https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ilovepizza2020&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-15T22:36:17Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_6&diff=147200 2018 AIME I Problems/Problem 6 2021-02-16T22:01:17Z <p>Ilovepizza2020: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;N&lt;/math&gt; be the number of complex numbers &lt;math&gt;z&lt;/math&gt; with the properties that &lt;math&gt;|z|=1&lt;/math&gt; and &lt;math&gt;z^{6!}-z^{5!}&lt;/math&gt; is a real number. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;a=z^{120}&lt;/math&gt;. This simplifies the problem constraint to &lt;math&gt;a^6-a \in \mathbb{R}&lt;/math&gt;. This is true if &lt;math&gt;\text{Im}(a^6)=\text{Im}(a)&lt;/math&gt;. Let &lt;math&gt;\theta&lt;/math&gt; be the angle &lt;math&gt;a&lt;/math&gt; makes with the positive x-axis. Note that there is exactly one &lt;math&gt;a&lt;/math&gt; for each angle &lt;math&gt;0\le\theta&lt;2\pi&lt;/math&gt;. This must be true for &lt;math&gt;12&lt;/math&gt; values of &lt;math&gt;a&lt;/math&gt; (it may help to picture the reference angle making one orbit from and to the positive x-axis; note every time &lt;math&gt;\sin\theta=\sin{6\theta}&lt;/math&gt;). For each of these solutions for &lt;math&gt;a&lt;/math&gt;, there are necessarily &lt;math&gt;120&lt;/math&gt; solutions for &lt;math&gt;z&lt;/math&gt;. Thus, there are &lt;math&gt;12*120=1440&lt;/math&gt; solutions for &lt;math&gt;z&lt;/math&gt;, yielding an answer of &lt;math&gt;\boxed{440}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The constraint mentioned in the problem is equivalent to the requirement that the imaginary part is equal to &lt;math&gt;0&lt;/math&gt;. Since &lt;math&gt;|z|=1&lt;/math&gt;, let &lt;math&gt;z=\cos \theta + i\sin \theta&lt;/math&gt;, then we can write the imaginary part of &lt;math&gt; \Im(z^{6!}-z^{5!})=\Im(z^{720}-z^{120})=\sin\left(720\theta\right)-\sin\left(120\theta\right)=0&lt;/math&gt;. Using the sum-to-product formula, we get &lt;math&gt;\sin\left(720\theta\right)-\sin\left(120\theta\right)=2\cos\left(\frac{720\theta+120\theta}{2}\right)\sin\left(\frac{720\theta-120\theta}{2}\right)=2\cos\left(\frac{840\theta}{2}\right)\sin\left(\frac{600\theta}{2}\right)\implies \cos\left(\frac{840\theta}{2}\right)=0&lt;/math&gt; or &lt;math&gt;\sin\left(\frac{600\theta}{2}\right)=0&lt;/math&gt;. The former yields &lt;math&gt;840&lt;/math&gt; solutions, and the latter yields &lt;math&gt;600&lt;/math&gt; solutions, giving a total of &lt;math&gt;840+600=1440&lt;/math&gt; solution, so our answer is &lt;math&gt;\boxed{440}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> As mentioned in solution one, for the difference of two complex numbers to be real, their imaginary parts must be equal. We use exponential form of complex numbers. Let &lt;math&gt;z = e^{i \theta}&lt;/math&gt;. We have two cases to consider. Either &lt;math&gt;z^{6!} = z^{5!}&lt;/math&gt;, or &lt;math&gt;z^{6!}&lt;/math&gt; and &lt;math&gt;z^{5!}&lt;/math&gt; are reflections across the imaginary axis.<br /> If &lt;math&gt;z^{6!} = z^{5!}&lt;/math&gt;, then &lt;math&gt;e^{6! \theta i} = e^{5! \theta i}&lt;/math&gt;. Thus, &lt;math&gt;720 \theta = 120 \theta&lt;/math&gt; or &lt;math&gt;600\theta = 0&lt;/math&gt;, giving us 600 solutions. (Equalities are taken modulo &lt;math&gt;2 \pi&lt;/math&gt;)<br /> For the second case, &lt;math&gt;e^{6! \theta i} = e^{(\pi - 5!\theta)i}&lt;/math&gt;. This means &lt;math&gt;840 \theta = \pi &lt;/math&gt;, giving us 840 solutions.<br /> Our total count is thus &lt;math&gt;1440&lt;/math&gt;, yielding a final answer of &lt;math&gt;\boxed{440}&lt;/math&gt;.<br /> <br /> == Solution 4 ==<br /> <br /> Because &lt;math&gt;|z| = 1,&lt;/math&gt; we know that &lt;math&gt;z\overline{z} = 1^2 = 1.&lt;/math&gt; Hence &lt;math&gt;\overline{z} = \frac 1 {z}.&lt;/math&gt; Because &lt;math&gt;z^{6!}-z^{5!}&lt;/math&gt; is real, it is equal to its complex conjugate. Hence &lt;math&gt;z^{6!}-z^{5!} = \overline{z^{6!}}-\overline{z^{5!}}.&lt;/math&gt; Substituting the expression we that we derived earlier, we get &lt;math&gt;z^{720}-z^{120} = \frac 1{z^{720}} - \frac 1{z^{120}}.&lt;/math&gt; This leaves us with a polynomial whose leading term is &lt;math&gt;z^{1440}.&lt;/math&gt; Hence our answer is &lt;math&gt;\boxed{440}&lt;/math&gt;.<br /> <br /> == Solution 5 ==<br /> Since &lt;math&gt;|z|=1&lt;/math&gt;, let &lt;math&gt;z=\cos \theta + i\sin \theta&lt;/math&gt;. For &lt;math&gt;z^{6!}-z^{5!}&lt;/math&gt; to be real, the imaginary parts of &lt;math&gt;z^{6!}&lt;/math&gt; and &lt;math&gt;z^{5!}&lt;/math&gt; must be equal, so &lt;math&gt;\sin 720\theta=\sin 120\theta&lt;/math&gt;. We need to find all solutions for &lt;math&gt;\theta&lt;/math&gt; in the interval &lt;math&gt;[0,2\pi)&lt;/math&gt;. This can be done by graphing &lt;math&gt;y=\sin 720\theta&lt;/math&gt; and &lt;math&gt;y=\sin 120\theta&lt;/math&gt; and finding their intersections. Since the period of &lt;math&gt;y=\sin 720\theta&lt;/math&gt; is &lt;math&gt;\frac{\pi}{360}&lt;/math&gt; and the period of &lt;math&gt;y=\sin 120\theta&lt;/math&gt; is &lt;math&gt;\frac{\pi}{60}&lt;/math&gt;, the common period of both graphs is &lt;math&gt;\frac{\pi}{60}&lt;/math&gt;. Therefore, we only graph the functions in the domain &lt;math&gt;[0, \frac{\pi}{60})&lt;/math&gt;. We can clearly see that there are twelve points of intersection. However, since we only graphed &lt;math&gt;\frac{1}{120}&lt;/math&gt; of the interval &lt;math&gt;[0,2\pi)&lt;/math&gt;, we need to multiply our answer by &lt;math&gt;120&lt;/math&gt;. The answer is &lt;math&gt;12 \cdot 120 = 1440 = \boxed{440} \pmod{1000}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> <br /> https://www.youtube.com/watch?v=iE8paW_ICxw<br /> <br /> == See also ==<br /> {{AIME box|year=2018|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_7&diff=147196 2004 AIME I Problems/Problem 7 2021-02-16T21:26:15Z <p>Ilovepizza2020: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt; C &lt;/math&gt; be the [[coefficient]] of &lt;math&gt; x^2 &lt;/math&gt; in the expansion of the product &lt;math&gt; (1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x). &lt;/math&gt; Find &lt;math&gt; |C|. &lt;/math&gt;<br /> <br /> __TOC__<br /> == Solutions ==<br /> === Solution 1 ===<br /> Let our [[polynomial]] be &lt;math&gt;P(x)&lt;/math&gt;.<br /> <br /> It is clear that the coefficient of &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;P(x)&lt;/math&gt; is &lt;math&gt;-1 + 2 - 3 + \ldots + 14 - 15 = -8&lt;/math&gt;, so &lt;math&gt;P(x) = 1 -8x + Cx^2 + Q(x)&lt;/math&gt;, where &lt;math&gt;Q(x)&lt;/math&gt; is some polynomial [[divisibility | divisible]] by &lt;math&gt;x^3&lt;/math&gt;.<br /> <br /> Then &lt;math&gt;P(-x) = 1 + 8x + Cx^2 + Q(-x)&lt;/math&gt; and so &lt;math&gt;P(x)\cdot P(-x) = 1 + (2C - 64)x^2 + R(x)&lt;/math&gt;, where &lt;math&gt;R(x)&lt;/math&gt; is some polynomial divisible by &lt;math&gt;x^3&lt;/math&gt;.<br /> <br /> However, we also know &lt;math&gt;P(x)\cdot P(-x) = (1 - x)(1 + x)(1 +2x)(1 - 2x) \cdots (1 - 15x)(1 + 15x)&lt;/math&gt; &lt;math&gt;= (1 - x^2)(1 - 4x^2)\cdots(1 - 225x^2)&lt;/math&gt; &lt;math&gt;= 1 - (1 + 4 + \ldots + 225)x^2 + R(x)&lt;/math&gt;.<br /> <br /> Equating coefficients, we have &lt;math&gt;2C - 64 = -(1 + 4 + \ldots + 225) = -1240&lt;/math&gt;, so &lt;math&gt;-2C = 1176&lt;/math&gt; and &lt;math&gt;|C| = \boxed{588}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;S&lt;/math&gt; be the [[set]] of integers &lt;math&gt;\{-1,2,-3,\ldots,14,-15\}&lt;/math&gt;. The coefficient of &lt;math&gt;x^2&lt;/math&gt; in the expansion is equal to the sum of the product of each pair of distinct terms, or &lt;math&gt;C = \sum_{1 \le i \neq j}^{15} S_iS_j&lt;/math&gt;. Also, we know that <br /> &lt;cmath&gt;\begin{align*}\left(\sum_{i=1}^{n} S_i\right)^2 &amp;= \left(\sum_{i=1}^{n} S_i^2\right) + 2\left(\sum_{1 \le i \neq j}^{15} S_iS_j\right)\\ (-8)^2 &amp;= \frac{15(15+1)(2\cdot 15+1)}{6} + 2C\end{align*}&lt;/cmath&gt;<br /> where the left-hand sum can be computed from:<br /> &lt;center&gt;&lt;math&gt;\sum_{i=1}^{15} S_i = S_{15} + \left(\sum_{i=1}^{7} S_{2i-1} + S_{2i}\right) = -15 + 7 = -8&lt;/math&gt;&lt;/center&gt;<br /> and the right-hand sum comes from the formula for the sum of the first &lt;math&gt;n&lt;/math&gt; perfect squares. Therefore, &lt;math&gt;|C| = \left|\frac{64-1240}{2}\right| = \boxed{588}&lt;/math&gt;.<br /> <br /> === Solution 3 (Bash)===<br /> <br /> Consider the set &lt;math&gt;[-1, 2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15]&lt;/math&gt;. Denote by &lt;math&gt;S&lt;/math&gt; all size 2 subsets of this set. Replace each element of &lt;math&gt;S&lt;/math&gt; by the product of the elements. Now, the quantity we seek is the sum of each element. Since consecutive elements add to &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;-1&lt;/math&gt;, we can simplify this to &lt;math&gt;|-1\cdot(-7)+2\cdot(-9)-3\cdot(-6)+4\cdot(-10)-5\cdot(-5)+\ldots+12\cdot(-14)-13\cdot(-1)+14\cdot(-15)|=|-588|=\boxed{588}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Let set &lt;math&gt;N&lt;/math&gt; be &lt;math&gt;\{-1, -3, \ldots -15\}&lt;/math&gt; and set &lt;math&gt;P&lt;/math&gt; be &lt;math&gt;\{2, 4, \ldots 14\}&lt;/math&gt;. The sum of the negative &lt;math&gt;x^2&lt;/math&gt; coefficients is the sum of the products of the elements in all two element sets such that one element is from &lt;math&gt;N&lt;/math&gt; and the other is from &lt;math&gt;P&lt;/math&gt;. Each summand is a term in the expansion of<br /> &lt;cmath&gt;(-1 - 3 - \ldots - 15)(2 + 4 + \ldots + 14)&lt;/cmath&gt;<br /> which equals &lt;math&gt;-56 * 64 = -(60^2 - 4^2) = -3584&lt;/math&gt;. The sum of the positive &lt;math&gt;x^2&lt;/math&gt; coefficients is the sum of the products of all two element sets such that the two elements are either both in &lt;math&gt;N&lt;/math&gt; or both in &lt;math&gt;P&lt;/math&gt;. By counting, the sum is &lt;math&gt;2992&lt;/math&gt;, so the sum of all &lt;math&gt;x^2&lt;/math&gt; coefficients is &lt;math&gt;-588&lt;/math&gt;. Thus, the answer is &lt;math&gt;\boxed{588}&lt;/math&gt;.<br /> <br /> <br /> == Solution 5==<br /> <br /> We can find out the coefficient of &lt;math&gt;x^2&lt;/math&gt; by multiplying every pair of two coefficients for &lt;math&gt;x&lt;/math&gt;. This means that we multiply &lt;math&gt;-1&lt;/math&gt; by &lt;math&gt;2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15&lt;/math&gt;. and etc. This sum can be easily simplified and is equal to &lt;math&gt;(-1)(-7)+(-3)(-6)+(-5)(-5)+(-7)(-4)+(-9)(-3)+(-11)(-2)+(-13)(-1)+2(-9)+4(-10)+6(-11)+8(-12)+10(-13)+12(-14)+14(-15)&lt;/math&gt; or &lt;math&gt;588&lt;/math&gt;.<br /> <br /> -David Camacho<br /> <br /> ==Solution 6==<br /> This is just another way of summing the subsets of 2 from &lt;math&gt;[-1, 2, -3, 4, -5, 6, -7, 8, -9, 10, -11, 12, -13, 14, -15]&lt;/math&gt;. Start from the right and multiply -15 to everything on its left. Use the distributive property and add all the 14 integers together to get 7. This gives us &lt;math&gt;-15 * 7&lt;/math&gt;. Doing this for 14 gives us &lt;math&gt;14 * -7&lt;/math&gt;, and for -13 we get &lt;math&gt;-13 * 6&lt;/math&gt;. This pattern repeats where every two integers will multiple 7, 6,... to 0. Combining and simplifying the pattern give us this: &lt;math&gt;-(29 * 7 + 25 * 6 + 21 * 5 + 17 * 4 + 13 * 3 + 9 * 2 + 5*1)&lt;/math&gt;. The expression gives us -588, or &lt;math&gt;C = \boxed{588}&lt;/math&gt;. This is a good solution because it guarantees we never add a product twice, and the pattern is simple to add by hand.<br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2004|n=I|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_18&diff=147183 2021 AMC 10B Problems/Problem 18 2021-02-16T17:41:47Z <p>Ilovepizza2020: </p> <hr /> <div>==Problem==<br /> <br /> A fair &lt;math&gt;6&lt;/math&gt;-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{1}{120} \qquad\textbf{(B)} ~\frac{1}{32} \qquad\textbf{(C)} ~\frac{1}{20} \qquad\textbf{(D)} ~\frac{3}{20} \qquad\textbf{(E)} ~\frac{1}{6}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> There is a &lt;math&gt;\frac36&lt;/math&gt; chance that the first number we choose is even.<br /> <br /> There is a &lt;math&gt;\frac{2}{5}&lt;/math&gt; chance that the next number that is distinct from the first is even.<br /> <br /> There is a &lt;math&gt;\frac{1}{4}&lt;/math&gt; chance that the next number distinct from the first two is even.<br /> <br /> &lt;math&gt;\frac{3}{6} \cdot \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{20}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) }\frac{1}{20}}.&lt;/math&gt;<br /> <br /> ~Tucker<br /> <br /> ==Solution 2==<br /> <br /> Every set of three numbers chosen from &lt;math&gt;\{1,2,3,4,5,6\}&lt;/math&gt; has an equal chance of being the first 3 distinct numbers rolled.<br /> <br /> Therefore, the probability that the first 3 distinct numbers are &lt;math&gt;\{2,4,6\}&lt;/math&gt; is &lt;math&gt;\frac{1}{{6 \choose 3}}=\boxed{(C)~\frac{1}{20}}&lt;/math&gt;<br /> <br /> ~kingofpineapplz<br /> <br /> ==Solution 3 (Quicksolve) ==<br /> <br /> Note that the problem is basically asking us to find the probability that in some permutation of &lt;math&gt;1,2,3,4,5,6&lt;/math&gt; that we get the three even numbers in the first three spots.<br /> <br /> There are &lt;math&gt;6!&lt;/math&gt; ways to order the &lt;math&gt;6&lt;/math&gt; numbers and &lt;math&gt;3!(3!)&lt;/math&gt; ways to order the evens in the first three spots and the odds in the next three spots.<br /> <br /> Therefore the probability is &lt;math&gt;\frac{3!(3!)}{6!} = \frac{1}{20} = \boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> <br /> --abhinavg0627<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;P_n&lt;/math&gt; denote the probability that the first odd number appears on roll &lt;math&gt;n&lt;/math&gt; and all our conditions are met. We now proceed with complementary counting. <br /> <br /> For &lt;math&gt;n \le 3&lt;/math&gt;, it's impossible to have all &lt;math&gt;3&lt;/math&gt; evens appear before an odd. Note that for &lt;math&gt;n \ge 4,&lt;/math&gt; &lt;cmath&gt;P_n = \frac {1}{2^{n}} - \frac {1}{2^{n}} \left(\frac{\binom{3}{2}(2^{n-1}-2)+\binom{3}{2}}{3^{n-1}}\right) = \frac {1}{2^{n}} - \left(\frac {3(2^{n-1})-3}{2^{n} \cdot 3^{n-1}}\right) = \frac {1}{2^{n}} - \left(\frac {1}{2 \cdot 3^{n-2}} - \frac{1}{2^{n} \cdot 3^{n-2}} \right).&lt;/cmath&gt;<br /> <br /> Summing for all &lt;math&gt;n&lt;/math&gt;, we get our answer of &lt;cmath&gt;\left (\frac {1}{2^{4}} + \frac {1}{2^{5}} + ... \right) - \left (\frac {1}{2 \cdot 3^{2}} + \frac {1}{2 \cdot 3^{3}} + ... \right) + \left (\frac {1}{2^{4} \cdot 3^{2}} + \frac {1}{2^{5} \cdot 3^{3}} + ... \right) = \left (\frac {1}{8} \right) - \left(\frac {\frac {1}{18}}{ \frac{2}{3}} \right) + \left(\frac {\frac {1}{144}}{\frac {5}{6}} \right) = \left (\frac {1}{8} \right) - \left(\frac {1}{12} \right) + \left (\frac{1}{120} \right) = \boxed{\textbf{(C) }\frac{1}{20}.}&lt;/cmath&gt;<br /> <br /> ~ike.chen<br /> <br /> ==Solution 5==<br /> Let &lt;math&gt; E_n &lt;/math&gt; be that probability that the condition in the problem is satisfied given that we need &lt;math&gt; n &lt;/math&gt; more distinct even numbers. Then, <br /> &lt;cmath&gt; E_1=\frac{1}{6}+\frac{1}{3}\cdot E_1+\frac{1}{2}\cdot 0, &lt;/cmath&gt;<br /> since there is a &lt;math&gt; \frac{1}{3} &lt;/math&gt; probability that we will roll an even number we already have rolled and will be in the same position again. Solving, we find that &lt;math&gt; E_1=\frac{1}{4} &lt;/math&gt;. <br /> <br /> We can apply the same concept for &lt;math&gt; E_2 &lt;/math&gt; and &lt;math&gt; E_3 &lt;/math&gt;. We find that &lt;cmath&gt; E_2=\frac{1}{3}\cdot E_1+\frac{1}{6}\cdot E_2+\frac{1}{2}\cdot 0, &lt;/cmath&gt; and so &lt;math&gt; E_2=\frac{1}{10} &lt;/math&gt;. Also, &lt;cmath&gt; E_3=\frac{1}{2}\cdot E_2+\frac{1}{2}\cdot 0, &lt;/cmath&gt; so &lt;math&gt; E_3=\frac{1}{20} &lt;/math&gt;. Since the problem is asking for &lt;math&gt; E_3 &lt;/math&gt;, our answer is &lt;math&gt; \boxed{\textbf{(C) }\frac{1}{20}} &lt;/math&gt;. -BorealBear<br /> <br /> == Video Solution by OmegaLearn (Conditional probability) ==<br /> https://youtu.be/IX-Y38KPxqs<br /> <br /> {{AMC10 box|year=2021|ab=B|num-b=17|num-a=19}}</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=Georgeooga-Harryooga_Theorem&diff=141489 Georgeooga-Harryooga Theorem 2021-01-04T14:22:51Z <p>Ilovepizza2020: </p> <hr /> <div>&lt;h1&gt;Overview&lt;/h1&gt;<br /> <br /> This is a very legit theorem and has many practical applications. Don't let the cool name fool you!<br /> <br /> &lt;h1&gt;Definition&lt;/h1&gt;<br /> The Georgeooga-Harryooga Theorem states that if you have &lt;math&gt;a&lt;/math&gt; distinguishable objects and &lt;math&gt;b&lt;/math&gt; are kept away from each other, then there are &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects.<br /> <br /> &lt;h1&gt;Proof&lt;/h1&gt;<br /> Let our group of &lt;math&gt;a&lt;/math&gt; objects be represented like so &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, ..., &lt;math&gt;a-1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;. Let the last &lt;math&gt;b&lt;/math&gt; objects be the ones we can't have together.<br /> <br /> Then we can organize our objects like so &lt;math&gt;\square1\square2\square3\square...\square a-b-1\square a-b\square&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;(a-b)!&lt;/math&gt; ways to arrange the objects in that list.<br /> <br /> Now we have &lt;math&gt;a-b+1&lt;/math&gt; blanks and &lt;math&gt;b&lt;/math&gt; other objects so we have &lt;math&gt;_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects we can't put together.<br /> <br /> By fundamental counting principal our answer is &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt;.<br /> <br /> <br /> Proof by [[User:Redfiretruck|RedFireTruck]]<br /> <br /> &lt;h1&gt;A side note by aryabhata000:&lt;/h1&gt;<br /> This can also be done by stars and bars like so:<br /> Let us call the &lt;math&gt;b&lt;/math&gt; people &lt;math&gt;1, 2, ... b&lt;/math&gt;<br /> <br /> Let the number of people before &lt;math&gt;1&lt;/math&gt; in line be &lt;math&gt;y_1&lt;/math&gt;, between &lt;math&gt;1, 2&lt;/math&gt; be &lt;math&gt;y_2&lt;/math&gt;, ... after &lt;math&gt;b&lt;/math&gt; b3 &lt;math&gt;y_{b+1}&lt;/math&gt;.<br /> We have &lt;cmath&gt;y_1 + y_2 + y_3 + \dots y_{b+1} = a-b&lt;/cmath&gt;<br /> <br /> The number of ways to determine &lt;math&gt;y_1, y_2, \dots&lt;/math&gt; is equivalent to the number of positive integer solutions to:<br /> &lt;cmath&gt;x_1 + x_2 + .. + x_{b+1}&lt;/cmath&gt; where &lt;math&gt;(x_2, ... x_b) = (y_2, ..., y_b) &lt;/math&gt; and &lt;math&gt;(x_1, x_{b+1}) = (y_1 +1, y_{b+1})&lt;/math&gt;.<br /> <br /> So, by stars and bars, the number of ways to determine &lt;math&gt;(y_2, ..., y_b) &lt;/math&gt; is &lt;cmath&gt;F(a,b) = \dbinom{a-b+1}{b} = \frac {(a-b+1)!}{b!(a-2b+1)!}&lt;/cmath&gt;<br /> <br /> Furthermore, after picking positions for the people, we have &lt;math&gt;(a-b)!&lt;/math&gt; ways to order the &lt;math&gt;(a-b)&lt;/math&gt; people who can be together, and &lt;math&gt;b!&lt;/math&gt; ways to order the &lt;math&gt;b&lt;/math&gt; people who cannot be together. So for each &lt;math&gt;(y_1, y_2, ... y_{b+1}&lt;/math&gt;, we have &lt;math&gt;b! (a-b)!&lt;/math&gt; orderings.<br /> <br /> Therefore, the final answer is &lt;cmath&gt;b! (a-b)! F(a,b) = \frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/cmath&gt;<br /> <br /> &lt;h1&gt;Application&lt;/h1&gt;<br /> <br /> Alice, Bob, Carl, David, Eric, Fred, George, and Harry want to stand in a line to buy ice cream. Fred and George are identical twins, so they are indistinguishable. Alice, Bob, and Carl had a serious disagreement in 6th grade, so none of them can be together in the line.<br /> With these conditions, how many different ways can you arrange these kids in a line?<br /> <br /> Problem by Math4Life2020<br /> <br /> &lt;h2&gt;Solution&lt;/h2&gt;<br /> <br /> If Eric and Fred were distinguishable we would have &lt;math&gt;\frac{(8-3)!(8-3+1)!}{(8-2\cdot3+1)!}=14400&lt;/math&gt; ways to arrange them by the Georgeooga-Harryooga Theorem. However, Eric and Fred are indistinguishable so we have to divide by &lt;math&gt;2!=2&lt;/math&gt;. Therefore, our answer is &lt;math&gt;\frac{14400}2=\boxed{7200}&lt;/math&gt;.<br /> <br /> <br /> Solution by [[User:Redfiretruck|RedFireTruck]]<br /> &lt;hr&gt;<br /> &lt;strong&gt;ALL THINGS ABOVE EXCEPT FOR THE OVERVIEW TAB AND THE PROBLEM IS MADE BY RedFireTruck &lt;/strong&gt;</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_12&diff=141030 2009 AIME I Problems/Problem 12 2020-12-30T16:04:44Z <p>Ilovepizza2020: </p> <hr /> <div>== Problem ==<br /> In right &lt;math&gt;\triangle ABC&lt;/math&gt; with hypotenuse &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;AC = 12&lt;/math&gt;, &lt;math&gt;BC = 35&lt;/math&gt;, and &lt;math&gt;\overline{CD}&lt;/math&gt; is the altitude to &lt;math&gt;\overline{AB}&lt;/math&gt;. Let &lt;math&gt;\omega&lt;/math&gt; be the circle having &lt;math&gt;\overline{CD}&lt;/math&gt; as a diameter. Let &lt;math&gt;I&lt;/math&gt; be a point outside &lt;math&gt;\triangle ABC&lt;/math&gt; such that &lt;math&gt;\overline{AI}&lt;/math&gt; and &lt;math&gt;\overline{BI}&lt;/math&gt; are both tangent to circle &lt;math&gt;\omega&lt;/math&gt;. The ratio of the perimeter of &lt;math&gt;\triangle ABI&lt;/math&gt; to the length &lt;math&gt;AB&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac {m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> <br /> == Solution 1==<br /> Let &lt;math&gt;O&lt;/math&gt; be center of the circle and &lt;math&gt;P&lt;/math&gt;,&lt;math&gt;Q&lt;/math&gt; be the two points of tangent such that &lt;math&gt;P&lt;/math&gt; is on &lt;math&gt;BI&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; is on &lt;math&gt;AI&lt;/math&gt;. We know that &lt;math&gt;AD:CD = CD:BD = 12:35&lt;/math&gt;.<br /> <br /> Since the ratios between corresponding lengths of two similar diagrams are equal, we can let &lt;math&gt;AD = 144, CD = 420&lt;/math&gt; and &lt;math&gt;BD = 1225&lt;/math&gt;. Hence &lt;math&gt;AQ = 144, BP = 1225, AB = 1369&lt;/math&gt; and the radius &lt;math&gt;r = OD = 210&lt;/math&gt;. <br /> <br /> Since we have &lt;math&gt;\tan OAB = \frac {35}{24}&lt;/math&gt; and &lt;math&gt;\tan OBA = \frac{6}{35}&lt;/math&gt; , we have &lt;math&gt;\sin {(OAB + OBA)} = \frac {1369}{\sqrt {(1801*1261)}},&lt;/math&gt;&lt;math&gt;\cos {(OAB + OBA)} = \frac {630}{\sqrt {(1801*1261)}}&lt;/math&gt;.<br /> <br /> Hence &lt;math&gt;\sin I = \sin {(2OAB + 2OBA)} = \frac {2*1369*630}{1801*1261}&lt;/math&gt;. let &lt;math&gt;IP = IQ = x&lt;/math&gt; , then we have Area&lt;math&gt;(IBC)&lt;/math&gt; = &lt;math&gt;(2x + 1225*2 + 144*2)*\frac {210}{2}&lt;/math&gt; = &lt;math&gt;(x + 144)(x + 1225)* \sin {\frac {I}{2}}&lt;/math&gt;. Then we get &lt;math&gt;x + 1369 = \frac {3*1369*(x + 144)(x + 1225)}{1801*1261}&lt;/math&gt;.<br /> <br /> Now the equation looks very complex but we can take a guess here. Assume that &lt;math&gt;x&lt;/math&gt; is a rational number<br /> (If it's not then the answer to the problem would be irrational which can't be in the form of &lt;math&gt;\frac {m}{n}&lt;/math&gt;) <br /> that can be expressed as &lt;math&gt;\frac {a}{b}&lt;/math&gt; such that &lt;math&gt;(a,b) = 1&lt;/math&gt;. Look at both sides; we can know that &lt;math&gt;a&lt;/math&gt; has to be a multiple of &lt;math&gt;1369&lt;/math&gt; and not of &lt;math&gt;3&lt;/math&gt; and it's reasonable to think that &lt;math&gt;b&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt; so that we can cancel out the &lt;math&gt;3&lt;/math&gt; on the right side of the equation.<br /> <br /> Let's see if &lt;math&gt;x = \frac {1369}{3}&lt;/math&gt; fits. Since &lt;math&gt;\frac {1369}{3} + 1369 = \frac {4*1369}{3}&lt;/math&gt;, and &lt;math&gt;\frac {3*1369*(x + 144)(x + 1225)}{1801*1261} = \frac {3*1369* \frac {1801}{3} * \frac {1261*4}{3}} {1801*1261} = \frac {4*1369}{3}&lt;/math&gt;. Amazingly it fits!<br /> <br /> Since we know that &lt;math&gt;3*1369*144*1225 - 1369*1801*1261 &lt; 0&lt;/math&gt;, the other solution of this equation is negative which can be ignored. Hence &lt;math&gt;x = 1369/3&lt;/math&gt;. <br /> <br /> Hence the perimeter is &lt;math&gt;1225*2 + 144*2 + \frac {1369}{3} *2 = 1369* \frac {8}{3}&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; is &lt;math&gt;1369&lt;/math&gt;. Hence &lt;math&gt;\frac {m}{n} = \frac {8}{3}&lt;/math&gt;, &lt;math&gt;m + n = 11&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> As in Solution &lt;math&gt;1&lt;/math&gt;, let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be the intersections of &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;BI&lt;/math&gt; and &lt;math&gt;AI&lt;/math&gt; respectively.<br /> <br /> First, by pythagorean theorem, &lt;math&gt;AB = \sqrt{12^2+35^2} = 37&lt;/math&gt;. Now the area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;1/2*12*35 = 1/2*37*CD&lt;/math&gt;, so &lt;math&gt;CD=\frac{420}{37}&lt;/math&gt; and the inradius of &lt;math&gt;\triangle ABI&lt;/math&gt; is &lt;math&gt;r=\frac{210}{37}&lt;/math&gt;.<br /> <br /> Now from &lt;math&gt;\triangle CDB \sim \triangle ACB&lt;/math&gt; we find that &lt;math&gt;\frac{BC}{BD} = \frac{AB}{BC}&lt;/math&gt; so &lt;math&gt;BD = BC^2/AB = 35^2/37&lt;/math&gt; and similarly, &lt;math&gt;AD = 12^2/37&lt;/math&gt;.<br /> <br /> Note &lt;math&gt;IP=IQ=x&lt;/math&gt;, &lt;math&gt;BP=BD&lt;/math&gt;, and &lt;math&gt;AQ=AD&lt;/math&gt;. So we have &lt;math&gt;AI = 144/37+x&lt;/math&gt;, &lt;math&gt;BI = 1225/37+x&lt;/math&gt;. Now we can compute the area of &lt;math&gt;\triangle ABI&lt;/math&gt; in two ways: by heron's formula and by inradius times semiperimeter, which yields<br /> <br /> &lt;math&gt;rs=210/37(37+x) = \sqrt{(37+x)(37-144/37)(37-1225/37)x}&lt;/math&gt;<br /> <br /> &lt;math&gt;210/37(37+x) = 12*35/37 \sqrt{x(37+x)}&lt;/math&gt;<br /> <br /> &lt;math&gt;37+x = 2 \sqrt{x(x+37)}&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2+74x+1369 = 4x^2 + 148x&lt;/math&gt;<br /> <br /> &lt;math&gt;3x^2 + 74x - 1369 = 0&lt;/math&gt;<br /> <br /> The quadratic formula now yields &lt;math&gt;x=37/3&lt;/math&gt;. Plugging this back in, the perimeter of &lt;math&gt;ABI&lt;/math&gt; is &lt;math&gt;2s=2(37+x)=2(37+37/3) = 37(8/3)&lt;/math&gt; so the ratio of the perimeter to &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;8/3&lt;/math&gt; and our answer is &lt;math&gt;8+3=\boxed{011}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> As in Solution &lt;math&gt;2&lt;/math&gt;, let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be the intersections of &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;BI&lt;/math&gt; and &lt;math&gt;AI&lt;/math&gt; respectively.<br /> <br /> Recall that the distance from a point outside a circle to that circle is the same along both tangent lines to the circle drawn from the point.<br /> <br /> Recall also that the length of the altitude to the hypotenuse of a right-angle triangle is the geometric mean of the two segments into which it cuts the hypotenuse.<br /> <br /> Let &lt;math&gt;x = \overline{AD} = \overline{AQ}&lt;/math&gt;. Let &lt;math&gt;y = \overline{BD} = \overline{BP}&lt;/math&gt;. Let &lt;math&gt;z = \overline{PI} = \overline{QI}&lt;/math&gt;. The semi-perimeter of &lt;math&gt;ABI&lt;/math&gt; is &lt;math&gt;x + y + z&lt;/math&gt;.<br /> Since the lengths of the sides of &lt;math&gt;ABI&lt;/math&gt; are &lt;math&gt;x + y&lt;/math&gt;, &lt;math&gt;y + z&lt;/math&gt; and &lt;math&gt;x + z&lt;/math&gt;, the square of its area by Heron's formula is &lt;math&gt;(x+y+z)xyz&lt;/math&gt;.<br /> <br /> The radius &lt;math&gt;r&lt;/math&gt; of &lt;math&gt;\omega&lt;/math&gt; is &lt;math&gt;\overline{CD}/2&lt;/math&gt;. Therefore &lt;math&gt;r^2 = xy/4&lt;/math&gt;. As &lt;math&gt;\omega&lt;/math&gt; is the in-circle of &lt;math&gt;ABI&lt;/math&gt;, the area of &lt;math&gt;ABI&lt;/math&gt; is also &lt;math&gt;r(x+y+z)&lt;/math&gt;, and so the square area is &lt;math&gt;r^2(x+y+z)^2&lt;/math&gt;.<br /> <br /> Therefore &lt;cmath&gt;(x+y+z)xyz = r^2(x+y+z)^2 = \frac{xy(x+y+z)^2}{4}&lt;/cmath&gt; Dividing both sides by &lt;math&gt;xy(x+y+z)/4&lt;/math&gt; we get: &lt;cmath&gt;4z = (x+y+z),&lt;/cmath&gt; and so &lt;math&gt;z = (x+y)/3&lt;/math&gt;. The semi-perimeter of &lt;math&gt;ABI&lt;/math&gt; is therefore &lt;math&gt;\frac{4}{3}(x+y)&lt;/math&gt; and the whole perimeter is &lt;math&gt;\frac{8}{3}(x+y)&lt;/math&gt;. Now &lt;math&gt;x + y = \overline{AB}&lt;/math&gt;, so the ratio of the perimeter of &lt;math&gt;ABI&lt;/math&gt; to the hypotenuse &lt;math&gt;\overline{AB}&lt;/math&gt; is &lt;math&gt;8/3&lt;/math&gt; and our answer is &lt;math&gt;8+3=\boxed{011}&lt;/math&gt;<br /> <br /> == Solution 4 ==<br /> <br /> We shall yet again let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be the intersections of &lt;math&gt;AI&lt;/math&gt; and &lt;math&gt;BI&lt;/math&gt; to &lt;math&gt;\omega&lt;/math&gt;, respectively. We want to find the perimeter of &lt;math&gt;ABI&lt;/math&gt;, which is &lt;math&gt;AD+BD+BQ+QI+IP+PA&lt;/math&gt;. We can easily find &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; using the fact that &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;ACD&lt;/math&gt;, and &lt;math&gt;BCD&lt;/math&gt; are all similar triangles. We get &lt;math&gt;AD=\frac{144}{37}&lt;/math&gt; and &lt;math&gt;\frac{1225}{37}&lt;/math&gt;. Since &lt;math&gt;AP&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; are tangents to &lt;math&gt;\omega&lt;/math&gt;, &lt;math&gt;AP=AD=\frac{144}{37}&lt;/math&gt;, and similarly &lt;math&gt;BQ=BD=\frac{1225}{37}&lt;/math&gt;. We now wish to find &lt;math&gt;IP&lt;/math&gt; and &lt;math&gt;IQ&lt;/math&gt;.<br /> <br /> Let the center of the given circle be &lt;math&gt;O&lt;/math&gt;. We know that &lt;math&gt;\angle AOP=\angle AOD&lt;/math&gt;, &lt;math&gt;\angle BOQ=\angle BOD&lt;/math&gt;, and &lt;math&gt;\angle IOQ=\angle IOP&lt;/math&gt;. Since all six angles sum to &lt;math&gt;360^{\circ}&lt;/math&gt;, &lt;math&gt;\angle AOP+\angle BOQ+\angle IOP=180^{\circ}&lt;/math&gt;. If we knew the radius of circle &lt;math&gt;\omega&lt;/math&gt; now, then we could find &lt;math&gt;\tan{\angle AOP}&lt;/math&gt; and &lt;math&gt;\tan{\angle BOQ}&lt;/math&gt;, and then we can use the sum (or difference) of tangents formula to find &lt;math&gt;\tan{\angle IOP}&lt;/math&gt;, which reveals &lt;math&gt;IP&lt;/math&gt;. This means we should find the radius of &lt;math&gt;\omega&lt;/math&gt;. We can easily see that the height of triangle &lt;math&gt;ABC&lt;/math&gt; from &lt;math&gt;C&lt;/math&gt; has length &lt;math&gt;\frac{12*35}{37}&lt;/math&gt;, so the radius of &lt;math&gt;\omega&lt;/math&gt; is &lt;math&gt;\frac{210}{37}&lt;/math&gt;. Now we shall proceed with the above plan.<br /> <br /> &lt;math&gt;\tan{\angle AOP}=\frac{144}{210}&lt;/math&gt;. &lt;math&gt;\tan{\angle BOQ}=\frac{1225}{210}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\tan{\angle IOP}=\tan{(180^{\circ}-\angle AOP-\angle BOQ)}=-\tan{(\angle AOP+\angle BOQ)}&lt;/math&gt;<br /> <br /> &lt;math&gt;=-\frac{\frac{144}{210}+\frac{1225}{210}}{1-\frac{144}{210}*\frac{1225}{210}}=-\frac{1369}{210-\frac{144*1225}{210}}=\frac{1369}{\frac{144*1225}{210}-210}=\frac{37*37}{35*18}&lt;/math&gt;.<br /> <br /> Therefore &lt;math&gt;OP=\frac{210}{37},IP=\frac{37}{3}&lt;/math&gt;, and the perimeter of &lt;math&gt;AIB&lt;/math&gt; is &lt;math&gt;2*\frac{37}{3}+2*\frac{144}{37}+2*\frac{1225}{37}=37*\frac{8}{3}&lt;/math&gt;. Since &lt;math&gt;AB=37&lt;/math&gt;, the desired ratio is &lt;math&gt;\frac{8}{3}&lt;/math&gt;, and &lt;math&gt;8+3=\boxed{011}&lt;/math&gt;.<br /> <br /> == Solution 5 ==<br /> <br /> This solution is not a real solution and is solving the problem with a ruler and compass.<br /> <br /> Draw &lt;math&gt;AC = 4.8, BC = 14, AB = 14.8&lt;/math&gt;. Then, drawing the tangents and intersecting them, we get that &lt;math&gt;IA&lt;/math&gt; is around &lt;math&gt;6.55&lt;/math&gt; and &lt;math&gt;IB&lt;/math&gt; is around &lt;math&gt;18.1&lt;/math&gt;. We then find the ratio to be around &lt;math&gt;\frac{39.45}{14.8}&lt;/math&gt;. Using long division, we find that this ratio is approximately 2.666, which you should recognize as &lt;math&gt;\frac{8}{3}&lt;/math&gt;. Since this seems reasonable, we find that the answer is &lt;math&gt;\boxed{11}&lt;/math&gt; ~ilp<br /> == See also ==<br /> {{AIME box|year=2009|n=I|num-b=11|num-a=13}}<br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_11&diff=140747 2011 AIME II Problems/Problem 11 2020-12-27T18:56:10Z <p>Ilovepizza2020: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;M_n&lt;/math&gt; be the &lt;math&gt;n \times n&lt;/math&gt; [[matrix]] with entries as follows: for &lt;math&gt;1 \le i \le n&lt;/math&gt;, &lt;math&gt;m_{i,i} = 10&lt;/math&gt;; for &lt;math&gt;1 \le i \le n - 1&lt;/math&gt;, &lt;math&gt;m_{i+1,i} = m_{i,i+1} = 3&lt;/math&gt;; all other entries in &lt;math&gt;M_n&lt;/math&gt; are zero. Let &lt;math&gt;D_n&lt;/math&gt; be the [[determinant]] of matrix &lt;math&gt;M_n&lt;/math&gt;. Then &lt;math&gt;\sum_{n=1}^{\infty} \frac{1}{8D_n+1}&lt;/math&gt; can be represented as &lt;math&gt;\frac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are [[relatively prime]] positive integers. Find &lt;math&gt;p + q&lt;/math&gt;.<br /> <br /> Note: The determinant of the &lt;math&gt;1 \times 1&lt;/math&gt; matrix &lt;math&gt;[a]&lt;/math&gt; is &lt;math&gt;a&lt;/math&gt;, and the determinant of the &lt;math&gt;2 \times 2&lt;/math&gt; matrix &lt;math&gt;\left[ {\begin{array}{cc}<br /> a &amp; b \\<br /> c &amp; d \\<br /> \end{array} } \right] = ad - bc&lt;/math&gt;; for &lt;math&gt;n \ge 2&lt;/math&gt;, the determinant of an &lt;math&gt;n \times n&lt;/math&gt; matrix with first row or first column &lt;math&gt;a_1&lt;/math&gt; &lt;math&gt;a_2&lt;/math&gt; &lt;math&gt;a_3&lt;/math&gt; &lt;math&gt;\dots&lt;/math&gt; &lt;math&gt;a_n&lt;/math&gt; is equal to &lt;math&gt;a_1C_1 - a_2C_2 + a_3C_3 - \dots + (-1)^{n+1}a_nC_n&lt;/math&gt;, where &lt;math&gt;C_i&lt;/math&gt; is the determinant of the &lt;math&gt;(n - 1) \times (n - 1)&lt;/math&gt; matrix formed by eliminating the row and column containing &lt;math&gt;a_i&lt;/math&gt;.<br /> <br /> ==Solution==<br /> &lt;cmath&gt;D_{1}=\begin{vmatrix}<br /> 10 <br /> \end{vmatrix} = 10, \quad<br /> D_{2}=\begin{vmatrix}<br /> 10 &amp; 3 \\<br /> 3 &amp; 10 \\ \end{vmatrix}<br /> =(10)(10) - (3)(3) = 91, \quad<br /> D_{3}=\begin{vmatrix}<br /> 10 &amp; 3 &amp; 0 \\<br /> 3 &amp; 10 &amp; 3 \\<br /> 0 &amp; 3 &amp; 10 \\<br /> \end{vmatrix}. &lt;/cmath&gt;<br /> Using the expansionary/recursive definition of determinants (also stated in the problem):<br /> <br /> &lt;math&gt;D_{3}=\left| {\begin{array}{ccc}<br /> 10 &amp; 3 &amp; 0 \\<br /> 3 &amp; 10 &amp; 3 \\<br /> 0 &amp; 3 &amp; 10 \\<br /> \end{array} } \right|=10\left| {\begin{array}{cc}<br /> 10 &amp; 3 \\<br /> 3 &amp; 10 \\<br /> \end{array} } \right| - 3\left| {\begin{array}{cc}<br /> 3 &amp; 3 \\<br /> 0 &amp; 10 \\<br /> \end{array} } \right| + 0\left| {\begin{array}{cc}<br /> 3 &amp; 10 \\<br /> 0 &amp; 3 \\<br /> \end{array} } \right| = 10D_{2} - 9D_{1} = 820&lt;/math&gt;<br /> <br /> This pattern repeats because the first element in the first row of &lt;math&gt;M_{n}&lt;/math&gt; is always 10, the second element is always 3, and the rest are always 0. The ten element directly expands to &lt;math&gt;10D_{n-1}&lt;/math&gt;. The three element expands to 3 times the determinant of the the matrix formed from omitting the second column and first row from the original matrix. Call this matrix &lt;math&gt;X_{n}&lt;/math&gt;. &lt;math&gt;X_{n}&lt;/math&gt; has a first column entirely of zeros except for the first element, which is a three. A property of matrices is that the determinant can be expanded over the rows instead of the columns (still using the recursive definition as given in the problem), and the determinant found will still be the same. Thus, expanding over this first column yields &lt;math&gt;3D_{n-2} + 0(\text{other things})=3D_{n-2}&lt;/math&gt;. Thus, the &lt;math&gt;3 \det(X_{n})&lt;/math&gt; expression turns into &lt;math&gt;9D_{n-2}&lt;/math&gt;. Thus, the equation &lt;math&gt;D_{n}=10D_{n-1}-9D_{n-2}&lt;/math&gt; holds for all n &gt; 2.<br /> <br /> This equation can be rewritten as &lt;math&gt;D_{n}=10(D_{n-1}-D_{n-2}) + D_{n-2}&lt;/math&gt;. This version of the equation involves the difference of successive terms of a recursive sequence. Calculating &lt;math&gt;D_{0}&lt;/math&gt; backwards from the recursive formula and &lt;math&gt;D_{4}&lt;/math&gt; from the formula yields &lt;math&gt;D_{0}=1, D_{4}=7381&lt;/math&gt;. Examining the differences between successive terms, a pattern emerges. <br /> &lt;math&gt;D_{0}=1=9^{0}&lt;/math&gt;, &lt;math&gt;D_{1}-D_{0}=10-1=9=9^{1}&lt;/math&gt;, &lt;math&gt;D_{2}-D_{1}=91-10=81=9^{2}&lt;/math&gt;, &lt;math&gt;D_{3}-D_{2}=820-91=729=9^{3}&lt;/math&gt;, &lt;math&gt;D_{4}-D_{3}=7381-820=6561=9^{4}&lt;/math&gt;.<br /> Thus, &lt;math&gt;D_{n}=D_{0} + 9^{1}+9^{2}+ . . . +9^{n}=\sum_{i=0}^{n}9^{i}=\frac{(1)(9^{n+1}-1)}{9-1}=\frac{9^{n+1}-1}{8}&lt;/math&gt;.<br /> <br /> Thus, the desired sum is &lt;math&gt;\sum_{n=1}^{\infty}\frac{1}{8\frac{9^{n+1}-1}{8}+1}=\sum_{n=1}^{\infty}\frac{1}{9^{n+1}-1+1} = \sum_{n=1}^{\infty}\frac{1}{9^{n+1}} &lt;/math&gt;<br /> <br /> This is an infinite [[geometric series]] with first term &lt;math&gt;\frac{1}{81}&lt;/math&gt; and common ratio &lt;math&gt;\frac{1}{9}&lt;/math&gt;. Thus, the sum is &lt;math&gt;\frac{\frac{1}{81}}{1-\frac{1}{9}}=\frac{\frac{1}{81}}{\frac{8}{9}}=\frac{9}{(81)(8)}=\frac{1}{(9)(8)}=\frac{1}{72}&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;p + q = 1 + 72 = \boxed{073}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We calculate the first &lt;math&gt;4&lt;/math&gt; terms of &lt;math&gt;8D_n+1&lt;/math&gt;. They are: &lt;math&gt;81, 729, 6561,&lt;/math&gt; and &lt;math&gt;59049&lt;/math&gt;. From here, we can tell that it is just &lt;math&gt;\frac{1}{9*8}=\frac{1}{72} \implies \boxed{073}&lt;/math&gt;<br /> ~Lcz<br /> <br /> ==Solution 3==<br /> We can calculate the first few determinants to be 10, 91, 820, 7381. These satisfy the pattern &lt;math&gt;D_n = 9D_{n-1}+1&lt;/math&gt;. So, we assume this works for all &lt;math&gt;n&lt;/math&gt;. We then calculate &lt;math&gt;8D_n+1&lt;/math&gt; to be &lt;math&gt;81,729,6561,&lt;/math&gt; and &lt;math&gt;59049&lt;/math&gt;. These look like powers of &lt;math&gt;9&lt;/math&gt;, so we assume it is that way. The sum is then &lt;math&gt;\frac{1}{81}+ \frac{1}{729}+\frac{1}{6561}... = \frac{\frac{1}{81}}{1-\frac{1}{9}} = \frac{1}{72}&lt;/math&gt;. Our answer is therefore &lt;math&gt;73&lt;/math&gt;. ~ilp<br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=Georgeooga-Harryooga_Theorem&diff=140016 Georgeooga-Harryooga Theorem 2020-12-20T13:44:04Z <p>Ilovepizza2020: </p> <hr /> <div>=Definition=<br /> The Georgeooga-Harryooga Theorem states that if you have &lt;math&gt;a&lt;/math&gt; distinguishable objects and &lt;math&gt;b&lt;/math&gt; are kept away from each other, then there are &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects.<br /> <br /> <br /> Created by George and Harry of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> =Proofs=<br /> ==Proof 1==<br /> Let our group of &lt;math&gt;a&lt;/math&gt; objects be represented like so &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, ..., &lt;math&gt;a-1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;. Let the last &lt;math&gt;b&lt;/math&gt; objects be the ones we can't have together.<br /> <br /> Then we can organize our objects like so &lt;math&gt;\square1\square2\square3\square...\square a-b-1\square a-b\square&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;(a-b)!&lt;/math&gt; ways to arrange the objects in that list.<br /> <br /> Now we have &lt;math&gt;a-b+1&lt;/math&gt; blanks and &lt;math&gt;b&lt;/math&gt; other objects so we have &lt;math&gt;_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects we can't put together.<br /> <br /> By fundamental counting principal our answer is &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt;.<br /> <br /> <br /> Proof by [[User:RedFireTruck|RedFireTruck]]<br /> <br /> =Applications=<br /> ==Application 1==<br /> ===Problem===<br /> Alice, Bob, Carl, David, Eric, Fred, George, and Harry want to stand in a line to buy ice cream.<br /> Fred and George are identical twins, so they are indistinguishable.<br /> Alice, Bob, and Carl had a serious disagreement in 6th grade, so none of them can be together in the line.<br /> <br /> With these conditions, how many different ways can you arrange these kids in a line?<br /> <br /> <br /> Problem by [https://artofproblemsolving.com/community/c4h2342517p18900597 Math4Life2020]<br /> <br /> ===Solutions===<br /> ====Solution 1====<br /> If Eric and Fred were distinguishable we would have &lt;math&gt;\frac{(8-3)!(8-3+1)!}{(8-2\cdot3+1)!}=14400&lt;/math&gt; ways to arrange them by the [[Georgeooga-Harryooga Theorem]]. However, Eric and Fred are indistinguishable so we have to divide by &lt;math&gt;2!=2&lt;/math&gt;. Therefore, our answer is &lt;math&gt;\frac{14400}2=\boxed{7200}&lt;/math&gt;.<br /> <br /> <br /> Solution by [[User:RedFireTruck|RedFireTruck]]<br /> <br /> ==Application 2==<br /> ===Problem===<br /> Zara has a collection of &lt;math&gt;4&lt;/math&gt; marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> <br /> Problem by [[2020 AMC 8 Problems/Problem 10|The Mathematical Association of America's American Mathematics Competitions]]<br /> <br /> ===Solutions===<br /> ====Solution 1====<br /> By the [[Georgeooga-Harryooga Theorem]] there are &lt;math&gt;\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\boxed{\textbf{(C) }12}&lt;/math&gt; way to arrange the marbles.<br /> <br /> <br /> Solution by [[User:RedFireTruck|RedFireTruck]]<br /> <br /> ====Solution 2====<br /> We can arrange our marbles like so &lt;math&gt;\square A\square B\square&lt;/math&gt;.<br /> <br /> To arrange the &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; we have &lt;math&gt;2!=2&lt;/math&gt; ways.<br /> <br /> To place the &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; in the blanks we have &lt;math&gt;_3P_2=6&lt;/math&gt; ways.<br /> <br /> By fundamental counting principle our final answer is &lt;math&gt;2\cdot6=\boxed{\textbf{(C) }12}&lt;/math&gt;<br /> <br /> <br /> Solution by [[User:Redfiretruck|RedFireTruck]]<br /> <br /> ====Solution 3====<br /> Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by &lt;math&gt;A,B,S,&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt;, respectively. If we ignore the constraint that &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; cannot be next to each other, we get a total of &lt;math&gt;4!=24&lt;/math&gt; ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; can be next to each other. If we place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. &lt;math&gt;ST\square\square, \square ST\square, \square\square ST&lt;/math&gt;). However, we could also have placed &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; in the opposite order (i.e. &lt;math&gt;TS\square\square, \square TS\square, \square\square TS&lt;/math&gt;). Thus there are 6 ways of placing &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; directly next to each other. Next, notice that for each of these placements, we have two open slots for placing &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. Specifically, we can place &lt;math&gt;A&lt;/math&gt; in the first open slot and &lt;math&gt;B&lt;/math&gt; in the second open slot or switch their order and place &lt;math&gt;B&lt;/math&gt; in the first open slot and &lt;math&gt;A&lt;/math&gt; in the second open slot. This gives us a total of &lt;math&gt;6\times 2=12&lt;/math&gt; ways to place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; next to each other. Subtracting this from the total number of arrangements gives us &lt;math&gt;24-12=12&lt;/math&gt; total arrangements &lt;math&gt;\implies\boxed{\textbf{(C) }12}&lt;/math&gt;.&lt;br&gt;<br /> <br /> We can also solve this problem directly by looking at the number of ways that we can place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; such that they are not directly next to each other. Observe that there are three ways to place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; (in that order) into the four slots so they are not next to each other (i.e. &lt;math&gt;S\square T\square, \square S\square T, S\square\square T&lt;/math&gt;). However, we could also have placed &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; in the opposite order (i.e. &lt;math&gt;T\square S\square, \square T\square S, T\square\square S&lt;/math&gt;). Thus there are 6 ways of placing &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. Specifically, we can place &lt;math&gt;A&lt;/math&gt; in the first open slot and &lt;math&gt;B&lt;/math&gt; in the second open slot or switch their order and place &lt;math&gt;B&lt;/math&gt; in the first open slot and &lt;math&gt;A&lt;/math&gt; in the second open slot. This gives us a total of &lt;math&gt;6\times 2=12&lt;/math&gt; ways to place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; such that they are not next to each other &lt;math&gt;\implies\boxed{\textbf{(C) }12}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ====Solution 4====<br /> Let's try complementary counting. There &lt;math&gt;4!&lt;/math&gt; ways to arrange the 4 marbles. However, there are &lt;math&gt;2\cdot3!&lt;/math&gt; arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus, &lt;cmath&gt;4!-2\cdot3!=\boxed{12 \textbf{(C)}}&lt;/cmath&gt;<br /> <br /> ====Solution 5====<br /> <br /> We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a &quot;super marble.&quot; There are &lt;math&gt;2!&lt;/math&gt; ways to arrange Steelie and Tiger within this &quot;super marble.&quot; Then there are &lt;math&gt;3!&lt;/math&gt; ways to arrange the &quot;super marble&quot; and Zara's two other marbles in a row. Since there are &lt;math&gt;4!&lt;/math&gt; ways to arrange the marbles without any restrictions, the answer is given by &lt;math&gt;4!-2!\cdot 3!=\textbf{(C) }12&lt;/math&gt;<br /> <br /> -franzliszt<br /> <br /> ====Solution 6====<br /> <br /> We will use the following<br /> <br /> &lt;math&gt;\textbf{Georgeooga-Harryooga Theorem:}&lt;/math&gt; The [[Georgeooga-Harryooga Theorem]] states that if you have &lt;math&gt;a&lt;/math&gt; distinguishable objects and &lt;math&gt;b&lt;/math&gt; of them cannot be together, then there are &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects.<br /> <br /> &lt;math&gt;\textit{Proof. (Created by AoPS user RedFireTruck)}&lt;/math&gt;<br /> <br /> Let our group of &lt;math&gt;a&lt;/math&gt; objects be represented like so &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, ..., &lt;math&gt;a-1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;. Let the last &lt;math&gt;b&lt;/math&gt; objects be the ones we can't have together.<br /> <br /> Then we can organize our objects like so &lt;math&gt;\square1\square2\square3\square...\square a-b-1\square a-b\square&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;(a-b)!&lt;/math&gt; ways to arrange the objects in that list.<br /> <br /> Now we have &lt;math&gt;a-b+1&lt;/math&gt; blanks and &lt;math&gt;b&lt;/math&gt; other objects so we have &lt;math&gt;_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects we can't put together.<br /> <br /> By fundamental counting principal our answer is &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt;.<br /> <br /> <br /> Proof by [[User:RedFireTruck|RedFireTruck]]<br /> <br /> <br /> Back to the problem. By the [[Georgeooga-Harryooga Theorem]], our answer is &lt;math&gt;\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\textbf{(C) }12&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ====Solution 7====<br /> https://youtu.be/pB46JzBNM6g<br /> <br /> ~savannahsolver<br /> <br /> =Testimonials=<br /> &quot;Thanks for rediscovering our theorem [[User:Redfiretruck|RedFireTruck]]&quot; - George and Harry of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> &quot;Wow! George and Harry are alive???&quot; ~ samrocksnature<br /> <br /> &quot;Hi&quot; ~ jasperE3<br /> <br /> &quot;I used this theorem on the AMC 8. Very useful!&quot; - [[User:Redfiretruck|RedFireTruck]]<br /> <br /> &quot;This is very complicated, but great.&quot; - Jiseop55406<br /> <br /> I used this theorem on the AMC 8 too! ~ ilp</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=Georgeooga-Harryooga_Theorem&diff=140015 Georgeooga-Harryooga Theorem 2020-12-20T13:42:04Z <p>Ilovepizza2020: Undo revision 140006 by Sugar rush (talk)</p> <hr /> <div>=Definition=<br /> The Georgeooga-Harryooga Theorem states that if you have &lt;math&gt;a&lt;/math&gt; distinguishable objects and &lt;math&gt;b&lt;/math&gt; are kept away from each other, then there are &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects.<br /> <br /> <br /> Created by George and Harry of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> =Proofs=<br /> ==Proof 1==<br /> Let our group of &lt;math&gt;a&lt;/math&gt; objects be represented like so &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, ..., &lt;math&gt;a-1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;. Let the last &lt;math&gt;b&lt;/math&gt; objects be the ones we can't have together.<br /> <br /> Then we can organize our objects like so &lt;math&gt;\square1\square2\square3\square...\square a-b-1\square a-b\square&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;(a-b)!&lt;/math&gt; ways to arrange the objects in that list.<br /> <br /> Now we have &lt;math&gt;a-b+1&lt;/math&gt; blanks and &lt;math&gt;b&lt;/math&gt; other objects so we have &lt;math&gt;_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects we can't put together.<br /> <br /> By fundamental counting principal our answer is &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt;.<br /> <br /> <br /> Proof by [[User:RedFireTruck|RedFireTruck]]<br /> <br /> =Applications=<br /> ==Application 1==<br /> ===Problem===<br /> Alice, Bob, Carl, David, Eric, Fred, George, and Harry want to stand in a line to buy ice cream.<br /> Fred and George are identical twins, so they are indistinguishable.<br /> Alice, Bob, and Carl had a serious disagreement in 6th grade, so none of them can be together in the line.<br /> <br /> With these conditions, how many different ways can you arrange these kids in a line?<br /> <br /> <br /> Problem by [https://artofproblemsolving.com/community/c4h2342517p18900597 Math4Life2020]<br /> <br /> ===Solutions===<br /> ====Solution 1====<br /> If Eric and Fred were distinguishable we would have &lt;math&gt;\frac{(8-3)!(8-3+1)!}{(8-2\cdot3+1)!}=14400&lt;/math&gt; ways to arrange them by the [[Georgeooga-Harryooga Theorem]]. However, Eric and Fred are indistinguishable so we have to divide by &lt;math&gt;2!=2&lt;/math&gt;. Therefore, our answer is &lt;math&gt;\frac{14400}2=\boxed{7200}&lt;/math&gt;.<br /> <br /> <br /> Solution by [[User:RedFireTruck|RedFireTruck]]<br /> <br /> ==Application 2==<br /> ===Problem===<br /> Zara has a collection of &lt;math&gt;4&lt;/math&gt; marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> <br /> Problem by [[2020 AMC 8 Problems/Problem 10|The Mathematical Association of America's American Mathematics Competitions]]<br /> <br /> ===Solutions===<br /> ====Solution 1====<br /> By the [[Georgeooga-Harryooga Theorem]] there are &lt;math&gt;\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\boxed{\textbf{(C) }12}&lt;/math&gt; way to arrange the marbles.<br /> <br /> <br /> Solution by [[User:RedFireTruck|RedFireTruck]]<br /> <br /> ====Solution 2====<br /> We can arrange our marbles like so &lt;math&gt;\square A\square B\square&lt;/math&gt;.<br /> <br /> To arrange the &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; we have &lt;math&gt;2!=2&lt;/math&gt; ways.<br /> <br /> To place the &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; in the blanks we have &lt;math&gt;_3P_2=6&lt;/math&gt; ways.<br /> <br /> By fundamental counting principle our final answer is &lt;math&gt;2\cdot6=\boxed{\textbf{(C) }12}&lt;/math&gt;<br /> <br /> <br /> Solution by [[User:Redfiretruck|RedFireTruck]]<br /> <br /> ====Solution 3====<br /> Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by &lt;math&gt;A,B,S,&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt;, respectively. If we ignore the constraint that &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; cannot be next to each other, we get a total of &lt;math&gt;4!=24&lt;/math&gt; ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; can be next to each other. If we place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. &lt;math&gt;ST\square\square, \square ST\square, \square\square ST&lt;/math&gt;). However, we could also have placed &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; in the opposite order (i.e. &lt;math&gt;TS\square\square, \square TS\square, \square\square TS&lt;/math&gt;). Thus there are 6 ways of placing &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; directly next to each other. Next, notice that for each of these placements, we have two open slots for placing &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. Specifically, we can place &lt;math&gt;A&lt;/math&gt; in the first open slot and &lt;math&gt;B&lt;/math&gt; in the second open slot or switch their order and place &lt;math&gt;B&lt;/math&gt; in the first open slot and &lt;math&gt;A&lt;/math&gt; in the second open slot. This gives us a total of &lt;math&gt;6\times 2=12&lt;/math&gt; ways to place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; next to each other. Subtracting this from the total number of arrangements gives us &lt;math&gt;24-12=12&lt;/math&gt; total arrangements &lt;math&gt;\implies\boxed{\textbf{(C) }12}&lt;/math&gt;.&lt;br&gt;<br /> <br /> We can also solve this problem directly by looking at the number of ways that we can place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; such that they are not directly next to each other. Observe that there are three ways to place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; (in that order) into the four slots so they are not next to each other (i.e. &lt;math&gt;S\square T\square, \square S\square T, S\square\square T&lt;/math&gt;). However, we could also have placed &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; in the opposite order (i.e. &lt;math&gt;T\square S\square, \square T\square S, T\square\square S&lt;/math&gt;). Thus there are 6 ways of placing &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. Specifically, we can place &lt;math&gt;A&lt;/math&gt; in the first open slot and &lt;math&gt;B&lt;/math&gt; in the second open slot or switch their order and place &lt;math&gt;B&lt;/math&gt; in the first open slot and &lt;math&gt;A&lt;/math&gt; in the second open slot. This gives us a total of &lt;math&gt;6\times 2=12&lt;/math&gt; ways to place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; such that they are not next to each other &lt;math&gt;\implies\boxed{\textbf{(C) }12}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ====Solution 4====<br /> Let's try complementary counting. There &lt;math&gt;4!&lt;/math&gt; ways to arrange the 4 marbles. However, there are &lt;math&gt;2\cdot3!&lt;/math&gt; arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus, &lt;cmath&gt;4!-2\cdot3!=\boxed{12 \textbf{(C)}}&lt;/cmath&gt;<br /> <br /> ====Solution 5====<br /> <br /> We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a &quot;super marble.&quot; There are &lt;math&gt;2!&lt;/math&gt; ways to arrange Steelie and Tiger within this &quot;super marble.&quot; Then there are &lt;math&gt;3!&lt;/math&gt; ways to arrange the &quot;super marble&quot; and Zara's two other marbles in a row. Since there are &lt;math&gt;4!&lt;/math&gt; ways to arrange the marbles without any restrictions, the answer is given by &lt;math&gt;4!-2!\cdot 3!=\textbf{(C) }12&lt;/math&gt;<br /> <br /> -franzliszt<br /> <br /> ====Solution 6====<br /> <br /> We will use the following<br /> <br /> &lt;math&gt;\textbf{Georgeooga-Harryooga Theorem:}&lt;/math&gt; The [[Georgeooga-Harryooga Theorem]] states that if you have &lt;math&gt;a&lt;/math&gt; distinguishable objects and &lt;math&gt;b&lt;/math&gt; of them cannot be together, then there are &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects.<br /> <br /> &lt;math&gt;\textit{Proof. (Created by AoPS user RedFireTruck)}&lt;/math&gt;<br /> <br /> Let our group of &lt;math&gt;a&lt;/math&gt; objects be represented like so &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, ..., &lt;math&gt;a-1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;. Let the last &lt;math&gt;b&lt;/math&gt; objects be the ones we can't have together.<br /> <br /> Then we can organize our objects like so &lt;math&gt;\square1\square2\square3\square...\square a-b-1\square a-b\square&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;(a-b)!&lt;/math&gt; ways to arrange the objects in that list.<br /> <br /> Now we have &lt;math&gt;a-b+1&lt;/math&gt; blanks and &lt;math&gt;b&lt;/math&gt; other objects so we have &lt;math&gt;_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects we can't put together.<br /> <br /> By fundamental counting principal our answer is &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt;.<br /> <br /> <br /> Proof by [[User:RedFireTruck|RedFireTruck]]<br /> <br /> <br /> Back to the problem. By the [[Georgeooga-Harryooga Theorem]], our answer is &lt;math&gt;\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\textbf{(C) }12&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ====Solution 7====<br /> https://youtu.be/pB46JzBNM6g<br /> <br /> ~savannahsolver<br /> <br /> =Testimonials=<br /> &quot;Thanks for rediscovering our theorem [[User:Redfiretruck|RedFireTruck]]&quot; - George and Harry of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> &quot;Wow! George and Harry are alive???&quot; ~ samrocksnature<br /> <br /> &quot;Hi&quot; ~ jasperE3<br /> <br /> &quot;I used this theorem on the AMC 8. Very useful!&quot; - [[User:Redfiretruck|RedFireTruck]]<br /> <br /> &quot;This is very complicated, but great.&quot; - Jiseop55406</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=Arrangement_Restriction_Theorem&diff=140014 Arrangement Restriction Theorem 2020-12-20T13:40:20Z <p>Ilovepizza2020: </p> <hr /> <div>The &lt;b&gt;Arrangement Restriction Theorem&lt;/b&gt; is discovered by [[User:aops-g5-gethsemanea2|aops-g5-gethsemanea2]] and is NOT an alternative to the [[Georgeooga-Harryooga Theorem]] because in this theorem the only situation that is not allowed is that all &lt;math&gt;k&lt;/math&gt; objects are together.<br /> <br /> ==Definition==<br /> If there are &lt;math&gt;n&lt;/math&gt; objects to be arranged and &lt;math&gt;k&lt;/math&gt; of them should not be beside each other &lt;b&gt;altogether&lt;/b&gt;, then the number of ways to arrange them is &lt;math&gt;n! - (n - k + 1)!k!&lt;/math&gt;.<br /> <br /> ==Proof/Derivation==<br /> <br /> If there are no restrictions, then we have &lt;math&gt;n!&lt;/math&gt;. But, if we put &lt;math&gt;k&lt;/math&gt; objects beside each other, we have &lt;math&gt;(n-k+1)!k!&lt;/math&gt; because we can count the &lt;math&gt;k&lt;/math&gt; objects as one object and just rearrange them.<br /> <br /> So, by complementary counting, we get &lt;math&gt;n! - (n - k + 1)!k!&lt;/math&gt;.<br /> <br /> ==Testimonials==<br /> <br /> I like this theorem, but not as much as the Georgeooga-Harryooga Theorem or the Wooga Looga Theorem ~ ilp</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=Wooga_Looga_Theorem&diff=138876 Wooga Looga Theorem 2020-12-01T15:07:56Z <p>Ilovepizza2020: </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 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 /> <br /> The Wooga Looga theorem is a direct application of Routh's Theorem when a=b=c.<br /> <br /> ==Proof 5==<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 /> 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 4==<br /> <br /> or this solution by ilovepizza2020:<br /> <br /> We use the &lt;math&gt;\mathbf{FUNDEMENTAL~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 tests 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.<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 /> 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 /> 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. - 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 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 /> &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 /> <br /> <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 [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<br /> <br /> [s]I ReAlLy don't get it - Senguamar[/s] 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</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=The_Devil%27s_Triangle&diff=136725 The Devil's Triangle 2020-11-06T14:53:55Z <p>Ilovepizza2020: </p> <hr /> <div>=Definition=<br /> For any triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, let &lt;math&gt;D, E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; be points on &lt;math&gt;BC, AC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; respectively. Devil's Triangle Theorem states that if &lt;math&gt;\frac{BD}{CD}=r, \frac{CE}{AE}=s&lt;/math&gt; and &lt;math&gt;\frac{AF}{BF}=t&lt;/math&gt;, then &lt;math&gt;\frac{[DEF]}{[ABC]}=1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}&lt;/math&gt;. <br /> <br /> =Proofs=<br /> ==Proof 1==<br /> Proof by CoolJupiter:<br /> <br /> We have the following ratios:<br /> &lt;math&gt;\frac{BD}{BC}=\frac{r}{r+1}, \frac{CD}{BC}=\frac{1}{r+1},\frac{CE}{AC}=\frac{s}{s+1}, \frac{AE}{AC}=\frac{1}{s+1},\frac{AF}{AB}=\frac{t}{t+1}, \frac{BF}{AB}=\frac{1}{t+1}&lt;/math&gt;.<br /> <br /> Now notice that &lt;math&gt;[DEF]=[ABC]-([BDF]+[CDE]+[AEF])&lt;/math&gt;.<br /> <br /> We attempt to find the area of each of the smaller triangles. <br /> <br /> <br /> Notice that &lt;math&gt;\frac{[BDF]}{[ABC]}=\frac{BF}{AB}\times \frac{BD}{BC}=\frac{r}{(r+1)(t+1)}&lt;/math&gt; using the ratios derived earlier.<br /> <br /> <br /> Similarly, &lt;math&gt;\frac{[CDE]}{[ABC]}=\frac{s}{(r+1)(s+1)}&lt;/math&gt; and &lt;math&gt;\frac{[AEF]}{[ABC]}=\frac{t}{(s+1)(t+1)}&lt;/math&gt;.<br /> <br /> <br /> Thus, &lt;math&gt;\frac{[BDF]+[CDE]+[AEF]}{[ABC]}=\frac{r}{(r+1)(t+1)}+\frac{s}{(r+1)(s+1)}+\frac{t}{(s+1)(t+1)}=\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}&lt;/math&gt;.<br /> <br /> Finally, we have &lt;math&gt;\frac{[DEF]}{[ABC]}=\boxed{1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}}&lt;/math&gt;.<br /> <br /> ~@CoolJupiter<br /> ==Proof 2==<br /> Proof 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; for &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y\in\mathbb{R}&lt;/math&gt;<br /> <br /> =Other Remarks=<br /> This theorem is a generalization of the Wooga Looga Theorem, which @RedFireTruck claims to have &quot;rediscovered&quot;. The link to the theorem can be found here:<br /> https://artofproblemsolving.com/wiki/index.php/Wooga_Looga_Theorem<br /> <br /> Essentially, Wooga Looga is a special case of this, specifically when &lt;math&gt;r=s=t&lt;/math&gt;.<br /> <br /> <br /> =Testimonials=<br /> The Ooga Booga Tribe would be proud of you. Amazing theorem - RedFireTruck<br /> <br /> This is Routh's theorem isn't it~ Ilovepizza2020</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=The_Devil%27s_Triangle&diff=136723 The Devil's Triangle 2020-11-06T14:53:29Z <p>Ilovepizza2020: </p> <hr /> <div>=Definition=<br /> For any triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, let &lt;math&gt;D, E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; be points on &lt;math&gt;BC, AC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; respectively. Devil's Triangle Theorem states that if &lt;math&gt;\frac{BD}{CD}=r, \frac{CE}{AE}=s&lt;/math&gt; and &lt;math&gt;\frac{AF}{BF}=t&lt;/math&gt;, then &lt;math&gt;\frac{[DEF]}{[ABC]}=1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}&lt;/math&gt;. <br /> <br /> =Proofs=<br /> ==Proof 1==<br /> Proof by CoolJupiter:<br /> <br /> We have the following ratios:<br /> &lt;math&gt;\frac{BD}{BC}=\frac{r}{r+1}, \frac{CD}{BC}=\frac{1}{r+1},\frac{CE}{AC}=\frac{s}{s+1}, \frac{AE}{AC}=\frac{1}{s+1},\frac{AF}{AB}=\frac{t}{t+1}, \frac{BF}{AB}=\frac{1}{t+1}&lt;/math&gt;.<br /> <br /> Now notice that &lt;math&gt;[DEF]=[ABC]-([BDF]+[CDE]+[AEF])&lt;/math&gt;.<br /> <br /> We attempt to find the area of each of the smaller triangles. <br /> <br /> <br /> Notice that &lt;math&gt;\frac{[BDF]}{[ABC]}=\frac{BF}{AB}\times \frac{BD}{BC}=\frac{r}{(r+1)(t+1)}&lt;/math&gt; using the ratios derived earlier.<br /> <br /> <br /> Similarly, &lt;math&gt;\frac{[CDE]}{[ABC]}=\frac{s}{(r+1)(s+1)}&lt;/math&gt; and &lt;math&gt;\frac{[AEF]}{[ABC]}=\frac{t}{(s+1)(t+1)}&lt;/math&gt;.<br /> <br /> <br /> Thus, &lt;math&gt;\frac{[BDF]+[CDE]+[AEF]}{[ABC]}=\frac{r}{(r+1)(t+1)}+\frac{s}{(r+1)(s+1)}+\frac{t}{(s+1)(t+1)}=\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}&lt;/math&gt;.<br /> <br /> Finally, we have &lt;math&gt;\frac{[DEF]}{[ABC]}=\boxed{1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}}&lt;/math&gt;.<br /> <br /> ~@CoolJupiter<br /> ==Proof 2==<br /> Proof by RedFireTruck:<br /> <br /> WLOG let &lt;math&gt;A=(0, 0), &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}$<br /> <br /> =Other Remarks=<br /> This theorem is a generalization of the Wooga Looga Theorem, which @RedFireTruck claims to have &quot;rediscovered&quot;. The link to the theorem can be found here:<br /> https://artofproblemsolving.com/wiki/index.php/Wooga_Looga_Theorem<br /> <br /> Essentially, Wooga Looga is a special case of this, specifically when &lt;math&gt;r=s=t&lt;/math&gt;.<br /> <br /> <br /> =Testimonials=<br /> The Ooga Booga Tribe would be proud of you. Amazing theorem - RedFireTruck<br /> This is Routh's theorem isn't it~ Ilovepizza2020</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=Wooga_Looga_Theorem&diff=136642 Wooga Looga Theorem 2020-11-06T00:03:57Z <p>Ilovepizza2020: </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 /> Created by the Ooga Booga Tribe of the Caveman Society, https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ<br /> =Proof=<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 /> =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=(\frac{7x+1}{8}, \frac{7y}{8})&lt;/math&gt;, &lt;math&gt;Y=(\frac{x}{8}, \frac{y}{8})&lt;/math&gt;, &lt;math&gt;Z=(\frac78, 0)&lt;/math&gt;. Then &lt;math&gt;[XYZ]=\frac12|\frac{43y}{64}|&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=(0,\tfrac 18,\tfrac 78),Y=(\tfrac 78,0,\tfrac 18),Z=(\tfrac18,\tfrac78,0)&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 /> or this solution by ilovepizza2020:<br /> <br /> We use the &lt;math&gt;\mathbf{FUNDEMENTAL~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 tests 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.<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 /> =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;72cm^2&lt;/math&gt;. If AD: DB = BE: EC =CF: FA =1: 5, find the area of triangle DEF<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 /> =Testimonials=<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 3rd 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)</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=Wooga_Looga_Theorem&diff=136641 Wooga Looga Theorem 2020-11-06T00:03:24Z <p>Ilovepizza2020: </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 /> Created by the Ooga Booga Tribe of the Caveman Society, https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ<br /> =Proof=<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 /> =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=(\frac{7x+1}{8}, \frac{7y}{8})&lt;/math&gt;, &lt;math&gt;Y=(\frac{x}{8}, \frac{y}{8})&lt;/math&gt;, &lt;math&gt;Z=(\frac78, 0)&lt;/math&gt;. Then &lt;math&gt;[XYZ]=\frac12|\frac{43y}{64}|&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=(0,\tfrac 18,\tfrac 78),Y=(\tfrac 78,0,\tfrac 18),Z=(\tfrac18,\tfrac78,0)&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 /> or this solution by ilovepizza2020:<br /> <br /> We use the &lt;math&gt;\mathbf{FUNDEMENTAL~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 tests 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.<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 /> =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;72cm^2&lt;/math&gt;. If AD: DB = BE: EC =CF: FA =1: 5, find the area of triangle DEF<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 /> =Testimonials=<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 3rd 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)</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=Wooga_Looga_Theorem&diff=136640 Wooga Looga Theorem 2020-11-06T00:02:48Z <p>Ilovepizza2020: </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 /> Created by the Ooga Booga Tribe of the Caveman Society, https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ<br /> =Proof=<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 /> =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=(\frac{7x+1}{8}, \frac{7y}{8})&lt;/math&gt;, &lt;math&gt;Y=(\frac{x}{8}, \frac{y}{8})&lt;/math&gt;, &lt;math&gt;Z=(\frac78, 0)&lt;/math&gt;. Then &lt;math&gt;[XYZ]=\frac12|\frac{43y}{64}|&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=(0,\tfrac 18,\tfrac 78),Y=(\tfrac 78,0,\tfrac 18),Z=(\tfrac18,\tfrac78,0)&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 /> or this solution by ilovepizza2020:<br /> <br /> We use the &lt;math&gt;\mathbf{FUNDEMENTAL~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 tests 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.<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 /> =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;72cm^2&lt;/math&gt;. If AD: DB = BE: EC =CF: FA =1: 5, find the area of triangle DEF<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 3=<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 /> =Testimonials=<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 3rd 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)</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=Wooga_Looga_Theorem&diff=136602 Wooga Looga Theorem 2020-11-05T17:19:51Z <p>Ilovepizza2020: </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 /> Created by the Ooga Booga Tribe of the Caveman Society, https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ<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=(\frac{7x+1}{8}, \frac{7y}{8})&lt;/math&gt;, &lt;math&gt;Y=(\frac{x}{8}, \frac{y}{8})&lt;/math&gt;, &lt;math&gt;Z=(\frac78, 0)&lt;/math&gt;. Then &lt;math&gt;[XYZ]=\frac12|\frac{43y}{64}|&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=(0,\tfrac 18,\tfrac 78),Y=(\tfrac 78,0,\tfrac 18),Z=(\tfrac18,\tfrac78,0)&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 /> or this solution by ilovepizza2020:<br /> <br /> We use the &lt;math&gt;\mathbf{FUNDEMENTAL~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 tests 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.<br /> <br /> ==Solution 6==<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 /> =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;72cm^2&lt;/math&gt;. If AD: DB = BE: EC =CF: FA =1: 5, find the area of triangle DEF<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 /> =Testimonials=<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 3rd 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! - Freezingpalmtree<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)</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=Wooga_Looga_Theorem&diff=136090 Wooga Looga Theorem 2020-10-30T00:10:07Z <p>Ilovepizza2020: </p> <hr /> <div>=Definition=<br /> If there is &lt;math&gt;\triangle ABC&lt;/math&gt; and points &lt;math&gt;X,Y,Z&lt;/math&gt; on the 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 mn&lt;/math&gt;, then the ratio &lt;math&gt;\frac{[XYZ]}{[ABC]}=\frac{m^2-m+n}{(m+n)^2}&lt;/math&gt;<br /> <br /> Created by the Ooga Booga Tribe of the Caveman Society, https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ<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=(\frac{7x+1}{8}, \frac{7y}{8})&lt;/math&gt;, &lt;math&gt;Y=(\frac{x}{8}, \frac{y}{8})&lt;/math&gt;, &lt;math&gt;Z=(\frac78, 0)&lt;/math&gt;. Then &lt;math&gt;[XYZ]=\frac12|\frac{43y}{64}|&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=(0,\tfrac 18,\tfrac 78),Y=(\tfrac 78,0,\tfrac 18),Z=(\tfrac18,\tfrac78,0)&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 /> or this solution by ilovepizza2020:<br /> <br /> We use the &lt;math&gt;\mathbf{FUNDEMENTAL~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 tests decided to ban it.)<br /> <br /> ==Solution 5==<br /> or this solution by eduD_looC:<br /> <br /> This is a perfect application of the Adhiytaha Anweopifanwpuiefhbavpwiuefnapveuihfnpvwheibfpanuwvfaw 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 /> =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 72cm2. If AD: DB = BE: EC =CF: FA =1: 5, find the area of triangle DEF<br /> <br /> ==Solution==<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 /> =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 /> =Testimonials=<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</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=Wooga_Looga_Theorem&diff=136080 Wooga Looga Theorem 2020-10-29T23:08:35Z <p>Ilovepizza2020: </p> <hr /> <div>=Definition=<br /> If there is &lt;math&gt;\triangle ABC&lt;/math&gt; and points &lt;math&gt;X,Y,Z&lt;/math&gt; on the 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 mn&lt;/math&gt;, then the ratio &lt;math&gt;\frac{[XYZ]}{[ABC]}=\frac{m^2-m+n}{(m+n)^2}&lt;/math&gt;<br /> <br /> Created by the Ooga Booga Tribe of the Caveman Society, https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ<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=(\frac{7x+1}{8}, \frac{7y}{8})&lt;/math&gt;, &lt;math&gt;Y=(\frac{x}{8}, \frac{y}{8})&lt;/math&gt;, &lt;math&gt;Z=(\frac78, 0)&lt;/math&gt;. Then &lt;math&gt;[XYZ]=\frac12|\frac{43y}{64}|&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=(0,\tfrac 18,\tfrac 78),Y=(\tfrac 78,0,\tfrac 18),Z=(\tfrac18,\tfrac78,0)&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 /> or this solution by ilovepizza2020:<br /> <br /> We use the &lt;math&gt;\mathbf{FUNDEMENTAL~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 tests decided to ban it.)<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 which area is 72cm2. If AD: DB = BE: EC =CF: FA =1: 5, find the area of triangle DEF<br /> <br /> ==Solution==<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 /> <br /> =Testimonials=<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 /> &quot;It is the best thing I have ever seen&quot; - Barack Obama, https://youtu.be/TSIAeHO3vxY</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=Wooga_Looga_Theorem&diff=136039 Wooga Looga Theorem 2020-10-29T15:50:17Z <p>Ilovepizza2020: </p> <hr /> <div>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 /> 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; and &lt;math&gt;X=(\frac{7x+1}{8}, \frac{7y}{8})&lt;/math&gt;, &lt;math&gt;Y=(\frac{x}{8}, \frac{y}{8})&lt;/math&gt;, &lt;math&gt;Z=(\frac78, 0)&lt;/math&gt;. Then &lt;math&gt;[XYZ]=\frac12|\frac{43y}{64}|&lt;/math&gt;. Therefore the answer is &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;.<br /> <br /> and that the solution to this problem by Matholic:<br /> <br /> The figure below shows a triangle ABC which area is 72cm2. If AD: DB = BE: EC =CF: FA =1: 5, find the area of triangle DEF<br /> <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 /> <br /> <br /> The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook.<br /> ~ilp2020</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_15&diff=133501 2012 AMC 10A Problems/Problem 15 2020-09-11T13:39:24Z <p>Ilovepizza2020: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=4;<br /> <br /> pair A=(0,0), B=(1,0); pair C=(0.8,-0.4);<br /> draw(A--(2,0)); draw((0,-1)--(2,-1)); draw((0,-2)--(1,-2));<br /> draw(A--(0,-2)); draw(B--(1,-2)); draw((2,0)--(2,-1));<br /> draw(A--(2,-1)); draw(B--(0,-2));<br /> <br /> pair[] ps={A,B,C};<br /> dot(ps);<br /> <br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,N);<br /> label(&quot;$C$&quot;,C,W);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}&lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> == Solution 1 ==<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=4;<br /> <br /> pair A=(0,0), B=(1,0); pair C=(0.8,-0.4);<br /> pair D=(1,-2), E=(0,-2);<br /> draw(A--(2,0)); draw((0,-1)--(2,-1)); draw(E--D);<br /> draw(A--E); draw(B--D); draw((2,0)--(2,-1));<br /> draw(A--(2,-1)); draw(B--E);<br /> <br /> pair[] ps={A,B,C,D,E};<br /> dot(ps);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,N);<br /> label(&quot;$C$&quot;,C,W);<br /> label(&quot;$D$&quot;,D,S);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$1$&quot;,(D--E),S);<br /> label(&quot;$1$&quot;,(A--B),N);<br /> label(&quot;$2$&quot;,(A--E),W);<br /> label(&quot;$\sqrt{5}$&quot;,(B--E),NW);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> &lt;math&gt;AC&lt;/math&gt; intersects &lt;math&gt;BC&lt;/math&gt; at a right angle, so &lt;math&gt;\triangle ABC \sim \triangle BED&lt;/math&gt;. The hypotenuse of right triangle BED is &lt;math&gt;\sqrt{1^2+2^2}=\sqrt{5}&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\frac{AC}{BC}=\frac{BD}{ED} \Rightarrow \frac{AC}{BC} = \frac21 \Rightarrow AC=2BC&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{AC}{AB}=\frac{BD}{BE} \Rightarrow \frac{AC}{1}=\frac{2}{\sqrt{5}} \Rightarrow AC=\frac{2}{\sqrt{5}}&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;AC=2BC&lt;/math&gt;, &lt;math&gt;BC=\frac{1}{\sqrt{5}}&lt;/math&gt;. &lt;math&gt;\triangle ABC&lt;/math&gt; is a right triangle so the area is just &lt;math&gt;\frac12 \cdot AC \cdot BC = \frac12 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \boxed{\textbf{(B)}\ \frac15}&lt;/math&gt;<br /> <br /> == Solution 2 (coordbash)==<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=4;<br /> <br /> pair A=(0,0), B=(1,0); pair C=(0.8,-0.4);<br /> pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0);<br /> draw(A--(2,0)); draw((0,-1)--F); draw(E--D);<br /> draw(A--E); draw(B--D); draw((2,0)--F);<br /> draw(A--F); draw(B--E); draw(C--G);<br /> <br /> pair[] ps={A,B,C,D,E,F,G};<br /> dot(ps);<br /> <br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,N);<br /> label(&quot;$C$&quot;,C,W);<br /> label(&quot;$D$&quot;,D,S);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$F$&quot;,F,E);<br /> label(&quot;$G$&quot;,G,N);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Let &lt;math&gt;\text{E}&lt;/math&gt; be the origin. Then,<br /> &lt;math&gt;\text{D}=(1, 0)&lt;/math&gt;<br /> &lt;math&gt;\text{A}=(0, 2)&lt;/math&gt;<br /> &lt;math&gt;\text{B}=(1, 2)&lt;/math&gt;<br /> &lt;math&gt;\text{F}=(2, 1)&lt;/math&gt;<br /> <br /> &lt;math&gt;{EB}&lt;/math&gt; can be represented by the line &lt;math&gt;y=2x&lt;/math&gt;<br /> Also, &lt;math&gt;{AF}&lt;/math&gt; can be represented by the line &lt;math&gt;y=-\frac{1}{2}x+2&lt;/math&gt;<br /> <br /> Subtracting the second equation from the first gives us &lt;math&gt;\frac{5}{2}x-2=0&lt;/math&gt;. <br /> Thus, &lt;math&gt;x=\frac{4}{5}&lt;/math&gt;.<br /> Plugging this into the first equation gives us &lt;math&gt;y=\frac{8}{5}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\text{C} (0.8, 1.6)&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt; is &lt;math&gt;(0.8, 2)&lt;/math&gt;,<br /> <br /> &lt;math&gt;{AB}=1&lt;/math&gt; and &lt;math&gt;{CG}=0.4&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot 0.4=0.2=\frac{1}{5}&lt;/math&gt;. The answer is &lt;math&gt;\boxed{\textbf{(B)}\ \frac15}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=4;<br /> <br /> pair A=(0,0), B=(1,0); pair C=(0.8,-0.4);<br /> pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0);<br /> pair H=(0,-1), I=(0.5,-1); <br /> draw(A--(2,0)); draw((0,-1)--F); draw(E--D);<br /> draw(A--E); draw(B--D); draw((2,0)--F);<br /> draw(A--F); draw(B--E); draw(C--G);<br /> draw(H--I);<br /> <br /> pair[] ps={A,B,C,D,E,F,G, H, I};<br /> dot(ps);<br /> <br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,N);<br /> label(&quot;$C$&quot;,C,W);<br /> label(&quot;$D$&quot;,D,S);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$F$&quot;,F,E);<br /> label(&quot;$G$&quot;,G,N);<br /> label(&quot;$H$&quot;,H,W);<br /> label(&quot;$I\$&quot;,I,E);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Triangle &lt;math&gt;EAB&lt;/math&gt; is similar to triangle &lt;math&gt;EHI&lt;/math&gt;; line &lt;math&gt;HI = 1/2&lt;/math&gt;<br /> <br /> Triangle &lt;math&gt;ACB&lt;/math&gt; is similar to triangle &lt;math&gt;FCI&lt;/math&gt; and the ratio of line &lt;math&gt;AB&lt;/math&gt; to line &lt;math&gt;IF = 1 : \frac{3}{2} = 2: 3&lt;/math&gt;.<br /> <br /> Based on similarity the length of the height of &lt;math&gt;GC&lt;/math&gt; is thus &lt;math&gt;\frac{2}{5}\cdot1 = \frac{2}{5}&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot \frac{2}{5}=\frac{1}{5}&lt;/math&gt;. <br /> The answer is &lt;math&gt;\boxed{\textbf{(B)}\ \frac15}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> <br /> <br /> <br /> <br /> <br /> <br /> Let &lt;math&gt;L&lt;/math&gt; be the point where the diagonal and the end of the unit square meet, on the right side of the diagram. Let &lt;math&gt;K&lt;/math&gt; be the top right corner of the top right unit square, where segment &lt;math&gt;ABL&lt;/math&gt; is 2 units in length.<br /> <br /> Because of the Pythagorean Theorem, since &lt;math&gt;AC = 2&lt;/math&gt; and &lt;math&gt;LK&lt;/math&gt; = 1, the diagonal of triangle &lt;math&gt;ALK&lt;/math&gt; is &lt;math&gt;\sqrt{5}&lt;/math&gt;. <br /> <br /> <br /> Triangle &lt;math&gt;ALK&lt;/math&gt; is clearly a similar triangle to triangle &lt;math&gt;ABC&lt;/math&gt;. Segment &lt;math&gt;AB&lt;/math&gt; is the hypotenuse of triangle &lt;math&gt;ABC&lt;/math&gt;. So, we can write down:<br /> <br /> &lt;cmath&gt;AK/AB = LK/BC&lt;/cmath&gt;, which is equal to:<br /> &lt;cmath&gt;\frac{\sqrt{5}}{1} = \frac{1}{BC}&lt;/cmath&gt; Solving this equation yields:<br /> <br /> &lt;cmath&gt;BC = \frac{1}{\sqrt{5}}&lt;/cmath&gt;<br /> <br /> By Pythagorean theorem, we can now find segment &lt;math&gt;AC&lt;/math&gt;<br /> &lt;cmath&gt;(\frac{1}{\sqrt{5}})^2 + AC^2 = 1&lt;/cmath&gt; Solving this yields:<br /> <br /> &lt;cmath&gt;AC^2 = \frac{4}{5}&lt;/cmath&gt;, so &lt;math&gt;AC = \frac{2}{\sqrt{5}}&lt;/math&gt;<br /> <br /> So then we can use &lt;cmath&gt;A = \frac{1}{2} * b * a.&lt;/cmath&gt;<br /> So &lt;cmath&gt;A = \frac{1}{2} * \frac{1}{\sqrt{5}} * \frac{2}{\sqrt{5}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \boxed{\textbf{(B)}\ \frac15}&lt;/cmath&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2012|ab=A|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_12&diff=132525 2004 AMC 12B Problems/Problem 12 2020-08-25T19:26:10Z <p>Ilovepizza2020: /* Solution 3 */</p> <hr /> <div>{{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #12]] and [[2004 AMC 10B Problems|2004 AMC 10B #19]]}}<br /> <br /> ==Problem==<br /> <br /> In the sequence &lt;math&gt;2001&lt;/math&gt;, &lt;math&gt;2002&lt;/math&gt;, &lt;math&gt;2003&lt;/math&gt;, &lt;math&gt;\ldots&lt;/math&gt; , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is &lt;math&gt;2001 + 2002 - 2003 = 2000&lt;/math&gt;. What is the<br /> &lt;math&gt;2004^\textrm{th}&lt;/math&gt; term in this sequence?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -2004 \qquad \mathrm{(B) \ } -2 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 4003 \qquad \mathrm{(E) \ } 6007 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> === Solution 1 ===<br /> <br /> We already know that &lt;math&gt;a_1=2001&lt;/math&gt;, &lt;math&gt;a_2=2002&lt;/math&gt;, &lt;math&gt;a_3=2003&lt;/math&gt;, and &lt;math&gt;a_4=2000&lt;/math&gt;. Let's compute the next few terms to get the idea how the sequence behaves. We get &lt;math&gt;a_5 = 2002+2003-2000 = 2005&lt;/math&gt;, &lt;math&gt;a_6=2003+2000-2005=1998&lt;/math&gt;, &lt;math&gt;a_7=2000+2005-1998=2007&lt;/math&gt;, and so on.<br /> <br /> We can now discover the following pattern: &lt;math&gt;a_{2k+1} = 2001+2k&lt;/math&gt; and &lt;math&gt;a_{2k}=2004-2k&lt;/math&gt;. This is easily proved by induction. It follows that &lt;math&gt;a_{2004}=a_{2\cdot 1002} = 2004 - 2\cdot 1002 = \boxed{0}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Note that the recurrence &lt;math&gt;a_n+a_{n+1}-a_{n+2}~=~a_{n+3}&lt;/math&gt; can be rewritten as &lt;math&gt;a_n+a_{n+1} ~=~ a_{n+2}+a_{n+3}&lt;/math&gt;. <br /> <br /> Hence we get that &lt;math&gt;a_1+a_2 ~=~ a_3+a_4 ~=~ a_5+a_6 ~= \cdots&lt;/math&gt; and also<br /> &lt;math&gt;a_2+a_3 ~=~ a_4+a_5 ~=~ a_6+a_7 ~= \cdots&lt;/math&gt;<br /> <br /> From the values given in the problem statement we see that &lt;math&gt;a_3=a_1+2&lt;/math&gt;. <br /> <br /> From &lt;math&gt;a_1+a_2 = a_3+a_4&lt;/math&gt; we get that &lt;math&gt;a_4=a_2-2&lt;/math&gt;.<br /> <br /> From &lt;math&gt;a_2+a_3 = a_4+a_5&lt;/math&gt; we get that &lt;math&gt;a_5=a_3+2&lt;/math&gt;.<br /> <br /> Following this pattern, we get &lt;math&gt;a_{2004} = a_{2002} - 2 = a_{2000} - 4 = \cdots = a_2 - 2002 = \boxed{0}&lt;/math&gt;.<br /> <br /> === Solution 3 (Sorta Overkill but whatever)===<br /> <br /> Our recurrence is &lt;math&gt;a_n+a_{n+1}-a_{n+2}~=~a_{n+3}&lt;/math&gt;, so we get &lt;math&gt;r^3+r^2-r-1 = 1&lt;/math&gt;, so &lt;math&gt;(r-1)(r+1)^2 = 1&lt;/math&gt;, so our formula for the recurrence is &lt;math&gt;a_n = A + (B + Cn)(-1)^n&lt;/math&gt;.<br /> <br /> Substituting our starting values gives us &lt;math&gt;a_n = 2002 + (2 - n)(-1)^n&lt;/math&gt;.<br /> <br /> So, &lt;math&gt;a_{2004} = 2002 - 2002 = 0.&lt;/math&gt;<br /> <br /> ~ ilovepizza2020<br /> <br /> == See also ==<br /> <br /> {{AMC12 box|year=2004|ab=B|num-b=11|num-a=13}}<br /> {{AMC10 box|year=2004|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_12&diff=132524 2004 AMC 12B Problems/Problem 12 2020-08-25T19:25:31Z <p>Ilovepizza2020: /* Solution 3 */</p> <hr /> <div>{{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #12]] and [[2004 AMC 10B Problems|2004 AMC 10B #19]]}}<br /> <br /> ==Problem==<br /> <br /> In the sequence &lt;math&gt;2001&lt;/math&gt;, &lt;math&gt;2002&lt;/math&gt;, &lt;math&gt;2003&lt;/math&gt;, &lt;math&gt;\ldots&lt;/math&gt; , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is &lt;math&gt;2001 + 2002 - 2003 = 2000&lt;/math&gt;. What is the<br /> &lt;math&gt;2004^\textrm{th}&lt;/math&gt; term in this sequence?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -2004 \qquad \mathrm{(B) \ } -2 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 4003 \qquad \mathrm{(E) \ } 6007 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> === Solution 1 ===<br /> <br /> We already know that &lt;math&gt;a_1=2001&lt;/math&gt;, &lt;math&gt;a_2=2002&lt;/math&gt;, &lt;math&gt;a_3=2003&lt;/math&gt;, and &lt;math&gt;a_4=2000&lt;/math&gt;. Let's compute the next few terms to get the idea how the sequence behaves. We get &lt;math&gt;a_5 = 2002+2003-2000 = 2005&lt;/math&gt;, &lt;math&gt;a_6=2003+2000-2005=1998&lt;/math&gt;, &lt;math&gt;a_7=2000+2005-1998=2007&lt;/math&gt;, and so on.<br /> <br /> We can now discover the following pattern: &lt;math&gt;a_{2k+1} = 2001+2k&lt;/math&gt; and &lt;math&gt;a_{2k}=2004-2k&lt;/math&gt;. This is easily proved by induction. It follows that &lt;math&gt;a_{2004}=a_{2\cdot 1002} = 2004 - 2\cdot 1002 = \boxed{0}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Note that the recurrence &lt;math&gt;a_n+a_{n+1}-a_{n+2}~=~a_{n+3}&lt;/math&gt; can be rewritten as &lt;math&gt;a_n+a_{n+1} ~=~ a_{n+2}+a_{n+3}&lt;/math&gt;. <br /> <br /> Hence we get that &lt;math&gt;a_1+a_2 ~=~ a_3+a_4 ~=~ a_5+a_6 ~= \cdots&lt;/math&gt; and also<br /> &lt;math&gt;a_2+a_3 ~=~ a_4+a_5 ~=~ a_6+a_7 ~= \cdots&lt;/math&gt;<br /> <br /> From the values given in the problem statement we see that &lt;math&gt;a_3=a_1+2&lt;/math&gt;. <br /> <br /> From &lt;math&gt;a_1+a_2 = a_3+a_4&lt;/math&gt; we get that &lt;math&gt;a_4=a_2-2&lt;/math&gt;.<br /> <br /> From &lt;math&gt;a_2+a_3 = a_4+a_5&lt;/math&gt; we get that &lt;math&gt;a_5=a_3+2&lt;/math&gt;.<br /> <br /> Following this pattern, we get &lt;math&gt;a_{2004} = a_{2002} - 2 = a_{2000} - 4 = \cdots = a_2 - 2002 = \boxed{0}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> Our recurrence is &lt;math&gt;a_n+a_{n+1}-a_{n+2}~=~a_{n+3}&lt;/math&gt;, so we get &lt;math&gt;r^3+r^2-r-1 = 1&lt;/math&gt;, so &lt;math&gt;(r-1)(r+1)^2 = 1&lt;/math&gt;, so our formula for the recurrence is &lt;math&gt;a_n = A + (B + Cn)(-1)^n&lt;/math&gt;.<br /> <br /> Substituting our starting values gives us &lt;math&gt;a_n = 2002 + (2 - n)(-1)^n&lt;/math&gt;.<br /> <br /> So, &lt;math&gt;a_{2004} = 2002 - 2002 = 0.&lt;/math&gt;<br /> <br /> ~ ilovepizza2020<br /> <br /> == See also ==<br /> <br /> {{AMC12 box|year=2004|ab=B|num-b=11|num-a=13}}<br /> {{AMC10 box|year=2004|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_12&diff=132523 2004 AMC 12B Problems/Problem 12 2020-08-25T19:25:10Z <p>Ilovepizza2020: /* Solution */</p> <hr /> <div>{{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #12]] and [[2004 AMC 10B Problems|2004 AMC 10B #19]]}}<br /> <br /> ==Problem==<br /> <br /> In the sequence &lt;math&gt;2001&lt;/math&gt;, &lt;math&gt;2002&lt;/math&gt;, &lt;math&gt;2003&lt;/math&gt;, &lt;math&gt;\ldots&lt;/math&gt; , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is &lt;math&gt;2001 + 2002 - 2003 = 2000&lt;/math&gt;. What is the<br /> &lt;math&gt;2004^\textrm{th}&lt;/math&gt; term in this sequence?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -2004 \qquad \mathrm{(B) \ } -2 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 4003 \qquad \mathrm{(E) \ } 6007 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> === Solution 1 ===<br /> <br /> We already know that &lt;math&gt;a_1=2001&lt;/math&gt;, &lt;math&gt;a_2=2002&lt;/math&gt;, &lt;math&gt;a_3=2003&lt;/math&gt;, and &lt;math&gt;a_4=2000&lt;/math&gt;. Let's compute the next few terms to get the idea how the sequence behaves. We get &lt;math&gt;a_5 = 2002+2003-2000 = 2005&lt;/math&gt;, &lt;math&gt;a_6=2003+2000-2005=1998&lt;/math&gt;, &lt;math&gt;a_7=2000+2005-1998=2007&lt;/math&gt;, and so on.<br /> <br /> We can now discover the following pattern: &lt;math&gt;a_{2k+1} = 2001+2k&lt;/math&gt; and &lt;math&gt;a_{2k}=2004-2k&lt;/math&gt;. This is easily proved by induction. It follows that &lt;math&gt;a_{2004}=a_{2\cdot 1002} = 2004 - 2\cdot 1002 = \boxed{0}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Note that the recurrence &lt;math&gt;a_n+a_{n+1}-a_{n+2}~=~a_{n+3}&lt;/math&gt; can be rewritten as &lt;math&gt;a_n+a_{n+1} ~=~ a_{n+2}+a_{n+3}&lt;/math&gt;. <br /> <br /> Hence we get that &lt;math&gt;a_1+a_2 ~=~ a_3+a_4 ~=~ a_5+a_6 ~= \cdots&lt;/math&gt; and also<br /> &lt;math&gt;a_2+a_3 ~=~ a_4+a_5 ~=~ a_6+a_7 ~= \cdots&lt;/math&gt;<br /> <br /> From the values given in the problem statement we see that &lt;math&gt;a_3=a_1+2&lt;/math&gt;. <br /> <br /> From &lt;math&gt;a_1+a_2 = a_3+a_4&lt;/math&gt; we get that &lt;math&gt;a_4=a_2-2&lt;/math&gt;.<br /> <br /> From &lt;math&gt;a_2+a_3 = a_4+a_5&lt;/math&gt; we get that &lt;math&gt;a_5=a_3+2&lt;/math&gt;.<br /> <br /> Following this pattern, we get &lt;math&gt;a_{2004} = a_{2002} - 2 = a_{2000} - 4 = \cdots = a_2 - 2002 = \boxed{0}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> Our recurrence is &lt;math&gt;a_n+a_{n+1}-a_{n+2}~=~a_{n+3}&lt;/math&gt;, so we get &lt;math&gt;r^3+r^2-r-1 = 1&lt;/math&gt;, so &lt;math&gt;(r-1)(r+1)^2 = 1&lt;/math&gt;, so our formula for the recurrence is &lt;math&gt;a_n = A + (B + Cn)(-1)^n&lt;/math&gt;.<br /> <br /> Substituting our starting values gives us &lt;math&gt;a_n = 2002 + (2 - n)(-1)^n&lt;/math&gt;.<br /> <br /> So, &lt;math&gt;a_{2004} = 2002 - 2002 = 0.&lt;/math&gt;<br /> <br /> == See also ==<br /> <br /> {{AMC12 box|year=2004|ab=B|num-b=11|num-a=13}}<br /> {{AMC10 box|year=2004|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Ilovepizza2020 https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=132150 User:Piphi 2020-08-19T16:26:18Z <p>Ilovepizza2020: /* User Count */</p> <hr /> <div>{{User:Piphi/Template:Header}}<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#dddddd&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;100px&quot;&gt;299&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;Piphi is legendary and made the USA IMO team in 2019.&lt;br&gt;<br /> <br /> Piphi is the creator of the [[User:Piphi/Games|AoPS Wiki Games by Piphi]], the future of games on AoPS.&lt;br&gt;<br /> <br /> Piphi started the signature trend at around May 2020.&lt;br&gt;<br /> <br /> Piphi is an extremely OP person - LJCoder619. &lt;br&gt;<br /> <br /> Piphi is OP --[[User:Aray10|Aray10]] ([[User talk:Aray10|talk]]) 23:22, 17 June 2020 (EDT) &lt;br&gt;<br /> <br /> Piphi has been very close to winning multiple [[Greed Control]] games, piphi placed 5th in game #18 and 2nd in game #19. Thanks to piphi, Greed Control games have started to be kept track of. Piphi made a spreadsheet that has all of Greed Control history [https://artofproblemsolving.com/community/c19451h2126208p15569802 here].&lt;br&gt;<br /> <br /> Piphi also found out who won [[Reaper]] games #1 and #2 as seen [https://artofproblemsolving.com/community/c19451h1826745p15526330 here].&lt;br&gt;<br /> <br /> Piphi has been called op by many AoPSers, including the legendary [[User:Radio2|Radio2]] himself [https://artofproblemsolving.com/community/c19451h1826745p15526800 here]. (note: Radio2 calls many users op.)&lt;br&gt;<br /> <br /> Piphi created the [[AoPS Administrators]] page, added most of the AoPS Admins to it, and created the scrollable table.&lt;br&gt;<br /> <br /> Piphi has also added a lot of the info that is in the [[Reaper Archives]].&lt;br&gt;<br /> <br /> Piphi has a side-project that is making the Wiki's [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]].&lt;br&gt;<br /> <br /> Piphi published Greed Control Game 19 statistics [https://artofproblemsolving.com/community/c19451h2126212 here].<br /> <br /> Piphi has a post that was made an announcement on a official AoPS Forum [https://artofproblemsolving.com/community/c68h2175116 here].<br /> <br /> Piphi is a proud member of [https://artofproblemsolving.com/community/c562043 The Interuniversal GMAAS Society].<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;<br /> You can check out more goals/statistics [[User:Piphi/Statistics|here]].<br /> <br /> A User Count of 330<br /> {{User:Piphi/Template:Progress_Bar|90.3|width=100%}}<br /> <br /> 200 subpages of [[User:Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|60.5|width=100%}}<br /> <br /> 200 signups for [[User:Piphi/Games|AoPS Wiki Games by Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|40|width=100%}}<br /> <br /> Make 10,000 edits<br /> {{User:Piphi/Template:Progress_Bar|18.45|width=100%}}&lt;/div&gt;<br /> &lt;/div&gt;</div> Ilovepizza2020