equiv 0\pmod {166}.&lt;/cmath&gt;<br /> Calculating and factoring out &lt;math&gt;83&lt;/math&gt;, we have that<br /> &lt;cmath&gt;83(1+83\cdot 83)\equiv 0\pmod {166}.&lt;/cmath&gt;<br /> As the right parenthesis is odd and &lt;math&gt;166=83\cdot 2&lt;/math&gt;, we know that this solution works, so we have another solution: &lt;math&gt;n=166-5=161&lt;/math&gt;.<br /> <br /> If &lt;math&gt;n+5=332&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{332}{2}\right)^3=83+166^3\equiv 0\pmod {322}.&lt;/cmath&gt;<br /> As the left hand side is odd, but all multiples of &lt;math&gt;322&lt;/math&gt; is even, this solution is therefore extraneous.<br /> <br /> If &lt;math&gt;n+5=664&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{664}{2}\right)^3=83+332^3\equiv 0\pmod {664}.&lt;/cmath&gt;<br /> Again, the left hand side is odd, and all multiples of &lt;math&gt;664&lt;/math&gt; are even, so this solution is extraneous.<br /> &lt;br&gt;&lt;br&gt;<br /> Therefore, our final answer is &lt;math&gt;78+161=\boxed{239}&lt;/math&gt;.<br /> &lt;br&gt;&lt;br&gt;<br /> ~ CoolCarsOnTheRun</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_10&diff=124277 2020 AIME II Problems/Problem 10 2020-06-07T20:07:47Z <p>CoolCarsOnTheRun: /* Solution 1 */</p> <hr /> <div>= Problem =<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that when &lt;math&gt;1^3+2^3+3^3+\cdots +n^3&lt;/math&gt; is divided by &lt;math&gt;n+5&lt;/math&gt;, the remainder is &lt;math&gt;17&lt;/math&gt;.<br /> <br /> == Solution 1 (If you don't remember formula for sum of cubes) ==<br /> We first note that since the remainder is &lt;math&gt;17&lt;/math&gt; and we are dividing by &lt;math&gt;n+5&lt;/math&gt;, &lt;math&gt;n+5&lt;/math&gt; must be greater than &lt;math&gt;17&lt;/math&gt;, meaning that &lt;math&gt;n&lt;/math&gt; has to be at least &lt;math&gt;13&lt;/math&gt;.<br /> <br /> We then notice that we can pair the &lt;math&gt;5^3&lt;/math&gt; term with the &lt;math&gt;n^3&lt;/math&gt; term to factor it into &lt;math&gt;(n+5)(n^2-5n+25)&lt;/math&gt; using the sum of cubes formula, which is divisible by &lt;math&gt;n+5&lt;/math&gt;. We can do the same for the &lt;math&gt;6^3&lt;/math&gt; term with the &lt;math&gt;(n-1)^3&lt;/math&gt; term, the &lt;math&gt;7^3&lt;/math&gt; term with the &lt;math&gt;(n-2)^3&lt;/math&gt;, and so on, which are all divisible by &lt;math&gt;n+5&lt;/math&gt;. However, when &lt;math&gt;n&lt;/math&gt; is odd, we will have a middle term that is not paired with any other terms, which is not necessarily divisible by &lt;math&gt;n+5&lt;/math&gt;. Thus, we have two cases:<br /> &lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;\textbf{CASE 1: } n&lt;/math&gt; is even &lt;br&gt;<br /> If &lt;math&gt;n&lt;/math&gt; is even, all terms that are greater than &lt;math&gt;4^3&lt;/math&gt; pair, as there are an even number of terms that are greater than &lt;math&gt;4^3&lt;/math&gt;. Therefore, all we need in order for the entire sequence to have a remainder of &lt;math&gt;17&lt;/math&gt; when divided by &lt;math&gt;n+5&lt;/math&gt; is &lt;math&gt;1^3+2^3+3^3+4^3&lt;/math&gt; to have a remainder of &lt;math&gt;17&lt;/math&gt; when divided by &lt;math&gt;n+5&lt;/math&gt;.<br /> <br /> Evaluating &lt;math&gt;1^3+2^3+3^3+4^3&lt;/math&gt; as &lt;math&gt;100&lt;/math&gt;, all we need to be true is <br /> &lt;cmath&gt;100\equiv 17\pmod {n+5},&lt;/cmath&gt;<br /> or that<br /> &lt;cmath&gt;83\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> Thus, &lt;math&gt;83&lt;/math&gt; will be divisible by &lt;math&gt;n+5&lt;/math&gt; where &lt;math&gt;n\geq 13&lt;/math&gt;. As &lt;math&gt;83&lt;/math&gt; is prime, &lt;math&gt;n+5&lt;/math&gt; must be equal to either &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;83&lt;/math&gt;. If &lt;math&gt;n+5=1&lt;/math&gt;, we have that &lt;math&gt;n=-4&lt;/math&gt;, which is not greater than or equal to &lt;math&gt;13&lt;/math&gt;, so that solution is extraneous.<br /> <br /> If &lt;math&gt;n+5=83&lt;/math&gt;, we have that &lt;math&gt;n=78&lt;/math&gt;, which is &lt;math&gt;\geq 13&lt;/math&gt;, so one of our solutions is &lt;math&gt;n=78&lt;/math&gt;, and we are done with our first case.<br /> &lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;\textbf{CASE 2: } n&lt;/math&gt; is odd<br /> If &lt;math&gt;n&lt;/math&gt; is odd, the only term that does not pair is the arithmetic mean of the numbers under the cube of the largest and smallest terms that would pair, or &lt;math&gt;(\frac{n+5}{2})^3&lt;/math&gt;. Therefore, as all other terms that are &lt;math&gt;\geq 5^3&lt;/math&gt; pair, the requirement that we have is<br /> &lt;cmath&gt;1^3+2^3+3^3+4^3+\left(\frac{n+5}{2}\right)^3\equiv 17\pmod {n+5}.&lt;/cmath&gt;<br /> Calculating and simplifying, we have that<br /> &lt;cmath&gt;83+\left(\frac{n+5}{2}\right)^3\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> Now, we multiply both sides by &lt;math&gt;8&lt;/math&gt;. However, since multiplication is not reversible in modular arithmetic, we need to check whether any solutions are extraneous after solving. The congruence that we now have is<br /> &lt;cmath&gt;8\cdot 83+(n+5)^3\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> As we know that &lt;math&gt;(n+5)^3&lt;/math&gt; is divisible by &lt;math&gt;n+5&lt;/math&gt;, what we need now is<br /> &lt;cmath&gt;8\cdot 83\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> We now check each solution to see whether it works.<br /> <br /> If &lt;math&gt;n+5=1, 2, 4, 8&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would be less than &lt;math&gt;13&lt;/math&gt;, so none of these solutions work. If &lt;math&gt;n+5=83&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would be even, so that solution does not work for this case. Therefore, the only three solutions we need to check for this case are when &lt;math&gt;n+5=166&lt;/math&gt;, &lt;math&gt;n+5=332&lt;/math&gt;, or &lt;math&gt;n+5=664&lt;/math&gt;. We plug these values into the congruence before we multiplied both sides by &lt;math&gt;8&lt;/math&gt; to check.<br /> <br /> If &lt;math&gt;n+5=166&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{166}{2}\right)^3\equiv 0\pmod {166}.&lt;/cmath&gt;<br /> Calculating and factoring out &lt;math&gt;83&lt;/math&gt;, we have that<br /> &lt;cmath&gt;83(1+83\cdot 83)\equiv 0\pmod {166}.&lt;/cmath&gt;<br /> As the right parenthesis is odd and &lt;math&gt;166=83\cdot 2&lt;/math&gt;, we know that this solution works, so we have another solution: &lt;math&gt;n=166-5=161&lt;/math&gt;.<br /> <br /> If &lt;math&gt;n+5=332&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{332}{2}\right)^3=83+166^3\equiv 0\pmod {322}.&lt;/cmath&gt;<br /> As the left hand side is odd, but all multiples of &lt;math&gt;322&lt;/math&gt; is even, this solution is therefore extraneous.<br /> <br /> If &lt;math&gt;n+5=664&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{664}{2}\right)^3=83+332^3\equiv 0\pmod {664}.&lt;/cmath&gt;<br /> Again, the left hand side is odd, and all multiples of &lt;math&gt;664&lt;/math&gt; are even, so this solution is extraneous.<br /> &lt;br&gt;&lt;br&gt;<br /> Therefore, our final answer is &lt;math&gt;78+161=\boxed{239}&lt;/math&gt;.<br /> &lt;br&gt;&lt;br&gt;<br /> ~ CoolCarsOnTheRun</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_10&diff=124268 2020 AIME II Problems/Problem 10 2020-06-07T19:15:16Z <p>CoolCarsOnTheRun: /* Problem */</p> <hr /> <div>= Problem =<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that when &lt;math&gt;1^3+2^3+3^3+\cdots +n^3&lt;/math&gt; is divided by &lt;math&gt;n+5&lt;/math&gt;, the remainder is &lt;math&gt;17&lt;/math&gt;.<br /> <br /> == Solution 1 ==<br /> We first note that since the remainder is &lt;math&gt;17&lt;/math&gt; and we are dividing by &lt;math&gt;n+5&lt;/math&gt;, &lt;math&gt;n+5&lt;/math&gt; must be greater than &lt;math&gt;17&lt;/math&gt;, meaning that &lt;math&gt;n&lt;/math&gt; has to be at least &lt;math&gt;13&lt;/math&gt;.<br /> <br /> We then notice that we can pair the &lt;math&gt;5^3&lt;/math&gt; term with the &lt;math&gt;n^3&lt;/math&gt; term to factor it into &lt;math&gt;(n+5)(n^2-5n+25)&lt;/math&gt; using the sum of cubes formula, which is divisible by &lt;math&gt;n+5&lt;/math&gt;. We can do the same for the &lt;math&gt;6^3&lt;/math&gt; term with the &lt;math&gt;(n-1)^3&lt;/math&gt; term, the &lt;math&gt;7^3&lt;/math&gt; term with the &lt;math&gt;(n-2)^3&lt;/math&gt;, and so on, which are all divisible by &lt;math&gt;n+5&lt;/math&gt;. However, when &lt;math&gt;n&lt;/math&gt; is odd, we will have a middle term that is not paired with any other terms, which is not necessarily divisible by &lt;math&gt;n+5&lt;/math&gt;. Thus, we have two cases:<br /> &lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;\textbf{CASE 1: } n&lt;/math&gt; is even &lt;br&gt;<br /> If &lt;math&gt;n&lt;/math&gt; is even, all terms that are greater than &lt;math&gt;4^3&lt;/math&gt; pair, as there are an even number of terms that are greater than &lt;math&gt;4^3&lt;/math&gt;. Therefore, all we need in order for the entire sequence to have a remainder of &lt;math&gt;17&lt;/math&gt; when divided by &lt;math&gt;n+5&lt;/math&gt; is &lt;math&gt;1^3+2^3+3^3+4^3&lt;/math&gt; to have a remainder of &lt;math&gt;17&lt;/math&gt; when divided by &lt;math&gt;n+5&lt;/math&gt;.<br /> <br /> Evaluating &lt;math&gt;1^3+2^3+3^3+4^3&lt;/math&gt; as &lt;math&gt;100&lt;/math&gt;, all we need to be true is <br /> &lt;cmath&gt;100\equiv 17\pmod {n+5},&lt;/cmath&gt;<br /> or that<br /> &lt;cmath&gt;83\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> Thus, &lt;math&gt;83&lt;/math&gt; will be divisible by &lt;math&gt;n+5&lt;/math&gt; where &lt;math&gt;n\geq 13&lt;/math&gt;. As &lt;math&gt;83&lt;/math&gt; is prime, &lt;math&gt;n+5&lt;/math&gt; must be equal to either &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;83&lt;/math&gt;. If &lt;math&gt;n+5=1&lt;/math&gt;, we have that &lt;math&gt;n=-4&lt;/math&gt;, which is not greater than or equal to &lt;math&gt;13&lt;/math&gt;, so that solution is extraneous.<br /> <br /> If &lt;math&gt;n+5=83&lt;/math&gt;, we have that &lt;math&gt;n=78&lt;/math&gt;, which is &lt;math&gt;\geq 13&lt;/math&gt;, so one of our solutions is &lt;math&gt;n=78&lt;/math&gt;, and we are done with our first case.<br /> &lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;\textbf{CASE 2: } n&lt;/math&gt; is odd<br /> If &lt;math&gt;n&lt;/math&gt; is odd, the only term that does not pair is the arithmetic mean of the numbers under the cube of the largest and smallest terms that would pair, or &lt;math&gt;(\frac{n+5}{2})^3&lt;/math&gt;. Therefore, as all other terms that are &lt;math&gt;\geq 5^3&lt;/math&gt; pair, the requirement that we have is<br /> &lt;cmath&gt;1^3+2^3+3^3+4^3+\left(\frac{n+5}{2}\right)^3\equiv 17\pmod {n+5}.&lt;/cmath&gt;<br /> Calculating and simplifying, we have that<br /> &lt;cmath&gt;83+\left(\frac{n+5}{2}\right)^3\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> Now, we multiply both sides by &lt;math&gt;8&lt;/math&gt;. However, since multiplication is not reversible in modular arithmetic, we need to check whether any solutions are extraneous after solving. The congruence that we now have is<br /> &lt;cmath&gt;8\cdot 83+(n+5)^3\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> As we know that &lt;math&gt;(n+5)^3&lt;/math&gt; is divisible by &lt;math&gt;n+5&lt;/math&gt;, what we need now is<br /> &lt;cmath&gt;8\cdot 83\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> We now check each solution to see whether it works.<br /> <br /> If &lt;math&gt;n+5=1, 2, 4, 8&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would be less than &lt;math&gt;13&lt;/math&gt;, so none of these solutions work. If &lt;math&gt;n+5=83&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would be even, so that solution does not work for this case. Therefore, the only three solutions we need to check for this case are when &lt;math&gt;n+5=166&lt;/math&gt;, &lt;math&gt;n+5=332&lt;/math&gt;, or &lt;math&gt;n+5=664&lt;/math&gt;. We plug these values into the congruence before we multiplied both sides by &lt;math&gt;8&lt;/math&gt; to check.<br /> <br /> If &lt;math&gt;n+5=166&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{166}{2}\right)^3\equiv 0\pmod {166}.&lt;/cmath&gt;<br /> Calculating and factoring out &lt;math&gt;83&lt;/math&gt;, we have that<br /> &lt;cmath&gt;83(1+83\cdot 83)\equiv 0\pmod {166}.&lt;/cmath&gt;<br /> As the right parenthesis is odd and &lt;math&gt;166=83\cdot 2&lt;/math&gt;, we know that this solution works, so we have another solution: &lt;math&gt;n=166-5=161&lt;/math&gt;.<br /> <br /> If &lt;math&gt;n+5=332&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{332}{2}\right)^3=83+166^3\equiv 0\pmod {322}.&lt;/cmath&gt;<br /> As the left hand side is odd, but all multiples of &lt;math&gt;322&lt;/math&gt; is even, this solution is therefore extraneous.<br /> <br /> If &lt;math&gt;n+5=664&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{664}{2}\right)^3=83+332^3\equiv 0\pmod {664}.&lt;/cmath&gt;<br /> Again, the left hand side is odd, and all multiples of &lt;math&gt;664&lt;/math&gt; are even, so this solution is extraneous.<br /> &lt;br&gt;&lt;br&gt;<br /> Therefore, our final answer is &lt;math&gt;78+161=\boxed{239}&lt;/math&gt;.<br /> &lt;br&gt;&lt;br&gt;<br /> ~ CoolCarsOnTheRun</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_10&diff=124264 2020 AIME II Problems/Problem 10 2020-06-07T19:12:09Z <p>CoolCarsOnTheRun: /* Problem */</p> <hr /> <div>= Problem =<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that when &lt;math&gt;1^3+2^3+3^3+\cdots +n^3&lt;/math&gt; is divided by &lt;math&gt;n+5&lt;/math&gt;, the remainder is &lt;math&gt;17&lt;/math&gt;.<br /> <br /> == Solution 1 ==<br /> We first note that since the remainder is &lt;math&gt;17&lt;/math&gt; and we are dividing by &lt;math&gt;n+5&lt;/math&gt;, &lt;math&gt;n+5&lt;/math&gt; must be greater than &lt;math&gt;17&lt;/math&gt;, meaning that &lt;math&gt;n&lt;/math&gt; has to be at least &lt;math&gt;13&lt;/math&gt;.<br /> <br /> We then notice that we can pair the &lt;math&gt;5^3&lt;/math&gt; term with the &lt;math&gt;n^3&lt;/math&gt; term to factor it into &lt;math&gt;(n+5)(n^2-5n+25)&lt;/math&gt; using the sum of cubes formula, which is divisible by &lt;math&gt;n+5&lt;/math&gt;. We can do the same for the &lt;math&gt;6^3&lt;/math&gt; term with the &lt;math&gt;(n-1)^3&lt;/math&gt; term, the &lt;math&gt;7^3&lt;/math&gt; term with the &lt;math&gt;(n-2)^3&lt;/math&gt;, and so on, which are all divisible by &lt;math&gt;n+5&lt;/math&gt;. However, when &lt;math&gt;n&lt;/math&gt; is odd, we will have a middle term that is not paired with any other terms, which is not necessarily divisible by &lt;math&gt;n+5&lt;/math&gt;. Thus, we have two cases:<br /> <br /> &lt;math&gt;\textbf{CASE 1: } n&lt;/math&gt; is even<br /> If &lt;math&gt;n&lt;/math&gt; is even, all terms that are greater than &lt;math&gt;4^3&lt;/math&gt; pair, as there are an even number of terms that are greater than &lt;math&gt;4^3&lt;/math&gt;. Therefore, all we need in order for the entire sequence to have a remainder of &lt;math&gt;17&lt;/math&gt; when divided by &lt;math&gt;n+5&lt;/math&gt; is &lt;math&gt;1^3+2^3+3^3+4^3&lt;/math&gt; to have a remainder of &lt;math&gt;17&lt;/math&gt; when divided by &lt;math&gt;n+5&lt;/math&gt;.<br /> <br /> Evaluating &lt;math&gt;1^3+2^3+3^3+4^3&lt;/math&gt; as &lt;math&gt;100&lt;/math&gt;, all we need to be true is <br /> &lt;cmath&gt;100\equiv 17\pmod {n+5},&lt;/cmath&gt;<br /> or that<br /> &lt;cmath&gt;83\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> Thus, &lt;math&gt;83&lt;/math&gt; will be divisible by &lt;math&gt;n+5&lt;/math&gt; where &lt;math&gt;n\geq 13&lt;/math&gt;. As &lt;math&gt;83&lt;/math&gt; is prime, &lt;math&gt;n+5&lt;/math&gt; must be equal to either &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;83&lt;/math&gt;. If &lt;math&gt;n+5=1&lt;/math&gt;, we have that &lt;math&gt;n=-4&lt;/math&gt;, which is not greater than or equal to &lt;math&gt;13&lt;/math&gt;, so that solution is extraneous.<br /> <br /> If &lt;math&gt;n+5=83&lt;/math&gt;, we have that &lt;math&gt;n=78&lt;/math&gt;, which is &lt;math&gt;\geq 13&lt;/math&gt;, so one of our solutions is &lt;math&gt;n=78&lt;/math&gt;, and we are done with our first case.<br /> <br /> &lt;math&gt;\textbf{CASE 2: } n&lt;/math&gt; is odd<br /> If &lt;math&gt;n&lt;/math&gt; is odd, the only term that does not pair is the arithmetic mean of the numbers under the cube of the largest and smallest terms that would pair, or &lt;math&gt;(\frac{n+5}{2})^3&lt;/math&gt;. Therefore, as all other terms that are &lt;math&gt;\geq 5^3&lt;/math&gt; pair, the requirement that we have is<br /> &lt;cmath&gt;1^3+2^3+3^3+4^3+\left(\frac{n+5}{2}\right)^3\equiv 17\pmod {n+5}.&lt;/cmath&gt;<br /> Calculating and simplifying, we have that<br /> &lt;cmath&gt;83+\left(\frac{n+5}{2}\right)^3\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> Now, we multiply both sides by &lt;math&gt;8&lt;/math&gt;. However, since multiplication is not reversible in modular arithmetic, we need to check whether any solutions are extraneous after solving. The congruence that we now have is<br /> &lt;cmath&gt;8\cdot 83+(n+5)^3\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> As we know that &lt;math&gt;(n+5)^3&lt;/math&gt; is divisible by &lt;math&gt;n+5&lt;/math&gt;, what we need now is<br /> &lt;cmath&gt;8\cdot 83\equiv 0\pmod {n+5}.&lt;/cmath&gt;<br /> We now check each solution to see whether it works.<br /> <br /> If &lt;math&gt;n+5=1, 2, 4, 8&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would be less than &lt;math&gt;13&lt;/math&gt;, so none of these solutions work. If &lt;math&gt;n+5=83&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would be even, so that solution does not work for this case. Therefore, the only three solutions we need to check for this case are when &lt;math&gt;n+5=166&lt;/math&gt;, &lt;math&gt;n+5=332&lt;/math&gt;, or &lt;math&gt;n+5=664&lt;/math&gt;. We plug these values into the congruence before we multiplied both sides by &lt;math&gt;8&lt;/math&gt; to check.<br /> <br /> If &lt;math&gt;n+5=166&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{166}{2}\right)^3\equiv 0\pmod {166}.&lt;/cmath&gt;<br /> Calculating and factoring out &lt;math&gt;83&lt;/math&gt;, we have that<br /> &lt;cmath&gt;83(1+83\cdot 83)\equiv 0\pmod {166}.&lt;/cmath&gt;<br /> As the right parenthesis is odd and &lt;math&gt;166=83\cdot 2&lt;/math&gt;, we know that this solution works, so we have another solution: &lt;math&gt;n=166-5=161&lt;/math&gt;.<br /> <br /> If &lt;math&gt;n+5=332&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{332}{2}\right)^3=83+166^3\equiv 0\pmod {322}.&lt;/cmath&gt;<br /> As the left hand side is odd, but all multiples of &lt;math&gt;322&lt;/math&gt; is even, this solution is therefore extraneous.<br /> <br /> If &lt;math&gt;n+5=664&lt;/math&gt;, we would need<br /> &lt;cmath&gt;83+\left(\frac{664}{2}\right)^3=83+332^3\equiv 0\pmod {664}.&lt;/cmath&gt;<br /> Again, the left hand side is odd, and all multiples of &lt;math&gt;664&lt;/math&gt; are even, so this solution is extraneous.<br /> <br /> Therefore, our final answer is &lt;math&gt;78+161=\boxed{239}&lt;/math&gt;.<br /> <br /> ~ CoolCarsOnTheRun</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_10&diff=124262 2020 AIME II Problems/Problem 10 2020-06-07T19:09:01Z <p>CoolCarsOnTheRun: </p> <hr /> <div>= Problem =<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that when &lt;math&gt;1^3+2^3+3^3+\cdots +n^3&lt;/math&gt; is divided by &lt;math&gt;n+5&lt;/math&gt;, the remainder is &lt;math&gt;17&lt;/math&gt;.</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_10&diff=124261 2020 AIME II Problems/Problem 10 2020-06-07T19:08:34Z <p>CoolCarsOnTheRun: Created page with &quot;=Problem= Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that when &lt;math&gt;1^3+2^3+3^3+\cdots +n^3&lt;/math&gt; is divided by &lt;math&gt;n+5&lt;/math&gt;, the remainder is &lt;math&gt;17&lt;/m...&quot;</p> <hr /> <div>=Problem=<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that when &lt;math&gt;1^3+2^3+3^3+\cdots +n^3&lt;/math&gt; is divided by &lt;math&gt;n+5&lt;/math&gt;, the remainder is &lt;math&gt;17&lt;/math&gt;.</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_15&diff=121074 1983 AIME Problems/Problem 15 2020-04-17T03:57:40Z <p>CoolCarsOnTheRun: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> The adjoining figure shows two intersecting chords in a circle, with &lt;math&gt;B&lt;/math&gt; on minor arc &lt;math&gt;AD&lt;/math&gt;. Suppose that the radius of the circle is &lt;math&gt;5&lt;/math&gt;, that &lt;math&gt;BC=6&lt;/math&gt;, and that &lt;math&gt;AD&lt;/math&gt; is bisected by &lt;math&gt;BC&lt;/math&gt;. Suppose further that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;. It follows that the sine of the central angle of minor arc &lt;math&gt;AB&lt;/math&gt; is a rational number. If this number is expressed as a fraction &lt;math&gt;\frac{m}{n}&lt;/math&gt; in lowest terms, what is the product &lt;math&gt;mn&lt;/math&gt;?<br /> <br /> &lt;asy&gt;size(100);<br /> defaultpen(linewidth(.8pt)+fontsize(11pt));<br /> dotfactor=1;<br /> pair O1=(0,0);<br /> pair A=(-0.91,-0.41);<br /> pair B=(-0.99,0.13);<br /> pair C=(0.688,0.728);<br /> pair D=(-0.25,0.97);<br /> path C1=Circle(O1,1);<br /> draw(C1);<br /> label(&quot;$A$&quot;,A,W);<br /> label(&quot;$B$&quot;,B,W);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,N);<br /> draw(A--D);<br /> draw(B--C);<br /> pair F=intersectionpoint(A--D,B--C);<br /> add(pathticks(A--F,1,0.5,0,3.5));<br /> add(pathticks(F--D,1,0.5,0,3.5));<br /> &lt;/asy&gt;<br /> &lt;!-- [[Image:1983_AIME-15.png|200px]] --&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> -Credit to Adamz for diagram-<br /> &lt;asy&gt;<br /> size(10cm);<br /> import olympiad;<br /> pair O = (0,0);dot(O);label(&quot;$O$&quot;,O,SW);<br /> pair M = (4,0);dot(M);label(&quot;$M$&quot;,M,SE);<br /> pair N = (4,2);dot(N);label(&quot;$N$&quot;,N,NE);<br /> draw(circle(O,5));<br /> pair B = (4,3);dot(B);label(&quot;$B$&quot;,B,NE);<br /> pair C = (4,-3);dot(C);label(&quot;$C$&quot;,C,SE);<br /> draw(B--C);draw(O--M);<br /> pair P = (1.5,2);dot(P);label(&quot;$P$&quot;,P,W);<br /> draw(circle(P,2.5));<br /> pair A=(3,4);dot(A);label(&quot;$A$&quot;,A,NE);<br /> draw(O--A);<br /> draw(O--B);<br /> pair Q = (1.5,0); dot(Q); label(&quot;$Q$&quot;,Q,S);<br /> pair R = (3,0); dot(R); label(&quot;$R$&quot;,R,S);<br /> draw(P--Q,dotted); draw(A--R,dotted);<br /> pair D=(5,0); dot(D); label(&quot;$D$&quot;,D,E);<br /> draw(A--D);<br /> &lt;/asy&gt;Let &lt;math&gt;A&lt;/math&gt; be any fixed point on circle &lt;math&gt;O&lt;/math&gt;, and let &lt;math&gt;AD&lt;/math&gt; be a chord of circle &lt;math&gt;O&lt;/math&gt;. The [[locus]] of midpoints &lt;math&gt;N&lt;/math&gt; of the chord &lt;math&gt;AD&lt;/math&gt; is a circle &lt;math&gt;P&lt;/math&gt;, with diameter &lt;math&gt;AO&lt;/math&gt;. Generally, the circle &lt;math&gt;P&lt;/math&gt; can intersect the chord &lt;math&gt;BC&lt;/math&gt; at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle &lt;math&gt;P&lt;/math&gt; is tangent to &lt;math&gt;BC&lt;/math&gt; at point &lt;math&gt;N&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;M&lt;/math&gt; be the midpoint of the chord &lt;math&gt;BC&lt;/math&gt;. From right triangle &lt;math&gt;OMB&lt;/math&gt;, we have &lt;math&gt;OM = \sqrt{OB^2 - BM^2} =4&lt;/math&gt;. This gives &lt;math&gt;\tan \angle BOM = \frac{BM}{OM} = \frac 3 4&lt;/math&gt;.<br /> <br /> Notice that the distance &lt;math&gt;OM&lt;/math&gt; equals &lt;math&gt;PN + PO \cos \angle AOM = r(1 + \cos \angle AOM)&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the radius of circle &lt;math&gt;P&lt;/math&gt;. <br /> <br /> Hence &lt;cmath&gt;\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5&lt;/cmath&gt; (where &lt;math&gt;R&lt;/math&gt; represents the radius, &lt;math&gt;5&lt;/math&gt;, of the large circle given in the question). Therefore, since &lt;math&gt;\angle AOM&lt;/math&gt; is clearly acute, we see that &lt;cmath&gt;\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3&lt;/cmath&gt;<br /> <br /> Next, notice that &lt;math&gt;\angle AOB = \angle AOM - \angle BOM&lt;/math&gt;. We can therefore apply the subtraction formula for &lt;math&gt;\tan&lt;/math&gt; to obtain &lt;cmath&gt;\tan \angle AOB =\frac{\tan \angle AOM - \tan \angle BOM}{1 + \tan \angle AOM \cdot \tan \angle BOM} =\frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}&lt;/cmath&gt; It follows that &lt;math&gt;\sin \angle AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}&lt;/math&gt;, such that the answer is &lt;math&gt;7 \cdot 25=\boxed{175}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> This solution, while similar to Solution 1, is far more motivated and less contrived.<br /> <br /> Firstly, we note the statement in the problem that &quot;&lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; and bisected by &lt;math&gt;BC&lt;/math&gt;&quot; &amp;ndash; what is its significance? What is the criterion for this statement to be true?<br /> <br /> We consider the locus of midpoints of the chords from &lt;math&gt;A&lt;/math&gt;. It is well-known that this is the circle with diameter &lt;math&gt;AO&lt;/math&gt;, where &lt;math&gt;O&lt;/math&gt; is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with scale factor &lt;math&gt;\frac{1}{2}&lt;/math&gt; and center &lt;math&gt;A&lt;/math&gt;. Thus, the locus is the result of the dilation with scale factor &lt;math&gt;\frac{1}{2}&lt;/math&gt; and centre &lt;math&gt;A&lt;/math&gt; of circle &lt;math&gt;O&lt;/math&gt;. Let the center of this circle be &lt;math&gt;P&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;AD&lt;/math&gt; is bisected by &lt;math&gt;BC&lt;/math&gt; if they cross at some point &lt;math&gt;N&lt;/math&gt; on the circle. Moreover, since &lt;math&gt;AD&lt;/math&gt; is the only chord, &lt;math&gt;BC&lt;/math&gt; must be tangent to the circle &lt;math&gt;P&lt;/math&gt;.<br /> <br /> The rest of this problem is straightforward.<br /> <br /> Our goal is to find &lt;math&gt;\sin \angle AOB = \sin{\left(\angle AOM - \angle BOM\right)}&lt;/math&gt;, where &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;. We have &lt;math&gt;BM=3&lt;/math&gt; and &lt;math&gt;OM=4&lt;/math&gt;.<br /> Let &lt;math&gt;R&lt;/math&gt; be the projection of &lt;math&gt;A&lt;/math&gt; onto &lt;math&gt;OM&lt;/math&gt;, and similarly let &lt;math&gt;Q&lt;/math&gt; be the projection of &lt;math&gt;P&lt;/math&gt; onto &lt;math&gt;OM&lt;/math&gt;. Then it remains to find &lt;math&gt;AR&lt;/math&gt; so that we can use the addition formula for &lt;math&gt;\sin&lt;/math&gt;.<br /> <br /> As &lt;math&gt;PN&lt;/math&gt; is a radius of circle &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;PN=2.5&lt;/math&gt;, and similarly, &lt;math&gt;PO=2.5&lt;/math&gt;. Since &lt;math&gt;OM=4&lt;/math&gt;, we have &lt;math&gt;OQ=OM-QM=OM-PN=4-2.5=1.5&lt;/math&gt;. Thus &lt;math&gt;PQ=\sqrt{2.5^2-1.5^2}=2&lt;/math&gt;.<br /> <br /> Further, we see that &lt;math&gt;\triangle OAR&lt;/math&gt; is a dilation of &lt;math&gt;\triangle OPQ&lt;/math&gt; about center &lt;math&gt;O&lt;/math&gt; with scale factor &lt;math&gt;2&lt;/math&gt;, so &lt;math&gt;AR=2PQ=4&lt;/math&gt;.<br /> <br /> Lastly, we apply the formula:<br /> &lt;cmath&gt; \sin{\left(\angle AOM - \angle BOM\right)} = \sin \angle AOM \cos \angle BOM - \sin \angle BOM \cos \angle AOM = \left(\frac{4}{5}\right)\left(\frac{4}{5}\right)-\left(\frac{3}{5}\right)\left(\frac{3}{5}\right)=\frac{7}{25}&lt;/cmath&gt;<br /> Thus the answer is &lt;math&gt;7\cdot25=\boxed{175}&lt;/math&gt;.<br /> <br /> === Solution 3 (coordinate geometry) ===<br /> [[File:Aime1983p15s2.png|500px|link=]]<br /> <br /> Let the circle have equation &lt;math&gt;x^2 + y^2 = 25&lt;/math&gt;, with centre &lt;math&gt;O(0,0)&lt;/math&gt;. Since &lt;math&gt;BC=6&lt;/math&gt;, we can calculate (by the Pythagorean Theorem) that the distance from &lt;math&gt;O&lt;/math&gt; to the line &lt;math&gt;BC&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;. Therefore, we can let &lt;math&gt;B=(3,4)&lt;/math&gt; and &lt;math&gt;C=(-3,4)&lt;/math&gt;. Now, assume that &lt;math&gt;A&lt;/math&gt; is any point on the major arc BC, and &lt;math&gt;D&lt;/math&gt; any point on the minor arc BC. We can write &lt;math&gt;A=(5 \cos \alpha, 5 \sin \alpha)&lt;/math&gt;, where &lt;math&gt;\alpha&lt;/math&gt; is the angle measured from the positive &lt;math&gt;x&lt;/math&gt; axis to the ray &lt;math&gt;OA&lt;/math&gt;. It will also be convenient to define &lt;math&gt;\angle XOB = \alpha_0&lt;/math&gt;. <br /> <br /> Firstly, since &lt;math&gt;B&lt;/math&gt; must lie in the minor arc &lt;math&gt;AD&lt;/math&gt;, we see that &lt;math&gt;\alpha \in \left(-\frac{\pi}{2}-\alpha_0,\alpha_0\right)&lt;/math&gt;. However, since the midpoint of &lt;math&gt;AD&lt;/math&gt; must lie on &lt;math&gt;BC&lt;/math&gt;, and the highest possible &lt;math&gt;y&lt;/math&gt;-coordinate of &lt;math&gt;D&lt;/math&gt; is &lt;math&gt;5&lt;/math&gt;, we see that the &lt;math&gt;y&lt;/math&gt;-coordinate cannot be lower than &lt;math&gt;3&lt;/math&gt;, that is, &lt;math&gt;\alpha \in \left[\sin^{-1}\frac{3}{5},\alpha_0\right)&lt;/math&gt;.<br /> <br /> Secondly, there is a theorem that says that, in a circle, if a chord is bisected by a radius, then they must be perpendicular. Therefore, suppose that &lt;math&gt;P&lt;/math&gt; is the intersection point of &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, so that by the theorem, &lt;math&gt;OP&lt;/math&gt; is perpendicular to &lt;math&gt;AD&lt;/math&gt;. So, if &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;, this means that &lt;math&gt;P&lt;/math&gt; is the only point on the chord &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;OP&lt;/math&gt; is perpendicular to &lt;math&gt;AD&lt;/math&gt;. Now suppose that &lt;math&gt;P=(p,4)&lt;/math&gt;, where &lt;math&gt;p \in (-3,3)&lt;/math&gt;. The fact that &lt;math&gt;OP&lt;/math&gt; must be perpendicular to &lt;math&gt;AD&lt;/math&gt; is equivalent to the following equation:<br /> <br /> &lt;cmath&gt; -1 = \left(\text{slope of } OP\right)\left(\text{slope of } AP\right)&lt;/cmath&gt;<br /> which becomes<br /> &lt;cmath&gt; -1 = \frac{4}{p} \cdot \frac{5\sin \alpha - 4}{5\cos \alpha - p}&lt;/cmath&gt;<br /> <br /> This rearranges to<br /> <br /> &lt;cmath&gt; p^2 - (5\cos \alpha)p + 16 - 20 \sin \alpha = 0&lt;/cmath&gt;<br /> <br /> Given that this equation must have only one real root &lt;math&gt;p\in (-3,3)&lt;/math&gt;, we study the following function:<br /> <br /> &lt;cmath&gt;f(x) = x^2 - (5\cos \alpha)x + 16 - 20 \sin \alpha&lt;/cmath&gt;<br /> <br /> First, by the fact that the equation &lt;math&gt;f(x)=0&lt;/math&gt; has real solutions, its discriminant &lt;math&gt;\Delta&lt;/math&gt; must be non-negative, so we calculate<br /> <br /> &lt;cmath&gt; \begin{split}\Delta &amp; = (5\cos \alpha)^2 - 4(16-20\sin \alpha) \\<br /> &amp; = 25 (1- \sin^2 \alpha) - 64 + 80 \sin \alpha \\<br /> &amp; = -25 \sin^2 \alpha + 80\sin \alpha - 39 \\<br /> &amp; = (13 - 5\sin \alpha)(5\sin \alpha - 3)\end{split}&lt;/cmath&gt;<br /> <br /> It is obvious that this is in fact non-negative. If it is actually zero, then &lt;math&gt;\sin \alpha = \frac{3}{5}&lt;/math&gt;, and &lt;math&gt;\cos \alpha = \frac{4}{5}&lt;/math&gt;. In this case, &lt;math&gt;p = \frac{5\cos \alpha}{2} = 2 \in (-3,3)&lt;/math&gt;, so we have found a possible solution. We thus calculate &lt;math&gt;\sin(\text{central angle of minor arc } AB) = \sin (\alpha_0 - \alpha) = \frac{4}{5}\cdot \frac{4}{5} - \frac{3}{5} \cdot \frac{3}{5} = \frac{7}{25}&lt;/math&gt; by the subtraction formula for &lt;math&gt;\sin&lt;/math&gt;. This means that the answer is &lt;math&gt;7 \cdot 25 = 175&lt;/math&gt;.<br /> <br /> === Addendum to Solution 3 ===<br /> Since this is an AIME problem, we can assume that we are done since we have found one possible case. However, in a full-solution contest, we could not assume that the answer is unique, but would need to prove that this is the unique solution. This can be proven as follows.<br /> <br /> Suppose that &lt;math&gt;\Delta &gt; 0&lt;/math&gt;, which would mean that there could be two real roots of &lt;math&gt;f(x)&lt;/math&gt;, one lying in the interval &lt;math&gt;(-3,3)&lt;/math&gt;, and another outside of it. We also see, by [[Vieta's Formulas]], that the average of the two roots is &lt;math&gt;\frac{5\cos \alpha}{2}&lt;/math&gt;, which is non-negative, so the root outside of &lt;math&gt;(-3,3)&lt;/math&gt; must be no less than &lt;math&gt;3&lt;/math&gt;. By considering the graph of &lt;math&gt;y=f(x)&lt;/math&gt;, which is a &quot;U-shaped&quot; parabola, it is now evident that &lt;math&gt;f(-3) &gt; 0&lt;/math&gt; and &lt;math&gt;f(3)\leq 0&lt;/math&gt;. We can just use the second inequality:<br /> <br /> &lt;cmath&gt;0 \geq f(3) = 25 - 15\cos \alpha - 20 \sin \alpha&lt;/cmath&gt;<br /> so<br /> &lt;cmath&gt; 3\cos \alpha + 4 \sin \alpha \geq 5 &lt;/cmath&gt;<br /> <br /> The only way for this inequality to be satisfied is when &lt;math&gt;A=B&lt;/math&gt; (by applying the Cauchy-Schwarz inequality, or just plotting the line &lt;math&gt;3x+4y=5&lt;/math&gt; to see that point &lt;math&gt;A&lt;/math&gt; can't be above this line), which does not make sense in the original problem statement. (For it would mean that the point &lt;math&gt;A&lt;/math&gt; lies in the half-plane above the line &lt;math&gt;3x+4y=5&lt;/math&gt;, inclusive, and the half-plane below the line &lt;math&gt;-3x+4y=5&lt;/math&gt;, exclusive. This can be seen to be impossible by drawing the lines and observing that the intersection of the two half-planes does not share any point with the circle.)<br /> ===Solution 4===<br /> Let the center of the circle be &lt;math&gt;O&lt;/math&gt;. Fix &lt;math&gt;B,C,&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt;. Then, as &lt;math&gt;D&lt;/math&gt; moves around the circle, the locus of the midpoints of &lt;math&gt;AD&lt;/math&gt; is clearly a circle. Since the problems gives that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; bisected by &lt;math&gt;BC&lt;/math&gt;, it follows that the circle with diameter &lt;math&gt;DO&lt;/math&gt; and &lt;math&gt;AO&lt;/math&gt; is tangent to &lt;math&gt;BC&lt;/math&gt;.<br /> <br /> Now, let the intersection of &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; be &lt;math&gt;E&lt;/math&gt; and let the midpoint of &lt;math&gt;AO&lt;/math&gt; (the center of the circle tangent to &lt;math&gt;BC&lt;/math&gt; that we described beforehand) be &lt;math&gt;F&lt;/math&gt;. Drop the altitude from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and call its intersection with &lt;math&gt;BC&lt;/math&gt; &lt;math&gt;K&lt;/math&gt;. Drop the perpendicular from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;KO&lt;/math&gt; and call its intersection with &lt;math&gt;KO&lt;/math&gt; &lt;math&gt;L&lt;/math&gt;. Clearly, &lt;math&gt;KO = \sqrt{OC^2-KC^2} = \sqrt{5^2-3^2} = 4&lt;/math&gt; and since &lt;math&gt;EF&lt;/math&gt; is radius, it equals &lt;math&gt;\frac{5}{2}&lt;/math&gt;. The same applies for &lt;math&gt;FO&lt;/math&gt;, which also equals &lt;math&gt;\frac{5}{2}&lt;/math&gt;. By the Pythagorean theorem, we deduce that &lt;math&gt;FL = 2&lt;/math&gt;, so &lt;math&gt;EK = 2&lt;/math&gt;. This is very important information! Now we know that &lt;math&gt;BE = 1&lt;/math&gt;, so by Power of a Point, &lt;math&gt;AE = ED = \sqrt{5}&lt;/math&gt;.<br /> <br /> We’re almost there! Since by the Pythagorean theorem, &lt;math&gt;ED^2 + EO^2 = 5&lt;/math&gt;, we deduce that &lt;math&gt;EO = 2\sqrt{5}&lt;/math&gt;. &lt;math&gt;EC=OC=5&lt;/math&gt;, so &lt;math&gt;\sin (CEO) = \frac{2\sqrt{5}}{5}&lt;/math&gt;. Furthermore, since &lt;math&gt;\sin (CEO) = \cos(DEC)&lt;/math&gt;, we know that &lt;math&gt;\cos (DEC) = \frac{2\sqrt{5}}{5}&lt;/math&gt;. By the law of cosines,<br /> &lt;cmath&gt;DC^2 = (\sqrt{5})^2 + 5^2 -10\sqrt{5} \cdot \frac{2\sqrt{5}}{5} = 10&lt;/cmath&gt;Therefore, &lt;math&gt;DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}&lt;/math&gt;. Now, drop the altitude from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;BA&lt;/math&gt; and call its intersection with &lt;math&gt;BA&lt;/math&gt; &lt;math&gt;Z&lt;/math&gt;. Then, by the Pythagorean theorem, &lt;math&gt;OZ = \frac{7\sqrt{2}}{2}&lt;/math&gt;. Thus, &lt;math&gt;\sin (BOZ) = \frac{\sqrt{2}}{10}&lt;/math&gt; and &lt;math&gt;\cos (BOZ) = \frac{7\sqrt{2}}{10}&lt;/math&gt;. As a result, &lt;math&gt;\sin (BOA) = \sin (2 BOZ) = 2\sin(BOZ)\cos(BOZ) = \frac{7}{25}&lt;/math&gt;. &lt;math&gt;7 \cdot 25 = \boxed{175}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> CoolCarsOnTheRun https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=121013 User:Piphi 2020-04-15T23:37:33Z <p>CoolCarsOnTheRun: /* User Count */</p> <hr /> <div>&lt;center&gt;[[File:Piphi-Avatar.png]]&lt;/center&gt;<br /> <br /> &lt;div style=&quot;border:2px solid black; background:#eeeeee;&quot;&gt;<br /> ::::&lt;font style=&quot;font-family: Verdana, sans-serif&quot;&gt;[[User:Piphi|Userpage]] | [[User talk:Piphi|Talk]] | [[Special:Contributions/Piphi|Contributions]]&lt;/font&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#dddddd; align:center&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: MV Boli, Verdana&quot;&gt;User Count&lt;/font&gt;==<br /> &lt;font color=&quot;black&quot;&gt;When you visit this page, edit it by incrementing the user count below by one.<br /> <br /> &lt;center&gt;&lt;font size=&quot;100px&quot;&gt;17&lt;/font&gt;&lt;/center&gt;<br /> &lt;/font&gt; <br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#cccccc; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: MV Boli, Verdana&quot;&gt;About Me&lt;/font&gt;==<br /> &lt;font color=&quot;black&quot;&gt;PM me if you want to find out about some cool things you can do with the AoPS wiki.<br /> <br /> My main project on the AoPS wiki is [[AoPS_Administrators#List_of_Admins | a list of all the AoPS admins]], everyone is welcome to add more admins to the list by clicking [https://artofproblemsolving.com/wiki/index.php?title=AoPS_Administrators&amp;action=edit&amp;section=1 here]. 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