https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Rocketsri&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-14T00:06:04Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_13&diff=150148 2021 AIME II Problems/Problem 13 2021-03-24T21:53:25Z <p>Rocketsri: /* Solution */</p> <hr /> <div>==Problem==<br /> Find the least positive integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;2^n + 5^n - n&lt;/math&gt; is a multiple of &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> 1000 divides this expression iff 8, 125 both divide it. It should be fairly obvious that &lt;math&gt;n \geq 3&lt;/math&gt;; so we may break up the initial condition into two sub-conditions.<br /> <br /> (1) &lt;math&gt;5^n \equiv n \pmod{8}&lt;/math&gt;. Notice that the square of any odd integer is 1 modulo 8 (why?), so the LHS of this expression goes &quot;5,1,5,1, ...&quot;, while the RHS goes &quot;1,2,3,4,5,6,7,8,1, ...&quot;. The cycle length of the LHS is 2, RHS is 8, so the cycle length of the solution is &lt;math&gt;lcm(2,8)=8&lt;/math&gt;. Indeed, the n that solve this equation are exactly those such that &lt;math&gt;n \equiv 5 \pmod{8}&lt;/math&gt;.<br /> <br /> (2) &lt;math&gt;2^n \equiv n \pmod{125}&lt;/math&gt;. This is extremely computationally intensive if you try to calculate all &lt;math&gt;2^1,2^2, ..., 2^{100} \pmod{125}&lt;/math&gt;, so instead, we take a divide-and-conquer approach. In order for this expression to be true &lt;math&gt;2^n \equiv n \pmod{5}&lt;/math&gt; is necessary; it shouldn't take too long for you to go through the 20 possible LHS-RHS combinations and convince yourself that &lt;math&gt;n \equiv 3 \pmod{20}&lt;/math&gt; or &lt;math&gt;n \equiv 17 \pmod{20}&lt;/math&gt;. <br /> <br /> With this in mind we consider &lt;math&gt;2^n \equiv n&lt;/math&gt; modulo 25. By the generalized Fermat's little theorem, &lt;math&gt;2^{20} \equiv 1 \pmod{25}&lt;/math&gt;, but we already have n modulo 20! Our calculation is greatly simplified. The LHS cycle length is 20, RHS cycle length is 25, the lcm is 100, in this step we need to test all the numbers between 1 to 100 that &lt;math&gt;n \equiv 3 \pmod{20}&lt;/math&gt; or &lt;math&gt;n \equiv 17 \pmod{17}&lt;/math&gt;. In the case that &lt;math&gt;n \equiv 3 \pmod{20}&lt;/math&gt;, &lt;math&gt;2^n \equiv 2^3 \equiv 8&lt;/math&gt;, and RHS goes &quot;3,23,43,63,83&quot;; clearly &lt;math&gt;n \equiv 83 \pmod{100}&lt;/math&gt;. In the case that &lt;math&gt;n \equiv 17 \pmod{20}&lt;/math&gt;, by a similar argument, &lt;math&gt;n \equiv 97 \pmod{100}&lt;/math&gt;. <br /> <br /> In the final step, we need to calculate &lt;math&gt;2^{97}, 2^{83}&lt;/math&gt; modulo 125. The former is simply &lt;math&gt;2^{-3}&lt;/math&gt;; because &lt;math&gt;8*47=376=1&lt;/math&gt; modulo 125, &lt;math&gt;2^{97} \equiv 47&lt;/math&gt;. &lt;math&gt;2^{83}&lt;/math&gt; is &lt;math&gt;2^{-17}=2^{-16}*2^{-1}&lt;/math&gt;. &lt;math&gt;2^{-1}=63,2^{-2}=63^2=3969=-31,2^{-4}=(-31)^2=961=-39,2^{-8}=1521=21, 2^{-16}=441,2^{-17}=63*441=7*{-31}=-217=33&lt;/math&gt;. This time, LHS cycle is 100, RHS cycle is 125, so we need to figure out what is n modulo 500. It should be &lt;math&gt;n \equiv 283,297 \pmod{500}&lt;/math&gt;.<br /> <br /> Put everything together. By the second subcondition, the only candidates &lt; 100 are &lt;math&gt;283,297,782,797&lt;/math&gt;. Apply the first subcondition, n=797 is the desired answer.<br /> <br /> -Ross Gao<br /> <br /> ==Solution 2==<br /> We have that &lt;math&gt;2^n + 5^n \equiv n\pmod{1000}&lt;/math&gt;, or &lt;math&gt;2^n + 5^n \equiv n \pmod{8}&lt;/math&gt; and &lt;math&gt;2^n + 5^n \equiv n \pmod{125}&lt;/math&gt; by CRT. It is easy to check &lt;math&gt;n &lt; 3&lt;/math&gt; don't work, so we have that &lt;math&gt;n \geq 3&lt;/math&gt;. Then, &lt;math&gt;2^n \equiv 0 \pmod{8}&lt;/math&gt; and &lt;math&gt;5^n \equiv 0 \pmod{125}&lt;/math&gt;, so we just have &lt;math&gt;5^n \equiv n \pmod{8}&lt;/math&gt; and &lt;math&gt;2^n \equiv n \pmod{125}&lt;/math&gt;. Let us consider both of these congruences separately.<br /> <br /> <br /> First, we look at &lt;math&gt;5^n \equiv n \pmod{8}&lt;/math&gt;. By Euler's Totient Theorem (ETT), we have &lt;math&gt;5^4 \equiv 1 \pmod{8}&lt;/math&gt;, so &lt;math&gt;5^5 \equiv 5 \pmod{8}&lt;/math&gt;. On the RHS of the congruence, the possible values of &lt;math&gt;n&lt;/math&gt; are all nonnegative integers less than &lt;math&gt;8&lt;/math&gt; and on the RHS the only possible values are &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;. However, for &lt;math&gt;5^n&lt;/math&gt; to be &lt;math&gt;1 \pmod{8}&lt;/math&gt; we must have &lt;math&gt;n \equiv 0 \pmod{4}&lt;/math&gt;, a contradiction. So, the only possible values of &lt;math&gt;n&lt;/math&gt; are when &lt;math&gt;n \equiv 5 \pmod{8} \implies n = 8k+5&lt;/math&gt;.<br /> <br /> Now we look at &lt;math&gt;2^n \equiv n \pmod{125}&lt;/math&gt;. Plugging in &lt;math&gt;n = 8k+5&lt;/math&gt;, we get &lt;math&gt;2^{8k+5} \equiv 8k+5 \pmod{125} \implies 2^{8k} \cdot 32 \equiv 8k+5 \pmod{125}&lt;/math&gt;. Note, for &lt;math&gt;2^n \equiv n\pmod{125}&lt;/math&gt; to be satisfied, we must have &lt;math&gt;2^n \equiv n \pmod{5}&lt;/math&gt; and &lt;math&gt;2^n \equiv n\pmod{25}&lt;/math&gt;. Since &lt;math&gt;2^{8k} \equiv 1\pmod{5}&lt;/math&gt; as &lt;math&gt;8k \equiv 0\pmod{4}&lt;/math&gt;, we have &lt;math&gt;2 \equiv -2k \pmod{5} \implies k = 5m-1&lt;/math&gt;. Then, &lt;math&gt;n = 8(5m-1) + 5 = 40m-3&lt;/math&gt;. Now, we get &lt;math&gt;2^{40m-3} \equiv 40m-3 \pmod{125}&lt;/math&gt;. Using the fact that &lt;math&gt;2^n \equiv n\pmod{25}&lt;/math&gt;, we get &lt;math&gt;2^{-3} \equiv 15m-3 \pmod{25}&lt;/math&gt;. The inverse of &lt;math&gt;2&lt;/math&gt; modulo &lt;math&gt;25&lt;/math&gt; is obviously &lt;math&gt;13&lt;/math&gt;, so &lt;math&gt;2^{-3} \equiv 13^3 \equiv 22 \pmod{25}&lt;/math&gt;, so &lt;math&gt;15m \equiv 0 \pmod{25} \implies m = 5s&lt;/math&gt;. Plugging in &lt;math&gt;m = 5s&lt;/math&gt;, we get &lt;math&gt;n = 200s - 3&lt;/math&gt;.<br /> <br /> Now, we are finally ready to plug &lt;math&gt;n&lt;/math&gt; into the congruence modulo &lt;math&gt;125&lt;/math&gt;. Plugging in, we get &lt;math&gt;2^{200s-3} \equiv 200s - 3 \pmod{125}&lt;/math&gt;. By ETT, we get &lt;math&gt;2^{100} \equiv 1 \pmod{125}&lt;/math&gt;, so &lt;math&gt;2^{200s- 3} \equiv 2^{-3} \equiv 47 \pmod{125}&lt;/math&gt;. Then, &lt;math&gt;200s \equiv 50 \pmod{125} \implies s \equiv 4 \pmod{5} \implies s = 5y+4&lt;/math&gt;. Plugging this in, we get &lt;math&gt;n = 200(5y+4) - 3 = 1000y+797&lt;/math&gt;, implying the smallest value of &lt;math&gt;n&lt;/math&gt; is simply &lt;math&gt;\boxed{797}&lt;/math&gt;. ~rocketsri<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=II|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_13&diff=150147 2021 AIME II Problems/Problem 13 2021-03-24T21:53:05Z <p>Rocketsri: /* Solution */</p> <hr /> <div>==Problem==<br /> Find the least positive integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;2^n + 5^n - n&lt;/math&gt; is a multiple of &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> 1000 divides this expression iff 8, 125 both divide it. It should be fairly obvious that &lt;math&gt;n \geq 3&lt;/math&gt;; so we may break up the initial condition into two sub-conditions.<br /> <br /> (1) &lt;math&gt;5^n \equiv n \pmod{8}&lt;/math&gt;. Notice that the square of any odd integer is 1 modulo 8 (why?), so the LHS of this expression goes &quot;5,1,5,1, ...&quot;, while the RHS goes &quot;1,2,3,4,5,6,7,8,1, ...&quot;. The cycle length of the LHS is 2, RHS is 8, so the cycle length of the solution is &lt;math&gt;lcm(2,8)=8&lt;/math&gt;. Indeed, the n that solve this equation are exactly those such that &lt;math&gt;n \equiv 5 \pmod{8}&lt;/math&gt;.<br /> <br /> (2) &lt;math&gt;2^n \equiv n \pmod{125}&lt;/math&gt;. This is extremely computationally intensive if you try to calculate all &lt;math&gt;2^1,2^2, ..., 2^{100} \pmod{125}&lt;/math&gt;, so instead, we take a divide-and-conquer approach. In order for this expression to be true &lt;math&gt;2^n \equiv n \pmod{5}&lt;/math&gt; is necessary; it shouldn't take too long for you to go through the 20 possible LHS-RHS combinations and convince yourself that &lt;math&gt;n \equiv 3 \pmod{20}&lt;/math&gt; or &lt;math&gt;n \equiv 17 \pmod{20}&lt;/math&gt;. <br /> <br /> With this in mind we consider &lt;math&gt;2^n \equiv n&lt;/math&gt; modulo 25. By the generalized Fermat's little theorem, &lt;math&gt;2^{20} \equiv 1 \pmod{25}&lt;/math&gt;, but we already have n modulo 20! Our calculation is greatly simplified. The LHS cycle length is 20, RHS cycle length is 25, the lcm is 100, in this step we need to test all the numbers between 1 to 100 that &lt;math&gt;n \equiv 3 \pmod{20}&lt;/math&gt; or &lt;math&gt;n \equiv 17 \pmod{17}&lt;/math&gt;. In the case that &lt;math&gt;n \equiv 3 \pmod{20}&lt;/math&gt;, &lt;math&gt;2^n \equiv 2^3 \equiv 8&lt;/math&gt;, and RHS goes &quot;3,23,43,63,83&quot;; clearly &lt;math&gt;n \equiv 83 \pmod{100}&lt;/math&gt;. In the case that &lt;math&gt;n \equiv 17 \pmod{20}&lt;/math&gt;, by a similar argument, &lt;math&gt;n \equiv 97 \pmod{100}&lt;/math&gt;. <br /> <br /> In the final step, we need to calculate &lt;math&gt;2^{97}, 2^{83}&lt;/math&gt; modulo 125. The former is simply &lt;math&gt;2^{-3}&lt;/math&gt;; because &lt;math&gt;8*47=376=1&lt;/math&gt; modulo 125, &lt;math&gt;2^{97} \equiv 47&lt;/math&gt;. &lt;math&gt;2^{83}&lt;/math&gt; is &lt;math&gt;2^{-17}=2^{-16}*2^{-1}&lt;/math&gt;. &lt;math&gt;2^{-1}=63,2^{-2}=63^2=3969=-31,2^{-4}=(-31)^2=961=-39,2^{-8}=1521=21, 2^{-16}=441,2^{-17}=63*441=7*{-31}=-217=33&lt;/math&gt;. This time, LHS cycle is 100, RHS cycle is 125, so we need to figure out what is n modulo 500. It should be &lt;math&gt;n \equiv 283,297 \pmod{500}&lt;/math&gt;.<br /> <br /> Put everything together. By the second subcondition, the only candidates &lt; 100 are &lt;math&gt;283,297,782,797&lt;/math&gt;. Apply the first subcondition, n=797 is the desired answer.<br /> <br /> -Ross Gao<br /> <br /> ==Solution 2==<br /> We have that &lt;math&gt;2^n + 5^n \equiv n\pmod{1000}&lt;/math&gt;, or &lt;math&gt;2^n + 5^n \equiv n \pmod{8}&lt;/math&gt; and &lt;math&gt;2^n + 5^n \equiv n \pmod{125}&lt;/math&gt; by CRT. It is easy to check &lt;math&gt;n &lt; 3&lt;/math&gt; don't work, so we have that &lt;math&gt;n \geq 3&lt;/math&gt;. Then, &lt;math&gt;2^n \equiv 0 \pmod{8}&lt;/math&gt; and &lt;math&gt;5^n \equiv 0 \pmod{125}&lt;/math&gt;, so we just have &lt;math&gt;5^n \equiv n \pmod{8}&lt;/math&gt; and &lt;math&gt;2^n \equiv n \pmod{125}&lt;/math&gt;. Let us consider both of these congruences separately.<br /> <br /> <br /> First, we look at &lt;math&gt;5^n \equiv n \pmod{8}&lt;/math&gt;. By Euler's Totient Theorem (ETT), we have &lt;math&gt;5^4 \equiv 1 \pmod{8}&lt;/math&gt;, so &lt;math&gt;5^5 \equiv 5 \pmod{8}&lt;/math&gt;. On the RHS of the congruence, the possible values of &lt;math&gt;n&lt;/math&gt; are all nonnegative integers less than &lt;math&gt;8&lt;/math&gt; and on the RHS the only possible values are &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;. However, for &lt;math&gt;5^n&lt;/math&gt; to be &lt;math&gt;1 \pmod{8}&lt;/math&gt; we must have &lt;math&gt;n \equiv 0 \pmod{4}&lt;/math&gt;, a contradiction. So, the only possible values of &lt;math&gt;n&lt;/math&gt; are when &lt;math&gt;n \equiv 5 \pmod{8} \implies n = 8k+5&lt;/math&gt;.<br /> <br /> Now we look at &lt;math&gt;2^n \equiv n \pmod{125}&lt;/math&gt;. Plugging in &lt;math&gt;n = 8k+5&lt;/math&gt;, we get &lt;math&gt;2^{8k+5} \equiv 8k+5 \pmod{125} \implies 2^{8k} \cdot 32 \equiv 8k+5 \pmod{125}&lt;/math&gt;. Note, for &lt;math&gt;2^n \equiv n\pmod{125}&lt;/math&gt; to be satisfied, we must have &lt;math&gt;2^n \equiv n \pmod{5}&lt;/math&gt; and &lt;math&gt;2^n \equiv n\pmod{25}&lt;/math&gt;. Since &lt;math&gt;2^{8k} \equiv 1\pmod{5}&lt;/math&gt; as &lt;math&gt;8k \equiv 0\pmod{4}&lt;/math&gt;, we have &lt;math&gt;2 \equiv -2k \pmod{5} \implies k = 5m-1&lt;/math&gt;. Then, &lt;math&gt;n = 8(5m-1) + 5 = 40m-3&lt;/math&gt;. Now, we get &lt;math&gt;2^{40m-3} \equiv 40m-3 \pmod{125}&lt;/math&gt;. Using the fact that &lt;math&gt;2^n \equiv n\pmod{25}&lt;/math&gt;, we get &lt;math&gt;2^{-3} \equiv 15m-3 \pmod{25}&lt;/math&gt;. The inverse of &lt;math&gt;2&lt;/math&gt; modulo &lt;math&gt;25&lt;/math&gt; is obviously &lt;math&gt;13&lt;/math&gt;, so &lt;math&gt;2^{-3} \equiv 13^3 \equiv 22 \pmod{25}&lt;/math&gt;, so &lt;math&gt;15m \equiv 0 \pmod{25} \implies m = 5s&lt;/math&gt;. Plugging in &lt;math&gt;m = 5s&lt;/math&gt;, we get &lt;math&gt;n = 200s - 3&lt;/math&gt;.<br /> <br /> Now, we are finally ready to plug &lt;math&gt;n&lt;/math&gt; into the congruence modulo &lt;math&gt;125&lt;/math&gt;. Plugging in, we get &lt;math&gt;2^{200s-3} \equiv 200s - 3 \pmod{125}&lt;/math&gt;. By ETT, we get &lt;math&gt;2^{100} \equiv 1 \pmod{125}&lt;/math&gt;, so &lt;math&gt;2^{200s- 3} \equiv 2^{-3} \equiv 47 \pmod{125}&lt;/math&gt;. Then, &lt;math&gt;200s \equiv 50 \pmod{125} \implies s \equiv 4 \pmod{5} \implies s = 5y+4&lt;/math&gt;. Plugging this in, we get &lt;math&gt;n = 200(5y+4) - 3 = 1000y+797&lt;/math&gt;, implying the smallest value of &lt;math&gt;n&lt;/math&gt; is simply &lt;math&gt;\boxed{797}&lt;/math&gt;. ~rocketsri<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=II|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_4&diff=149233 2021 AIME I Problems/Problem 4 2021-03-12T15:22:08Z <p>Rocketsri: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> Find the number of ways &lt;math&gt;66&lt;/math&gt; identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.<br /> <br /> ==Solution 1==<br /> Suppose we have &lt;math&gt;1&lt;/math&gt; coin in the first pile. Then &lt;math&gt;(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)&lt;/math&gt; all work for a total of &lt;math&gt;31&lt;/math&gt; piles. Suppose we have &lt;math&gt;2&lt;/math&gt; coins in the first pile, then &lt;math&gt;(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)&lt;/math&gt; all work, for a total of &lt;math&gt;29&lt;/math&gt;. Continuing this pattern until &lt;math&gt;21&lt;/math&gt; coins in the first pile, we have the sum &lt;math&gt;31+29+28+26+25+\ldots+4+2+1=(31+28+25+22+\ldots+1)+(29+26+23+\ldots+2)=176+155=\boxed{331}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let the three piles have &lt;math&gt;a, b, c&lt;/math&gt; coins respectively. If we disregard order, then we just need to divide by &lt;math&gt;3! = 6&lt;/math&gt; at the end.<br /> <br /> We know &lt;math&gt;a + b + c = 66&lt;/math&gt;. Since &lt;math&gt;a, b, c&lt;/math&gt; are positive integers, there are &lt;math&gt;\binom{65}{2}&lt;/math&gt; ways from Stars and Bars.<br /> <br /> However, we must discard the cases where &lt;math&gt;a = b&lt;/math&gt; or &lt;math&gt;a = c&lt;/math&gt; or &lt;math&gt;b = c&lt;/math&gt;. The three cases are symmetric, so we just take the first case and multiply by 3. We have &lt;math&gt;2a + c = 66 \implies a = 1, 2, \dots 32&lt;/math&gt; for 32 solutions. Multiplying by 3, we will subtract 96 from our total.<br /> <br /> But we undercounted where &lt;math&gt;a = b = c = 22&lt;/math&gt;. This is first counted 1 time, then we subtract it 3 times, so we add it back twice. There is clearly only 1 way, for a total of 2.<br /> <br /> Hence, the answer is &lt;math&gt;\frac{\binom{65}{2} - 96 + 2}{6} = \boxed{331}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Let the piles have &lt;math&gt;a, b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; coins, with &lt;math&gt;0 &lt; a &lt; b &lt; c&lt;/math&gt;. Then, let &lt;math&gt;b = a + k_1&lt;/math&gt;, and &lt;math&gt;c = b + k_2&lt;/math&gt;, such that each &lt;math&gt;k_i \geq 1&lt;/math&gt;. The sum is then &lt;math&gt;a + a+k_1 + a+k_1+k_2 = 66 \implies 3a+2k_1 + k_2 = 66&lt;/math&gt;. This is simply the number of positive solutions to the equation &lt;math&gt;3x+2y+z = 66&lt;/math&gt;. Now, we take cases on &lt;math&gt;a&lt;/math&gt;. <br /> <br /> If &lt;math&gt;a = 1&lt;/math&gt;, then &lt;math&gt;2k_1 + k_2 = 63 \implies 1 \leq k_1 \leq 31&lt;/math&gt;. Each value of &lt;math&gt;k_1&lt;/math&gt; corresponds to a unique value of &lt;math&gt;k_2&lt;/math&gt;, so there are &lt;math&gt;31&lt;/math&gt; solutions in this case. Similarly, if &lt;math&gt;a = 2&lt;/math&gt;, then &lt;math&gt;2k_1 + k_2 = 60 \implies 1 \leq k_1 \leq 29&lt;/math&gt;, for a total of &lt;math&gt;29&lt;/math&gt; solutions in this case. If &lt;math&gt;a = 3&lt;/math&gt;, then &lt;math&gt;2k_1 + k_1 = 57 \implies 1 \leq k_1 \leq 28&lt;/math&gt;, for a total of &lt;math&gt;28&lt;/math&gt; solutions. In general, the number of solutions is just all the numbers that aren't a multiple of &lt;math&gt;3&lt;/math&gt;, that are less than or equal to &lt;math&gt;31&lt;/math&gt;. <br /> <br /> We then add our cases to get &lt;cmath&gt;1 + 2 + 4 + 5 + \cdots + 31 = 1 + 2 + 3 + \cdots + 31 - 3(1 + 2 + 3 + \cdots + 10) = \frac{31(32)}{2} - 3(55) = 31 \cdot 16 - 165 = 496 - 165 = \boxed{331}&lt;/cmath&gt; as our answer.<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_4&diff=149232 2021 AIME I Problems/Problem 4 2021-03-12T15:20:48Z <p>Rocketsri: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> Find the number of ways &lt;math&gt;66&lt;/math&gt; identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.<br /> <br /> ==Solution 1==<br /> Suppose we have &lt;math&gt;1&lt;/math&gt; coin in the first pile. Then &lt;math&gt;(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)&lt;/math&gt; all work for a total of &lt;math&gt;31&lt;/math&gt; piles. Suppose we have &lt;math&gt;2&lt;/math&gt; coins in the first pile, then &lt;math&gt;(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)&lt;/math&gt; all work, for a total of &lt;math&gt;29&lt;/math&gt;. Continuing this pattern until &lt;math&gt;21&lt;/math&gt; coins in the first pile, we have the sum &lt;math&gt;31+29+28+26+25+\ldots+4+2+1=(31+28+25+22+\ldots+1)+(29+26+23+\ldots+2)=176+155=\boxed{331}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let the three piles have &lt;math&gt;a, b, c&lt;/math&gt; coins respectively. If we disregard order, then we just need to divide by &lt;math&gt;3! = 6&lt;/math&gt; at the end.<br /> <br /> We know &lt;math&gt;a + b + c = 66&lt;/math&gt;. Since &lt;math&gt;a, b, c&lt;/math&gt; are positive integers, there are &lt;math&gt;\binom{65}{2}&lt;/math&gt; ways from Stars and Bars.<br /> <br /> However, we must discard the cases where &lt;math&gt;a = b&lt;/math&gt; or &lt;math&gt;a = c&lt;/math&gt; or &lt;math&gt;b = c&lt;/math&gt;. The three cases are symmetric, so we just take the first case and multiply by 3. We have &lt;math&gt;2a + c = 66 \implies a = 1, 2, \dots 32&lt;/math&gt; for 32 solutions. Multiplying by 3, we will subtract 96 from our total.<br /> <br /> But we undercounted where &lt;math&gt;a = b = c = 22&lt;/math&gt;. This is first counted 1 time, then we subtract it 3 times, so we add it back twice. There is clearly only 1 way, for a total of 2.<br /> <br /> Hence, the answer is &lt;math&gt;\frac{\binom{65}{2} - 96 + 2}{6} = \boxed{331}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Let the piles have &lt;math&gt;a, b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; coins, with &lt;math&gt;0 &lt; a &lt; b &lt; c&lt;/math&gt;. Then, let &lt;math&gt;b = a + k_1&lt;/math&gt;, and &lt;math&gt;c = b + k_2&lt;/math&gt;, such that each &lt;math&gt;k_i \geq 1&lt;/math&gt;. The sum is then &lt;math&gt;a + a+k_1 + a+k_1+k_2 = 66 \implies 3a+2k_1 + k_2 = 66&lt;/math&gt;. This is simply the number of positive solutions to the equation &lt;math&gt;3x+2y+z = 66&lt;/math&gt;. Now, we take cases on &lt;math&gt;a&lt;/math&gt;. <br /> <br /> If &lt;math&gt;a = 1&lt;/math&gt;, then &lt;math&gt;2k_1 + k_2 = 63 \implies 1 \leq k_1 \leq 31&lt;/math&gt;. Each value of &lt;math&gt;k_1&lt;/math&gt; corresponds to a unique value of &lt;math&gt;k_2&lt;/math&gt;, so there are &lt;math&gt;31&lt;/math&gt; solutions in this case. Similarly, if &lt;math&gt;a = 2&lt;/math&gt;, then &lt;math&gt;2k_1 + k_2 = 60 \implies 1 \leq k_1 \leq 29&lt;/math&gt;, for a total of &lt;math&gt;29&lt;/math&gt; solutions in this case. If &lt;math&gt;a = 3&lt;/math&gt;, then &lt;math&gt;2k_1 + k_1 = 57 \implies 1 \leq k_1 \leq 28&lt;/math&gt;, for a total of &lt;math&gt;28&lt;/math&gt; solutions. In general, the number of solutions is just all the numbers that aren't a multiple of &lt;math&gt;3&lt;/math&gt;, that are less than or equal to &lt;math&gt;31&lt;/math&gt;. <br /> <br /> We then add our cases to get&lt;math&gt;1 + 2 + 4 + 5 + \cdots + 31 = 1 + 2 + 3 + \cdots + 31 - 3(1 + 2 + 3 + \cdots + 10) = \frac{31(32)}{2} - 3(55) = 31 \cdot 16 - 165 = 496 - 165 = \boxed{331}&lt;/math&gt; solutions.<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_4&diff=149224 2021 AIME I Problems/Problem 4 2021-03-12T14:56:42Z <p>Rocketsri: </p> <hr /> <div>==Problem==<br /> Find the number of ways &lt;math&gt;66&lt;/math&gt; identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.<br /> <br /> ==Solution 1==<br /> Suppose we have &lt;math&gt;1&lt;/math&gt; coin in the first pile. Then &lt;math&gt;(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)&lt;/math&gt; all work for a total of &lt;math&gt;31&lt;/math&gt; piles. Suppose we have &lt;math&gt;2&lt;/math&gt; coins in the first pile, then &lt;math&gt;(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)&lt;/math&gt; all work, for a total of &lt;math&gt;29&lt;/math&gt;. Continuing this pattern until &lt;math&gt;21&lt;/math&gt; coins in the first pile, we have the sum &lt;math&gt;31+29+28+26+25+\ldots+4+2+1=(31+28+25+22+\ldots+1)+(29+26+23+\ldots+2)=176+155=\boxed{331}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let the three piles have &lt;math&gt;a, b, c&lt;/math&gt; coins respectively. If we disregard order, then we just need to divide by &lt;math&gt;3! = 6&lt;/math&gt; at the end.<br /> <br /> We know &lt;math&gt;a + b + c = 66&lt;/math&gt;. Since &lt;math&gt;a, b, c&lt;/math&gt; are positive integers, there are &lt;math&gt;\binom{65}{2}&lt;/math&gt; ways from Stars and Bars.<br /> <br /> However, we must discard the cases where &lt;math&gt;a = b&lt;/math&gt; or &lt;math&gt;a = c&lt;/math&gt; or &lt;math&gt;b = c&lt;/math&gt;. The three cases are symmetric, so we just take the first case and multiply by 3. We have &lt;math&gt;2a + c = 66 \implies a = 1, 2, \dots 32&lt;/math&gt; for 32 solutions. Multiplying by 3, we will subtract 96 from our total.<br /> <br /> But we undercounted where &lt;math&gt;a = b = c = 22&lt;/math&gt;. This is first counted 1 time, then we subtract it 3 times, so we add it back twice. There is clearly only 1 way, for a total of 2.<br /> <br /> Hence, the answer is &lt;math&gt;\frac{\binom{65}{2} - 96 + 2}{6} = \boxed{331}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_1&diff=149218 2021 AIME I Problems/Problem 1 2021-03-12T14:51:57Z <p>Rocketsri: /* Solution 2 (Casework but Bashier) */</p> <hr /> <div>==Problem==<br /> Zou and Chou are practicing their 100-meter sprints by running &lt;math&gt;6&lt;/math&gt; races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is &lt;math&gt;\frac23&lt;/math&gt; if they won the previous race but only &lt;math&gt;\frac13&lt;/math&gt; if they lost the previous race. The probability that Zou will win exactly &lt;math&gt;5&lt;/math&gt; of the &lt;math&gt;6&lt;/math&gt; races is &lt;math&gt;\frac mn&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> ==Solution 1(Casework)==<br /> For the last five races, Zou wins four and loses one. There are five possible outcome sequences, and we will proceed by casework:<br /> <br /> &lt;b&gt;&lt;u&gt;Case (1): Zou does not lose the last race.&lt;/u&gt;&lt;/b&gt;<br /> <br /> The probability that Zou loses a race is &lt;math&gt;\frac13,&lt;/math&gt; and the probability that he wins the following race is &lt;math&gt;\frac13.&lt;/math&gt; For each of the three other races, the probability that Zou wins is &lt;math&gt;\frac23.&lt;/math&gt;<br /> <br /> There are four such outcome sequences. The probability of one such sequence is &lt;math&gt;\left(\frac13\right)^2\left(\frac23\right)^3.&lt;/math&gt;<br /> <br /> &lt;b&gt;&lt;u&gt;Case (2): Zou loses the last race.&lt;/u&gt;&lt;/b&gt;<br /> <br /> The probability that Zou loses a race is &lt;math&gt;\frac13.&lt;/math&gt; For each of the four other races, the probability that Zou wins is &lt;math&gt;\frac23.&lt;/math&gt;<br /> <br /> There is one such outcome sequence. The probability is &lt;math&gt;\left(\frac13\right)^1\left(\frac23\right)^4.&lt;/math&gt;<br /> <br /> &lt;b&gt;&lt;u&gt;Answer&lt;/u&gt;&lt;/b&gt;<br /> <br /> The requested probability is &lt;cmath&gt;4\left(\frac13\right)^2\left(\frac23\right)^3+\left(\frac13\right)^1\left(\frac23\right)^4=\frac{32}{243}+\frac{16}{243}=\frac{48}{243}=\frac{16}{81},&lt;/cmath&gt; and the answer is &lt;math&gt;16+81=\boxed{097}.&lt;/math&gt;<br /> <br /> ~MRENTHUSIASM<br /> <br /> == Solution 2 (Casework but Bashier) ==<br /> We have &lt;math&gt;5&lt;/math&gt; cases, depending on which race Zou lost. Let &lt;math&gt;W&lt;/math&gt; denote a won race, and &lt;math&gt;L&lt;/math&gt; denote a lost race for Zou. The possible cases are &lt;math&gt;WWWWL, WWWLW, WWLWW, WLWWW, LWWWW&lt;/math&gt;. The first case has probability &lt;math&gt;\left(\frac{2}{3} \right)^4 \cdot \frac{1}{3} = \frac{16}{3^5}&lt;/math&gt;. The second case has probability &lt;math&gt;\left( \frac{2}{3} \right)^3 \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{8}{3^5}&lt;/math&gt;. The third has probability &lt;math&gt;\left( \frac{2}{3} \right)^2 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{8}{3^5}&lt;/math&gt;. The fourth has probability &lt;math&gt;\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^2 = \frac{8}{3^5}&lt;/math&gt;. Lastly, the fifth has probability &lt;math&gt;\frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^3 = \frac{8}{3^5}&lt;/math&gt;. Adding these up, the total probability is &lt;math&gt;\frac{16 + 8 \cdot 4}{3^5} = \frac{16 \cdot 3}{3^5} = \frac{16}{81}&lt;/math&gt;, so &lt;math&gt;m+n = \boxed{97}&lt;/math&gt;. ~rocketsri<br /> <br /> This is for if you're paranoid like me, and like to sometimes write out all of the cases if there are only a few.<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtube.com/watch?v=H17E9n2nIyY<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|before=First problem|num-a=2}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_1&diff=149217 2021 AIME I Problems/Problem 1 2021-03-12T14:43:57Z <p>Rocketsri: </p> <hr /> <div>==Problem==<br /> Zou and Chou are practicing their 100-meter sprints by running &lt;math&gt;6&lt;/math&gt; races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is &lt;math&gt;\frac23&lt;/math&gt; if they won the previous race but only &lt;math&gt;\frac13&lt;/math&gt; if they lost the previous race. The probability that Zou will win exactly &lt;math&gt;5&lt;/math&gt; of the &lt;math&gt;6&lt;/math&gt; races is &lt;math&gt;\frac mn&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> ==Solution 1(Casework)==<br /> For the last five races, Zou wins four and loses one. There are five possible outcome sequences, and we will proceed by casework:<br /> <br /> &lt;b&gt;&lt;u&gt;Case (1): Zou does not lose the last race.&lt;/u&gt;&lt;/b&gt;<br /> <br /> The probability that Zou loses a race is &lt;math&gt;\frac13,&lt;/math&gt; and the probability that he wins the following race is &lt;math&gt;\frac13.&lt;/math&gt; For each of the three other races, the probability that Zou wins is &lt;math&gt;\frac23.&lt;/math&gt;<br /> <br /> There are four such outcome sequences. The probability of one such sequence is &lt;math&gt;\left(\frac13\right)^2\left(\frac23\right)^3.&lt;/math&gt;<br /> <br /> &lt;b&gt;&lt;u&gt;Case (2): Zou loses the last race.&lt;/u&gt;&lt;/b&gt;<br /> <br /> The probability that Zou loses a race is &lt;math&gt;\frac13.&lt;/math&gt; For each of the four other races, the probability that Zou wins is &lt;math&gt;\frac23.&lt;/math&gt;<br /> <br /> There is one such outcome sequence. The probability is &lt;math&gt;\left(\frac13\right)^1\left(\frac23\right)^4.&lt;/math&gt;<br /> <br /> &lt;b&gt;&lt;u&gt;Answer&lt;/u&gt;&lt;/b&gt;<br /> <br /> The requested probability is &lt;cmath&gt;4\left(\frac13\right)^2\left(\frac23\right)^3+\left(\frac13\right)^1\left(\frac23\right)^4=\frac{32}{243}+\frac{16}{243}=\frac{48}{243}=\frac{16}{81},&lt;/cmath&gt; and the answer is &lt;math&gt;16+81=\boxed{097}.&lt;/math&gt;<br /> <br /> ~MRENTHUSIASM<br /> <br /> == Solution 2 (Casework but Bashier) ==<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtube.com/watch?v=H17E9n2nIyY<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|before=First problem|num-a=2}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_1&diff=149216 2021 AIME I Problems/Problem 1 2021-03-12T14:43:21Z <p>Rocketsri: /* Solution (Casework) */</p> <hr /> <div>==Problem==<br /> Zou and Chou are practicing their 100-meter sprints by running &lt;math&gt;6&lt;/math&gt; races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is &lt;math&gt;\frac23&lt;/math&gt; if they won the previous race but only &lt;math&gt;\frac13&lt;/math&gt; if they lost the previous race. The probability that Zou will win exactly &lt;math&gt;5&lt;/math&gt; of the &lt;math&gt;6&lt;/math&gt; races is &lt;math&gt;\frac mn&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> ==Solution 1(Casework)==<br /> For the last five races, Zou wins four and loses one. There are five possible outcome sequences, and we will proceed by casework:<br /> <br /> &lt;b&gt;&lt;u&gt;Case (1): Zou does not lose the last race.&lt;/u&gt;&lt;/b&gt;<br /> <br /> The probability that Zou loses a race is &lt;math&gt;\frac13,&lt;/math&gt; and the probability that he wins the following race is &lt;math&gt;\frac13.&lt;/math&gt; For each of the three other races, the probability that Zou wins is &lt;math&gt;\frac23.&lt;/math&gt;<br /> <br /> There are four such outcome sequences. The probability of one such sequence is &lt;math&gt;\left(\frac13\right)^2\left(\frac23\right)^3.&lt;/math&gt;<br /> <br /> &lt;b&gt;&lt;u&gt;Case (2): Zou loses the last race.&lt;/u&gt;&lt;/b&gt;<br /> <br /> The probability that Zou loses a race is &lt;math&gt;\frac13.&lt;/math&gt; For each of the four other races, the probability that Zou wins is &lt;math&gt;\frac23.&lt;/math&gt;<br /> <br /> There is one such outcome sequence. The probability is &lt;math&gt;\left(\frac13\right)^1\left(\frac23\right)^4.&lt;/math&gt;<br /> <br /> &lt;b&gt;&lt;u&gt;Answer&lt;/u&gt;&lt;/b&gt;<br /> <br /> The requested probability is &lt;cmath&gt;4\left(\frac13\right)^2\left(\frac23\right)^3+\left(\frac13\right)^1\left(\frac23\right)^4=\frac{32}{243}+\frac{16}{243}=\frac{48}{243}=\frac{16}{81},&lt;/cmath&gt; and the answer is &lt;math&gt;16+81=\boxed{097}.&lt;/math&gt;<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtube.com/watch?v=H17E9n2nIyY<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|before=First problem|num-a=2}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=148209 AMC historical results 2021-03-02T17:54:40Z <p>Rocketsri: /* AMC 12A */</p> <hr /> <div>&lt;!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --&gt;<br /> This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br /> ==2021==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: 103.5<br /> *Distinction: <br /> *Distinguished Honor Roll:<br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: <br /> *Distinction: <br /> *Distinguished Honor Roll:<br /> <br /> ===AMC 12A===<br /> *Average score:<br /> *AIME floor: 93<br /> *Distinction: <br /> *Distinguished Honor Roll:<br /> <br /> ===AMC 12B===<br /> <br /> *Average score: <br /> *AIME floor: <br /> *Distinction: <br /> *Distinguished Honor Roll:<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: <br /> *USAJMO cutoff:<br /> ===AIME II===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: <br /> *USAJMO cutoff:<br /> ===AMC 8===<br /> *Average score:<br /> *Honor Roll: <br /> *DHR:<br /> <br /> ==2020==<br /> ===AMC 10A===<br /> *Average score: 64.29<br /> *AIME floor: 103.5<br /> *Distinction: 105<br /> *Distinguished Honor Roll: 124.5<br /> <br /> ===AMC 10B===<br /> *Average score: 61.22<br /> *AIME floor: 102<br /> *Distinction: 103.5 <br /> *Distinguished Honor Roll: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 61.42<br /> *AIME floor: 87<br /> *Distinction: 100.5<br /> *Distinguished Honor Roll: 123<br /> <br /> ===AMC 12B===<br /> *Average score: 60.47<br /> *AIME floor: 87<br /> *Distinction: 97.5<br /> *Distinguished Honor Roll: 120<br /> <br /> ===AIME I===<br /> *Average score: 5.69<br /> *Median score: 6<br /> *USAMO cutoff: 233.5 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 229.5 (AMC 10A), 230 (AMC 10B)<br /> <br /> ===AIME II===<br /> Due to COVID-19, the 2020 AIME II was administered online and referred to as the AOIME.<br /> *Average score: 6.13<br /> *Median score: 6<br /> *USAMO cutoff: 234 (AMC 12A), 234.5 (AMC 12B)<br /> *USAJMO cutoff: 233.5 (AMC 10A), 229.5 (AMC 10B)<br /> <br /> ===AMC 8===<br /> *Average score: 10.00<br /> *Honor Roll: 18<br /> *DHR: 21<br /> <br /> ==2019==<br /> ===AMC 10A===<br /> *Average score: 51.66<br /> *Honor roll: 96<br /> *AIME floor: 103.5<br /> *DHR: 123<br /> <br /> ===AMC 10B===<br /> *Average score: 58.42<br /> *Honor roll: 102<br /> *AIME floor: 108<br /> *Distinguished Honor Roll: 121.5<br /> <br /> ===AMC 12A===<br /> *Average score: 49.22<br /> *AIME floor: 84<br /> *DHR: 121.5<br /> <br /> ===AMC 12B===<br /> *Average score: 56.73<br /> *AIME floor: 94.5 <br /> *DHR: 123<br /> <br /> ===AIME I===<br /> *Average score: 5.88<br /> *Median score: 6<br /> *USAMO cutoff: 220 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 209.5 (AMC 10A), 216 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 6.47<br /> *Median score: 6<br /> *USAMO cutoff: 230.5 (AMC 12A), 236 (AMC 12B)<br /> *USAJMO cutoff: 216.5 (AMC 10A), 220.5 (AMC 10B)<br /> <br /> ===AMC 8===<br /> *Average score: 9.43<br /> *Honor roll: 19<br /> *DHR: 23<br /> <br /> ==2018==<br /> ===AMC 10A===<br /> *Average score: 53.84<br /> *Honor roll: 100.5<br /> *AIME floor: 111<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 57.81<br /> *Honor roll: 97.5<br /> *AIME floor: 108<br /> *DHR: 123<br /> <br /> ===AMC 12A===<br /> *Average score: 56.36<br /> *AIME floor: 93<br /> *DHR: 120<br /> <br /> ===AMC 12B===<br /> *Average score: 57.85<br /> *AIME floor: 99<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.09<br /> *Median score: 5<br /> *USAMO cutoff: 215 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.48<br /> *Median score: 5<br /> *USAMO cutoff: 216 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br /> <br /> ===AMC 8===<br /> *Average score: 8.51<br /> *Honor roll: 15<br /> *DHR: 19<br /> <br /> ==2017==<br /> ===AMC 10A===<br /> *Average score: 59.33<br /> *AIME floor: 112.5<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 66.56<br /> *AIME floor: 120<br /> *DHR: 136.5<br /> <br /> ===AMC 12A===<br /> *Average score: 59.66<br /> *AIME floor: 96<br /> *DHR: 115.5<br /> <br /> ===AMC 12B===<br /> *Average score: 58.35<br /> *AIME floor: 100<br /> *DHR: 129<br /> <br /> ===AIME I===<br /> *Average score: 5.69<br /> *Median score: 5<br /> *USAMO cutoff: 225 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 224.5 (AMC 10A), 233 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.64<br /> *Median score: 5<br /> *USAMO cutoff: 221 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 219 (AMC 10A), 225 (AMC 10B)<br /> <br /> ===AMC 8===<br /> *Average score: 8.96<br /> *Honor roll: 17<br /> *DHR: 20<br /> <br /> ==2016==<br /> ===AMC 10A===<br /> *Average score: 65.31<br /> *AIME floor: 110<br /> *DHR: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 65.40<br /> *AIME floor: 110<br /> *DHR: 124.5<br /> <br /> ===AMC 12A===<br /> *Average score: 60.32<br /> *AIME floor: 93<br /> *DHR: 111<br /> <br /> ===AMC 12B===<br /> *Average score: 68.65<br /> *AIME floor: 100.5<br /> *DHR: 127.5<br /> <br /> ===AIME I===<br /> *Average score: 5.83<br /> *Median score: 6<br /> *USAMO cutoff: 220<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 4.43<br /> *Median score: 4<br /> *USAMO cutoff: 205<br /> *USAJMO cutoff: 200<br /> <br /> ===AMC 8===<br /> *Average score: 9.36<br /> *Honor roll: 18<br /> *DHR: 22<br /> <br /> ==2015==<br /> ===AMC 10A===<br /> *Average score: 73.39<br /> *AIME floor: 106.5<br /> *DHR: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.09<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 69.90<br /> *AIME floor: 99<br /> *DHR: 117<br /> <br /> ===AMC 12B===<br /> *Average score: 66.88<br /> *AIME floor: 100.5<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.29<br /> *Median score: 5<br /> *USAMO cutoff: 219.0<br /> *USAJMO cutoff: 213.0<br /> <br /> ===AIME II===<br /> *Average score: 6.63<br /> *Median score: 6<br /> *USAMO cutoff: 229.0<br /> *USAJMO cutoff: 223.5<br /> <br /> ===AMC 8===<br /> *Average score: 8.55<br /> *Honor roll: 16<br /> *DHR: 21<br /> <br /> ==2014==<br /> ===AMC 10A===<br /> *Average score: 63.34<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 10B===<br /> *Average score: 71.42<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 63.60<br /> *AIME floor: 93<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.12<br /> *AIME floor: 100.5<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 4.88<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AIME II===<br /> *Average score: 5.49<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AMC 8===<br /> *Average score: 11.43<br /> *Honor roll: 19<br /> *DHR: 23<br /> <br /> ==2013==<br /> ===AMC 10A===<br /> *Average score: 72.50<br /> *AIME floor: 108<br /> *DHR: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 72.62<br /> *AIME floor: 120<br /> *DHR: 129<br /> <br /> ===AMC 12A===<br /> *Average score: 65.06<br /> *AIME floor: 88.5<br /> *DHR: 106.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.21<br /> *AIME floor: 93<br /> *DHR: 108<br /> <br /> ===AIME I===<br /> *Average score: 4.69<br /> *Median score: 4<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 6.56<br /> *Median score: 6<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AMC 8===<br /> *Average score: 10.69<br /> *Honor roll: 18<br /> *DHR: 22<br /> <br /> ==2012==<br /> ===AMC 10A===<br /> *Average score: 72.51<br /> *AIME floor: 115.5<br /> *DHR: 121.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.59<br /> *AIME floor: 120<br /> *DHR: 133.5<br /> <br /> ===AMC 12A===<br /> *Average score: 64.62<br /> *AIME floor: 94.5<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 70.08<br /> *AIME floor: 99<br /> *DHR: 114<br /> <br /> ===AIME I===<br /> *Average score: 5.13<br /> *Median score: 5<br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AIME II===<br /> *Average score: 4.94<br /> *Median score: 5<br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AMC 8===<br /> *Average score: 10.67<br /> *Honor roll: 18<br /> *DHR: 22<br /> <br /> ==2011==<br /> ===AMC 10A===<br /> *Average score: 64.24<br /> *AIME floor: 117<br /> *DHR: 129<br /> <br /> ===AMC 10B===<br /> *Average score: 71.78<br /> *AIME floor: 117<br /> *DHR: 133.5<br /> <br /> ===AMC 12A===<br /> *Average score: 65.38<br /> *AIME floor: 93<br /> *DHR: 112.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.71<br /> *AIME floor: 97.5<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 2.23<br /> *Median score: 2<br /> *USAMO cutoff: 188<br /> *USAJMO cutoff: 179<br /> <br /> ===AIME II===<br /> *Average score: 5.47<br /> *Median score: 5<br /> *USAMO cutoff: 215.5<br /> *USAJMO cutoff: 196.5<br /> <br /> ===AMC 8===<br /> *Average score: 10.75<br /> *Honor roll: 17<br /> *DHR: 22<br /> <br /> ==2010==<br /> ===AMC 10A===<br /> *Average score: 68.11<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 68.57<br /> *AIME floor: 118.5<br /> <br /> ===AMC 12A===<br /> *Average score: 61.02<br /> *AIME floor: 88.5<br /> *DHR cutoff: 108<br /> <br /> ===AMC 12B===<br /> *Average score: 59.98<br /> *AIME floor: 88.5<br /> *DHR cutoff: 109.5<br /> <br /> ===AIME I===<br /> *Average score: 5.90<br /> *Median score: 6<br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AIME II===<br /> *Average score: 3.39<br /> *Median score: 3<br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AMC 8===<br /> *Average score: 9.59<br /> *Honor roll: 17<br /> *DHR: 22<br /> <br /> ==2009==<br /> ===AMC 10A===<br /> *Average score: 67.41<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 74.73<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 66.37<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 71.88<br /> *AIME floor: 100 (Top 5% (1.00))<br /> <br /> ===AIME I===<br /> *Average score: 4.17<br /> *Median score: 4<br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score: 3.27<br /> *Median score: 3<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Average score: 10.28<br /> *Honor roll: 17<br /> *DHR: 20<br /> <br /> ==2008==<br /> ===AMC 10A===<br /> *Average score: 60.25<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 65.6<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.9<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: 4.77<br /> *Median score: 4<br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score: 5.27<br /> *Median score: 5<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Average score: 11.45<br /> *Honor roll: 19<br /> *DHR: 22<br /> <br /> ==2007==<br /> <br /> ===AMC 10A===<br /> *Average score: 67.9<br /> *AIME floor: 117<br /> <br /> ===AMC 10B=== <br /> *Average score: 61.5<br /> *AIME floor: 115.5<br /> <br /> ===AMC 12A===<br /> *Average score: 66.8<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 73.1<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5<br /> *Median score: 3<br /> *USAMO floor: 6<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Average score: 9.87<br /> *Honor roll: 17<br /> *DHR: 21<br /> <br /> ==2006==<br /> ===AMC 10A===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 68.5<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 85.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 85.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score: <br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Honor roll: 17<br /> *DHR: 21<br /> <br /> ==2005==<br /> ===AMC 10A===<br /> *Average score: 74.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 78.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 83.4<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Honor roll: 16<br /> *DHR: 20<br /> <br /> ==2004==<br /> ===AMC 10A===<br /> *Average score: 69.1<br /> *AIME floor: 110<br /> <br /> ===AMC 10B===<br /> *Average score: 80.4<br /> *AIME floor: 115<br /> <br /> ===AMC 12A===<br /> *Average score: 73.9<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 84.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Honor roll: 17<br /> *DHR: 21<br /> <br /> ==2003==<br /> ===AMC 10A===<br /> *Average score: 74.4<br /> *AIME floor: 119<br /> <br /> ===AMC 10B===<br /> *Average score: 79.6<br /> *AIME floor: 121<br /> <br /> ===AMC 12A===<br /> *Average score: 77.8<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 76.6<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Honor roll: 18<br /> *DHR: 22<br /> <br /> ==2002==<br /> ===AMC 10A===<br /> *Average score: 68.5<br /> *AIME floor: 115<br /> <br /> ===AMC 10B===<br /> *Average score: 74.9<br /> *AIME floor: 118<br /> <br /> ===AMC 12A===<br /> *Average score: 72.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 80.8<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2001==<br /> ===AMC 10===<br /> *Average score: 67.8<br /> *AIME floor: 116<br /> <br /> ===AMC 12===<br /> *Average score: 56.6<br /> *AIME floor: 84<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2000==<br /> ===AMC 10===<br /> *Average score: &lt;math&gt;64.2&lt;/math&gt;<br /> *AIME floor: &lt;math&gt;110&lt;/math&gt;<br /> <br /> ===AMC 12===<br /> *Average score: &lt;math&gt;64.9&lt;/math&gt;<br /> *AIME floor: &lt;math&gt;92&lt;/math&gt;<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1999==<br /> ===AHSME===<br /> *Average score: &lt;math&gt;68.8&lt;/math&gt;<br /> *AIME floor:<br /> <br /> ===AIME===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_24&diff=146471 2021 AMC 10A Problems/Problem 24 2021-02-13T18:56:29Z <p>Rocketsri: /* Solution 3 (Geometry) */</p> <hr /> <div>==Problem==<br /> The interior of a quadrilateral is bounded by the graphs of &lt;math&gt;(x+ay)^2 = 4a^2&lt;/math&gt; and &lt;math&gt;(ax-y)^2 = a^2&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; a positive real number. What is the area of this region in terms of &lt;math&gt;a&lt;/math&gt;, valid for all &lt;math&gt;a &gt; 0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{8a^2}{(a+1)^2}\qquad\textbf{(B)} ~\frac{4a}{a+1}\qquad\textbf{(C)} ~\frac{8a}{a+1}\qquad\textbf{(D)} ~\frac{8a^2}{a^2+1}\qquad\textbf{(E)} ~\frac{8a}{a^2+1}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The conditions &lt;math&gt;(x+ay)^2 = 4a^2&lt;/math&gt; and &lt;math&gt;(ax-y)^2 = a^2&lt;/math&gt; give &lt;math&gt;|x+ay| = |2a|&lt;/math&gt; and &lt;math&gt;|ax-y| = |a|&lt;/math&gt; or &lt;math&gt;x+ay = \pm 2a&lt;/math&gt; and &lt;math&gt;ax-y = \pm a&lt;/math&gt;. The slopes here are perpendicular, so the quadrilateral is a rectangle.<br /> Plug in &lt;math&gt;a=1&lt;/math&gt; and graph it. We quickly see that the area is &lt;math&gt;2\sqrt{2} \cdot \sqrt{2} = 4&lt;/math&gt;, so the answer can't be &lt;math&gt;A&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt; by testing the values they give (test it!). Now plug in &lt;math&gt;a=2&lt;/math&gt;. We see using a ruler that the sides of the rectangle are about &lt;math&gt;\frac74&lt;/math&gt; and &lt;math&gt;\frac72&lt;/math&gt;. So the area is about &lt;math&gt;\frac{49}8 = 6.125&lt;/math&gt;. Testing &lt;math&gt;C&lt;/math&gt; we get &lt;math&gt;\frac{16}3&lt;/math&gt; which is clearly less than &lt;math&gt;6&lt;/math&gt;, so it is out. Testing &lt;math&gt;D&lt;/math&gt; we get &lt;math&gt;\frac{32}5&lt;/math&gt; which is near our answer, so we leave it. Testing &lt;math&gt;E&lt;/math&gt; we get &lt;math&gt;\frac{16}5&lt;/math&gt;, way less than &lt;math&gt;6&lt;/math&gt;, so it is out. So, the only plausible answer is &lt;math&gt;\boxed{D}&lt;/math&gt; ~firebolt360<br /> <br /> ==Solution 2 (Casework)==<br /> For the equation &lt;math&gt;(x+ay)^2 = 4a^2,&lt;/math&gt; the cases are <br /> <br /> (1) &lt;math&gt;x+ay=2a.&lt;/math&gt; This is a line with &lt;math&gt;x&lt;/math&gt;-intercept &lt;math&gt;2a&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;-intercept &lt;math&gt;2&lt;/math&gt;, and slope &lt;math&gt;-\frac 1a.&lt;/math&gt;<br /> <br /> (2) &lt;math&gt;x+ay=-2a.&lt;/math&gt; This is a line with &lt;math&gt;x&lt;/math&gt;-intercept &lt;math&gt;-2a&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;-intercept &lt;math&gt;-2&lt;/math&gt;, and slope &lt;math&gt;-\frac 1a.&lt;/math&gt;<br /> <br /> For the equation &lt;math&gt;(ax-y)^2 = a^2,&lt;/math&gt; the cases are<br /> <br /> (1)* &lt;math&gt;ax-y=a.&lt;/math&gt; This is a line with &lt;math&gt;x&lt;/math&gt;-intercept &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;-intercept &lt;math&gt;-a&lt;/math&gt;, and slope &lt;math&gt;a.&lt;/math&gt;<br /> <br /> (2)* &lt;math&gt;ax-y=-a.&lt;/math&gt; This is a line with &lt;math&gt;x&lt;/math&gt;-intercept &lt;math&gt;-1&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;-intercept &lt;math&gt;a&lt;/math&gt;, and slope &lt;math&gt;a.&lt;/math&gt;<br /> <br /> Since the slopes of the intersecting lines (from the 4 above equations) are negative reciprocals, the quadrilateral is a rectangle.<br /> <br /> Plugging &lt;math&gt;a=2&lt;/math&gt; into the 4 above equations and solving systems of equations for intersecting lines [(1) and (1)*, (1) and (2)*, (2) and (1)*, (2) and (2)*], we get the respective solutions &lt;cmath&gt;(x,y)=\left(\frac 85, \frac 65\right), (0,2), (0,-2), \left(-\frac 85, -\frac 65\right).&lt;/cmath&gt; <br /> Finally, by the Distance Formula, the length and width of the rectangle are &lt;math&gt;\frac{8\sqrt5}{5}&lt;/math&gt; and &lt;math&gt;\frac{4\sqrt5}{5}.&lt;/math&gt; The area we seek is &lt;cmath&gt;\left(\frac{8\sqrt5}{5}\right)\left(\frac{4\sqrt5}{5}\right)=\frac{32}{5}.&lt;/cmath&gt;<br /> Plugging &lt;math&gt;a=2&lt;/math&gt; into the choices gives <br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{32}{9}\qquad\textbf{(B)} ~\frac{8}{3}\qquad\textbf{(C)} ~\frac{16}{3}\qquad\textbf{(D)} ~\frac{32}{5}\qquad\textbf{(E)} ~\frac{16}{5}&lt;/math&gt;<br /> <br /> The answer is &lt;math&gt;\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.&lt;/math&gt;<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 3 (Geometry)==<br /> Similar to Solution 2, we will use the equations of the four cases:<br /> <br /> (1) &lt;math&gt;x+ay=2a.&lt;/math&gt; This is a line with &lt;math&gt;x&lt;/math&gt;-intercept &lt;math&gt;2a&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;-intercept &lt;math&gt;2&lt;/math&gt;, and slope &lt;math&gt;-\frac 1a.&lt;/math&gt;<br /> <br /> (2) &lt;math&gt;x+ay=-2a.&lt;/math&gt; This is a line with &lt;math&gt;x&lt;/math&gt;-intercept &lt;math&gt;-2a&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;-intercept &lt;math&gt;-2&lt;/math&gt;, and slope &lt;math&gt;-\frac 1a.&lt;/math&gt;<br /> <br /> (3)* &lt;math&gt;ax-y=a.&lt;/math&gt; This is a line with &lt;math&gt;x&lt;/math&gt;-intercept &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;-intercept &lt;math&gt;-a&lt;/math&gt;, and slope &lt;math&gt;a.&lt;/math&gt;<br /> <br /> (4)* &lt;math&gt;ax-y=-a.&lt;/math&gt; This is a line with &lt;math&gt;x&lt;/math&gt;-intercept &lt;math&gt;-1&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;-intercept &lt;math&gt;a&lt;/math&gt;, and slope &lt;math&gt;a.&lt;/math&gt;<br /> <br /> The area of the rectangle created by the four equations can be written as &lt;math&gt;2a\cdot \sin A\cdot4\sin A&lt;/math&gt;<br /> <br /> = &lt;math&gt;8a(\sin A)^2&lt;/math&gt;<br /> <br /> = &lt;math&gt;8a(~\frac{1}{\sqrt(a^2+1)})^2&lt;/math&gt;<br /> <br /> = &lt;math&gt;\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.&lt;/math&gt;<br /> <br /> (Note: &lt;math&gt;\tan A=&lt;/math&gt; slope &lt;math&gt;a&lt;/math&gt;)<br /> <br /> -fnothing4994<br /> <br /> == Video Solution by OmegaLearn (System of Equations and Shoelace Formula) ==<br /> https://youtu.be/2iohPYkZpkQ<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_23&diff=146469 2021 AMC 10B Problems/Problem 23 2021-02-13T18:26:51Z <p>Rocketsri: /* Solution */</p> <hr /> <div>==Problem==<br /> A square with side length &lt;math&gt;8&lt;/math&gt; is colored white except for &lt;math&gt;4&lt;/math&gt; black isosceles right triangular regions with legs of length &lt;math&gt;2&lt;/math&gt; in each corner of the square and a black diamond with side length &lt;math&gt;2\sqrt{2}&lt;/math&gt; in the center of the square, as shown in the diagram. A circular coin with diameter &lt;math&gt;1&lt;/math&gt; is dropped onto the square and lands in a random location where the coin is completely contained within the square. The probability that the coin will cover part of the black region of the square can be written as &lt;math&gt;\frac{1}{196}(a+b\sqrt{2}+\pi)&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b&lt;/math&gt;?<br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> draw((0,0)--(8,0)--(8,8)--(0,8)--(0,0));<br /> fill((2,0)--(0,2)--(0,0)--cycle, black);<br /> fill((6,0)--(8,0)--(8,2)--cycle, black);<br /> fill((8,6)--(8,8)--(6,8)--cycle, black);<br /> fill((0,6)--(2,8)--(0,8)--cycle, black);<br /> fill((4,6)--(2,4)--(4,2)--(6,4)--cycle, black);<br /> filldraw(circle((2.6,3.31),0.47),gray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~64 \qquad\textbf{(B)} ~66 \qquad\textbf{(C)} ~68 \qquad\textbf{(D)} ~70 \qquad\textbf{(E)} ~72&lt;/math&gt;<br /> <br /> <br /> == Solution ==<br /> <br /> To find the probability, we look at the &lt;math&gt;\frac{\text{success region}}{\text{total possible region}}&lt;/math&gt;. For the coin to be completely contained within the square, we must have the distance from the center of the coin to a side of the square to be at least &lt;math&gt;\frac{1}{2}&lt;/math&gt;, as it's the radius of the coin. This implies the &lt;math&gt;\text{total possible region}&lt;/math&gt; is a square with side length &lt;math&gt;8 - \frac{1}{2} - \frac{1}{2} = 7&lt;/math&gt;, with an area of &lt;math&gt;49&lt;/math&gt;. Now, we consider cases on where needs to land to partially cover a black region.<br /> <br /> &lt;b&gt; Near The Center Square &lt;/b&gt;<br /> <br /> We can have the center of the coin land within &lt;math&gt;\frac{1}{2}&lt;/math&gt; of the center square, or inside of the center square. We have that the center lands either outside of the square, or inside. So, we have a region with &lt;math&gt;\frac{1}{2}&lt;/math&gt; emanating from every point on the exterior of the square, forming &lt;math&gt;4&lt;/math&gt; quarter circles and &lt;math&gt;4&lt;/math&gt; rectangles. The &lt;math&gt;4&lt;/math&gt; quarter circles combine to make a full circle, with radius of &lt;math&gt;\frac{1}{2}&lt;/math&gt;, so that has an area of &lt;math&gt;\frac{\pi}{4}&lt;/math&gt;. The area of a rectangle is &lt;math&gt;2 \sqrt 2 \cdot \frac{1}{2} = \sqrt 2&lt;/math&gt;, so &lt;math&gt;4&lt;/math&gt; of them combine to an area of &lt;math&gt;4 \sqrt 2&lt;/math&gt;. The area of the black square is simply &lt;math&gt;(2\sqrt 2)^2 = 8&lt;/math&gt;. So, for this case, we have a combined total of &lt;math&gt;8 + 4\sqrt 2 + \frac{\pi}{4}&lt;/math&gt;. Onto the second (and last) case.<br /> <br /> &lt;b&gt; Near A Triangle &lt;/b&gt;<br /> <br /> We can also have the coin land within &lt;math&gt;\frac{1}{2}&lt;/math&gt; of one of the triangles. By symmetry, we can just find the successful region for one of them, then multiply by &lt;math&gt;4&lt;/math&gt;. Consider this diagram. We can draw in an altitude from the bottom corner of the square to hit the hypotenuse of the blue triangle. The length of this when passing through the black region is &lt;math&gt;\sqrt 2&lt;/math&gt;, and when passing through the white region (while being contained in the blue triangle) is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. However, we have to subtract off when it doesn't pass through the red square. Then, it's the hypotenuse of a small isosceles right triangle with side lengths of &lt;math&gt;\frac{1}{2}&lt;/math&gt;, or &lt;math&gt;\frac{\sqrt 2}{2}&lt;/math&gt;. So, our altitude of the blue triangle is &lt;math&gt;\sqrt 2 + \frac{1}{2} - \frac{\sqrt 2}{2} = \frac{\sqrt 2 + 1}{2}&lt;/math&gt;. Then, recall, the area of an isosceles right triangle is &lt;math&gt;h^2&lt;/math&gt;, where &lt;math&gt;h&lt;/math&gt; is the altitude from the right angle. So, squaring this, we get &lt;math&gt;\frac{3 + 2\sqrt 2}{4}&lt;/math&gt;. Now, we have to multiply this by &lt;math&gt;4&lt;/math&gt; to account for all of the black triangles, to get &lt;math&gt;3 + 2\sqrt 2&lt;/math&gt; as the final area for this case. <br /> <br /> <br /> &lt;b&gt; Finishing &lt;/b&gt;<br /> <br /> Then, to have the coin touching a black region, we add up the area of our successful regions, or &lt;math&gt;8 + 4\sqrt 2 + \frac{\pi}{4} + 3 + 2\sqrt 2 = 11 + 6\sqrt 2 + \frac{\pi}{4} = \frac{44 + 24\sqrt 2 + \pi}{4}&lt;/math&gt;. The total region is &lt;math&gt;49&lt;/math&gt;, so our probability is &lt;math&gt;\frac{\frac{44 + 24\sqrt 2 + \pi}{4}}{49} = \frac{44 + 24\sqrt 2 + \pi}{196}&lt;/math&gt;, which implies &lt;math&gt;a+b = 44+24 = 68&lt;/math&gt;. This corresponds to answer choice &lt;math&gt;\boxed{\textbf{(A)} ~68}&lt;/math&gt;. ~rocketsri<br /> <br /> == Video Solution by OmegaLearn (Similar Triangles and Area Calculations) ==<br /> https://youtu.be/-UvivZ0UA1U<br /> <br /> ~ pi_is_3.14<br /> <br /> <br /> <br /> {{AMC10 box|year=2021|ab=B|num-b=22|num-a=24}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=File:2021_AMC_10B_-23_Solution_2.png&diff=146468 File:2021 AMC 10B -23 Solution 2.png 2021-02-13T18:25:01Z <p>Rocketsri: </p> <hr /> <div></div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_23&diff=146467 2021 AMC 10B Problems/Problem 23 2021-02-13T18:20:45Z <p>Rocketsri: /* Solution */</p> <hr /> <div>==Problem==<br /> A square with side length &lt;math&gt;8&lt;/math&gt; is colored white except for &lt;math&gt;4&lt;/math&gt; black isosceles right triangular regions with legs of length &lt;math&gt;2&lt;/math&gt; in each corner of the square and a black diamond with side length &lt;math&gt;2\sqrt{2}&lt;/math&gt; in the center of the square, as shown in the diagram. A circular coin with diameter &lt;math&gt;1&lt;/math&gt; is dropped onto the square and lands in a random location where the coin is completely contained within the square. The probability that the coin will cover part of the black region of the square can be written as &lt;math&gt;\frac{1}{196}(a+b\sqrt{2}+\pi)&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b&lt;/math&gt;?<br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> draw((0,0)--(8,0)--(8,8)--(0,8)--(0,0));<br /> fill((2,0)--(0,2)--(0,0)--cycle, black);<br /> fill((6,0)--(8,0)--(8,2)--cycle, black);<br /> fill((8,6)--(8,8)--(6,8)--cycle, black);<br /> fill((0,6)--(2,8)--(0,8)--cycle, black);<br /> fill((4,6)--(2,4)--(4,2)--(6,4)--cycle, black);<br /> filldraw(circle((2.6,3.31),0.47),gray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~64 \qquad\textbf{(B)} ~66 \qquad\textbf{(C)} ~68 \qquad\textbf{(D)} ~70 \qquad\textbf{(E)} ~72&lt;/math&gt;<br /> <br /> <br /> == Solution ==<br /> <br /> To find the probability, we look at the &lt;math&gt;\frac{\text{success region}}{\text{total possible region}}&lt;/math&gt;. For the coin to be completely contained within the square, we must have the distance from the center of the coin to a side of the square to be at least &lt;math&gt;\frac{1}{2}&lt;/math&gt;, as it's the radius of the coin. This implies the &lt;math&gt;\text{total possible region}&lt;/math&gt; is a square with side length &lt;math&gt;8 - \frac{1}{2} - \frac{1}{2} = 7&lt;/math&gt;, with an area of &lt;math&gt;49&lt;/math&gt;. Now, we consider cases on where needs to land to partially cover a black region.<br /> <br /> &lt;b&gt; Near The Center Square &lt;/b&gt;<br /> <br /> We can have the center of the coin land within &lt;math&gt;\frac{1}{2}&lt;/math&gt; of the center square, or inside of the center square. We have that the center lands either outside of the square, or inside. So, we have a region with &lt;math&gt;\frac{1}{2}&lt;/math&gt; emanating from every point on the exterior of the square, forming &lt;math&gt;4&lt;/math&gt; quarter circles and &lt;math&gt;4&lt;/math&gt; rectangles. &lt;img&gt; https://latex.artofproblemsolving.com/3/6/d/36ddfcdad3ad425c1a9c62d2380136463179dff1.png &lt;/img&gt; The &lt;math&gt;4&lt;/math&gt; quarter circles combine to make a full circle, with radius of &lt;math&gt;\frac{1}{2}&lt;/math&gt;, so that has an area of &lt;math&gt;\frac{\pi}{4}&lt;/math&gt;. The area of a rectangle is &lt;math&gt;2 \sqrt 2 \cdot \frac{1}{2} = \sqrt 2&lt;/math&gt;, so &lt;math&gt;4&lt;/math&gt; of them combine to an area of &lt;math&gt;4 \sqrt 2&lt;/math&gt;. The area of the black square is simply &lt;math&gt;(2\sqrt 2)^2 = 8&lt;/math&gt;. So, for this case, we have a combined total of &lt;math&gt;8 + 4\sqrt 2 + \frac{\pi}{4}&lt;/math&gt;. Onto the second (and last) case.<br /> <br /> &lt;b&gt; Near A Triangle &lt;/b&gt;<br /> <br /> We can also have the coin land within &lt;math&gt;\frac{1}{2}&lt;/math&gt; of one of the triangles. By symmetry, we can just find the successful region for one of them, then multiply by &lt;math&gt;4&lt;/math&gt;. Consider this diagram. [img] https://latex.artofproblemsolving.com/e/7/1/e7194733f1299176a89a56fb28ba3cf247e736c5.png [/img] We can draw in an altitude from the bottom corner of the square to hit the hypotenuse of the blue triangle. The length of this when passing through the black region is &lt;math&gt;\sqrt 2&lt;/math&gt;, and when passing through the white region (while being contained in the blue triangle) is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. However, we have to subtract off when it doesn't pass through the red square. Then, it's the hypotenuse of a small isosceles right triangle with side lengths of &lt;math&gt;\frac{1}{2}&lt;/math&gt;, or &lt;math&gt;\frac{\sqrt 2}{2}&lt;/math&gt;. So, our altitude of the blue triangle is &lt;math&gt;\sqrt 2 + \frac{1}{2} - \frac{\sqrt 2}{2} = \frac{\sqrt 2 + 1}{2}&lt;/math&gt;. Then, recall, the area of an isosceles right triangle is &lt;math&gt;h^2&lt;/math&gt;, where &lt;math&gt;h&lt;/math&gt; is the altitude from the right angle. So, squaring this, we get &lt;math&gt;\frac{3 + 2\sqrt 2}{4}&lt;/math&gt;. Now, we have to multiply this by &lt;math&gt;4&lt;/math&gt; to account for all of the black triangles, to get &lt;math&gt;3 + 2\sqrt 2&lt;/math&gt; as the final area for this case. <br /> <br /> <br /> &lt;b&gt; Finishing &lt;/b&gt;<br /> <br /> Then, to have the coin touching a black region, we add up the area of our successful regions, or &lt;math&gt;8 + 4\sqrt 2 + \frac{\pi}{4} + 3 + 2\sqrt 2 = 11 + 6\sqrt 2 + \frac{\pi}{4} = \frac{44 + 24\sqrt 2 + \pi}{4}&lt;/math&gt;. The total region is &lt;math&gt;49&lt;/math&gt;, so our probability is &lt;math&gt;\frac{\frac{44 + 24\sqrt 2 + \pi}{4}}{49} = \frac{44 + 24\sqrt 2 + \pi}{196}&lt;/math&gt;, which implies &lt;math&gt;a+b = 44+24 = 68&lt;/math&gt;. This corresponds to answer choice &lt;math&gt;\boxed{\textbf{(A)} ~68}&lt;/math&gt;. ~rocketsri (diagrams from the MathJam)<br /> <br /> == Video Solution by OmegaLearn (Similar Triangles and Area Calculations) ==<br /> https://youtu.be/-UvivZ0UA1U<br /> <br /> ~ pi_is_3.14<br /> <br /> <br /> <br /> {{AMC10 box|year=2021|ab=B|num-b=22|num-a=24}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_23&diff=146466 2021 AMC 10B Problems/Problem 23 2021-02-13T18:06:11Z <p>Rocketsri: /* Solution */</p> <hr /> <div>==Problem==<br /> A square with side length &lt;math&gt;8&lt;/math&gt; is colored white except for &lt;math&gt;4&lt;/math&gt; black isosceles right triangular regions with legs of length &lt;math&gt;2&lt;/math&gt; in each corner of the square and a black diamond with side length &lt;math&gt;2\sqrt{2}&lt;/math&gt; in the center of the square, as shown in the diagram. A circular coin with diameter &lt;math&gt;1&lt;/math&gt; is dropped onto the square and lands in a random location where the coin is completely contained within the square. The probability that the coin will cover part of the black region of the square can be written as &lt;math&gt;\frac{1}{196}(a+b\sqrt{2}+\pi)&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b&lt;/math&gt;?<br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> draw((0,0)--(8,0)--(8,8)--(0,8)--(0,0));<br /> fill((2,0)--(0,2)--(0,0)--cycle, black);<br /> fill((6,0)--(8,0)--(8,2)--cycle, black);<br /> fill((8,6)--(8,8)--(6,8)--cycle, black);<br /> fill((0,6)--(2,8)--(0,8)--cycle, black);<br /> fill((4,6)--(2,4)--(4,2)--(6,4)--cycle, black);<br /> filldraw(circle((2.6,3.31),0.47),gray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~64 \qquad\textbf{(B)} ~66 \qquad\textbf{(C)} ~68 \qquad\textbf{(D)} ~70 \qquad\textbf{(E)} ~72&lt;/math&gt;<br /> <br /> <br /> == Solution ==<br /> To find the probability, we look at the &lt;math&gt;\frac{\text{success region}}{\text{total possible region}}&lt;/math&gt;. For the coin to be completely contained within the square, we must have the distance from the center of the coin to a side of the square to be at least &lt;math&gt;\frac{1}{2}&lt;/math&gt;, as it's the radius of the coin. This implies the &lt;math&gt;\text{total possible region}&lt;/math&gt; is a square with side length &lt;math&gt;8 - \frac{1}{2} - \frac{1}{2} = 7&lt;/math&gt;, with an area of &lt;math&gt;49&lt;/math&gt;. Now, we consider cases on where needs to land to partially cover a black region.<br /> <br /> &lt;b&gt; Near The Center Square &lt;/b&gt;<br /> <br /> We can have the center of the coin land within &lt;math&gt;\frac{1}{2}&lt;/math&gt; of the center square, or inside of the center square. We have that the center lands either outside of the square, or inside. So, we have a region with &lt;math&gt;\frac{1}{2}&lt;/math&gt; emanating from every point on the exterior of the square, forming &lt;math&gt;4&lt;/math&gt; quarter circles and &lt;math&gt;4&lt;/math&gt; rectangles. &lt;img&gt; https://latex.artofproblemsolving.com/3/6/d/36ddfcdad3ad425c1a9c62d2380136463179dff1.png &lt;/img&gt; The &lt;math&gt;4&lt;/math&gt; quarter circles combine to make a full circle, with radius of &lt;math&gt;\frac{1}{2}&lt;/math&gt;, so that has an area of &lt;math&gt;\frac{\pi}{4}&lt;/math&gt;. The area of a rectangle is &lt;math&gt;2 \sqrt 2 \cdot \frac{1}{2} = \sqrt 2&lt;/math&gt;, so &lt;math&gt;4&lt;/math&gt; of them combine to an area of &lt;math&gt;4 \sqrt 2&lt;/math&gt;. The area of the black square is simply &lt;math&gt;(2\sqrt 2)^2 = 8&lt;/math&gt;. So, for this case, we have a combined total of &lt;math&gt;8 + 4\sqrt 2 + \frac{\pi}{4}&lt;/math&gt;. Onto the second (and last) case.<br /> <br /> &lt;b&gt; Near A Triangle &lt;/b&gt;<br /> <br /> We can also have the coin land within &lt;math&gt;\frac{1}{2}&lt;/math&gt; of one of the triangles. By symmetry, we can just find the successful region for one of them, then multiply by &lt;math&gt;4&lt;/math&gt;. Consider this diagram. &lt;img&gt; https://latex.artofproblemsolving.com/e/7/1/e7194733f1299176a89a56fb28ba3cf247e736c5.png &lt;/img&gt; We can draw in an altitude from the bottom corner of the square to hit the hypotenuse of the blue triangle. The length of this when passing through the black region is &lt;math&gt;\sqrt 2&lt;/math&gt;, and when passing through the white region (while being contained in the blue triangle) is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. However, we have to subtract off when it doesn't pass through the red square. Then, it's the hypotenuse of a small isosceles right triangle with side lengths of &lt;math&gt;\frac{1}{2}&lt;/math&gt;, or &lt;math&gt;\frac{\sqrt 2}{2}&lt;/math&gt;. So, our altitude of the blue triangle is &lt;math&gt;\sqrt 2 + \frac{1}{2} - \frac{\sqrt 2}{2} = \frac{\sqrt 2 + 1}{2}&lt;/math&gt;. Then, recall, the area of an isosceles right triangle is &lt;math&gt;h^2&lt;/math&gt;, where &lt;math&gt;h&lt;/math&gt; is the altitude from the right angle. So, squaring this, we get &lt;math&gt;\frac{3 + 2\sqrt 2}{4}&lt;/math&gt;. Now, we have to multiply this by &lt;math&gt;4&lt;/math&gt; to account for all of the black triangles, to get &lt;math&gt;3 + 2\sqrt 2&lt;/math&gt; as the final area for this case. <br /> <br /> <br /> &lt;b&gt; Finishing &lt;/b&gt;<br /> <br /> Then, to have the coin touching a black region, we add up the area of our successful regions, or &lt;math&gt;8 + 4\sqrt 2 + \frac{\pi}{4} + 3 + 2\sqrt 2 = 11 + 6\sqrt 2 + \frac{\pi}{4} = \frac{44 + 24\sqrt 2 + \pi}{4}&lt;/math&gt;. The total region is &lt;math&gt;49&lt;/math&gt;, so our probability is &lt;math&gt;\frac{\frac{44 + 24\sqrt 2 + \pi}{4}}{49} = \frac{44 + 24\sqrt 2 + \pi}{196}&lt;/math&gt;, which implies &lt;math&gt;a+b = 44+24 = 68&lt;/math&gt;. This corresponds to answer choice &lt;math&gt;\boxed{\textbf{(A)} ~68}&lt;/math&gt;. ~rocketsri (diagrams from the MathJam)<br /> <br /> == Video Solution by OmegaLearn (Similar Triangles and Area Calculations) ==<br /> https://youtu.be/-UvivZ0UA1U<br /> <br /> ~ pi_is_3.14<br /> <br /> <br /> <br /> {{AMC10 box|year=2021|ab=B|num-b=22|num-a=24}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_23&diff=146465 2021 AMC 10B Problems/Problem 23 2021-02-13T18:05:29Z <p>Rocketsri: </p> <hr /> <div>==Problem==<br /> A square with side length &lt;math&gt;8&lt;/math&gt; is colored white except for &lt;math&gt;4&lt;/math&gt; black isosceles right triangular regions with legs of length &lt;math&gt;2&lt;/math&gt; in each corner of the square and a black diamond with side length &lt;math&gt;2\sqrt{2}&lt;/math&gt; in the center of the square, as shown in the diagram. A circular coin with diameter &lt;math&gt;1&lt;/math&gt; is dropped onto the square and lands in a random location where the coin is completely contained within the square. The probability that the coin will cover part of the black region of the square can be written as &lt;math&gt;\frac{1}{196}(a+b\sqrt{2}+\pi)&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b&lt;/math&gt;?<br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> draw((0,0)--(8,0)--(8,8)--(0,8)--(0,0));<br /> fill((2,0)--(0,2)--(0,0)--cycle, black);<br /> fill((6,0)--(8,0)--(8,2)--cycle, black);<br /> fill((8,6)--(8,8)--(6,8)--cycle, black);<br /> fill((0,6)--(2,8)--(0,8)--cycle, black);<br /> fill((4,6)--(2,4)--(4,2)--(6,4)--cycle, black);<br /> filldraw(circle((2.6,3.31),0.47),gray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~64 \qquad\textbf{(B)} ~66 \qquad\textbf{(C)} ~68 \qquad\textbf{(D)} ~70 \qquad\textbf{(E)} ~72&lt;/math&gt;<br /> <br /> <br /> == Solution ==<br /> To find the probability, we look at the &lt;math&gt;\frac{\text{success region}}{\text{total possible region}}&lt;/math&gt;. For the coin to be completely contained within the square, we must have the distance from the center of the coin to a side of the square to be at least &lt;math&gt;\frac{1}{2}&lt;/math&gt;, as it's the radius of the coin. This implies the &lt;math&gt;\text{total possible region}&lt;/math&gt; is a square with side length &lt;math&gt;8 - \frac{1}{2} - \frac{1}{2} = 7&lt;/math&gt;, with an area of &lt;math&gt;49&lt;/math&gt;. Now, we consider cases on where needs to land to partially cover a black region.<br /> <br /> &lt;b&gt; Near The Center Square &lt;/b&gt;<br /> We can have the center of the coin land within &lt;math&gt;\frac{1}{2}&lt;/math&gt; of the center square, or inside of the center square. We have that the center lands either outside of the square, or inside. So, we have a region with &lt;math&gt;\frac{1}{2}&lt;/math&gt; emanating from every point on the exterior of the square, forming &lt;math&gt;4&lt;/math&gt; quarter circles and &lt;math&gt;4&lt;/math&gt; rectangles. &lt;img&gt; https://latex.artofproblemsolving.com/3/6/d/36ddfcdad3ad425c1a9c62d2380136463179dff1.png &lt;/img&gt; The &lt;math&gt;4&lt;/math&gt; quarter circles combine to make a full circle, with radius of &lt;math&gt;\frac{1}{2}&lt;/math&gt;, so that has an area of &lt;math&gt;\frac{\pi}{4}&lt;/math&gt;. The area of a rectangle is &lt;math&gt;2 \sqrt 2 \cdot \frac{1}{2} = \sqrt 2&lt;/math&gt;, so &lt;math&gt;4&lt;/math&gt; of them combine to an area of &lt;math&gt;4 \sqrt 2&lt;/math&gt;. The area of the black square is simply &lt;math&gt;(2\sqrt 2)^2 = 8&lt;/math&gt;. So, for this case, we have a combined total of &lt;math&gt;8 + 4\sqrt 2 + \frac{\pi}{4}&lt;/math&gt;. Onto the second (and last) case.<br /> <br /> &lt;b&gt; Near A Triangle &lt;/b&gt;<br /> We can also have the coin land within &lt;math&gt;\frac{1}{2}&lt;/math&gt; of one of the triangles. By symmetry, we can just find the successful region for one of them, then multiply by &lt;math&gt;4&lt;/math&gt;. Consider this diagram. &lt;img&gt; https://latex.artofproblemsolving.com/e/7/1/e7194733f1299176a89a56fb28ba3cf247e736c5.png &lt;/img&gt; We can draw in an altitude from the bottom corner of the square to hit the hypotenuse of the blue triangle. The length of this when passing through the black region is &lt;math&gt;\sqrt 2&lt;/math&gt;, and when passing through the white region (while being contained in the blue triangle) is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. However, we have to subtract off when it doesn't pass through the red square. Then, it's the hypotenuse of a small isosceles right triangle with side lengths of &lt;math&gt;\frac{1}{2}&lt;/math&gt;, or &lt;math&gt;\frac{\sqrt 2}{2}&lt;/math&gt;. So, our altitude of the blue triangle is &lt;math&gt;\sqrt 2 + \frac{1}{2} - \frac{\sqrt 2}{2} = \frac{\sqrt 2 + 1}{2}&lt;/math&gt;. Then, recall, the area of an isosceles right triangle is &lt;math&gt;h^2&lt;/math&gt;, where &lt;math&gt;h&lt;/math&gt; is the altitude from the right angle. So, squaring this, we get &lt;math&gt;\frac{3 + 2\sqrt 2}{4}&lt;/math&gt;. Now, we have to multiply this by &lt;math&gt;4&lt;/math&gt; to account for all of the black triangles, to get &lt;math&gt;3 + 2\sqrt 2&lt;/math&gt; as the final area for this case. <br /> <br /> &lt;b&gt; Finishing &lt;/b&gt;<br /> Then, to have the coin touching a black region, we add up the area of our successful regions, or &lt;math&gt;8 + 4\sqrt 2 + \frac{\pi}{4} + 3 + 2\sqrt 2 = 11 + 6\sqrt 2 + \frac{\pi}{4} = \frac{44 + 24\sqrt 2 + \pi}{4}&lt;/math&gt;. The total region is &lt;math&gt;49&lt;/math&gt;, so our probability is &lt;math&gt;\frac{\frac{44 + 24\sqrt 2 + \pi}{4}}{49} = \frac{44 + 24\sqrt 2 + \pi}{196}&lt;/math&gt;, which implies &lt;math&gt;a+b = 44+24 = 68&lt;/math&gt;. This corresponds to answer choice &lt;math&gt;\boxed{\textbf{(A)} ~68}&lt;/math&gt;. ~rocketsri (diagrams from the MathJam)<br /> <br /> == Video Solution by OmegaLearn (Similar Triangles and Area Calculations) ==<br /> https://youtu.be/-UvivZ0UA1U<br /> <br /> ~ pi_is_3.14<br /> <br /> <br /> <br /> {{AMC10 box|year=2021|ab=B|num-b=22|num-a=24}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_23&diff=146242 2021 AMC 12A Problems/Problem 23 2021-02-12T23:50:01Z <p>Rocketsri: /* Solution 6 (Casework) */</p> <hr /> <div>==Problem==<br /> Frieda the frog begins a sequence of hops on a &lt;math&gt;3 \times 3&lt;/math&gt; grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she &quot;wraps around&quot; and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops &quot;up&quot;, the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}&lt;/math&gt;<br /> <br /> ==Solution 1 (complementary counting)==<br /> We will use complementary counting.<br /> First, the frog can go left with probability &lt;math&gt;\frac14&lt;/math&gt;. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since &lt;math&gt;4\cdot \frac14=1&lt;/math&gt;, we will ignore the leading probability.<br /> <br /> From the left, she either goes left to another edge (&lt;math&gt;\frac14&lt;/math&gt;) or back to the center (&lt;math&gt;\frac14&lt;/math&gt;). Time for some casework.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; She goes back to the center.<br /> <br /> Now, she can go in any 4 directions, and then has 2 options from that edge. This gives &lt;math&gt;\frac12&lt;/math&gt;. --End case 1<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; She goes to another edge (rightmost).<br /> <br /> Subcase 1: She goes back to the left edge. She now has 2 places to go, giving &lt;math&gt;\frac12&lt;/math&gt;<br /> <br /> Subcase 2: She goes to the center. Now any move works.<br /> <br /> &lt;math&gt;\frac14 \cdot \frac12 + \frac14 \cdot 1=\frac18 + \frac 14=\frac38&lt;/math&gt; for this case. --End case 2<br /> <br /> She goes back to the center in Case 1 with probability &lt;math&gt;\frac14&lt;/math&gt;, and to the right edge with probability &lt;math&gt;\frac14&lt;/math&gt;<br /> <br /> So, our answer is &lt;math&gt;\frac14 \cdot \frac12 + \frac14 \cdot \frac38=\frac14 (\frac12+\frac38)=\frac14 \cdot \frac78 = \frac7{32}&lt;/math&gt;<br /> <br /> But, don't forget complementary counting. So, we get &lt;math&gt;1-\frac7{32}=\frac{25}{32} \implies \boxed{D}&lt;/math&gt;. ~ firebolt360<br /> <br /> Video Solution for those who prefer: https://youtu.be/ude2rzO1cmk ~ firebolt360<br /> <br /> ==Solution 2 (direct counting and probability states)==<br /> We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have &lt;math&gt;1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{D}&lt;/math&gt;. ~IceWolf10<br /> <br /> ==Solution 3 (Similar to Solution 2, but Finds the Numerator and Denominator Separately)==<br /> &lt;b&gt;Denominator&lt;/b&gt;<br /> <br /> There are &lt;math&gt;4^4=256&lt;/math&gt; ways to make 4 hops without restrictions.<br /> <br /> <br /> &lt;b&gt;Numerator (Casework)&lt;/b&gt;<br /> <br /> Suppose Frieda makes 4 hops without stopping. We perform casework on which hop reaches a corner for the first time.<br /> <br /> <br /> (1) Hop #2 (Hops #3 and #4 have no restrictions)<br /> <br /> The 4 independent hops have 4, 2, 4, 4 options, respectively. So, this case has &lt;math&gt;4\cdot2\cdot4\cdot4=128&lt;/math&gt; ways.<br /> <br /> <br /> (2) Hop #3 (Hop #4 has no restriction)<br /> <br /> No matter which direction the first hop takes, the second hop must &quot;wrap around&quot;.<br /> <br /> The 4 independent hops have 4, 1, 2, 4 options, respectively. So, this case has &lt;math&gt;4\cdot1\cdot2\cdot4=32&lt;/math&gt; ways.<br /> <br /> <br /> (3) Hop #4<br /> <br /> Two sub-cases:<br /> <br /> (3.1) The second hop &quot;wraps around&quot;. It follows that the third hop also &quot;wraps around&quot;.<br /> <br /> The 4 independent hops have 4, 1, 1, 2 options, respectively. So, this sub-case has &lt;math&gt;4\cdot1\cdot1\cdot2=8&lt;/math&gt; ways.<br /> <br /> (3.2) The second hop backs to the center.<br /> <br /> The 4 independent hops have 4, 1, 4, 2 options, respectively. So, this sub-case has &lt;math&gt;4\cdot1\cdot4\cdot2=32&lt;/math&gt; ways.<br /> <br /> Together, Case (3) has &lt;math&gt;8+32=40&lt;/math&gt; ways.<br /> <br /> <br /> The numerator is &lt;math&gt;128+32+40=200.&lt;/math&gt;<br /> <br /> <br /> &lt;b&gt;Probability&lt;/b&gt;<br /> <br /> &lt;math&gt;\frac{200}{256}=\boxed{\textbf{(D)} ~\frac{25}{32}}.&lt;/math&gt;<br /> <br /> This problem is quite similar to 1995 AIME Problem 3: https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3<br /> <br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;C_n&lt;/math&gt; be the probability that Frieda is on the central square after n moves, &lt;math&gt;E_n&lt;/math&gt; be the probability that Frieda is on one of the four squares on the middle of the edges after n moves, and &lt;math&gt;V_n&lt;/math&gt; (V for vertex) be the probability that Frieda is on a corner after n moves. The only way to reach the center is by moving in &lt;math&gt;1&lt;/math&gt; specific direction out of &lt;math&gt;4&lt;/math&gt; total directions from the middle of an edge, so &lt;math&gt;C_{n+1}=\frac{E_n}{4}&lt;/math&gt;. The ways to reach the middle of an edge are by moving in any direction from the center or by moving in &lt;math&gt;1&lt;/math&gt; specific direction from the middle of an edge, so &lt;math&gt;E_{n+1}=C_n+\frac{E_n}{4}&lt;/math&gt;. The ways to reach a corner are by simply staying there after reaching there in a previous move or by moving in &lt;math&gt;2&lt;/math&gt; specific directions from the middle of an edge, so &lt;math&gt;V_{n+1}=V_n+\frac{E_n}{2}&lt;/math&gt;. Since Frieda always start from the center, &lt;math&gt;C_0=1&lt;/math&gt;, &lt;math&gt;E_0=0&lt;/math&gt;, and &lt;math&gt;V_0=0&lt;/math&gt;. We use the previous formulas to work out &lt;math&gt;V_4&lt;/math&gt; and find it to be &lt;math&gt;\boxed{\textbf{(D)} ~\frac{25}{32}}&lt;/math&gt;.<br /> <br /> -SmileKat32<br /> <br /> ==Solution 5==<br /> Imagine an infinite grid of &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; squares such that there is a &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; square centered at &lt;math&gt;(3x, 3y)&lt;/math&gt; for all ordered pairs of integers &lt;math&gt;(x, y).&lt;/math&gt; <br /> <br /> &lt;asy&gt;<br /> dot((0,0)); draw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle); draw((-4,-1)--(-4,1)--(-2,1)--(-2,-1)--cycle); draw((2,-1)--(2,1)--(4,1)--(4,-1)--cycle); draw((-1,2)--(-1,4)--(1,4)--(1,2)--cycle); draw((-1,-4)--(-1,-2)--(1,-2)--(1,-4)--cycle); draw((-4,2)--(-2,2)--(-2,4)--(-4,4)--cycle); draw((4,-2)--(2,-2)--(2,-4)--(4,-4)--cycle); draw((4,2)--(2,2)--(2,4)--(4,4)--cycle); draw((-4,-2)--(-2,-2)--(-2,-4)--(-4,-4)--cycle); draw((-3,-3)--(3,-3)--(3,3)--(-3,3)--cycle); draw((-4,0)--(4,0)); draw((0,4)--(0,-4)); <br /> &lt;/asy&gt;<br /> <br /> It is easy to see that the problem is equivalent to Frieda moving left, right, up, or down on this infinite grid starting at &lt;math&gt;(0, 0)&lt;/math&gt;. (minus the teleportations) Since counting the complement set is easier, we'll count the number of &lt;math&gt;4&lt;/math&gt;-step paths such that Frieda never reaches a corner point. <br /> <br /> In other words, since the reachable corner points are &lt;math&gt;(\pm 1, \pm 1), (\pm 1, \pm 2), (\pm 2, \pm 1), &lt;/math&gt; and &lt;math&gt;(\pm 2, \pm 2),&lt;/math&gt; Frieda can only travel along the collection of points included in &lt;math&gt;S&lt;/math&gt;, where &lt;math&gt;S&lt;/math&gt; is all points on &lt;math&gt;x=0&lt;/math&gt; and &lt;math&gt;y=0&lt;/math&gt; such that &lt;math&gt;|y|&lt;4&lt;/math&gt; and &lt;math&gt;|x|&lt;4&lt;/math&gt;, respectively, plus all points on the big square with side length &lt;math&gt;6&lt;/math&gt; centered at &lt;math&gt;(0, 0).&lt;/math&gt; We then can proceed with casework:<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: Frieda never reaches &lt;math&gt;(0, \pm 3)&lt;/math&gt; nor &lt;math&gt;(\pm 3, 0).&lt;/math&gt;<br /> <br /> When Frieda only moves horizontally or vertically for her four moves, she can do so in &lt;math&gt;2^4 - 4 = 12&lt;/math&gt; ways for each case . Thus, &lt;math&gt;12 \cdot 2&lt;/math&gt; total paths for the subcase of staying in one direction. (For instance, all length &lt;math&gt;4&lt;/math&gt; combinations of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; except &lt;math&gt;FFFF&lt;/math&gt;, &lt;math&gt;BBBB&lt;/math&gt;, &lt;math&gt;FFFB&lt;/math&gt;, and &lt;math&gt;BBBF&lt;/math&gt; for the horizontal direction.) <br /> <br /> There is another subcase where she changes directions during her path. There are four symmetric cases for this subcase depending on which of the four quadrants Frieda hugs. For the first quadrant, the possible paths are &lt;math&gt;FBUD&lt;/math&gt;, &lt;math&gt;FBUU&lt;/math&gt;, &lt;math&gt;UDFB&lt;/math&gt;, and &lt;math&gt;UDFF.&lt;/math&gt; Thus, a total of &lt;math&gt;4 \cdot 4 = 16&lt;/math&gt; ways for this subcase. <br /> <br /> Total for Case &lt;math&gt;1&lt;/math&gt;: &lt;math&gt;24 + 16 = 40&lt;/math&gt;<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: Frieda reaches &lt;math&gt;(0, \pm 3)&lt;/math&gt; or &lt;math&gt;(\pm 3, 0)&lt;/math&gt;.<br /> <br /> Once Frieda reaches one of the points listed above (by using three moves), she has four choices for her last move. Thus, a total of &lt;math&gt;4 \cdot 4 = 16&lt;/math&gt; paths for this case.<br /> <br /> Our total number of paths never reaching coroners is thus &lt;math&gt;16+40=56,&lt;/math&gt; making for an answer of &lt;cmath&gt;\frac{4^4-56}{4^4} = \boxed{\textbf{(D)} ~\frac{25}{32}}.&lt;/cmath&gt;<br /> <br /> -fidgetboss_4000<br /> <br /> == Solution 6 (Casework) ==<br /> We take cases on the number of hops needed to reach a corner. For simplicity, denote &lt;math&gt;E&lt;/math&gt; as a move that takes Frieda to an edge, &lt;math&gt;W&lt;/math&gt; as wrap-around move and &lt;math&gt;C&lt;/math&gt; as a corner move. Also, denote &lt;math&gt;O&lt;/math&gt; as a move that takes us to the center.<br /> <br /> <br /> &lt;b&gt; 2 Hops &lt;/b&gt;<br /> <br /> Then, Frieda will have to &lt;math&gt;(E, C)&lt;/math&gt; as her set of moves. There are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, and &lt;math&gt;2&lt;/math&gt; corners to move to, for a total of &lt;math&gt;4 \cdot 2 = 8&lt;/math&gt; cases here. Then, there are &lt;math&gt;4&lt;/math&gt; choices for each move, for a probability of &lt;math&gt;\frac{8}{4 \cdot 4} = \frac{1}{2}&lt;/math&gt;. <br /> <br /> &lt;b&gt; 3 Hops &lt;/b&gt;<br /> <br /> In this case, Frieda must wrap-around. There's only one possible combination, just &lt;math&gt;(E, W, C)&lt;/math&gt;. There are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, &lt;math&gt;1&lt;/math&gt; way to wrap-around (you must continue in the same direction) and &lt;math&gt;2&lt;/math&gt; corners, for a total of &lt;math&gt;4 \cdot 1 \cdot 2 = 8&lt;/math&gt; cases here. Then, there are &lt;math&gt;4&lt;/math&gt; choices for each move, for a probability of &lt;math&gt;\frac{8}{4 \cdot 4 \cdot 4} = \frac{1}{8}&lt;/math&gt;. <br /> <br /> &lt;b&gt; 4 Hops &lt;/b&gt; <br /> <br /> Lastly, there are two cases we must consider here. The first case is &lt;math&gt;(E, O, E, C)&lt;/math&gt;, and the second is &lt;math&gt;(E, W, W, C)&lt;/math&gt;. For the first case, there are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, &lt;math&gt;1&lt;/math&gt; way to return to the center, &lt;math&gt;4&lt;/math&gt; ways to move to an edge once again, and &lt;math&gt;2&lt;/math&gt; ways to move to a corner. Hence, there is a total of &lt;math&gt;4 \cdot 1 \cdot 4 \cdot 2 = 32&lt;/math&gt; cases here. Then, for the second case, there are &lt;math&gt;4&lt;/math&gt; ways to move to a corner, &lt;math&gt;1&lt;/math&gt; way to wrap-around, &lt;math&gt;1&lt;/math&gt; way to wrap-around again, and &lt;math&gt;2&lt;/math&gt; ways to move to a corner. This implies there are &lt;math&gt;4 \cdot 1 \cdot 1 \cdot 2 = 8&lt;/math&gt; cases here. Then, there is a total of &lt;math&gt;8+32 = 40&lt;/math&gt; cases, out of a total of &lt;math&gt;4^4 = 256&lt;/math&gt; cases, for a probability of &lt;math&gt;\frac{40}{256} = \frac{5}{32}&lt;/math&gt;. <br /> <br /> <br /> Then, the total probability that Frieda ends up on a corner is &lt;math&gt;\frac{1}{2} + \frac{1}{8} + \frac{5}{32} = \frac{25}{32}&lt;/math&gt;, corresponding to choice &lt;math&gt;\boxed{\textbf{(D)} ~\frac{25}{32}}&lt;/math&gt;. ~rocketsri<br /> <br /> == Video Solution by OmegaLearn (Using Probability States) ==<br /> https://youtu.be/V_Sn30N2q50<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2021|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_21&diff=146188 2021 AMC 10B Problems/Problem 21 2021-02-12T21:58:47Z <p>Rocketsri: /* Problem */</p> <hr /> <div>==Problem==<br /> A square piece of paper has side length &lt;math&gt;1&lt;/math&gt; and vertices &lt;math&gt;A,B,C,&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; in that order. As shown in the figure, the paper is folded so that vertex &lt;math&gt;C&lt;/math&gt; meets edge &lt;math&gt;\overline{AD}&lt;/math&gt; at point &lt;math&gt;C'&lt;/math&gt;, and edge &lt;math&gt;\overline{AB}&lt;/math&gt; at point &lt;math&gt;E&lt;/math&gt;. Suppose that &lt;math&gt;C'D = \frac{1}{3}&lt;/math&gt;. What is the perimeter of triangle &lt;math&gt;\bigtriangleup AEC' ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{136} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}&lt;/math&gt;<br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> pair A=(0,1);<br /> pair CC=(0.666666666666,1);<br /> pair D=(1,1);<br /> pair F=(1,0.62);<br /> pair C=(1,0);<br /> pair B=(0,0);<br /> pair G=(0,0.25);<br /> pair H=(-0.13,0.41);<br /> pair E=(0,0.5);<br /> dot(A^^CC^^D^^C^^B^^E);<br /> draw(E--A--D--F);<br /> draw(G--B--C--F, dashed);<br /> fill(E--CC--F--G--H--E--CC--cycle, gray);<br /> draw(E--CC--F--G--H--E--CC);<br /> label(&quot;A&quot;,A,NW);<br /> label(&quot;B&quot;,B,SW);<br /> label(&quot;C&quot;,C,SE);<br /> label(&quot;D&quot;,D,NE);<br /> label(&quot;E&quot;,E,NW);<br /> label(&quot;C'&quot;,CC,N);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> <br /> We can set the point on &lt;math&gt;CD&lt;/math&gt; where the fold occurs as point &lt;math&gt;F&lt;/math&gt;. Then, we can set &lt;math&gt;FD&lt;/math&gt; as &lt;math&gt;x&lt;/math&gt;, and &lt;math&gt;CF&lt;/math&gt; as &lt;math&gt;1-x&lt;/math&gt; because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for &lt;math&gt;x&lt;/math&gt;, we get, <br /> <br /> &lt;cmath&gt;x^2 + (\frac{1}{3})^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}&lt;/cmath&gt;<br /> <br /> We know this is a 3-4-5 triangle because the side lengths are &lt;math&gt;\frac{3}{9}, \frac{4}{9}, \frac{5}{9}&lt;/math&gt;. We also know that &lt;math&gt;EAC'&lt;/math&gt; is similar to &lt;math&gt;C'DF&lt;/math&gt; because angle &lt;math&gt;C'&lt;/math&gt; is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of &lt;math&gt;C'DF * \frac{AC'}{DF}&lt;/math&gt;. Thats just &lt;math&gt;\frac{4}{3} * \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} * \frac{3}{2} = 2&lt;/math&gt;. Therefore, the final answer is &lt;math&gt;\boxed{A}&lt;/math&gt;<br /> <br /> ~Tony_Li2007<br /> <br /> ==Solution 2==<br /> Let line we're reflecting over be &lt;math&gt;\ell&lt;/math&gt;, and let the points where it hits &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, be &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt;, respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line &lt;math&gt;\ell&lt;/math&gt;. The segment &lt;math&gt;CC'&lt;/math&gt; has slope &lt;math&gt;\frac{0 - 1}{1 - 2/3} = -3&lt;/math&gt;, implying line &lt;math&gt;\ell&lt;/math&gt; has a slope of &lt;math&gt;\frac{1}{3}&lt;/math&gt;. Also, the midpoint of segment &lt;math&gt;CC'&lt;/math&gt; is &lt;math&gt;\left( \frac{5}{6}, \frac{1}{2} \right)&lt;/math&gt;, so line &lt;math&gt;\ell&lt;/math&gt; passes through this point. Then, we get the equation of line &lt;math&gt;\ell&lt;/math&gt; is simply &lt;math&gt;y = \frac{1}{3} x + \frac{2}{9}&lt;/math&gt;. Then, if the point where &lt;math&gt;B&lt;/math&gt; is reflected over line &lt;math&gt;\ell&lt;/math&gt; is &lt;math&gt;B'&lt;/math&gt;, then we get &lt;math&gt;BB'&lt;/math&gt; is the line &lt;math&gt;y = -3x&lt;/math&gt;. The intersection of &lt;math&gt;\ell&lt;/math&gt; and segment &lt;math&gt;BB'&lt;/math&gt; is &lt;math&gt;\left( - \frac{1}{15}, \frac{1}{5} \right)&lt;/math&gt;. So, we get &lt;math&gt;B' = \left(- \frac{2}{15}, \frac{2}{5} \right)&lt;/math&gt;. Then, line segment &lt;math&gt;B'C'&lt;/math&gt; has equation &lt;math&gt;y = \frac{3}{4} x + \frac{1}{2}&lt;/math&gt;, so the point &lt;math&gt;E&lt;/math&gt; is the &lt;math&gt;y&lt;/math&gt;-intercept, or &lt;math&gt;\left(0, \frac{1}{2} \right)&lt;/math&gt;. This implies that &lt;math&gt;AE = \frac{1}{2}, AC' = \frac{2}{3}&lt;/math&gt;, and by the Pythagorean Theorem, &lt;math&gt;EC' = \frac{5}{6}&lt;/math&gt; (or you could notice &lt;math&gt;\triangle AEC'&lt;/math&gt; is a &lt;math&gt;3-4-5&lt;/math&gt; right triangle). Then, the perimeter is &lt;math&gt;\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{\textbf{(A)} ~2}&lt;/math&gt;. ~rocketsri<br /> <br /> ==Solution 3 (Fakesolve):==<br /> Assume that E is the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt;. Then, &lt;math&gt;\overline{AE}=\frac{1}{2}&lt;/math&gt; and since &lt;math&gt;C'D=\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\overline{AC'}=\frac{2}{3}&lt;/math&gt;. By the Pythagorean Theorem, &lt;math&gt;\overline{EC'}=\frac{5}{6}&lt;/math&gt;. It easily follows that our desired perimeter is &lt;math&gt;2 \rightarrow \boxed{A}&lt;/math&gt; ~samrocksnature<br /> <br /> == Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) ==<br /> https://youtu.be/cagzLmdbqYQ<br /> <br /> ~ pi_is_3.14<br /> <br /> {{AMC10 box|year=2021|ab=B|num-b=20|num-a=22}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_21&diff=146186 2021 AMC 10B Problems/Problem 21 2021-02-12T21:19:38Z <p>Rocketsri: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> A square piece of paper has side length &lt;math&gt;1&lt;/math&gt; and vertices &lt;math&gt;A,B,C,&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; in that order. As shown in the figure, the paper is folded so that vertex &lt;math&gt;C&lt;/math&gt; meets edge &lt;math&gt;\overline{AD}&lt;/math&gt; at point &lt;math&gt;C'&lt;/math&gt;, and edge &lt;math&gt;\overline{AB}&lt;/math&gt; at point &lt;math&gt;E&lt;/math&gt;. Suppose that &lt;math&gt;C'D = \frac{1}{3}&lt;/math&gt;. What is the perimeter of triangle &lt;math&gt;\bigtriangleup AEC' ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{13}{6} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}&lt;/math&gt;<br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> pair A=(0,1);<br /> pair CC=(0.666666666666,1);<br /> pair D=(1,1);<br /> pair F=(1,0.62);<br /> pair C=(1,0);<br /> pair B=(0,0);<br /> pair G=(0,0.25);<br /> pair H=(-0.13,0.41);<br /> pair E=(0,0.5);<br /> dot(A^^CC^^D^^C^^B^^E);<br /> draw(E--A--D--F);<br /> draw(G--B--C--F, dashed);<br /> fill(E--CC--F--G--H--E--CC--cycle, gray);<br /> draw(E--CC--F--G--H--E--CC);<br /> label(&quot;A&quot;,A,NW);<br /> label(&quot;B&quot;,B,SW);<br /> label(&quot;C&quot;,C,SE);<br /> label(&quot;D&quot;,D,NE);<br /> label(&quot;E&quot;,E,NW);<br /> label(&quot;C'&quot;,CC,N);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> <br /> We can set the point on &lt;math&gt;CD&lt;/math&gt; where the fold occurs as point &lt;math&gt;F&lt;/math&gt;. Then, we can set &lt;math&gt;FD&lt;/math&gt; as &lt;math&gt;x&lt;/math&gt;, and &lt;math&gt;CF&lt;/math&gt; as &lt;math&gt;1-x&lt;/math&gt; because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for &lt;math&gt;x&lt;/math&gt;, we get, <br /> <br /> &lt;cmath&gt;x^2 + (\frac{1}{3})^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}&lt;/cmath&gt;<br /> <br /> We know this is a 3-4-5 triangle because the side lengths are &lt;math&gt;\frac{3}{9}, \frac{4}{9}, \frac{5}{9}&lt;/math&gt;. We also know that &lt;math&gt;EAC'&lt;/math&gt; is similar to &lt;math&gt;C'DF&lt;/math&gt; because angle &lt;math&gt;C'&lt;/math&gt; is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of &lt;math&gt;C'DF * \frac{AC'}{DF}&lt;/math&gt;. Thats just &lt;math&gt;\frac{4}{3} * \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} * \frac{3}{2} = 2&lt;/math&gt;. Therefore, the final answer is &lt;math&gt;\boxed{A}&lt;/math&gt;<br /> <br /> ~Tony_Li2007<br /> <br /> ==Solution 2==<br /> Let line we're reflecting over be &lt;math&gt;\ell&lt;/math&gt;, and let the points where it hits &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, be &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt;, respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line &lt;math&gt;\ell&lt;/math&gt;. The segment &lt;math&gt;CC'&lt;/math&gt; has slope &lt;math&gt;\frac{0 - 1}{1 - 2/3} = -3&lt;/math&gt;, implying line &lt;math&gt;\ell&lt;/math&gt; has a slope of &lt;math&gt;\frac{1}{3}&lt;/math&gt;. Also, the midpoint of segment &lt;math&gt;CC'&lt;/math&gt; is &lt;math&gt;\left( \frac{5}{6}, \frac{1}{2} \right)&lt;/math&gt;, so line &lt;math&gt;\ell&lt;/math&gt; passes through this point. Then, we get the equation of line &lt;math&gt;\ell&lt;/math&gt; is simply &lt;math&gt;y = \frac{1}{3} x + \frac{2}{9}&lt;/math&gt;. Then, if the point where &lt;math&gt;B&lt;/math&gt; is reflected over line &lt;math&gt;\ell&lt;/math&gt; is &lt;math&gt;B'&lt;/math&gt;, then we get &lt;math&gt;BB'&lt;/math&gt; is the line &lt;math&gt;y = -3x&lt;/math&gt;. The intersection of &lt;math&gt;\ell&lt;/math&gt; and segment &lt;math&gt;BB'&lt;/math&gt; is &lt;math&gt;\left( - \frac{1}{15}, \frac{1}{5} \right)&lt;/math&gt;. So, we get &lt;math&gt;B' = \left(- \frac{2}{15}, \frac{2}{5} \right)&lt;/math&gt;. Then, line segment &lt;math&gt;B'C'&lt;/math&gt; has equation &lt;math&gt;y = \frac{3}{4} x + \frac{1}{2}&lt;/math&gt;, so the point &lt;math&gt;E&lt;/math&gt; is the &lt;math&gt;y&lt;/math&gt;-intercept, or &lt;math&gt;\left(0, \frac{1}{2} \right)&lt;/math&gt;. This implies that &lt;math&gt;AE = \frac{1}{2}, AC' = \frac{2}{3}&lt;/math&gt;, and by the Pythagorean Theorem, &lt;math&gt;EC' = \frac{5}{6}&lt;/math&gt; (or you could notice &lt;math&gt;\triangle AEC'&lt;/math&gt; is a &lt;math&gt;3-4-5&lt;/math&gt; right triangle). Then, the perimeter is &lt;math&gt;\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{\textbf{(A)} ~2}&lt;/math&gt;. ~rocketsri<br /> <br /> ==Solution 3 (Fakesolve):==<br /> Assume that E is the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt;. Then, &lt;math&gt;\overline{AE}=\frac{1}{2}&lt;/math&gt; and since &lt;math&gt;C'D=\frac{1}{3}&lt;/math&gt;, &lt;math&gt;\overline{AC'}=\frac{2}{3}&lt;/math&gt;. By the Pythagorean Theorem, &lt;math&gt;\overline{EC'}=\frac{5}{6}&lt;/math&gt;. It easily follows that our desired perimeter is &lt;math&gt;2 \rightarrow \boxed{A}&lt;/math&gt; ~samrocksnature<br /> <br /> == Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) ==<br /> https://youtu.be/cagzLmdbqYQ<br /> <br /> ~ pi_is_3.14<br /> <br /> {{AMC10 box|year=2021|ab=B|num-b=20|num-a=22}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_21&diff=146185 2021 AMC 10B Problems/Problem 21 2021-02-12T21:19:09Z <p>Rocketsri: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> A square piece of paper has side length &lt;math&gt;1&lt;/math&gt; and vertices &lt;math&gt;A,B,C,&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; in that order. As shown in the figure, the paper is folded so that vertex &lt;math&gt;C&lt;/math&gt; meets edge &lt;math&gt;\overline{AD}&lt;/math&gt; at point &lt;math&gt;C'&lt;/math&gt;, and edge &lt;math&gt;\overline{AB}&lt;/math&gt; at point &lt;math&gt;E&lt;/math&gt;. Suppose that &lt;math&gt;C'D = \frac{1}{3}&lt;/math&gt;. What is the perimeter of triangle &lt;math&gt;\bigtriangleup AEC' ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{13}{6} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}&lt;/math&gt;<br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> pair A=(0,1);<br /> pair CC=(0.666666666666,1);<br /> pair D=(1,1);<br /> pair F=(1,0.62);<br /> pair C=(1,0);<br /> pair B=(0,0);<br /> pair G=(0,0.25);<br /> pair H=(-0.13,0.41);<br /> pair E=(0,0.5);<br /> dot(A^^CC^^D^^C^^B^^E);<br /> draw(E--A--D--F);<br /> draw(G--B--C--F, dashed);<br /> fill(E--CC--F--G--H--E--CC--cycle, gray);<br /> draw(E--CC--F--G--H--E--CC);<br /> label(&quot;A&quot;,A,NW);<br /> label(&quot;B&quot;,B,SW);<br /> label(&quot;C&quot;,C,SE);<br /> label(&quot;D&quot;,D,NE);<br /> label(&quot;E&quot;,E,NW);<br /> label(&quot;C'&quot;,CC,N);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> <br /> We can set the point on &lt;math&gt;CD&lt;/math&gt; where the fold occurs as point &lt;math&gt;F&lt;/math&gt;. Then, we can set &lt;math&gt;FD&lt;/math&gt; as &lt;math&gt;x&lt;/math&gt;, and &lt;math&gt;CF&lt;/math&gt; as &lt;math&gt;1-x&lt;/math&gt; because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for &lt;math&gt;x&lt;/math&gt;, we get, <br /> <br /> &lt;cmath&gt;x^2 + (\frac{1}{3})^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}&lt;/cmath&gt;<br /> <br /> We know this is a 3-4-5 triangle because the side lengths are &lt;math&gt;\frac{3}{9}, \frac{4}{9}, \frac{5}{9}&lt;/math&gt;. We also know that &lt;math&gt;EAC'&lt;/math&gt; is similar to &lt;math&gt;C'DF&lt;/math&gt; because angle &lt;math&gt;C'&lt;/math&gt; is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of &lt;math&gt;C'DF * \frac{AC'}{DF}&lt;/math&gt;. Thats just &lt;math&gt;\frac{4}{3} * \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} * \frac{3}{2} = 2&lt;/math&gt;. Therefore, the final answer is &lt;math&gt;\boxed{A}&lt;/math&gt;<br /> <br /> ~Tony_Li2007<br /> <br /> ==Solution 2==<br /> Let line we're reflecting over be &lt;math&gt;\ell&lt;/math&gt;, and let the points where it hits &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, be &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt;, respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line &lt;math&gt;\ell&lt;/math&gt;. The segment &lt;math&gt;CC'&lt;/math&gt; has slope &lt;math&gt;\frac{0 - 1}{1 - 2/3} = -3&lt;/math&gt;, implying line &lt;math&gt;\ell&lt;/math&gt; has a slope of &lt;math&gt;\frac{1}{3}&lt;/math&gt;. Also, the midpoint of segment &lt;math&gt;CC'&lt;/math&gt; is &lt;math&gt;\left( \frac{5}{6}, \frac{1}{2} \right)&lt;/math&gt;, so line &lt;math&gt;\ell&lt;/math&gt; passes through this point. Then, we get the equation of line &lt;math&gt;\ell&lt;/math&gt; is simply &lt;math&gt;y = \frac{1}{3} x + \frac{2}{9}&lt;/math&gt;. Then, if the point where &lt;math&gt;B&lt;/math&gt; is reflected over line &lt;math&gt;\ell&lt;/math&gt; is &lt;math&gt;B'&lt;/math&gt;, then we get &lt;math&gt;BB'&lt;/math&gt; is the line &lt;math&gt;y = -3x&lt;/math&gt;. The intersection of &lt;math&gt;\ell&lt;/math&gt; and segment &lt;math&gt;BB'&lt;/math&gt; is &lt;math&gt;\left( - \frac{1}{15}, \frac{1}{5} \right)&lt;/math&gt;. So, we get &lt;math&gt;B' = \left(- \frac{2}{15}, \frac{2}{5} \right)&lt;/math&gt;. Then, line segment &lt;math&gt;B'C'&lt;/math&gt; has equation &lt;math&gt;y = \frac{3}{4} x + \frac{1}{2}&lt;/math&gt;, so the point &lt;math&gt;E&lt;/math&gt; is the &lt;math&gt;y&lt;/math&gt;-intercept, or &lt;math&gt;\left(0, \frac{1}{2} \right)&lt;/math&gt;. This implies that &lt;math&gt;AE = \frac{1}{2}&lt;/math&gt;, AC' = \frac{2}{3}&lt;math&gt;, and by the Pythagorean Theorem, &lt;/math&gt;EC' = \frac{5}{6}&lt;math&gt; (or you could notice &lt;/math&gt;\triangle AEC'&lt;math&gt; is a &lt;/math&gt;3-4-5&lt;math&gt; right triangle). Then, the perimeter is &lt;/math&gt;\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2&lt;math&gt;, so our answer is &lt;/math&gt;\boxed{\textbf{(A)} ~2}&lt;math&gt;. ~rocketsri<br /> <br /> ==Solution 3 (Fakesolve):==<br /> Assume that E is the midpoint of &lt;/math&gt;\overline{AB}&lt;math&gt;. Then, &lt;/math&gt;\overline{AE}=\frac{1}{2}&lt;math&gt; and since &lt;/math&gt;C'D=\frac{1}{3}&lt;math&gt;, &lt;/math&gt;\overline{AC'}=\frac{2}{3}&lt;math&gt;. By the Pythagorean Theorem, &lt;/math&gt;\overline{EC'}=\frac{5}{6}&lt;math&gt;. It easily follows that our desired perimeter is &lt;/math&gt;2 \rightarrow \boxed{A}$~samrocksnature<br /> <br /> == Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) ==<br /> https://youtu.be/cagzLmdbqYQ<br /> <br /> ~ pi_is_3.14<br /> <br /> {{AMC10 box|year=2021|ab=B|num-b=20|num-a=22}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_21&diff=146184 2021 AMC 10B Problems/Problem 21 2021-02-12T21:18:34Z <p>Rocketsri: </p> <hr /> <div>==Problem==<br /> A square piece of paper has side length &lt;math&gt;1&lt;/math&gt; and vertices &lt;math&gt;A,B,C,&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; in that order. As shown in the figure, the paper is folded so that vertex &lt;math&gt;C&lt;/math&gt; meets edge &lt;math&gt;\overline{AD}&lt;/math&gt; at point &lt;math&gt;C'&lt;/math&gt;, and edge &lt;math&gt;\overline{AB}&lt;/math&gt; at point &lt;math&gt;E&lt;/math&gt;. Suppose that &lt;math&gt;C'D = \frac{1}{3}&lt;/math&gt;. What is the perimeter of triangle &lt;math&gt;\bigtriangleup AEC' ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{13}{6} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}&lt;/math&gt;<br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> pair A=(0,1);<br /> pair CC=(0.666666666666,1);<br /> pair D=(1,1);<br /> pair F=(1,0.62);<br /> pair C=(1,0);<br /> pair B=(0,0);<br /> pair G=(0,0.25);<br /> pair H=(-0.13,0.41);<br /> pair E=(0,0.5);<br /> dot(A^^CC^^D^^C^^B^^E);<br /> draw(E--A--D--F);<br /> draw(G--B--C--F, dashed);<br /> fill(E--CC--F--G--H--E--CC--cycle, gray);<br /> draw(E--CC--F--G--H--E--CC);<br /> label(&quot;A&quot;,A,NW);<br /> label(&quot;B&quot;,B,SW);<br /> label(&quot;C&quot;,C,SE);<br /> label(&quot;D&quot;,D,NE);<br /> label(&quot;E&quot;,E,NW);<br /> label(&quot;C'&quot;,CC,N);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> <br /> We can set the point on &lt;math&gt;CD&lt;/math&gt; where the fold occurs as point &lt;math&gt;F&lt;/math&gt;. Then, we can set &lt;math&gt;FD&lt;/math&gt; as &lt;math&gt;x&lt;/math&gt;, and &lt;math&gt;CF&lt;/math&gt; as &lt;math&gt;1-x&lt;/math&gt; because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for &lt;math&gt;x&lt;/math&gt;, we get, <br /> <br /> &lt;cmath&gt;x^2 + (\frac{1}{3})^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}&lt;/cmath&gt;<br /> <br /> We know this is a 3-4-5 triangle because the side lengths are &lt;math&gt;\frac{3}{9}, \frac{4}{9}, \frac{5}{9}&lt;/math&gt;. We also know that &lt;math&gt;EAC'&lt;/math&gt; is similar to &lt;math&gt;C'DF&lt;/math&gt; because angle &lt;math&gt;C'&lt;/math&gt; is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of &lt;math&gt;C'DF * \frac{AC'}{DF}&lt;/math&gt;. Thats just &lt;math&gt;\frac{4}{3} * \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} * \frac{3}{2} = 2&lt;/math&gt;. Therefore, the final answer is &lt;math&gt;\boxed{A}&lt;/math&gt;<br /> <br /> ~Tony_Li2007<br /> <br /> ==Solution 2==<br /> Let line we're reflecting over be &lt;math&gt;\ell&lt;/math&gt;, and let the points where it hits &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, be &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt;, respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line &lt;math&gt;\ell&lt;/math&gt;. The segment &lt;math&gt;CC'&lt;/math&gt; has slope &lt;math&gt;\frac{0 - 1}{1 - 2/3} = \frac{-3}&lt;/math&gt;, implying line &lt;math&gt;\ell&lt;/math&gt; has a slope of &lt;math&gt;\frac{1}{3}&lt;/math&gt;. Also, the midpoint of segment &lt;math&gt;CC'&lt;/math&gt; is &lt;math&gt;\left( \frac{5}{6}, \frac{1}{2} \right)&lt;/math&gt;, so line &lt;math&gt;\ell&lt;/math&gt; passes through this point. Then, we get the equation of line &lt;math&gt;\ell&lt;/math&gt; is simply &lt;math&gt;y = \frac{1}{3} x + \frac{2}{9}&lt;/math&gt;. Then, if the point where &lt;math&gt;B&lt;/math&gt; is reflected over line &lt;math&gt;\ell&lt;/math&gt; is &lt;math&gt;B'&lt;/math&gt;, then we get &lt;math&gt;BB'&lt;/math&gt; is the line &lt;math&gt;y = -3x&lt;/math&gt;. The intersection of &lt;math&gt;\ell&lt;/math&gt; and segment &lt;math&gt;BB'&lt;/math&gt; is &lt;math&gt;\left( - \frac{1}{15}, \frac{1}{5} \right)&lt;/math&gt;. So, we get &lt;math&gt;B' = \left(- \frac{2}{15}, \frac{2}{5} \right)&lt;/math&gt;. Then, line segment &lt;math&gt;B'C'&lt;/math&gt; has equation &lt;math&gt;y = \frac{3}{4} x + \frac{1}{2}&lt;/math&gt;, so the point &lt;math&gt;E&lt;/math&gt; is the &lt;math&gt;y&lt;/math&gt;-intercept, or &lt;math&gt;\left(0, \frac{1}{2} \right)&lt;/math&gt;. This implies that &lt;math&gt;AE = \frac{1}{2}&lt;/math&gt;, AC' = \frac{2}{3}&lt;math&gt;, and by the Pythagorean Theorem, &lt;/math&gt;EC' = \frac{5}{6}&lt;math&gt; (or you could notice &lt;/math&gt;\triangle AEC'&lt;math&gt; is a &lt;/math&gt;3-4-5&lt;math&gt; right triangle). Then, the perimeter is &lt;/math&gt;\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2&lt;math&gt;, so our answer is &lt;/math&gt;\boxed{\textbf{(A)} ~2}&lt;math&gt;. ~rocketsri<br /> <br /> ==Solution 3 (Fakesolve):==<br /> Assume that E is the midpoint of &lt;/math&gt;\overline{AB}&lt;math&gt;. Then, &lt;/math&gt;\overline{AE}=\frac{1}{2}&lt;math&gt; and since &lt;/math&gt;C'D=\frac{1}{3}&lt;math&gt;, &lt;/math&gt;\overline{AC'}=\frac{2}{3}&lt;math&gt;. By the Pythagorean Theorem, &lt;/math&gt;\overline{EC'}=\frac{5}{6}&lt;math&gt;. It easily follows that our desired perimeter is &lt;/math&gt;2 \rightarrow \boxed{A}$ ~samrocksnature<br /> <br /> == Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) ==<br /> https://youtu.be/cagzLmdbqYQ<br /> <br /> ~ pi_is_3.14<br /> <br /> {{AMC10 box|year=2021|ab=B|num-b=20|num-a=22}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_20&diff=146179 2021 AMC 12B Problems/Problem 20 2021-02-12T20:40:47Z <p>Rocketsri: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;Q(z)&lt;/math&gt; and &lt;math&gt;R(z)&lt;/math&gt; be the unique polynomials such that&lt;cmath&gt;z^{2021}+1=(z^2+z+1)Q(z)+R(z)&lt;/cmath&gt;and the degree of &lt;math&gt;R&lt;/math&gt; is less than &lt;math&gt;2.&lt;/math&gt; What is &lt;math&gt;R(z)?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }-z \qquad \textbf{(B) }-1 \qquad \textbf{(C) }2021\qquad \textbf{(D) }z+1 \qquad \textbf{(E) }2z+1&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that<br /> &lt;cmath&gt;z^3-1\equiv 0\pmod{z^2+z+1}&lt;/cmath&gt;<br /> so if &lt;math&gt;F(z)&lt;/math&gt; is the remainder when dividing by &lt;math&gt;z^3-1&lt;/math&gt;,<br /> &lt;cmath&gt;F(z)\equiv R(z)\pmod{z^2+z+1}.&lt;/cmath&gt;<br /> Now,<br /> &lt;cmath&gt;z^{2021}+1= (z^3-1)(z^{2018} + z^{2015} + \cdots + z^2) + z^2+1&lt;/cmath&gt;<br /> So &lt;math&gt;F(z) = z^2+1&lt;/math&gt;, and<br /> &lt;cmath&gt;R(z)\equiv F(z) \equiv -z\pmod{z^2+z+1}&lt;/cmath&gt;<br /> The answer is &lt;math&gt;\boxed{\textbf{(A) }-z}.&lt;/math&gt;<br /> <br /> ==Solution 1b (More Thorough Version of 1)==<br /> Instead of dealing with a nasty &lt;math&gt;z^2+z+1&lt;/math&gt;, we can instead deal with the nice &lt;math&gt;z^3 - 1&lt;/math&gt;, as &lt;math&gt;z^2+z+1&lt;/math&gt; is a factor of &lt;math&gt;z^3-1&lt;/math&gt;. Then, we try to see what &lt;math&gt;\frac{z^{2021} + 1}{z^3 - 1}&lt;/math&gt; is. Of course, we will need a &lt;math&gt;z^{2018}&lt;/math&gt;, getting &lt;math&gt;z^{2021} - z^{2018}&lt;/math&gt;. Then, we've gotta get rid of the &lt;math&gt;z^{2018}&lt;/math&gt; term, so we add a &lt;math&gt;z^{2015}&lt;/math&gt;, to get &lt;math&gt;z^{2021} - z^{2015}&lt;/math&gt;. This pattern continues, until we add a &lt;math&gt;z^2&lt;/math&gt; to get rid of &lt;math&gt;z^5&lt;/math&gt;, and end up with &lt;math&gt;z^{2021} - z^2&lt;/math&gt;. We can't add anything more to get rid of the &lt;math&gt;z^2&lt;/math&gt;, so our factor is &lt;math&gt;z^{2018} + z^{2015} + z^{2012} + \cdots + z^2&lt;/math&gt;. Then, to get rid of the &lt;math&gt;z^2&lt;/math&gt;, we must have a remainder of &lt;math&gt;+z^2&lt;/math&gt;, and to get the &lt;math&gt;+1&lt;/math&gt; we have to also have a &lt;math&gt;+1&lt;/math&gt; in the remainder. So, our product is &lt;cmath&gt;z^{2021}+1= (z^3-1)(z^{2018} + z^{2015} + \cdots + z^2) + z^2+1.&lt;/cmath&gt; Then, our remainder is &lt;math&gt;z^2+1&lt;/math&gt;. The remainder when dividing by &lt;math&gt;z^3-1&lt;/math&gt; must be the same when dividing by &lt;math&gt;z^2+z+1&lt;/math&gt;, modulo &lt;math&gt;z^2+z+1&lt;/math&gt;. So, we have that &lt;math&gt;z^2 + 1 \equiv R(z) \pmod{z^2+z+1}&lt;/math&gt;, or &lt;math&gt;R(z) \equiv -z\pmod{z^2+z+1}&lt;/math&gt;. This corresponds to answer choice &lt;math&gt;\boxed{\textbf{(A)} ~ -z}&lt;/math&gt;. ~rocketsri<br /> <br /> ==Solution 2 (Complex numbers)==<br /> One thing to note is that &lt;math&gt;R(z)&lt;/math&gt; takes the form of &lt;math&gt;Az + B&lt;/math&gt; for some constants A and B.<br /> Note that the roots of &lt;math&gt;z^2 + z + 1&lt;/math&gt; are part of the solutions of &lt;math&gt;z^3 -1 = 0&lt;/math&gt;<br /> They can be easily solved with roots of unity:<br /> &lt;cmath&gt;z^3 = 1&lt;/cmath&gt;<br /> &lt;cmath&gt;z^3 = e^{i 0}&lt;/cmath&gt;<br /> &lt;cmath&gt;z = e^{i 0}, e^{i \frac{2\pi}{3}}, e^{i -\frac{2\pi}{3}}&lt;/cmath&gt; &lt;cmath&gt;\newline&lt;/cmath&gt;<br /> Obviously the right two solutions are the roots of &lt;math&gt;z^2 + z + 1 = 0&lt;/math&gt;<br /> We substitute &lt;math&gt;e^{i \frac{2\pi}{3}}&lt;/math&gt; into the original equation, and &lt;math&gt;z^2 + z + 1&lt;/math&gt; becomes 0. Using De Moivre's theorem, we get:<br /> &lt;cmath&gt;e^{i\frac{4042\pi}{3}} + 1 = A \cdot e^{i \frac{2\pi}{3}} + B&lt;/cmath&gt; <br /> &lt;cmath&gt;e^{i\frac{4\pi}{3}} + 1 = A \cdot e^{i \frac{2\pi}{3}} + B&lt;/cmath&gt;<br /> Expanding into rectangular complex number form:<br /> &lt;cmath&gt;\frac{1}{2} - \frac{\sqrt{3}}{2} i = (-\frac{1}{2}A + B) + \frac{\sqrt{3}}{2} i A&lt;/cmath&gt;<br /> Comparing the real and imaginary parts, we get:<br /> &lt;cmath&gt;A = -1, B = 0&lt;/cmath&gt;<br /> The answer is &lt;math&gt;\boxed{\textbf{(A) }-z}&lt;/math&gt;. ~Jamess2022(burntTacos;-;)<br /> <br /> == Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-solving) ==<br /> https://youtu.be/nnjr17q7fS0<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_9&diff=146177 2021 AMC 12A Problems/Problem 9 2021-02-12T20:23:34Z <p>Rocketsri: /* Solution 4 (Engineer's Induction) */</p> <hr /> <div>==Problem==<br /> Which of the following is equivalent to<br /> &lt;cmath&gt;(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)} ~3^{127} + 2^{127} \qquad\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} \qquad\textbf{(C)} ~3^{128}-2^{128} \qquad\textbf{(D)} ~3^{128} + 3^{128} \qquad\textbf{(E)} ~5^{127}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> All you need to do is multiply the entire equation by &lt;math&gt;(3-2)&lt;/math&gt;. Then all the terms will easily simplify by difference of squares and you will get &lt;math&gt;3^{128}-2^{128}&lt;/math&gt; or &lt;math&gt;\boxed{C}&lt;/math&gt; as your final answer. Notice you don't need to worry about &lt;math&gt;3-2&lt;/math&gt; because that's equal to &lt;math&gt;1&lt;/math&gt;.<br /> <br /> -Lemonie<br /> <br /> ==Solution 2==<br /> <br /> If you weren't able to come up with the &lt;math&gt;(3 - 2)&lt;/math&gt; insight, then you could just notice that the answer is divisible by <br /> &lt;math&gt;(2 + 3) = 5&lt;/math&gt;, and &lt;math&gt;(2^2 + 3^2) = 13&lt;/math&gt;. We can then use Fermat's Little Theorem for &lt;math&gt;p = 5, 13&lt;/math&gt; on the answer choices to determine which of the answer choices are divisible by both &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;13&lt;/math&gt;. This is &lt;math&gt;\boxed{C}&lt;/math&gt;.<br /> <br /> -MEWTO<br /> <br /> ==Solution 3==<br /> <br /> After expanding the first few terms, the result after each term appears to be &lt;math&gt;2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}&lt;/math&gt; where n is the number of terms expanded. We can prove this using mathematical induction. The base step is trivial. When expanding another term, all of the previous terms multiplied by &lt;math&gt;2^{2^{n-1}}&lt;/math&gt; would give &lt;math&gt;2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^{2^{n-1}+1}\cdot{3^{2^{n-1}-1}} + 2^{2^{n-1}}\cdot{3^{2^{n-1}}}&lt;/math&gt;, and all the previous terms multiplied by &lt;math&gt;3^{2^{n-1}}&lt;/math&gt; would give &lt;math&gt;3^{2^n-1} + 3^{2^n-2}\cdot{2^1} + 3^{2^n-3}\cdot{2^2} + ... + 3^{2^{n-1}+1}\cdot{2^{2^{n-1}-1}} + 3^{2^{n-1}}\cdot{2^{2^{n-1}}}&lt;/math&gt;. Their sum is equal to &lt;math&gt;2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}&lt;/math&gt;, so the proof is complete. Since &lt;math&gt;\frac{3^{2^n}-2^{2^n}}{3-2}&lt;/math&gt; is equal to &lt;math&gt;2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}&lt;/math&gt;, the answer is &lt;math&gt;\frac{3^{2^7}-2^{2^7}}{3-2}=\boxed{C}&lt;/math&gt;.<br /> <br /> -SmileKat32<br /> <br /> == Solution 4 (Engineer's Induction) ==<br /> We can compute some of the first few partial products, and notice that &lt;math&gt;\prod_{k = 0}^{2^n} (2^{2^n}+3^{2^n}) = 3^{2^{n+1}} - 2^{2^{n+1}}&lt;/math&gt;. As we don't have to prove this, we get the product is &lt;math&gt;3^{2^7} - 2^{2^7} = 3^{128} - 2^{128}&lt;/math&gt;, and smugly click &lt;math&gt;\boxed{\textbf{(C)} ~3^{128} - 2^{128}}&lt;/math&gt;. ~rocketsri<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=P5al76DxyHY<br /> <br /> == Video Solution by OmegaLearn(Factorizations/Telescoping&amp; Meta-solving) ==<br /> https://youtu.be/H34IFMlq7Lk<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=9|num-a=11}}<br /> {{AMC12 box|year=2021|ab=A|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_25&diff=146176 2021 AMC 12A Problems/Problem 25 2021-02-12T20:23:14Z <p>Rocketsri: /* Solution 3 (Only Use If Low On Time) */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;d(n)&lt;/math&gt; denote the number of positive integers that divide &lt;math&gt;n&lt;/math&gt;, including &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;d(1)=1,d(2)=2,&lt;/math&gt; and &lt;math&gt;d(12)=6&lt;/math&gt;. (This function is known as the divisor function.) Let&lt;cmath&gt;f(n)=\frac{d(n)}{\sqrt n}.&lt;/cmath&gt;There is a unique positive integer &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;f(N)&gt;f(n)&lt;/math&gt; for all positive integers &lt;math&gt;n\ne N&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;N?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> Consider the prime factorization &lt;cmath&gt;n={p_1}^{e_1}{p_2}^{e_2}{p_3}^{e_3}\cdots{p_k}^{e_k}.&lt;/cmath&gt; By the Multiplication Principle, &lt;cmath&gt;d(n)=(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1).&lt;/cmath&gt; Now, we rewrite &lt;math&gt;f(n)&lt;/math&gt; as &lt;cmath&gt;f(n)=\frac{d(n)}{\sqrt n}=\frac{(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1)}{{p_1}^{{e_1}/3}{p_2}^{{e_2}/3}{p_3}^{{e_3}/3}\cdots{p_k}^{{e_k}/3}}=\left(\frac{e_1+1}{{p_1}^{{e_1}/3}}\right)\left(\frac{e_2+1}{{p_2}^{{e_2}/3}}\right)\left(\frac{e_3+1}{{p_3}^{{e_3}/3}}\right)\cdots\left(\frac{e_k+1}{{p_k}^{{e_k}/3}}\right).&lt;/cmath&gt; As &lt;math&gt;f(n)&gt;0&lt;/math&gt; for all positive integers &lt;math&gt;n,&lt;/math&gt; it follows that for all positive integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;f(a)&gt;f(b)&lt;/math&gt; if and only if &lt;math&gt;f(a)^3&gt;f(b)^3.&lt;/math&gt; So, &lt;math&gt;f(n)&lt;/math&gt; is maximized if and only if &lt;cmath&gt;f(n)^3=\left(\frac{(e_1+1)^3}{{p_1}^{e_1}}\right)\left(\frac{(e_2+1)^3}{{p_2}^{e_2}}\right)\left(\frac{(e_3+1)^3}{{p_3}^{e_3}}\right)\cdots\left(\frac{(e_k+1)^3}{{p_k}^{e_k}}\right)&lt;/cmath&gt; is maximized.<br /> <br /> For every factor &lt;math&gt;\frac{(e_i+1)^3}{{p_i}^{e_i}}&lt;/math&gt; with a fixed &lt;math&gt;p_i&lt;/math&gt; where &lt;math&gt;1\leq i\leq k,&lt;/math&gt; the denominator grows faster than the numerator, as exponential functions grow faster than polynomial functions. For each prime &lt;math&gt;p_i=2,3,5,7,\cdots,&lt;/math&gt; we look for the &lt;math&gt;e_i&lt;/math&gt; for which &lt;math&gt;\frac{(e_i+1)^3}{{p_i}^{e_i}}&lt;/math&gt; is a relative maximum:<br /> &lt;cmath&gt;\begin{tabular}{ c c c c }<br /> pi &amp; ei &amp; fraction &amp; choose? \\ <br /> \hline<br /> 2 &amp; 0 &amp; 1 &amp; \\ <br /> 2 &amp; 1 &amp; 4 &amp; \\<br /> 2 &amp; 2 &amp; 27/4 &amp;\\<br /> 2 &amp; 3 &amp; 8 &amp; yes\\<br /> 2 &amp; 4 &amp; 125/16 &amp; \\<br /> \hline<br /> 3 &amp; 0 &amp; 1 &amp;\\<br /> 3 &amp; 1 &amp; 8/3 &amp; \\<br /> 3 &amp; 2 &amp; 3 &amp; yes\\<br /> 3 &amp; 3 &amp; 64/27 &amp; \\<br /> \hline<br /> 5 &amp; 0 &amp; 1 &amp; \\<br /> 5 &amp; 1 &amp; 8/5 &amp; yes\\<br /> 5 &amp; 2 &amp; 27/25 &amp; \\<br /> \hline<br /> 7 &amp; 0 &amp; 1 &amp; \\<br /> 7 &amp; 1 &amp; 8/7 &amp; yes\\<br /> 7 &amp; 2 &amp; 27/49 &amp; \\<br /> \hline<br /> 11 &amp; 0 &amp; 1 &amp; yes \\<br /> 11 &amp; 1 &amp; 8/11 &amp; \\<br /> \hline<br /> ,,, &amp; ... &amp; ... &amp; <br /> \end{tabular}&lt;/cmath&gt;<br /> <br /> Finally, the number we seek is &lt;math&gt;N=2^3 3^2 5^1 7^1 = 2520.&lt;/math&gt; The sum of its digits is &lt;math&gt;2+5+2+0=\boxed{\textbf{(E) }9}&lt;/math&gt;<br /> <br /> ~MRENTHUSIASM<br /> <br /> ===Solution 2===<br /> A cube root seems bad, so we should just cube it. It seems that if the number is a multiple of 3, there are only two choices. If the number is a multiple of 9, there is one choice. We can prove that for all k is indivisible by 3, f(9k) &gt; f(3k) &gt; f(k). The divisors of 3k contain the divisors of k and the divisors of k multiplied by 3. The divisors of 9k contain the divisors of k, the divisors of k multiplied by 3, and the divisors of k multiplied by 9. &lt;cmath&gt;\frac{27}{9}d(k)^3 &gt; \frac{8}{3}d(k)^3 &gt; d(k)^3&lt;/cmath&gt; so &lt;math&gt;f(9k) &gt; f(3k) &gt; f(k)&lt;/math&gt; and since &lt;math&gt;\boxed{\textbf{(E) }9}&lt;/math&gt; is the only possible answer choice, it is the answer.<br /> <br /> ==Solution 3 (Only Use If Low On Time)==<br /> We can guess that &lt;math&gt;N&lt;/math&gt; would be divisible by &lt;math&gt;9&lt;/math&gt;. Recall, for a number to be divisible by &lt;math&gt;9&lt;/math&gt;, the sum of its digits must also be divisible by &lt;math&gt;9&lt;/math&gt;. Since there's only one choice where it's divisible by &lt;math&gt;9&lt;/math&gt;, we get &lt;math&gt;\boxed{\textbf{(E)} ~9}&lt;/math&gt; as our answer. ~rocketsri<br /> <br /> == Video Solution by OmegaLearn (Multiplicative function properties + Meta-solving ) ==<br /> https://youtu.be/6P-0ZHAaC_A<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=24|after=Last problem}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_25&diff=146175 2021 AMC 12A Problems/Problem 25 2021-02-12T20:22:57Z <p>Rocketsri: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;d(n)&lt;/math&gt; denote the number of positive integers that divide &lt;math&gt;n&lt;/math&gt;, including &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;d(1)=1,d(2)=2,&lt;/math&gt; and &lt;math&gt;d(12)=6&lt;/math&gt;. (This function is known as the divisor function.) Let&lt;cmath&gt;f(n)=\frac{d(n)}{\sqrt n}.&lt;/cmath&gt;There is a unique positive integer &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;f(N)&gt;f(n)&lt;/math&gt; for all positive integers &lt;math&gt;n\ne N&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;N?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> Consider the prime factorization &lt;cmath&gt;n={p_1}^{e_1}{p_2}^{e_2}{p_3}^{e_3}\cdots{p_k}^{e_k}.&lt;/cmath&gt; By the Multiplication Principle, &lt;cmath&gt;d(n)=(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1).&lt;/cmath&gt; Now, we rewrite &lt;math&gt;f(n)&lt;/math&gt; as &lt;cmath&gt;f(n)=\frac{d(n)}{\sqrt n}=\frac{(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1)}{{p_1}^{{e_1}/3}{p_2}^{{e_2}/3}{p_3}^{{e_3}/3}\cdots{p_k}^{{e_k}/3}}=\left(\frac{e_1+1}{{p_1}^{{e_1}/3}}\right)\left(\frac{e_2+1}{{p_2}^{{e_2}/3}}\right)\left(\frac{e_3+1}{{p_3}^{{e_3}/3}}\right)\cdots\left(\frac{e_k+1}{{p_k}^{{e_k}/3}}\right).&lt;/cmath&gt; As &lt;math&gt;f(n)&gt;0&lt;/math&gt; for all positive integers &lt;math&gt;n,&lt;/math&gt; it follows that for all positive integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;f(a)&gt;f(b)&lt;/math&gt; if and only if &lt;math&gt;f(a)^3&gt;f(b)^3.&lt;/math&gt; So, &lt;math&gt;f(n)&lt;/math&gt; is maximized if and only if &lt;cmath&gt;f(n)^3=\left(\frac{(e_1+1)^3}{{p_1}^{e_1}}\right)\left(\frac{(e_2+1)^3}{{p_2}^{e_2}}\right)\left(\frac{(e_3+1)^3}{{p_3}^{e_3}}\right)\cdots\left(\frac{(e_k+1)^3}{{p_k}^{e_k}}\right)&lt;/cmath&gt; is maximized.<br /> <br /> For every factor &lt;math&gt;\frac{(e_i+1)^3}{{p_i}^{e_i}}&lt;/math&gt; with a fixed &lt;math&gt;p_i&lt;/math&gt; where &lt;math&gt;1\leq i\leq k,&lt;/math&gt; the denominator grows faster than the numerator, as exponential functions grow faster than polynomial functions. For each prime &lt;math&gt;p_i=2,3,5,7,\cdots,&lt;/math&gt; we look for the &lt;math&gt;e_i&lt;/math&gt; for which &lt;math&gt;\frac{(e_i+1)^3}{{p_i}^{e_i}}&lt;/math&gt; is a relative maximum:<br /> &lt;cmath&gt;\begin{tabular}{ c c c c }<br /> pi &amp; ei &amp; fraction &amp; choose? \\ <br /> \hline<br /> 2 &amp; 0 &amp; 1 &amp; \\ <br /> 2 &amp; 1 &amp; 4 &amp; \\<br /> 2 &amp; 2 &amp; 27/4 &amp;\\<br /> 2 &amp; 3 &amp; 8 &amp; yes\\<br /> 2 &amp; 4 &amp; 125/16 &amp; \\<br /> \hline<br /> 3 &amp; 0 &amp; 1 &amp;\\<br /> 3 &amp; 1 &amp; 8/3 &amp; \\<br /> 3 &amp; 2 &amp; 3 &amp; yes\\<br /> 3 &amp; 3 &amp; 64/27 &amp; \\<br /> \hline<br /> 5 &amp; 0 &amp; 1 &amp; \\<br /> 5 &amp; 1 &amp; 8/5 &amp; yes\\<br /> 5 &amp; 2 &amp; 27/25 &amp; \\<br /> \hline<br /> 7 &amp; 0 &amp; 1 &amp; \\<br /> 7 &amp; 1 &amp; 8/7 &amp; yes\\<br /> 7 &amp; 2 &amp; 27/49 &amp; \\<br /> \hline<br /> 11 &amp; 0 &amp; 1 &amp; yes \\<br /> 11 &amp; 1 &amp; 8/11 &amp; \\<br /> \hline<br /> ,,, &amp; ... &amp; ... &amp; <br /> \end{tabular}&lt;/cmath&gt;<br /> <br /> Finally, the number we seek is &lt;math&gt;N=2^3 3^2 5^1 7^1 = 2520.&lt;/math&gt; The sum of its digits is &lt;math&gt;2+5+2+0=\boxed{\textbf{(E) }9}&lt;/math&gt;<br /> <br /> ~MRENTHUSIASM<br /> <br /> ===Solution 2===<br /> A cube root seems bad, so we should just cube it. It seems that if the number is a multiple of 3, there are only two choices. If the number is a multiple of 9, there is one choice. We can prove that for all k is indivisible by 3, f(9k) &gt; f(3k) &gt; f(k). The divisors of 3k contain the divisors of k and the divisors of k multiplied by 3. The divisors of 9k contain the divisors of k, the divisors of k multiplied by 3, and the divisors of k multiplied by 9. &lt;cmath&gt;\frac{27}{9}d(k)^3 &gt; \frac{8}{3}d(k)^3 &gt; d(k)^3&lt;/cmath&gt; so &lt;math&gt;f(9k) &gt; f(3k) &gt; f(k)&lt;/math&gt; and since &lt;math&gt;\boxed{\textbf{(E) }9}&lt;/math&gt; is the only possible answer choice, it is the answer.<br /> <br /> ==Solution 3 (Only Use If Low On Time)==<br /> We can guess that &lt;math&gt;N&lt;/math&gt; would be divisible by &lt;math&gt;9&lt;/math&gt;. Recall, for a number to be divisible by &lt;math&gt;9&lt;/math&gt;, the sum of its digits must also be divisible by &lt;math&gt;9&lt;/math&gt;. Since there's only one choice where it's divisible by &lt;math&gt;9&lt;/math&gt;, we get &lt;math&gt;\boxed{\textbf{(E)} ~9}&lt;/math&gt; as our answer<br /> <br /> == Video Solution by OmegaLearn (Multiplicative function properties + Meta-solving ) ==<br /> https://youtu.be/6P-0ZHAaC_A<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=24|after=Last problem}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_9&diff=146174 2021 AMC 12A Problems/Problem 9 2021-02-12T20:22:18Z <p>Rocketsri: /* Solution 4 (Engineer's Induction) */</p> <hr /> <div>==Problem==<br /> Which of the following is equivalent to<br /> &lt;cmath&gt;(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)} ~3^{127} + 2^{127} \qquad\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} \qquad\textbf{(C)} ~3^{128}-2^{128} \qquad\textbf{(D)} ~3^{128} + 3^{128} \qquad\textbf{(E)} ~5^{127}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> All you need to do is multiply the entire equation by &lt;math&gt;(3-2)&lt;/math&gt;. Then all the terms will easily simplify by difference of squares and you will get &lt;math&gt;3^{128}-2^{128}&lt;/math&gt; or &lt;math&gt;\boxed{C}&lt;/math&gt; as your final answer. Notice you don't need to worry about &lt;math&gt;3-2&lt;/math&gt; because that's equal to &lt;math&gt;1&lt;/math&gt;.<br /> <br /> -Lemonie<br /> <br /> ==Solution 2==<br /> <br /> If you weren't able to come up with the &lt;math&gt;(3 - 2)&lt;/math&gt; insight, then you could just notice that the answer is divisible by <br /> &lt;math&gt;(2 + 3) = 5&lt;/math&gt;, and &lt;math&gt;(2^2 + 3^2) = 13&lt;/math&gt;. We can then use Fermat's Little Theorem for &lt;math&gt;p = 5, 13&lt;/math&gt; on the answer choices to determine which of the answer choices are divisible by both &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;13&lt;/math&gt;. This is &lt;math&gt;\boxed{C}&lt;/math&gt;.<br /> <br /> -MEWTO<br /> <br /> ==Solution 3==<br /> <br /> After expanding the first few terms, the result after each term appears to be &lt;math&gt;2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}&lt;/math&gt; where n is the number of terms expanded. We can prove this using mathematical induction. The base step is trivial. When expanding another term, all of the previous terms multiplied by &lt;math&gt;2^{2^{n-1}}&lt;/math&gt; would give &lt;math&gt;2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^{2^{n-1}+1}\cdot{3^{2^{n-1}-1}} + 2^{2^{n-1}}\cdot{3^{2^{n-1}}}&lt;/math&gt;, and all the previous terms multiplied by &lt;math&gt;3^{2^{n-1}}&lt;/math&gt; would give &lt;math&gt;3^{2^n-1} + 3^{2^n-2}\cdot{2^1} + 3^{2^n-3}\cdot{2^2} + ... + 3^{2^{n-1}+1}\cdot{2^{2^{n-1}-1}} + 3^{2^{n-1}}\cdot{2^{2^{n-1}}}&lt;/math&gt;. Their sum is equal to &lt;math&gt;2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}&lt;/math&gt;, so the proof is complete. Since &lt;math&gt;\frac{3^{2^n}-2^{2^n}}{3-2}&lt;/math&gt; is equal to &lt;math&gt;2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}&lt;/math&gt;, the answer is &lt;math&gt;\frac{3^{2^7}-2^{2^7}}{3-2}=\boxed{C}&lt;/math&gt;.<br /> <br /> -SmileKat32<br /> <br /> == Solution 4 (Engineer's Induction) ==<br /> We can compute some of the first few partial products, and notice that &lt;math&gt;\prod_{k = 0}^{2^n} (2^{2^n}+3^{2^n}) = 3^{2^{n+1}} - 2^{2^{n+1}}&lt;/math&gt;. As we don't have to prove this, we get the product is &lt;math&gt;3^{2^7} - 2^{2^7} = 3^{128} - 2^{128}&lt;/math&gt;, and smugly click &lt;math&gt;\boxed{\textbf{(C)} 3^{128} - 2^{128}}&lt;/math&gt;.<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=P5al76DxyHY<br /> <br /> == Video Solution by OmegaLearn(Factorizations/Telescoping&amp; Meta-solving) ==<br /> https://youtu.be/H34IFMlq7Lk<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=9|num-a=11}}<br /> {{AMC12 box|year=2021|ab=A|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_9&diff=146168 2021 AMC 12A Problems/Problem 9 2021-02-12T20:13:55Z <p>Rocketsri: </p> <hr /> <div>==Problem==<br /> Which of the following is equivalent to<br /> &lt;cmath&gt;(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)} ~3^{127} + 2^{127} \qquad\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} \qquad\textbf{(C)} ~3^{128}-2^{128} \qquad\textbf{(D)} ~3^{128} + 3^{128} \qquad\textbf{(E)} ~5^{127}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> All you need to do is multiply the entire equation by &lt;math&gt;(3-2)&lt;/math&gt;. Then all the terms will easily simplify by difference of squares and you will get &lt;math&gt;3^{128}-2^{128}&lt;/math&gt; or &lt;math&gt;\boxed{C}&lt;/math&gt; as your final answer. Notice you don't need to worry about &lt;math&gt;3-2&lt;/math&gt; because that's equal to &lt;math&gt;1&lt;/math&gt;.<br /> <br /> -Lemonie<br /> <br /> ==Solution 2==<br /> <br /> If you weren't able to come up with the &lt;math&gt;(3 - 2)&lt;/math&gt; insight, then you could just notice that the answer is divisible by <br /> &lt;math&gt;(2 + 3) = 5&lt;/math&gt;, and &lt;math&gt;(2^2 + 3^2) = 13&lt;/math&gt;. We can then use Fermat's Little Theorem for &lt;math&gt;p = 5, 13&lt;/math&gt; on the answer choices to determine which of the answer choices are divisible by both &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;13&lt;/math&gt;. This is &lt;math&gt;\boxed{C}&lt;/math&gt;.<br /> <br /> -MEWTO<br /> <br /> ==Solution 3==<br /> <br /> After expanding the first few terms, the result after each term appears to be &lt;math&gt;2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}&lt;/math&gt; where n is the number of terms expanded. We can prove this using mathematical induction. The base step is trivial. When expanding another term, all of the previous terms multiplied by &lt;math&gt;2^{2^{n-1}}&lt;/math&gt; would give &lt;math&gt;2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^{2^{n-1}+1}\cdot{3^{2^{n-1}-1}} + 2^{2^{n-1}}\cdot{3^{2^{n-1}}}&lt;/math&gt;, and all the previous terms multiplied by &lt;math&gt;3^{2^{n-1}}&lt;/math&gt; would give &lt;math&gt;3^{2^n-1} + 3^{2^n-2}\cdot{2^1} + 3^{2^n-3}\cdot{2^2} + ... + 3^{2^{n-1}+1}\cdot{2^{2^{n-1}-1}} + 3^{2^{n-1}}\cdot{2^{2^{n-1}}}&lt;/math&gt;. Their sum is equal to &lt;math&gt;2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}&lt;/math&gt;, so the proof is complete. Since &lt;math&gt;\frac{3^{2^n}-2^{2^n}}{3-2}&lt;/math&gt; is equal to &lt;math&gt;2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}&lt;/math&gt;, the answer is &lt;math&gt;\frac{3^{2^7}-2^{2^7}}{3-2}=\boxed{C}&lt;/math&gt;.<br /> <br /> -SmileKat32<br /> <br /> == Solution 4 (Engineer's Induction) ==<br /> We can compute some of the first few partial products, and notice that &lt;math&gt;\prod_{k = 0}^{2^n} (2^{2^n}+3^{2^n}) = 3^{2^{n+1}} - 2^{2^{n+1}}&lt;/math&gt;. As we don't have to prove this, we get the product is &lt;math&gt;3^{2^7} - 2^{2^7} = 3^{128} - 2^{128}&lt;/math&gt;, and smugly circle &lt;math&gt;\boxed{\textbf{(C)} 3^{128} - 2^{128}}&lt;/math&gt;.<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=P5al76DxyHY<br /> <br /> == Video Solution by OmegaLearn(Factorizations/Telescoping&amp; Meta-solving) ==<br /> https://youtu.be/H34IFMlq7Lk<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=9|num-a=11}}<br /> {{AMC12 box|year=2021|ab=A|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_23&diff=146161 2021 AMC 12A Problems/Problem 23 2021-02-12T19:24:06Z <p>Rocketsri: /* Solution 6 (Casework) */</p> <hr /> <div>==Problem==<br /> Frieda the frog begins a sequence of hops on a &lt;math&gt;3 \times 3&lt;/math&gt; grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she &quot;wraps around&quot; and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops &quot;up&quot;, the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}&lt;/math&gt;<br /> <br /> ==Solution 1 (complementary counting)==<br /> We will use complementary counting.<br /> First, the frog can go left with probability &lt;math&gt;\frac14&lt;/math&gt;. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since &lt;math&gt;4\cdot \frac14=1&lt;/math&gt;, we will ignore the leading probability.<br /> <br /> From the left, she either goes left to another edge (&lt;math&gt;\frac14&lt;/math&gt;) or back to the center (&lt;math&gt;\frac14&lt;/math&gt;). Time for some casework.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; She goes back to the center.<br /> <br /> Now, she can go in any 4 directions, and then has 2 options from that edge. This gives &lt;math&gt;\frac12&lt;/math&gt;. --End case 1<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; She goes to another edge (rightmost).<br /> <br /> Subcase 1: She goes back to the left edge. She now has 2 places to go, giving &lt;math&gt;\frac12&lt;/math&gt;<br /> <br /> Subcase 2: She goes to the center. Now any move works.<br /> <br /> &lt;math&gt;\frac14 \cdot \frac12 + \frac14 \cdot 1=\frac18 + \frac 14=\frac38&lt;/math&gt; for this case. --End case 2<br /> <br /> She goes back to the center in Case 1 with probability &lt;math&gt;\frac14&lt;/math&gt;, and to the right edge with probability &lt;math&gt;\frac14&lt;/math&gt;<br /> <br /> So, our answer is &lt;math&gt;\frac14 \cdot \frac12 + \frac14 \cdot \frac38=\frac14 (\frac12+\frac38)=\frac14 \cdot \frac78 = \frac7{32}&lt;/math&gt;<br /> <br /> But, don't forget complementary counting. So, we get &lt;math&gt;1-\frac7{32}=\frac{25}{32} \implies \boxed{D}&lt;/math&gt;. ~ firebolt360<br /> <br /> Video Solution for those who prefer: https://youtu.be/ude2rzO1cmk ~ firebolt360<br /> <br /> ==Solution 2 (direct counting and probability states)==<br /> We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have &lt;math&gt;1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{D}&lt;/math&gt;. ~IceWolf10<br /> <br /> ==Solution 3 (Similar to Solution 2, but Finds the Numerator and Denominator Separately)==<br /> &lt;b&gt;Denominator&lt;/b&gt;<br /> <br /> There are &lt;math&gt;4^4=256&lt;/math&gt; ways to make 4 hops without restrictions.<br /> <br /> <br /> &lt;b&gt;Numerator (Casework)&lt;/b&gt;<br /> <br /> Suppose Frieda makes 4 hops without stopping. We perform casework on which hop reaches a corner for the first time.<br /> <br /> <br /> (1) Hop #2 (Hops #3 and #4 have no restrictions)<br /> <br /> The 4 independent hops have 4, 2, 4, 4 options, respectively. So, this case has &lt;math&gt;4\cdot2\cdot4\cdot4=128&lt;/math&gt; ways.<br /> <br /> <br /> (2) Hop #3 (Hope #4 has no restriction)<br /> <br /> No matter which direction the first hop takes, the second hop must &quot;wrap around&quot;.<br /> <br /> The 4 independent hops have 4, 1, 2, 4 options, respectively. So, this case has &lt;math&gt;4\cdot1\cdot2\cdot4=32&lt;/math&gt; ways.<br /> <br /> <br /> (3) Hop #4<br /> <br /> Two sub-cases:<br /> <br /> (3.1) The second hop &quot;wraps around&quot;. It follows that the third hop also &quot;wraps around&quot;.<br /> <br /> The 4 independent hops have 4, 1, 1, 2 options, respectively. So, this sub-case has &lt;math&gt;4\cdot1\cdot1\cdot2=8&lt;/math&gt; ways.<br /> <br /> (3.2) The second hop backs to the center.<br /> <br /> The 4 independent hops have 4, 1, 4, 2 options, respectively. So, this sub-case has &lt;math&gt;4\cdot1\cdot4\cdot2=32&lt;/math&gt; ways.<br /> <br /> Together, Case (3) has &lt;math&gt;8+32=40&lt;/math&gt; ways.<br /> <br /> <br /> The numerator is &lt;math&gt;128+32+40=200.&lt;/math&gt;<br /> <br /> <br /> &lt;b&gt;Probability&lt;/b&gt;<br /> <br /> &lt;math&gt;\frac{200}{256}=\boxed{\textbf{(D)} ~\frac{25}{32}}.&lt;/math&gt;<br /> <br /> <br /> This problem is quite similar to 1995 AIME Problem 3: https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3<br /> <br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;C_n&lt;/math&gt; be the probability that Frieda is on the central square after n moves, &lt;math&gt;E_n&lt;/math&gt; be the probability that Frieda is on one of the four squares on the middle of the edges after n moves, and &lt;math&gt;V_n&lt;/math&gt; (V for vertex) be the probability that Frieda is on a corner after n moves. The only way to reach the center is by moving in &lt;math&gt;1&lt;/math&gt; specific direction out of &lt;math&gt;4&lt;/math&gt; total directions from the middle of an edge, so &lt;math&gt;C_{n+1}=\frac{E_n}{4}&lt;/math&gt;. The ways to reach the middle of an edge are by moving in any direction from the center or by moving in &lt;math&gt;1&lt;/math&gt; specific direction from the middle of an edge, so &lt;math&gt;E_{n+1}=C_n+\frac{E_n}{4}&lt;/math&gt;. The ways to reach a corner are by simply staying there after reaching there in a previous move or by moving in &lt;math&gt;2&lt;/math&gt; specific directions from the middle of an edge, so &lt;math&gt;V_{n+1}=V_n+\frac{E_n}{2}&lt;/math&gt;. Since Frieda always start from the center, &lt;math&gt;C_0=1&lt;/math&gt;, &lt;math&gt;E_0=0&lt;/math&gt;, and &lt;math&gt;V_0=0&lt;/math&gt;. We use the previous formulas to work out &lt;math&gt;V_4&lt;/math&gt; and find it to be &lt;math&gt;\boxed{\textbf{(D)} ~\frac{25}{32}}&lt;/math&gt;.<br /> <br /> -SmileKat32<br /> <br /> ==Solution 5==<br /> Imagine an infinite grid of &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; squares such that there is a &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; square centered at &lt;math&gt;(3x, 3y)&lt;/math&gt; for all ordered pairs of integers &lt;math&gt;(x, y).&lt;/math&gt; It is easy to see that the problem is equivalent to Frieda moving left, right, up, or down on this infinite grid starting at &lt;math&gt;(0, 0)&lt;/math&gt;. (minus the teleportations) Since counting the complement set is easier, we'll count the number of &lt;math&gt;4&lt;/math&gt;-step paths such that Frieda never reaches a corner point. <br /> <br /> In other words, since the reachable corner points are &lt;math&gt;(\pm 1, \pm 1), (\pm 1, \pm 2), (\pm 2, \pm 1), &lt;/math&gt; and &lt;math&gt;(\pm 2, \pm 2),&lt;/math&gt; Frieda can only travel along the collection of points included in &lt;math&gt;S&lt;/math&gt;, where &lt;math&gt;S&lt;/math&gt; is all points on &lt;math&gt;x=0&lt;/math&gt; and &lt;math&gt;y=0&lt;/math&gt; such that &lt;math&gt;|y|&lt;4&lt;/math&gt; and &lt;math&gt;|x|&lt;4&lt;/math&gt;, respectively, plus all points on the big square with side length &lt;math&gt;6&lt;/math&gt; centered at &lt;math&gt;(0, 0).&lt;/math&gt; We then can proceed with casework:<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: Frieda never reaches &lt;math&gt;(0, \pm 3)&lt;/math&gt; nor &lt;math&gt;(\pm 3, 0).&lt;/math&gt;<br /> <br /> When Frieda only moves horizontally or vertically for her four moves, she can do so in &lt;math&gt;2^4 - 4 = 12&lt;/math&gt; ways for each case . Thus, &lt;math&gt;12 \cdot 2&lt;/math&gt; total paths for the subcase of staying in one direction. (For instance, all length &lt;math&gt;4&lt;/math&gt; combinations of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; except &lt;math&gt;FFFF&lt;/math&gt;, &lt;math&gt;BBBB&lt;/math&gt;, &lt;math&gt;FFFB&lt;/math&gt;, and &lt;math&gt;BBBF&lt;/math&gt; for the horizontal direction.) <br /> <br /> There is another subcase where she changes directions during her path. There are four symmetric cases for this subcase depending on which of the four quadrants Frieda hugs. For the first quadrant, the possible paths are &lt;math&gt;FBUD&lt;/math&gt;, &lt;math&gt;FBUU&lt;/math&gt;, &lt;math&gt;UDFB&lt;/math&gt;, and &lt;math&gt;UDFF.&lt;/math&gt; Thus, a total of &lt;math&gt;4 \cdot 4 = 16&lt;/math&gt; ways for this subcase. <br /> <br /> Total for Case &lt;math&gt;1&lt;/math&gt;: &lt;math&gt;24 + 16 = 40&lt;/math&gt;<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: Frieda reaches &lt;math&gt;(0, \pm 3)&lt;/math&gt; or &lt;math&gt;(\pm 3, 0)&lt;/math&gt;.<br /> <br /> Once Frieda reaches one of the points listed above (by using three moves), she has four choices for her last move. Thus, a total of &lt;math&gt;4 \cdot 4 = 16&lt;/math&gt; paths for this case.<br /> <br /> Our total number of paths never reaching coroners is thus &lt;math&gt;16+40=56,&lt;/math&gt; making for an answer of &lt;cmath&gt;\frac{4^4-56}{4^4} = \boxed{\textbf{(D)} ~\frac{25}{32}}.&lt;/cmath&gt;<br /> <br /> -fidgetboss_4000<br /> <br /> == Solution 6 (Casework) ==<br /> We take cases on the number of hops needed to reach a corner. For simplicity, denote &lt;math&gt;E&lt;/math&gt; as a move that takes Freida to an edge, &lt;math&gt;W&lt;/math&gt; as wrap-around move and &lt;math&gt;C&lt;/math&gt; as a corner move. Also, denote &lt;math&gt;O&lt;/math&gt; as a move that takes us to the center.<br /> <br /> <br /> &lt;b&gt; 2 Hops &lt;/b&gt;<br /> <br /> Then, Freida will have to &lt;math&gt;(E, C)&lt;/math&gt; as her set of moves. There are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, and &lt;math&gt;2&lt;/math&gt; corners to move to, for a total of &lt;math&gt;4 \cdot 2 = 8&lt;/math&gt; cases here. Then, there are &lt;math&gt;4&lt;/math&gt; choices for each move, for a probability of &lt;math&gt;\frac{8}{4 \cdot 4} = \frac{1}{2}&lt;/math&gt;. <br /> <br /> &lt;b&gt; 3 Hops &lt;/b&gt;<br /> <br /> In this case, Freida must wrap-around. There's only one possible combination, just &lt;math&gt;(E, W, C)&lt;/math&gt;. There are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, &lt;math&gt;1&lt;/math&gt; way to wrap-around (you must continue in the same direction) and &lt;math&gt;2&lt;/math&gt; corners, for a total of &lt;math&gt;4 \cdot 1 \cdot 2 = 8&lt;/math&gt; cases here. Then, there are &lt;math&gt;4&lt;/math&gt; choices for each move, for a probability of &lt;math&gt;\frac{8}{4 \cdot 4 \cdot 4} = \frac{1}{8}&lt;/math&gt;. <br /> <br /> &lt;b&gt; 4 Hops &lt;/b&gt; <br /> <br /> Lastly, there are two cases we must consider here. The first case is &lt;math&gt;(E, O, E, C)&lt;/math&gt;, and the second is &lt;math&gt;(E, W, W, C)&lt;/math&gt;. For the first case, there are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, &lt;math&gt;1&lt;/math&gt; way to return to the center, &lt;math&gt;4&lt;/math&gt; ways to move to an edge once again, and &lt;math&gt;2&lt;/math&gt; ways to move to a corner. Hence, there is a total of &lt;math&gt;4 \cdot 1 \cdot 4 \cdot 2 = 32&lt;/math&gt; cases here. Then, for the second case, there are &lt;math&gt;4&lt;/math&gt; ways to move to a corner, &lt;math&gt;1&lt;/math&gt; way to wrap-around, &lt;math&gt;1&lt;/math&gt; way to wrap-around again, and &lt;math&gt;2&lt;/math&gt; ways to move to a corner. This implies there are &lt;math&gt;4 \cdot 1 \cdot 1 \cdot 2 = 8&lt;/math&gt; cases here. Then, there is a total of &lt;math&gt;8+32 = 40&lt;/math&gt; cases, out of a total of &lt;math&gt;4^4 = 256&lt;/math&gt; cases, for a probability of &lt;math&gt;\frac{40}{256} = \frac{5}{32}&lt;/math&gt;. <br /> <br /> <br /> Then, the total probability that Freida ends up on a corner is &lt;math&gt;\frac{1}{2} + \frac{1}{8} + \frac{5}{32} = \frac{25}{32}&lt;/math&gt;, corresponding to choice &lt;math&gt;\boxed{\textbf{(D)} ~\frac{25}{32}}&lt;/math&gt;. ~rocketsri<br /> <br /> == Video Solution by OmegaLearn (Using Probability States) ==<br /> https://youtu.be/V_Sn30N2q50<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2021|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_23&diff=146159 2021 AMC 12A Problems/Problem 23 2021-02-12T19:23:47Z <p>Rocketsri: /* Solution 6 (Casework) */</p> <hr /> <div>==Problem==<br /> Frieda the frog begins a sequence of hops on a &lt;math&gt;3 \times 3&lt;/math&gt; grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she &quot;wraps around&quot; and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops &quot;up&quot;, the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}&lt;/math&gt;<br /> <br /> ==Solution 1 (complementary counting)==<br /> We will use complementary counting.<br /> First, the frog can go left with probability &lt;math&gt;\frac14&lt;/math&gt;. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since &lt;math&gt;4\cdot \frac14=1&lt;/math&gt;, we will ignore the leading probability.<br /> <br /> From the left, she either goes left to another edge (&lt;math&gt;\frac14&lt;/math&gt;) or back to the center (&lt;math&gt;\frac14&lt;/math&gt;). Time for some casework.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; She goes back to the center.<br /> <br /> Now, she can go in any 4 directions, and then has 2 options from that edge. This gives &lt;math&gt;\frac12&lt;/math&gt;. --End case 1<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; She goes to another edge (rightmost).<br /> <br /> Subcase 1: She goes back to the left edge. She now has 2 places to go, giving &lt;math&gt;\frac12&lt;/math&gt;<br /> <br /> Subcase 2: She goes to the center. Now any move works.<br /> <br /> &lt;math&gt;\frac14 \cdot \frac12 + \frac14 \cdot 1=\frac18 + \frac 14=\frac38&lt;/math&gt; for this case. --End case 2<br /> <br /> She goes back to the center in Case 1 with probability &lt;math&gt;\frac14&lt;/math&gt;, and to the right edge with probability &lt;math&gt;\frac14&lt;/math&gt;<br /> <br /> So, our answer is &lt;math&gt;\frac14 \cdot \frac12 + \frac14 \cdot \frac38=\frac14 (\frac12+\frac38)=\frac14 \cdot \frac78 = \frac7{32}&lt;/math&gt;<br /> <br /> But, don't forget complementary counting. So, we get &lt;math&gt;1-\frac7{32}=\frac{25}{32} \implies \boxed{D}&lt;/math&gt;. ~ firebolt360<br /> <br /> Video Solution for those who prefer: https://youtu.be/ude2rzO1cmk ~ firebolt360<br /> <br /> ==Solution 2 (direct counting and probability states)==<br /> We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have &lt;math&gt;1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{D}&lt;/math&gt;. ~IceWolf10<br /> <br /> ==Solution 3 (Similar to Solution 2, but Finds the Numerator and Denominator Separately)==<br /> &lt;b&gt;Denominator&lt;/b&gt;<br /> <br /> There are &lt;math&gt;4^4=256&lt;/math&gt; ways to make 4 hops without restrictions.<br /> <br /> <br /> &lt;b&gt;Numerator (Casework)&lt;/b&gt;<br /> <br /> Suppose Frieda makes 4 hops without stopping. We perform casework on which hop reaches a corner for the first time.<br /> <br /> <br /> (1) Hop #2 (Hops #3 and #4 have no restrictions)<br /> <br /> The 4 independent hops have 4, 2, 4, 4 options, respectively. So, this case has &lt;math&gt;4\cdot2\cdot4\cdot4=128&lt;/math&gt; ways.<br /> <br /> <br /> (2) Hop #3 (Hope #4 has no restriction)<br /> <br /> No matter which direction the first hop takes, the second hop must &quot;wrap around&quot;.<br /> <br /> The 4 independent hops have 4, 1, 2, 4 options, respectively. So, this case has &lt;math&gt;4\cdot1\cdot2\cdot4=32&lt;/math&gt; ways.<br /> <br /> <br /> (3) Hop #4<br /> <br /> Two sub-cases:<br /> <br /> (3.1) The second hop &quot;wraps around&quot;. It follows that the third hop also &quot;wraps around&quot;.<br /> <br /> The 4 independent hops have 4, 1, 1, 2 options, respectively. So, this sub-case has &lt;math&gt;4\cdot1\cdot1\cdot2=8&lt;/math&gt; ways.<br /> <br /> (3.2) The second hop backs to the center.<br /> <br /> The 4 independent hops have 4, 1, 4, 2 options, respectively. So, this sub-case has &lt;math&gt;4\cdot1\cdot4\cdot2=32&lt;/math&gt; ways.<br /> <br /> Together, Case (3) has &lt;math&gt;8+32=40&lt;/math&gt; ways.<br /> <br /> <br /> The numerator is &lt;math&gt;128+32+40=200.&lt;/math&gt;<br /> <br /> <br /> &lt;b&gt;Probability&lt;/b&gt;<br /> <br /> &lt;math&gt;\frac{200}{256}=\boxed{\textbf{(D)} ~\frac{25}{32}}.&lt;/math&gt;<br /> <br /> <br /> This problem is quite similar to 1995 AIME Problem 3: https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3<br /> <br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;C_n&lt;/math&gt; be the probability that Frieda is on the central square after n moves, &lt;math&gt;E_n&lt;/math&gt; be the probability that Frieda is on one of the four squares on the middle of the edges after n moves, and &lt;math&gt;V_n&lt;/math&gt; (V for vertex) be the probability that Frieda is on a corner after n moves. The only way to reach the center is by moving in &lt;math&gt;1&lt;/math&gt; specific direction out of &lt;math&gt;4&lt;/math&gt; total directions from the middle of an edge, so &lt;math&gt;C_{n+1}=\frac{E_n}{4}&lt;/math&gt;. The ways to reach the middle of an edge are by moving in any direction from the center or by moving in &lt;math&gt;1&lt;/math&gt; specific direction from the middle of an edge, so &lt;math&gt;E_{n+1}=C_n+\frac{E_n}{4}&lt;/math&gt;. The ways to reach a corner are by simply staying there after reaching there in a previous move or by moving in &lt;math&gt;2&lt;/math&gt; specific directions from the middle of an edge, so &lt;math&gt;V_{n+1}=V_n+\frac{E_n}{2}&lt;/math&gt;. Since Frieda always start from the center, &lt;math&gt;C_0=1&lt;/math&gt;, &lt;math&gt;E_0=0&lt;/math&gt;, and &lt;math&gt;V_0=0&lt;/math&gt;. We use the previous formulas to work out &lt;math&gt;V_4&lt;/math&gt; and find it to be &lt;math&gt;\boxed{\textbf{(D)} ~\frac{25}{32}}&lt;/math&gt;.<br /> <br /> -SmileKat32<br /> <br /> ==Solution 5==<br /> Imagine an infinite grid of &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; squares such that there is a &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; square centered at &lt;math&gt;(3x, 3y)&lt;/math&gt; for all ordered pairs of integers &lt;math&gt;(x, y).&lt;/math&gt; It is easy to see that the problem is equivalent to Frieda moving left, right, up, or down on this infinite grid starting at &lt;math&gt;(0, 0)&lt;/math&gt;. (minus the teleportations) Since counting the complement set is easier, we'll count the number of &lt;math&gt;4&lt;/math&gt;-step paths such that Frieda never reaches a corner point. <br /> <br /> In other words, since the reachable corner points are &lt;math&gt;(\pm 1, \pm 1), (\pm 1, \pm 2), (\pm 2, \pm 1), &lt;/math&gt; and &lt;math&gt;(\pm 2, \pm 2),&lt;/math&gt; Frieda can only travel along the collection of points included in &lt;math&gt;S&lt;/math&gt;, where &lt;math&gt;S&lt;/math&gt; is all points on &lt;math&gt;x=0&lt;/math&gt; and &lt;math&gt;y=0&lt;/math&gt; such that &lt;math&gt;|y|&lt;4&lt;/math&gt; and &lt;math&gt;|x|&lt;4&lt;/math&gt;, respectively, plus all points on the big square with side length &lt;math&gt;6&lt;/math&gt; centered at &lt;math&gt;(0, 0).&lt;/math&gt; We then can proceed with casework:<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: Frieda never reaches &lt;math&gt;(0, \pm 3)&lt;/math&gt; nor &lt;math&gt;(\pm 3, 0).&lt;/math&gt;<br /> <br /> When Frieda only moves horizontally or vertically for her four moves, she can do so in &lt;math&gt;2^4 - 4 = 12&lt;/math&gt; ways for each case . Thus, &lt;math&gt;12 \cdot 2&lt;/math&gt; total paths for the subcase of staying in one direction. (For instance, all length &lt;math&gt;4&lt;/math&gt; combinations of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; except &lt;math&gt;FFFF&lt;/math&gt;, &lt;math&gt;BBBB&lt;/math&gt;, &lt;math&gt;FFFB&lt;/math&gt;, and &lt;math&gt;BBBF&lt;/math&gt; for the horizontal direction.) <br /> <br /> There is another subcase where she changes directions during her path. There are four symmetric cases for this subcase depending on which of the four quadrants Frieda hugs. For the first quadrant, the possible paths are &lt;math&gt;FBUD&lt;/math&gt;, &lt;math&gt;FBUU&lt;/math&gt;, &lt;math&gt;UDFB&lt;/math&gt;, and &lt;math&gt;UDFF.&lt;/math&gt; Thus, a total of &lt;math&gt;4 \cdot 4 = 16&lt;/math&gt; ways for this subcase. <br /> <br /> Total for Case &lt;math&gt;1&lt;/math&gt;: &lt;math&gt;24 + 16 = 40&lt;/math&gt;<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: Frieda reaches &lt;math&gt;(0, \pm 3)&lt;/math&gt; or &lt;math&gt;(\pm 3, 0)&lt;/math&gt;.<br /> <br /> Once Frieda reaches one of the points listed above (by using three moves), she has four choices for her last move. Thus, a total of &lt;math&gt;4 \cdot 4 = 16&lt;/math&gt; paths for this case.<br /> <br /> Our total number of paths never reaching coroners is thus &lt;math&gt;16+40=56,&lt;/math&gt; making for an answer of &lt;cmath&gt;\frac{4^4-56}{4^4} = \boxed{\textbf{(D)} ~\frac{25}{32}}.&lt;/cmath&gt;<br /> <br /> -fidgetboss_4000<br /> <br /> == Solution 6 (Casework) ==<br /> We take cases on the number of hops needed to reach a corner. For simplicity, denote &lt;math&gt;E&lt;/math&gt; as a move that takes Freida to an edge, &lt;math&gt;W&lt;/math&gt; as wrap-around move and &lt;math&gt;C&lt;/math&gt; as a corner move. Also, denote &lt;math&gt;O&lt;/math&gt; as a move that takes us to the center.<br /> <br /> <br /> &lt;b&gt; 2 Hops &lt;/b&gt;<br /> <br /> Then, Freida will have to &lt;math&gt;(E, C)&lt;/math&gt; as her set of moves. There are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, and &lt;math&gt;2&lt;/math&gt; corners to move to, for a total of &lt;math&gt;4 \cdot 2 = 8&lt;/math&gt; cases here. Then, there are &lt;math&gt;4&lt;/math&gt; choices for each move, for a probability of &lt;math&gt;\frac{8}{4 \cdot 4} = \frac{1}{2}&lt;/math&gt;. <br /> <br /> &lt;b&gt; 3 Hops &lt;/b&gt;<br /> <br /> In this case, Freida must wrap-around. There's only one possible combination, just &lt;math&gt;(E, W, C)&lt;/math&gt;. There are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, &lt;math&gt;1&lt;/math&gt; way to wrap-around (you must continue in the same direction) and &lt;math&gt;2&lt;/math&gt; corners, for a total of &lt;math&gt;4 \cdot 1 \cdot 2 = 8&lt;/math&gt; cases here. Then, there are &lt;math&gt;4&lt;/math&gt; choices for each move, for a probability of &lt;math&gt;\frac{8}{4 \cdot 4 \cdot 4} = \frac{1}{8}&lt;/math&gt;. <br /> <br /> &lt;b&gt; 4 Hops &lt;/b&gt; <br /> <br /> Lastly, there are two cases we must consider here. The first case is &lt;math&gt;(E, O, E, C)&lt;/math&gt;, and the second is &lt;math&gt;(E, W, W, C)&lt;/math&gt;. For the first case, there are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, &lt;math&gt;1&lt;/math&gt; way to return to the center, &lt;math&gt;4&lt;/math&gt; ways to move to an edge once again, and &lt;math&gt;2&lt;/math&gt; ways to move to a corner. Hence, there is a total of &lt;math&gt;4 \cdot 1 \cdot 4 \cdot 2 = 32&lt;/math&gt; cases here. Then, for the second case, there are &lt;math&gt;4&lt;/math&gt; ways to move to a corner, &lt;math&gt;1&lt;/math&gt; way to wrap-around, &lt;math&gt;1&lt;/math&gt; way to wrap-around again, and &lt;math&gt;2&lt;/math&gt; ways to move to a corner. This implies there are &lt;math&gt;4 \cdot 1 \cdot 1 \cdot 2 = 8&lt;/math&gt; cases here. Then, there is a total of &lt;math&gt;8+32 = 40&lt;/math&gt; cases, out of a total of &lt;math&gt;4^4 = 256&lt;/math&gt; cases, for a probability of &lt;math&gt;\frac{40}{256} = \frac{5}{32}&lt;/math&gt;. <br /> <br /> <br /> Then, the total probability that Freida ends up on a corner is &lt;math&gt;\frac{1}{2} + \frac{1}{8} + \frac{5}{32} = \frac{25}{32}&lt;/math&gt;, corresponding to choice &lt;math&gt;\boxed{\textbf{(D)} \frac{25}{32}}&lt;/math&gt;. ~rocketsri<br /> <br /> == Video Solution by OmegaLearn (Using Probability States) ==<br /> https://youtu.be/V_Sn30N2q50<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2021|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_23&diff=146157 2021 AMC 12A Problems/Problem 23 2021-02-12T19:23:30Z <p>Rocketsri: /* Solution 6 (Casework) */</p> <hr /> <div>==Problem==<br /> Frieda the frog begins a sequence of hops on a &lt;math&gt;3 \times 3&lt;/math&gt; grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she &quot;wraps around&quot; and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops &quot;up&quot;, the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}&lt;/math&gt;<br /> <br /> ==Solution 1 (complementary counting)==<br /> We will use complementary counting.<br /> First, the frog can go left with probability &lt;math&gt;\frac14&lt;/math&gt;. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since &lt;math&gt;4\cdot \frac14=1&lt;/math&gt;, we will ignore the leading probability.<br /> <br /> From the left, she either goes left to another edge (&lt;math&gt;\frac14&lt;/math&gt;) or back to the center (&lt;math&gt;\frac14&lt;/math&gt;). Time for some casework.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; She goes back to the center.<br /> <br /> Now, she can go in any 4 directions, and then has 2 options from that edge. This gives &lt;math&gt;\frac12&lt;/math&gt;. --End case 1<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; She goes to another edge (rightmost).<br /> <br /> Subcase 1: She goes back to the left edge. She now has 2 places to go, giving &lt;math&gt;\frac12&lt;/math&gt;<br /> <br /> Subcase 2: She goes to the center. Now any move works.<br /> <br /> &lt;math&gt;\frac14 \cdot \frac12 + \frac14 \cdot 1=\frac18 + \frac 14=\frac38&lt;/math&gt; for this case. --End case 2<br /> <br /> She goes back to the center in Case 1 with probability &lt;math&gt;\frac14&lt;/math&gt;, and to the right edge with probability &lt;math&gt;\frac14&lt;/math&gt;<br /> <br /> So, our answer is &lt;math&gt;\frac14 \cdot \frac12 + \frac14 \cdot \frac38=\frac14 (\frac12+\frac38)=\frac14 \cdot \frac78 = \frac7{32}&lt;/math&gt;<br /> <br /> But, don't forget complementary counting. So, we get &lt;math&gt;1-\frac7{32}=\frac{25}{32} \implies \boxed{D}&lt;/math&gt;. ~ firebolt360<br /> <br /> Video Solution for those who prefer: https://youtu.be/ude2rzO1cmk ~ firebolt360<br /> <br /> ==Solution 2 (direct counting and probability states)==<br /> We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have &lt;math&gt;1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{D}&lt;/math&gt;. ~IceWolf10<br /> <br /> ==Solution 3 (Similar to Solution 2, but Finds the Numerator and Denominator Separately)==<br /> &lt;b&gt;Denominator&lt;/b&gt;<br /> <br /> There are &lt;math&gt;4^4=256&lt;/math&gt; ways to make 4 hops without restrictions.<br /> <br /> <br /> &lt;b&gt;Numerator (Casework)&lt;/b&gt;<br /> <br /> Suppose Frieda makes 4 hops without stopping. We perform casework on which hop reaches a corner for the first time.<br /> <br /> <br /> (1) Hop #2 (Hops #3 and #4 have no restrictions)<br /> <br /> The 4 independent hops have 4, 2, 4, 4 options, respectively. So, this case has &lt;math&gt;4\cdot2\cdot4\cdot4=128&lt;/math&gt; ways.<br /> <br /> <br /> (2) Hop #3 (Hope #4 has no restriction)<br /> <br /> No matter which direction the first hop takes, the second hop must &quot;wrap around&quot;.<br /> <br /> The 4 independent hops have 4, 1, 2, 4 options, respectively. So, this case has &lt;math&gt;4\cdot1\cdot2\cdot4=32&lt;/math&gt; ways.<br /> <br /> <br /> (3) Hop #4<br /> <br /> Two sub-cases:<br /> <br /> (3.1) The second hop &quot;wraps around&quot;. It follows that the third hop also &quot;wraps around&quot;.<br /> <br /> The 4 independent hops have 4, 1, 1, 2 options, respectively. So, this sub-case has &lt;math&gt;4\cdot1\cdot1\cdot2=8&lt;/math&gt; ways.<br /> <br /> (3.2) The second hop backs to the center.<br /> <br /> The 4 independent hops have 4, 1, 4, 2 options, respectively. So, this sub-case has &lt;math&gt;4\cdot1\cdot4\cdot2=32&lt;/math&gt; ways.<br /> <br /> Together, Case (3) has &lt;math&gt;8+32=40&lt;/math&gt; ways.<br /> <br /> <br /> The numerator is &lt;math&gt;128+32+40=200.&lt;/math&gt;<br /> <br /> <br /> &lt;b&gt;Probability&lt;/b&gt;<br /> <br /> &lt;math&gt;\frac{200}{256}=\boxed{\textbf{(D)} ~\frac{25}{32}}.&lt;/math&gt;<br /> <br /> <br /> This problem is quite similar to 1995 AIME Problem 3: https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3<br /> <br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;C_n&lt;/math&gt; be the probability that Frieda is on the central square after n moves, &lt;math&gt;E_n&lt;/math&gt; be the probability that Frieda is on one of the four squares on the middle of the edges after n moves, and &lt;math&gt;V_n&lt;/math&gt; (V for vertex) be the probability that Frieda is on a corner after n moves. The only way to reach the center is by moving in &lt;math&gt;1&lt;/math&gt; specific direction out of &lt;math&gt;4&lt;/math&gt; total directions from the middle of an edge, so &lt;math&gt;C_{n+1}=\frac{E_n}{4}&lt;/math&gt;. The ways to reach the middle of an edge are by moving in any direction from the center or by moving in &lt;math&gt;1&lt;/math&gt; specific direction from the middle of an edge, so &lt;math&gt;E_{n+1}=C_n+\frac{E_n}{4}&lt;/math&gt;. The ways to reach a corner are by simply staying there after reaching there in a previous move or by moving in &lt;math&gt;2&lt;/math&gt; specific directions from the middle of an edge, so &lt;math&gt;V_{n+1}=V_n+\frac{E_n}{2}&lt;/math&gt;. Since Frieda always start from the center, &lt;math&gt;C_0=1&lt;/math&gt;, &lt;math&gt;E_0=0&lt;/math&gt;, and &lt;math&gt;V_0=0&lt;/math&gt;. We use the previous formulas to work out &lt;math&gt;V_4&lt;/math&gt; and find it to be &lt;math&gt;\boxed{\textbf{(D)} ~\frac{25}{32}}&lt;/math&gt;.<br /> <br /> -SmileKat32<br /> <br /> ==Solution 5==<br /> Imagine an infinite grid of &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; squares such that there is a &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; square centered at &lt;math&gt;(3x, 3y)&lt;/math&gt; for all ordered pairs of integers &lt;math&gt;(x, y).&lt;/math&gt; It is easy to see that the problem is equivalent to Frieda moving left, right, up, or down on this infinite grid starting at &lt;math&gt;(0, 0)&lt;/math&gt;. (minus the teleportations) Since counting the complement set is easier, we'll count the number of &lt;math&gt;4&lt;/math&gt;-step paths such that Frieda never reaches a corner point. <br /> <br /> In other words, since the reachable corner points are &lt;math&gt;(\pm 1, \pm 1), (\pm 1, \pm 2), (\pm 2, \pm 1), &lt;/math&gt; and &lt;math&gt;(\pm 2, \pm 2),&lt;/math&gt; Frieda can only travel along the collection of points included in &lt;math&gt;S&lt;/math&gt;, where &lt;math&gt;S&lt;/math&gt; is all points on &lt;math&gt;x=0&lt;/math&gt; and &lt;math&gt;y=0&lt;/math&gt; such that &lt;math&gt;|y|&lt;4&lt;/math&gt; and &lt;math&gt;|x|&lt;4&lt;/math&gt;, respectively, plus all points on the big square with side length &lt;math&gt;6&lt;/math&gt; centered at &lt;math&gt;(0, 0).&lt;/math&gt; We then can proceed with casework:<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: Frieda never reaches &lt;math&gt;(0, \pm 3)&lt;/math&gt; nor &lt;math&gt;(\pm 3, 0).&lt;/math&gt;<br /> <br /> When Frieda only moves horizontally or vertically for her four moves, she can do so in &lt;math&gt;2^4 - 4 = 12&lt;/math&gt; ways for each case . Thus, &lt;math&gt;12 \cdot 2&lt;/math&gt; total paths for the subcase of staying in one direction. (For instance, all length &lt;math&gt;4&lt;/math&gt; combinations of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; except &lt;math&gt;FFFF&lt;/math&gt;, &lt;math&gt;BBBB&lt;/math&gt;, &lt;math&gt;FFFB&lt;/math&gt;, and &lt;math&gt;BBBF&lt;/math&gt; for the horizontal direction.) <br /> <br /> There is another subcase where she changes directions during her path. There are four symmetric cases for this subcase depending on which of the four quadrants Frieda hugs. For the first quadrant, the possible paths are &lt;math&gt;FBUD&lt;/math&gt;, &lt;math&gt;FBUU&lt;/math&gt;, &lt;math&gt;UDFB&lt;/math&gt;, and &lt;math&gt;UDFF.&lt;/math&gt; Thus, a total of &lt;math&gt;4 \cdot 4 = 16&lt;/math&gt; ways for this subcase. <br /> <br /> Total for Case &lt;math&gt;1&lt;/math&gt;: &lt;math&gt;24 + 16 = 40&lt;/math&gt;<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: Frieda reaches &lt;math&gt;(0, \pm 3)&lt;/math&gt; or &lt;math&gt;(\pm 3, 0)&lt;/math&gt;.<br /> <br /> Once Frieda reaches one of the points listed above (by using three moves), she has four choices for her last move. Thus, a total of &lt;math&gt;4 \cdot 4 = 16&lt;/math&gt; paths for this case.<br /> <br /> Our total number of paths never reaching coroners is thus &lt;math&gt;16+40=56,&lt;/math&gt; making for an answer of &lt;cmath&gt;\frac{4^4-56}{4^4} = \boxed{\textbf{(D)} ~\frac{25}{32}}.&lt;/cmath&gt;<br /> <br /> -fidgetboss_4000<br /> <br /> == Solution 6 (Casework) ==<br /> We take cases on the number of hops needed to reach a corner. For simplicity, denote &lt;math&gt;E&lt;/math&gt; as a move that takes Freida to an edge, &lt;math&gt;W&lt;/math&gt; as wrap-around move and &lt;math&gt;C&lt;/math&gt; as a corner move. Also, denote &lt;math&gt;O&lt;/math&gt; as a move that takes us to the center.<br /> <br /> <br /> &lt;b&gt; 2 Hops &lt;/b&gt;<br /> <br /> Then, Freida will have to &lt;math&gt;(E, C)&lt;/math&gt; as her set of moves. There are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, and &lt;math&gt;2&lt;/math&gt; corners to move to, for a total of &lt;math&gt;4 \cdot 2 = 8&lt;/math&gt; cases here. Then, there are &lt;math&gt;4&lt;/math&gt; choices for each move, for a probability of &lt;math&gt;\frac{8}{4 \cdot 4} = \frac{1}{2}&lt;/math&gt;. <br /> <br /> &lt;b&gt; 3 Hops &lt;/b&gt;<br /> <br /> In this case, Freida must wrap-around. There's only one possible combination, just &lt;math&gt;(E, W, C)&lt;/math&gt;. There are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, &lt;math&gt;1&lt;/math&gt; way to wrap-around (you must continue in the same direction) and &lt;math&gt;2&lt;/math&gt; corners, for a total of &lt;math&gt;4 \cdot 1 \cdot 2 = 8&lt;/math&gt; cases here. Then, there are &lt;math&gt;4&lt;/math&gt; choices for each move, for a probability of &lt;math&gt;\frac{8}{4 \cdot 4 \cdot 4} = \frac{1}{8}&lt;/math&gt;. <br /> <br /> &lt;b&gt; 4 Hops &lt;/b&gt; <br /> <br /> Lastly, there are two cases we must consider here. The first case is &lt;math&gt;(E, O, E, C)&lt;/math&gt;, and the second is &lt;math&gt;(E, W, W, C)&lt;/math&gt;. For the first case, there are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, &lt;math&gt;1&lt;/math&gt; way to return to the center, &lt;math&gt;4&lt;/math&gt; ways to move to an edge once again, and &lt;math&gt;2&lt;/math&gt; ways to move to a corner. Hence, there is a total of &lt;math&gt;4 \cdot 1 \cdot 4 \cdot 2 = 32&lt;/math&gt; cases here. Then, for the second case, there are &lt;math&gt;4&lt;/math&gt; ways to move to a corner, &lt;math&gt;1&lt;/math&gt; way to wrap-around, &lt;math&gt;1&lt;/math&gt; way to wrap-around again, and &lt;math&gt;2&lt;/math&gt; ways to move to a corner. This implies there are &lt;math&gt;4 \cdot 1 \cdot 1 \cdot 2 = 8&lt;/math&gt; cases here. Then, there is a total of &lt;math&gt;8+32 = 40&lt;/math&gt; cases, out of a total of &lt;math&gt;4^4 = 256&lt;/math&gt; cases, for a probability of &lt;math&gt;\frac{40}{256} = \frac{5}{32}&lt;/math&gt;. <br /> <br /> <br /> Then, the total probability that Freida ends up on a corner is &lt;math&gt;\frac{1}{2} + \frac{1}{8} + \frac{5}{32} = \frac{25}{32}&lt;/math&gt;, corresponding to choice &lt;math&gt;\boxed{\textbf{(D)} \frac{25}{32}}&lt;/math&gt;. ~rocketsri<br /> <br /> == Video Solution by OmegaLearn (Using Probability States) ==<br /> https://youtu.be/V_Sn30N2q50<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2021|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_23&diff=146155 2021 AMC 12A Problems/Problem 23 2021-02-12T19:22:50Z <p>Rocketsri: /* Solution 6 (Casework) */</p> <hr /> <div>==Problem==<br /> Frieda the frog begins a sequence of hops on a &lt;math&gt;3 \times 3&lt;/math&gt; grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she &quot;wraps around&quot; and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops &quot;up&quot;, the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}&lt;/math&gt;<br /> <br /> ==Solution 1 (complementary counting)==<br /> We will use complementary counting.<br /> First, the frog can go left with probability &lt;math&gt;\frac14&lt;/math&gt;. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since &lt;math&gt;4\cdot \frac14=1&lt;/math&gt;, we will ignore the leading probability.<br /> <br /> From the left, she either goes left to another edge (&lt;math&gt;\frac14&lt;/math&gt;) or back to the center (&lt;math&gt;\frac14&lt;/math&gt;). Time for some casework.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; She goes back to the center.<br /> <br /> Now, she can go in any 4 directions, and then has 2 options from that edge. This gives &lt;math&gt;\frac12&lt;/math&gt;. --End case 1<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; She goes to another edge (rightmost).<br /> <br /> Subcase 1: She goes back to the left edge. She now has 2 places to go, giving &lt;math&gt;\frac12&lt;/math&gt;<br /> <br /> Subcase 2: She goes to the center. Now any move works.<br /> <br /> &lt;math&gt;\frac14 \cdot \frac12 + \frac14 \cdot 1=\frac18 + \frac 14=\frac38&lt;/math&gt; for this case. --End case 2<br /> <br /> She goes back to the center in Case 1 with probability &lt;math&gt;\frac14&lt;/math&gt;, and to the right edge with probability &lt;math&gt;\frac14&lt;/math&gt;<br /> <br /> So, our answer is &lt;math&gt;\frac14 \cdot \frac12 + \frac14 \cdot \frac38=\frac14 (\frac12+\frac38)=\frac14 \cdot \frac78 = \frac7{32}&lt;/math&gt;<br /> <br /> But, don't forget complementary counting. So, we get &lt;math&gt;1-\frac7{32}=\frac{25}{32} \implies \boxed{D}&lt;/math&gt;. ~ firebolt360<br /> <br /> Video Solution for those who prefer: https://youtu.be/ude2rzO1cmk ~ firebolt360<br /> <br /> ==Solution 2 (direct counting and probability states)==<br /> We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have &lt;math&gt;1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{D}&lt;/math&gt;. ~IceWolf10<br /> <br /> ==Solution 3 (Similar to Solution 2, but Finds the Numerator and Denominator Separately)==<br /> &lt;b&gt;Denominator&lt;/b&gt;<br /> <br /> There are &lt;math&gt;4^4=256&lt;/math&gt; ways to make 4 hops without restrictions.<br /> <br /> <br /> &lt;b&gt;Numerator (Casework)&lt;/b&gt;<br /> <br /> Suppose Frieda makes 4 hops without stopping. We perform casework on which hop reaches a corner for the first time.<br /> <br /> <br /> (1) Hop #2 (Hops #3 and #4 have no restrictions)<br /> <br /> The 4 independent hops have 4, 2, 4, 4 options, respectively. So, this case has &lt;math&gt;4\cdot2\cdot4\cdot4=128&lt;/math&gt; ways.<br /> <br /> <br /> (2) Hop #3 (Hope #4 has no restriction)<br /> <br /> No matter which direction the first hop takes, the second hop must &quot;wrap around&quot;.<br /> <br /> The 4 independent hops have 4, 1, 2, 4 options, respectively. So, this case has &lt;math&gt;4\cdot1\cdot2\cdot4=32&lt;/math&gt; ways.<br /> <br /> <br /> (3) Hop #4<br /> <br /> Two sub-cases:<br /> <br /> (3.1) The second hop &quot;wraps around&quot;. It follows that the third hop also &quot;wraps around&quot;.<br /> <br /> The 4 independent hops have 4, 1, 1, 2 options, respectively. So, this sub-case has &lt;math&gt;4\cdot1\cdot1\cdot2=8&lt;/math&gt; ways.<br /> <br /> (3.2) The second hop backs to the center.<br /> <br /> The 4 independent hops have 4, 1, 4, 2 options, respectively. So, this sub-case has &lt;math&gt;4\cdot1\cdot4\cdot2=32&lt;/math&gt; ways.<br /> <br /> Together, Case (3) has &lt;math&gt;8+32=40&lt;/math&gt; ways.<br /> <br /> <br /> The numerator is &lt;math&gt;128+32+40=200.&lt;/math&gt;<br /> <br /> <br /> &lt;b&gt;Probability&lt;/b&gt;<br /> <br /> &lt;math&gt;\frac{200}{256}=\boxed{\textbf{(D)} ~\frac{25}{32}}.&lt;/math&gt;<br /> <br /> <br /> This problem is quite similar to 1995 AIME Problem 3: https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3<br /> <br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;C_n&lt;/math&gt; be the probability that Frieda is on the central square after n moves, &lt;math&gt;E_n&lt;/math&gt; be the probability that Frieda is on one of the four squares on the middle of the edges after n moves, and &lt;math&gt;V_n&lt;/math&gt; (V for vertex) be the probability that Frieda is on a corner after n moves. The only way to reach the center is by moving in &lt;math&gt;1&lt;/math&gt; specific direction out of &lt;math&gt;4&lt;/math&gt; total directions from the middle of an edge, so &lt;math&gt;C_{n+1}=\frac{E_n}{4}&lt;/math&gt;. The ways to reach the middle of an edge are by moving in any direction from the center or by moving in &lt;math&gt;1&lt;/math&gt; specific direction from the middle of an edge, so &lt;math&gt;E_{n+1}=C_n+\frac{E_n}{4}&lt;/math&gt;. The ways to reach a corner are by simply staying there after reaching there in a previous move or by moving in &lt;math&gt;2&lt;/math&gt; specific directions from the middle of an edge, so &lt;math&gt;V_{n+1}=V_n+\frac{E_n}{2}&lt;/math&gt;. Since Frieda always start from the center, &lt;math&gt;C_0=1&lt;/math&gt;, &lt;math&gt;E_0=0&lt;/math&gt;, and &lt;math&gt;V_0=0&lt;/math&gt;. We use the previous formulas to work out &lt;math&gt;V_4&lt;/math&gt; and find it to be &lt;math&gt;\boxed{\textbf{(D)} ~\frac{25}{32}}&lt;/math&gt;.<br /> <br /> -SmileKat32<br /> <br /> ==Solution 5==<br /> Imagine an infinite grid of &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; squares such that there is a &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; square centered at &lt;math&gt;(3x, 3y)&lt;/math&gt; for all ordered pairs of integers &lt;math&gt;(x, y).&lt;/math&gt; It is easy to see that the problem is equivalent to Frieda moving left, right, up, or down on this infinite grid starting at &lt;math&gt;(0, 0)&lt;/math&gt;. (minus the teleportations) Since counting the complement set is easier, we'll count the number of &lt;math&gt;4&lt;/math&gt;-step paths such that Frieda never reaches a corner point. <br /> <br /> In other words, since the reachable corner points are &lt;math&gt;(\pm 1, \pm 1), (\pm 1, \pm 2), (\pm 2, \pm 1), &lt;/math&gt; and &lt;math&gt;(\pm 2, \pm 2),&lt;/math&gt; Frieda can only travel along the collection of points included in &lt;math&gt;S&lt;/math&gt;, where &lt;math&gt;S&lt;/math&gt; is all points on &lt;math&gt;x=0&lt;/math&gt; and &lt;math&gt;y=0&lt;/math&gt; such that &lt;math&gt;|y|&lt;4&lt;/math&gt; and &lt;math&gt;|x|&lt;4&lt;/math&gt;, respectively, plus all points on the big square with side length &lt;math&gt;6&lt;/math&gt; centered at &lt;math&gt;(0, 0).&lt;/math&gt; We then can proceed with casework:<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: Frieda never reaches &lt;math&gt;(0, \pm 3)&lt;/math&gt; nor &lt;math&gt;(\pm 3, 0).&lt;/math&gt;<br /> <br /> When Frieda only moves horizontally or vertically for her four moves, she can do so in &lt;math&gt;2^4 - 4 = 12&lt;/math&gt; ways for each case . Thus, &lt;math&gt;12 \cdot 2&lt;/math&gt; total paths for the subcase of staying in one direction. (For instance, all length &lt;math&gt;4&lt;/math&gt; combinations of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; except &lt;math&gt;FFFF&lt;/math&gt;, &lt;math&gt;BBBB&lt;/math&gt;, &lt;math&gt;FFFB&lt;/math&gt;, and &lt;math&gt;BBBF&lt;/math&gt; for the horizontal direction.) <br /> <br /> There is another subcase where she changes directions during her path. There are four symmetric cases for this subcase depending on which of the four quadrants Frieda hugs. For the first quadrant, the possible paths are &lt;math&gt;FBUD&lt;/math&gt;, &lt;math&gt;FBUU&lt;/math&gt;, &lt;math&gt;UDFB&lt;/math&gt;, and &lt;math&gt;UDFF.&lt;/math&gt; Thus, a total of &lt;math&gt;4 \cdot 4 = 16&lt;/math&gt; ways for this subcase. <br /> <br /> Total for Case &lt;math&gt;1&lt;/math&gt;: &lt;math&gt;24 + 16 = 40&lt;/math&gt;<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: Frieda reaches &lt;math&gt;(0, \pm 3)&lt;/math&gt; or &lt;math&gt;(\pm 3, 0)&lt;/math&gt;.<br /> <br /> Once Frieda reaches one of the points listed above (by using three moves), she has four choices for her last move. Thus, a total of &lt;math&gt;4 \cdot 4 = 16&lt;/math&gt; paths for this case.<br /> <br /> Our total number of paths never reaching coroners is thus &lt;math&gt;16+40=56,&lt;/math&gt; making for an answer of &lt;cmath&gt;\frac{4^4-56}{4^4} = \boxed{\textbf{(D)} ~\frac{25}{32}}.&lt;/cmath&gt;<br /> <br /> -fidgetboss_4000<br /> <br /> == Solution 6 (Casework) ==<br /> We take cases on the number of hops needed to reach a corner. For simplicity, denote &lt;math&gt;E&lt;/math&gt; as a move that takes Freida to an edge, &lt;math&gt;W&lt;/math&gt; as wrap-around move and &lt;math&gt;C&lt;/math&gt; as a corner move. Also, denote &lt;math&gt;O&lt;/math&gt; as a move that takes us to the center.<br /> <br /> <br /> &lt;b&gt; 2 Hops &lt;/b&gt;<br /> <br /> Then, Freida will have to &lt;math&gt;(E, C)&lt;/math&gt; as her set of moves. There are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, and &lt;math&gt;2&lt;/math&gt; corners to move to, for a total of &lt;math&gt;4 \cdot 2 = 8&lt;/math&gt; cases here. Then, there are &lt;math&gt;4&lt;/math&gt; choices for each move, for a probability of &lt;math&gt;\frac{8}{4 \cdot 4} = \frac{1}{2}&lt;/math&gt;. <br /> <br /> &lt;b&gt; 3 Hops &lt;/b&gt;<br /> <br /> In this case, Freida must wrap-around. There's only one possible combination, just &lt;math&gt;(E, W, C)&lt;/math&gt;. There are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, &lt;math&gt;1&lt;/math&gt; way to wrap-around (you must continue in the same direction) and &lt;math&gt;2&lt;/math&gt; corners, for a total of &lt;math&gt;4 \cdot 1 \cdot 2 = 8&lt;/math&gt; cases here. Then, there are &lt;math&gt;4&lt;/math&gt; choices for each move, for a probability of &lt;math&gt;\frac{8}{4 \cdot 4 \cdot 4} = \frac{1}{8}&lt;/math&gt;. <br /> <br /> &lt;b&gt; 4 Hops &lt;/b&gt; <br /> <br /> Lastly, there are two cases we must consider here. The first case is &lt;math&gt;(E, O, E, C)&lt;/math&gt;, and the second is &lt;math&gt;(E, W, W, C)&lt;/math&gt;. For the first case, there are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, &lt;math&gt;1&lt;/math&gt; way to return to the center, &lt;math&gt;4&lt;/math&gt; ways to move to an edge once again, and &lt;math&gt;2&lt;/math&gt; ways to move to a corner. Hence, there is a total of &lt;math&gt;4 \cdot 1 \cdot 4 \cdot 2 = 32&lt;/math&gt; cases here. Then, for the second case, there are &lt;math&gt;4&lt;/math&gt; ways to move to a corner, &lt;math&gt;1&lt;/math&gt; way to wrap-around, &lt;math&gt;1&lt;/math&gt; way to wrap-around again, and &lt;math&gt;2&lt;/math&gt; ways to move to a corner. This implies there are &lt;math&gt;4 \cdot 1 \cdot 1 \cdot 2 = 8&lt;/math&gt; cases here. Then, there is a total of &lt;math&gt;8+32 = 40&lt;/math&gt; cases, out of a total of &lt;math&gt;4^4 = 256&lt;/math&gt; cases, for a probability of &lt;math&gt;\frac{40}{256} = \frac{5}{32}&lt;/math&gt;. <br /> <br /> <br /> Then, the total probability that Freida ends up on a corner is &lt;math&gt;\frac{1}{2} + \frac{1}{8} + \frac{5}{32} = \frac{25}{32}&lt;/math&gt;, corresponding to choice &lt;math&gt;\boxed{&lt;b&gt; (D) &lt;/b&gt; = \frac{25}{32}}&lt;/math&gt;. ~rocketsri<br /> <br /> == Video Solution by OmegaLearn (Using Probability States) ==<br /> https://youtu.be/V_Sn30N2q50<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2021|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_23&diff=146154 2021 AMC 12A Problems/Problem 23 2021-02-12T19:21:46Z <p>Rocketsri: /* Solution 6 */</p> <hr /> <div>==Problem==<br /> Frieda the frog begins a sequence of hops on a &lt;math&gt;3 \times 3&lt;/math&gt; grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she &quot;wraps around&quot; and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops &quot;up&quot;, the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}&lt;/math&gt;<br /> <br /> ==Solution 1 (complementary counting)==<br /> We will use complementary counting.<br /> First, the frog can go left with probability &lt;math&gt;\frac14&lt;/math&gt;. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since &lt;math&gt;4\cdot \frac14=1&lt;/math&gt;, we will ignore the leading probability.<br /> <br /> From the left, she either goes left to another edge (&lt;math&gt;\frac14&lt;/math&gt;) or back to the center (&lt;math&gt;\frac14&lt;/math&gt;). Time for some casework.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; She goes back to the center.<br /> <br /> Now, she can go in any 4 directions, and then has 2 options from that edge. This gives &lt;math&gt;\frac12&lt;/math&gt;. --End case 1<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; She goes to another edge (rightmost).<br /> <br /> Subcase 1: She goes back to the left edge. She now has 2 places to go, giving &lt;math&gt;\frac12&lt;/math&gt;<br /> <br /> Subcase 2: She goes to the center. Now any move works.<br /> <br /> &lt;math&gt;\frac14 \cdot \frac12 + \frac14 \cdot 1=\frac18 + \frac 14=\frac38&lt;/math&gt; for this case. --End case 2<br /> <br /> She goes back to the center in Case 1 with probability &lt;math&gt;\frac14&lt;/math&gt;, and to the right edge with probability &lt;math&gt;\frac14&lt;/math&gt;<br /> <br /> So, our answer is &lt;math&gt;\frac14 \cdot \frac12 + \frac14 \cdot \frac38=\frac14 (\frac12+\frac38)=\frac14 \cdot \frac78 = \frac7{32}&lt;/math&gt;<br /> <br /> But, don't forget complementary counting. So, we get &lt;math&gt;1-\frac7{32}=\frac{25}{32} \implies \boxed{D}&lt;/math&gt;. ~ firebolt360<br /> <br /> Video Solution for those who prefer: https://youtu.be/ude2rzO1cmk ~ firebolt360<br /> <br /> ==Solution 2 (direct counting and probability states)==<br /> We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have &lt;math&gt;1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{D}&lt;/math&gt;. ~IceWolf10<br /> <br /> ==Solution 3 (Similar to Solution 2, but Finds the Numerator and Denominator Separately)==<br /> &lt;b&gt;Denominator&lt;/b&gt;<br /> <br /> There are &lt;math&gt;4^4=256&lt;/math&gt; ways to make 4 hops without restrictions.<br /> <br /> <br /> &lt;b&gt;Numerator (Casework)&lt;/b&gt;<br /> <br /> Suppose Frieda makes 4 hops without stopping. We perform casework on which hop reaches a corner for the first time.<br /> <br /> <br /> (1) Hop #2 (Hops #3 and #4 have no restrictions)<br /> <br /> The 4 independent hops have 4, 2, 4, 4 options, respectively. So, this case has &lt;math&gt;4\cdot2\cdot4\cdot4=128&lt;/math&gt; ways.<br /> <br /> <br /> (2) Hop #3 (Hope #4 has no restriction)<br /> <br /> No matter which direction the first hop takes, the second hop must &quot;wrap around&quot;.<br /> <br /> The 4 independent hops have 4, 1, 2, 4 options, respectively. So, this case has &lt;math&gt;4\cdot1\cdot2\cdot4=32&lt;/math&gt; ways.<br /> <br /> <br /> (3) Hop #4<br /> <br /> Two sub-cases:<br /> <br /> (3.1) The second hop &quot;wraps around&quot;. It follows that the third hop also &quot;wraps around&quot;.<br /> <br /> The 4 independent hops have 4, 1, 1, 2 options, respectively. So, this sub-case has &lt;math&gt;4\cdot1\cdot1\cdot2=8&lt;/math&gt; ways.<br /> <br /> (3.2) The second hop backs to the center.<br /> <br /> The 4 independent hops have 4, 1, 4, 2 options, respectively. So, this sub-case has &lt;math&gt;4\cdot1\cdot4\cdot2=32&lt;/math&gt; ways.<br /> <br /> Together, Case (3) has &lt;math&gt;8+32=40&lt;/math&gt; ways.<br /> <br /> <br /> The numerator is &lt;math&gt;128+32+40=200.&lt;/math&gt;<br /> <br /> <br /> &lt;b&gt;Probability&lt;/b&gt;<br /> <br /> &lt;math&gt;\frac{200}{256}=\boxed{\textbf{(D)} ~\frac{25}{32}}.&lt;/math&gt;<br /> <br /> <br /> This problem is quite similar to 1995 AIME Problem 3: https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3<br /> <br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;C_n&lt;/math&gt; be the probability that Frieda is on the central square after n moves, &lt;math&gt;E_n&lt;/math&gt; be the probability that Frieda is on one of the four squares on the middle of the edges after n moves, and &lt;math&gt;V_n&lt;/math&gt; (V for vertex) be the probability that Frieda is on a corner after n moves. The only way to reach the center is by moving in &lt;math&gt;1&lt;/math&gt; specific direction out of &lt;math&gt;4&lt;/math&gt; total directions from the middle of an edge, so &lt;math&gt;C_{n+1}=\frac{E_n}{4}&lt;/math&gt;. The ways to reach the middle of an edge are by moving in any direction from the center or by moving in &lt;math&gt;1&lt;/math&gt; specific direction from the middle of an edge, so &lt;math&gt;E_{n+1}=C_n+\frac{E_n}{4}&lt;/math&gt;. The ways to reach a corner are by simply staying there after reaching there in a previous move or by moving in &lt;math&gt;2&lt;/math&gt; specific directions from the middle of an edge, so &lt;math&gt;V_{n+1}=V_n+\frac{E_n}{2}&lt;/math&gt;. Since Frieda always start from the center, &lt;math&gt;C_0=1&lt;/math&gt;, &lt;math&gt;E_0=0&lt;/math&gt;, and &lt;math&gt;V_0=0&lt;/math&gt;. We use the previous formulas to work out &lt;math&gt;V_4&lt;/math&gt; and find it to be &lt;math&gt;\boxed{\textbf{(D)} ~\frac{25}{32}}&lt;/math&gt;.<br /> <br /> -SmileKat32<br /> <br /> ==Solution 5==<br /> Imagine an infinite grid of &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; squares such that there is a &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; square centered at &lt;math&gt;(3x, 3y)&lt;/math&gt; for all ordered pairs of integers &lt;math&gt;(x, y).&lt;/math&gt; It is easy to see that the problem is equivalent to Frieda moving left, right, up, or down on this infinite grid starting at &lt;math&gt;(0, 0)&lt;/math&gt;. (minus the teleportations) Since counting the complement set is easier, we'll count the number of &lt;math&gt;4&lt;/math&gt;-step paths such that Frieda never reaches a corner point. <br /> <br /> In other words, since the reachable corner points are &lt;math&gt;(\pm 1, \pm 1), (\pm 1, \pm 2), (\pm 2, \pm 1), &lt;/math&gt; and &lt;math&gt;(\pm 2, \pm 2),&lt;/math&gt; Frieda can only travel along the collection of points included in &lt;math&gt;S&lt;/math&gt;, where &lt;math&gt;S&lt;/math&gt; is all points on &lt;math&gt;x=0&lt;/math&gt; and &lt;math&gt;y=0&lt;/math&gt; such that &lt;math&gt;|y|&lt;4&lt;/math&gt; and &lt;math&gt;|x|&lt;4&lt;/math&gt;, respectively, plus all points on the big square with side length &lt;math&gt;6&lt;/math&gt; centered at &lt;math&gt;(0, 0).&lt;/math&gt; We then can proceed with casework:<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: Frieda never reaches &lt;math&gt;(0, \pm 3)&lt;/math&gt; nor &lt;math&gt;(\pm 3, 0).&lt;/math&gt;<br /> <br /> When Frieda only moves horizontally or vertically for her four moves, she can do so in &lt;math&gt;2^4 - 4 = 12&lt;/math&gt; ways for each case . Thus, &lt;math&gt;12 \cdot 2&lt;/math&gt; total paths for the subcase of staying in one direction. (For instance, all length &lt;math&gt;4&lt;/math&gt; combinations of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; except &lt;math&gt;FFFF&lt;/math&gt;, &lt;math&gt;BBBB&lt;/math&gt;, &lt;math&gt;FFFB&lt;/math&gt;, and &lt;math&gt;BBBF&lt;/math&gt; for the horizontal direction.) <br /> <br /> There is another subcase where she changes directions during her path. There are four symmetric cases for this subcase depending on which of the four quadrants Frieda hugs. For the first quadrant, the possible paths are &lt;math&gt;FBUD&lt;/math&gt;, &lt;math&gt;FBUU&lt;/math&gt;, &lt;math&gt;UDFB&lt;/math&gt;, and &lt;math&gt;UDFF.&lt;/math&gt; Thus, a total of &lt;math&gt;4 \cdot 4 = 16&lt;/math&gt; ways for this subcase. <br /> <br /> Total for Case &lt;math&gt;1&lt;/math&gt;: &lt;math&gt;24 + 16 = 40&lt;/math&gt;<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: Frieda reaches &lt;math&gt;(0, \pm 3)&lt;/math&gt; or &lt;math&gt;(\pm 3, 0)&lt;/math&gt;.<br /> <br /> Once Frieda reaches one of the points listed above (by using three moves), she has four choices for her last move. Thus, a total of &lt;math&gt;4 \cdot 4 = 16&lt;/math&gt; paths for this case.<br /> <br /> Our total number of paths never reaching coroners is thus &lt;math&gt;16+40=56,&lt;/math&gt; making for an answer of &lt;cmath&gt;\frac{4^4-56}{4^4} = \boxed{\textbf{(D)} ~\frac{25}{32}}.&lt;/cmath&gt;<br /> <br /> -fidgetboss_4000<br /> <br /> == Solution 6 (Casework) ==<br /> We take cases on the number of hops needed to reach a corner. For simplicity, denote &lt;math&gt;E&lt;/math&gt; as a move that takes Freida to an edge, &lt;math&gt;W&lt;/math&gt; as wrap-around move and &lt;math&gt;C&lt;/math&gt; as a corner move. Also, denote &lt;math&gt;O&lt;/math&gt; as a move that takes us to the center.<br /> <br /> <br /> &lt;b&gt; 2 Hops &lt;/b&gt;<br /> <br /> Then, Freida will have to &lt;math&gt;(E, C)&lt;/math&gt; as her set of moves. There are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, and &lt;math&gt;2&lt;/math&gt; corners to move to, for a total of &lt;math&gt;4 \cdot 2 = 8&lt;/math&gt; cases here. Then, there are &lt;math&gt;4&lt;/math&gt; choices for each move, for a probability of &lt;math&gt;\frac{8}{4 \cdot 4} = \frac{1}{2}&lt;/math&gt;. <br /> <br /> &lt;b&gt; 3 Hops &lt;/b&gt;<br /> <br /> In this case, Freida must wrap-around. There's only one possible combination, just &lt;math&gt;(E, W, C)&lt;/math&gt;. There are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, &lt;math&gt;1&lt;/math&gt; way to wrap-around (you must continue in the same direction) and &lt;math&gt;2&lt;/math&gt; corners, for a total of &lt;math&gt;4 \cdot 1 \cdot 2 = 8&lt;/math&gt; cases here. Then, there are &lt;math&gt;4&lt;/math&gt; choices for each move, for a probability of &lt;math&gt;\frac{8}{4 \cdot 4 \cdot 4} = \frac{1}{8}&lt;/math&gt;. <br /> <br /> &lt;b&gt; 4 Hops &lt;/b&gt; <br /> <br /> Lastly, there are two cases we must consider here. The first case is &lt;math&gt;(E, O, E, C)&lt;/math&gt;, and the second is &lt;math&gt;(E, W, W, C)&lt;/math&gt;. For the first case, there are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, &lt;math&gt;1&lt;/math&gt; way to return to the center, &lt;math&gt;4&lt;/math&gt; ways to move to an edge once again, and &lt;math&gt;2&lt;/math&gt; ways to move to a corner. Hence, there is a total of &lt;math&gt;4 \cdot 1 \cdot 4 \cdot 2 = 32&lt;/math&gt; cases here. Then, for the second case, there are &lt;math&gt;4&lt;/math&gt; ways to move to a corner, &lt;math&gt;1&lt;/math&gt; way to wrap-around, &lt;math&gt;1&lt;/math&gt; way to wrap-around again, and &lt;math&gt;2&lt;/math&gt; ways to move to a corner. This implies there are &lt;math&gt;4 \cdot 1 \cdot 1 \cdot 2 = 8&lt;/math&gt; cases here. Then, there is a total of &lt;math&gt;8+32 = 40&lt;/math&gt; cases, out of a total of &lt;math&gt;4^4 = 256&lt;/math&gt; cases, for a probability of &lt;math&gt;\frac{40}{256} = \frac{5}{32}&lt;/math&gt;. <br /> <br /> <br /> Then, the total probability that Freida ends up on a corner is &lt;math&gt;\frac{1}{2} + \frac{1}{8} + \frac{5}{32} = \frac{25}{32}&lt;/math&gt;, corresponding to choice &lt;math&gt;\boxed{D}&lt;/math&gt;. ~rocketsri<br /> <br /> == Video Solution by OmegaLearn (Using Probability States) ==<br /> https://youtu.be/V_Sn30N2q50<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2021|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_23&diff=146153 2021 AMC 12A Problems/Problem 23 2021-02-12T19:21:24Z <p>Rocketsri: </p> <hr /> <div>==Problem==<br /> Frieda the frog begins a sequence of hops on a &lt;math&gt;3 \times 3&lt;/math&gt; grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she &quot;wraps around&quot; and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops &quot;up&quot;, the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}&lt;/math&gt;<br /> <br /> ==Solution 1 (complementary counting)==<br /> We will use complementary counting.<br /> First, the frog can go left with probability &lt;math&gt;\frac14&lt;/math&gt;. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since &lt;math&gt;4\cdot \frac14=1&lt;/math&gt;, we will ignore the leading probability.<br /> <br /> From the left, she either goes left to another edge (&lt;math&gt;\frac14&lt;/math&gt;) or back to the center (&lt;math&gt;\frac14&lt;/math&gt;). Time for some casework.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; She goes back to the center.<br /> <br /> Now, she can go in any 4 directions, and then has 2 options from that edge. This gives &lt;math&gt;\frac12&lt;/math&gt;. --End case 1<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; She goes to another edge (rightmost).<br /> <br /> Subcase 1: She goes back to the left edge. She now has 2 places to go, giving &lt;math&gt;\frac12&lt;/math&gt;<br /> <br /> Subcase 2: She goes to the center. Now any move works.<br /> <br /> &lt;math&gt;\frac14 \cdot \frac12 + \frac14 \cdot 1=\frac18 + \frac 14=\frac38&lt;/math&gt; for this case. --End case 2<br /> <br /> She goes back to the center in Case 1 with probability &lt;math&gt;\frac14&lt;/math&gt;, and to the right edge with probability &lt;math&gt;\frac14&lt;/math&gt;<br /> <br /> So, our answer is &lt;math&gt;\frac14 \cdot \frac12 + \frac14 \cdot \frac38=\frac14 (\frac12+\frac38)=\frac14 \cdot \frac78 = \frac7{32}&lt;/math&gt;<br /> <br /> But, don't forget complementary counting. So, we get &lt;math&gt;1-\frac7{32}=\frac{25}{32} \implies \boxed{D}&lt;/math&gt;. ~ firebolt360<br /> <br /> Video Solution for those who prefer: https://youtu.be/ude2rzO1cmk ~ firebolt360<br /> <br /> ==Solution 2 (direct counting and probability states)==<br /> We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have &lt;math&gt;1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{D}&lt;/math&gt;. ~IceWolf10<br /> <br /> ==Solution 3 (Similar to Solution 2, but Finds the Numerator and Denominator Separately)==<br /> &lt;b&gt;Denominator&lt;/b&gt;<br /> <br /> There are &lt;math&gt;4^4=256&lt;/math&gt; ways to make 4 hops without restrictions.<br /> <br /> <br /> &lt;b&gt;Numerator (Casework)&lt;/b&gt;<br /> <br /> Suppose Frieda makes 4 hops without stopping. We perform casework on which hop reaches a corner for the first time.<br /> <br /> <br /> (1) Hop #2 (Hops #3 and #4 have no restrictions)<br /> <br /> The 4 independent hops have 4, 2, 4, 4 options, respectively. So, this case has &lt;math&gt;4\cdot2\cdot4\cdot4=128&lt;/math&gt; ways.<br /> <br /> <br /> (2) Hop #3 (Hope #4 has no restriction)<br /> <br /> No matter which direction the first hop takes, the second hop must &quot;wrap around&quot;.<br /> <br /> The 4 independent hops have 4, 1, 2, 4 options, respectively. So, this case has &lt;math&gt;4\cdot1\cdot2\cdot4=32&lt;/math&gt; ways.<br /> <br /> <br /> (3) Hop #4<br /> <br /> Two sub-cases:<br /> <br /> (3.1) The second hop &quot;wraps around&quot;. It follows that the third hop also &quot;wraps around&quot;.<br /> <br /> The 4 independent hops have 4, 1, 1, 2 options, respectively. So, this sub-case has &lt;math&gt;4\cdot1\cdot1\cdot2=8&lt;/math&gt; ways.<br /> <br /> (3.2) The second hop backs to the center.<br /> <br /> The 4 independent hops have 4, 1, 4, 2 options, respectively. So, this sub-case has &lt;math&gt;4\cdot1\cdot4\cdot2=32&lt;/math&gt; ways.<br /> <br /> Together, Case (3) has &lt;math&gt;8+32=40&lt;/math&gt; ways.<br /> <br /> <br /> The numerator is &lt;math&gt;128+32+40=200.&lt;/math&gt;<br /> <br /> <br /> &lt;b&gt;Probability&lt;/b&gt;<br /> <br /> &lt;math&gt;\frac{200}{256}=\boxed{\textbf{(D)} ~\frac{25}{32}}.&lt;/math&gt;<br /> <br /> <br /> This problem is quite similar to 1995 AIME Problem 3: https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3<br /> <br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;C_n&lt;/math&gt; be the probability that Frieda is on the central square after n moves, &lt;math&gt;E_n&lt;/math&gt; be the probability that Frieda is on one of the four squares on the middle of the edges after n moves, and &lt;math&gt;V_n&lt;/math&gt; (V for vertex) be the probability that Frieda is on a corner after n moves. The only way to reach the center is by moving in &lt;math&gt;1&lt;/math&gt; specific direction out of &lt;math&gt;4&lt;/math&gt; total directions from the middle of an edge, so &lt;math&gt;C_{n+1}=\frac{E_n}{4}&lt;/math&gt;. The ways to reach the middle of an edge are by moving in any direction from the center or by moving in &lt;math&gt;1&lt;/math&gt; specific direction from the middle of an edge, so &lt;math&gt;E_{n+1}=C_n+\frac{E_n}{4}&lt;/math&gt;. The ways to reach a corner are by simply staying there after reaching there in a previous move or by moving in &lt;math&gt;2&lt;/math&gt; specific directions from the middle of an edge, so &lt;math&gt;V_{n+1}=V_n+\frac{E_n}{2}&lt;/math&gt;. Since Frieda always start from the center, &lt;math&gt;C_0=1&lt;/math&gt;, &lt;math&gt;E_0=0&lt;/math&gt;, and &lt;math&gt;V_0=0&lt;/math&gt;. We use the previous formulas to work out &lt;math&gt;V_4&lt;/math&gt; and find it to be &lt;math&gt;\boxed{\textbf{(D)} ~\frac{25}{32}}&lt;/math&gt;.<br /> <br /> -SmileKat32<br /> <br /> ==Solution 5==<br /> Imagine an infinite grid of &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; squares such that there is a &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; square centered at &lt;math&gt;(3x, 3y)&lt;/math&gt; for all ordered pairs of integers &lt;math&gt;(x, y).&lt;/math&gt; It is easy to see that the problem is equivalent to Frieda moving left, right, up, or down on this infinite grid starting at &lt;math&gt;(0, 0)&lt;/math&gt;. (minus the teleportations) Since counting the complement set is easier, we'll count the number of &lt;math&gt;4&lt;/math&gt;-step paths such that Frieda never reaches a corner point. <br /> <br /> In other words, since the reachable corner points are &lt;math&gt;(\pm 1, \pm 1), (\pm 1, \pm 2), (\pm 2, \pm 1), &lt;/math&gt; and &lt;math&gt;(\pm 2, \pm 2),&lt;/math&gt; Frieda can only travel along the collection of points included in &lt;math&gt;S&lt;/math&gt;, where &lt;math&gt;S&lt;/math&gt; is all points on &lt;math&gt;x=0&lt;/math&gt; and &lt;math&gt;y=0&lt;/math&gt; such that &lt;math&gt;|y|&lt;4&lt;/math&gt; and &lt;math&gt;|x|&lt;4&lt;/math&gt;, respectively, plus all points on the big square with side length &lt;math&gt;6&lt;/math&gt; centered at &lt;math&gt;(0, 0).&lt;/math&gt; We then can proceed with casework:<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: Frieda never reaches &lt;math&gt;(0, \pm 3)&lt;/math&gt; nor &lt;math&gt;(\pm 3, 0).&lt;/math&gt;<br /> <br /> When Frieda only moves horizontally or vertically for her four moves, she can do so in &lt;math&gt;2^4 - 4 = 12&lt;/math&gt; ways for each case . Thus, &lt;math&gt;12 \cdot 2&lt;/math&gt; total paths for the subcase of staying in one direction. (For instance, all length &lt;math&gt;4&lt;/math&gt; combinations of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; except &lt;math&gt;FFFF&lt;/math&gt;, &lt;math&gt;BBBB&lt;/math&gt;, &lt;math&gt;FFFB&lt;/math&gt;, and &lt;math&gt;BBBF&lt;/math&gt; for the horizontal direction.) <br /> <br /> There is another subcase where she changes directions during her path. There are four symmetric cases for this subcase depending on which of the four quadrants Frieda hugs. For the first quadrant, the possible paths are &lt;math&gt;FBUD&lt;/math&gt;, &lt;math&gt;FBUU&lt;/math&gt;, &lt;math&gt;UDFB&lt;/math&gt;, and &lt;math&gt;UDFF.&lt;/math&gt; Thus, a total of &lt;math&gt;4 \cdot 4 = 16&lt;/math&gt; ways for this subcase. <br /> <br /> Total for Case &lt;math&gt;1&lt;/math&gt;: &lt;math&gt;24 + 16 = 40&lt;/math&gt;<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: Frieda reaches &lt;math&gt;(0, \pm 3)&lt;/math&gt; or &lt;math&gt;(\pm 3, 0)&lt;/math&gt;.<br /> <br /> Once Frieda reaches one of the points listed above (by using three moves), she has four choices for her last move. Thus, a total of &lt;math&gt;4 \cdot 4 = 16&lt;/math&gt; paths for this case.<br /> <br /> Our total number of paths never reaching coroners is thus &lt;math&gt;16+40=56,&lt;/math&gt; making for an answer of &lt;cmath&gt;\frac{4^4-56}{4^4} = \boxed{\textbf{(D)} ~\frac{25}{32}}.&lt;/cmath&gt;<br /> <br /> -fidgetboss_4000<br /> <br /> == Solution 6 ==<br /> We take cases on the number of hops needed to reach a corner. For simplicity, denote &lt;math&gt;E&lt;/math&gt; as a move that takes Freida to an edge, &lt;math&gt;W&lt;/math&gt; as wrap-around move and &lt;math&gt;C&lt;/math&gt; as a corner move. Also, denote &lt;math&gt;O&lt;/math&gt; as a move that takes us to the center.<br /> <br /> <br /> &lt;b&gt; 2 Hops &lt;/b&gt;<br /> <br /> Then, Freida will have to &lt;math&gt;(E, C)&lt;/math&gt; as her set of moves. There are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, and &lt;math&gt;2&lt;/math&gt; corners to move to, for a total of &lt;math&gt;4 \cdot 2 = 8&lt;/math&gt; cases here. Then, there are &lt;math&gt;4&lt;/math&gt; choices for each move, for a probability of &lt;math&gt;\frac{8}{4 \cdot 4} = \frac{1}{2}&lt;/math&gt;. <br /> <br /> &lt;b&gt; 3 Hops &lt;/b&gt;<br /> <br /> In this case, Freida must wrap-around. There's only one possible combination, just &lt;math&gt;(E, W, C)&lt;/math&gt;. There are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, &lt;math&gt;1&lt;/math&gt; way to wrap-around (you must continue in the same direction) and &lt;math&gt;2&lt;/math&gt; corners, for a total of &lt;math&gt;4 \cdot 1 \cdot 2 = 8&lt;/math&gt; cases here. Then, there are &lt;math&gt;4&lt;/math&gt; choices for each move, for a probability of &lt;math&gt;\frac{8}{4 \cdot 4 \cdot 4} = \frac{1}{8}&lt;/math&gt;. <br /> <br /> &lt;b&gt; 4 Hops &lt;/b&gt; <br /> <br /> Lastly, there are two cases we must consider here. The first case is &lt;math&gt;(E, O, E, C)&lt;/math&gt;, and the second is &lt;math&gt;(E, W, W, C)&lt;/math&gt;. For the first case, there are &lt;math&gt;4&lt;/math&gt; ways to move to an edge, &lt;math&gt;1&lt;/math&gt; way to return to the center, &lt;math&gt;4&lt;/math&gt; ways to move to an edge once again, and &lt;math&gt;2&lt;/math&gt; ways to move to a corner. Hence, there is a total of &lt;math&gt;4 \cdot 1 \cdot 4 \cdot 2 = 32&lt;/math&gt; cases here. Then, for the second case, there are &lt;math&gt;4&lt;/math&gt; ways to move to a corner, &lt;math&gt;1&lt;/math&gt; way to wrap-around, &lt;math&gt;1&lt;/math&gt; way to wrap-around again, and &lt;math&gt;2&lt;/math&gt; ways to move to a corner. This implies there are &lt;math&gt;4 \cdot 1 \cdot 1 \cdot 2 = 8&lt;/math&gt; cases here. Then, there is a total of &lt;math&gt;8+32 = 40&lt;/math&gt; cases, out of a total of &lt;math&gt;4^4 = 256&lt;/math&gt; cases, for a probability of &lt;math&gt;\frac{40}{256} = \frac{5}{32}&lt;/math&gt;. <br /> <br /> <br /> Then, the total probability that Freida ends up on a corner is &lt;math&gt;\frac{1}{2} + \frac{1}{8} + \frac{5}{32} = \frac{25}{32}&lt;/math&gt;, corresponding to choice &lt;math&gt;\boxed{D}&lt;/math&gt;. ~rocketsri<br /> <br /> == Video Solution by OmegaLearn (Using Probability States) ==<br /> https://youtu.be/V_Sn30N2q50<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2021|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2020_CIME_I_Problems/Problem_7&diff=144649 2020 CIME I Problems/Problem 7 2021-02-02T16:05:54Z <p>Rocketsri: /* Solution */</p> <hr /> <div>==Problem 7==<br /> For every positive integer &lt;math&gt;n&lt;/math&gt;, define &lt;cmath&gt;f(n)=\frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1)}.&lt;/cmath&gt; Suppose that the sum &lt;math&gt;f(1)+f(2)+\cdots+f(2020)&lt;/math&gt; can be expressed as &lt;math&gt;\frac{p}{q}&lt;/math&gt; for relatively prime integers &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;. Find the remainder when &lt;math&gt;p&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;S(n) = f(1)+f(2)+\cdots+f(n)&lt;/math&gt;. We claim that &lt;cmath&gt;S(n) = \frac{1}{2} - \frac{1}{2 \cdot (2n+1)!!}.&lt;/cmath&gt; We show this using induction. Suppose this is true for some &lt;math&gt;n = k&lt;/math&gt;. Then, it must be true for &lt;math&gt;k+1&lt;/math&gt;. The base case when &lt;math&gt;k = 1&lt;/math&gt; is trivial. Then, we have that &lt;cmath&gt;\begin{align*} S(k+1) &amp;= f(k) + S(k) \\ {} &amp;= \frac{n}{(2n+1)!!} + \frac{1}{2} - \frac{1}{2 \cdot (2n+1)!!} \\ {} &amp;= \frac{1}{2} - \frac{1}{2 \cdot (2n+3)!!}. \end{align*}&lt;/cmath&gt; Hence, this completes the induction therefore proving the claim. So, the numerator &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;\frac{1}{2}(4041!! - 1)&lt;/math&gt;. <br /> <br /> We proceed using Euler's Theorem combined with Chinese Remainder Theorem. It is obvious that &lt;math&gt;4041!! \equiv 0 \pmod{125}&lt;/math&gt; so &lt;math&gt;p \equiv 62\pmod{125}&lt;/math&gt;. Also, instead of considering &lt;math&gt;p&lt;/math&gt; modulo &lt;math&gt;8&lt;/math&gt;, we consider it modulo &lt;math&gt;16&lt;/math&gt;. Then, we get &lt;cmath&gt;4041!! \equiv 3^{253} \cdot 5^{253} \cdot 7^{253} \cdot 9^{252} \cdot 11^{252} \cdot 13^{252} \cdot 15^{252} \equiv 3 \cdot 5 \cdot 7 \cdot 9 \equiv 1\pmod{16},&lt;/cmath&gt; by Euler's Totient Theorem as &lt;math&gt;\phi(16) = 8&lt;/math&gt;. This implies that &lt;math&gt;\frac{1}{2}(4041!! - 1) \equiv 0\pmod{8}&lt;/math&gt;, so &lt;math&gt;p \equiv 0\pmod{8}&lt;/math&gt;. Solving the system of congruences, we get &lt;math&gt;p \equiv \boxed{312}\pmod{1000}&lt;/math&gt;. ~rocketsri (based off of official solution)<br /> <br /> ==See also==<br /> {{CIME box|year=2020|n=I|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAC Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2020_CIME_I_Problems/Problem_7&diff=144647 2020 CIME I Problems/Problem 7 2021-02-02T16:05:24Z <p>Rocketsri: /* Solution */</p> <hr /> <div>==Problem 7==<br /> For every positive integer &lt;math&gt;n&lt;/math&gt;, define &lt;cmath&gt;f(n)=\frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1)}.&lt;/cmath&gt; Suppose that the sum &lt;math&gt;f(1)+f(2)+\cdots+f(2020)&lt;/math&gt; can be expressed as &lt;math&gt;\frac{p}{q}&lt;/math&gt; for relatively prime integers &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;. Find the remainder when &lt;math&gt;p&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;S(n) = f(1)+f(2)+\cdots+f(n)&lt;/math&gt;. We claim that &lt;cmath&gt;S(n) = \frac{1}{2} - \frac{1}{2 \cdot (2n+1)!!}.&lt;/cmath&gt; We show this using induction. Suppose this is true for some &lt;math&gt;n = k&lt;/math&gt;. Then, it must be true for &lt;math&gt;k+1&lt;/math&gt;. The base case when &lt;math&gt;k = 1&lt;/math&gt; is trivial. Then, we have that &lt;cmath&gt;\begin{align*} S(k+1) &amp;= f(k) + S(k) \\ {} &amp;= \frac{n}{(2n+1)!!} + \frac{1}{2} - \frac{1}{2 \cdot (2n+1)!!} \\ {} &amp;= \frac{1}{2} - \frac{1}{2 \cdot (2n+3)!!}. \end{align*}&lt;/cmath&gt; Hence, this completes the induction therefore proving the claim. So, the numerator &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;\frac{1}{2}(4041!! - 1)&lt;/math&gt;. We proceed using Euler's Theorem combined with Chinese Remainder Theorem. It is obvious that &lt;math&gt;4041!! \equiv 0 \pmod{125}&lt;/math&gt; so &lt;math&gt;p \equiv 62\pmod{125}&lt;/math&gt;. Also, instead of considering &lt;math&gt;p&lt;/math&gt; modulo &lt;math&gt;8&lt;/math&gt;, we consider it modulo &lt;math&gt;16&lt;/math&gt;. Then, we get &lt;cmath&gt;4041!! \equiv 3^{253} \cdot 5^{253} \cdot 7^{253} \cdot 9^{252} \cdot 11^{252} \cdot 13^{252} \cdot 15^{252} \equiv 3 \cdot 5 \cdot 7 \cdot 9 \equiv 1\pmod{16},&lt;/cmath&gt; by Euler's Totient Theorem as &lt;math&gt;\phi(16) = 8&lt;/math&gt;. This implies that &lt;math&gt;\frac{1}{2}(4041!! - 1) \equiv 0\pmod{8}&lt;/math&gt;, so &lt;math&gt;p \equiv 0\pmod{8}&lt;/math&gt;. Solving the system of congruences, we get &lt;math&gt;p \equiv \boxed{312}\pmod{1000}&lt;/math&gt;. ~rocketsri (based off of official solution)<br /> <br /> ==See also==<br /> {{CIME box|year=2020|n=I|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAC Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2020_CIME_I_Problems/Problem_10&diff=144639 2020 CIME I Problems/Problem 10 2021-02-02T15:47:28Z <p>Rocketsri: /* Solution */</p> <hr /> <div>==Problem 10==<br /> Let &lt;math&gt;1=d_1&lt;d_2&lt;\cdots&lt;d_k=n&lt;/math&gt; be the divisors of a positive integer &lt;math&gt;n&lt;/math&gt;. Let &lt;math&gt;S&lt;/math&gt; be the sum of all positive integers &lt;math&gt;n&lt;/math&gt; satisfying &lt;cmath&gt;n=d_1^1+d_2^2+d_3^3+d_4^4.&lt;/cmath&gt; Find the remainder when &lt;math&gt;S&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Note that &lt;math&gt;n&lt;/math&gt; is even, as if &lt;math&gt;n&lt;/math&gt; is odd then the RHS is even and the LHS is odd. This implies that the first two divisors of &lt;math&gt;n&lt;/math&gt; are &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;. Now, there are three cases (note that &lt;math&gt;p_i&lt;/math&gt; represents a prime): <br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; When &lt;math&gt;\{ d_3, d_4 \} = \{ p_1, p_2 \}&lt;/math&gt;. Then, &lt;math&gt;1+2^2+p_1^3+p_2^4&lt;/math&gt; is always odd, so this is a contradiction. <br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; When &lt;math&gt;\{ d_3, d_4 \} = \{ p_1, 2p_1 \}&lt;/math&gt;. This implies that &lt;math&gt;p_1 | 1+2^2&lt;/math&gt; or &lt;math&gt;p_1 = 5&lt;/math&gt;. Then, &lt;math&gt;n = 1+2^2+5^3+10^4 = 10130&lt;/math&gt;. <br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; When &lt;math&gt;\{ d_3, d_4 \} = \{ 4, p_1 \}&lt;/math&gt;. If &lt;math&gt;p_1 &lt; 4 \implies p_1 = 3&lt;/math&gt; then &lt;math&gt;n = 288&lt;/math&gt;. If &lt;math&gt;p_1 \geq 5&lt;/math&gt; then &lt;math&gt;4p_1 | 1+2^3+4^3+p_1^4 \implies p_1 | 1+2^2+4^3 = 69&lt;/math&gt;. This means &lt;math&gt;p_1 = 23&lt;/math&gt; as &lt;math&gt;p_1 \geq 5&lt;/math&gt;, however, this fails. <br /> <br /> <br /> So, the sum of all possible values of &lt;math&gt;n&lt;/math&gt; are &lt;math&gt;10130 + 288 = 10418&lt;/math&gt;, so the remainder is &lt;math&gt;\boxed{418}&lt;/math&gt;. ~rocketsri<br /> <br /> ==See also==<br /> {{CIME box|year=2020|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAC Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_12&diff=141966 2020 AMC 12A Problems/Problem 12 2021-01-12T21:30:06Z <p>Rocketsri: /* Solution 7 (Complex Numbers) */</p> <hr /> <div>== Problem ==<br /> <br /> Line &lt;math&gt;l&lt;/math&gt; in the coordinate plane has equation &lt;math&gt;3x-5y+40=0&lt;/math&gt;. This line is rotated &lt;math&gt;45^{\circ}&lt;/math&gt; counterclockwise about the point &lt;math&gt;(20,20)&lt;/math&gt; to obtain line &lt;math&gt;k&lt;/math&gt;. What is the &lt;math&gt;x&lt;/math&gt;-coordinate of the &lt;math&gt;x&lt;/math&gt;-intercept of line &lt;math&gt;k?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> The slope of the line is &lt;math&gt;\frac{3}{5}&lt;/math&gt;. We must transform it by &lt;math&gt;45^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;45^{\circ}&lt;/math&gt; creates an isosceles right triangle since the sum of the angles of the triangle must be &lt;math&gt;180^{\circ}&lt;/math&gt; and one angle is &lt;math&gt;90^{\circ}&lt;/math&gt; which means the last leg angle must also be &lt;math&gt;45^{\circ}&lt;/math&gt;. <br /> <br /> In the isosceles right triangle, the two legs are congruent. We can, therefore, construct an isosceles right triangle with a line of &lt;math&gt;\frac{3}{5}&lt;/math&gt; slope on graph paper. That line with &lt;math&gt;\frac{3}{5}&lt;/math&gt; slope starts at &lt;math&gt;(0,0)&lt;/math&gt; and will go to &lt;math&gt;(5,3)&lt;/math&gt;, the vector &lt;math&gt;&lt;5,3&gt;&lt;/math&gt;. <br /> <br /> Construct another line from &lt;math&gt;(0,0)&lt;/math&gt; to &lt;math&gt;(3,-5)&lt;/math&gt;, the vector &lt;math&gt;&lt;3,-5&gt;&lt;/math&gt;. This is &lt;math&gt;\perp&lt;/math&gt; and equal to the original line segment. The difference between the two vectors is &lt;math&gt;&lt;2,8&gt;&lt;/math&gt;, which is the slope &lt;math&gt;4&lt;/math&gt;, and that is the slope of line &lt;math&gt;k&lt;/math&gt;. <br /> <br /> Furthermore, the equation &lt;math&gt;3x-5y+40=0&lt;/math&gt; passes straight through &lt;math&gt;(20,20)&lt;/math&gt; since &lt;math&gt;3(20)-5(20)+40=60-100+40=0&lt;/math&gt;, which means that any rotations about &lt;math&gt;(20,20)&lt;/math&gt; would contain &lt;math&gt;(20,20)&lt;/math&gt;. We can create a line of slope &lt;math&gt;4&lt;/math&gt; through &lt;math&gt;(20,20)&lt;/math&gt;. The &lt;math&gt;x&lt;/math&gt;-intercept is therefore &lt;math&gt;20-\frac{20}{4} = \boxed{\textbf{(B) } 15.}&lt;/math&gt;~lopkiloinm<br /> <br /> ==Solution 2==<br /> Since the slope of the line is &lt;math&gt;\frac{3}{5}&lt;/math&gt;, and the angle we are rotating around is x, then &lt;math&gt;\tan x = \frac{3}{5}&lt;/math&gt;<br /> &lt;math&gt;\tan(x+45^{\circ}) = \frac{\tan x + \tan(45^{\circ})}{1-\tan x*\tan(45^{\circ})} = \frac{0.6+1}{1-0.6} = \frac{1.6}{0.4} = 4&lt;/math&gt;<br /> <br /> Hence, the slope of the rotated line is &lt;math&gt;4&lt;/math&gt;. Since we know the line intersects the point &lt;math&gt;(20,20)&lt;/math&gt;, then we know the line is &lt;math&gt;y=4x-60&lt;/math&gt;. Set &lt;math&gt;y=0&lt;/math&gt; to find the x-intercept, and so &lt;math&gt;x=\boxed{15}&lt;/math&gt;<br /> <br /> ~Solution by IronicNinja<br /> <br /> ==Solution 3==<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(20, 0)--(20, 20)--(0, 20)--cycle);<br /> draw((20, 20)--(0, 8));<br /> draw((15, 0)--(20, 20));<br /> <br /> dot(&quot;$P$&quot;, (20, 20));<br /> dot(&quot;$A$&quot;, (0, 8), dir(75));<br /> dot(&quot;$B$&quot;, (15, 0), dir(45));<br /> dot(&quot;$X$&quot;, (20, 0));<br /> dot(&quot;$Y$&quot;, (0, 20), dir(50));<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;P&lt;/math&gt; be &lt;math&gt;(20, 20)&lt;/math&gt; and &lt;math&gt;X, Y&lt;/math&gt; be &lt;math&gt;(20, 0)&lt;/math&gt; and &lt;math&gt;(0, 20)&lt;/math&gt; respectively. Since the slope of the line is &lt;math&gt;3/5&lt;/math&gt; we know that &lt;math&gt;\tan{\angle{YPA}} = 3/5.&lt;/math&gt; Segments &lt;math&gt;\overline{PA}&lt;/math&gt; and &lt;math&gt;\overline{PB}&lt;/math&gt; represent the before and after of rotating &lt;math&gt;l&lt;/math&gt; by 45 counterclockwise. Thus, &lt;math&gt;\angle{XPB} = 45 - \angle{YPA}&lt;/math&gt; and &lt;cmath&gt;BX = 20 \tan{\angle{XPB}} = 20 \cdot \frac{1 - 3/5}{1 + 3/5} = 5&lt;/cmath&gt; by tangent addition formula. Since &lt;math&gt;BX&lt;/math&gt; is 5 and the sidelength of the square is 20 the answer is &lt;math&gt;20 - 5 \implies \boxed{\textbf{B}}.&lt;/math&gt;<br /> <br /> ==Solution 4 (Cheap)==<br /> <br /> Using the protractor you brought, carefully graph the equation and rotate the given line &lt;math&gt;45^{\circ}&lt;/math&gt; counter-clockwise about the point &lt;math&gt;(20,20)&lt;/math&gt;. Scaling everything down by a factor of 5 makes this process easier.<br /> <br /> It should then become fairly obvious that the x intercept is &lt;math&gt;x=\boxed{15}&lt;/math&gt; (only use this as a last resort). <br /> <br /> ~Silverdragon<br /> <br /> ==Solution 5 (Rotation Matrix)==<br /> First note that the given line goes through &lt;math&gt;(20,20)&lt;/math&gt; with a slope of &lt;math&gt;\frac{3}{5}&lt;/math&gt;. This means that &lt;math&gt;(25,23)&lt;/math&gt; is on the line. Now consider translating the graph so that &lt;math&gt;(20,20)&lt;/math&gt; goes to the origin, then &lt;math&gt;(25,23)&lt;/math&gt; becomes &lt;math&gt;(5,3)&lt;/math&gt;. We now rotate the line &lt;math&gt;45^\circ&lt;/math&gt; about the origin using a rotation matrix. This maps &lt;math&gt;(5,3)&lt;/math&gt; to <br /> &lt;cmath&gt;\begin{bmatrix} \frac{\sqrt{2}}{2} &amp; -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} &amp; \frac{\sqrt{2}}{2}\end{bmatrix}\begin{bmatrix} 5 \\ 3\end{bmatrix}=\begin{bmatrix}\sqrt{2} \\ 4\sqrt{2}\end{bmatrix}&lt;/cmath&gt;<br /> The line through the origin and &lt;math&gt;(\sqrt{2}, 4\sqrt{2})&lt;/math&gt; has slope &lt;math&gt;4&lt;/math&gt;. Translating this line so that the origin is mapped to &lt;math&gt;(20,20)&lt;/math&gt;, we find that the equation for the new line is &lt;math&gt;4x-60&lt;/math&gt;, meaning that the &lt;math&gt;x&lt;/math&gt;-intercept is &lt;math&gt;x=\frac{60}{4}=\boxed{\textbf{(B) }15}&lt;/math&gt;.<br /> <br /> ==Solution 6 (Angle Bisector)==<br /> Note &lt;math&gt;(20,20)&lt;/math&gt; is on the line. Construct the perpendicular line &lt;math&gt;5x+3y-160=0&lt;/math&gt;. This creates a right triangle that intersects the x-axis at &lt;math&gt;(\frac{-40}{3},0)&lt;/math&gt; and &lt;math&gt;(32,0)&lt;/math&gt; a distance of &lt;math&gt;136&lt;/math&gt; apart. The &lt;math&gt;45^\circ&lt;/math&gt; transformation will bisect the right angle.<br /> The angle bisector theorem tells us the &lt;math&gt;136&lt;/math&gt; will split in ratio to the lengths of the sides.<br /> These are &lt;math&gt;\sqrt(12^2+20^2)&lt;/math&gt; and &lt;math&gt;\sqrt(\frac{100}{3}^2 + 20^2) = 4\sqrt(34)&lt;/math&gt; and &lt;math&gt;\frac{20}{3}\sqrt(34)&lt;/math&gt;. Thus the x intercept will split the line from &lt;math&gt;\frac{-40}{3}&lt;/math&gt; to &lt;math&gt;32&lt;/math&gt; into a ratio of &lt;math&gt;5:3&lt;/math&gt; making the x-intercept &lt;math&gt;15&lt;/math&gt;.<br /> <br /> ==Solution 7 (Complex Numbers)==<br /> Converting to the complex plane, we can see that two numbers on the line are &lt;math&gt;8i&lt;/math&gt; and &lt;math&gt;5+11i&lt;/math&gt;. Translating &lt;math&gt;20+20i&lt;/math&gt; to the origin, we get &lt;math&gt;8i-20-2i = -20+6i&lt;/math&gt; and &lt;math&gt;5+11i-20-20i = -15-9i&lt;/math&gt;. Multiplying each of them by &lt;math&gt;e^{\pi i/4}&lt;/math&gt;, we get &lt;math&gt;-4\sqrt 2 - 16 \sqrt2 i&lt;/math&gt; and &lt;math&gt;-3\sqrt 2 - 12 \sqrt 2 i&lt;/math&gt;. This line has a slope of &lt;math&gt;4&lt;/math&gt;. Now, back to the cartesian plane. We have a line passing through &lt;math&gt;(20, 20)&lt;/math&gt; with slope &lt;math&gt;4&lt;/math&gt; which gives the equation as &lt;math&gt;y = 4x-60&lt;/math&gt; which implies the &lt;math&gt;x&lt;/math&gt; coordinate of the &lt;math&gt;x&lt;/math&gt; intercept is &lt;math&gt;60/4 = 15&lt;/math&gt;. ~rocketsri<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_12&diff=141965 2020 AMC 12A Problems/Problem 12 2021-01-12T21:29:07Z <p>Rocketsri: </p> <hr /> <div>== Problem ==<br /> <br /> Line &lt;math&gt;l&lt;/math&gt; in the coordinate plane has equation &lt;math&gt;3x-5y+40=0&lt;/math&gt;. This line is rotated &lt;math&gt;45^{\circ}&lt;/math&gt; counterclockwise about the point &lt;math&gt;(20,20)&lt;/math&gt; to obtain line &lt;math&gt;k&lt;/math&gt;. What is the &lt;math&gt;x&lt;/math&gt;-coordinate of the &lt;math&gt;x&lt;/math&gt;-intercept of line &lt;math&gt;k?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> The slope of the line is &lt;math&gt;\frac{3}{5}&lt;/math&gt;. We must transform it by &lt;math&gt;45^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;45^{\circ}&lt;/math&gt; creates an isosceles right triangle since the sum of the angles of the triangle must be &lt;math&gt;180^{\circ}&lt;/math&gt; and one angle is &lt;math&gt;90^{\circ}&lt;/math&gt; which means the last leg angle must also be &lt;math&gt;45^{\circ}&lt;/math&gt;. <br /> <br /> In the isosceles right triangle, the two legs are congruent. We can, therefore, construct an isosceles right triangle with a line of &lt;math&gt;\frac{3}{5}&lt;/math&gt; slope on graph paper. That line with &lt;math&gt;\frac{3}{5}&lt;/math&gt; slope starts at &lt;math&gt;(0,0)&lt;/math&gt; and will go to &lt;math&gt;(5,3)&lt;/math&gt;, the vector &lt;math&gt;&lt;5,3&gt;&lt;/math&gt;. <br /> <br /> Construct another line from &lt;math&gt;(0,0)&lt;/math&gt; to &lt;math&gt;(3,-5)&lt;/math&gt;, the vector &lt;math&gt;&lt;3,-5&gt;&lt;/math&gt;. This is &lt;math&gt;\perp&lt;/math&gt; and equal to the original line segment. The difference between the two vectors is &lt;math&gt;&lt;2,8&gt;&lt;/math&gt;, which is the slope &lt;math&gt;4&lt;/math&gt;, and that is the slope of line &lt;math&gt;k&lt;/math&gt;. <br /> <br /> Furthermore, the equation &lt;math&gt;3x-5y+40=0&lt;/math&gt; passes straight through &lt;math&gt;(20,20)&lt;/math&gt; since &lt;math&gt;3(20)-5(20)+40=60-100+40=0&lt;/math&gt;, which means that any rotations about &lt;math&gt;(20,20)&lt;/math&gt; would contain &lt;math&gt;(20,20)&lt;/math&gt;. We can create a line of slope &lt;math&gt;4&lt;/math&gt; through &lt;math&gt;(20,20)&lt;/math&gt;. The &lt;math&gt;x&lt;/math&gt;-intercept is therefore &lt;math&gt;20-\frac{20}{4} = \boxed{\textbf{(B) } 15.}&lt;/math&gt;~lopkiloinm<br /> <br /> ==Solution 2==<br /> Since the slope of the line is &lt;math&gt;\frac{3}{5}&lt;/math&gt;, and the angle we are rotating around is x, then &lt;math&gt;\tan x = \frac{3}{5}&lt;/math&gt;<br /> &lt;math&gt;\tan(x+45^{\circ}) = \frac{\tan x + \tan(45^{\circ})}{1-\tan x*\tan(45^{\circ})} = \frac{0.6+1}{1-0.6} = \frac{1.6}{0.4} = 4&lt;/math&gt;<br /> <br /> Hence, the slope of the rotated line is &lt;math&gt;4&lt;/math&gt;. Since we know the line intersects the point &lt;math&gt;(20,20)&lt;/math&gt;, then we know the line is &lt;math&gt;y=4x-60&lt;/math&gt;. Set &lt;math&gt;y=0&lt;/math&gt; to find the x-intercept, and so &lt;math&gt;x=\boxed{15}&lt;/math&gt;<br /> <br /> ~Solution by IronicNinja<br /> <br /> ==Solution 3==<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(20, 0)--(20, 20)--(0, 20)--cycle);<br /> draw((20, 20)--(0, 8));<br /> draw((15, 0)--(20, 20));<br /> <br /> dot(&quot;$P$&quot;, (20, 20));<br /> dot(&quot;$A$&quot;, (0, 8), dir(75));<br /> dot(&quot;$B$&quot;, (15, 0), dir(45));<br /> dot(&quot;$X$&quot;, (20, 0));<br /> dot(&quot;$Y$&quot;, (0, 20), dir(50));<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;P&lt;/math&gt; be &lt;math&gt;(20, 20)&lt;/math&gt; and &lt;math&gt;X, Y&lt;/math&gt; be &lt;math&gt;(20, 0)&lt;/math&gt; and &lt;math&gt;(0, 20)&lt;/math&gt; respectively. Since the slope of the line is &lt;math&gt;3/5&lt;/math&gt; we know that &lt;math&gt;\tan{\angle{YPA}} = 3/5.&lt;/math&gt; Segments &lt;math&gt;\overline{PA}&lt;/math&gt; and &lt;math&gt;\overline{PB}&lt;/math&gt; represent the before and after of rotating &lt;math&gt;l&lt;/math&gt; by 45 counterclockwise. Thus, &lt;math&gt;\angle{XPB} = 45 - \angle{YPA}&lt;/math&gt; and &lt;cmath&gt;BX = 20 \tan{\angle{XPB}} = 20 \cdot \frac{1 - 3/5}{1 + 3/5} = 5&lt;/cmath&gt; by tangent addition formula. Since &lt;math&gt;BX&lt;/math&gt; is 5 and the sidelength of the square is 20 the answer is &lt;math&gt;20 - 5 \implies \boxed{\textbf{B}}.&lt;/math&gt;<br /> <br /> ==Solution 4 (Cheap)==<br /> <br /> Using the protractor you brought, carefully graph the equation and rotate the given line &lt;math&gt;45^{\circ}&lt;/math&gt; counter-clockwise about the point &lt;math&gt;(20,20)&lt;/math&gt;. Scaling everything down by a factor of 5 makes this process easier.<br /> <br /> It should then become fairly obvious that the x intercept is &lt;math&gt;x=\boxed{15}&lt;/math&gt; (only use this as a last resort). <br /> <br /> ~Silverdragon<br /> <br /> ==Solution 5 (Rotation Matrix)==<br /> First note that the given line goes through &lt;math&gt;(20,20)&lt;/math&gt; with a slope of &lt;math&gt;\frac{3}{5}&lt;/math&gt;. This means that &lt;math&gt;(25,23)&lt;/math&gt; is on the line. Now consider translating the graph so that &lt;math&gt;(20,20)&lt;/math&gt; goes to the origin, then &lt;math&gt;(25,23)&lt;/math&gt; becomes &lt;math&gt;(5,3)&lt;/math&gt;. We now rotate the line &lt;math&gt;45^\circ&lt;/math&gt; about the origin using a rotation matrix. This maps &lt;math&gt;(5,3)&lt;/math&gt; to <br /> &lt;cmath&gt;\begin{bmatrix} \frac{\sqrt{2}}{2} &amp; -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} &amp; \frac{\sqrt{2}}{2}\end{bmatrix}\begin{bmatrix} 5 \\ 3\end{bmatrix}=\begin{bmatrix}\sqrt{2} \\ 4\sqrt{2}\end{bmatrix}&lt;/cmath&gt;<br /> The line through the origin and &lt;math&gt;(\sqrt{2}, 4\sqrt{2})&lt;/math&gt; has slope &lt;math&gt;4&lt;/math&gt;. Translating this line so that the origin is mapped to &lt;math&gt;(20,20)&lt;/math&gt;, we find that the equation for the new line is &lt;math&gt;4x-60&lt;/math&gt;, meaning that the &lt;math&gt;x&lt;/math&gt;-intercept is &lt;math&gt;x=\frac{60}{4}=\boxed{\textbf{(B) }15}&lt;/math&gt;.<br /> <br /> ==Solution 6 (Angle Bisector)==<br /> Note &lt;math&gt;(20,20)&lt;/math&gt; is on the line. Construct the perpendicular line &lt;math&gt;5x+3y-160=0&lt;/math&gt;. This creates a right triangle that intersects the x-axis at &lt;math&gt;(\frac{-40}{3},0)&lt;/math&gt; and &lt;math&gt;(32,0)&lt;/math&gt; a distance of &lt;math&gt;136&lt;/math&gt; apart. The &lt;math&gt;45^\circ&lt;/math&gt; transformation will bisect the right angle.<br /> The angle bisector theorem tells us the &lt;math&gt;136&lt;/math&gt; will split in ratio to the lengths of the sides.<br /> These are &lt;math&gt;\sqrt(12^2+20^2)&lt;/math&gt; and &lt;math&gt;\sqrt(\frac{100}{3}^2 + 20^2) = 4\sqrt(34)&lt;/math&gt; and &lt;math&gt;\frac{20}{3}\sqrt(34)&lt;/math&gt;. Thus the x intercept will split the line from &lt;math&gt;\frac{-40}{3}&lt;/math&gt; to &lt;math&gt;32&lt;/math&gt; into a ratio of &lt;math&gt;5:3&lt;/math&gt; making the x-intercept &lt;math&gt;15&lt;/math&gt;.<br /> <br /> ==Solution 7 (Complex Numbers)==<br /> Converting to the complex plane, we can see that two numbers on the line are &lt;math&gt;8i&lt;/math&gt; and &lt;math&gt;5+11i&lt;/math&gt;. Translating &lt;math&gt;20+20i&lt;/math&gt; to the origin, we get &lt;math&gt;8i-20-2i = -20-12i&lt;/math&gt; and &lt;math&gt;5+11i-20-20i = -15-9i&lt;/math&gt;. Multiplying each of them by &lt;math&gt;e^{\pi i/4}&lt;/math&gt;, we get &lt;math&gt;-4\sqrt 2 - 16 \sqrt2 i&lt;/math&gt; and &lt;math&gt;-3\sqrt 2 - 12 \sqrt 2 i&lt;/math&gt;. This line has a slope of &lt;math&gt;4&lt;/math&gt;. Now, back to the cartesian plane. We have a line passing through &lt;math&gt;(20, 20)&lt;/math&gt; with slope &lt;math&gt;4&lt;/math&gt; which gives the equation as &lt;math&gt;y = 4x-60&lt;/math&gt; which implies the &lt;math&gt;x&lt;/math&gt; coordinate of the &lt;math&gt;x&lt;/math&gt; intercept is &lt;math&gt;60/4 = 15&lt;/math&gt;. ~rocketsri<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=User:Edud_looc&diff=141911 User:Edud looc 2021-01-11T21:26:22Z <p>Rocketsri: /* User Count */</p> <hr /> <div>==&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;102px”&gt;1 (reset as some people just can't stop)<br /> &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;&lt;br&gt;<br /> <br /> eduD looC is 11 years old.<br /> He wants to qualify for the 2021 AIME I and get 8+ (got 5 on 2020 AIME II)<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 /> Unknown<br /> &lt;/div&gt;<br /> &lt;/div&gt;</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=Wooga_Looga_Theorem&diff=140041 Wooga Looga Theorem 2020-12-20T21:22:08Z <p>Rocketsri: /* Testimonials */</p> <hr /> <div>=Definition=<br /> If there is &lt;math&gt;\triangle ABC&lt;/math&gt; and points &lt;math&gt;D,E,F&lt;/math&gt; on the sides &lt;math&gt;BC,CA,AB&lt;/math&gt; respectively such that &lt;math&gt;\frac{DB}{DC}=\frac{EC}{EA}=\frac{FA}{FB}=r&lt;/math&gt;, then the ratio &lt;math&gt;\frac{[DEF]}{[ABC]}=\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 /> One solution is this one 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 /> <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. Great for math contests with proofs.<br /> <br /> ==Solution 6==<br /> or this solution by CoolJupiter:<br /> <br /> Wow. All of your solutions are slow, compared to my sol:<br /> <br /> By math, we have &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;.<br /> <br /> ~CoolJupiter<br /> ^<br /> |<br /> EVERYONE USE THIS SOLUTION IT'S BRILLIANT <br /> ~bsu1<br /> Yes, very BRILLIANT!<br /> ~ TheAoPSLebron<br /> <br /> ==The Best Solution==<br /> <br /> By the 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;\boxed7&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> By franzliszt<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Then &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. We can also find that &lt;math&gt;D=(0,\tfrac 14,\tfrac 34),E=(\tfrac 34,0,\tfrac 14),F=(\tfrac 14,\tfrac 34,0)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[DEF]}{[ABC]}=\begin{vmatrix} 0&amp;\tfrac 14&amp;\tfrac 34\\ \tfrac 34&amp;0&amp;\tfrac 14\\ \tfrac 14&amp;\tfrac 34&amp;0 \end{vmatrix}=\frac{7}{16}.&lt;/cmath&gt;So the answer is &lt;math&gt;\boxed{7}&lt;/math&gt;.<br /> <br /> =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 /> 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<br /> <br /> If only I knew this on some contests that I had done previously... - JacobJB<br /> <br /> The Wooga Looga Theorem is so pr0 that it needs to be nerfed. - rocketsri</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=Wooga_Looga_Theorem&diff=139937 Wooga Looga Theorem 2020-12-18T22:01:20Z <p>Rocketsri: /* Testimonials */</p> <hr /> <div>=Definition=<br /> If there is &lt;math&gt;\triangle ABC&lt;/math&gt; and points &lt;math&gt;D,E,F&lt;/math&gt; on the sides &lt;math&gt;BC,CA,AB&lt;/math&gt; respectively such that &lt;math&gt;\frac{DB}{DC}=\frac{EC}{EA}=\frac{FA}{FB}=r&lt;/math&gt;, then the ratio &lt;math&gt;\frac{[DEF]}{[ABC]}=\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 /> One solution is this one 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 /> <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. Great for math contests with proofs.<br /> <br /> ==Solution 6==<br /> or this solution by CoolJupiter:<br /> <br /> Wow. All of your solutions are slow, compared to my sol:<br /> <br /> By math, we have &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;.<br /> <br /> ~CoolJupiter<br /> ^<br /> |<br /> EVERYONE USE THIS SOLUTION IT'S BRILLIANT <br /> ~bsu1<br /> Yes, very BRILLIANT!<br /> ~ TheAoPSLebron<br /> <br /> ==The Best Solution==<br /> <br /> By the 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;\boxed7&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> By franzliszt<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Then &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. We can also find that &lt;math&gt;D=(0,\tfrac 14,\tfrac 34),E=(\tfrac 34,0,\tfrac 14),F=(\tfrac 14,\tfrac 34,0)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[DEF]}{[ABC]}=\begin{vmatrix} 0&amp;\tfrac 14&amp;\tfrac 34\\ \tfrac 34&amp;0&amp;\tfrac 14\\ \tfrac 14&amp;\tfrac 34&amp;0 \end{vmatrix}=\frac{7}{16}.&lt;/cmath&gt;So the answer is &lt;math&gt;\boxed{7}&lt;/math&gt;.<br /> <br /> =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 /> 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<br /> <br /> If only I knew this on some contests that I had done previously... - JacobJB<br /> <br /> The Wooga Looga Theorem is so pr0 that test writers have banned it from comps - rocketsri</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=Lagrange_Multipliers&diff=138849 Lagrange Multipliers 2020-12-01T01:06:24Z <p>Rocketsri: /* Example */</p> <hr /> <div>This article discusses Lagrange multipliers, a topic of multivariable calculus.<br /> __TOC__<br /> <br /> == Definition ==<br /> Useful in optimization, Lagrange multipliers, based on a calculus approach, can be used to find local minimums and maximums of a function given a constraint. Suppose there is a continuous function &lt;math&gt;f(x,y)&lt;/math&gt; and there exists a continuous constraint function on the values of the function &lt;math&gt;c = g(x,y)&lt;/math&gt;. The method of Lagrange multipliers states that, to find the minimum or maximum satisfying both requirements (&lt;math&gt;\lambda&lt;/math&gt; is a constant):<br /> <br /> &lt;cmath&gt;\dfrac{\partial f}{\partial x} = \lambda \dfrac{\partial g}{\partial x}&lt;/cmath&gt;<br /> &lt;cmath&gt;\dfrac{\partial f}{\partial y} = \lambda \dfrac{\partial g}{\partial y}&lt;/cmath&gt;<br /> <br /> The method can be extended to multiple variables, as well as multiple constraints. If we had a continuous function &lt;math&gt;f(x, y, z)&lt;/math&gt; and the constraint &lt;math&gt;c = g(x, y, z)&lt;/math&gt;, we would just have another equation relating the partial derivatives with respect to &lt;math&gt;z&lt;/math&gt; through a factor of &lt;math&gt;\lambda&lt;/math&gt;. Similarly, if we have another constraint &lt;math&gt;d = h(x, y, z)&lt;/math&gt;, then we would add the partial with respect to each variable to their respective equation with another factor &lt;math&gt;\mu&lt;/math&gt;. Thus, we have the general form for Lagrange multipliers for the function &lt;math&gt;f(x_1, x_2,...,x_n)&lt;/math&gt; bounded by a certain number of constraints &lt;math&gt;c_1 = g_1(x_1, x_2,..., x_n)&lt;/math&gt;, &lt;math&gt;c_2 = g_2(x_1, x_2,..., x_n)&lt;/math&gt;,...,&lt;math&gt;c_m = g_m(x_1, x_2,..., x_n)&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2 + ... + \lambda_m \nabla g_m&lt;/cmath&gt;<br /> <br /> Where &lt;math&gt;\nabla&lt;/math&gt; is the del operator, which when applied to a scalar function &lt;math&gt;f&lt;/math&gt; results in a vector with each component the partial in that component's direction, representing total change.<br /> <br /> It is important to remember that sometimes there maybe a function that is bounded on some interval. In such cases, Lagrange multipliers may give a result, but the answer may not be the one this method results. After using this method, it is imperative to check the endpoints and also plug back in the numbers resulted from this method to verify the final answer. Also, sometimes the function may be over-constrained and there maybe no point that satisfies all requirements.<br /> <br /> == Example ==<br /> This method can become extremely difficult to use when there are several variables and constraints, but it is very effective and useful in some situations, including some contest math problems. Consider the following question:<br /> <br /> A rectangular prism lies on the &lt;math&gt;xy&lt;/math&gt; plane with one vertex at the origin. The vertex that does not share a face with this vertex at the origin lies on the plane bounded by the points &lt;math&gt;(0, 0, 4)&lt;/math&gt;, &lt;math&gt;(0, 8, 0)&lt;/math&gt;, &lt;math&gt;(16, 0, 0)&lt;/math&gt;. Find the maximum volume of the box.<br /> <br /> <br /> Solution:<br /> The volume of the box is given by the equation &lt;math&gt;V(x, y, z) = xyz&lt;/math&gt;. Because the way the box is described, the point that lies on the plane given determines the volume of the box. Thus we need only need to consider that point. The constraint we are given for that point is given by the plane &lt;math&gt;x + 2y + 4z = 16&lt;/math&gt;, since that is the plane that goes through the three points mentioned. Let us call that function &lt;math&gt;16 = g(x, y, z)&lt;/math&gt;. Now we can begin to find our partial derivatives:<br /> <br /> &lt;cmath&gt;\dfrac{\partial V}{\partial x} = yz&lt;/cmath&gt; &lt;cmath&gt;\dfrac{\partial V}{\partial y} = xz&lt;/cmath&gt; &lt;cmath&gt;\dfrac{\partial V}{\partial z} = xy&lt;/cmath&gt;<br /> &lt;cmath&gt;\dfrac{\partial g}{\partial x} = 1&lt;/cmath&gt; &lt;cmath&gt;\dfrac{\partial g}{\partial y} = 2&lt;/cmath&gt; &lt;cmath&gt;\dfrac{\partial g}{\partial z} = 4&lt;/cmath&gt;<br /> <br /> We equate the corresponding partials through a constant factor, and we also use our original constraint:<br /> &lt;cmath&gt;x + 2y + 4z = 16&lt;/cmath&gt;<br /> &lt;cmath&gt;yz = \left(\lambda \right)1&lt;/cmath&gt; &lt;cmath&gt;xz = \left(\lambda \right)2&lt;/cmath&gt; &lt;cmath&gt;xy = \left(\lambda \right)4&lt;/cmath&gt;<br /> <br /> Multiplying each side of the last three equations by the variable not in the equation, we have:<br /> &lt;cmath&gt;xyz = \left(\lambda \right)1x&lt;/cmath&gt; &lt;cmath&gt;xyz = \left(\lambda \right)2y&lt;/cmath&gt; &lt;cmath&gt;xyz = \left(\lambda \right)4z&lt;/cmath&gt;<br /> <br /> We can equate these to result:<br /> &lt;math&gt;x = 2y = 4z&lt;/math&gt;<br /> <br /> Plugging back into the constraint, we have:<br /> &lt;cmath&gt;x + x + x = 16&lt;/cmath&gt;<br /> &lt;cmath&gt;x = \dfrac{16}{3}&lt;/cmath&gt; &lt;cmath&gt;y = \dfrac{8}{3}&lt;/cmath&gt; &lt;cmath&gt;z = \dfrac{4}{3}&lt;/cmath&gt;<br /> <br /> Thus, the maximum volume, given by &lt;math&gt;xyz&lt;/math&gt;, is:<br /> &lt;math&gt;\boxed{\dfrac{512}{9}}&lt;/math&gt;<br /> <br /> ==Problems==<br /> '''Introductory''' <br /> <br /> Find the maximum area of a rectangular prism with a surface area of &lt;math&gt;624&lt;/math&gt;.<br /> <br /> '''Olympiad'''<br /> <br /> Prove that &lt;cmath&gt;0\le yz+zx+xy-2xyz\le \frac{7}{27}&lt;/cmath&gt;, where &lt;math&gt;x,y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; are non-negative real numbers satisfying &lt;cmath&gt;x+y+z=1&lt;/cmath&gt;.</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=Lagrange_Multipliers&diff=138848 Lagrange Multipliers 2020-12-01T01:05:05Z <p>Rocketsri: /* Example */</p> <hr /> <div>This article discusses Lagrange multipliers, a topic of multivariable calculus.<br /> __TOC__<br /> <br /> == Definition ==<br /> Useful in optimization, Lagrange multipliers, based on a calculus approach, can be used to find local minimums and maximums of a function given a constraint. Suppose there is a continuous function &lt;math&gt;f(x,y)&lt;/math&gt; and there exists a continuous constraint function on the values of the function &lt;math&gt;c = g(x,y)&lt;/math&gt;. The method of Lagrange multipliers states that, to find the minimum or maximum satisfying both requirements (&lt;math&gt;\lambda&lt;/math&gt; is a constant):<br /> <br /> &lt;cmath&gt;\dfrac{\partial f}{\partial x} = \lambda \dfrac{\partial g}{\partial x}&lt;/cmath&gt;<br /> &lt;cmath&gt;\dfrac{\partial f}{\partial y} = \lambda \dfrac{\partial g}{\partial y}&lt;/cmath&gt;<br /> <br /> The method can be extended to multiple variables, as well as multiple constraints. If we had a continuous function &lt;math&gt;f(x, y, z)&lt;/math&gt; and the constraint &lt;math&gt;c = g(x, y, z)&lt;/math&gt;, we would just have another equation relating the partial derivatives with respect to &lt;math&gt;z&lt;/math&gt; through a factor of &lt;math&gt;\lambda&lt;/math&gt;. Similarly, if we have another constraint &lt;math&gt;d = h(x, y, z)&lt;/math&gt;, then we would add the partial with respect to each variable to their respective equation with another factor &lt;math&gt;\mu&lt;/math&gt;. Thus, we have the general form for Lagrange multipliers for the function &lt;math&gt;f(x_1, x_2,...,x_n)&lt;/math&gt; bounded by a certain number of constraints &lt;math&gt;c_1 = g_1(x_1, x_2,..., x_n)&lt;/math&gt;, &lt;math&gt;c_2 = g_2(x_1, x_2,..., x_n)&lt;/math&gt;,...,&lt;math&gt;c_m = g_m(x_1, x_2,..., x_n)&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2 + ... + \lambda_m \nabla g_m&lt;/cmath&gt;<br /> <br /> Where &lt;math&gt;\nabla&lt;/math&gt; is the del operator, which when applied to a scalar function &lt;math&gt;f&lt;/math&gt; results in a vector with each component the partial in that component's direction, representing total change.<br /> <br /> It is important to remember that sometimes there maybe a function that is bounded on some interval. In such cases, Lagrange multipliers may give a result, but the answer may not be the one this method results. After using this method, it is imperative to check the endpoints and also plug back in the numbers resulted from this method to verify the final answer. Also, sometimes the function may be over-constrained and there maybe no point that satisfies all requirements.<br /> <br /> == Example ==<br /> This method can become extremely difficult to use when there are several variables and constraints, but it is very effective and useful in some situations, including some contest math problems. Consider the following question:<br /> <br /> A rectangular prism lies on the &lt;math&gt;xy&lt;/math&gt; plane with one vertex at the origin. The vertex that does not share a face with this vertex at the origin lies on the plane bounded by the points &lt;math&gt;(0, 0, 4)&lt;/math&gt;, &lt;math&gt;(0, 8, 0)&lt;/math&gt;, &lt;math&gt;(16, 0, 0)&lt;/math&gt;. Find the maximum volume of the box.<br /> <br /> <br /> Solution:<br /> The volume of the box is given by the equation &lt;math&gt;V(x, y, z) = xyz&lt;/math&gt;. Because the way the box is described, the point that lies on the plane given determines the volume of the box. Thus we need only need to consider that point. The constraint we are given for that point is given by the plane &lt;math&gt;x + 2y + 4z = 16&lt;/math&gt;, since that is the plane that goes through the three points mentioned. Let us call that function &lt;math&gt;16 = g(x, y, z)&lt;/math&gt;. Now we can begin to find our partial derivatives:<br /> <br /> &lt;cmath&gt;\dfrac{\partial V}{\partial x} = yz&lt;/cmath&gt; &lt;cmath&gt;\dfrac{\partial V}{\partial y} = xz&lt;/cmath&gt; &lt;cmath&gt;\dfrac{\partial V}{\partial z} = xy&lt;/cmath&gt;<br /> &lt;cmath&gt;\dfrac{\partial g}{\partial x} = 1&lt;/cmath&gt; &lt;cmath&gt;\dfrac{\partial g}{\partial y} = 2&lt;/cmath&gt; &lt;cmath&gt;\dfrac{\partial g}{\partial z} = 4&lt;/cmath&gt;<br /> <br /> We equate the corresponding partials through a constant factor, and we also use our original constraint:<br /> &lt;cmath&gt;x + 2y + 4z = 16&lt;/cmath&gt;<br /> &lt;cmath&gt;yz = \left(\lambda \right)1&lt;/cmath&gt; &lt;cmath&gt;xz = \left(\lambda \right)2&lt;/cmath&gt; &lt;cmath&gt;xy = \left(\lambda \right)4&lt;/cmath&gt;<br /> <br /> Multiplying each side of the last three equations by the variable not in the equation, we have:<br /> &lt;cmath&gt;xyz = \left(\lambda \right)1x&lt;/cmath&gt; &lt;cmath&gt;xyz = \left(\lambda \right)2y&lt;/cmath&gt; &lt;cmath&gt;xyz = \left(\lambda \right)4z&lt;/cmath&gt;<br /> <br /> We can equate these to result:<br /> &lt;math&gt;x = 2y = 4z&lt;/math&gt;<br /> <br /> Plugging back into the constraint, we have:<br /> &lt;cmath&gt;x + x + x = 16&lt;/cmath&gt;<br /> &lt;cmath&gt;x = \dfrac{16}{3}&lt;/cmath&gt; &lt;cmath&gt;y = \dfrac{8}{3}&lt;/cmath&gt; &lt;cmath&gt;z = \dfrac{4}{3}&lt;/cmath&gt;<br /> <br /> Thus, the maximum volume, given by &lt;math&gt;xyz&lt;/math&gt;, is:<br /> &lt;math&gt;\box{\dfrac{512}{9}}&lt;/math&gt;<br /> <br /> ==Problems==<br /> '''Introductory''' <br /> <br /> Find the maximum area of a rectangular prism with a surface area of &lt;math&gt;624&lt;/math&gt;.<br /> <br /> '''Olympiad'''<br /> <br /> Prove that &lt;cmath&gt;0\le yz+zx+xy-2xyz\le \frac{7}{27}&lt;/cmath&gt;, where &lt;math&gt;x,y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; are non-negative real numbers satisfying &lt;cmath&gt;x+y+z=1&lt;/cmath&gt;.</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=Lagrange_Multipliers&diff=138847 Lagrange Multipliers 2020-12-01T01:04:50Z <p>Rocketsri: /* Example */</p> <hr /> <div>This article discusses Lagrange multipliers, a topic of multivariable calculus.<br /> __TOC__<br /> <br /> == Definition ==<br /> Useful in optimization, Lagrange multipliers, based on a calculus approach, can be used to find local minimums and maximums of a function given a constraint. Suppose there is a continuous function &lt;math&gt;f(x,y)&lt;/math&gt; and there exists a continuous constraint function on the values of the function &lt;math&gt;c = g(x,y)&lt;/math&gt;. The method of Lagrange multipliers states that, to find the minimum or maximum satisfying both requirements (&lt;math&gt;\lambda&lt;/math&gt; is a constant):<br /> <br /> &lt;cmath&gt;\dfrac{\partial f}{\partial x} = \lambda \dfrac{\partial g}{\partial x}&lt;/cmath&gt;<br /> &lt;cmath&gt;\dfrac{\partial f}{\partial y} = \lambda \dfrac{\partial g}{\partial y}&lt;/cmath&gt;<br /> <br /> The method can be extended to multiple variables, as well as multiple constraints. If we had a continuous function &lt;math&gt;f(x, y, z)&lt;/math&gt; and the constraint &lt;math&gt;c = g(x, y, z)&lt;/math&gt;, we would just have another equation relating the partial derivatives with respect to &lt;math&gt;z&lt;/math&gt; through a factor of &lt;math&gt;\lambda&lt;/math&gt;. Similarly, if we have another constraint &lt;math&gt;d = h(x, y, z)&lt;/math&gt;, then we would add the partial with respect to each variable to their respective equation with another factor &lt;math&gt;\mu&lt;/math&gt;. Thus, we have the general form for Lagrange multipliers for the function &lt;math&gt;f(x_1, x_2,...,x_n)&lt;/math&gt; bounded by a certain number of constraints &lt;math&gt;c_1 = g_1(x_1, x_2,..., x_n)&lt;/math&gt;, &lt;math&gt;c_2 = g_2(x_1, x_2,..., x_n)&lt;/math&gt;,...,&lt;math&gt;c_m = g_m(x_1, x_2,..., x_n)&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2 + ... + \lambda_m \nabla g_m&lt;/cmath&gt;<br /> <br /> Where &lt;math&gt;\nabla&lt;/math&gt; is the del operator, which when applied to a scalar function &lt;math&gt;f&lt;/math&gt; results in a vector with each component the partial in that component's direction, representing total change.<br /> <br /> It is important to remember that sometimes there maybe a function that is bounded on some interval. In such cases, Lagrange multipliers may give a result, but the answer may not be the one this method results. After using this method, it is imperative to check the endpoints and also plug back in the numbers resulted from this method to verify the final answer. Also, sometimes the function may be over-constrained and there maybe no point that satisfies all requirements.<br /> <br /> == Example ==<br /> This method can become extremely difficult to use when there are several variables and constraints, but it is very effective and useful in some situations, including some contest math problems. Consider the following question:<br /> <br /> A rectangular prism lies on the &lt;math&gt;xy&lt;/math&gt; plane with one vertex at the origin. The vertex that does not share a face with this vertex at the origin lies on the plane bounded by the points &lt;math&gt;(0, 0, 4)&lt;/math&gt;, &lt;math&gt;(0, 8, 0)&lt;/math&gt;, &lt;math&gt;(16, 0, 0)&lt;/math&gt;. Find the maximum volume of the box.<br /> <br /> <br /> Solution:<br /> The volume of the box is given by the equation &lt;math&gt;V(x, y, z) = xyz&lt;/math&gt;. Because the way the box is described, the point that lies on the plane given determines the volume of the box. Thus we need only need to consider that point. The constraint we are given for that point is given by the plane &lt;math&gt;x + 2y + 4z = 16&lt;/math&gt;, since that is the plane that goes through the three points mentioned. Let us call that function &lt;math&gt;16 = g(x, y, z)&lt;/math&gt;. Now we can begin to find our partial derivatives:<br /> <br /> &lt;cmath&gt;\dfrac{\partial V}{\partial x} = yz&lt;/cmath&gt; &lt;cmath&gt;\dfrac{\partial V}{\partial y} = xz&lt;/cmath&gt; &lt;cmath&gt;\dfrac{\partial V}{\partial z} = xy&lt;/cmath&gt;<br /> &lt;cmath&gt;\dfrac{\partial g}{\partial x} = 1&lt;/cmath&gt; &lt;cmath&gt;\dfrac{\partial g}{\partial y} = 2&lt;/cmath&gt; &lt;cmath&gt;\dfrac{\partial g}{\partial z} = 4&lt;/cmath&gt;<br /> <br /> We equate the corresponding partials through a constant factor, and we also use our original constraint:<br /> &lt;cmath&gt;x + 2y + 4z = 16&lt;/cmath&gt;<br /> &lt;cmath&gt;yz = \left(\lambda \right)1&lt;/cmath&gt; &lt;cmath&gt;xz = \left(\lambda \right)2&lt;/cmath&gt; &lt;cmath&gt;xy = \left(\lambda \right)4&lt;/cmath&gt;<br /> <br /> Multiplying each side of the last three equations by the variable not in the equation, we have:<br /> &lt;cmath&gt;xyz = \left(\lambda \right)1x&lt;/cmath&gt; &lt;cmath&gt;xyz = \left(\lambda \right)2y&lt;/cmath&gt; &lt;cmath&gt;xyz = \left(\lambda \right)4z&lt;/cmath&gt;<br /> <br /> We can equate these to result:<br /> &lt;math&gt;x = 2y = 4z&lt;/math&gt;<br /> <br /> Plugging back into the constraint, we have:<br /> &lt;cmath&gt;x + x + x = 16&lt;/cmath&gt;<br /> &lt;cmath&gt;x = \dfrac{16}{3}&lt;/cmath&gt; &lt;cmath&gt;y = \dfrac{8}{3}&lt;/cmath&gt; &lt;cmath&gt;z = \dfrac{4}{3}&lt;/cmath&gt;<br /> <br /> Thus, the maximum volume, given by &lt;math&gt;xyz&lt;/math&gt;, is:<br /> &lt;math&gt;box{\dfrac{512}{9}}&lt;/math&gt;<br /> <br /> ==Problems==<br /> '''Introductory''' <br /> <br /> Find the maximum area of a rectangular prism with a surface area of &lt;math&gt;624&lt;/math&gt;.<br /> <br /> '''Olympiad'''<br /> <br /> Prove that &lt;cmath&gt;0\le yz+zx+xy-2xyz\le \frac{7}{27}&lt;/cmath&gt;, where &lt;math&gt;x,y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; are non-negative real numbers satisfying &lt;cmath&gt;x+y+z=1&lt;/cmath&gt;.</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=136669 User:Piphi 2020-11-06T03:30:29Z <p>Rocketsri: /* User Count */</p> <hr /> <div>{{User:Piphi/Template:Header}}<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;100px&quot;&gt;375&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#B1B2B3;-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 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 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 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 in 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 /> <br /> Piphi has won 2 gold trophies at the [https://artofproblemsolving.com/community/c1124279 Asymptote Competition] and is now part of the staff.<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#727373;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;#f0f2f3&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;#f0f2f3&quot;&gt;<br /> You can check out more goals/statistics [[User:Piphi/Statistics|here]].<br /> <br /> A User Count of 500<br /> {{User:Piphi/Template:Progress_Bar|74.4|width=100%}}<br /> <br /> 200 subpages of [[User:Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|63.5|width=100%}}<br /> <br /> 200 signups for [[User:Piphi/Games|AoPS Wiki Games by Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|47.5|width=100%}}<br /> <br /> Make 10,000 edits<br /> {{User:Piphi/Template:Progress_Bar|20.09|width=100%}}&lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=1996_USAMO_Problems/Problem_1&diff=131505 1996 USAMO Problems/Problem 1 2020-08-11T16:01:36Z <p>Rocketsri: /* Solution 3 -hashtagmath */</p> <hr /> <div>==Problem==<br /> Prove that the average of the numbers &lt;math&gt; n\sin n^{\circ}\; (n = 2,4,6,\ldots,180) &lt;/math&gt; is &lt;math&gt;\cot 1^\circ&lt;/math&gt;.<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> First, as &lt;math&gt;180\sin{180^\circ}=0,&lt;/math&gt; we omit that term. Now, we multiply by &lt;math&gt;\sin 1^\circ&lt;/math&gt; to get, after using product to sum, &lt;math&gt;(\cos 1^\circ-\cos 3^\circ)+2(\cos 3^\circ-\cos 5^\circ)+\cdots +89(\cos 177^\circ-\cos 179^\circ)&lt;/math&gt;. <br /> This simplifies to &lt;math&gt;\cos 1^\circ+\cos 3^\circ +\cos 5^\circ+\cos 7^\circ+...+\cos 177^\circ-89\cos 179^\circ&lt;/math&gt;. Since &lt;math&gt;\cos x=-\cos(180-x),&lt;/math&gt; this simplifies to &lt;math&gt;90\cos 1^\circ&lt;/math&gt;. We multiplied by &lt;math&gt;\sin 1^\circ&lt;/math&gt; in the beginning, so we must divide by it now, and thus the sum is just &lt;math&gt;90\cot 1^\circ&lt;/math&gt;, so the average is &lt;math&gt;\cot 1^\circ&lt;/math&gt;, as desired.<br /> <br /> &lt;math&gt;\Box&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> Notice that for every &lt;math&gt;n\sin n^\circ&lt;/math&gt; there exists a corresponding pair term &lt;math&gt;(180^\circ - n)\sin{180^\circ - n} = (180^\circ - n)\sin n^\circ&lt;/math&gt;, for &lt;math&gt;n&lt;/math&gt; not &lt;math&gt;90^\circ&lt;/math&gt;. Pairing gives the sum of all &lt;math&gt;n\sin n^\circ&lt;/math&gt; terms to be &lt;math&gt;90(\sin 2^\circ + \sin 4^\circ + ... + \sin 178^\circ)&lt;/math&gt;, and thus the average is &lt;cmath&gt;S = (\sin 2^\circ + \sin 4^\circ + ... + \sin 178^\circ). (*)&lt;/cmath&gt; We need to show that &lt;math&gt;S = \cot 1^\circ&lt;/math&gt;. Multiplying (*) by &lt;math&gt;2\sin 1^\circ&lt;/math&gt; and using sum-to-product and telescoping gives &lt;math&gt;2\sin 1^\circ S = \cos 1^\circ - \cos 179^\circ = 2\cos 1^\circ&lt;/math&gt;. Thus, &lt;math&gt;S = \frac{\cos 1^\circ}{\sin 1^\circ} = \cot 1^\circ&lt;/math&gt;, as desired.<br /> <br /> &lt;math&gt;\Box&lt;/math&gt;<br /> <br /> ===Solution 3 -hashtagmath===<br /> We know that the average of a list of numbers is the sum of all the terms, divided by the number of terms. So we will set up an average. This average will average to &lt;math&gt;\cot 1^\circ&lt;/math&gt;. So we can set it equal to &lt;math&gt;\frac{\cos 1^\circ}{\sin 1^\circ}&lt;/math&gt;. Doing so, gives us &lt;cmath&gt;\frac{2 \sin 2^\circ + 4 \sin 4^\circ + 6 \sin 6^\circ + 8 \sin 8^\circ + 10 \sin 10^\circ + 12 \sin 12^\circ ... + 180 \sin 180^\circ}{90} = \frac{\cos 1^\circ}{\sin 1^\circ}.(1)&lt;/cmath&gt;<br /> <br /> We should try to simplify the numerator as it looks pretty messy.<br /> <br /> Now we know that &lt;math&gt;\sin x = \sin (180-x)&lt;/math&gt;, so maybe we can use this to clear things up a bit. Applying this to &lt;math&gt;(1)&lt;/math&gt; gives us &lt;cmath&gt;\frac{2 \sin 178^\circ + 4 \sin 176^\circ + 6 \sin 174^\circ + 8 \sin 172^\circ + 10 \sin 170^\circ + 12 \sin 168^\circ ... + 180 \sin 0^\circ}{90} = \frac{\cos 1^\circ}{\sin 1^\circ}. (2)&lt;/cmath&gt;<br /> <br /> Now, we know that &lt;math&gt;\sin 2^\circ = \sin 178^\circ&lt;/math&gt;, and &lt;math&gt;\sin 4^\circ = \sin 176^\circ&lt;/math&gt; and so on. We also know that &lt;math&gt;\sin 0^\circ = 0&lt;/math&gt;, so we can omit the last term. Also, we know that &lt;math&gt;\sin 90 = 1&lt;/math&gt;. Thus &lt;math&gt;90 \sin 90 = 90&lt;/math&gt;. So we will use all these when we further simplify. Another thing that we can simplify is to notice that when we reach the term &lt;math&gt;92\sin 88^\circ&lt;/math&gt;, we can rewrite it as &lt;math&gt;92\sin 92^\circ&lt;/math&gt;. Thus we can rewrite everything in the form &lt;math&gt;x\sin y^\circ&lt;/math&gt; as &lt;math&gt;x \sin x^\circ&lt;/math&gt;, starting with the term &lt;math&gt;92\sin 88^\circ&lt;/math&gt;, then &lt;math&gt;94\sin 86^\circ&lt;/math&gt;, and so on. <br /> <br /> Thus we can further simplify our equation. Doing so, gives us &lt;cmath&gt;\frac{2 \sin 178^\circ + 4 \sin 176^\circ + 6 \sin 174^\circ + 8 \sin 172^\circ + 10 \sin 170^\circ + 12 \sin 168^\circ ...+ 176 \sin 176^\circ + 178 \sin 178^\circ + 90}{90} = \frac{\cos 1^\circ}{\sin 1^\circ}. (3)&lt;/cmath&gt;<br /> <br /> Then, to make things easier, we can rewrite all the numbers with degrees less than &lt;math&gt;90^\circ&lt;/math&gt; and combine like terms. Doing so gives us &lt;cmath&gt;\frac{180 \sin 2^\circ + 180 \sin 4^\circ + 180 \sin 6^\circ +180 \sin 8^\circ +180 \sin 10^\circ ... +180 \sin 88^\circ + (90)}{90} = \frac{\cos 1^\circ}{\sin 1^\circ}. (4)&lt;/cmath&gt;<br /> <br /> Now we can factor out &lt;math&gt;180&lt;/math&gt; to get &lt;cmath&gt;\frac{180(\sin 2^\circ+ \sin 4^\circ + \sin 6^\circ + \sin 8^\circ + \sin 10^\circ ... + \sin 88^\circ) + 90}{90} = \frac{\cos 1^\circ}{\sin 1^\circ}. (5)&lt;/cmath&gt;<br /> <br /> Then we can simplify it by getting rid of the denominator to get to &lt;cmath&gt;2(\sin 2^\circ+ \sin 4^\circ + \sin 6^\circ + \sin 8^\circ + \sin 10^\circ ... + \sin 88^\circ) + 1 = \frac{\cos 1^\circ}{\sin 1^\circ}. (6)&lt;/cmath&gt;<br /> <br /> Now we notice that we can multiply both sides by &lt;math&gt;\sin 1^\circ&lt;/math&gt;. Doing so gives us &lt;cmath&gt; \sin 1^\circ(2 (\sin 2^\circ+ \sin 4^\circ + \sin 6^\circ + \sin 8^\circ + \sin 10^\circ ... + \sin 88^\circ) + 1) = \cos 1^\circ. (7)&lt;/cmath&gt;<br /> <br /> Now, simplifying &lt;math&gt;\sin 2^\circ+ \sin 4^\circ + \sin 6^\circ + \sin 8^\circ + \sin 10^\circ ... + \sin 88^\circ&lt;/math&gt; doesn't look too promising. So maybe if we expand again, we can maybe somehow use our product to sum formulas. <br /> <br /> Doing so, gives us &lt;cmath&gt; (\sin 1^\circ)(2 \sin 2^\circ)+(\sin 1^\circ)(2\sin 4^\circ) +(\sin 1^\circ)(2\sin 6^\circ) + (\sin 1^\circ)(2\sin 8^\circ) +(\sin 1^\circ)(2 \sin 10^\circ) ... + (\sin 1^\circ)(2\sin 88^\circ) + \sin 1^\circ = \cos 1^\circ. (8)&lt;/cmath&gt;<br /> <br /> Now we can use our product of sines to sum formula and see if we can find a pattern. <br /> <br /> We will start with expanding &lt;math&gt;(\sin 1^\circ)(2 \sin 2^\circ)&lt;/math&gt;. Because the format of our formula for the product of sines is &lt;math&gt;\sin \alpha + \sin \beta&lt;/math&gt;, we can factor out the &lt;math&gt;2&lt;/math&gt; and find the product, then multiply by &lt;math&gt;2&lt;/math&gt;. So, we are at &lt;math&gt;2(\sin 1^\circ)(\sin 2^\circ)&lt;/math&gt;. Our formula for the product of sines is &lt;math&gt;\sin \alpha + \sin \beta = \frac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha + \beta)]&lt;/math&gt;. Plugging in our values into the formula and simplifying gives us &lt;math&gt;2\left(\frac{1}{2}(\cos 1^\circ - \cos 3^\circ)\right)&lt;/math&gt;. We know that &lt;math&gt;2\left(\frac{1}{2}\right)&lt;/math&gt; will just cancel out to &lt;math&gt;1&lt;/math&gt;. So we are left with &lt;math&gt;\cos 1^\circ - \cos 3^\circ&lt;/math&gt;. &lt;math&gt;(9)&lt;/math&gt;<br /> <br /> Next we expand &lt;math&gt;(\sin 1^\circ)(2\sin 4^\circ)&lt;/math&gt;. We use the same steps or strategy as we did above and get &lt;math&gt;\cos 3^\circ - \cos 5^\circ&lt;/math&gt;. &lt;math&gt;(10)&lt;/math&gt;<br /> <br /> We may be noticing a pattern. Just to make sure it is true, we will expand &lt;math&gt;(\sin 1^\circ)(2\sin 6^\circ)&lt;/math&gt;. When we expand this using the same strategy above, we get &lt;math&gt;\cos 5^\circ - \cos 7^\circ&lt;/math&gt;. &lt;math&gt;(11)&lt;/math&gt; So our pattern is that &lt;math&gt;(\sin 1^\circ)(2\sin \mu^\circ) = \cos (\mu-1)^\circ - \cos (\mu+1)^\circ. (12)&lt;/math&gt;<br /> <br /> Now we notice that when we add &lt;math&gt;(9)&lt;/math&gt; and &lt;math&gt;(10)&lt;/math&gt;, &lt;math&gt;-\cos 3^\circ&lt;/math&gt; and &lt;math&gt;+\cos 3^\circ&lt;/math&gt; cancel themselves out to just &lt;math&gt;\cos 1^\circ - \cos 5^\circ&lt;/math&gt;. Then, when we add &lt;math&gt;(11)&lt;/math&gt; to it, &lt;math&gt;+ \cos 5^\circ&lt;/math&gt; and &lt;math&gt;- \cos 5^\circ&lt;/math&gt; also cancel themselves out. <br /> <br /> Now, we need to figure out when we should stop canceling out. We can use formula &lt;math&gt;(12)&lt;/math&gt; and replace &lt;math&gt;88&lt;/math&gt; with &lt;math&gt;\mu&lt;/math&gt; since &lt;math&gt;88&lt;/math&gt; is the last one in &lt;math&gt;(8)&lt;/math&gt;. Applying this formula and simplifying gives us &lt;math&gt;\cos 87^\circ - \cos 89&lt;/math&gt;, so the last term would be &lt;math&gt;- \cos 89&lt;/math&gt;.<br /> <br /> Thus we know that everything we cancel themselves out except the first and the last term. So, we are left with &lt;cmath&gt;(\cos 1^\circ - \cos 89^\circ) + \sin 1^\circ = \cos 1^\circ. (13)&lt;/cmath&gt;<br /> <br /> Now we notice that we have &lt;math&gt;\cos 1^\circ&lt;/math&gt; on both sides. Thus we can subtract &lt;math&gt;\cos 1^\circ&lt;/math&gt; from both sides to get &lt;math&gt;- \cos 89^\circ + \sin 1^\circ = 0&lt;/math&gt;. Now we can add &lt;math&gt;\cos 89^\circ&lt;/math&gt; to both sides and get &lt;math&gt;\sin 1^\circ = \cos 89^\circ&lt;/math&gt;. Now we also know that &lt;math&gt;\sin (\omega)^\circ = \cos(90-\omega)&lt;/math&gt;. Thus we know that &lt;math&gt;\sin 1^\circ = \cos 89^\circ&lt;/math&gt;. Hence, our proof is complete. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{USAMO newbox|year=1996|before=First Question|num-a=2}}<br /> {{MAA Notice}}<br /> [[Category:Olympiad Algebra Problems]]</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=User:Rusczyk&diff=131347 User:Rusczyk 2020-08-10T16:46:53Z <p>Rocketsri: /* User Count */</p> <hr /> <div>Rusczyk's Page:<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#4EC284&quot;&gt;<br /> ==&lt;font color=&quot;white&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;white&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;&lt;font color=&quot;white&quot;&gt;<br /> &lt;center&gt;&lt;font size=&quot;101px&quot;&gt;56&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#4EC284;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;white&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;white&quot;&gt;Rusczyk is currently borderline AIME.&lt;br&gt;<br /> <br /> Rusczyk just turned 13 years old.&lt;br&gt;<br /> <br /> Rusczyk scored 44/46 when mocking the 2018 MATHCOUNTS State test, and got silver on the 2020 online MATHCOUNTS State held on AoPS.&lt;br&gt;<br /> <br /> Rusczyk is a pro at maths and physics<br /> <br /> Rusczyk has come world and country #1 in various international tournaments and competitions starting from 2017<br /> <br /> Rusczyk is better than me at maths<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#4EC284;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;white&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;white&quot;&gt; A User Count of 500<br /> <br /> Make AIME 2021 (Currently borderline)<br /> <br /> Pass AP Calculus AB, BC and AP Physics exam<br /> <br /> Get in the Alcumus HoF in the next 6 months<br /> <br /> Get &lt;math&gt;\color{white}{2 \times}&lt;/math&gt; medals this year as compared to what they did last year. That is &lt;math&gt;\color{white}{2 \times 14 = \boxed{28}}&lt;/math&gt; which is nearly impossible.<br /> &lt;/div&gt;<br /> &lt;/div&gt;</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=User:Wuwang2002&diff=127255 User:Wuwang2002 2020-07-02T03:20:21Z <p>Rocketsri: /* Somewhat accurate visit count */</p> <hr /> <div>Hello! I am wuwang2002!<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 This Place&lt;/div&gt;&lt;/font&gt;==<br /> <br /> What you saw above was made by piphi.<br /> <br /> Uh, well...I have nothing else here to say.<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;Links you should click NOW&lt;/div&gt;&lt;/font&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;My own links from AoPS&lt;/div&gt;&lt;/font&gt;===<br /> https://artofproblemsolving.com/community/c1132787_wuwang2002s_games_collection<br /> <br /> Check that out: it is my own collection of my own fora.<br /> <br /> https://artofproblemsolving.com/community/c1194307_the_land_of_empires<br /> <br /> Check that out if you want.<br /> <br /> And lastly: [[Clickbait|A last link]]<br /> <br /> Thanks!<br /> <br /> Sorry if this is not allowed @admins. I read that article, but I wasn't sure if this classified.<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;Outside links&lt;/div&gt;&lt;/font&gt;===<br /> <br /> &lt;font color=&quot;red&quot;&gt;[NOT PUT HERE BY WUWANG2002 BECAUSE OF POSSIBLE VIOLATIONS OF THE WIKI RULES]&lt;/font&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;More info&lt;/div&gt;&lt;/font&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;Somewhat accurate visit count&lt;/div&gt;&lt;/font&gt;===<br /> Add 1 to this number whenever you visit: <br /> &lt;center&gt;&lt;font size=&quot;150 pts&quot;&gt;3&lt;/center&gt;&lt;/font size&gt;<br /> <br /> Please put your name under this sentence when it is your first time visiting:<br /> *wuwang2002<br /> *Lcz<br /> *rocketsri<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;My Talk&lt;/div&gt;&lt;/font&gt;===<br /> <br /> [[User talk:Wuwang2002|Click here for my talk]]<br /> <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;LaTeX 'n bbcode&lt;/div&gt;&lt;/font&gt;==<br /> <br /> C'mon! Ask me for any LaTeX and then I will give the code!<br /> Same with bbcode!<br /> Thanks!<br /> <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;Games&lt;/div&gt;&lt;/font&gt;==<br /> <br /> So far, my games are all in forums. But however, I hope to someday make some own games like Go! and Gomuku, which I don't have a translation for!<br /> <br /> EDIT: I am making a collaboration with [[User:Piphi|piphi]] to make more games.<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;Some Facts&lt;/div&gt;&lt;/font&gt;==<br /> <br /> 1. wuwang2002 is the best!...or not<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;Testing Stuff&lt;/div&gt;&lt;/font&gt;==<br /> <br /> This is free to anyone to edit.<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;Test Level 1&lt;/div&gt;&lt;/font&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;Test level 2&lt;/div&gt;&lt;/font&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;Test level 3&lt;/div&gt;&lt;/font&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;Last level&lt;/div&gt;&lt;/font&gt;======<br /> <br /> Go more!<br /> <br /> Oh, you can't.<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;Stories about bots&lt;/div&gt;&lt;/font&gt;==<br /> <br /> To be made<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;Math&lt;/div&gt;&lt;/font&gt;==<br /> <br /> Sorry this came so late, but...<br /> Put your opinions with your name under the latest post in the appropriate section. Feel free to add examples.<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;Arithmetic&lt;/div&gt;&lt;/font&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;Opinions&lt;/div&gt;&lt;/font&gt;====<br /> Bashy, annoying, drill-and-kill, and not that helpful or interesting. -wuwang2002<br /> <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;Examples&lt;/div&gt;&lt;/font&gt;====<br /> <br /> &lt;math&gt;9+\dfrac{8}{7}\cdot5-\dfrac{19}{7}&lt;/math&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;Algebra&lt;/div&gt;&lt;/font&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;Opinions&lt;/div&gt;&lt;/font&gt;====<br /> Much better than arithmetic, but sometimes bashy. -wuwang2002<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;Examples&lt;/div&gt;&lt;/font&gt;====<br /> <br /> &lt;math&gt;\dfrac{9}{4}x+32=50&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2-8x+12=-3&lt;/math&gt;</div> Rocketsri https://artofproblemsolving.com/wiki/index.php?title=Simon%27s_Favorite_Factoring_Trick&diff=112234 Simon's Favorite Factoring Trick 2019-11-27T19:04:34Z <p>Rocketsri: /* Awesome Practice Problems */</p> <hr /> <div><br /> ==About==<br /> '''Dr. Simon's Favorite Factoring Trick''' (abbreviated '''SFFT''') is a special factorization first popularized by [[AoPS]] user [[user:ComplexZeta | Simon Rubinstein-Salzedo]].<br /> <br /> ==The General Statement==<br /> The general statement of SFFT is: &lt;math&gt;{xy}+{xk}+{yj}+{jk}=(x+j)(y+k)&lt;/math&gt;. Two special common cases are: &lt;math&gt;xy + x + y + 1 = (x+1)(y+1)&lt;/math&gt; and &lt;math&gt;xy - x - y +1 = (x-1)(y-1)&lt;/math&gt;.<br /> <br /> The act of adding &lt;math&gt;{jk}&lt;/math&gt; to &lt;math&gt;{xy}+{xk}+{yj}&lt;/math&gt; in order to be able to factor it could be called &quot;completing the rectangle&quot; in analogy to the more familiar &quot;completing the square.&quot;<br /> <br /> == Applications ==<br /> This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are variables and &lt;math&gt;j,k&lt;/math&gt; are known constants. Also, it is typically necessary to add the &lt;math&gt;jk&lt;/math&gt; term to both sides to perform the factorization.<br /> <br /> == Fun Practice Problems ==<br /> ===Introductory===<br /> *Two different [[prime number]]s between &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;18&lt;/math&gt; are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 22 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 } &lt;/math&gt;<br /> <br /> ([[2000 AMC 12/Problem 6|Source]])<br /> <br /> ===Intermediate===<br /> *&lt;math&gt;m, n&lt;/math&gt; are integers such that &lt;math&gt;m^2 + 3m^2n^2 = 30n^2 + 517&lt;/math&gt;. Find &lt;math&gt;3m^2n^2&lt;/math&gt;.<br /> <br /> ([[1987 AIME Problems/Problem 5|Source]])<br /> <br /> ===Olympiad===<br /> <br /> *The integer &lt;math&gt;N&lt;/math&gt; is positive. There are exactly 2005 ordered pairs &lt;math&gt;(x, y)&lt;/math&gt; of positive integers satisfying:<br /> <br /> &lt;cmath&gt;\frac 1x +\frac 1y = \frac 1N&lt;/cmath&gt;<br /> <br /> Prove that &lt;math&gt;N&lt;/math&gt; is a perfect square. (British Mathematical Olympiad Round 2, 2005)<br /> <br /> Enjoy the problems!<br /> <br /> == See More==<br /> * [[Algebra]]<br /> * [[Factoring]]<br /> <br /> [[Category:Elementary algebra]]<br /> [[Category:Theorems]]</div> Rocketsri